Solucionario faires

SECTION 1– DESIGN FOR SIMPLE STRESSES

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TENSION, COMPRESSION, SHEAR

DESIGN PROBLEMS

1. The link shown, made of AISI C1045 steel, as rolled, is subjected to a tensile load

of 8000 lb. Let bh 5.1= . If the load is repeated but not reversed, determine the

dimensions of the section with the design based on (a) ultimate strength, (b) yield

strength. (c) If this link, which is 15 in. long., must not elongate more than 0.005

in., what should be the dimensions of the cross section?

Problems 1 – 3.

Solution:

For AISI C1045 steel, as rolled (Table AT 7)

ksisu 96=

ksisy 59=

psiE 61030×=

A

Fsd =

where

lbF 8000=

bhA =

but

bh 5.1=

therefore 25.1 bA =

(a) Based on ultimate strength

N = factor of safety = 6 for repeated but not reversed load (Table 1.1)

A

F

N

ss u

d ==

25.1

8000

6

000,96

b=

inb 577.0= say in8

5.


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inbh16

155.1 ==

(b) Based on yield strength


A

F

N

ss u

d ==

25.1

8000

3

000,59

b=

inb 521.0= say in16

9.

inbh32

275.1 ==

(c) Elongation = AE

FL=δ

where,

in005.0=δ

lbF 8000=

psiE 61030×=

inL 15= 25.1 bA =

then,

AE

FL=δ

( )( )( )( )62 10305.1

158000005.0

×=

b

inb 730.0= say in4

3.

inbh8

115.1 ==

2. The same as 1 except that the material is malleable iron, ASTM A47-52, grade 35

018.

Solution:

For malleable iron, ASTM A47-52, grade 35 018(Table AT 6)

ksisu 55=

ksisy 5.36=

psiE 61025×=


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A

Fsd =

where

lbF 8000=

bhA =

but

bh 5.1=

therefore 25.1 bA =



A

F

N

ss u

d ==

25.1

8000

6

000,55

b=

inb 763.0= say in8

7.

inbh16

515.1 ==



A

F

N

ss u

d ==

25.1

8000

3

500,36

b=

inb 622.0= say in16

11.

inbh32

115.1 ==

(c) Elongation = AE

FL=δ

where,

in005.0=δ

lbF 8000=

psiE 61025×=

inL 15= 25.1 bA =

then,


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AE

FL=δ

( )( )( )( )62 10255.1

158000005.0

×=

b

inb 8.0= say in8

7.

inbh16

515.1 ==

3. The same as 1 except that the material is gray iron, ASTM 30.

Solution:

For ASTM 30 (Table AT 6)

ksisu 30= , no ys

psiE 6105.14 ×=

Note: since there is no ys for brittle materials. Solve only for (a) and (c)

A

Fsd =

where

lbF 8000=

bhA =

but

bh 5.1=

therefore 25.1 bA =


N = factor of safety = 7 ~ 8 say 7.5 (Table 1.1)

A

F

N

ss u

d ==

25.1

8000

5.7

000,30

b=

inb 1547.1= say in16

31 .

inbh32

2515.1 ==

(c) Elongation = AE

FL=δ

where,

in005.0=δ

lbF 8000=

psiE 6105.14 ×=


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inL 15= 25.1 bA =

then,

AE

FL=δ

( )( )( )( )62 105.145.1

158000005.0

×=

b

inb 050.1= say in16

11 .

inbh32

1915.1 ==

4. A piston rod, made of AISI 3140 steel, OQT 1000 F (Fig. AF 2), is subjected to a

repeated, reversed load. The rod is for a 20-in. air compressor, where the

maximum pressure is 125 psig. Compute the diameter of the rod using a design

factor based on (a) ultimate strength, (b) yield strength.

Solution:

From Fig. AF 2 for AISI 3140, OQT 1000 F

ksisu 5.152=

ksisy 5.132=

( ) ( ) kipslbforceF 27.39270,39125204

2====

π

From Table 1.1, page 20

8=uN

4=yN


u

u

s

FNA =

( )( )5.152

27.398

4

2 =dπ

ind 62.1= say in8

51


y

y

s

FNA =

( )( )5.132

27.394

4

2 =dπ


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ind 23.1= say in4

11

5. A hollow, short compression member, of normalized cast steel (ASTM A27-58,

65 ksi), is to support a load of 1500 kips with a factor of safety of 8 based on the

ultimate strength. Determine the outside and inside diameters if io DD 2= .

Solution:

ksisu 65=

8=uN

kipsF 1500=

( ) ( )4

34

44

22222 iiiio

DDDDDA

πππ=−=−=

( )( )65

15008

4

3 2

===u

ui

s

FNDA

π

inDi 85.8= say in8

78

inDD io4

317

8

7822 =

==

6. A short compression member with io DD 2= is to support a dead load of 25 tons.

The material is to be 4130 steel, WQT 1100 F. Calculate the outside and inside

diameters on the basis of (a) yield strength, (b) ultimate strength.

Solution:

From Table AT 7 for 4130, WQT 1100 F

ksisu 127=

ksisy 114=

From Table 1.1 page 20, for dead load

4~3=uN , say 4

2~5.1=yN , say 2

Area, ( ) ( )4

34

44

22222 iiiio

DDDDDA

πππ=−=−=

kipstonsF 5025 ==

(a) Based on yield strength

( )( )114

502

4

3 2

===y

yi

s

FNDA

π


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inDi 61.0= say in8

5

inDD io4

11

8

522 =

==

(b) Based on ultimate strength

( )( )127

504

4

3 2

===u

ui

s

FNDA

π

inDi 82.0= say in8

7

inDD io4

31

8

722 =

==

7. A round, steel tension member, 55 in. long, is subjected to a maximum load of

7000 lb. (a) What should be its diameter if the total elongation is not to exceed

0.030 in? (b) Choose a steel that would be suitable on the basis of yield strength if

the load is gradually applied and repeated (not reversed).

Solution:

(a) AE

FL=δ or

E

FLA

δ=

where,

lbF 7000=

inL 55=

in030.0=δ

psiE 61025×=

( )( )( )( )6

2

1030030.0

557000

4 ×== dA

π

ind 74.0= say in4

3

(b) For gradually applied and repeated (not reversed) load

3=yN

( )( )

( )psi

A

FNs

y

y 534,47

75.04

70003

2

===π

ksisy 48≈

say C1015 normalized condition ( ksisy 48= )

8. A centrifuge has a small bucket, weighing 0.332 lb. with contents, suspended on a

manganese bronze pin (B138-A, ½ hard) at the end of a horizontal arm. If the pin

is in double shear under the action of the centrifugal force, determine the diameter


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needed for 10,000 rpm of the arm. The center of gravity of the bucket is 12 in.

from the axis of rotation.

Solution:

From Table AT 3, for B138-A, ½ hard

ksisus 48=

rg

WF 2ω=

where

lbW 332.0= 22.32 fpsg =

( )sec1047

60

000,102

60

2rad

n===

ππω

inr 12=

( ) ( ) kipslbrg

WF 3.11300,1111047

2.32

332.0 22 ==== ω

From Table 1.1, page 20

4~3=N , say 4

u

u

s

FNA =

( )( )48

3.114

42 2 =

d

π for double shear

ind 774.0= say in32

25

CHECK PROBLEMS

9. The link shown is made of AISIC1020 annealed steel, with inb4

3= and

inh2

11= . (a) What force will cause breakage? (b) For a design factor of 4 based

on the ultimate strength, what is the maximum allowable load? (c) If 5.2=N

based on the yield strength, what is the allowable load?

Problem 9.


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Solution:

For AISI C1020 annealed steel, from Table AT 7

ksisu 57=

ksisy 42=

(a) AsF u=

2125.12

11

4

3inbhA =

==

( )( ) kipsF 64125.157 ==

(b) u

u

N

AsF =

4=uN

2125.12

11

4

3inbhA =

==

( )( )kipsF 16

4

125.157==

(c) y

y

N

AsF =

5.2=yN

2125.12

11

4

3inbhA =

==

( )( )kipsF 9.18

2

125.142==

10. A ¾-in.bolt, made of cold-finished B1113, has an effective stress area of 0.334 sq.

in. and an effective grip length of 5 in. The bolt is to be loaded by tightening until

the tensile stress is 80 % of the yield strength, as determined by measuring the

total elongation. What should be the total elongation?

Solution:

E

sL=δ

from Table AT 7 for cold-finished B1113

ksisy 72=

then, ( ) ksiss y 6.57728.080.0 ===

ksipsiE 000,301030 6 =×=

( )( )in

E

sL0096.0

000,30

56.57===δ


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11. A 4-lb. weight is attached by a 3/8-in. bolt to a rotating arm 14-in. from the center

of rotation. The axis of the bolts is normal to the plane in which the centrifugal

force acts and the bolt is in double shear. At what speed will the bolt shear in two

if it is made of AISI B1113, cold finish?

Solution:

From Table AT 7, psiksisus 000,6262 ==

( ) 2

2

2209.08

3

4

12 inA =

= π

Asrg

WF us== 2ω

( ) ( )( )2209.0000,62142.32

4 2 =ω

sec74.88 rad=ω

74.8860

2==

nπω

rpmn 847=

12. How many ¾-in. holes could be punched in one stroke in annealed steel plate of

AISI C1040, 3/16-in. thick, by a force of 60 tons?

Solution:

For AISI C1040, from Figure AF 1

ksisu 80=

( ) ksiksiss uus 608075.075.0 ===

tdA π=

kipstonsF 12060 ==

n = number of holes

( )( )9

602209.0

120===

usAs

Fn holes

13. What is the length of a bearing for a 4-in. shaft if the load on the bearing is 6400

lb. and the allowable bearing pressure is 200 psi of the projected area?

Solution:

WpDL =

where

psip 200=

inD 4=

lbW 6400=


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( )( ) 64004200 =L

inL 8=

BENDING STRESSES

DESIGN PROBLEMS

14. A lever keyed to a shaft is inL 15= long and has a rectangular cross section of

th 3= . A 2000-lb load is gradually applied and reversed at the end as shown; the

material is AISI C1020, as rolled. Design for both ultimate and yield strengths. (a)

What should be the dimensions of a section at ina 13= ? (b) at inb 4= ? (c) What

should be the size where the load is applied?

Problem 14.

Solution:

For AISI C1020, as rolled, Table AT 7

ksisu 65=

ksisy 49=

Design factors for gradually applied and reversed load

8=uN

4=yN

12

3th

I = , moment of inertial

but th 3=

36

4h

I =

Moment Diagram (Load Upward)


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Based on ultimate strength

u

u

N

ss =

(a) I

Fac

I

Mcs ==

2

hc =

kipslbsF 22000 ==

( )( )

==

36

2132

8

654

h

h

s

inh 86.3=

inh

t 29.13

86.3

3===

say

ininh2

145.4 ==

inint2

115.1 ==

(b) I

Fbc

I

Mcs ==

2

hc =

kipslbsF 22000 ==

( )( )

==

36

242

8

654

h

h

s

inh 61.2=

inh

t 87.03

61.2

3===

say

inh 3=

int 1=

(c)


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413

35.4

4

3

−

−=

− h

inh 33.2=

413

15.1

4

1

−

−=

− t

int 78.0=

say

inh 625.2= or inh8

52=

15. A simple beam 54 in. long with a load of 4 kips at the center is made of cast steel,

SAE 080. The cross section is rectangular (let bh 3≈ ). (a) Determine the

dimensions for 3=N based on the yield strength. (b) Compute the maximum

deflection for these dimensions. (c) What size may be used if the maximum

deflection is not to exceed 0.03 in.?

Solution:

For cast steel, SAE 080 (Table AT 6)

ksisy 40=

psiE 61030×=


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From Table AT 2

Max. ( )( )

inkipsFL

M −=== 544

544

4

12

3bh

I =

but bh 3=

36

4h

I =

(a) I

Mc

N

ss

y

y==

2

hc =

( )

=

36

254

3

404

h

h

inh 18.4=

inh

b 39.13

18.4

3===

say inh2

14= , inin

hb

2

115.1

3

5.4

3====

(b) ( )( )

( ) ( )( )in

EI

FL0384.0

12

5.45.1103048

544000

48 3

6

33

=

×

==δ

(c)

=

3648

4

3

hE

FLδ

( )( ) ( )( )( )46

3

103048

3654400003.0

h×=

inh 79.4=

inh

b 60.13

79.4

3===

say ininh4

1525.5 == , inin

hb

4

3175.1

3

25.5

3====


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16. The same as 15, except that the beam is to have a circular cross section.

Solution:

(a) I

Mc

N

ss

y

y==

64

4d

Iπ

=

2

dc =

34

32

64

2

d

M

d

dM

sππ

=

=

( )3

5432

3

40

dπ=

ind 46.3=

say ind2

13=

(b) EI

FL

48

3

=δ

64

4d

Iπ

=

( )( )( )

( )( )( )in

dE

FL0594.0

5.3103048

54400064

48

6446

3

4

3

=×

==ππ

δ

(c) ( )4

3

48

64

dE

FL

πδ =

( )( )( )( ) 46

3

103048

5440006403.0

dπ×=

ind 15.4=

say ind4

14=

17. A simple beam, 48 in. long, with a static load of 6000 lb. at the center, is made of

C1020 structural steel. (a) Basing your calculations on the ultimate strength,

determine the dimensions of the rectangular cross section for bh 2= . (b)

Determine the dimensions based on yield strength. (c) Determine the dimensions

using the principle of “limit design.”


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Solution:

From Table AT 7 and Table 1.1

ksisu 65=

ksisy 48=

4~3=uN , say 4

2~5.1=yN , say 2

( )( )kipsin

FLM −=== 72

4

486

4

I

Mcs =

2

hc =

12

3bh

I =

but 2

hb =

24

4h

I =

34

12

24

2

h

M

h

hM

s =

=


3

12

h

M

N

ss

u

u ==

( )3

7212

4

65

h=

inh 76.3=


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inh

b 88.12

76.3

2===

say ininh4

3375.3 == , inin

hb

8

71875.1

2

75.3

2====


3

12

h

M

N

ss

y

y==

( )3

7212

2

48

h=

inh 30.3=

inh

b 65.12

30.3

2===

say ininh2

135.3 == , inin

hb

4

3175.1

2

5.3

2====

(c) Limit design (Eq. 1.6)

4

2bh

sM y=

( )4

24872

2h

h

=

inh 29.2=

inh

b 145.12

29.2

2===

say ininh2

125.2 == , inin

hb

4

1125.1

2

5.2

2====

18. The bar shown is subjected to two vertical loads, 1F and 2F , of 3000 lb. each, that

are inL 10= apart and 3 in. ( a , d ) from the ends of the bar. The design factor is 4

based on the ultimate strength; bh 3= . Determine the dimensions h and b if the

bar is made of (a) gray cast iron, SAE 111; (b) malleable cast iron, ASTM A47-

52, grade 35 018; (c) AISI C1040, as rolled (Fig. AF 1). Sketch the shear and

moment diagrams approximately to scale.


of 131

Problems18, 19.

Solution:

lbRRFF 30002121 ====

Moment Diagram

( )( ) inkipsinlbsaRM −=−=== 99000330001

N = factor of safety = 4 based on us

12

3bh

I =

2

hc =

3612

34

3

hh

h

I =

=

(a) For gray cast iron, SAE 111

ksisu 30= , Table AT 6

34

18

36

2

h

M

h

hM

I

Mc

N

ss u =

===

( )3

918

4

30

hs ==

inh 78.2=

inh

b 93.03

78.2

3===

say inh 5.3= , inb 1=

(b) For malleable cast iron, ASTM A47-52, grade 35 018


of 131


34

18

36

2

h

M

h

hM

I

Mc

N

ss u =

===

( )3

918

4

55

hs ==

inh 28.2=

inh

b 76.03

28.2

3===

say inh4

12= , inb

4

3=

(c) For AISI C1040, as rolled

ksisu 90= , Fig. AF 1

34

18

36

2

h

M

h

hM

I

Mc

N

ss u =

===

( )3

918

4

90

hs ==

inh 93.1=

inh

b 64.03

93.1

3===

say inh8

71= , inb

8

5=

19. The same as 18, except that 1F acts up ( 2F acts down).

Solution:

[ ]∑ = 0AM lbRR 187521 ==


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Shear Diagram

Moment Diagram

=M maximum moment = 5625 lb-in = 5.625 kips-in

(a) For gray cast iron

3

18

h

M

N

ss u ==

( )3

625.518

4

30

h=

inh 38.2=

inh

b 79.03

38.2

3===

say inh4

12= , inb

4

3=

(b) For malleable cast iron

3

18

h

M

N

ss u ==

( )3

625.518

4

55

h=

inh 95.1=

inh

b 65.03

95.1

3===

say inh8

71= , inb

8

5=


of 131


3

18

h

M

N

ss u ==

( )3

625.518

4

90

h=

inh 65.1=

inh

b 55.03

65.1

3===

say inh2

11= , inb

2

1=

20. The bar shown, supported at A and B , is subjected to a static load F of 2500 lb.

at 0=θ . Let ind 3= , inL 10= and bh 3= . Determine the dimensions of the

section if the bar is made of (a) gray iron, SAE 110; (b) malleable cast iron,

ASTM A47-52, grade 32 510; (c) AISI C1035 steel, as rolled. (d) For economic

reasons, the pins at A, B, and C are to be the same size. What should be their

diameter if the material is AISI C1035, as rolled, and the mounting is such that

each is in double shear? Use the basic dimensions from (c) as needed. (e) What

sectional dimensions would be used for the C1035 steel if the principle of “limit

design” governs in (c)?

Problems 20, 21.

Solution:


of 131

[ ]∑ = 0AM ( )2500133 =BR

lbRB 833,10=

[ ]∑ = 0BM ( )2500103 =AR

lbRA 8333=

Shear Diagram

Moment Diagram

( )( ) inkipsinlbM −=−== 25000,25102500

bh 3=

12

3bh

I =

36

4h

I =

2

hc =

34

18

36

2

h

M

h

hM

I

Mcs =

==



6~5=N , say 6 for cast iron, dead load

3

18

h

M

N

ss u ==

( )3

2518

6

20

h=


of 131

inh 13.5=

inh

b 71.13

==

say inh4

15= , inb

4

31=

(b) For malleable cast iron, ASTM A47-32 grade 32510

ksisu 52= , ksisy 34=

4~3=N , say 4 for ductile, dead load

3

18

h

M

N

ss u ==

( )3

2518

4

52

h=

inh 26.3=

inh

b 09.13

==

say inh4

33= , inb

4

11=



4=N , based on ultimate strength

3

18

h

M

N

ss u ==

( )3

2518

4

85

h=

inh 77.2=

inh

b 92.03

==

say inh 3= , inb 1=

(d) For AISI C1035, as rolled

ksissu 64=

4=N , kipsRB 833.10=

A

R

N

ss Bsu

s ==

22

242 DDA

ππ=

=

2

2

833.10

4

64

D

ss π==

inD 657.0=


of 131

say inD16

11=

(e) Limit Design

4

2bh

sM y=

For AISI C1035 steel, ksisy 55=

3

hb =

( )4

35525

2h

h

M

==

inh 76.1=

inh

b 59.03

==

say ininh8

71875.1 == , inb

8

5=

21. The same as 20, except that o30=θ . Pin B takes all the horizontal thrust.

Solution:

θcosFFV =

[ ]∑ = 0AM VB FR 133 =

( ) 30cos2500133 =BR

lbRB 9382=

[ ]∑ = 0BM VA FR 103 =

( ) 30cos2500103 =AR

lbRA 7217=

Shear Diagram


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Moment Diagram

( )( ) inkipsinlbM −=−== 65.21650,21102165

3

18

h

Ms =



6~5=N , say 6 for cast iron, dead load

3

18

h

M

N

ss u ==

( )3

65.2118

6

20

h=

inh 89.4=

inh

b 63.13

==

say inh4

15= , inb

4

31=

(b) For malleable cast iron, ASTM A47-32 grade 32510


4~3=N , say 4 for ductile, dead load

3

18

h

M

N

ss u ==

( )3

65.2118

4

52

h=

inh 11.3=

inh

b 04.13

==

say inh 3= , inb 1=





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3

18

h

M

N

ss u ==

( )3

65.2118

4

85

h=

inh 64.2=

inh

b 88.03

==

say inh8

52= , inb

8

7=

(d) For AISI C1035, as rolled

ksissu 64=

4=N , lbRBV 9382=

lbFFR HBH 125030sin2500sin ==== θ

( ) ( )22222 12509382 +=+= BHBVB RRR

lbRB 9465=

A

R

N

ss Bsu

s ==

22

242 DDA

ππ=

=

2

2

465.9

4

64

D

ss π==

inD 614.0=

say inD8

5=

(e) Limit Design

4

2bh

sM y=

For AISI C1035 steel, ksisy 55=

3

hb =

( )4

35565.21

2h

h

M

==

inh 68.1=

inh

b 56.03

==

say ininh8

71875.1 == , inb

8

5=


of 131

22. A cast-iron beam, ASTM 50, as shown, is 30 in. long and supports two gradually

applied, repeated loads (in phase), one of 2000 lb. at ine 10= from the free end,

and one of 1000 lb at the free end. (a) Determine the dimensions of the cross

section if acb 3≈= . (b) The same as (a) except that the top of the tee is below.

Problem 22.

Solution:

For cast iron, ASTM 50

ksisu 50= , ksisuc 164=

For gradually applied, repeated load

8~7=N , say 8

( )edFdFM ++= 21

where:

lbF 20001 =

lbF 10002 =

ind 201030 =−=

ined 30=+

( )( ) ( )( ) inkipsinlbM −=−=+= 70000,70301000202000

I

Mcs =

Solving for I , moment of inertia

( )( ) ( )( ) ( )( ) ( )( )[ ]yaaaaa

aaa

aa 332

53

23 +=

+

2

3ay =


of 131

( )( ) ( )( )( ) ( )( ) ( )( )( )2

173

12

33

12

3 42

3

2

3a

aaaaa

aaaaa

I =+++=

(a)

2

3act =

2

5acc =

Based on tension

I

Mc

N

ss tu

t ==

( )

=

2

17

2

370

8

504

a

a

ina 255.1=

Based on compression

I

Mc

N

ss cuc

c ==

( )

=

2

17

2

570

8

1644

a

a

ina 001.1=

Therefore ina 255.1=

Or say ina4

11=

And ( ) inacb 75.325.133 ====


of 131

Or incb4

33==

(b) If the top of the tee is below

2

5act =

2

3acc =

2

17 4a

I =

inkipsM −= 70

Based on tension

I

Mc

N

ss tu

t ==

( )

=

2

17

2

570

8

504

a

a

ina 488.1=

Based on compression

I

Mc

N

ss cuc

c ==

( )

=

2

17

2

370

8

1644

a

a

ina 845.0=

Therefore ina 488.1=

Or say ina2

11=

And inacb2

143 ===

CHECK PROBLEMS


of 131

23. An I-beam is made of structural steel, AISI C1020, as rolled. It has a depth of 3

in. and is subjected to two loads; 1F and 12 2FF = ; 1F is 5 in. from one end and

2F is 5 in. from the other ends. The beam is 25 in. long; flange width is

inb 509.2= ; 49.2 inI x = . Determine (a) the approximate values of the load to

cause elastic failure, (b) the safe loads for a factor of safety of 3 based on the yield

strength, (c) the safe load allowing for flange buckling (i1.24), (f) the maximum

deflection caused by the safe loads.

Problems 23 – 25.

Solution:

[ ]∑ = 0AM ( ) BRFF 252205 11 =+

18.1 FRB =

[ ]∑ = 0VF BA RRFF +=+ 11 2

111 2.18.13 FFFRA =−=

Shear Diagram

Moment Diagram


of 131

19FM = = maximum moment

For AISI C1020, as rolled

ksisy 48=

(a) I

Mcsy =

where ind

c 5.12

3

2===

( )( )9.2

5.1948 1F

sy ==

kipsF 31.101 =

kipsFF 62.202 12 ==

(b) I

Mc

N

ss

y==

( )( )9.2

5.19

3

48 1Fs ==

kipsF 44.31 =

kipsFF 88.62 12 ==

(c) 1596.9509.2

25<==

b

L (page 34)

ksisc 20= ( page 34, i1.24)

I

Mcsc =

( )( )9.2

5.1920 1F

=

kipsF 30.41 =

kipsFF 60.82 12 ==

(d) For maximum deflection,

by method of superposition, Table AT 2

( ) 2

3

max33

′+′=

bLa

EIL

bFy , ba ′>

or

( ) 2

3

max33

+′=

aLb

EIL

Fay , ab >′


of 131

maxy caused by 1F

( ) 2

3

1111max

331

+′=

aLb

EIL

aFy , 11 ab >′

where ksiE 000,30=

ina 51 =

inb 201 =′

inL 25=

49.2 inI =

( )( )( )( )

( )1

2

3

1max 0022.0

3

52520

259.2000,303

51

FF

y =

+=

maxy caused by 2F

( ) 2

3

2222max

332

′+′=

bLa

EIL

bFy , 22 ba ′>

where inb 52 =′

ina 202 =

( )( )( )( )

( )1

2

3

1max 0043.0

3

52520

259.2000,303

522

FF

y =

+=

Total deflection = δ

111maxmax 0065.00043.0022.021

FFFyy =+=+=δ

Deflection caused by the safe loads in (a)

( ) ina 067.031.100065.0 ==δ

Deflection caused by the safe loads in (b)

( ) inb 022.044.30065.0 ==δ

Deflection caused by the safe loads in (c)

( ) inc 028.030.40065.0 ==δ

24. The same as 23, except that the material is aluminum alloy, 2024-T4, heat treated.

Solution:

For aluminum alloy, 2024-T4, heat treated

ksisy 47=

(a) I

Mcsy =


of 131

( )( )9.2

5.1947 1F

sy ==

kipsF 10.101 =

kipsFF 20.202 12 ==

(b) I

Mc

N

ss

y==

( )( )9.2

5.19

3

47 1Fs ==

kipsF 36.31 =

kipsFF 72.62 12 ==

(c) 1596.9509.2

25<==

b

L (page 34)

ksisc 20= ( page 34, i1.24)

I

Mcsc =

( )( )9.2

5.1920 1F

=

kipsF 30.41 =

kipsFF 60.82 12 ==

(d) Total deflection = δ

111maxmax 0065.00043.0022.021

FFFyy =+=+=δ

Deflection caused by the safe loads in (a)

( ) ina 066.010.100065.0 ==δ

Deflection caused by the safe loads in (b)

( ) inb 022.036.30065.0 ==δ

Deflection caused by the safe loads in (c)

( ) inc 028.030.40065.0 ==δ

25. A light I-beam is 80 in. long, simply supported, and carries a static load at the

midpoint. The cross section has a depth of ind 4= , a flange width of inb 66.2= ,

and 40.6 inI x = (see figure). (a) What load will the beam support if it is made of

C1020, as-rolled steel, and flange buckling (i1.24) is considered? (b) Consider the

stress owing to the weight of the beam, which is 7.7 lb/ft, and decide whether or

not the safe load should be less.


of 131

Solution:

(a) For C1020, as rolled, ksisu 65=

Consider flange buckling

3066.2

80==

b

L

since 4015 <<b

L

( )ksi

b

Lsc 15

1800

301

5.22

18001

5.2222

=

+

=

+

=

I

Mcs =

ind

c 22

4

2===

From Table AT 2

( )F

FFLM 20

4

80

4===

I

Mcss c ==

( )( )6

22015

F=

kipsF 25.2= , safe load

(b) Considering stress owing to the weight of the beam

add’l 8

2wL

M = (Table AT 2)

where ftlbw 7.7=


of 131

add’l ( )

inkipsinlbwL

M −=−=

== 513.0513

8

80

12

7.7

8

22

513.020 += FM = total moment

I

Mcss c ==

( )( )6

2513.02015

+=

F

kipsF 224.2=

Therefore, the safe load should be less.

26. What is the stress in a band-saw blade due to being bent around a 13 ¾-in. pulley?

The blade thickness is 0.0265 in. (Additional stresses arise from the initial tension

and forces of sawing.)

Solution:

int

c 01325.00265.02

===

inr 76325.1301325.075.13 =+=

Using Eq. (1.4) page 11 (Text)

r

Ecs =

where psiE 61030×=

( )( )psis 881,28

76325.13

01325.01030 6

=×

=

27. A cantilever beam of rectangular cross section is tapered so that the depth varies

uniformly from 4 in. at the fixed end to 1 in. at the free end. The width is 2 in. and

the length 30 in. What safe load, acting repeated with minor shock, may be

applied to the free end? The material is AISI C1020, as rolled.

Solution:


ksisu 65= (Table AT 7)

Designing based on ultimate strength,

6=N , for repeated, minor shock load


of 131

ksiN

ss u 8.10

6

65===

Loading Diagram

x

h 1

30

14 −=

−

110.0 += xh

12

3wh

I =

2

hc =

FxM =

( )

( )2223 110.0

33

2

6

12

2

+===

==x

Fx

h

Fx

h

Fx

wh

hFx

I

Mcs

Differentiating with respect to x then equate to zero to solve for x giving maximum

stress.

( ) ( ) ( )( )( )( )

0110.0

10.0110.021110.03

4

2

=

+

+−+=

x

xxxF

dx

ds

( ) 010.02110.0 =−+ xx

inx 10=

( ) inh 211010.0 =+=

2

3

h

Fx

N

ss u ==

( )( )22

1038.10

F=

kipsF 44.1=

TORSIONAL STRESSES

DESIGN PROBLEMS


of 131

28. A centrifugal pump is to be driven by a 15-hp electric motor at 1750 rpm. What

should be the diameter of the pump shaft if it is made of AISI C1045 as rolled?

Consider the load as gradually repeated.

Solution:

For C1045 as rolled,

ksisy 59=

ksisus 72=

Designing based on ultimate strength

N

ss us= , 6=N (Table 1.1)

ksis 126

72==

Torque, ( )

( )kipsinlbinlbft

n

hpT −=−=−=== 540.054045

17502

15000,33

2

000,33

ππ

For diameter,

3

16

d

Ts

π=

( )3

540.01612

dπ=

ind 612.0=

say ind8

5=

29. A shaft in torsion only is to transmit 2500 hp at 570 rpm with medium shocks. Its

material is AISI 1137 steel, annealed. (a) What should be the diameter of a solid

shaft? (b) If the shaft is hollow, io DD 2= , what size is required? (c) What is the

weight per foot of length of each of these shafts? Which is the lighter? By what

percentage? (d) Which shaft is the more rigid? Compute the torsional deflection

of each for a length of 10 ft.

Solution:

( )( )

kipsinlbftn

hpT −=−=== 276036,23

5702

2500000,33

2

000,33

ππ

For AISI 1137, annealed

ksisy 50= (Table AT 8)

ksiss yys 306.0 ==

Designing based on yield strength

3=N for medium shock, one direction


of 131

Design stress

ksiN

ss

ys10

3

30===

(a) Let D = shaft diameter

J

Tcs =

32

4D

Jπ

=

2

Dc =

3

16

D

Ts

π=

( )3

2761610

Dπ=

inD 20.5=

say inD4

15=

(b) ( ) ( )[ ]

32

15

32

2

32

44444

iiiio DDDDDJ

πππ=

−=

−=

iio D

DDc ===

2

2

2

34 15

32

32

15 ii

i

D

T

D

TDs

ππ=

=

( )315

2763210

iDπ=

inDi 66.2=

inDD io 32.52 ==

say

inDi8

52=

inDo4

15=

(c) Density, 3284.0 inlb=ρ (Table AT 7)


of 131

For solid shaft

=w weight per foot of length

( )( ) ftlbDDw 8.7325.5284.0334

12222 ===

= ππρ

πρ

For hollow shaft

( ) ( ) ( ) ( ) ( )[ ] ftlbDDDDw ioio 3.55625.225.5284.0334

12222222 =−=−=−

= ππρ

πρ

Therefore hollow shaft is lighter

Percentage lightness = ( ) %5.33%1003.55

3.558.73=

−

(d) Torsional Deflection

JG

TL=θ

where

inftL 12010 ==

ksiG 3105.11 ×=

For solid shaft, 32

4D

Jπ

=

( )( )

( ) ( )( ) o2.2

180039.0039.0

105.1125.532

120276

34

=

==

×

=

ππθ rad

For hollow shaft, ( )

32

44

io DDJ

−=

π

( )( )

( ) ( )[ ]( )( ) o4.2

180041.0041.0

105.11625.225.532

120276

344

=

==

×−

=

ππθ rad

Therefore, solid shaft is more rigid, oo 4.22.2 <

30. The same as 29, except that the material is AISI 4340, OQT 1200 F.

Solution:

( )( )

kipsinlbftn

hpT −=−=== 276036,23

5702

2500000,33

2

000,33

ππ

For AISI 4340, OQT 1200 F

ksisy 130=

( ) ksiss yys 781306.06.0 ===

Designing based on yield strength


of 131

3=N for mild shock

Design stress

ksiN

ss

ys26

3

78===

(a) Let D = shaft diameter

J

Tcs =

32

4D

Jπ

=

2

Dc =

3

16

D

Ts

π=

( )3

2761626

Dπ=

inD 78.3=

say inD4

33=

(b) ( ) ( )[ ]

32

15

32

2

32

44444

iiiio DDDDDJ

πππ=

−=

−=

iio D

DDc ===

2

2

2

34 15

32

32

15 ii

i

D

T

D

TDs

ππ=

=

( )315

2763226

iDπ=

inDi 93.1=

inDD io 86.32 ==

say

inDi 2=

inDo 4=

(c) Density, 3284.0 inlb=ρ (Table AT 7)


of 131

For solid shaft

=w weight per foot of length

( )( ) ftlbDDw 6.3775.3284.0334

12222 ===

= ππρ

πρ

For hollow shaft

( ) ( ) ( ) ( ) ( )[ ] ftlbDDDDw ioio 1.3224284.0334

12222222 =−=−=−

= ππρ

πρ

Therefore hollow shaft is lighter

Percentage lightness = ( ) %1.17%1001.32

1.326.37=

−

(d) Torsional Deflection

JG

TL=θ

where

inftL 12010 ==

ksiG 3105.11 ×=

For solid shaft, 32

4D

Jπ

=

( )( )

( ) ( )( ) o48.8

180148.0148.0

105.1175.332

120276

34

=

==

×

=

ππθ rad

For hollow shaft, ( )

32

44

io DDJ

−=

π

( )( )

( ) ( )[ ]( )( ) o99.6

180122.0122.0

105.112432

120276

344

=

==

×−

=

ππθ rad

Therefore, hollow shaft is more rigid, oo 48.899.6 < .

31. A steel shaft is transmitting 40 hp at 500 rpm with minor shock. (a) What should

be its diameter if the deflection is not to exceed 1o in D20 ? (b) If deflection is

primary what kind of steel would be satisfactory?

Solution:

(a) ( )

( )kipsinlbft

n

hpT −=−=== 04.5420

5002

40000,33

2

000,33

ππ

ksiG 3105.11 ×=

DL 20=


of 131

rad180

1π

θ == o

JG

TL=θ

( )( )

( )34

105.1132

2004.5

180×

=

D

D

π

π

inD 72.1=

say inD4

31=

(b) ( )

( )ksi

D

Ts 8.4

75.1

04.5161633

===ππ

Based on yield strength

3=N

( )( ) ksiNssys 4.148.43 ===

ksis

sys

y 246.0

4.14

6.0===

Use C1117 normalized steel ksisy 35=

32. A square shaft of cold-finish AISI 1118 transmits a torsional moment of 1200 in-

lb. For medium shock, what should be its size?

Solution:

For AISI 1118 cold-finish

ksisy 75=

ksiss yys 456.0 ==

3=N for medium shock

Z

T

N

ss

ys

′==

where, bh =

9

2

9

2 32bhb

Z ==′ (Table AT 1)

kipsinlbinT −=−= 2.11200

( )32

92.1

3

45

bs ==

inhb 71.0==

say inhb4

3==


of 131

CHECK PROBLEMS

33. A punch press is designed to exert a force sufficient to shear a 15/16-in. hole in a

½-in. steel plate, AISI C1020, as rolled. This force is exerted on the shaft at a

radius of ¾-in. (a) Compute the torsional stress in the 3.5-in. shaft (bending

neglected). (b) What will be the corresponding design factor if the shaft is made

of cold-rolled AISI 1035 steel (Table AT 10)? Considering the shock loading that

is characteristics of this machine, do you thick the design is safe enough?

Solution:


ksisus 49=

( )DtsF us π=

where inD16

15=

int2

1=

( ) kipsF 2.722

1

16

1549 =

= π

FrT =

where inr4

3=

( ) kipsinT −=

= 2.54

4

32.72

(a) 3

16

d

Ts

π=

where ind 5.3=

( )( )

ksis 44.65.3

2.54163

==π

(b) For AISI 1035 steel, ksisus 64=

for shock loading, traditional factor of safety, 15~10=N

Design factor , 94.944.6

64===

s

sN us , the design is safe ( 10≈N )

34. The same as 33, except that the shaft diameter is 2 ¾ in.

Solution:


of 131

ind 75.2=

(a) 3

16

d

Ts

π=

( )( )

ksis 3.1375.2

2.54163

==π

(b) For AISI 1035 steel, ksisus 64=

for shock loading, traditional factor of safety, 15~10=N

Design factor , 8.43.13

64===

s

sN us , the design is not safe ( 10<N )

35. A hollow annealed Monel propeller shaft has an external diameter of 13 ½ in. and

an internal diameter of 6 ½ in.; it transmits 10,000 hp at 200 rpm. (a) Compute the

torsional stress in the shaft (stress from bending and propeller thrust are not

considered). (b) Compute the factor of safety. Does it look risky?

Solution:

For Monel shaft,

ksisus 98= (Table AT 3)

4~3=N , for dead load, based on ultimate strength

(a) J

Tcs =

( ) ( ) ( )[ ] 4

4444

308632

5.65.13

32in

DDJ io =

−=

−=

ππ

inD

c o 75.62

5.13

2===

( )( )

kipsinlbftn

hpT −=−=== 3152606,262

2002

000,10000,33

2

000,33

ππ

( )( )ksis 9.6

3086

75.63152==

(b) Factor of safety,

2.149.6

98===

s

sN us , not risky


of 131

STRESS ANALYSIS

DESIGN PROBLEMS

36. A hook is attached to a plate as shown and supports a static load of 12,000 lb. The

material is to be AISI C1020, as rolled. (a) Set up strength equations for

dimensions d , D , h , and t . Assume that the bending in the plate is negligible.

(b) Determine the minimum permissible value of these dimensions. In estimating

the strength of the nut, let dD 2.11 = . (c) Choose standard fractional dimensions

which you think would be satisfactory.

Problems 36 – 38.

Solution:

s = axial stress

ss = shear stress

(a)

22

4

4

1 d

F

d

Fs

ππ==

Equation (1) s

Fd

π

4=


of 131

( ) ( ) ( )[ ] ( )22222

1

22

1

2 44.1

4

2.1

44

4

1 dD

F

dD

F

DD

F

DD

Fs

−=

−=

−=

−

=ππππ

Equation (2) 244.14

ds

FD +=

π

dh

F

hD

Fss

ππ 2.11

==

Equation (3) sds

Fh

π2.1=

Dt

Fss

π=

Equation (4) sDs

Ft

π=

(b) Designing based on ultimate strength,

Table AT 7, AISI C1020, as rolled

ksisu 65=

ksisus 49=

4~3=N say 4, design factor for static load

ksiN

ss u 16

4

65===

ksiN

ss us

s 124

49===

kipslbF 12000,12 ==

From Equation (1)

( )( )

ins

Fd 98.0

16

1244===

ππ

From Equation (2)

( )( )

( ) inds

FD 53.198.044.1

16

12444.1

4 22 =+=+=ππ

From Equation (3)

( )( )in

ds

Fh

s

27.01298.02.1

12

2.1===

ππ

From Equation (4)

( )( )in

Ds

Ft

s

21.01253.1

12===

ππ


of 131

(c) Standard fractional dimensions

ind 1=

inD2

11=

inh4

1=

int4

1=

37. The same as 36, except that a shock load of 4000 lb. is repeatedly applied.

Solution:

(a) Same as 36.

(b) 15~10=N for shock load, based on ultimate strength

say 15=N , others the same.

ksiN

ss u 4

15

65===

ksiN

ss us

s 315

49===

kipslbF 44000 ==

From Equation (1)

( )( )

ins

Fd 13.1

4

444===

ππ

From Equation (2)

( )( )

( ) inds

FD 76.113.144.1

4

4444.1

4 22 =+=+=ππ

From Equation (3)

( )( )in

ds

Fh

s

31.0313.12.1

4

2.1===

ππ

From Equation (4)

( )( )in

Ds

Ft

s

24.0376.1

4===

ππ


of 131

(c) Standard fractional dimensions

ind8

11=

inD4

31=

inh8

3=

int4

1=

38. The connection between the plate and hook, as shown, is to support a load F .

Determine the value of dimensions D , h , and t in terms of d if the connection

is to be as strong as the rod of diameter d . Assume that dD 2.11 = , uus ss 75.0= ,

and that bending in the plate is negligible.

Solution:

2

4

1d

Fs

π=

sdF2

4

1π=

(1)

=

N

sdF u2

4

1π


of 131

( ) ( )222

1

2 44.14

1

4

1dD

F

DD

Fs

−

=

−

=

ππ

( )sdDF22 44.1

4

1−= π

(2) ( )

−=

N

sdDF u22 44.1

4

1π

dh

F

hD

Fss

ππ 2.11

==

sdhsF π2.1=

=

=

N

sdh

N

sdhF uus 75.0

2.12.1 ππ

(3)

=

N

sdhF u5

9.0 π

Dt

Fss

π=

sDtsF π=

=

=

N

sDt

N

sDtF uus 75.0

ππ

(4)

=

N

sDtF uπ75.0

Equate (2) and (1)

( )

=

−=

N

sd

N

sdDF uu 222

4

144.1

4

1ππ

22 44.2 dD =

dD 562.1=

Equate (3) and (1)

=

=

N

sd

N

sdhF uu 2

4

19.0 ππ

( )d

dh 278.0

9.04==

Equate (4) and (1)

=

=

N

sd

N

sDtF uu 2

4

175.0 ππ

( )( )

=

=

N

sd

N

stdF uu 2

4

1562.175.0 ππ

( )( )d

dt 214.0

562.175.04==


of 131

39. (a) For the connection shown, set up strength equations representing the various

methods by which it might fail. Neglect bending effects. (b) Design this

connection for a load of 2500 lb. Both plates and rivets are of AISI C1020, as

rolled. The load is repeated and reversed with mild shock. Make the connection

equally strong on the basis of yield strengths in tension, shear, and compression.

Problems 39, 40

Solution:

(a)

=

2

4

15 D

Fss

π

Equation (1) ss

FD

π5

4=

( )Dbt

Fs

2−=

Equation (2) Dts

Fb 2+=

Dt

Fs

5=

Equation (3) Ds

Ft

5=

(b) For AISI C1020, as rolled


ksiss yys 286.0 ==

4=N for repeated and reversed load (mild shock) based on yield strength

ksis 124

48==

ksiss 74

28==

From Equation (1)


of 131

ss

FD

π5

4=

where

kipslbF 5.22500 ==

( )( )

ins

FD

s

30.075

5.24

5

4===

ππ say in

16

5

From Equation (3)

( )in

Ds

Ft 13.0

1216

55

5.2

5=

== say in

32

5

From Equation (2)

( )inD

ts

Fb 96.1

16

52

1232

5

5.22 =

+

=+= say in2

40. The same as 39, except that the material is 2024-T4, aluminum alloy.

Solution:

(a) Same as 39.

(b) ) For 2024-T4, aluminum alloy


ksiss yys 2555.0 ==

4=N for repeated and reversed load (mild shock) based on yield strength

ksis 124

47==

ksiss 64

25==

From Equation (1)

ss

FD

π5

4=

where

kipslbF 5.22500 ==

( )( )

ins

FD

s

33.065

5.24

5

4===

ππ say in

8

3

From Equation (3)


of 131

( )in

Ds

Ft 11.0

128

35

5.2

5=

== say in

8

1

From Equation (2)

( )inD

ts

Fb 42.2

8

32

128

1

5.22 =

+

=+= say in

2

12

41. (a) For the connection shown, set up strength equations representing the various

methods by which it might fail. (b) Design this connection for a load of 8000 lb.

Use AISI C1015, as rolled, for the rivets, and AISI C1020, as rolled, for the

plates. Let the load be repeatedly applied with minor shock in one direction and

make the connection equally strong on the basis of ultimate strengths in tension,

shear, and compression.

Problem 41.

Solution:

(a)

( )Dbt

FsP

−= or

( )Dbt

F

sP2

4

3

−= Equation (1)

( )24

14 2

=

D

FssR

π

Equation (2)


of 131

Dt

FsR

4= Equation (3)

(b) For AISI C1015, as rolled

ksisuR 61= , ksiss uRusR 4575.0 ==


ksisuP 65=


ksiN

ss uP

P 8.106

65===

ksiN

ss uR

R 1.106

61===

ksiN

ss usR

sR 5.76

45===

kipslbF 88000 ==

Solving for D

22 D

FssR

π=

( )in

s

FD

sR

412.05.72

8

2===

ππ say in

16

7

Solving for t

Dt

FsR

4=

( )in

Ds

Ft

R

453.0

1.1016

74

8

4=

== say in

2

1

Solving for b

Using ( )Dbt

FsP

−=

( )inD

ts

Fb

P

92.116

7

8.102

1

8=+

=+= say in2

Using ( )Dbt

F

sP2

4

3

−=


of 131

( )

( )inD

ts

Fb

P

99.116

72

8.102

14

832

4

3=

+

=+= say in2

Therefore

inb 2=

inD16

7=

int2

1=

42. Give the strength equations for the connection shown, including that for the shear

of the plate by the cotter.

Problems 42 – 44.

Solution:

Axial Stresses

2

12

1

4

4

1 D

F

D

Fs

ππ== Equation (1)

( )eDL

Fs

2−= Equation (2)


of 131

eD

Fs

2

= Equation (3)

( ) ( )2

2

22

2

2

4

4

1 Da

F

Da

Fs

−=

−

=ππ

Equation (4)

eDD

F

eDD

Fs

2

2

22

2

2

4

4

4

1 −=

−

=ππ

Equation (5)

Shear Stresses

eb

Fss

2= Equation (6)

( )teDL

Fss

+−=

22 Equation (7)


of 131

at

Fss

π= Equation (8)

mD

Fss

1π= Equation (9)

hD

Fss

22= Equation (10)

43. A steel rod, as-rolled AISI C1035, is fastened to a 7/8-in., as-rolled C1020 plate

by means of a cotter that is made of as-rolled C1020, in the manner shown. (a)

Determine all dimensions of this joint if it is to withstand a reversed shock load

kipsF 10= , basing the design on yield strengths. (b) If all fits are free-running

fits, decide upon tolerances and allowances.

Solution: (See figure of Prob. 42)

inint 875.08

7== , ysy ss 6.0=

For steel rod, AISI C1035, as rolled

ksisy 551

=

ksissy 331

=

For plate and cotter, AISI C1020, as rolled

ksisy 482

=

ksissy 282

=

7~5=N based on yield strength

say 7=N

From Equation (1) (Prob. 42)

2

1

41

D

F

N

ss

y

π==

( )2

1

104

7

55

Dπ=

inD 27.11 =

say inD4

111 =


of 131

From Equation (9)

mD

F

N

ss

sy

s

1

1

π==

m

=

4

11

10

7

33

π

inm 54.0=

say inm16

9=

From Equation (3)

eD

F

N

ss

y

2

1 ==

eDs

2

10

7

55==

273.12 =eD

From Equation (5)

eDD

F

N

ss

y

2

2

2 4

41

−==

π

( )( )273.14

104

7

552

2 −=

Dπ

inD 80.12 =

say inD4

312 =

and 273.12 =eD

273.14

31 =

e

ine 73.0=

say ine4

3=

By further adjustment

Say inD 22 = , ine8

5=

From Equation (8)

at

F

N

ss

sy

sπ

== 2

( )875.0

10

7

28

aπ=

ina 91.0=

say ina 1=


of 131

From Equation (4)

( )2

2

2

42

Da

F

N

ss

y

−==

π

( )( )22 2

104

7

48

−=

aπ

ina 42.2=

say ina2

12=

use ina2

12=

From Equation (7)

( )teDL

F

N

ss

sy

s+−

==22

2

( )875.08

522

10

7

28

+−

=

L

inL 80.2=

say inL 3=

From Equation (6)

eb

F

N

ss

sy

s2

2 ==

b

=

8

52

10

7

28

inb 2=

From Equation (10)

hD

F

N

ss

sy

s

22

2 ==

( )h22

10

7

28=

ininh8

5625.0 ==

Summary of Dimensions

inL 3=

inh8

5=

inb 2=

int8

7=


of 131

inm16

9=

ina2

12=

inD4

111 =

inD 22 =

ine8

5=

(b) Tolerances and allowances, No fit, tolerance = in010.0±

inL 010.03±=

inh 010.0625.0 ±=

int 010.0875.0 ±=

inm 010.05625.0 ±=

ina 010.0500.2 ±=

inD 010.025.11 ±=

For Free Running Fits (RC 7) Table 3.1

Female Male

inb0000.0

0030.00.2

−

+= inb

0058.0

0040.00.2

−

−=

allowance = 0.0040 in

inD0000.0

0030.00.22

−

+= inD

0058.0

0040.00.22

−

−=


ine0000.0

0016.0625.0

−

+= ine

0030.0

0020.0625.0

−

−=


44. A 1-in. ( 1D ) steel rod (as-rolled AISI C1035) is to be anchored to a 1-in. steel

plate (as-rolled C1020) by means of a cotter (as rolled C1035) as shown. (a)

Determine all the dimensions for this connection so that all parts have the same

ultimate strength as the rod. The load F reverses direction. (b) Decide upon

tolerances and allowances for loose-running fits.

Solution: (Refer to Prob. 42)

(a) For AISI C1035, as rolled

ksisu 851

=

ksisus 641

=



of 131

ksisu 652

=

ksisus 482

=

Ultimate strength

Use Equation (1)

( ) ( ) kipsDsF uu 8.6614

185

4

1 22

11=

=

= ππ

Equation (9)

mDsF usu 11π=

( )( )( )m1648.66 π=

inm 33.0=

say inm8

3=

From Equation (3)

eDsF uu 21=

( ) eD2858.66 =

7859.02 =eD

From Equation (5)

−= eDDsF uu 2

2

24

11

π

( )

−= 7859.0

4

1858.66 2

2Dπ

inD 42.12 =

say inD8

312 =

7859.08

312 =

= eeD

ine 57.0=

say ine16

9=

From Equation (4)

( )

−= 2

2

2

4

12

DasF uu π

( )

−

=

2

2

8

31

4

1658.66 aπ

ina 79.1=

say ina4

31=

From Equation (8)


of 131

atsF usu π2

=

( )( )( )( )1488.66 aπ=

ina 44.0=

say ina2

1=

use ina4

31=

From Equation (2)

( )eDLsF uu 22−=

( )

−=

16

9

8

31658.66 L

inL 20.3=

say inL4

13=

From Equation (7)

( )teDLsF usu −−= 222

( ) ( )116

9

8

314828.66

−−= L

inL 51.1=

say inL2

11=

use inL4

13=

From Equation (6)

ebsF usu 12=

( ) b

=

16

96428.66

inb 93.0=

say inb 1=

From Equation (10)

hDsF usu 212=

( ) h

=

8

316428.66

inh 38.0=

say inh8

3=

Dimensions

inL4

13=


of 131

inh8

3=

inb 1=

int 1=

inm8

3=

ina4

31=

inD 11 =

inD8

312 =

ine16

9=

(b) Tolerances and allowances, No fit, tolerance = in010.0±

inL 010.025.3 ±=

inh 010.0375.0 ±=

int 010.0000.1 ±=

inm 010.0375.0 ±=

ina 010.075.1 ±=

inD 010.0000.11 ±=

For Loose Running Fits (RC 8) Table 3.1

Female Male

inb0000.0

0035.00.1

−

+= inb

0065.0

0045.00.1

−

−=


inD0000.0

0040.0375.12

−

+= inD

0075.0

0050.0375.12

−

−=


ine0000.0

0028.05625.0

−

+= ine

0051.0

0035.05625.0

−

−=


45. Give all the simple strength equations for the connection shown. (b) Determine

the ratio of the dimensions a , b , c , d , m , and n to the dimension D so that the

connection will be equally strong in tension, shear, and compression. Base the

calculations on ultimate strengths and assume uus ss 75.0= .


of 131

Problems 45 – 47.

Solution:

(a) Neglecting bending

Equation (1):

= 2

4

1DsF π

Equation (2):

= 2

4

12 csF s π

Equation (3): ( )bcsF 2=

Equation (4): ( )acsF =

Equation (5): ( )[ ]bcdsF −= 2

Equation (6): ( )mbsF s 4=

Equation (7): ( )nbsF s 2=

Equation (8): ( )acdsF −=

(b) N

ss u= and

N

ss us

s =

Therefore

sss 75.0=

Equate (2) and (1)

=

= 22

4

1

4

12 DscsF s ππ

=

22

4

1

2

175.0 Dscs

Dc 8165.0=

Equate (3) and (1)

( )

== 2

4

12 DsbcsF π

( ) 2

4

18165.02 DDb π=

Db 4810.0=


of 131

Equate (4) and (1)

== 2

4

1DssacF π

( ) 2

4

18165.0 DDa π=

Da 9619.0=

Equate (5) and (1)

( )[ ]

=−= 2

4

12 DsbcdsF π

( )( ) 2

4

14810.08165.02 DDd π=−

Dd 6329.1=

Equate (6) and (1)

( )

== 2

4

14 DsmbsF s π

( )( ) 2

4

14810.0475.0 DDm π=

Dm 5443.0=

Equate (7) and (1)

( )

== 2

4

12 DsnbsF s π

( )( ) 2

4

14810.0275.0 DDn π=

Dn 0886.1=

Equate (8) and (1)

( )

=−= 2

4

1DsacdsF π

( ) 2

4

18165.06329.1 DaDD π=−−

Da 9620.0=

Summary

Da 9620.0=

Db 4810.0=

Dc 8165.0=

Dd 6329.1=

Dm 5443.0=

Dn 0886.1=

46. The same as 45, except that the calculations are to be based on yield strengths. Let

ysy ss 6.0= .


of 131

Solution: (Refer to Prob. 45)

(a) Neglecting bending

Equation (1):

= 2

4

1DsF π

Equation (2):

= 2

4

12 csF s π

Equation (3): ( )bcsF 2=

Equation (4): ( )acsF =

Equation (5): ( )[ ]bcdsF −= 2

Equation (6): ( )mbsF s 4=

Equation (7): ( )nbsF s 2=

Equation (8): ( )acdsF −=

(b) N

ss

y= and

N

ss

sy

s =

Therefore

sss 6.0=

Equate (2) and (1)

=

= 22

4

1

4

12 DscsF s ππ

=

22

4

1

2

16.0 Dscs

Dc 9129.0=

Equate (3) and (1)

( )

== 2

4

12 DsbcsF π

( ) 2

4

19129.02 DDb π=

Db 4302.0=

Equate (4) and (1)

== 2

4

1DssacF π

( ) 2

4

19129.0 DDa π=

Da 8603.0=

Equate (5) and (1)

( )[ ]

=−= 2

4

12 DsbcdsF π

SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS

Page 1 of 62

VARYING STRESSES – NO CONCENTRATION

DESIGN PROBLEMS

141. The maximum pressure of air in a 20-in. cylinder (double-acting air compressor)

is 125 psig. What should be the diameter of the piston rod if it is made of AISI

3140, OQT at 1000 F, and if there are no stress raisers and no column action? Let

75.1=N ; indefinite life desired. How does your answer compare with that

obtained for 4?

Solution:


ksisu 153=

ksisy 134=

( ) ksiss un 5.761535.05.0 ===

For axial loading, with size factor

( )( )( ) ksiss un 525.7685.08.05.0 ===

Soderberg line

n

a

y

m

s

s

s

s

N+=

1

For double-acting

( ) ( ) kipslbpAFF 27.39270,39204

1252

max ==

===

π

kipsFF 27.39min −=−=

0=ms

( )222

5027.3944

ddd

Fsa ===

ππ

52

50

075.1

11 2

+==d

N

ind 2972.1=

say ind16

51=

comparative to Problem 4.

142. A link as shown is to be made of AISI 2330, WQT 1000 F. The load kipsF 5=

is repeated and reversed. For the time being, ignore stress concentrations. (a) If

its surface is machined, what should be its diameter for 40.1=N . (b) The same

as (a), except that the surface is mirror polished. What would be the percentage

saving in weight? (c) The same as (a), except that the surface is as forged.


Page 2 of 62

Prob. 142 – 144

Solution:

For AISI 2330, WQT 1000 F

ksisu 105=

ksisy 85=

( ) ksiss un 5.521055.05.0 ===

0=ms

( )222

20544

ddd

Fsa

πππ===

Soderberg line

n

a

y

m

s

s

s

s

N+=

1

n

a

s

s

N+= 0

1

N

ss n

a =

Size factor = 0.85

Factor for axial loading = 0.80

(a) Machined surface

Surface factor = 0.85 (Fig. AF 5)

( )( )( )( ) ksiksiss un 345.305.5285.085.080.05.0 ===

4.1

345.30202

==D

saπ

inD 542.0=

say inD16

9=

(b) Mirror polished surface


( )( )( )( ) ksiksiss un 7.355.5200.185.080.05.0 ===

4.1

7.35202

==D

saπ


Page 3 of 62

inD 5.0=

Savings in weight = ( ) %21%100

16

9

2

1

16

9

2

22

=

−

(c) As forged surface


( )( )( )( ) ksiksiss un 28.145.5240.085.080.05.0 ===

4.1

28.14202

==D

saπ

inD 79.0=

say inD4

3=

143. The same as 142, except that, because of a corrosive environment, the link is

made from cold-drawn silicon bronze B and the number of reversals of the load

is expected to be less than 3 x 107.

Solution:

For cold-drawn silicon bronze, Type B.

ksisn 30= at 3 x 108

ksisy 69=

ksisu 75.93=

ns at 3 x 107 ( ) ksi5.36

103

10330

085.0

7

8

=

×

×=

( )( )( ) ksisn 82.245.3685.080.0 ==

4.1

82.24202

==D

saπ

inD 60.0=

say inD8

5=

144. The same as 142, except that the link is made of aluminum alloy 2024-T4 with a

minimum life of 107 cycles.

Solution:

For AA 2024-T4

ksisy 47=

ksisu 68=

ksisn 20= at 5 x108


Page 4 of 62

ns at 107 ( ) ksi9.27

10

10520

085.0

7

8

=

×

( )( )( ) ksisn 199.2785.080.0 ==

4.1

19202

==D

saπ

inD 685.0=

say inD16

11=

145. A shaft supported as a simple beam, 18 in. long, is made of carburized AISI 3120

steel (Table AT 10). With the shaft rotating, a steady load of 2000 lb. is appliled

midway between the bearings. The surfaces are ground. Indefinite life is desired

with 6.1=N based on endurance strength. What should be its diameter if there

are no surface discontinuities?

Solution:

For AISI 3120 steel, carburized

ksisn 90=

ksisy 100=

ksisu 141=

Size Factor = 0.85

Surface factor (ground) = 0.88

( )( )( ) ksisn 32.679088.085.0 ==

0=ms

3

32

D

Msa

π=

( )( )kipsinlbin

FLM −=−=== 0.99000

4

182000

4

Soderberg line

n

a

y

m

s

s

s

s

N+=

1


Page 5 of 62

n

a

s

s

N+= 0

1

N

ss n

a =

( )6.1

32.679323

=Dπ

inD 2964.1=

say inD4

11=

146. (a) A lever as shown with a rectangular section is to be designed for indefinite

life and a reversed load of lbF 900= . Find the dimensions of a section without

discontinuity where tb 8.2= and inL 14= . for a design factor of 2=N . The

material is AISI C1020, as rolled, with an as-forged surface. (b) compute the

dimensions at a section where ine 4= .

Problems 146, 147

Solution:


ksisu 65=

ksisy 48=

ksiss un 5.325.0 ==

Surface factor (as forged) = 0.55

(a) 0=ms

I

Mcsa =

( ) 4

33

8293.112

8.2

12t

tttbI ===

ttb

c 4.12

8.2

2===

( )( ) kipsinlbinFLM −=−=== 6.12600,1214900

( )( )34

643.9

8293.1

4.16.12

tt

tsa ==

( )( )( ) ksisn 20.155.3255.085.0 ==


Page 6 of 62

Soderberg line

n

a

y

m

s

s

s

s

N+=

1

n

a

s

s

N+= 0

1

N

ss n

a =

2

20.15643.93

=t

int 08.1=

( ) intb 0.308.18.28.2 ===

say int16

11= , inb 0.3=

(b) ( )( ) kipsinlbinFeM −=−=== 6.3600,34900

( )( )34

755.2

18293

4.16.3

tt

tsa ==

2

20.15755.23

=t

int 713.0=

( ) intb 996.1713.08.28.2 ===

say int32

23= , inb 2=

147. The same as 146, except that the reversal of the load are not expected to exceed

105 (Table AT 10).

Solution:

ksisn 5.32=

ns at 105 ( ) ksi5.39

10

105.32

085.0

5

6

=

=

( )( )( ) ksisn 5.185.3955.085.0 ==

(a) N

ss n

a =

2

5.18643.93

=t


Page 7 of 62

int 014.1=

( ) intb 839.2014.18.28.2 ===

say int 1= , inb16

132=

(b) N

ss n

a =

2

5.18755.23

=t

int 6678.0=

( ) intb 870.16678.08.28.2 ===

say int16

11= , inb

8

71=

148. A shaft is to be subjected to a maximum reversed torque of 15,000 in-lb. It is

machined from AISI 3140 steel, OQT 1000 F (Fig. AF 2). What should be its

diameter for 75.1=N ?

Solution:

For AISI 3140 steel, OQT 1000 F

ksisu 152=

ksisy 134=

ksiss un 765.0 ==

For machined surface,

Surface factor = 0.78

Size factor = 0.85

( )( )( )( ) ksisns 3.5313478.085.06.0 ==

( ) ksiss yys 4.801346.06.0 ===

ns

as

ys

ms

s

s

s

s

N+=

1

0=mss

3

16

D

Tsas

π=

kipsinT −= 15

( )33

2401516

DDsas

ππ==

ns

as

s

s

N+= 0

1

N

ss ns

as =


Page 8 of 62

75.1

3.532403

=Dπ

inD 3587.1=

say inD8

31=

149. The same as 148, except that the shaft is hollow with the outside diameter twice

the inside diameter.

Solution:

io DD 2=

( )( )( )

( )[ ] 34444

32

2

2151616

iii

i

io

oas

DDD

D

DD

TDs

πππ=

−=

−=

N

ss ns

as =

75.1

3.53323

=iDπ

inDi 694.0=

say inDi16

11= , inDo

8

31=

150. The link shown is machined from AISI 1035 steel, as rolled, and subjected to a

repeated tensile load that varies from zero to 10 kips; bh 2= . (a) Determine these

dimensions for 40.1=N (Soderberg) at a section without stress concentration.

(b) How much would these dimensions be decreased if the surfaces of the link

were mirror polished?

Problems 150, 151, 158.

Solution:

For AISI 1035, steel as rolled

ksisu 85=

ksisy 55=

ksiss un 5.425.0 ==


Page 9 of 62

( ) kipsFm 50102

1=+=

( ) kipsFa 50102

1=−=

22 3

10

5.1

5

bbbh

Fs m

m ===

22 3

10

5.1

5

bbbh

Fs a

a ===

(a) Soderberg line

n

a

y

m

s

s

s

s

N+=

1

For machined surface,

Factor = 0.88

Size factor = 0.85

( )( )( )( ) ksisn 4.255.4288.085.080.0 ==

( ) ( )4.253

10

553

10

40.1

122

bb+=

inb 5182.0=

say inb16

9=

inbh32

275.1 ==

(b) Mirror polished,

Factor = 1.00

Size factor = 0.85

( )( )( )( ) ksisn 9.285.4200.185.080.0 ==

( ) ( )9.283

10

553

10

40.1

122

bb+=

inb 4963.0=

say inb2

1=

inbh4

35.1 ==

151. The same as 150, except that the link operates in brine solution. (Note: The

corroding effect of the solution takes precedence over surface finish.)


Page 10 of 62

Solution:

Table AT 10, in brine, AISI 1035,

ksisn 6.24=

ksisy 58=

( )( )( ) ksisn 73.166.2485.080.0 ==

( ) ( )73.163

10

553

10

40.1

122

bb+=

inb 60.0=

say inb8

5=

inbh16

155.1 ==

152. The simple beam shown, 30-in. long ( dLa ++= ), is made of AISI C1022 steel,

as rolled, left a forged. At ina 10= , .30001 lbF = is a dead load. At

ind 10= , .24002 lbF = is repeated, reversed load. For 5.1=N , indefinite life,

and bh 3= , determine b and h . (Ignore stress concentration).

Problem 152, 153

Solution:


ksisu 72=

ksisy 52=

ksiss un 365.0 ==

For as forged surface

Figure AF 5, factor = 0.52

Size factor = 0.85

( )( )( ) ksisn 163652.085.0 ==

Loading:


Page 11 of 62

∑ = 0AM

( ) ( ) 230240020300010 R=+

lbR 26002 =

∑ = 0VF

2121 FFRR +=+

2400300026001 +=+R

lbR 28001 =

Shear Diagram

( )( ) kipsinlbinMC −=−== 28000,28102800

1

( )( ) kipsinlbinM D −=−== 26000,261026001

Then

Loading

∑ = 0AM

( ) ( )24002030300010 2 =+ R

lbR 6002 =

∑ = 0VF

2121 RFFR +=+

600300024001 +=+R

lbR 12001 =


Page 12 of 62

Shear Diagram

( )( ) kipsinlbinMC −=−== 12000,121012002

( )( ) kipsinlbinM D −=−== 6000,6106002

Then using

kipsinMM C −== 281max

kipsinMM C −== 122min

( ) ( ) kipsinMMMm −=+=+= 2012282

1

2

1minmax

( ) ( ) kipsinMMM a −=−=−= 812282

1

2

1minmax

I

cMs m

m = , I

cMs a

a =

( ) 4

33

25.212

3

12b

bbbhI ===

bh

c 5.12

==

35.1 b

Ms m

m = ,35.1 b

Ms a

a =

n

a

y

m

s

s

s

s

N+=

1

16

5.1

8

52

5.1

20

5.1

1 33

+

=bb

inb 96.0=

say inb 1=

inbh 33 ==

153. The same as 152, except that the cycles of 2F will not exceed 100,000 and all

surfaces are machined.

Solution:


Page 13 of 62

ns at 105 cycles ( ) ksi8.43

10

1036

085.0

5

6

=

=

ksisu 72=

Machined surface, factor = 0.90

( )( )( ) ksisn 5.338.4390.085.0 ==

5.33

5.1

8

52

5.1

20

5.1

1 33

+

=bb

inb 8543.0=

say inb8

7=

inbh8

523 ==

154. A round shaft, made of cold-finished AISI 1020 steel, is subjected to a variable

torque whose maximum value is 6283 in-lb. For 5.1=N on the Soderberg

criterion, determine the diameter if (a) the torque is reversed, (b) the torque varies

from zero to a maximum, (c) the torque varies from 3141 in-lb to maximum.

Solution:

For AISI 1020, cold-finished

ksisu 78=

ksisy 66=

ksiss un 395.0 ==

size factor = 0.85

( )( )( ) ksisns 203985.06.0 ==

( ) ksiss yys 40666.06.0 ===

ns

as

ys

ms

s

s

s

s

N+=

1

(a) Reversed torque

0=mss

3

16

D

Tsas

π=

lbinT −= 6283

( )ksi

Dpsi

DDsas 333

32000,32628316===

π

ns

as

s

s

N+= 0

1


Page 14 of 62

20

32

05.1

1 3

+=D

inD 34.1=

say inD8

31=

(b) 0min =T , lbinT −= 6283max

( ) lbinTm −== 314162832

1

( ) lbinTa −== 314162832

1

( )ksi

Dpsi

DDsms 333

16000,16314116===

π

( )ksi

Dpsi

DDsas 333

16000,16314116===

π

20

16

40

16

5.1

1 33

+

=DD

inD 22.1=

say inD4

11=

(c) lbinT −= 3141min , lbinT −= 6283max

( ) lbinTm −=+= 4712314162832

1

( ) lbinTa −=−= 1571314162832

1

( )ksi

Dpsi

DDsms 333

24000,24471216===

π

( )ksi

Dpsi

DDsas 333

8000,8157116===

π

20

8

40

24

5.1

1 33

+

=DD

inD 145.1=

say inD32

51=


Page 15 of 62

CHECK PROBLEMS

155. A simple beam 2 ft. long is made of AISI C1045 steel, as rolled. The dimensions

of the beam, which is set on edge, are 1 in. x 3 in. At the midpoint is a repeated,

reversed load of 4000 lb. What is the factor of safety?

Solution:


ksisu 96=

ksisy 59=

( ) ksiss un 48965.05.0 ===

size factor = 0.85

( )( ) ksisn 8.404885.0 ==

n

a

y

m

s

s

s

s

N+=

1

0=ms

2

6

bh

Msa =

inh 3=

inb 1=

( )( )kipsinlbin

FLM −=−=== 24000,24

4

244000

4

( )( )( )

ksisa 1631

2462

==

8.40

160

1+=

N

55.2=N

156. The same as 155, except that the material is normalized and tempered cast steel,

SAE 080.

Solution:

Table AT 6

ksisn 35=′

ksisy 40=

( )( ) ksisn 75.293585.0 ==

75.29

160

1+=

N

86.1=N

157. A 1 ½-in. shaft is made of AISI 1045 steel, as rolled. For 2=N , what repeated

and reversed torque can the shaft sustain indefinitely?


Page 16 of 62

Solution:

For AISI 1045, as rolled

ksisu 96=

ksisy 59=

( ) ksiss un 48965.05.0 ===′

( )( )( ) ksisns 48.244885.06.0 ==

( )( ) ksiss yys 4.35596.06.0 ===

ns

as

ys

ms

s

s

s

s

N+=

1

0=mss

48.240

2

1 ass+=

ksisas 24.12=

24.1216

3==

D

Tsas

π

kipsinT −= 8

VARIABLE STRESSES WITH STRESS CONCENTRATIONS

DESIGN PROBLEMS

158. The load on the link shown (150) is a maximum of 10 kips, repeated and

reversed. The link is forged from AISI C020, as rolled, and it has a ¼ in-hole

drilled on the center line of the wide side. Let bh 2= and 5.1=N . Determine b

and h at the hole (no column action) (a) for indefinite life, (b) for 50,000

repetitions (no reversal) of the maximum load, (c) for indefinite life but with a

ground and polished surface. In this case, compute the maximum stress.

Solution:


ksisu 65=

ksisy 48=

( ) ksiss un 5.32655.05.0 ===



Size factor = 0.85

( )( )( )( ) ksisn 2.125.3255.085.080.0 ==

n

af

y

m

s

sK

s

s

N+=

1


Page 17 of 62

Fig. AF 8, 1>hb

Assume 5.3=tK

Figure AF 7, inind

r 125.08

1

2===

ina 01.0=

926.0

125.0

01.01

1

1

1=

+

=

+

=

r

aq

( ) ( ) 3.3115.3926.011 =+−=+−= tf KqK

0=ms

( ) ( )25.02

10

−=

−=

bbdhb

Fsa

(a) n

af

s

sK

N+= 0

1

( )( )( )( )2.1225.02

103.30

5.1

1

−+=

bb

06.425.02 2 =− bb

003.2125.02 =−− bb

inb 489.1=

say inb2

11= , inbh 32 ==

(b) For 50,000 repetitions or 50,000 cycles

( ) ksisn 74.15105

102.12

085.0

4

6

=

×=

( ) ( )( )

0.210

105

103.3log

33.3log4

log

3log

=×

==f

f

K

K

fl

nK

n

afl

s

sK

N=

1

( )( )( )( )74.1525.02

100.2

5.1

1

−=

bb

906.125.02 2 =− bb

0953.0125.02 =−− bb

inb 04.1=

say inb16

11= , inbh

8

122 ==

(c) For indefinite life, ground and polished surface



Page 18 of 62

( )( )( )( ) ksisn 205.3290.085.080.0 ==

n

af

s

sK

N=

1

( )( )( )( )2025.02

103.3

5.1

1

−=

bb

02375.1125.02 =−− bb

inb 18.1=

say inb16

31= , inbh

8

322 ==

Maximum stress = ( )dhb

FK f

−

1>hb , 105.0375.225.0 ==hd

Figure AF 8

5.3=tK

( ) ( ) 315.3115.3926.011 =+−=+−= tf KqK

( )( )( )

ksis 14.1325.0375.21875.1

10315.3max =

−=

159. A connecting link as shown, except that there is a 1/8-in. radial hole drilled

through it at the center section. It is machined from AISI 2330, WQT 1000 F, and

it is subjected to a repeated, reversed axial load whose maximum value is 5 kips.

For 5.1=N , determine the diameter of the link at the hole (a) for indefinite life;

(b) for a life of 105 repetitions (no column action). (c) In the link found in (a)

what is the maximum tensile stress?

Problem 159

Solution:

For AISI 2330, WQT 1000 F

ksisu 135=

ksisy 126=

( ) ksiss un 5.671355.05.0 ===

For machined surface, Fig. AF 7, surface factor = 0.80

Size factor = 0.85

( )( )( )( ) ksisn 72.365.6780.085.080.0 ==


Page 19 of 62

n

af

y

m

s

sK

s

s

N+=

1

Fig. AF 8, 1>hb

Assume 5.2=tK

Figure AF 7, inind

r 0625.016

1

2===

ina 0025.0=

96.0

0625.0

0025.01

1

1

1=

+

=

+

=

r

aq

( ) ( ) 44.2115.296.011 =+−=+−= tf KqK

(a) Indefinite life, 44.2=fK

ksisn 72.36=

0=ms

( )DD

DDDdD

F

DdD

Fsa

5.0

20

8

14

54

4

4

4

22

22−

=

−

=−

=

−

=π

πππ

n

af

s

sK

N+= 0

1

( )( )( )DD 5.072.36

2044.2

5.1

12 −

=π

00.25.02 =− DDπ

inD 88.0=

say inD8

7=

(b) For a life of 105 repetitions or cycles

( ) ksisn 66.4410

1072.36

085.0

5

6

=

=

( ) ( )( )

81.110

10

1044.2log

34.2log5

log

3log

===f

f

K

K

fl

nK

n

afl

s

sK

N=

1

( )( )( )DD 5.066.44

2081.1

5.1

12 −

=π

216.15.02 =− DDπ

inD 71.0=


Page 20 of 62

say inD4

3=

(c) DD

FKs

f

5.0

42max

−=

π

inD8

7= , 14.0

875.0

125.0==

D

d

Figure AF 8

6.2=tK

( ) ( ) 54.2116.296.011 =+−=+−= tf KqK

( )( )ksis 82.25

8

75.0

8

7

554.242max =

−

=

π

160. A machine part of uniform thickness 5.2bt = is shaped as shown and machined

all over from AISI C1020, as rolled. The design is for indefinite life for a load

repeated from 1750 lb to 3500 lb. Let bd = . (a) For a design factor of 1.8

(Soderberg), what should be the dimensions of the part? (b) What is the

maximum tensile stress in the part designed?

Problems 160, 161

Solution:


ksisu 65=

ksisy 48=

( ) ksiss un 5.32655.05.0 ===′

For machined surface


Size factor = 0.85

( )( )( )( ) ksisn 205.3290.085.080.0 ==

n

af

y

m

s

sK

s

s

N+=

1

(a) For flat plate with fillets

Figure AF 9

33

dbr ==


Page 21 of 62

333.03

1==

d

r

22

==b

b

d

h

65.1=tK

ina 01.0=

0.1

1

1≈

+

=

r

aq

65.1=≈ tf KK

bt

Fs m

m =

bt

Fs a

a =

5.2

bt =

( ) lbFm 2625175035002

1=+=

( ) lbFa 875175035002

1=−=

2

5.6562

5.2

2625

bbb

sm =

=

2

5.2187

5.2

875

bbb

sa =

=

( )( )22 000,20

5.218765.1

000,48

5.6562

8.1

1

bb+=

inb 7556.0=

or inb 75.0=

inb

t 3.05.2

75.0

5.2===

For flat plate with central hole

Fig. AF 8, 1>hb , 212 == bbhd

Assume 9.2=≈ tf KK

( ) ( ) bt

F

tbb

F

tdh

Fs mmm

m =−

=−

=2

( ) ( ) bt

F

tbb

F

tdh

Fs aaa

a =−

=−

=2


Page 22 of 62

2

5.6562

5.2

2625

bbb

sm =

=

2

5.2187

5.2

875

bbb

sa =

=

( )( )22 000,20

5.21879.2

000,48

5.6562

8.1

1

bb+=

inb 904.0=

or ininb16

159375.0 ==

inb

t8

3

5.2==

inbd16

15==

use inb16

15= , int

8

3= , ind

16

15=

(b) afm sKss +=max

ind

r32

15

2==

98.0

32

15

01.01

1=

+

=q

9.2=tK

( ) ( ) 86.2119.298.011 =+−=+−= tf KqK

psibbt

Fs m

m 7467

16

15

5.65625.656222

=

===

psibbt

Fs a

a 2489

16

15

5.21875.218722

=

===

( )( ) psisKss afm 586,14248986.27467max =+=+=

162. The beam shown has a circular cross section and supports a load F that

varies from 1000 lb to 3000 lb; it is machined from AISI C1020 steel, as

rolled. Determine the diameter D if Dr 2.0= and 2=N ; indefinite life.


Page 23 of 62

Problems 162 – 164.

Solution:


ksisu 65=

ksisy 48=

( ) ksiss un 5.32655.05.0 ===′



Size factor = 0.85

( )( )( ) ksisn 86.245.3290.085.0 ==

∑ = 0AM

BF 2412 =

BF 2=

2

FB =

2

FBA ==

At discontinuity

FF

M 32

6==

( ) kipsinlbinlbinM −=−=−= 9900030003max

( ) kipsinlbinlbinM −=−=−= 3300010003min

( ) kipsinM m −=+= 6392

1

( ) kipsinM a −=−= 3392

1

3

32

D

Ms

π=

Figure AF 12

5.15.1 == dddD

2.02.0 == dddr

42.1=tK

assume 42.1=≈ tf KK


Page 24 of 62

n

af

y

m

s

sK

s

s

N+=

1

( )( ) ( )( )( )33 86.24

33242.1

48

632

2

1

DD ππ+=

inD 821.1=

say inD16

131=

At maximum moment

FF

M 62

12==



( ) kipsinM m −=+= 126182

1

( ) kipsinM a −=−= 66182

1

3

32

D

Ms

π=

00.1=fK

n

af

y

m

s

sK

s

s

N+=

1

( )( ) ( )( )( )33 86.24

6320.1

48

1232

2

1

DD ππ+=

inD 4368.1=

Therefore use inD16

131=

164. The shaft shown is machined from C1040, OQT 1000 F (Fig. AF 1). It is

subjected to a torque that varies from zero to 10,000 in-lb. ( 0=F ). Let Dr 2.0=

and 2=N . Compute D . What is the maximum torsional stress in the shaft?

Solution:

For C1040, OQT 1000 F

ksisu 104=

ksisy 72=


Page 25 of 62

( ) ksiss un 521045.05.0 ===′



Size factor = 0.85

( )( )( )( ) ksisns 5.225285.085.060.0 ==

( ) kipsinlbinTT ma −=−=== 55000000,102

1

( ) ksiss yys 2.43726.06.0 ===

3

16

D

Tss asms

π==

ns

asfs

ys

ms

s

sK

s

s

N+=

1

Figure AF 12

5.15.1 == dddD

2.02.0 == dddr

2.1=tsK

assume 2.1=≈ tsfs KK

( )( ) ( )( )( )33 5.22

5162.1

2.43

516

2

1

DD ππ+=

inD 5734.1=

say inD16

91=

afm sKss +=max

( )( ) ( )( )( )ksis 686.14

16

91

5162.1

16

91

51633max =

+

=

ππ

165. An axle (nonrotating) is to be machined from AISI 1144, OQT 1000 F, to the

proportions shown, with a fillet radius Dr 25.0≈ ; F varies from 400 lb to 1200

lb.; the supports are to the left of BB not shown. Let 2=N (Soderberg line). (a)

At the fillet, compute D and the maximum tensile stress. (b) Compute D at

section BB. (c) Specify suitable dimensions keeping the given proportions, would

a smaller diameter be permissible if the fillet were shot-peened?

Problems 165 – 167


Page 26 of 62

Solution:


ksisu 118=

ksisy 83=

ksiss un 595.0 ==′



Size factor = 0.85

( )( )( ) ksisn 62.415983.085.0 ==

(a) At the fillet

5.15.1 == dddD

25.025.0 == dddr

35.1=tK

assume 35.1=≈ tf KK

FM 6=

( ) kipsinlbinlbinM −=−=−= 2.7720012006max

( ) kipsinlbinlbinM −=−=−= 4.224004006min

( ) kipsinM m −=+= 8.44.22.72

1

( ) kipsinM a −=−= 4.24.22.72

1

3

32

D

Ms

π=

n

af

y

m

s

sK

s

s

N+=

1

( )( ) ( )( )( )33 62.41

4.23235.1

83

8.432

2

1

DD ππ+=

inD 4034.1=

say inD16

71=

(b) At section BB,

FM 30=



( ) kipsinM m −=+= 8.44.22.72

1

( ) kipsinM a −=−= 4.24.22.72

1


Page 27 of 62

3

32

D

Ms

π=

0.1=fK

n

af

y

m

s

sK

s

s

N+=

1

( )( )( )

( )( )( )( )33

5.162.41

12320.1

5.183

3632

2

1

DD ππ+=

inD 6335.1=

say inD16

111=

(c) Specified dimension:

inD 2= , inD 35.1 =

A smaller diameter is permissible if the fillet were shot-peened because of increased

fatigue strength.

166. A pure torque varying from 5 in-kips to 15 in-kips is applied at section C.

( 0=F ) of the machined shaft shown. The fillet radius 8Dr = and the torque

passes through the profile keyway at C. The material is AISI 1050, OQT 1100 F,

and 6.1=N . (a) What should be the diameter? (b) If the fillet radius were

increased to 4D would it be reasonable to use a smaller D ?

Solution:

kipsinT −= 15max

kipsinT −= 5min

( ) kipsinTm −=+= 105152

1

( ) kipsinTa −=−= 55152

1


ksisu 101=

ksisy 5.58=

( ) ksiss un 5.501015.05.0 ===


Page 28 of 62



Size factor = 0.85

( )( )( )( ) ksisns 9.215.5085.085.060.0 ==

(a) At the fillet

81=== Drdr

5.1=dD

3.1=tsK

assume 3.1=≈ tsfs KK

At the key profile

6.1=fsK

use 6.1=fsK

( ) ksiss yys 1.355.586.06.0 ===

ns

asfs

ys

ms

s

sK

s

s

N+=

1

( )( ) ( )( )( )33 9.21

5166.1

1.35

1016

6.1

1

DD ππ+=

inD 7433.1=

say inD4

31=

(b) 4Dr =

25.0=Dr

5.1=dD

Figure AF 12

18.1=tsK

6.118.1 <=≈ tsfs KK

Therefore, smaller D is not reasonable.

170. The beam shown is made of AISI C1020 steel, as rolled; ine 8= . The load F is

repeated from zero to a maximum of 1400 lb. Assume that the stress

concentration at the point of application of F is not decisive. Determine the

depth h and width t if th 4≈ ; 1.05.1 ±=N for Soderberg line. Iteration is

necessary because fK depends on the dimensions. Start by assuming a logical

fK for a logical h (Fig. AF 11), with a final check of fK . Considerable

estimation inevitable.


Page 29 of 62

Problem 170

Solution:

FBA2

1==

At the hole

( ) FF

eBM 42

8 =

==

FM 4max =

0min =M

( ) ( ) kipsinFFM m −==== 8.24.12242

1

( ) ( ) kipsinFFM a −==== 8.24.12242

1

I

Mcs =

( )12

23tdh

I−

=

inind 5.02

1==

inc 75.12

1

2

1

2

11 =

+=


ksisu 65=

ksisy 48=

( ) ksiss un 5.32655.05.0 ===

Size factor = 0.85

( )( ) ksisn 62.275.3285.0 ==

Fig. AF 7, 5.05.35.075.1 >==dc


Page 30 of 62

Assume 5.3=tK

inr 25.02

1

2

1=

=

ina 010.0=

962.0

25.0

010.01

1

1

1=

+

=

+

=

r

aq

( ) ( ) 4.3115.3962.011 =+−=+−= tf KqK

n

af

y

m

s

sK

s

s

N+=

1

( )( )( )

( )( )( )( )( ) tdhtdh

33262.27

75.18.2124.3

248

75.18.212

5.1

1

−+

−=

( ) 70.1223

=− tdh

( )[ ] 70.1250.023

=− th

( ) 70.12143

=− tt

int 8627.0=

say int8

7=

inth 5.34 ==

inh2

1

2

11

2

11 ++>

inh 5.3>

Figure AF 11, 10>dh

( ) indh 550.01010 ===

5.0

2

11

2

52

1

=

−

=b

d

Therefore 5.3=tK , 4.3=fK

Use inh 5= , int4

11=

171. Design a crank similar to that shown with a design factor of 16.06.1 ± based on

the modified Goodman line. The crank is to be forged with certain surfaces

milled as shown and two ¼-in. holes. It is estimated that the material must be of

the order of AISI 8630, WQT 1100 F. The length .17 inL = , .5 ina = , and the

load varies form + 15 kips to –9 kips. (a) Compute the dimensions at section AB

with bh 3= . Check the safety of the edges (forged surfaces). (Iteration involves;

one could first make calculations for forged surfaces and then check safety at

holes.) (b) Without redesigning but otherwise considering relevant factors ,


Page 31 of 62

quantitatively discuss actions that might be taken to reduce the size; holes must

remain as located.

Problems 171-174.

Solution:

(a) AISI 8630, WQT 1100 F

ksisu 96=

( ) ksiss un 48965.05.0 ===

Size factor = 0.85

As-forged surface (Fig. AF I)


( )( )( ) ksisn 174842.085.0 ==

Milled surface (Machined)


( )( )( ) ksisn 68.344885.085.0 ==

At AB, machined

n

af

u

m

s

sK

s

s

N+=

1

Figure AF 11

ininb 5.02

1==

inind 25.04

1==

5.05.0

25.0== in

b

d

Assume 50.3=fK

998.0=q

( ) ( ) 495.3115.3998.011 =+−=+−= tf KqK

I

Mcs =

( )12

23bdh

I−

=


Page 32 of 62

( ) ( )348

1144

8

1

4

11

2

1

4

1

2

1

2

1

2−=+−=

+−=

+−= hhh

hc

bh 3=

( )

12

4

12

348

1

3

bh

hM

s

−

−

=

( )

( ) bb

bM

s3

5.03

3122

3

−

−=

( )( ) bb

bMs

35.03

145.4

−

−=

( )aLFM −=

( )( ) kipsinM −=−= 18051715max

( )( ) kipsinM −=−−= 1085179min

( ) kipsinMm −=−

= 36108180

2

1

( ) kipsinM a −=+

= 144108180

2

1

n

af

u

m

s

sK

s

s

N+=

1

( )( )( )

( )( )( )( )( ) bb

b

bb

b33

5.0368.34

141445.4495.3

5.0396

14365.4

6.1

1

−

−+

−

−=

( )( ) 2.107

1

5.03

143

=−

−

bb

b

( )( )

2.10714

5.033

=−

−

b

bb

inb 6.2=

say inb8

52=

inbh8

773 ==

Checking at the edges (as forged)

( )( ) kipsinM −== 2551715max

( )( ) kipsinM −−=−= 153179min


= 51153255

2

1


Page 33 of 62

( ) kipsinM a −=+

= 204153255

2

1

332 3

2

9

66

b

M

b

M

bh

Ms ===

0.1≈fK

n

af

u

m

s

sK

s

s

N+=

1

( )( )

( )( )( )( )173

20420.1

963

512

6.1

133

bb+=

inb 373.2=

say inb8

32=

since ininb8

32

8

52 >= , ∴ safe.

(c) Action: reduce number of repetitions of load.

CHECK PROBLEMS

173. For the crank shown, inL 15= , ina 3= , ind 5.4= , inb 5.1= . It is as forged

from AISI 8630, WQT 1100 F, except for machined areas indicated. The load F

varies from +5 kips to –3 kips. The crank has been designed without detailed

attention to factors that affect its endurance strength. In section AB only,

compute the factor of safety by the Soderberg criterion. Suppose it were desired

to improve the margin of safety, with significant changes of dimensions

prohibited, what various steps could be taken? What are your particular

recommendations?

Solution:


ksisn 17=


ksisn 68.34=

ksisn 72=

In section AB, machined


Page 34 of 62

( )aLFM −=

( )( ) kipsinM −=−+= 603155max

( )( ) kipsinM −−=−−= 363153min


= 123660

2

1

( ) kipsinM a −=+

= 483660

2

1

inhd 5.4== , inb 5.1=

3=b

h

( )( ) bb

bMs

35.03

145.4

−

−=

( ) ( )[ ]( )[ ] ( )

ksism 8125.25.15.05.13

15.14125.43

=−

−=

( ) ( )[ ]( )[ ] ( )

ksisa 25.115.15.05.13

15.14485.43

=−

−=

n

af

y

m

s

sK

s

s

N+=

1

495.3=fK from Problem 171.

( )( )68.34

25.11495.3

72

8125.21+=

N

185.0 <=N , unsafe

To increase the margin of safety

1. reduce the number of repetitions of loads

2. shot-peening

3. good surface roughness

Recommendation:

No. 1, reducing the number of repetitions of loads.

175. The link shown is made of AISI C1020, as rolled, machined all over. It is loaded

in tension by pins in the inD8

3= holes in the ends; ina

16

9= , int

16

5= ,

inh8

11= . Considering sections at A, B, and C, determine the maximum safe

axial load for 2=N and indefinite life (a) if it is repeated and reversed; (b) if it

is repeated varying from zero to maximum; (c) if it is repeatedly varies or

WF −= to WF 3= . (d) Using the results from (a) and (b), determine the ratio of

the endurance strength for a repeated load to that for a reversed load (Soderberg

line).


Page 35 of 62

Problems 175 - 178

Solution:


ksisu 65=

ksisy 48=

( ) ksiss un 5.32655.05.0 ===

Size factor = 0.85

For machined all over


( )( )( )( ) ksisn 205.3280.090.085.0 ==

n

af

y

m

s

sK

s

s

N+=

1

at A, Figure AF 8

inb16

9=

inh8

11=

inDd8

3==

int16

5=

33.0

8

11

8

3

==h

d

5.0

8

11

16

9

==h

b

6.3=tAK

ind

r16

3

2==

ina 01.0=


Page 36 of 62

95.0

16

3

01.01

1

1

1=

+

=

+

=

r

aq

( ) ( ) 47.3116.395.011 =+−=+−= tAfA kqk

( ) 15

64

16

5

8

3

8

11

FF

tdh

Fs =

−

=−

=

( )( )( )2015

6447.3

4815

64

2

1 am FF+=

am FF 48.145

81 += at A

At B Figure AF 9

inad16

9==

inh8

11=

inr16

3=

int16

5=

33.0

16

916

3

==d

r

2

16

98

11

==d

h

63.1=tBK

ina 01.0=

95.0

16

3

01.01

1

1

1=

+

=

+

=

r

aq

( ) ( ) 6.11163.195.011 =+−=+−= tBfB kqk

45

256

16

5

16

9

FF

dt

Fs =

==


Page 37 of 62

( )( )

( )2045

2566.1

4845

256

2

1 am FF+=

am FF 455.0135

321 += at B

at C, Figure AF 8, 1>h

b

inD8

1=

inah16

9==

22.0

16

98

1

==h

d

5.3=tCK

ind

r16

1

2==

ina 01.0=

862.0

16

1

01.01

1

1

1=

+

=

+

=

r

aq

( ) ( ) 2.3115.3862.011 =+−=+−= tCfC kqk

( ) 35

256

16

5

8

1

16

9

FF

tdh

Fs =

−

=−

=

( )( )

( )2035

2562.3

4835

256

2

1 am FF+=

am FF 17.1105

321 += at C

Equations

At A, am FF 48.145

81 +=

At B, am FF 455.0135

321 +=

At C, am FF 17.1105

321 +=


Page 38 of 62

(a) Repeated and reversed load

0=mF

FFa =

use at A

am FF 48.145

81 +=

( ) aF48.1045

81 +=

kipF 676.0=

(b) FFF am ==

at A, FF 48.145

81 +=

kipF 603.0=

at B, FF 455.0135

321 +=

kipsF 480.1=

at C, FF 17.1105

321 +=

kipF 678.0=

use kipF 603.0=

(c) WF −=min , WF 3max =

( ) WWWFm =−= 32

1

( ) WWWFa 232

1=+=

at A, ( )WW 248.145

81 +=

kipW 319.0=

at B, ( )WW 2455.0135

321 +=

kipW 884.0=

at C, ( )WW 217.1105

321 +=

kipW 378.0=

use kipW 319.0=

kipF 957.0max =


Page 39 of 62

(d) ( )( )

892.0676.0

603.0===

aF

bFRatio

179. A steel rod shown, AISI 2320, hot rolled, has been machined to the following

dimensions: .1 inD = , .4

3inc = , .

8

1ine = A semicircular groove at the

midsection has .8

1inr = ; for radial hole, .

4

1ina = An axial load of 5 kips is

repeated and reversed ( 0=M ). Compute the factor of safety (Soderberg) and

make a judgement on its suitability (consider statistical variations of endurance

strength – i4.4). What steps may be taken to improve the design factor?

Problems 179-183

Solution:

AISI 2320 hot-rolled (Table AT 10)

ksisu 96=

ksisy 51=

ksisn 48=

Size factor = 0.85

Surface factor = 0.85 (machined)

( )( )( )( ) ksisn 74.274885.085.080.0 ==

n

af

y

m

s

sK

s

s

N+=

1

0=ms , reversed

ssa =

n

af

s

sK

N=

1

f

na

NK

ss =

at the fillet, Figure AF 12

iner8

1==


Page 40 of 62

incd4

3==

inD 1=

17.0

4

38

1

==d

r

3.1

4

3

1==

d

D

55.1=tK

ina 010.0=

926.0

8

1

010.01

1

1

1=

+

=

+

=

r

aq

( ) ( ) 51.11155.1926.011 =+−=+−= tf KqK

( )ksissa 32.11

4

3

542

=

==

π

( )( )62.1

51.132.11

74.27===

fa

n

Ks

sN

At the groove, Figure AF 14

inininrDbd4

3

8

1212 =

−=−==

inD 1=

inr8

1=

17.0

4

38

1

==d

r

3.1

4

3

1==

d

D

75.1=tK

ina 010.0=


Page 41 of 62

926.0

8

1

010.01

1

1

1=

+

=

+

=

r

aq

( ) ( ) 7.11175.1926.011 =+−=+−= tf KqK

( )ksi

d

Fssa 32.11

4

3

54422

=

===

ππ

( )( )44.1

7.132.11

74.27===

fa

n

Ks

sN

At the hole, Figure AF8

inhD 1==

inad4

1==

25.014

1

==h

D

44.2=tK

ina 010.0=

926.0

8

1

010.01

1

1

1=

+

=

+

=

r

aq

( ) ( ) 33.21144.2926.011 =+−=+−= tf KqK

( ) ( )ksi

DdD

Fssa 34.9

4

11

4

1

5

4

22=

−

=

−

==ππ

( )( )27.1

33.234.9

74.27===

fa

n

Ks

sN

Factor of safety is 1.27

From i4.4

nss 76.0=

27.1min32.176.0

>==n

n

s

sN

Therefore, dimensions are not suitable.

Steps to be taken:

1. Reduce number of cycle to failure


Page 42 of 62

2. Good surface condition

3. Presetting

186. A stock stud that supports a roller follower on a needle bearing for a cam is

made as shown, where ina8

5= , inb

16

7= , inc

4

3= . The nature of the junction

of the diameters at B is not defined. Assume that the inside corner is sharp. The

material of the stud is AISI 2317, OQT 1000 F. Estimate the safe, repeated load

F for 2=N . The radial capacity of the needle bearing is given as 1170 lb. at

2000 rpm for a 2500-hr life. See Fig. 20.9, p. 532, Text.

Problem 186

Solution:

AISI 2317, OQT 1000 F

ksisu 106=

ksisy 71=

ksiss un 535.0 ==

Size factor = 0.85

( )( ) ksisn 455385.0 ==

Figure AF 12

inad8

5==

incD4

3==

0≈dr , sharp corner

2.1

8

54

3

==d

D

Assume 7.2=tK

7.2=≈ tf KK

3

32

a

Ms

π=

FFFbM 4375.016

7=

==


Page 43 of 62

inina 625.08

5==

( )( )

FF

s 25.18625.0

4375.0323

==π

Fsss am 25.18===

n

af

y

m

s

sK

s

s

N+=

1

( )( )45

25.187.2

71

25.18

2

1 FF+=

lbkipF 370370.0 == < less than radial capacity of the needle bearing. Ok.

187. The link shown is made of AISI C1035 steel, as rolled, with the following

dimensions .8

3ina = , .

8

7inb = , .1 inc = , .

2

1ind = , .12 inL = , .

16

1inr = The

axial load F varies from 3000 lb to 5000 lb and is applied by pins in the holes.

(a) What are the factors of safety at points A, B, and C if the link is machined all

over? What are the maximum stresses at these points?

Problems 187, 188

Solution:

AISI C1035, as rolled

ksisu 85=

ksisy 55=

ksiss un 5.425.0 ==

size factor = 0.85

( )( )( ) ksisn 68.215.4285.06.0 ==

n

af

y

m

s

sK

s

s

N+=

1

( ) kipsFm 4352

1=+=

( ) kipFa 1352

1=−=


Page 44 of 62

(a) at A, Figure AF 9

inr16

1=

inad8

3==

inbh8

7==

17.0

8

316

1

==d

r

33.2

8

38

7

==d

h

9.1=tK

ina 010.0=

862.0

16

1

010.01

1

1

1=

+

=

+

=

r

aq

( ) ( ) 78.1119.1862.011 =+−=+−= tf KqK

ac

Fs =

( )ksism 67.10

18

3

4=

=

( )ksisa 67.2

18

3

1=

=

( )( )68.21

67.278.1

55

67.101+=

N

42.2=N

At B, same as A, 78.1=fK

( )cab

Fs

−=

( )ksism 8

18

3

8

7

4=

−

=


Page 45 of 62

( )ksisa 2

18

3

8

7

1=

−

=

( )( )68.21

278.1

55

81+=

N

23.3=N

At C, Figure AF 8

ind2

1=

inch 1==

1>hb

5.012

1

==h

d

2.2=tK

ina 010.0=

inind

r 25.04

1

2===

962.0

25.0

010.01

1

1

1=

+

=

+

=

r

aq

( ) ( ) 15.2112.2962.011 =+−=+−= tf KqK

( )( )dcab

Fs

−−=

ksism 16

2

11

8

3

8

7

4=

−

−

=

ksism 4

2

11

8

3

8

7

1=

−

−

=

( )( )68.21

415.2

55

161+=

N

45.1=N

(b) Maximum stresses

at A

( ) ksisKss afmA 42.1567.278.167.10 =+=+=

at B

( ) ksisKss afmB 56.11278.18 =+=+=


Page 46 of 62

at C

( ) ksisKss afmC 6.24415.216 =+=+=

IMPACT PROBLEMS

189. A wrought-iron bar is 1in. in diameter and 5 ft. long. (a) What will be the stress

and elongation if the bar supports a static load of 5000 lb? Compute the stress

and elongation if a 5000 lb. weight falls freely 0.05 in. and strikes a stop at the

end of the bar. (b) The same as (a), except that the bar is aluminum alloy 3003-

H14.

Solution:

.1 inD = , ftL 5=

For wrought iron,

psiE 61028×=

(a) elongation

lbF 5000=

( )( )( )

( ) ( )in

AE

FL01364.0

102814

1255000

62

=

×

==π

δ

Stress and elongation

inh 05.0=

lbW 5000=

inftL 605 ==

2

1

21

++=

LW

hEA

A

W

A

Ws

( ) ( )

( )( ) ( )

( )( )psis 741,24

500060

14

102805.02

1

14

5000

14

5000

2

1

26

22

=

×

++=

π

ππ

( )( )in

E

sL053.0

1028

60741,246

=×

==δ

(b) Aluminum alloy 3003-H14

psiE 61010×=

lbF 5000=

( )( )( )

( ) ( )in

AE

FL038.0

101014

1255000

62

=

×

==π

δ

Stress and elongation

inh 05.0=


Page 47 of 62

lbW 5000=

inftL 605 ==

2

1

21

++=

LW

hEA

A

W

A

Ws

( ) ( )

( )( ) ( )

( )( )psis 475,18

500060

14

101005.02

1

14

5000

14

5000

2

1

26

22

=

×

++=

π

ππ

( )( )in

E

sL111.0

1010

60475,186

=×

==δ

190. What should be the diameter of a rod 5 ft. long, made of an aluminum alloy

2024-T4, if it is to resist the impact of a weight of lbW 500= dropped through a

distance of 2 in.? The maximum computed stress is to be 20 ksi.

Solution:

For aluminum alloy, 2024-T4

psiE 6106.10 ×=

lbW 500=

inh 2=

inftL 605 ==

psiksis 000,2020 ==

2

1

21

++=

LW

hEA

A

W

A

Ws

( )( )( )( )

2

16

50060

106.10221

50005000000,20

×++=

A

AA

( )2

1

14131140 AA ++=

9332.04

2

==D

Aπ

inD 09.1= , say inD16

11=

191. A rock drill has the heads of the cylinder bolted on by 7/8-in. bolts somewhat as

shown. The grip of the bolt is 4 in. (a) If the shank of the bolt is turned down to

the minor diameter of the coarse-thread screw, 0.7387 in., what energy may each

bolt absorb if the stress is not to exceed 25 ksi? (b) Short bolts used as described

above sometimes fail under repeated shock loads. It was found in one instance

that if long bolts, running from head to head, were used, service failures were

eliminated. How much more energy will the bolt 21 in. long absorb for a stress of


Page 48 of 62

25 ksi. That the bolt 4 in. long? As before, let the bolt be turned down to the

minor diameter. The effect of the threads on the strength is to be neglected.

Problem 191

Solution:

( )E

ALsAL

E

sU

22

22

==

(a) 4

2D

Aπ

=

inL 4=

inD 7387.0=

psiE 61030×=

psiksis 000,2525 ==

( ) ( ) ( )

( )lbinU −=

×

= 86.1710302

47387.04

000,25

6

22 π

(b) inL 21=

( ) ( ) ( )

( )lbinU −=

×

= 75.9310302

217387.04

000,25

6

22 π

lbinU −=−=∆ 89.7586.1775.93

192. As seen in the figure, an 8.05-lb body A moving down with a constant

acceleration of 12 fps2, having started from rest at point C. If A is attached to a

steel wire, W & M gage 8 (0.162 in. diameter) and if for some reason the sheave

D is instantly stopped, what stress is induced in the wire?

Problems 192, 193

Solution:

E

ALsU

2

2

=

( ) maLmahahmmvU ==== 22

1

2

1 2


Page 49 of 62

maLE

ALs=

2

2

gA

WaE

A

maEs

222 ==

lbW 05.8= 212 fpsa = 232 fpsg =

212 fpsb =

psiE 61030×=

4

2D

Aπ

=

( )( )( )( ) ( )32162.0

10301205.8882

6

2

2

ππ

×==

gD

WaEs

psis 741,93=

193. The hoist A shown, weighing 5000 lb. and moving at a constant fpsv 4= is

attached to a 2 in. wire rope that has a metal area of 1.6 sq. in. and a modulus

psiE 61012×= . When fth 100= , the sheave D is instantly stopped by a brake

(since this is impossible, it represents the worst conceivable condition).

Assuming that the stretching is elastic, compute the maximum stress in the rope.

Solution:

E

ALsU

2

2

=

22

22

1v

g

WmvU ==

22

22v

g

W

E

ALs=

gAL

EWvs

22 =

lbW 5000=


Page 50 of 62

fpsv 4=

psiE 61012×= 26.1 inA =

fthL 100== 232 fpsg =

( )( ) ( )( )( )( )1006.132

101245000 62

2 ×=s

psis 693,13=

194. A coarse-thread steel bolt, ¾ in. in diameter, with 2 in. of threaded and 3 in. of

unthreaded shank, receives an impact caused by a falling 500-lb weight. The area

at the root of the thread is 0.334 sq. in. and the effects of threads are to be

neglected. (a) What amount of energy in in-lb. could be absorbed if the maximum

calculated stress is 10 ksi? (b) From what distance h could the weight be

dropped for this maximum stress? (c) How much energy could be absorbed at the

same maximum stress if the unthreaded shank were turned down to the root

diameter.

Solution:

E

ALsU

2

2

=

(a) 21 UUU +=

E

LAsU

2

11

2

11 =

E

LAsU

2

22

2

22 =

2

1 334.0 inA =

( ) 2

2 442.075.04

inA ==π

psis 000,101 =

( )( )psi

A

Ass 7556

442.0

334.0000,10

2

112 ===

inL 21 =

inL 32 =

psiE 61030 ×=

( ) ( )( )( )

lbinU −=×

= 113.110302

2334.0000,106

2

1

( ) ( )( )( )

lbinU −=×

= 262.110302

3442.075566

2

2


Page 51 of 62

lbinUUU −=+=+= 375.2262.1113.121

(b)

++=

2

1

211

LW

hEA

A

Ws

+

++=

2

1

2

2

1

11

211

A

L

A

LW

hE

A

Ws

( )

+++=

2

1

2112

21

1

211

LALAW

AhEA

A

Ws

lbW 500= 2

1 334.0 inA = 2

2 442.0 inA =

inL 21 =

inL 32 =

psiE 61030×=

psis 000,10=

( )( )( )( )( ) ( )( )[ ]

2

16

3334.02442.0500

442.0334.01030211

334.0

500000,10

+

×++=

h

inh 0033.0=

(c) E

ALsU

2

2

=

2334.0 inA =

inL 5=

psiE 61030×=

psis 000,10=

( ) ( )( )( )

lbinU −=×

= 783.210302

5334.0000,106

2

196. A part of a machine that weighs 1000 lb. raised and lowered by 1 ½-in. steel rod

that has Acme threads on one end (see i8.18 Text, for minor diameter). The

length of the rod is 10 ft. and the upper 4 ft are threaded. As the part being


Page 52 of 62

lowered it sticks, then falls freely a distance of 1/8 in. (a) Compute the maximum

stress in the rod. (b) What would be the maximum stress in the rod if the lower

end had been turned down to the root diameter?

Solution:

++=

2

1

211

LW

hEA

A

Ws

+

++=

2

1

2

2

1

11

211

A

L

A

LW

hE

A

Ws

( )

+++=

2

1

2112

21

1

211

LALAW

AhEA

A

Ws

see i8.18 , inD2

112 = , inD 25.11 =

( ) 2

2

1 227.14

25.1inA ==

π

( ) 2

2

2 767.14

5.1inA ==

π

inL 41 =

inL 62 =

ininh 125.08

1==

lbW 1000=

psiE 61030×=

( )( )( )( )( )( ) ( )( )[ ]

psis 186,286227.14767.11000

767.1227.11030125.0211

227.1

1000 2

16

=

+

×++=

(b)

++=

2

1

211

LW

hEA

A

Ws

2

1 227.1 inAA ==

inLLL 1021 =+=


Page 53 of 62

( )( )( )( )

psis 552,25100010

227.11030125.0211

227.1

1000 2

16

=

×

++=

197. A weight W of 50 lb is moving on a smooth horizontal surface with a velocity of

2 fps when it strikes head-on the end of a ¾-in. round steel rod, 6 ft. long.

Compute the maximum stress in the rod. What design factor based on yield

strength is indicated for AISI 1010, cold drawn?

Solution:

2

1

2

1

+

=

W

WALg

EWvs

eo

3

be

WW =

ALWb ρ=

3284.0 inlb=ρ

2

2

442.04

3

4inA =

=

π

inftL 726 ==

( )( )( ) lbWb 038.972442.0284.0 ==

lbWe 013.33

038.9==

lbW 50=

fpsv 2= 232 fpsgo =

psiE 61030×=

ftL 6=

( )( ) ( )

( )( )( )psis 8166

50

013.316442.032

1030250

2

1

62

=

+

×=

For AISI 1010, cold drawn

psiksisy 000,5555 ==

74.68166

000,55===

s

sN

y


Page 54 of 62

199. A rigid weight of 100 lb is dropped a distance of 25 in. upon the center of a 12

in., 50-lb. I-beam ( 46.301 inI x = ) that is simply supported on supports 10 ft

apart. Compute the maximum stress in the I-beam both with and without

allowing for the beam’s weight.

Solution:

Without beams weight

st

sty

yss =

EI

FLy

48

3

=

3

48

L

EI

y

Fk ==

++==

2

1

211

W

hk

k

Wy δ

psiE 61030×=

inftL 12010 == 46.301 inI =

( )( )( )

inlbk 333,251120

6.3011030483

6

=×

=

lbW 100=

inh 25=

( )( )iny 1415.0

100

333,25125211

333,251

100 2

1

=

++

=

( )( )( )( )

inEI

WLyst 0004.0

6.301103048

120100

48 6

33

=×

==

I

Mcsst =

( )( )lbin

WLM −=== 3000

4

120100

4

inh

c 62

12

2===

( )( )psisst 68.59

6.301

63000==

( ) psis 112,210004.0

1415.068.59 =

=


Page 55 of 62

with mass of beam

2

1

21

++=

st

ststy

hyyy

h - correction factor =

W

We+1

1

35

17 be

WW =

( )( ) lbftftlbWb 5001050 ==

( )lbWe 243

35

50017==

h - correction factor = 292.0

100

2431

1=

+

( )( )iny 0764.0

0004.0

292.0252110004.0

2

1

=

++=

( ) psiy

yss

st

st 400,110004.0

0764.068.59 =

==

201. A 3000 lb. automobile (here considered rigid) strikes the midpoint of a guard rail

that is an 8-in. 23-lb. I-beam, 40 ft. long; 42.64 inI = . Made of AISI C1020, as

rolled, the I-beam is simply supported on rigid posts at its ends. (a) What level

velocity of the automobile results in stressing the I-beam to the tensile yield

strength? Compare results observed by including and neglecting the beam’s

mass.

Solution:


psiksisy 000,4848 ==

og

WvF

22

2

=δ

3

48

L

EIFk ==

δ

I

FLc

I

Mcs

4==

Lc

IsF

4=

( ) 2

2

22

32232

696

16

962 Ec

ILs

EIcL

LsI

EI

LFF===

δ


Page 56 of 62

neglecting mass of beam

og

Wv

Ec

ILsF

262

2

2

2

==δ

ILg

EcWvs

o2

3 222 =

lbW 3000= 232 fpsgo =

inh

c 42

8

2===

psiE 61030×= 42.64 inI =

ftL 40=

psiksiss y 000,4848 ===

( ) ( ) ( )( )( )( )402.6432

4103030003

2

3000,48

2622222 ×

===v

ILg

EcWvs

o

fpsv 62.6=

Including mass of beam

+

=

W

WILg

EcWvs

eo 1

1

2

3 222

35

17 be

WW =

( )( ) lbftftlbWb 9204023 ==

( )lbWe 447

35

92017==

( ) ( ) ( )( )( )( )

+

×===

3000

4471

1

402.6432

4103030003

2

3000,48

2622222 v

ILg

EcWvs

o

fpsv 10.7=

DATA LACKING – DESIGNER’S DECISIONS

202. A simple beam is struck midway between supports by a 32.2-lb. weight that has

fallen 20 in. The length of the beam is 12 ft. If the stress is not to exceed 20 ksi,

what size I-beam should be used?

Solution:


Page 57 of 62

2

1

21

++=

st

ststy

hyyy

st

sty

yss =

inh 20=

psis 000,20=

EI

WLyst

48

3

=

2

1

3

9611

++=

WL

EIh

y

y

st

with correction factor

2

1

3

1

19611

+

++=

W

WWL

EIh

y

y

est

I

WLd

I

Mcsst

8==

35

17wLWe =

+

++=

2

1

3

35

171

19611

8

W

wLWL

EIh

I

WLds

lbW 2.32=

inh 20=

inftL 14412 ==

psiE 61030×=

( )( ) ( )( )( )( )( ) ( )( )

( )

+

×++=

2

1

3

6

2.3235

12171

1

1442.32

2010309611

8

1442.32

w

I

I

ds

+++=

2

1

181.01

159911

6.579

wI

I

ds

From The Engineer’s Manual

By Ralph G. Hudson, S.B.


Page 58 of 62

Use 3”, 5.7 lb, 45.2 inI =

( ) ( )( )

psipsis 000,20600,197.5181.01

15.259911

5.2

36.579 2

1

<=

+++=

Therefore use 3-in depth, 5.7-lb I-beam ( 45.2 inI = )

204. A 10-in., 25.4-lb.., I-bean, AISI 1020, as rolled, is 10 ft. long and is simply

supported at the ends shown. There is a static load of kipsF 101 = , 4 ft from the

left end, and a repeated reversed load of kipsF 102 = , 3 ft from the right end. It is

desired to make two attachments to the beam through holes as shown. No

significant load is supported by these attachments, but the holes cause stress

concentration. Will it be safe to make these attachments as planned? Determine

the factor of safety at the point of maximum moment and at points of stress

concentration.

Problem 204

Solution:

Mass of beam negligible


ksisy 48=

ksisu 65=

( )∑ = 0AM

( ) BFF 103104 21 =−+

( )21 7410

1FFB +=

( )∑ = 0BM

( ) AFF 104103 12 =−+

( )21 3610

1FFA +=

kipsF 101 =

kipstoF 10102 −=

( ) ( )[ ] kipsB 310710410

1min −=−+=


Page 59 of 62

( ) ( )[ ] kipsB 1110710410

1max =+=

( ) ( )[ ] kipsA 310310610

1min =−+=

( ) ( )[ ] kipsA 930710610

1max =+=

Figure AF 11,

ine2

11= ,

ind4

1=

inc 625.14

12

2

11 =

+=

inh 10=

ineh

b 5.32

11

2

10

2=−=−=

07.05.3

25.0==

b

d

5.0625.0

50.1>==

d

e

Use 0.3=tK

926.0

8

1

010.01

1=

+

=q

( ) ( ) 85.2113926.011 =+−=+−= tf KqK

( ) ksiss un 5.32655.05.0 ===

size factor = 0.85

( ) ksisn 6.275.3285.0 ==

left hole, ( )AM 2=

( ) kipsftM −== 1892max

( ) kipsftM −== 632min

I

Mcs =

( ) kipsinkipsftMm −=−=+= 144126182

1

( ) kipsinkipsftM a −=−=−= 7266182

1

inc 625.1= 41.122 inI = (Tables)


Page 60 of 62

( )( )ksism 92.1

1.122

625.1144==

( )( )ksisa 96.0

1.122

625.172==

n

af

y

m

s

sK

s

s

N+=

1

( )( )6.27

96.085.2

48

92.11+=

N

2.7=N

right hole , ( )BM 5.1=

( ) kipsftM −== 5.16115.1max

( ) kipsftM −−=−= 5.435.1min

I

Mcs =

( ) kipsinkipsftMm −=−=−= 7265.45.162

1

( ) kipsinkipsftM a −=−=+= 1265.105.45.162

1

inc 625.1= 41.122 inI = (Tables)

( )( )ksism 96.0

1.122

625.172==

( )( )ksisa 68.1

1.122

625.1126==

n

af

y

m

s

sK

s

s

N+=

1

( )( )6.27

68.185.2

48

96.01+=

N

67.5=N

at maximum moment, or at , 2F

( ) kipsftM −== 33113max

( ) kipsftM −−=−= 933min

I

Mcs =

( ) kipsinkipsftMm −=−=−= 144129332

1

( ) kipsinkipsftM a −=−=+= 252219332

1


Page 61 of 62

inc 52

10==

41.122 inI = (Tables)

( )( )ksism 90.5

1.122

5144==

( )( )ksisa 32.10

1.122

5252==

0.1=fK

n

af

y

m

s

sK

s

s

N+=

1

( )( )6.27

32.100.1

48

90.51+=

N

2=N

Since the design factor at the holes is much larger than at the point of maximum moment,

it is safe to make these attachment as planned.

205. The runway of a crane consists of .20 ftL = lengths of 15-in., 42.9-lb. I-beams,

as shown, each section being supported at its ends; AISI C1020, as rolled. The

wheels of the crane are 9 ft apart, and the maximum load expected is

lbF 000,10= on each wheel. Neglecting the weight of the beam, find the design

factor (a) based on variable stresses for 105 cycles, (b) based on the ultimate

strength. (Hint. Since the maximum moment will occur under the wheel, assume

the wheels at some distance x from the point of support, and determine the

reaction, 1R as a function of x ; 0=dx

dM gives position for a maximum bending

moment.)

Problem 205.

Solution:

( )∑ = 02RM

( ) ( ) 1LRFaxLFxL =−−+−

( )L

FaxLR

−−=

221

( )FaxLL

xxRM −−== 221


Page 62 of 62

( ) ( )[ ] 0222 =−+−−= xaxLL

F

dx

dM

0222 =−−− xaxL

−=

22

1 aLx

L

Fa

L

Faa

LLL

aL

M2

2

222

2

max

−

=

−

−−

−=

inftL 24020 ==

infta 1089 ==

kipslbF 10000,10 ==

( )

( )kipsinM −=

−

= 75.7202402

102

108240

2

max

For 15-in., 42.9 lb, I-beam 48.441 inI =

inc 5.72

15==

( )( )ksi

I

Mcs 24.12

8.441

5.775.720max ===


ksisu 65=

ksiss un 5.325.0 ==

size factor = 0.85

( ) ksisn 6.275.3285.0 ==

(a) at 105

cycles

ksisn 3410

106.27

085.0

5

6

=

=

724.12

34===

s

sN n

(b) 31.524.12

65===

s

sN u

- end -

SECTION 3 – SCREW FASTENINGS

Page 1 of 42

SIMPLE TENSION INCLUDING TIGHTENING STRESSES

DESIGN PROBLEMS

221. A 5000-lb. gear box is provided with a steel (as rolled B1113) eyebolt for use

in moving it. What size bolt should be used: (a) if UNC threads are used? (b)

If UNF threads are used? (c) If the 8-thread series is used? Explain the basis

of your choice of design factor.

Solution:

B1113, as rolled

ksisy 45= (Table AT-7)

lbFe 5000=

i5.6, ( ) 21

6s

y

d As

s =

< inD

4

3

For inD4

3=

..35.0 insqAs ≈

( ) 21

35.06

y

d

ss =

10

y

d

ss =

use design factor = 10

10

000,45 psisd =

psisd 4500=

..111.14500

5000insq

s

FA

d

es ===

Table AT 14 and Table 5.1

(a) UNC Threads

Use inD8

31= , ..155.1 insqAs =


Page 2 of 42

(b) UNF Threads

Use inD8

31= , ..155.1 insqAs =

(c) 8-Thread Series

Use inD8

31= , ..233.1 insqAs =

222. A motor weighing 2 tons is lifted by a wrought-iron eye bolt which is screwed

into the frame. Decide upon a design factor and determine the size of the

eyebolt if (a) UNC threads are used, (b) UNF threads are used. Note: Fine

threads are not recommended for brittle materials.

Solution:

Table AT-7

Wrought iron, ksisy 25=

Assume design factor = 10

10

y

d

ss =

10

000,25 psisd =

psisd 2500=

( )..60.1

2500

20002insq

s

FA

d

es ===

Table AT 17

(a) UNC Threads

Use inD4

31= , ..90.1 insqAs =

(b) UNF Threads

Use inD2

11= , ..581.1 insqAs =


Page 3 of 42

224. A wall bracket, Fig. 8-13, Text, is loaded so that the two top bolts that fasten it

to the wall are each subjected to a tensile load of 710 lb. The bolts are to be

cold forged from AISI C1020 steel with UNC threads, Neglecting the effect of

shearing stresses, determine the diameter of these bolts if they are well

tightened.

Figure 8-13

Solution:

cold forged, AISI C1020

ksisy 66= (Table AT-7)

lbFe 710=

( ) 23

6s

y

e As

F =

< inD

4

3

( ) 23

6

000,66710 sA=

..161.0 insqAs = , inD4

3<

Table AT 14 , UNC Threads

Use inD16

9= , ..1820.0 insqAs =


Page 4 of 42

225. A connection similar to Fig. 5.9, Text, is subjected to an external load eF of

1250 lb. The bolt is made from cold-finished AISI B1113 steel with UNC

threads. (a) Determine the diameter of the bolt if it is well tightened. (b)

Compute the initial tension and corresponding approximate tightening torque

if yi ss 85.0= (i5.8).

Figure 5.9

Solution:

Cold-finished AISI B1113

Table A-7, ksisy 72=

lbFe 1250=

(a) ( ) 23

6s

y

e As

F =

( ) 23

6

000,721250 sA=

..2214.0 insqAs = , inD4

3<


Use inD8

5= , ..2260.0 insqAs =

(b) ( ) psiss yi 200,61000,7285.085.0 ===

Initial Tension

( )( ) lbAsF sii 831,132260.0200,61 ===

Tightening torque

iCDFT =


Page 5 of 42

( ) lbinDFT i −=

== 1729831,13

8

52.02.0

226. The cylinder head of a 10 x 18 in. Freon compressor is attached by 10 stud

bolts made of SAE Grade 5. The cylinder pressure is 200 psi. (a) What size

bolts should be used? (b) What approximate tightening torque should be

needed to induce a tightening stress is of 0.9 times the proof stress?

Solution:

Table 5.2

SAE Grade 5

Assume ksisy 88=

(a)

( )lbFe 1571

10

104

2002

=

=

π

( ) 23

6s

y

e As

F = , inD4

3<

( ) 23

6

000,881571 sA=

..2255.0 insqAs = , inD4

3<


Use inD8

5= , ..2260.0 insqAs =

(b) iCDFT =

2.0=C

pi ss 9.0=

ksisp 85= , (Table 5.2)

( ) psisi 500,76000,859.0 ==

( )( ) lbAsF sii 289,172260.0500,76 ===

Tightening torque

( ) lbinDFT i −=

== 2161289,17

8

52.02.0

227. The American Steel Flange Standard specifies that 8 bolts are to be used on

flanges for 4-in. pipe where the steam or water pressure is 1500 psi. It is also

specified that, in calculating the bolt load, the outside diameter of the gasket,

which is 6 3/16 in., should be used. Determine (a) the diameter of the UNC bolts

if they are well-tightened and made of ASTM 354 BD (Table 5-2), (b) the

approximate torque to tighten the nuts if the initial stress is 90 % of the proof


Page 6 of 42

stress. The Standard specifies that 1 1/4 –in. bolts with 8 th./in. be used (these

bolts are also subjected to bending). How does your answer compare?

Solution:

Table 5.2, ASTM 354 BD

ksisp 120=

ksisy 125=

lbFe 56388

16

36

41500

2

=

=

π

(a) ( ) 23

6s

y

e As

F = , inD4

3<

( ) 23

6

000,1255638 sA=

..4184.0 insqAs = , inD4

3<


Use inD8

7= , ..4620.0 insqAs =

inD4

3>

use

( ) ( ) psiss yd 750,18000,12515.085.01 ==−=

..3007.0750,18

5638insq

s

FA

d

es ===


Use inD8

3= , ..334.0 insqAs =

(b) iDFT 2.0=

pi ss 9.0=

( ) psisi 000,108000,1209.0 ==

( )( ) lbAsF sii 072,363340.0000,108 ===

Tightening torque

( ) lbinDFT i −=

== 5411072,36

4

32.02.0


Page 7 of 42

inD4

11< as specified by the standard.

CHECK PROBLEMS

228. A cap screw, ¾ in.-10-UNC-2, with a hexagonal head that is 9/16 in. thick,

carries a tensile load of 3000 lb. If the material is AISI 1015, cold drawn, find

the factor of safety based on ultimate strengths of (a) the threaded shank, (b)

the head against being sheared off, and (c) the bearing surface under the head.

(d) Is there any need to consider the strength of standard cap-screw heads in

design?

Solution:

For ¾ in. UNC, Table AT 14,

..334.0 insqAs =

Head:

.8

11 inA =


ksisu 77= , ksisus 58=

(a) psiA

Fs

s

8982334.0

3000===

57.88982

000,77===

d

u

s

sN

(b) Dt

Fss

π=

int16

9=

psiss 2264

16

9

4

3

3000=

=

π

6.252264

000,58===

s

us

s

sN


Page 8 of 42

(c) oo

3012

360==θ

( ) ( ) ..096.130tan2

8

11

2

126tan

222

126

2

insqAA

Area =

=

= θ

psiAArea

Fs

b

b 4586

4

3

4096.1

30002

=

−

=−

=π

8.164586

000,77===

b

u

s

sN

(d) No need to consider the strength of standard cap-screw heads since its factor of

safety is very much higher than for the threaded shank.

229. A bolt, 1 1/8 in.-7-UNC-2, is subjected to a tensile load of 10,000 lb. The head

has a thickness of ¾ in. and the nut a thickness of 1 in. If the material is SAE

grade 2 (Table 5.2), find the design factor as based on ultimate stresses (a) of

the threaded shank, (b) of the head against being sheared off, and (c) of the

bearing surface under the head. The bolt head is finished. (d) Is there any need

to consider the strength of standard bolt heads in design?

Solution:

For SAE grade 2 (Table 5.2), inD8

11=

ksisu 55= , uus ss 75.0=

For 1 1/8 in.-7-UNC-2 (Table AT 14)

..763.0 insqAs =

inA16

111=

.000,10 lbF =


Page 9 of 42

(a) psiA

Fs

s

106,13763.0

000,10===

2.4106,13

000,55===

d

u

s

sN

(b) Dt

Fss

π=

int4

3=

psiss 3773

4

3

8

11

000,10=

=

π

( )11

3773

000,5575.0===

s

us

s

sN

(c) oo

3012

360==θ

( ) ( ) ..4661.230tan2

16

111

2

126tan

222

126

2

insqAA

Area =

=

= θ

psiAArea

Fs

b

b 6793

8

11

44661.2

000,102

=

−

=−

=π

1.86793

000,55===

b

u

s

sN

(d) No need to consider the strength of standard bolt head in design since its factor of

safety is higher than for the threaded shank.

230. An axial force is applied to a regular nut which of course tends to shear the

threads on the screw. (a) What is the ratio of the force necessary to shear the

threads (all threads initially in intimate contact) to the force necessary to pull

the bolt in two? Use coarse threads, a 1 ½ -in. bolt, and assume that

uus ss 75.0= . The head thickness is 1 in. and the nut thickness is 1 5/16 in. (b)

Is failure of the thread by shear likely in this bolt?


Page 10 of 42

Solution:

1 ½ - in. UNC

..405.1 insqAs =

(a) sF = shear force = Dtsusπ

.2

11 inD =

.16

51 int =

uus ss 75.0=

( )( ) uus ssF 6388.416

51

2

1175.0 =

= π

usu sAsF 405.1==

Ratio = 3.3405.1

6388.4=

u

u

s

s

(b) Ratio > 1, failure by shear is not likely to occur.

231. For bolted structural joints, specifications suggest that ½-in. bolts (high-

strength material) be tightened to an initial tension of lbFi 500,12= . What

should be the approximate tightening torque? How does your answer compare

with lbftT −= 90 ., which is the value in the specification?

Solution:

( ) lbinDFT i −=

== 1250500,12

2

12.02.0

lbinlbinlbftT −<−=−= 1250108090 o.k.

232. One method of estimating the initial tensile stress in a tightened bolt is to turn

the nut until it is snug, but with no significant stress in the bolt. Then the nut is

turned through a predetermined angle that induces a certain unit strain

corresponding to the desired stress. A ¾ - in. bolt of the type shown in Fig.

5.4, Text, is turned down until, for practical purposes, the diameter of the

entire shank is the minor diameter. The material is AISI 4140, OQT 1200 oF.

The grip is 5 in. and the effective strain length is estimated to be 5.3 in. If the

initial tensile stress at the root diameter is to be about 75 % of the yield

strength, through what angle should the nut be turned after it is just snug? The

threads are UNC and the parts being bolted are assumed to be rigid.

Solution:


Page 11 of 42

For ¾ in., UNC

inDr 6273.0=

..334.0 insqAs =

10. =inTh

AISI 4140, OQT 1200 oF

ksisy 115=

( ) ksis 25.8611575.0 ==

E

sL=δ

inL 3.5=

pitch, ininp 10.010

1==

( )o

p360

δθ =

( )o

pE

sL360=θ

( )( )( )( )

( ) oo 55360103010.0

3.5250,866

=×

=θ

233. When both ends of a bolt are accessible for micrometer measurements, the

total elongation δ caused by tightening can be determined by measuring

lengths before and after tightening. In order to reduce this total elongation to

unit elongation, thence to stress, the effective strain length for the bolt must be

known. For a 1 ¼-in steel bolt, threaded for its full length, 8-thread series, the

effective strain length has been found by experiment to be

.1.197.0 inGLe += , where G is the grip (by W.A. McDonald, North

Carolina State College). Let the bolt material be AISI 8742, OQT 1000 oF. (a)

It is desired that the initial tensile stress be about ys7.0 . What total elongation

should be obtained for a grip length of 4.8 in.? (b) Investigate the approximate

tightening torque for the specified condition. How could this torque be

obtained?

Solution:

1 ¼ in., 8-thread series

Table 5.1


Page 12 of 42

inDr 0966.1=

..000.1 insqAs =

8. =inTh


ksisy 147=

(a) yi ss 70.0=

( ) psiksisi 900,1029.10214770.0 ===

E

Ls ei=δ

.1.197.0 inGLe +=

inG 8.4=

( ) ininLe 756.5.1.18.497.0 =+=

( )( )in

E

Ls ei 01975.01030

756.5900,1026

=×

==δ

(b) GD

TL

p r

4

64

π

δθ ==

psiG 6105.11 ×=

ininp 125.08

1==

( )( ) ( )64

105.110966.1

756.564

125.0

01975.0

×==

πθ

T

lbinT −= 408,22

ELASTIC CONSIDERATIONS

235. The member C shown is part of a swivel connection that is to be clamped by a

1-in. bolt D to the member B, which has large dimensions in the plane

perpendicular to the paper. Both B and C are aluminum alloy 2024-T4, HT

aged. The bolt is made of AISI C1113, cold-drawn steel; consider the

unthreaded shank to be 2 in. long; it is well tightened with a torque of 250 ft-

lb.; UNC threads, unlubricated. (a) Estimate the initial tension by equation

(5.2), assume elastic action, and compute the bolt elongation and the total

deformation of B and C. Let the effective strain length be 2 in. (b) After

tightening an external axial force eF of 5000 lb. is applied to member C.

Determine the total normal stresses in the bolt and in B and C. (c) Determine

the load required to “open” the connection. Draw a diagram similar to Fig.

5.6, Text, locating points A, B, D and M.


Page 13 of 42

Prob. 235, 236

Solution:

For aluminum alloy, 2024-T4 HT aged,

psiE 6106.10 ×=

ksisy 47=

For AISI C1113, cold-drawn steel,

psiE 61030×=

ksisy 72=

(a) iDFT 2.0=

.1 inD =

lbinlbftT −=−= 3000250

lbFi 000,15=

Deformations: .2 inL =

Table AT 14, 1-in. UNC Bolt,

..66.0 insqAs =

( ) ..785.014

2insqAb ==

π

Bolt:

( )( )( )( )

inEA

LF

ib

i

i 00127.01030785.0

2000,156

=×

==δ

Member B and C

cc

ic

EA

LF=δ

22

44DDA ec

ππ−=

eD = (Nut or head width across flats) + 2

h

Table AT 14

inA2

11=


Page 14 of 42

.2

12

2

2

2

11 inDe =+=

22

44DDA ec

ππ−=

( ) ( )[ ] ..1234.415.24

2

2insqAc =−=

π

( )( )( )( )

.000686.0106.101234.4

2000,156

inEA

LF

cc

ic =

×==δ

(b) lbFe 5000=

+=

+=

+=∆

00127.0000686.0

000686.05000

cb

be

cb

beb F

kk

kFF

δδ

δ

lbFb 1754=∆

Bolt:

lbFFF bit 754,161754000,15 =+=∆+=

psiA

Fs

s

tb 132,29

606.0

754,16===

Member B and C

+−=

cb

ceic

kk

kFFF

+−=

cb

ceic FFF

δδ

δ

lbFc 754,11000686.000127.0

00127.05000000,15 =

+−=

psiA

Fs

c

cc 2851

1234.4

754,11===

(c) oF = opening load

lbFFi

ciio 102,23

00127.0

000686.000127.0000,15 =

+=

+=

δ

δδ

Fig. 5.6


Page 15 of 42

237. A 1-in. steel bolt is used to clamp two aluminum (2014-T6, HT aged) plates

together as shown by Fig. 5.9, Text. The aluminum plates have a total

thickness of 2 in. and an equivalent diameter of 2 in. The bolt is heated to a

temperature of 200 oF, the inserted in the aluminum plates, which are at 80

oF,

and tightened so as to have a tensile tightening stress of 30 ksi in the

unthreaded shank while steel at 200 oF. What is the tensile stress in the bolt

after assembly has cooled to 80 oF? The deformations are elastic.

Figure 5.9

Solution:

For aluminum 2014-T6

psiE 6106.10 ×=

psisb 000,30=

( ) ( ) lbAsF bbi 562,2314

000,302

=

==

π

Steel bolt. psiEb

61030×=

( )( ).002.0

1030

2000,306

inE

Ls

b

b

i =×

==δ

.cc

i

cEA

LF=δ

22

44DDA ec

ππ−=

( ) ( )[ ] ..3562.2124

22insqAc =−=

π


Page 16 of 42

psiEc

6106.10 ×=

( )( )( )( )

inEA

LF

cc

ic 001887.0

106.103562.2

2562,236

=×

==δ

.998113.1001887.02 inLL c =−=−=′ δ

tLL ∆′=∆ α

( )Finin −= ..000007.0α for steel

( )( )( ) .001678.020080998113.1000007.0 inL −=−=∆

.000322.0001678.0002.0 inLii =−=∆+=′ δδ

b

bi

E

Ls′=′δ

( )61030

2000322.0

×

′= bs

psisb 4830=′

238. A 1 1/8-in. steel bolt A passes through a yellow brass (B36-8) tube B as

shown. The length of the tube is 30 in. (virtually the unthreaded bolt length),

the threads on the bolt are UNC, and the tube’s cross-sectional area is 2 sq. in.

After the nut is snug it is tightened ¼ turn. (a) What normal stresses will be

produced in the bolt and in the tube? Assume that washers, nut, and head are

rigid. (b) What are the stresses if an axial load of 5 kips is now applied to the

bolts end? Compute the bolt load that just results in a zero stress in the tube.

Prob. 238

Solution:

For Yellow brass, B36-8,

psiE 61015×=

Steel bolt

psiE 61030×=

Table AT 14, 1 1/8 in., UNC

.9497.0 inDr =

..763.0 insqAs =

7=inTh


Page 17 of 42

.30 inL =

p

iδθ =

.7

1inp =

4

1=θ turn

.28

1

7

1

4

1ini =

=δ

bb

ii

EA

LF=δ

( )

( )6

2

10308

11

4

30

28

1

×

=

π

iF

lbFi 500,35=

(a) Bolt: psiA

Fs

s

ib 527,46

763.0

500,35===

Tube: c

ic

A

Fs =

..2 insqAc =

psiA

Fs

c

ic 750,17

2

500,35===

(b) lbFe 5000=

( )( )inlb

L

EAk cc

c 000,000,130

10152 6

=×

==

( )inlb

L

EAk bb

b 000,99430

10308

11

4

6

2

=

×

==

π

Bolts:

e

cb

bit F

kk

kFF

++=

( ) lbFt 000,385000000,000,1000,994

000,994500,35 =

++=

psiA

Fs

s

tt 800,49

763.0

000,38===


Page 18 of 42

Tube:

e

cb

cic F

kk

kFF

+−=

( ) lbFc 000,335000000,000,1000,994

000,000,1500,35 =

+−=

psiA

Fs

c

cc 500,16

2

000,33===

For zero stress in the tube

( ) lbFk

kkF i

c

cbo 787,70500,35

000,000,1

000,000,1000,994=

+=

+=

ENDURANCE STRENGTH

DESIGN PROBLEMS

239. As shown diagrammatically, a bearing is supported in a pillow block attached

to an overhead beam by two cap screws, each of which, it may be assumed,

carried half the total bearing load. This load acts vertically downward, varying

from 0 to 1500 lb. The screws are to be made of AISI C1118, as rolled, and

they are tightened to give an initial stress of about yi ss 5.0= . The pillow

block is made of class-20 cast iron. Assume that the effective length of screw

is equal to the thickness t , as shown, and that the head and beam are rigid

(overly conservative?). The equivalent diameter of the compression area may

be taken as twice the bolt diameter. For a design factor of 1.75, determine the

size of the screw: (a) from the Soderberg line, (b) from the modified Goodman

line. (c) What size do you recommend using?

Problem 239

Solution:


ksisy 46=

ksisu 75=

yi ss 5.0=


Page 19 of 42

( ) psiksisi 000,2323465.0 ===

lbAkipAAsF sssii 000,2323 ===

e

cb

bb F

kk

kF

+=∆

b

bbb

L

EAk =

psiEb

61030×= (steel)

tLb =

c

ccc

L

EAk =

For cast-iron class 20

psiEc

6106.9 ×=

tLc =

22

44DDA ec

ππ−=

2

4DAb

π=

DDe 2=

( ) bc ADDDA 34

3

42

4

222==−=

πππ

b

bbb

L

EAk =

( )t

Ak b

b

61030×=

( )t

Ak b

c

6106.93 ×=

( ) 588

300

106.931030

103066

6

=×+×

×=

+ cb

b

kk

k

01 =∆ bF

lbFkk

kF e

cb

bb 383

2

1500

588

3002 =

=

+=∆

( ) ( ) lbAAFFFF ssbbim 192000,2303832

1000,23

2

112 +=++=∆+∆+=

( ) ( ) lbFFF bba 19203832

1

2

112 =−=∆−∆=


Page 20 of 42

( ) psipsiss un 500,37000,755.05.0 ===

For axial loading with size factor

( )( )( ) psipsisn 500,25500,3785.08.0 ==

75.1=N

sss

s

s

mm

AAA

A

A

Fs

192000,23

192000,23+=+==

ss

aa

AA

Fs

192==

Table AT 12, 8.1=fK

(a) Soderberg line

n

af

y

m

s

sK

s

s

N+=

1

( )

500,25

1928.1

000,46

192000,23

75.1

1

+

+

= ss AA

..2482.0 insqAs =

Table AT 14, UNC

Use .4

3inD = , ..334.0 insqAs =

(b) Modifies Goodman line

n

af

u

m

s

sK

s

s

N+=

1

( )

500,25

1928.1

000,75

192000,23

75.1

1

+

+

= ss AA

..0609.0 insqAs =

Table AT 14, UNC

Use .8

3inD = , ..0775.0 insqAs =


Page 21 of 42

(c) Recommended, UNCinD −= .4

3

240. A connection similar to Fig. 5.9, Text, is subjected to an external load that

varied from 0 to 1250 lb. The bolt is cold forged from AISI B1113 steel; UNC

threads.The aluminum parts C (3003 H14) have a total thickness of 1 ½ in.

and an external diameter of D2 . It is desired that the connection not open for

an external load of eF5.1 . Determine (a) the initial tensile load on the bolt, (b)

the bolt diameter for 2=N based on the Soderberg line.

Fig. 5.9

Solution:

(a) lbkk

kQFF

cb

cei

+=

5.1=Q

b

bbb

L

EAk =

2

4DAb

π=

psiEb

61030×=

.2

11 inLb =

c

ccc

L

EAk =

22

44DDA ec

ππ−=

DDe 2=

( ) bc ADDDA 34

3

42

4

222==−=

πππ

psiEc

61010×= (3003-H14 aluminum)


Page 22 of 42

.2

11 inLc =

( )

×=

2

11

1030 6

bb

Ak

( )

×=

2

11

10103 6

bc

Ak

( )( )

5.0101031030

1010366

6

=×+×

×=

+ cb

c

kk

k

lbFe 1250=

lbkk

kQFF

cb

cei

+=

( )( )( ) lbFi 5.9375.012505.1 ==

(b) For AISI B1113 steel, cold forged

ksisu 83=

ksisy 72=

( ) psiksiss un 500,415.41835.05.0 ====


( )( )( ) psipsisn 220,28500,4185.08.0 ==

e

cb

bb F

kk

kF

+=∆

01 =∆ bF

( )( ) lbF

kk

kF e

cb

bb 6251250

101031030

103066

6

2 =

×+×

×=

+=∆

( ) ( ) lbFFFF bbim 125006252

15.937

2

112 =++=∆+∆+=

( ) ( ) lbFFF bba 5.31206252

1

2

112 =−=∆−∆=

ss

mm

AA

Fs

1250==


Page 23 of 42

ss

aa

AA

Fs

5.312==

Soderberg line, 8.1=fK Table AT 12

n

af

y

m

s

sK

s

s

N+=

1

( )( )

ss AA 220,28

5.3128.1

000,72

1250

2

1+=

..07459.0 insqAs =

Table AT 14, UNC

Use .8

3inD = , ..0775.0 insqAs =

243. This problem concerns the Freon compressor of 226: size, 10 x 18 in.; 10

studs, UNC; made of C1118, as rolled; 200 psi gas pressure. The initial

tension in the bolts, assumed to be equally loaded, is such that a cylinder

pressure of 300 psi is required for the joint to be on the opening. The bolted

parts are cast steel and for the first calculations, it will be satisfactorily to

assume the equivalent diameter of the compressed parts to be twice the bolt

size. (a) For 2=N on the Soderberg criterion, what bolt size is required? (b)

Compute the torque required for the specified initial tension.

Solution:

( ).2356

10

10

4300

2

lbFo =

=

π

+=

cb

coi

kk

kFF

b

bbb

L

EAk =

2

4DAb

π=

psiEb

61030×=

LLb =

c

ccc

L

EAk =

22

44DDA ec

ππ−=

DDe 2=


Page 24 of 42

( ) bc ADDDA 34

3

42

4

222==−=

πππ

Cast Steel, psiEc

61030×=

LLc =

( )L

Ak b

b

61030×=

( )b

bc k

L

Ak 3

10303 6

=×

=

( ) lbkk

k

kk

kFF

bb

b

cb

coi 1767

3

32356 =

+=

+=

(a) e

cb

bb F

kk

kF

+=∆

01 =∆ bF

( ) ( )lb

kk

kF

kk

kF

bb

be

cb

bb 393

10

20010

43

2

2 =

+=

+=∆

π

( ) ( ) lbFFFF bbim 196403932

11767

2

112 =++=∆+∆+=

( ) ( ) lbFFF bba 19603932

1

2

112 =−=∆−∆=

ss

mm

AA

Fs

1964==

ss

aa

AA

Fs

196==

For C1118, as rolled

ksisu 75=

ksisy 46=

( ) psiksiss un 500,375.37755.05.0 ====


( )( )( ) psipsisn 500,25500,3785.08.0 ==

8.1=fK Table AT 12

n

af

y

m

s

sK

s

s

N+=

1

( )( )

ss AA 500,25

1968.1

000,46

1964

2

1+=

..1131.0 insqAs =


Page 25 of 42

Table AT 14, UNC

Use .2

1inD = , ..1419.0 insqAs =

(b) iDFT 2.0=

( ) .7.17617672

12.0 lbinT −=

=

245. A cast-iron (class 35) Diesel-engine cylinder head is held on 8 stud bolts with

UNC threads. These bolts are made of AISI 3140 steel, OQT 1000 oF (Fig.

AF2). Assume that the compressed material has an equivalent diameter twice

the bolt size. The maximum cylinder pressure is 750 psi and the bore of the

engine is 8 in. Let the initial bolt load be such that a cylinder pressure of 1500

psi brings the joint to the point of opening. For a design factor of 2, determine

the bolt diameter (a) using the Soderberg equation, (b) using the Goodman

equation. (c) What approximate torque will be required to induce the desired

initial stress? (d) Determine the ratio of the initial stress to the yield strength.

Considering the lessons of experience (i5.8), what initial stress would you

recommend? Using this value, what factor of safety is computed from the

Soderberg equation?

Solution:

( ).9425

8

8

41500

2

lbFo =

=

π

+=

cb

coi

kk

kFF

b

bbb

L

EAk =

2

4DAb

π=

psiEb

61030×=

LLb =

c

ccc

L

EAk =

22

44DDA ec

ππ−=

DDe 2=

( ) bc ADDDA 34

3

42

4

222==−=

πππ


Page 26 of 42

psiEc

6105.14 ×= , for cast-iron (class 35)

LLc =

( )L

Ak b

b

61030×=

( )L

Ak b

c

6105.143 ×=

( ) ( )( )

lbkk

kFF

cb

coi 5578

105.1431030

105.1439425

66

6

=

×+×

×=

+=

e

cb

bb F

kk

kF

+=∆

01 =∆ bF

( )( ) ( )

lbFkk

kF e

cb

bb 1923

8

7508

4105.1431030

10302

66

6

2 =

×+×

×=

+=∆

π

( ) ( ) lbFFFF bbim 6540019232

15578

2

112 =++=∆+∆+=

( ) ( ) lbFFF bba 962019232

1

2

112 =−=∆−∆=

ss

mm

AA

Fs

6540==

ss

aa

AA

Fs

962==

(a) For AISI 3140 steel, OQT 1000 oF

ksisu 153=

ksisy 134=

( ) psiksiss un 500,765.761535.05.0 ====


( )( )( ) psipsisn 000,52500,7685.08.0 ==

Table AT 12, 3.3=fK (hardened)

Soderberg Equation

n

af

y

m

s

sK

s

s

N+=

1

( )( )

ss AA 000,52

9623.3

000,134

6540

2

1+=

..2197.0 insqAs =


Page 27 of 42

Table AT 14, UNC

Use .8

5inD = , ..226.0 insqAs =

(b) Goodman Equation

n

af

u

m

s

sK

s

s

N+=

1

( )( )

ss AA 000,52

9623.3

000,153

6540

2

1+=

..2076.0 insqAs =

Table AT 14, UNC

Use .8

5inD = , ..226.0 insqAs =

(c) iDFT 2.0=

( ) .69755788

52.0 lbinT −=

=

(d) psiA

Fs

s

ii 681,24

226.0

5578===

Ratio = 184.0000,134

681,24==

y

i

s

s

i5.8 ( ) psiss yi 900,113000,13485.085.0 ===

Factor of safety

( )( ) lbAsF sii 742,25226.0900,113 ===

( ) lbFm 704,2619232

1742,25 =+=

( ) lbFa 96219232

1==

psiA

Fs

s

mm 159,118

226.0

704,26===

psiA

Fs

s

aa 4257

226.0

962===

Soderberg Equation


Page 28 of 42

n

af

y

m

s

sK

s

s

N+=

1

( )( )000,52

42573.3

000,134

159,1181+=

N

87.0=N

246. A 30,000-lb. body is to be mounted on a shaker (vibrator). The shaker will

exert a harmonic force of .2sin000,30 lbftF π= on the body where f cps is

the frequency and t sec. is the time. The frequency can be varied from 5 to

10,000 cps. The harmonic force will exert a tensile load on the bolts that

attach the body to the shaker when F is positive. Determine the minimum

number of ½-in.-UNF bolts that must be used for 2=N based on Soderberg

line. The material of the bolts is to be AISI 8630, WQT 1100 oF; the material

of the body that is to be vibrated is aluminum alloy, 2014-T6 and the joint is

not to open for an external force that is 1.25 times the maximum force exerted

by the shaker. It may be assumed that the equivalent diameter of the material

in compression is twice the bolt diameter.

Solution:

0min =eF

lbFe 000,30max =

+=

cb

cei

kk

kQFF

25.1=Q

b

bbb

L

EAk =

2

4DAb

π=

psiEb

61030×=

LLb =


Page 29 of 42

c

ccc

L

EAk =

22

44DDA ec

ππ−=

DDe 2=

( ) bc ADDDA 34

3

42

4

222==−=

πππ

psiEc

6106.10 ×= , (Aluminum 2014-T6)

LLc =

( )L

A

L

EAk b

b

bbb

61030×==

( )L

A

L

EAk b

c

ccc

6106.103 ×==

+=

cb

cei

kk

kQFF

( )( ) ( )( )

lbFi 296,19106.1031030

106.103000,3025.1

66

6

=

×+×

×=

e

cb

bb F

kk

kF

+=∆

01 =∆ bF

( )( ) lbF

kk

kF e

cb

bb 563,14000,30

106.1031030

103066

6

2 =

×+×

×=

+=∆

( ) ( ) lbFFFF bbim 578,260563,142

1296,19

2

112 =++=∆+∆+=

( ) ( ) lbFFF bba 72820563,142

1

2

112 =−=∆−∆=

s

mm

nA

Fs =

s

a

anA

Fs =

For ½-in.-UNF (Table AT 14)

..1419.0 insqAs =

nnnA

Fs

s

mm

300,187

1419.0

578,26===


Page 30 of 42

nnnA

Fs

s

aa

318,51

1419.0

7282===

For AISI 8630, WQT 1100 oF

3.3=fK

ksisu 137=

ksisy 125=

( ) psiksiss un 500,685.681375.05.0 ====


( )( )( ) psipsisn 580,46500,6885.08.0 ==

Soderberg Equation, 2=N

n

af

y

m

s

sK

s

s

N+=

1

( )( )nn 580,46

318,513.3

000,125

300,187

2

1+=

3.10=n

Minimum number of bolts = 10 bolts

248. The maximum external load on the cap bolts of an automotive connecting rod

end, imposed by inertia forces at top dead center, is taken to be 4000 lb.; the

minimum load is zero at bottom dead center. The material is AISI 4140, OQT

1100oF (qualifying for SAE grade 5); assume that un ss 45.0=′ . The grip for

through bolts is 1.5 in. For design purposed, let each bolt take half the load,

and use an equivalent .8

31 inDe = for the connected parts. The threads extend

a negligible amount into the grip. For the initial computation, use an opening

load eo FF 75.1= . Considering the manner in which the bolt is loaded, we

decide that a design factor of 1.4 (Soderberg) should be quite adequate. (a)

Does a 5/16-24 UNF satisfy this situation? If not, what size do you

recommend? (b) Experience suggests that, in situations such as this, an initial

stress of the order suggested in i5.8, Text, is good insurance against fatigue

failure. Decide upon such an is and recomputed N . How does it change?

Would you be concerned about the safety in this case? Consider the variation

of is as a consequences of the use of torque wrench and also the stress

relaxation with time (due to seating and other factors), and discuss. Compute

the required tightening torque for each is .


Page 31 of 42

Solution:

( ) lbFF eo 7000400075.175.1 ===

+=

cb

cei

kk

kFF

b

bbb

L

EAk =

2

4DAb

π=

psiEb

61030×=

.5.1 inLb =

c

ccc

L

EAk =

22

44DDA ec

ππ−=

bc ADDA −=−=−

= 485.1

4485.1

48

31

4

22

2πππ

psiEc

6106.10 ×= , (Aluminum 2014-T6)

.5.1 inLc =

bs AA ≈

( )5.1

1030 6×== b

b

bbb

A

L

EAk

( )( )5.1

1030485.1 6×−== s

c

ccc

A

L

EAk

ss

i AA

F 47147000485.1

485.17000 −=

−=

e

cb

bb F

kk

kF

+=∆

01 =∆ bF

( )s

se

cb

bb A

AF

kk

kF 2694000,4

485.12 =

=

+=∆

( ) ( ) sssbbim AAAFFFF 33677000026942

147147000

2

112 −=++−=∆+∆+=

( ) ( ) ssbba AAFFF 1347026942

1

2

112 =−=∆−∆=


Page 32 of 42

Table 5.2

ksisu 120=

ksisy 88=

( ) psiksiss un 000,545412045.045.0 ====′

33677000

−==ss

mm

AA

Fs

1347==s

aa

A

Fs

3.3=fK (hardened, Table AT 12)

Soderberg Equation, 4.1=N

n

af

y

m

s

sK

s

s

N+=

1

( )( )000,54

13473.3

000,88

3367

000,88

7000

4.1

1+−=

sA

..1187.0 insqAs =

Table At14, we inD16

7= , ..1187.0 insqAs =

(a) 5/16-24 UNF will not satisfy the situation. Instead use inD16

7= ,

..1187.0 insqAs =

(b) i5.8, Text

( ) psiksiss yi 800,748.748885.085.0 ====

( )( ) lbAsF sii 88791187.0800,74 ===

( ) ( ) ssbbim AAFFFF 13478879026942

18879

2

112 +=++=∆+∆+=

( ) ( ) ssbba AAFFF 1347026942

1

2

112 =−=∆−∆=

13478879

+==ss

mm

AA

Fs

1347==s

aa

A

Fs

n

af

y

m

s

sK

s

s

N+=

1


Page 33 of 42

( )( )000,54

13473.3

000,88

13471187.0

8879

1+

+

=N

06.1=N , it decreases

1>N , therefore, safe.

Considering variation of is , is tends to exceeds the limiting stress therefore reduces the

factor of safety. While stress relaxation tends to reduce the limiting stress approaching

the is and causing lower design factor.

(c) ( ) lbAF si 64401187.04714700047147000 =−=−=

( ) lbinDFT i −=

== 5646440

16

72.02.0

at lbFi 8879=

( ) lbinDFT i −=

== 7778879

16

72.02.0

CHECK PROBLEMS

249. A 1-in. steel bolt A (normalized AISI 1137, cold-rolled threads) passes

through a yellow brass tube B (B36-8, ½ hard) as shown. The tube length is

30 in., its cross-sectional area is 2 sq. in. and the UNC bolt threads extend a

negligible amount below the nut. The steel washers are ¼ in. thick and are

assumed not to bend (clearances are exaggerated). The nut is turned ¼ turn.

(a) If an external tensile axial load, varying from 0 to 5 kips, is repeatedly

applied to the bolt, what is the factor of safety of the bolt by the Soderberg

criterion? (b) What is the external load on the bolt at the instant that the load

on the tube becomes zero.

Problem 249, 250

Solution:

For 1-in. UNC

..606.0 insqAs =


Page 34 of 42

.8466.0 inDr =

8. =inTh

.8

11in

inThp ==

4

1=

p

δ

.32

1

8

1

4

1in=

=δ

.5.304

1230 inLb =

+=

psiEb

61030×=

( )

( ) ( )6210301

4

5.30

32

1

×

==π

δ iF

lbFi 141,24=

+=∆

cb

beb

kk

kFF

( ) ( )525,772

5.30

103014

62

=×

==

π

b

bbb

L

EAk

c

ccc

L

EAk =

..2 insqAc =

.30 inLc =

psiEc

61015×= (Yellow Brass)

( )000,000,1

30

10152 6

=×

=ck

(a) 01 =∆ bF

( ) lbFb 2179000,000,1525,772

525,77250002 =

+=∆

( ) ( ) lbFFFF bbim 230,25021792

1141,24

2

112 =++=∆+∆+=

( ) ( ) lbFFF bba 1090021792

1

2

112 =−=∆−∆=

psiA

Fs

s

mm 227,38

606.0

230,25===


Page 35 of 42

psiA

Fs

s

aa 1799

606.0

1090===

For normalized AISI 1137, cold-rolled thread

ksisu 98=

ksisy 58=

( ) psiksiss un 000,4949985.05.0 ====

4.1=fK (Table AT 12)

For axial loading, ( ) psisn 200,39000,498.0 ==

n

af

y

m

s

sK

s

s

N+=

1

( )( )200,39

17994.1

000,58

227,381+=

N

38.1=N

(b)

+=

cb

coi

kk

kFF

+=

000,000,1525,772

000,000,1141,24 oF

lbFo 790,42=

250. A ¾-in. fine-thread bolt, made of AISI 1117, cold drawn, with rolled threads,

passes through a yellow brass tube and two steel washers, as shown. The tube

is 4 in. long, 7/8 in. internal diameter, 1 ¼-in. external diameter. The washers

are each ¼-in. thick. The unthreaded part of the bolt is 3 in. long. Assume that

there is no stretching of the bolt inside the nut in finding its k . The

unlubricated bolt is tightened by a torque of 1800 in-lb. The external load,

varying from 0 to 4 kips, is axially applied to the washers an indefinite

number of times. (a) Compute the factor of safety of the bolt by the Soderberg

criterion. Is there any danger of failure of the bolt? (b) What pull must be

exerted by the washers to remove all load from the brass tube?

Solution:

iDFT 2.0=

iF

=

4

32.01800

lbFi 000,12=


Page 36 of 42

+=∆

cb

beb

kk

kFF

21

111

bbb kkk+=

1

1

b

bbb

L

EAk =

.31 inLb =

..4418.04

3

4

2

insqAb =

=

π

psiEb

61030×=

( )( )000,418,4

3

10304418.0 6

1 =×

=bk

2

2

b

bsb

L

EAk =

For ¾-in. UNF (Table AT 14)

..373.0 insqAs =

.5.132

1242 inLb =−

+=

( )( )000,460,7

5.1

1030373.0 6

2 =×

=bk

21

111

bbb kkk+=

000,460,7

1

000,418,4

11+=

bk

733,774,2=bk

c

ccc

L

EAk =

..6259.08

7

4

11

4

22

insqAc =

−

=

π

psiEc

61015×=

.4 inLc =

( )( )125,347,2

4

10156259.0 6

=×

=ck

01 =∆ bF

( ) lbFb 2167125,347,2733,774,2

733,774,240002 =

+=∆


Page 37 of 42

( ) ( ) lbFFFF bbim 084,13021672

1000,12

2

112 =++=∆+∆+=

( ) ( ) lbFFF bba 1084021672

1

2

112 =−=∆−∆=

psiA

Fs

s

mm 078,35

373.0

084,13===

psiA

Fs

s

aa 2906

373.0

1084===

For AISI 111, cold drawn, rolled threads

ksisn 40=

ksisy 68=

4.1=fK

( ) psiksisn 000,3232408.0 === , axial loading

(a) n

af

y

m

s

sK

s

s

N+=

1

( )( )000,32

29064.1

000,68

078,351+=

N

56.1=N

(b)

+=

cb

coi

kk

kFF

+=

125,347,2733,774,2

125,347,2000,12 oF

lbFo 186,26=

251. A coupling bolt (i5.13, Text) is used to connect two parts made of cast-iron,

class 35. The diameter of the coarse-thread bolt is ½-in.; its grip is 2 in., which

is also nearly the unthreaded length. The bolt tightened to have an initial

tension of 4000 lb. The parts support an external load eF that tends to separate

them and it varies from zero to 5000 lb. What is the factor of safety,

(Soderberg)?

Solution:

lbFi 4000=

+=∆

cb

beb

kk

kFF


Page 38 of 42

b

bbb

L

EAk =

..19635.02

1

4

2

insqAb =

=

π (unthreaded length)

psiEb

61030×=

.2 inLb =

( )( )250,945,2

2

103019635.0 6

=×

=bk

Table AT 14, UNC

.2

1inD =

..1419.0 insqAs =

.4

3inA =

2

hADe +=

.2 inh =

.4

31

2

2

4

3inDe =+=

c

ccc

L

EAk =

..209.22

1

4

31

444

22

22insqDDA ec =

−

=−=

πππ

psiEc

6105.14 ×= , (Cast iron, class 35)

.2 inLc =

( )( )250,015,16

2

105.14209.2 6

=×

=ck

01 =∆ bF

( ) lbFb 777250,015,16250,945,2

250,945,250002 =

+=∆

( ) ( ) lbFFFF bbim 438907772

14000

2

112 =++=∆+∆+=

( ) ( ) lbFFF bba 38907772

1

2

112 =−=∆−∆=

psiA

Fs

s

mm 930,30

1419.0

4389===

psiA

Fs

s

aa 2741

1419.0

389===


Page 39 of 42

For ASTM 354 BC (Table 5.2), .2

1inD =

ksisu 125=

ksisy 109=

un ss 5.0=

For axial loading

( )( )( ) psiksisn 000,50501255.08.0 ===

8.1=fK

Soderberg Line

n

af

y

m

s

sK

s

s

N+=

1

( )( )000,50

27418.1

000,109

930,301+=

N

6.2=N

252. The cap on the end of a connecting rod (automotive engine) is held on by two

5/16-in. bolts that are forged integrally with the main connecting rod. These

bolts have UNF threads with a 5/8-in. on an unthreaded length of virtually 5/8

in. The nuts are to be tightened with a torque of 20 ft-lb. and the maximum

external load on one bolt is expected to be 2330 lb. Let the equivalent

diameter of the connected parts be ¾ in. (a) Estimate the maximum force on

the bolt. (b) Compute the opening load. Is this satisfactory? (c) If the bolt

material is AISI 4140, OQT 1000 oF, what is the factor of safety based on the

Soderberg criterion?

Solution:

lbinlbftT −=−= 24020

iDFT 2.0=

iF

=

16

52.0240

lbFi 3840=

+=∆

cb

beb

kk

kFF

b

bbb

L

EAk =

..0767.016

5

4

2

insqAb =

=

π (unthreaded length)

psiEb

61030×=

.8

5inLb =


Page 40 of 42

( )( )600,681,3

8

5

10300767.0 6

=

×=bk

c

ccc

L

EAk =

..3651.016

5

4

3

444

22

22insqDDA ec =

−

=−=

πππ

psiEc

61030×= , (Cast iron, class 35)

.8

5inLc =

( )( )800,524,17

8

5

10303651.0 6

=

×=ck

( ) lbFb 405800,524,17600,681,3

600,681,32330 =

+=∆

(a) lbFFF bi 42454053840max =+=∆+=

(b)

+=

cb

coi

kk

kFF

+=

80,524,17600,681,3

800,524,173840 oF

max4647 FlbFo <=

(c) lbF

FF bim 4042

2

4053840

2=+=

∆+=

lbF

F ba 202

2

405

2==

∆=

For AISI 4140, OQT 1000 oF

ksisu 170=

ksisy 155=

Table AT 12, 6.2=fK

un ss 5.0=

For axial loading

( )( )( ) psiksisn 000,68681705.08.0 ===

Soderberg Line

n

af

y

m

s

sK

s

s

N+=

1


Page 41 of 42

For 5/16-in.-UNF, Table AT 14, ..0580.0 insqAs =

psiA

Fs

s

mm 690,69

0580.0

4042===

psiA

Fs

s

aa 3843

0580.0

202===

n

af

y

m

s

sK

s

s

N+=

1

( )( )000,68

34836.2

000,155

690,691+=

N

72.1=N

SET SCREWS

254. A 6-in. pulley is fastened to a 1 ¼ in. shaft by a set screw. If a net tangential

force of 75 lb, is applied to the surface of the pulley, what size screw should

be used when the load is steady?

Solution:

Tangential force = ( ) lblb 36525.1

675 =

Assume tangential force = holding force

Table 5.3, use Screw size 8, Holding force = 385 lb.

255. An eccentric is to be connected to a 3-in. shaft by a setscrew. The center of the

eccentric is 1 ¼ in. from the center of the shaft when a tensile force of 1000

lb. is applied to the eccentric rod perpendicular to the line of centers. What

size set screw should be used for a deign factor of 6?

Solution:

Tangential force = ( ) lblb 83323

25.11000 =

Holding force = ( )( ) lb50008336 =

Table 5.3, use Screw size ¾ in.

256. A lever 16 in. long is to be fastened to a 2-in. shaft. A load of 40 lb. is to be

applied normal to the lever at its end. What size of set screw should be used

for a design factor of 5?

Solution:

Torque = ( )( ) lbin −= 6404016

Tangential force = ( )

lb6402

6402=


Page 42 of 42

Holding force = ( )( ) lb32006405 =

Table 5.3, use Screw size 9/16 in.

257. A 12-in. gear is mounted on a 2-in. shaft and is held in place by a 7/16 in.

setscrew. For a design factor of 3, what would be the tangential load that

could be applied to the teeth and what horsepower could be transmitted by the

screw.

Solution:

Table 5.3, 7/16 in.

Holding force = 2500 lb

Tangential force = lb8333

2500=

Tangential load on gear = lb13912

2833 =

Assume fpmvm 4500=

Hp transmitted = ( )( )

hp19000,33

4500139=

- end -

SECTION 4 - SPRINGS

Page 1 of 70

HELICAL COMPRESSION SPRINGS

DESIGN – LIGHT, MEDIUM SERVICE

271. A solenoid brake (Fig. 18.2, Text) is to be actuated by a helical compression

spring. The spring should have a free length of approximately 18 in. and is to

exert a maximum force of 2850 lb. when compressed to a length of 15 in. The

outside diameter must not exceed 7 in. Using oil-tempered wire, design a

spring for this brake, (wire diameter, coil diameter, number of active coils,

pitch, pitch angle, “solid stress”). General Electric used a spring made of 1 in.

wire, with an outside diameter of 6 in., and 11 ½ free coils for a similar

application.

Solution:

For oil tempered wire, Table AT 17

ksiD

s

w

u 19.0

146= , [ ]5.0032.0 << wD

“solid stress” = us6.0

design stress, (average service)

usd ss 324.0=

( )ksi

DDs

ww

sd 19.019.0

304.47146324.0==

7≤+ mw DD

kipslbF 85.22850 ==

19.03

304.478

ww

ms

DD

FDKs =

=

π

say 3.1=K

( )( )19.03

304.47785.283.1

ww

ws

DD

Ds =

−=

π

ininDw 5.0062.1 >=

use ( )

ksiksissd 545.0

304.4719.0

==

( )( )54

785.283.1

3=

−=

w

ws

D

Ds

π

inDw 015.1=

say inDw 0.1=

( )( )

541

85.283.1

3=

=

πm

s

Ds

inDm 72.5=

SECTION 4 - SPRINGS

Page 2 of 70

say inDm 0.5=

.760.10.5 ininDDOD wm <=+=+=

50.1

0.5===

w

m

D

DC

δ = Free length – Compressed length = 18 in – 15 in = 3 in.

w

c

GD

NFC38

=δ

ksiG 500,10= , inDw8

3>

( )( )( )( )1500,10

585.283

3

cN==δ

05.11=cN

say 5.11=cN

( )( ) ( )( )( )

in12.31500,10

5.11585.283

==δ

Free length = 15 + 3.12 = 18.12 in

At 2.85 kips

=

3

8

w

ms

D

FDKs

π

5=C ( )( )

3105.15

615.0

454

154615.0

44

14=+

−

−=+

−

−=

CC

CK

( )( )( )

ksiss 55.471

585.283105.1

3=

=

π

Permissible solid stress

( )( )

ksiksiss uso 93.995.0

1466.06.0

19.0===

using δ

Fk =

or let Tδ = Free length – Solid height

Tδ

93.99

12.3

55.47=

inT 56.6=δ

Tδ = Free length – Solid height = ( ) cw NDP −

( )( )5.11156.6 −= P inP 570.1=

SECTION 4 - SPRINGS

Page 3 of 70

use inP2

11=

Pitch angle,

( )oo 125.5

5

5.1tantan 11 <=

== −−

ππλ

D

P, o.k.

For actual solid stress

( )( ) .75.55.1115.1 inT =−=δ

75.512.3

55.47 sos=

ksiksisso 93.9963.87 <= , ok

Summary of answer:

wD = wire diameter = 1 in.

mD = coil diameter = 5 in.

cN = no. of active coils = 11 1/2

P = pitch = 1 ½ in.

γ = pitch angle = 5.5o

sos = solid stress = 87.63 ksi

272. A coil spring is to be used for the front spring of a automobile. The spring is

to have a rate of 400 lb./in., an inside diameter of 4 3/64 in., and a free length

of 14 1/8 in., with squared-and-ground ends. The material is to be oil-

tempered chrome vanadium steel. Decide upon the diameter of the wire and

the number of free coils for a design load of lbF 1500= . Be sure “solid

stress” is all right. How much is the pitch angle?

Solution:

Table AT 17 Cr-V steel

ksiD

s

w

u 166.0

168= , [ ]437.0032.0 wD<

average service

usd ss 324.0=

( )ksi

DDs

ww

sd 166.0166.0

432.54168324.0==

Max “solid stress” = us6.0

SECTION 4 - SPRINGS

Page 4 of 70

ininDDID wm 046875.464

34 ==−=

inDD wm 046875.4+=

sd

w

ms s

D

FDKs =

=

3

8

π

Assume 3.1=K

kipslbF 5.11500 ==

( )( )

+==

3166.0

046875.45.183.1

432.54

w

w

w

sdD

D

Ds

π

ininDw 437.0747.0 >=

use

( )ksiksissd 45.62

437.0

432.5419.0

==

( )( )45.62

046875.45.183.1

3=

+=

w

wsd

D

Ds

π

inDw 724.0=

use inDw4

3=

inDm64

514

64

34

4

3=+=

=

3

8

w

ms

D

FDKs

π

CC

CK

615.0

44

14+

−

−=

4.6

4

3

64

514

≈

==w

m

D

DC

( )( )

235.14.6

615.0

44.64

14.64=+

−

−=K

( )ksiksiss 45.6264.53

4

3

64

5145.18

235.13

<=

=

π

, (o.k.)

w

c

GD

NFC38

=δ

SECTION 4 - SPRINGS

Page 5 of 70

ksiG 500,10= , inDw8

3>

ink

F75.3

400

1500===δ

( )( )

( )

==

4

3500,10

4.65.1875.3

3

cNδ

4.9=cN

Table AT 16, Total coils = 4.1124.92 =+=+cN for square and grounded end.

Summary of answer:

wD = wire diameter = ¾ in.

No. of free coils = 11.4

To check for solid stress.

Permissible solid stress = ( )

( )ksi65.115

437.0

1686.0166.0

=

Free length = wc DPN 2+

Solid height = ( ) ( ) inND cw 55.84.114

32 =

=+

Solid stress = ( ) ksiksi 65.11574.7875.3

55.88

114

64.53 <=

−

(safe)

Pitch:

inDPN wc8

1142 =+

( )8

114

4

324.9 =

+P

ininP32

111343.1 ==

Pitch angle,

oo 121.5

64

514

32

111

tantan 11 <=

== −−

ππ

λD

P, o.k.

SECTION 4 - SPRINGS

Page 6 of 70

273. A coiled compression spring is to fit inside a cylinder 5/8 in. in diameter. For

one position of the piston, the spring is to exert a pressure on the piston

equivalent to 5 psi of piston area, and in this position, the overall length of the

spring must not exceed (but may be less than) 2 in. A pressure of 46 psi on the

piston is to compress the spring ¾ in. from the position described above.

Design a spring for medium service. Specify the cheapest suitable material,

number of total and active coils for square-and-ground ends, and investigate

the pitch angle, and “solid stress”.

Solution:

=

3

8

w

ms

D

FDKs

π

28

5 wwm

DinDDOD −=+=

inDD wm8

55.1 =+

( ) lbF 534.18

5

45

2

1 =

=

π

( ) lbF 647.158

5

4546

2

2 =

+=

π

Using hard-drawn spring wire, Cost Index = 1

( )85.0324.0 usd ss =

ksiD

sw

u 19.0

140=

, [ ]625.0028.0 << wD

Max “solid stress” = ksiDw

19.0

70

( )19.019.0

556.3814085.0324.0

ww

sdDD

s ==

psiD

ksiDD

FCKs

www

s 19.019.02

556,38556.388==

=

π

( ) 81.1556,38647.158

wDC

K =

π ( ) 81.1556,38845.39 wDCK =

625.05.1 =+ wm DD

625.05.1 =+ ww DCD

5.1

625.0

+=

CDw

SECTION 4 - SPRINGS

Page 7 of 70

CC

CK

615.0

44

14+

−

−=

( )81.1

5.1

625.0556,38845.39

615.0

44

14

+=

+

−

−

CC

CC

C

( ) 3.4135.1615.0

44

14 81.1=+

+

−

−CC

CC

C

035.7=C

inC

Dw 0732.05.1035.7

625.0

5.1

625.0=

+=

+=

Table AT 15, inDw 0720.0= , W & M 15

( ) inDm 5065.00720.0035.7 ==

For cN

( )

w

c

GD

NCFF3

1212

8 −=−δδ

psiG 6105.11 ×=

( )( )( )( )0720.0105.11

035.7534.1647.158

4

36

3

12×

−==− cN

δδ

8.15=cN

Table AT 16,

Total coils = 8.1728.152 =+=+cN

Solid height = ( ) ( )( ) inDN wc 28.10720.028.152 =+=+


Free length = 12 δ+

( )

w

c

GD

NCF3

11

8=δ

( )( ) ( )( )( )

.082.00720.0105.11

8.15035.7534.186

3

1 in=×

=δ

Free length = in082.2082.02 =+

( )

w

c

GD

NCF3

22

8=δ

( )( ) ( )( )( )

.832.00720.0105.11

8.15035.7647.1586

3

2 in=×

=δ

Solid Height ≤ Free Length - 2δ

Solid Height ≤ in832.0082.2 −

Solid Height ≤ in25.1

But Solid Height > 1.25 in.

Therefore change material to Oil-tempered spring wire, Cost Index = 1.5

SECTION 4 - SPRINGS

Page 8 of 70

Table AT 17

ksiD

s

w

u 19.0

146= , 5.0028.0 << wD

Max “solid stress” = ksiD

w

19.0

5.87

19.019.0

304.47146324.0

ww

sdDD

s ==

psiD

ksiDD

FCKs

www

s 19.019.02

304,47304.478==

=

π

( ) 81.1304,47647.158

wDC

K =

π ( ) 81.1304,47845.39 wDCK =

5.1

625.0

+=

CDw

CC

CK

615.0

44

14+

−

−=

( )81.1

5.1

625.0304,47845.39

615.0

44

14

+=

+

−

−

CC

CC

C

( ) 1.5075.1615.0

44

14 81.1=+

+

−

−CC

CC

C

684.7=C

inC

Dw 0680.05.1684.7

625.0

5.1

625.0=

+=

+=

Table AT 15, inDw 0625.0= , W & M 16

( ) inDm 48025.00625.0684.7 ==

say inDm 46875.032

15==

5.70625.0

46875.0===

w

m

D

DC

=

2

8

w

sD

FCKs

π

( )( )

1974.15.7

615.0

45.74

15.74615.0

44

14=+

−

−=+

−

−=

CC

CK

( )( )( )

ksipsiss 6.91600,910625.0

5.7647.1581974.1

2==

=

π

For cN

SECTION 4 - SPRINGS

Page 9 of 70

( )

w

c

GD

NCFF3

1212

8 −=−δδ

psiG 6105.11 ×=

( )( )( )( )0625.0105.11

5.7534.1647.158

4

36

3

12×

−==− cN

δδ

32.11=cN

Table AT 16, squared and ground ends

Total coils = 32.13232.112 =+=+cN

Solid height = ( ) ( )( ) inDN wc 8325.00625.0232.112 =+=+


Free length = 12 δ+

( )

w

c

GD

NCF3

11

8=δ

( )( ) ( )( )( )

.082.00625.0105.11

32.115.7534.186

3

1 in=×

=δ

Free length = ( ) ( )0625.0232.11082.2082.02 +==+ Pin

ininP64

111729.0 ≈=

Pitch angle,

( )oo 127.6

46875.0

1729.0tantan 11 <=

== −−

ππλ

D

P, o.k.

Solid stress

( ) ksisso 6.14275.0

8325.026.91 =

−=

Permissible solid stress = ( )

ksiksi 5.1378.1480625.0

5.8719.0

>= , safe.

Summary of answer:

Suitable material = Oil-Tempered Spring Wire

Total Coils = 13.32

Active Coils, 32.11=cN

274. A helical spring is to fit about a 11/16-in. rod with a free length of 2 ¾ in. or

less. A maximum load of 8 lb. is to produce a deflection of 1 ¾ in. The spring

is expected to be compressed less than 5000 times during its life, but is

subjected to relatively high temperatures and corrosive atmosphere. Select a

material and determine the necessary wire size, mean coil diameter, and

number of coils. Meet all conditions advised by Text.

SECTION 4 - SPRINGS

Page 10 of 70

Solution:

For 5000 cycles < 104 cycles, use light service

Use stainless steel, type 302 (Cr-Ni), ASTM A313 – for relative high temperature and

corrosive atmosphere, Table AT 17.

usd ss 32.0= (i)

ksiD

sw

u 14.0

170= , [ ]13.001.0 << wD

ksiD

sw

u 41.0

97= , [ ]375.013.0 << wD

Maximum “solid” uo ss 47.0=

=

3

8

w

ms

D

FDKs

π

lbF 8=

216

11 wwm

DDD +=−

inDD wm 6875.05.1 =−

6875.05.1 =− ww DCD

5.1

6875.0

−=

CDw

CC

CK

615.0

44

14+

−

−=

assume ksiD

sw

u 14.0

170=

( )psi

Dksi

DDs

www

sd 14.014.014.0

400,544.5417032.0===

( ) 86.1400,5488615.0

44

14w

DC

CC

C=

+

−

−

π 86.1

5.1

6875.0400,54

64615.0

44

14

−=

+

−

−

C

C

CC

C

π

( ) 13305.1615.0

44

14 86.1=−

+

−

−CC

CC

C

919.12=C

inDw 0602.05.1919.12

6875.0=

−=

Use Table AT 15, inDw 0625.0= , 16 W & M

( ) inDm 8074.00602.0919.12 ==

say ininDm 78125.032

25==

SECTION 4 - SPRINGS

Page 11 of 70

16

11>− wm DD

6875.00625.078125.0 >−

6875.071875.0 >

203125.06875.071875.0 wD

==− , o.k.

5.120625.0

71875.0===

w

m

D

DC

[ ]13.00625.0 < , therefore, ksiD

sw

u 14.0

170= is o.k.

=

2

8

w

sD

FCKs

π

( )( )

1144.15.12

615.0

45.124

15.124=+

−

−=K

( )( )( )

psiss 648,720625.0

5.12881144.1

2=

=

π

( )

w

c

GD

NCF38

=δ psiG 6106.10 ×=

( )( )( )( )0625.0106.10

5.1288

4

31

6

3

×== cN

δ

3.9=cN

To check for solid stress and pitch

Minimum solid height = ( )( ) inND cw 58125.03.90625.0 ==

Solid stress =

( )ksipsi 90000,90

4

31

58125.04

32648,72

==

−

Permissible solid stress = ( )( )( )

ksiksi 908.1170625.0

17047.014.0

>= , o.k.

Free length = cPN , minimum

( )4

323.9 =P

inP 2957.0=

Pitch angle,

( )oo 125.7

71825.0

2957.0tantan 11 <=

== −−

ππλ

D

P, o.k.

Summary of answer

Material, Stainless Steel, Cr-Ni. ASTM A313

SECTION 4 - SPRINGS

Page 12 of 70

inDw 0625.0= , 16 W & M

inDm32

25=

3.9=cN

275. In order to isolate vibrations, helical compression springs are used to support a

machine. The static load on each spring is 3500 lb., under which the deflection

should be about 0.5 in. The solid deflection should be about 1 in. and the

outside coil diameter should not exceed 6 in. Recommend a spring for this

application; include scale, wire size, static stress, material, number of coils,

solid stress, and pitch of coils.

Solution:

Use Music wire (The best material)

Table AT 17

ksiD

sw

u 154.0

190= , [ ]192.0004.0 << wD

Maximum “solid” uso ss 5.0=

Light service, usd ss 405.0=

( )psi

Dksi

DDs

www

sd 154.0154.0154.0

950,7695.76190405.0===

=

2

8

w

sD

FCKs

π

lbF 3500= inDDOD wm 6=+=

( ) 61 =+ wDC

1

6

+=

CDw

( )154.0

1

6

950,76

1

6

35008615.0

44

14

+

=

+

+

−

−=

CC

C

CC

Css

π

( )[ ] 9.2351615.0

44

14 846.1=+

+

−

−CC

CC

C

635.5=C

ininDw 192.09043.01635.5

6>=

+=

use ( )

psiss 216,99192.0

950.76154.0

==

SECTION 4 - SPRINGS

Page 13 of 70

( )216,99

1

6

35008615.0

44

14=

+

+

−

−=

C

C

CC

Css

π

( ) 8.4001615.0

44

14 2=+

+

−

−CC

CC

C

205.6=C

inDw 8328.01205.6

6=

+=

Say ininDw 8125.016

13==

( )( ) inDm 042.58125.0205.6 ==

Say inDm 5=

154.68125.0

5===

w

m

D

DC

( )( )

2455.1154.6

615.0

4154.64

1154.64=+

−

−=K

=

2

8

w

sD

FCKs

π

( )( )( )

psipsiss 216,99481,1038125.0

154.6350082455.1

2>=

=

π, not o.k.

Use inDm 5.4=

5385.58125.0

5.4===

w

m

D

DC

( )

2763.15385.5

615.0

45385.45

15385.54=+

−

−=K

( )( )( )

psipsiss 216,99435,958125.0

5385.5350082763.1

2>=

=

π, o.k.

To check for solid stress

Permissible solid stress = ( )( )( )

psiksi 488,122488.122192.0

1905.0154.0

==

Solid stress = ( ) psipsi 488,122870,190435,955.0

1>=

, not ok

Use

.

psissd 244,611

5.0488,122 =

=

SECTION 4 - SPRINGS

Page 14 of 70

( )244,61

1

6

35008615.0

44

14=

+

+

−

−=

C

C

CC

Css

π

( ) 4.2471615.0

44

14 2=+

+

−

−CC

CC

C

1.5=C

inDw 9836.011.5

6=

+=

Say inDw 0.1=

( )( ) inDm 1.50.11.5 ==

Say inDm 5=

51

5===

w

m

D

DC

=

2

8

w

sD

FCKs

π

( )( )

3105.15

615.0

454

154=+

−

−=K

( )( )( )

psipsiss 244,61400,580.1

5350083105.1

2>=

=

πo.k.

Use inDw 0.1= , inDm 5=

Solid stress = ( ) psipsi 488,122800,116400,585.0

1<=

, o.k.

( )

w

c

GD

NCF38

=δ

(Table AT 17)

psiG 61012×=

( )( )( )( )0.11012

5350085.0

6

3

×== cN

δ

7143.1=cN

say 75.1=cN

Free length – Solid length = Solid Deflection

inNDPN cwc 1=−

( ) ( )( ) 175.1175.1 =−P

ininP16

915714.1 ≈=

Pitch angle,

SECTION 4 - SPRINGS

Page 15 of 70

( )oo 1268.5

5

16

91

tantan 11 <=

== −−

ππλ

D

P, o.k.

Summary of answer.

Scale, inlbF

k 70005.0

3500===

δ

Wire size, inDw 0.1=

Material = Music Wire

Solid sress = 116,800 psi

Pitch of stress = inP16

91=

CHECK PROBLEMS – LIGHT, MEDIUM SERVICE

276. The front spring of an automobile has a total of 9 ½ coils, 7 3/8 active coils

(square-and-ground ends), an inside diameter of 4 3/64 in., and a free length

of 14 ¼ in. It is made of SAE 9255 steel wire, OQT 1000oF, with a diameter

of 43/64 in. Compute (a) the rate (scale) of the spring; (b) the “solid stress”

and compare with a permissible value (is a stop needed to prevent solid

compression?). (c) Can 95 % of the solid stress be repeated 105 times without

danger of failure? Would you advise shot peening of the spring?

Solution:

(a) w

c

GD

NFC38

=δ

ininDw8

3

64

43>=

psiG 6105.10 ×=

w

m

D

DC =

IDDD wm =−

inDm64

34

64

43=−

inDm32

234=

0233.7

64

4332

234

==C

SECTION 4 - SPRINGS

Page 16 of 70

8

37=cN

( )

( )inlb

NC

GDFratek

c

w 345

8

370233.78

64

43105.10

8 3

6

3=

×

====δ

(b) “Solid Stress”

Solid height = ( )( ) inCoilsTotalDN wc 3828.62

19

64

43=

==

Solid deflection = Free length – Solid height = 14 ¼ - 6.3828 = 7.8672 in.

Solid Force = ( ) lbFso 27143458672.7 ==

Solid Stress =

2

8

w

so

D

CFK

π

CC

CK

615.0

44

14+

−

−=

( )( )

212.10233.7

615.0

40233.74

10233.74=+

−

−=K

( )( )psiss 322,130

64

43

0233.727148212.1

2=

=

π

Permissible value, syyss ss 6.0== , [ ]inDw 5.0>

SAE 9255, OQT 1000 oF

ksisy 160=,

ksisu 180=

( ) psipsiksisys 322,130000,96961606.0 <===

Therefore a stop is needed to prevent solid compression.

(c) usd ss 324.0= (105 cycles)

( ) ksissd 32.58180324.0 ==

( ) ksiksipsisso 32.588.123800,123322,13095.095.0 >===

There is a danger of failure, shot peening is advisable

( )soys spsis 95.0000,120000,9625.1 ≈==

277. An oil-tempered steel helical compression spring has a wire size of No. 3 W

& M, a spring index of 4.13, 30 active coils, a pitch of 0.317 in., ground-and-

squared ends; medium service. (a) What maximum load is permitted if the

recommended stress is not exceeded (static approach)? Compute (b) the

SECTION 4 - SPRINGS

Page 17 of 70

corresponding deflection, (c) “solid stress,”. (d) pitch angle, (e) scale, (f) the

energy absorbed by the spring from a deflection of 0.25 in. to that of the

working load. (g) Is there any danger of this spring buckling? (h) What

maximum load could be used if the spring were shot peened?

Solution:

Table AT 17, oil-tempered

ksiD

sw

u 19.0

146= , [ ]5.0032.0 << wD

Maximum “solid” ksiD

sw

so 19.0

5.87=

usd ss 324.0= (medium service)

Table AT 15, No. 3 W & M

inDw 2437.0=

13.4=C ( ) inCDD wm 0.12437.013.4 ===

(a)

=

2

8

w

ss

D

CFKs

π

CC

CK

615.0

44

14+

−

−=

( )( )

3885.113.4

615.0

413.44

113.44=+

−

−=K

( )

( )psiksiss sds 858,61858.61

2437.0

146324.019.0

====

( )( )( )

==

22437.0

13.483885.1858,61

π

Fss

lbF 252=

w

c

GD

NFC38

=δ

psiG 6105.11 ×= 30=cN

( )( ) ( )( )( )

in52.12437.0105.11

3013.425286

3

=×

=δ

(c) For solid stress . Square-and-ground end)

Free length = ( )( ) ( ) inDPN wc 9974.92437.0230317.02 =+=+

SECTION 4 - SPRINGS

Page 18 of 70

Solid height = ( )( ) inDND wcw 7984.72437.02302 =+=+

Solid deflection = 9.9974 – 7.7984 = 2.199 in.

Solid stress = ( ) psi491,8952.1

199.2858,61 =

Maximum “solid” ( )

ksiksiksiksiD

sw

so 491.894.1142437.0

5.875.8719.019.0

>=== , o.k. safe

(d) ( )

oo 1276.51

317.0tantan 11 <=

== −−

ππλ

D

P, o.k.

(e) inlbF

kscale 16652.1

252====

δ

(f) ( )2

1

2

22

1δδ −= kU s

inlbk 166= in25.01 =δ in52.12 =δ

( ) ( ) ( )[ ] lbinU s −=−= 6.18625.052.11662

1 22

(g) i 6.18 Free length = 9.9974 in

Mean Diameter = inDm 0.1=

49974.90.1

9974.9>==

DiameterMean

lengthFree

There is a danger for spring buckling

(h) Shot peened, Table AT 17

( )( ) psissd 322,7725.1858,61 ==

( )( )( )

==

22437.0

13.483885.1322,77

π

Fss

lbF 314=

280. It is desired to isolate a furnace, weighing 47,300 lb., from the surroundings

by mounting it on helical springs. Under the weight, the springs should deflect

approximately 1 in., and at least 2 in. before becoming solid. It has been

decided to use springs having a wire diameter of 1 in., an outside diameter of

5 3/8 in., 4.3 free coils. Determine (a) the number of springs to be used, (b)

the stress caused by the weight, (c) the “solid stress”. (d) What steel should be

used?

Solution:

SECTION 4 - SPRINGS

Page 19 of 70

inDw 1=

inDD wm8

35=+

inDm8

34=

375.41

8

34

===w

m

D

DC

(a) w

c

GD

NFC38

=δ

Assume 3.4=cN

psiG 6105.10 ×= , inDw8

3>

( ) ( )( )( )1105.10

3.4375.480.1

6

3

×==

Fδ

lbF 3645=

No. of springs = 133645

300,47==

F

W

(b) lbW

F 363813

300,47

13===

=

2

8

w

ss

D

CFKs

π

CC

CK

615.0

44

14+

−

−=

( )( )

3628.1375.4

615.0

4375.44

1375.44=+

−

−=K

( )( )( )

psiss 235,550.1

375.4363883628.1

2=

=

π

(c) “Solid Stress” = psiss 470,1101

2235,55 =

=

(d) psisys 470,110≈

ksipsis

sys

y 117.184117,1846.0

470,110

6.0====

From Table AT 7,

Use AISI 8760, OQT 800 oF, ksisy 200=

SECTION 4 - SPRINGS

Page 20 of 70

VARYING STRESS APPROACH

DESIGN PROBLEMS

282. A spring, subjected to a load varying from 100 lb. to 250 lb., is to be made of oil-

tempered, cold-wound wire. Determine the diameter of the wire and the mean

diameter of the coil for a design factor of 1.25 based on Wahl’s line. The spring

index is to be at least 5. Conform to good practice, showing checks for all

significant parameters. Let the free length be between 6 and 8.

Solution:

lbF 250max =

lbF 100min =

( ) ( ) kiplbFFFm 175.01751002502

1

2

1minmax ==+=+=

( ) ( ) kiplbFFFa 075.0751002502

1

2

1minmax ==−=−=

Wahl’s line

no

as

ys

asms

s

s

s

ss

N

21+

−=

23

88

w

a

w

maas

D

CKF

D

DKFs

ππ==

23

88

wc

m

wc

mmms

DK

CKF

DK

DKFs

ππ==

5=C

CC

CK

615.0

44

14+

−

−=

( )( )

31.15

615.0

454

154=+

−

−=K

Fig. AF 15, 5=C

19.1=cK

For oil-tempered wire,

19.0

5.87

w

ysD

s =

, [ ]5.0032.0 << wD

1.0

47

w

noD

s =

, [ ]15.0041.0 << wD

34.0

30

w

noD

s =

, [ ]625.015.0 << wD

25.1=N

SECTION 4 - SPRINGS

Page 21 of 70

( )( )( )22

251.15075.031.18

ww

asDD

s ==π

( )( )( )

( ) 22

453.2

19.1

5175.031.18

ww

msDD

s ==π

say ksiD

sw

no 34.0

30=

no

as

ys

asms

s

s

s

ss

N

21+

−=

+

−

=

34.0

2

19.0

2

30

251.12

5.87

251.1453.2

25.1

1

w

w

w

w

D

D

D

D

66.181.1 99.11

1

8.72

1

25.1

1

ww DD+=

ininDw 15.02857.0 >=

Table AT 15, use No. 1, W & M

inDw 2830.0=

( ) inCDD wm 415.12830.05 ===

say inDm16

71=

Check for Free length

6 in < Free length < 8 in

Free length = inDm 75.516

7144 =

=

Increase mD

inDm2

11=

Free length = inDm 62

1144 =

= , o.k.

Summary of answer

inDw 2830.0=

inDm2

11=

283. A carbon-steel spring is to be subjected to a load that varies from 500 to 1200 lb.

The outside diameter should be between 3.5 and 4 in., the spring index between 5

to 10; approximate scale of 500 lb./in. Choose a steel and for a design factor of

1.4 by the Wahl line, find the wire diameter. Also determine the number of active

SECTION 4 - SPRINGS

Page 22 of 70

coils and the free length for squared-and-ground ends. Conform to the general

conditions specified in the Text.

Solution:

For carbon steel, Table AT 17

ksiD

sw

ys 1.0

91= , [ ]25.0093.0 << wD

ksiD

sw

no 15.0

49= , [ ]25.0093.0 << wD

lbF 1200max =

lbF 500min =

( ) ( ) kiplbFFFm 85.085050012002

1

2

1minmax ==+=+=

( ) ( ) kiplbFFFa 35.035050012002

1

2

1minmax ==−=−=

inOD 0.4~5.3=

10~5=C Wahl’s line

no

as

ys

asms

s

s

s

ss

N

21+

−=

Figure AF 15, 10~5=C

Assume 2.1=K , 125.1=cK

3

8

w

maas

D

DKFs

π=

3

8

wc

mmms

DK

DKFs

π=

inOD 75.3≈

wm DD −= 75.3

( )( )( ) ( )33

75.30695.175.335.02.18

w

w

w

was

D

D

D

Ds

−=

−=

π

( )( )( )( )

( )33

75.33088.2

125.1

75.385.02.18

w

w

w

wms

D

D

D

Ds

−=

−=

π

( ) ( )

−

+

−−

=

15.0

3

1.0

3

49

75.30695.12

91

75.30695.13088.2

4.1

1

w

w

w

w

w

w

D

D

D

D

D

D

85.29.2 9079.22

75.3

4285.73

75.3

4.1

1

w

w

w

w

D

D

D

D −+

−=

ininDw 25.06171.0 >=

Use

SECTION 4 - SPRINGS

Page 23 of 70

( )ksisys 53.104

25.0

911.0

==

( )ksisno 33.60

25.0

4915.0

==

( ) ( )

33.60

75.30695.12

53.104

75.30695.13088.2

4.1

133

−

+

−−

= w

w

w

w

D

D

D

D

33 205.28

75.3

346.84

75.3

4.1

1

w

w

w

w

D

D

D

D −+

−=

3137.21

75.3

4.1

1

w

w

D

D−=

inDw 5935.0=

use

inDw32

19=

inDD wm4

33≈+

inDm4

33

32

19=+

inDm32

53=

316.5

32

19

32

53

=

==w

m

D

DC

. o.k.

Wire Diameter inDw32

19= , Carbon Steel

Number of coils:

w

c

GD

NFC38

=δ

ksipsiG 500,10105.10 6 =×= , inDw8

3>

c

w

NC

GDk

F38

==δ

( )

( )cN

3

6

316.58

32

19105.10

500

×

=

4.10=cN

Table AT 16, square-and-ground ends

SECTION 4 - SPRINGS

Page 24 of 70


Solid height = wcw DND 2+

Total Coils = 2+cN

Solid height = ( ) inDND wcw 3625.732

1924.102 =

+=+

ink

F4.2

500

1200===δ

Min. Free length = 2.4 + 7.3625 in = 9.7625 in

Use Free length = 10 in

To check for pitch angle.


( ) 1032

1924.10 =

+P

inP 8474.0=

oo 12885.4

32

53

8474.0tantan 11 <=

== −−

ππ

λmD

P, o.k.

Solid stress:

indeflectionsolidT 6375.23625.710 =−==δ ( )( ) lbkF T 13196375.2500 === δ

CC

CK

615.0

44

14+

−

−=

( )( )

29.1316.5

615.0

4316.54

1316.54=+

−

−=K

( )( )( )ksisksipsi

D

KFDs ys

w

ms 53.104033.23033,23

32

19

32

53131929.18

833

=<==

==

ππ

284. A helical compression spring, made of oil-tempered, cold-wound carbon steel, is

to be subjected to a working load varying from 100 to 300 lb. for an indefinite

time (severe). A mean coil diameter of 2 in. should be satisfactory. (a) Using the

static approach, compute a wire diameter. (b) For this wire size, compute the

factor of safety as given by the Wahl line.

Solution:

SECTION 4 - SPRINGS

Page 25 of 70

Table AT 16,

For carbon steel,

ksiD

sw

u 1.0

182= , [ ]25.0093.0 << wD

Max. “solid” ksiD

sw

ys 1.0

91=

ksiD

sw

ys 1.0

91=

ksiD

sw

no 15.0

49= , [ ]25.0093.0 << wD

.2 inDm = lbF 300max =

lbF 100min =

(a) kiplbF 3.0300 ==

severe service, ( )( )

ksiDD

ssww

usd 1.01.0

866.47182263.0263.0 ===

=

2

8

w

ss

D

CFKs

π

CC

CK

615.0

44

14+

−

−=

w

m

D

DC =

CC

DD m

w

2==

( )( )1.03

2

866.47

2

23.08615.0

44

14

=

+

−

−=

CC

CC

Css

π

84.233615.0

44

14 9.2 =

+

−

−C

CC

C

075.6=C

ininDw 25.03292.0075.6

2>==

Therefore use ( )

ksissd 984.5425.0

866.471.0

==

SECTION 4 - SPRINGS

Page 26 of 70

( )( )984.54

2

23.08615.0

44

143

=

+

−

−=

C

CC

Css

π

9.287615.0

44

14 3 =

+

−

−C

CC

C

136.6=C

inDw 3259.0136.6

2==

say inDw64

21=

(b) ( )

ksisys 53.10425.0

911.0

==

( )ksisno 33.60

25.0

4915.0

==

( ) ( ) kiplbFFFm 2.02001003002

1

2

1minmax ==+=+=

( ) ( ) kiplbFFFa 1.01001003002

1

2

1minmax ==−=−=

095.6

64

21

2=

==

w

m

D

DC

Figure AF 15

15.1=cK

25.1=K

=

3

8

w

mm

c

msD

DF

K

Ks

π

( )( )ksisms 34.31

64

21

22.08

15.1

25.13

=

=

π

3

8

w

maas

D

DKFs

π=

SECTION 4 - SPRINGS

Page 27 of 70

( )( )ksisas 02.18

64

21

21.0825.1

3=

=

π

Wahl’s line

no

as

ys

asms

s

s

s

ss

N

21+

−=

( )

33.60

02.182

53.104

02.1834.311+

−=

N 38.1=N

285. A helical spring of hard-drawn wire with a mean diameter of 1 ½ in. and square-

and-ground ends is to be subjected to a maximum load of 325 lb. (a) Compute the

wire diameter for average service. (b) How many total coils are required if the

scale is 800 lb./in.? (c) For a minimum load of 100 lb., what is the factor of safety

according to Wahl line? Would it be safe for an indefinite life?

Solution:

Table AT 17,

Hard-drawn wire,

ksiD

sw

u 19.0

140= , [ ]625.0028.0 << wD

Maximum “solid” ksiD

ssw

yss 19.0

70==

( )( )ksi

Ds

w

no 1.0

479.0= , [ ]15.0041.0 << wD

( )( )ksi

Ds

w

no 34.0

309.0= , [ ]625.015.0 << wD

Average service

(a) ( ) ( )ksi

DDsss

ww

uusd 19.019.0

556.381402754.02754.0324.085.0 ====

kiplbF 325.0325 ==

inDm2

11=

=

3

8

w

ms

D

FDKs

π

SECTION 4 - SPRINGS

Page 28 of 70

CC

CK

615.0

44

14+

−

−=

CDw

5.1=

( )( )19.03

5.1

556.38

5.1

5.1325.08615.0

44

14

=

+

−

−=

CC

CC

Css

π

05.97615.0

44

14 81.2 =

+

−

−C

CC

C

586.4=C

ininC

Dw 625.03271.0586.4

5.15.1<===

inDw64

21=

(b) 57.4

64

21

5.1=

==

w

m

D

DC

( )( )

345.157.4

615.0

457.44

157.44=+

−

−=K

w

c

GD

NFC38

=δ

c

w

NC

GDk

F38

==δ

inkipinlbk 8.0800 ==

( )

cN357.84

64

21500,11

8.0

=

2.6=cN

(c) ksisys 5.86

64

21

7019.0

=

=

( )( )ksisno 44.39

64

21

309.019.0

=

= , inDw 15.0>

( ) kiplbFm 2125.05.2121003252

1==+=

SECTION 4 - SPRINGS

Page 29 of 70

( ) kiplbFa 1125.05.1121003252

1==−=

212.1=cK , Fig. AF 15

345.1=K

( )( )ksi

D

DF

K

Ks

w

mm

c

ms 5.25

64

21

5.12125.08

212.1

345.1833

=

=

=

ππ

( )( )ksi

D

DFKs

w

mmas 36.16

64

21

5.11125.08345.1

833

=

=

=

ππ

( )44.39

36.162

5.86

36.165.2521+

−=+

−=

no

as

ys

asms

s

s

s

ss

N

[ ]min15.107.1 NN <=

Not safe for indefinite life.

286. A helical spring is to be subjected to a maximum load of 200 lb. (a) Determine

the wire size suitable for medium service if the material is carbon steel ASTM

A230; 6=C . Determine the factor of safety of this spring according to the Wahl

line (b) If the minimum force is 150 lb., (c) if the minimum force is 100 lb., (d) if

the minimum force is 25 lb.

Solution:

For carbon steel ASTM A230

Table AT 17

ksiD

sw

u 1.0

182= , [ ]25.0093.0 << wD

ksiD

sw

ys 1.0

91= , [ ]25.0093.0 << wD

ksiD

sw

no 15.0

49= , [ ]25.0093.0 << wD

Medium Service

usd ss 324.0=

(a) psiD

ksiDD

swww

sd 1.01.01.0

968,58968.58182324.0 ==

=

SECTION 4 - SPRINGS

Page 30 of 70

=

3

8

w

ms

D

FDKs

π

CC

CK

615.0

44

14+

−

−=

( )( )

2525.16

615.0

464

164=+

−

−=K

lbF 200=

( )( )1.02

968,58620082525.1

ww

sDD

s =

=

π

inDw 2371.0=

Table At 15, use inDw 2437.0= , No. 3 W & M

ininDw 25.02437.0 <= , o.k.

Factor of safety.

( )ksiksiksi

Ds

w

ys 8.1042437.0

91911.01.0

===

( )ksiksiksi

Ds

w

no 56.602437.0

494915.0

===

(a) ( ) kiplbFm 175.01751502002

1==+=

( ) kiplbFa 025.0251502002

1==−=

Figure AF 15, 156.1=cK

( )( )( )

ksiD

CF

K

Ks

w

m

c

ms 8.482437.0

6175.08

156.1

2525.1822

=

=

=

ππ

( )( )( )

ksiD

CFKs

w

aas 1.8

2437.0

6025.082525.1

822

=

=

=

ππ

no

as

ys

asms

s

s

s

ss

N

21+

−=

( )56.60

1.82

8.104

1.88.481+

−=

N

525.1=N

(b) ( ) kiplbFm 15.01501002002

1==+=

( ) kiplbFa 05.0501002002

1==−=

SECTION 4 - SPRINGS

Page 31 of 70


( )( )( )

ksiD

CF

K

Ks

w

m

c

ms 8.412437.0

615.08

156.1

2525.1822

=

=

=

ππ

( )( )( )

ksiD

CFKs

w

aas 11.16

2437.0

605.082525.1

822

=

=

=

ππ

no

as

ys

asms

s

s

s

ss

N

21+

−=

( )56.60

11.162

8.104

11.168.411+

−=

N

287.1=N

(c) ( ) kiplbFm 1125.05.112252002

1==+=

( ) kiplbFa 0875.05.87252002

1==−=


( )( )( )

ksiD

CF

K

Ks

w

m

c

ms 36.312437.0

61125.08

156.1

2525.1822

=

=

=

ππ

( )( )( )

ksiD

CFKs

w

aas 2.28

2437.0

60875.082525.1

822

=

=

=

ππ

no

as

ys

asms

s

s

s

ss

N

21+

−=

( )56.60

20.282

8.104

20.2836.311+

−=

N

04.1=N

CHECK PROBLEMS

A Diesel valve spring is made of 3/8-in. chrome-vanadium steel wire, shot-peened; inside

diameter is 3 in., 7 active coils, free length is 7 3/8 in., solid length is 4 1/8 in., length

with valve closed, 6 ¼ in., length when open, 5 1/8 in. (a) Compute the spring constant

and the factor of safety as defined by the Wahl criterion (see § 6.13, Text). (b) Is there

any danger of damage to the spring if it is compressed solid? (c) What is the natural

frequency? If this spring is used on a 4-stroke Diesel engine at 450 rpm, is there any

danger of surge? Compute the change of stored energy between working lengths.

Solution:

For chrome-vanadium steel wire, shot-peened, Table AT 17

SECTION 4 - SPRINGS

Page 32 of 70

( )( )ksi

Ds

w

u 166.0

16825.1= , [ ]437.0032.0 << wD

( )( )ksi

Ds

w

ys 166.0

10025.1= , [ ]437.0032.0 << wD

( )( )ksi

Ds

w

no 15.0

4925.1= , [ ]5.0028.0 << wD

ininDw 375.08

3==

( )( )( )

ksiksisys 1.147375.0

10025.1166.0

==

( )( )( )

ksiksisno 96.70375.0

4925.115.0

==

(a) w

c

GD

NFC38

=δ

c

w

NC

GDk

F38

==δ

psiG 6105.11 ×=

7=cN

inDw 375.0=

inIDDD wm 3==−

inDm 375.3=

9375.0

375.3===

w

m

D

DC

nstantcospringk =

( )( )( ) ( )

inlbNC

GDk

c

w 64.105798

375.0105.11

83

6

3=

×==

in25.38

14

8

371 =−=δ

( )( ) lbkF 33.34325.364.10511 === δ

in125.14

16

8

372 =−=δ

( )( ) lbkF 85.118125.164.10522 === δ

( ) kiplbFm 231.009.23185.11833.3432

1==+=

( ) kiplbFa 11224.024.11285.11833.3432

1==−=

=

2

8

w

m

c

msD

CF

K

Ks

π

SECTION 4 - SPRINGS

Page 33 of 70

CC

CK

615.0

44

14+

−

−=

( )( )

162.19

615.0

494

194=+

−

−=K


( )( )( )

ksiD

CF

K

Ks

w

m

c

ms 8.39375.0

9231.08

10.1

162.1822

=

=

=

ππ

( )( )( )

ksiD

CFKs

w

aas 3.21

375.0

911224.08162.1

822

=

=

=

ππ

no

as

ys

asms

s

s

s

ss

N

21+

−=

( )96.70

3.212

1.147

3.218.391+

−=

N

377.1=N

(b) max. “solid” ksiss yss 1.147==

Min. Solid Height = ( )( ) inND cw 625.27375.0 ==

Solid deflection = .75.4625.28

37 in=−

( )( ) kiplbkF 5018.08.50175.464.105 ==== δ

Solid stress = ( )( )

( )ksiksi

D

FCKs

w

s 1.14795375.0

95018.08162.1

822

<=

=

=

ππ

There is no danger of damage

(c) Natural frequency

For steel

cpsDN

D

mc

w

2

050,14=φ

cpsDCN wc

2

050,14=φ

( )( ) ( )cpscps 66

375.097

050,142

==φ

For 450 rpm, cps4760

2450 =

=

πφ

124.147

66<= , there is danger of surging.

SECTION 4 - SPRINGS

Page 34 of 70

(d) ( ) ( ) ( ) ( )[ ] lbinkU s −=−=−= 491125.125.364.1052

1

2

1 222

2

2

1 δδ

289. A helical spring is hot wound from 5/8-in. carbon-steel wire with an outside

diameter of 3 ¼ in. A force of 3060 lb. is required to compress the spring 1 ¾

in to the solid heigh. In service the spring is compressed so that its

deformation varies form ½ in. to1 1/8 in. (a) What is the factor of safety by

the Wahl criterion? (b) Is the “solid stress” safe? Compute (c) the pitch angle,

(d) the change of stored energy between the working lengths, (e) the factor of

safety if the spring is peened?

Solution:

For hot-wound carbon steel wire

inDw8

5=

Table AT 17

ksiD

sw

ys 1.0

91= , [ ]25.0093.0 << wD

( )ksiksisys 5.104

25.0

911.0

== , .25.0 inDw >

ksiD

sw

no 15.0

49= , [ ]25.0093.0 << wD

( )ksiksisno 33.60

25.0

4915.0

== , .25.0 inDw >

Permissible solid stress = ksiD

sw

s 31.0

117= , [ ].375.0 inDw > § 6.3

( )ksiksiss 4.35

625.0

11731.0

==

(a) inlbF

k 6.174875.1

3060===

δ

( ) lbkF 3.8742

16.174811 =

== δ

( ) lbkF 2.19678

116.174822 =

== δ

( ) kiplbFm 4207.17.14203.8742.19672

1==+=

( ) kiplbFa 5464.04.5463.8742.19672

1==−=

SECTION 4 - SPRINGS

Page 35 of 70

ininDw 625.08

5==

inDD wm4

13=+

inDm 625.2=

2.4625.0

625.2===

w

m

D

DC

CC

CK

615.0

44

14+

−

−=

( )( )

3808.12.4

615.0

42.44

12.44=+

−

−=K

234.1=cK

( )( )( )

ksiD

CF

K

Ks

w

m

c

ms 5.43625.0

2.44207.18

234.1

3808.1822

=

=

=

ππ

( )( )( )

ksiD

CFKs

w

aas 7.20

625.0

2.45464.083808.1

822

=

=

=

ππ

no

as

ys

asms

s

s

s

ss

N

21+

−=

( )33.60

7.202

5.104

7.205.431+

−=

N

106.1=N

(b) Permissible solid stress = 135.4 ksi

kipF 060.3=

Solid stress, ( )( )

( )ksiksi

D

FCKs

w

s 4.1357.115625.0

2.4060.383808.1

822

<=

=

=

ππ, safe

(c) Solid deflection = in4

31

( ) inNDP cw 75.1=−

w

c

GD

NFC38

=δ

psiG 6105.10 ×= , hot-wound

c

w

NC

GDFk

38==

δ

SECTION 4 - SPRINGS

Page 36 of 70

( )( )( )

cN3

6

2.48

625.0105.106.1748

×=

332.6=cN

( )( ) 75.1332.6625.0 =−P

inP 9014.0=

Pitch angle

wm CD

P

D

P

ππλ ==tan

( )( )o24.6

625.02.4

9014.0tantan 11 =

== −−

ππλ

wCD

P

(d) ( ) ( ) ( ) ( )[ ] lbinkU s −=−=−= 8885.0125.16.17482

1

2

1 222

2

2

1 δδ

(e) When peened

( ) ksisys 6.1305.1045.12 ==

( ) ksisno 4.7533.6025.1 ==

( )4.75

7.202

6.130

7.205.431+

−=

N

38.1=N

ENERGY STORAGE

293. A 10-lb. body falls 10 in. and then strikes a helical spring. Design a hard-

drawn carbon steel spring that will absorb this shock occasionally without

permanent damage. Determine appropriate values of wire diameter, coil

diameter, pitch, free length, closed length, and the maximum stress under the

specified conditions, and scale. Let 7=C .

Solution:

For hard-drawn carbon steel, Table AT 17

ksiD

sw

u 1.0

182= , [ ]25.0093.0 << wD


sw

s 1.0

91= , [ ]25.0093.0 << wD

( )( ) ksiD

ssw

usd 1.0

855.36405.050.0 ==

GK

VsU s

s 2

2

4=

SECTION 4 - SPRINGS

Page 37 of 70

CC

CK

615.0

44

14+

−

−=

( )( )

213.17

615.0

474

174=+

−

−=K

( )cm

w NDD

V ππ

≈

4

2

4

22

cmw NDDV

π=

wm CDD =

4

32

cwNCDV

π=

w

c

GD

NFC38

=δ

( )δ+= hWU s

=

2

8

w

sD

FCKs

π

KC

DsF ws

8

2π=

=

w

cws

GD

NC

KC

Ds32 8

8

πδ

KG

NCDs cws

2πδ =

GK

NCDs

KG

NCDshWU

ccwss

ws

2

3222

16

ππ=

+=

KG

WCDs

GK

CDs

WhN

wsws

c 2

2

222

16

ππ−

=

when ksiD

sw

s 1.0

855.36=

( )KG

WCD

GK

CD

WhN

ww

c 29.0

2

8.222 855.36

16

855.36 ππ−

=

( )( )

( ) ( )( ) ( )

( ) ( )( )( )500,11213.1

010.07855.36

500,11213.116

7855.36

10010.029.0

2

8.222

wDD

N

w

cππ

−

=

9.08.2 004067.03466.0

10.0

wDD

Nw

c−

=

SECTION 4 - SPRINGS

Page 38 of 70

combination of wD and cN

Gage No. W & M wD cN cwND

12 0.1055 991.2 105

11 0.1205 312.1 37.6

10 0.1350 166.1 22.4

9 0.1483 108.0 16.0

8 0.1620 75.2 12.2

7 0.1770 53.7 9.5

6 0.1920 40.2 7.7

5 0.2070 31 6.4

4 0.2253 23.4 5.3

3 0.2437 18.1 4.4

Use ininDw 25.02437.0 <= , 1.18=cN

( ) ininDD wm64

4517059.12437.077 ====

( ) ( ) ( )( )( )

inKG

NCD

KG

NCDs ccws w 066.2500,11213.1

1.1872437.0855.36855.36 29.029.02

====πππ

δ

( )ksiss 44.42

2437.0

855.361.0

==

( )ksi

Ds

w

so 8.1042437.0

91911.01.0

===

Solid deflection

( ) in1.5066.244.42

8.104=

=

( ) 1.5=− cw NDP

( )( ) 1.51.182437.0 =−P

inP 5255.0=

say inP 53125.032

17==

Minimum Solid Height = ( )( ) inND cw 41.41.182437.0 ==

Assume squared and ground end

Solid height = ( )( ) ( ) inDND wcw 0.52437.021.182437.02 =+=+

Solid deflection = ( )( ) in2.51.182437.053125.0 =−

Free length = 5.0 in + 5.2 in = 10.2in

Summary of answer:

inDw 2437.0= , No. 3 W & M

SECTION 4 - SPRINGS

Page 39 of 70

inDm64

451=

inP32

17=

Free length = 10.2 in

Closed length = 5 in

Maximum stress = 42.44 ksi

294. A helical spring, of hard-drawn steel wire, is to absorb 75 in-lb of energy

without being stressed beyond the recommended value of average service. Let

6=C . Decide upon satisfactory dimensions; wD , mD , cN , free length, pitch

angle, solid stress, volume of metal, possibility of spring buckling.

Solution:

For hard-drawn steel wire, shock load, average service

ksiD

sw

u 19.0

140= , [ ]625.0028.0 << wD


sw

s 19.0

70= , [ ]625.0028.0 << wD

( )( )( ) ( )( )( ) ksiDD

ssww

usd 19.019.0

278.19140324.085.050.0324.085.050.0 =

==

GK

CNDs

GK

VsU cwss

s 2

322

2

2

164

π==

6=C

CC

CK

615.0

44

14+

−

−=

( )( )

2525.16

615.0

464

164=+

−

−=K

kipinlbinU s −=−= 075.075

( )( ) ( )500,112525.116

6278.19075.0

2

32

19.0

cw

w

s

ND

DU

π

==

cw ND62.29837.0 =

Table AT-15

W & M wD cN cwND

SECTION 4 - SPRINGS

Page 40 of 70

9 0.1483 146 21.65

8 0.1620 116 18.79

7 0.1770 92 16.28

6 0.1920 74 14.21

5 0.207 61 12.63

4 0.2253 49 11.04

3 0.2437 40 9.75

2 0.2625 32.7 8.58

1 0.2830 26.9 7.61

0 0.3065 21.8 6.68

2-0 0.3310 17.8 5.89

3-0 0.3625 14.0 5.075

4-0 0.3938 11.3 4.45

5-0 0.4305 8.95 3.85

Use inDw 4305.0= , 5-0 W & M

9≈cN

( ) ininDm16

92583.24305.06 ≈==

( )ksiss 63.22

4305.0

278.1919.0

==

Max. Solid Stress = ( )

ksisso 16.824305.0

7019.0

==

( )( )( )( ) ( )( )( )

inKG

NCDs cws 6885.0500,112525.1

964305.063.2222

===ππ

δ

Solid deflection ( ) in5.26885.063.22

16.82=

=

( ) 5.2=− cw NDP

( )( ) 5.294305.0 =−P

inP 7083.0=

say inP 703125.064

45==

Solid deflection = ( )( ) in453625.294305.0703125.0 =−

Solid stress ksi65.806885.0

453625.263.22 =

=

Minimum Solid Height = ( )( ) ininND cw8

738745.394305.0 ≈==

Minimum Free Length = ( ) ininPNc64

216328125.69

64

45≈=

=

Pitch Angle

SECTION 4 - SPRINGS

Page 41 of 70

oo 125

16

92

64

45

tantan 11 <=

== −−

ππ

λmD

P

Volume

( ) ( ) ( ) 3

22

55.10916

92

4

4305.0

4inND

DV cm

w =

=

≈ π

ππ

π

Summary of answer:

inDw 4305.0= , No. 5-0 W & M

inDm16

92=

9=cN

Free length = in64

216

Pitch Angle = o5=λ

Solid Stress = 80.65 ksi

Volume of metal = 10.55 in3

Possibility of spring buckling

447.2

16

92

64

216

<= , no possibility

CONCENTRIC HELICAL SPRINGS

297. Two concentric helical springs are to be subjected to a load that varies from a

maximum of 235 lb. to a maximum of 50 lb. They are to fit inside a 1 5/8 in.

cylinder. The maximum deflection is to be ¾ in., and the deflection when

compressed solid is to be approximately 1 in. Using the “static approach” for

severe service (maximum load), determine the wire diameter, mean coil

diameter, number of coils, solid length, and free length of both springs. (Start

with oil-tempered wire and assume a diametral clearance between the outer

spring and the cylinder of 2

wD, assume a similar clearance between springs.

Search for a suitable spring index and wire size.)

Solution:

For oil-tempered wire

Table AT 17

SECTION 4 - SPRINGS

Page 42 of 70

ksiD

sw

u 19.0

146= , [ ]5.0032.0 << wD


sw

s 19.0

5.87= , [ ]5.0032.0 << wD

Severe service

( )ksi

DDss

ww

usd 19.019.0

398.38146263.0263.0 ===

kiplbF 235.0235 ==

io δδ =

wi

iii

wo

cooo

GD

NCF

GD

NCF33 88

=

Assume, io CC =

co

woo

NC

GDF

332

3=

ci

wii

NC

GDF

332

3=

=

2

8

wo

oso

D

CFKs

π

=

2

8

wi

isi

D

CFKs

π

wi

mi

wo

mo

D

D

D

DC ==

CC

CK

615.0

44

14+

−

−=

wimiwi

womo DDD

DD +=−−2

wiwomimo DDDD 5.1+=−

womowo DD

D+=−

2625.1

625.15.1 =+ womo DD

625.15.1 =+ wowo DCD

5.1

625.1

+=

CDwo

5.1

625.1

+=

C

CDmo

wiwowiwo DDCDCD 5.1+=−

( ) ( ) wiwo DCDC 5.11 +=−

SECTION 4 - SPRINGS

Page 43 of 70

( )( )2

5.1

1625.1

+

−=

C

CDwi

( )( )2

5.1

1625.1

+

−=

C

CCDmi

ksiDD

CFKs

wowo

oso 19.02

398.388=

=

π

KC

DF wo

o

81.108.15=

ksiDD

CFKs

wiwi

isi 19.02

398.388=

=

π

KC

DF wi

i

81.108.15=

kipFFF io 235.0==+

235.008.1508.15 81.181.1

=+KC

D

KC

D wiwo

KCDD wiwo 235.008.1508.15 81.181.1 =+

( )KC

C

C

C235.0

5.1

1625.108.15

5.1

625.108.15

81.181.1

=

+

−+

+

( )( )

( )C

CC

C

C

C

C

+

−

−=

+

−+

+

615.0

44

14235.0

5.1

1

5.1

152.154

62.3

81.1

81.1

328.5=C

( )( )

inDwi 1509.05.1328.5

1328.5625.12

=+

−=

inDwo 2380.05.1328.5

625.1=

+=

Table AT 15, use inDwi 1620.0= , No. 8 W & M and inDwo 2625.0= , No. 2 W & M

( )( ) ininCDD womo32

1313986.12625.0328.5 ≈===

( )( ) ininCDD wimi8

78631.01620.0328.5 ≈===

401.51620.0

8

7

=

==wi

mii

D

DC

357.52625.0

32

131

=

==wo

moo

D

DC

SECTION 4 - SPRINGS

Page 44 of 70

oo

woo

CK

DF

81.108.15=

( )( )

287.1357.5

615.0

1357.54

1357.54=+

−

−=oK

( )( )( )

kipFo 194.0357.5287.1

2625.008.1581.1

==

ii

wii

CK

DF

81.108.15=

( )( )

2843.1401.5

615.0

1401.54

1401.54=+

−

−=iK

( )( )( )

kipFi 081.0401.52843.1

1620.008.15==

kipkipFF io 235.0275.0071.0194.0 >=+=+ , ok

co

woo

NC

GDF

332

3=

( )( )( )

coN3

357.532

2625.0500,113194.0 =

5.9=coN

ci

wii

NC

GDF

332

3=

( )( )( )

ciN3

401.532

1620.0500,113071.0 =

6.15=ciN

Max. solid stress, ksiD

sw

ss 19.0

5.87= ,

( )ksissso 82.112

2625.0

5.8719.0

==

( )ksisssi 65.123

1620.0

5.8719.0

==

Stress

( )ksissi 26.54

1620.0

398.3819.0

==

( )ksisso 51.49

2625.0

398.3819.0

==

Solid stress

ksiksisso 82.11201.6675.0

151.49 <=

=

SECTION 4 - SPRINGS

Page 45 of 70

ksiksissi 65.12335.7275.0

126.54 <=

=

Solid length

( )( ) inND cowo 5.25.92625.0 ==

( )( ) inND ciwi 53.26.151620.0 ==

assume solid length = 3 in

( ) ( )( ) inxxND iiciwi 36.151620.0 =+=+

92.2=ix

Total coils = 15.6 + 2.92 = 18.52

( ) ( )( ) inxxND oocowo 35.92625.0 =+=+

93.1=ox

Total coils = 9.5 + 1.93 = 11.43

Free Length = 3 in + 1 in = 4 in

Summary of answer:

Outside wire.

inDwo 2625.0= , No. 2 W & M

inDmo32

131=

43.11=toN

Solid length = 3 in

Free length = 4 in

Inside wire.

inDwi 1620.0= , No. 8 W & M

inDmi8

7=

52.18=tiN

Solid length = 3 in

Free length = 4 in

298. Two concentric, helical compression springs are used on a freight car. The

larger spring has an outside diameter of 7 in., a free length of 7 1/8 in., and is

made of a 1 ¾ in. steel bar. The smaller has an outside diameter of 4 1/8 in., a

free length of 6 13/16 in. , and is made of 7/8 in. steel bar. The solid height of

each spring is 5 ¼ in. and the forces required to compress them solid are

15,530 lb. and 7,000 lb., respectively. The working load on the two springs is

11,350 lb. Determine (a) the number of free coils in each spring, (b) the stress

in each spring when compressed solid, (c) the stresses induced by the working

SECTION 4 - SPRINGS

Page 46 of 70

load. Notice that the outer spring deflects 5/16 in. before the inner one takes a

load. (d) What energy is absorbed while changing deflection from that at the

working load to that when the springs are compressed “solid”?

Solution:

inODo 7=

inDwo8

31=

inFLo8

17=

inODi8

14=

inDwi8

7=

inFLi16

136=

(a) Solid height = inND Tw4

15=

82.3375.1

25.5==ToN

6875.0

25.5==TiN

(b) lbFo 530,15=

lbFi 7000=

=

2

8

w

sD

FCKs

π

inDmo 625.58

317 =−=

091.4375.1

625.5===

wo

moo

D

DC

( )( )

393.1091.4

615.0

4091.44

1091.44=+

−

−=oK

inDmi 25.38

7

8

14 =−=

714.3875.0

25.3===

wi

mii

D

DC

( )( )

442.1714.3

615.0

4714.34

1714.34=+

−

−=iK

SECTION 4 - SPRINGS

Page 47 of 70

Solid stress

( )( )( )

psisso 203,119375.1

091.4530,158393.1

2=

=

π

( )( )( )

psissi 689,124875.0

714.370008442.1

2=

=

π

(b) Stresses induced by working load

lbFF oi 350,11=+

inlbko 8283

4

15

8

17

530,15=

−

=

inlbko 4480

4

15

16

136

7000=

−

=

inio 3125.016

5==−δδ

iiii kF δδ 4480==

( )iooo kF δδ +== 3125.08283

( ) lbkFF iioooi 350,113125.082834480 =++==+ δδδ

ini 6865.0=δ

ino 9990.06865.03125.0 =+=δ

( )( ) lbFi 30766865.04480 ==

( )( ) lbFo 82759990.08283 ==

Stresses

( )( )( )

psisso 516,63375.1

091.482758393.1

2=

=

π

( )( )( )

psissi 792,54875.0

714.330768442.1

2=

=

π

(d) Energy

( )2

1

2

22

1oooso kU δδ −=

ino 875.14

15

8

172 =−=δ

ino 9990.01 =δ

( ) ( ) ( )[ ] lbinU so −=−= 427,10999.0875.182832

1 22

SECTION 4 - SPRINGS

Page 48 of 70

( )2

1

2

22

1iiisi kU δδ −=

ini 5625.14

15

16

1362 =−=δ

ino 6865.01 =δ

( ) ( ) ( )[ ] lbinU si −=−= 413,46865.05625.144802

1 22

TORSION-BAR SPRINGS

299. A torsion-bar similar to that shown is to be used for the front spring of an

automobile. Its rate should be 400 lb./in. of deflection of the end of the arm

which is .10 ine = long. It is made of AISI 9261,OQT 900 oF, and the

maximum repeated load is 1500 lb. perpendicular to the centerline of the arm.

The support is such that bending of the bar is negligible. (a) Determine its

diameter and length so that no permanent set occurs due to a 30 % overload

(limited by a stop). Use yys ss 6.0= , but check with equation (c) § 6.3, Text, if

appropriate. (b) Determine the factor of safety according to the Soderberg

criterion if the load varies from 1200 lb. to 1500 lb.; minimum 1.0=dr ,

3=dD . (c) The same as (b) except that the bar is shot-peened all over. What

other steps may be taken to improve the fatigue strength?

Problem 299, 300

Solution:

ine 10=

For AISI 9261, OQT 900 oF

ksisy 192=

ksisu 215=

psiksiss yys 200,1152.1156.0 ===

SECTION 4 - SPRINGS

Page 49 of 70

(a) 3

16

d

Tss

π=

( )( )( ) lbinFeT −=== 500,191015003.1

( )3

500,1916200,115

dπ=

ind 95.0=

use ind 1=

§ 6.3 ( ) ys

w

s sksiD

s ≈=== 1171

1171173.03.0

(b) Soderberg Criterion

ns

asf

ys

ms

s

sK

s

s

N+=

1

( )( )( ) ksisns 5.642155.06.0 ==

Figure AF 12, 1.0=dr , 3=dD

45.1=tK

45.1=≈ tf KK

( ) lbFm 1350120015002

1=+=

( )( ) kipsinlbinTm −=−== 5.13500,13101350

( )( )

ksisms 8.681

5.13163

==π

( ) lbFa 150120015002

1=−=

( )( ) kipsinlbinTa −=−== 5.1150010150

( )( )

ksisas 64.71

5.1163

==π

( )( )5.64

64.745.1

2.115

8.681+=

N

30.1=N

(c) Shot-peened

( ) ksisys 1442.11525.1 ==

( ) ksisns 6.805.6425.1 ==

( )( )6.80

64.745.1

144

8.681+=

N

625.1=N

SECTION 4 - SPRINGS

Page 50 of 70

300. A solid steel torsion bar is loaded through a 10 in. arm as shown. The load F

perpendicular to the center-line of the arm varies from 500 to 1000 lb.,

200,000 cycles. The bar is .8

7ind = in diameter and 30 in. long; let 3=dD ;

1.0=dr ; (a) Determine the maximum stress in the bar, the angular

deflection, and the scale (lb./in.) where F is applied. The support is such that

bending of the bar is negligible. (b) Select a material and heat treatment for

this bar for a minimum 2.1=N , Soderberg criterion.

Problem 299, 300

Solution:

Fig. AF 12, 45.1=fK

( ) lbFm 75050010002

1=+=

( ) lbFa 25050010002

1=−=

( )( ) kipsinlbinTm −=−== 5.7750010750

( )( ) kipsinlbinTa −=−== 5.2250010250

( ) ( )( )

33.145.1

000,200345.1log3log

==f

K

flK

nK

f

3

16

d

Tss

π=

( )ksisms 57

8

7

5.7163

=

=

π

( )ksisas 19

8

7

5.2163

=

=

π

(a) ( )( ) ksisKss asflms 27.821933.157max =+=+=

SECTION 4 - SPRINGS

Page 51 of 70

Gd

TL

JG

TL4

64

πθ ==

( )( )( )

( )rad4533.0

105.118

7

301050064min

6

4=

×

=

π

θ

( )( )( )

( )rad9066.0

105.118

7

3010100064max

6

4=

×

=

π

θ

( )( )inlb

e

Fscale 3.110

109066.0

1000===

θ

(c) ns

asf

ys

ms

s

sK

s

s

N+=

1

( )( )uuns sss 344.0

000,200

105.06.0

085.06

=

=

yys ss 6.0=

( )( )

uy ss 344.0

1933.1

6.0

57

2.1

1+=

Use AISI 8760, OQT 800 oF

ksisy 200=

ksisu 220=

24.1=N

HELICAL SPRINGS – NON CIRCULAR SECTION

301. A spring is to be designed of square oil-tempered steel wire and subjected to a

repeated maximum load of 325 lb.; mean coil diameter, 1 ½ in.; deflection,

13/32 in. Determine (a) the wire size for average service, (b) the required

number of active coils, (c) the solid height, free length, and pitch (the ends are

squared and ground, the “solid stress” must be satisfactory, and the pitch angle

not excessive). (d) What amount of energy is stored when the load is 325 lb.?

Express in in-lb. and Btu.

Solution:

For oil-tempered wire,

19.0

146

w

uD

s = , [ ]5.0032.0 << wD

Max. “solid” us ss 6.0=

SECTION 4 - SPRINGS

Page 52 of 70

(a) average service,

wDb = , bt =

( )322

4.2

2

8.13

b

FDK

tb

tbFDKs m

q

mq

s =+

=

usd ss 324.0= , average service

( )19.03

146324.0

2

4.2

bb

FDK m

q =

kipF 325.0=

inDm2

11=

25.1=qK (assumed)

( )( ) ( )19.03

146324.05.1325.04.225.1

bb=

inb 2902.0=

Table AT 15, use inb 313.0= , # 1 wire size

8.4313.0

5.1===

b

DC m

Figure AF 15, 275.1=qK

(b) ( ) 4

3

3

3

44.0

45.2

56.0

45.2

Gb

NFD

tbGt

NFD cmcm =−

=δ

( )( )( )( )4

3

313.0500,1144.0

5.1325.045.2

32

13 cN=

34.7=cN

(c) Solid height = ( ) ( ) inNb c 92.2234.7313.02 =+=+

Free length = bPNc 2+

lbF 325.0=

( )( )( )

ksib

FDKs m

qs 65.48313.0

5.1325.04.2275.1

4.233

=

=

=

solid stress = ( )

( )ksi2.109

313.0

1466.019.0

==

solid deflection = in91.032

13

65.48

2.109=

=

( ) 91.0=− cNbP

( )( ) 91.034.7313.0 =−P

inP 437.0=

SECTION 4 - SPRINGS

Page 53 of 70

use inP16

7=

Free length = ( ) ( ) ininbPNc32

273837.3313.0234.7

16

72 ≈=+

=+

( )5.1

16

7

tanππ

λ

==mD

P

oo 103.5 <=λ

(d) ( ) lbinkipinFkU s −=−=

=== 66066.0

32

13325.0

2

1

2

1

2

1 2 δδ

BtuU s 085.0778

66==

302. A coil spring, of hard-drawn carbon steel, is to deflect 1 in. under a load of

100 lb. The outside coil diameter is to be 1 in. Compute the number of active

coils, (a) if the wire is round, 5/32 in. in diameter, (b) if the wire is square,

5/32 in. on the side, (c) if the wire is rectangular 1/8 x 3/16 in., long

dimension parallel to the axis, (d) If the wire is rectangular 3/16 x 1/8 in.,

short dimension parallel to the axis. (e) What is the maximum stress in each of

the above springs under the 100-lb load? (f) What is the ratio of the

approximate volumes, square- or rectangular-wire to round wire spring?

Solution:

inDD wm 1=+

(a) inDw32

5=

inDm32

27

32

51 =−=

4.5

32

5

32

27

=

==w

m

D

DC

w

c

GD

NFC38

=δ

( )( )

( )

×

=

32

5105.11

4.510081

6

3

cN

3.14=cN

SECTION 4 - SPRINGS

Page 54 of 70

(b) Square, inb32

5=

inDm32

27

32

51 =−=

4

3

44.0

45.2

Gb

NFD cm

=δ

( )

( )4

6

3

32

5105.1144.0

32

2710045.2

1

×

=cN

5.20=cN

(c) inb16

3= , int

8

1=

intDm8

7

8

111 =−=−=

( )tbGt

NFD cm

56.0

45.23

3

−=δ

( )

( )

−

×

=

8

156.0

16

3

8

1105.11

8

710045.2

13

6

3

cN

1.16=cN

(d) inb8

1= , int

16

3=

intDm16

13

16

311 =−=−=

( )tbGt

NFD cm

56.0

45.23

3

−=δ

( )

( )

−

×

=

16

356.0

8

1

16

3105.11

16

1310045.2

13

6

3

cN

5.11=cN

(e) Maximum Stress

SECTION 4 - SPRINGS

Page 55 of 70

For (a) 3

8

w

sD

FCKs

π=

( )( )

284.14.5

615.0

44.54

14.54=+

−

−=K

( )( )psiss 320,72

32

5

4.51008284.1

3=

=

π

For (b) ( )

322

4.2

2

8.13

b

FDK

tb

tbFDKs m

qm

qs =+

=

4.5

32

5

32

27

=

==b

DC m

25.1=qK

( )psiss 355,66

32

5

32

271004.2

25.13

=

=

For (c) ( )

222

8.13

tb

tbFDKs m

qs

+=

7

8

1

8

7

=

==t

DC m

1.1=qK

( )psiss 992,68

8

18.1

16

33

8

1

16

32

8

7100

1.122

=

+

=

For (d) ( )

222

8.13

tb

tbFDKs m

qs

+=

33.4

16

3

16

13

=

==t

DC m

SECTION 4 - SPRINGS

Page 56 of 70

2.1=qK

( )psiss 232,63

16

38.1

8

13

16

3

8

12

16

13100

2.122

=

+

=

(e) Ratio of the approximate volumes

For (a) Round wire

( ) cmwa NDDV ππ

= 2

4

( ) 3

2

727.03.1432

27

32

5

4inVa =

= π

π

For (b) Square wire

( ) cmb NDbV π2=

( ) 3

2

327.15.2032

27

32

5inVb =

= π

For (c) rectangular wire

( ) cmc NDbtV π=

( ) 3037.11.168

7

8

1

16

3inVc =

= π

For (d) rectangular wire

( ) cmd NDbtV π=

( ) 3688.05.1116

13

16

3

8

1inVd =

= π

Ratio of volume

Square to round wire

825.1727.0

327.1===

a

b

V

V

Rectangular to round wire (long dimension parallel to the axis)

426.1727.0

037.1===

a

c

V

V

Rectangular to round wire (short dimension parallel to the axis)

946.0727.0

688.0===

a

d

V

V

SECTION 4 - SPRINGS

Page 57 of 70

TENSION SPRINGS

305. Design two tension springs for a spring balance with a capacity of 200 lb.

Each spring supports a maximum load of 100 lb. The outside diameter must

not exceed 1 ¼ in. and the total length including end loops must not exceed 9

½ in. Select a material and determine the dimension, including wire diameter,

number of coils, and free length.

Solution:

Table AT 17, assume oil tempered wire

ksiD

sw

u 19.0

146=

ksiD

sw

ys 19.0

5.87=

( )ksi

DDs

ww

sd 19.019.0

705.878.0== , [ ]5.0032.0 << wD

δkFF i +=

3

8

w

mcs

D

FDKs

π=

w

m

w

m

D

D

D

rC ==

2

w

ca

DG

NCF38

=δ

NC

DGk w

38=

a

w

caw FGD

NCF

NC

DGk =

=

3

3

8

8δ

kiplbFa 10.0100 ==

Figure AF 15, assume 2.1=cK

33

88

w

mac

w

mics

D

DFK

D

DFKs

ππ+=

2

8

w

acics

D

CFKsKs

π+=

inDDOD wm 25.1=+=

1

25.1

+=

CDw

§ 6.21, assume ksisi 18=

ssd ss =

SECTION 4 - SPRINGS

Page 58 of 70

( )( ) ( )( )219.0

1.02.18182.1

70

ww D

C

D π+=

( )( )

( )( ) ( )( )2

2

19.0

19.0

25.1

11.02.186.21

25.1

170

π

++=

+ CCC

( ) ( )219.011956.06.2111.67 ++=+ CCC

( ) ( ) 6.2111956.011.67219.0

=+−+ CCC

7.6=C

inC

Dw 1623.017.6

25.1

1

25.1=

+=

+=

Table AT 15, use inDw 1620.0= , 8 W & M

( )( ) inCDD wm 085.11620.07.6 ===

say inDm 0.1=

17.61620.0

0.1===

w

m

D

DC

ksisi 7.17=

To check, Fig. AF 15, 15.1=cK

( ) ( )( )( )( )

ksiss 20.891620.0

17.610.015.187.1715.1

2=+=

π

( )ksiksissd 20.8992.98

1620.0

7019.0

>== , o.k.

Total length = ( )wmcw DDND ++ 2

( ) ( )162.00.12162.05.9 ++= cN

coilsNc 3.44=

Free length = ( )( ) inND cw 18.73.441620.0 ==

Summary of answer:

Material, oil-tempered wire

inDw 1620.0= , 8 W & M

coilsNc 3.44=

Free length = 7.18 in.

306. Two helical tension springs are to be used in scales for weighing milk. The

capacity of the scales is 30 lb., each spring carries 15 lb. with a deflection of 3

9/16 in. The springs are made of No. 14, W & M steel wire, outside diameter,

29/32 in. (a) how many coils should each spring have? (b) What is the

maximum stress in the wire? What material should be used?

SECTION 4 - SPRINGS

Page 59 of 70

Solution:

lbk 15=δ

in16

93=δ

(a) Table AT 15, No. 14 W &M

inDw 0800.0=

inDODD wm 82625.00800.032

29=−=−=

328.100800.0

82625.0===

w

m

D

DC

( )

w

c

GD

NCk38 δ

δ =

( )( )( )( )080.0105.11

328.10158

16

93

6

3

×= cN

8.24=cN

(b) δkFF i +=

m

wii

D

DsF

8

3π=

§ 6.21, 328.10=C

psisi 272,11=

( )( )( )

lbFi 743.282625.08

08.0272,113

==π

lbF 743.1715743.2 =+=


( )( )( )( )

psiD

FDKs

w

mcs 476,79

080.0

82625.0743.1709.18833

===ππ

ksipsis

s sys 345.99345,99

8.0

476,79

8.0===≈

Table AT 17, use Hard drawn wire

( )ksiksi

Ds

w

ys 345.99113080.0

707019.019.0

>===

307. A tension spring for a gas-control lever is made of inDw 078.0= steel wire;

inside diameter, 0.609 in.; number of coils, 55; free length including end

loops, 5 9/16 in. When the spring is extended to a length of 6 5/16 in., it must

exert a force 5 ½ lb.; it must extend to (a) the initial tension, (b) the stress in

the spring caused by the initial tension (compare with the recommended

SECTION 4 - SPRINGS

Page 60 of 70

maximum values), (c) the stress caused by the 5 ½-lb load, (d) the maximum

stress. What material should be used? (e) What energy is absorbed from the

point where the load is the initial tension until the spring’s length is 6 5/16 in.?

(Data courtesy Worthington Corporation.)

Solution:

inDw 078.0=

inDD wm 609.0=−

inDm 687.0078.0609.0 =+=

8.8078.0

687.0===

w

m

D

DC

55=cN

w

c

GD

NFC38

=δ

lbF2

15=

in75.016

95

16

56 =−=δ

( )( ) ( )( )( )078.0105.11

58.8875.0

6

3

×==

δδ

k

lbk 244.2=δ

(a) lbkFFi 256.3244.25.5 =−=−= δ

(b) ( )( )

( )psi

D

CFs

w

ii 000,12

078.0

8.8256.38822

===ππ

§ 6.21, 8.8=C

psipsisi 000,12300,13 >= , ok

(c) lbF 5.5=

2

8

w

cs

D

FCKs

π=

8.8=C

Figure AF 15

1.1=cK

( )( )( )( )

psiss 284,22078.0

8.85.51.182

==π

(d) maximum stress

inlbk

k 992.275.0

244.2===

δ

δ

SECTION 4 - SPRINGS

Page 61 of 70

δ ′= kF

in75.316

95

16

59 =−=′δ

( )( ) lbkFF i 476.1475.3992.2256.3 =+=′+= δ

( )( )( )( )

psiD

FCKs

w

cs 651,58

078.0

8.8476.141.18822

===ππ

Table AT 16

ksipsis

s sys 3.73300,73

8.0

651,58

8.0===≈

Table AT 17, use Hard drawn wire

( )ksiksi

Ds

w

ys 3.73658.113078.0

707019.019.0

>===

(e) ( )( ) lbinkU s −=== 8415.075.0992.22

1

2

1 22δ

TORSION SPRINGS

308. A carbon-steel (ASTM A230) torsion spring is to resist a force of 55 lb. at a

radius of 2 in.; the mean diameter is to be 2 ½ in. Compute (a) the diameter of

the wire for average service, (b) the number of coils for a deflection of 180o

under the given torque, (c) the energy the spring has absorbed when the force

is 55 lb.

Solution:

FaMT ==

lbF 55=

ina 2=

( )( ) lbinMT −=== 110255

inDm 5.2=

Table AT 17, ksiD

sw

u 1.0

182= , [ ]25.0093.0 << wD

Average service

( )( ) psiD

ksiDD

sswww

ud 1.01.01.0

936,117936.117182648.0405.06.1 ==

==

(a) I

KMcss =

For round wire, assume 08.1== cic KK , Table AT 18

2

wDc =

SECTION 4 - SPRINGS

Page 62 of 70

32

3

wD

c

I π=

( )( )( )1.03

936,1173211008.1

ww

sDD

s ==π

ininDw 25.02060.0 <=

Table AT 15, use inDw 2070.0= , No. 5 W & M

To check: 966.92070.0

2>===

w

m

D

D

c

r, ok

Table AT 18, 08.1=K

( )( )( )( )

psiss 430,1362070.0

3211008.13

==π

( )psipsissd 430,136054,138

2070.0

936,1171.0

>==

Therefore, use No. 5 W & M, inDw 2070.0=

(b) EI

NDM cmπθ =

psiE 61030×=

64

4

wDI

π=

πθ == o180

4

64

w

cm

ED

NMD=θ

( )( )( )( )46 2070.01030

211064

×= cN

π

29.12=cN

(c) ( )( ) lbinTU s −=== 8.1721102

1

2

1πθ

312. A pivoted roller follower is held in contact with the cam by a torsion spring.

The moment exerted by the spring varies from 20 lb-in to 50 lb-in. as the

follower oscillates through 30o. The spring is made of AISI 6152 steel, OQT

1000 oF. What should be the value of wD , mD , and cN if the factor of safety

is 1.75 based on the Soderberg line? Would this be a conservative or risky

approach?

Solution:


ksisu 184=

SECTION 4 - SPRINGS

Page 63 of 70

ksisy 173=

ksiss un 925.0 ==

( ) inlbM m −=+= 3520502

1

( ) inlbM a −=−= 1520502

1

assume 08.1=K

( )( )psi

DDD

KMs

www

mm 333

3853508.13232===

ππ

( )( )psi

DDD

KMs

www

aa 333

1651508.13232===

ππ

n

a

y

m

s

s

s

s

N+=

1

33 000,92

165

000,173

385

75.1

1

ww DD+=

inDw 1916.0=

Table AT 15, use inDw 1920.0= , No. 6 W & M

To solve for K

( )( )

psiKK

sm 369,501920.0

35323

==π

( )( )

psiKK

sa 587,211920.0

15323

==π

000,92

587,20

000,173

369,50

75.1

1 KK+=

0868.1=K

Table AT 18 0868.1== KKci

932.9 >==w

m

D

D

c

r, ok

( ) inDm 7894.11920.032.9 ==

use ininDm 875.18

71 ==

4

64

w

cmcm

ED

NMD

EI

NDM ∆=

∆=∆

πθ

( )( )( )( )46 1920.01030

875.1205064

180

30

×

−= cNπ

93.5=cN

Summary of answer:

inDw 1920.0= , No. 6 W & M

SECTION 4 - SPRINGS

Page 64 of 70

inDm8

71=

93.5=cN , 4.1>N , therefore conservative.

FLAT AND LEAF SPRINGS

315. A cantilever flat spring of uniform strength, Fig. 6.20, Text, is to absorb an

energy impact of 500 ft-lb. Let the thickness of the steel, AISI 1095, OQT 900 oF, be ½ in. and let the maximum stress be half of the yield strength. (a) Find

the width b of the spring at the widest point in terms of the length L .

Determine values of b for lengths of 36 in., 48 in., 60 in., and 72 in. (b)

Determine the deflection of the spring for each set of values found in (a).

Solution.

Fig. 6/20

2

6

bh

FLsB =

3

36

Ebh

FL=δ

AISI 1095, OQT 900 oF, ksisy 104= , Table AT 9

( ) psiksiss yB 000,52521045.05.0 ====

δFU s2

1=

L

bhsF B

6

2

=

Eh

Ls

Ebh

L

L

bhs BB

2

3

32

66 =

=δ

=

=

E

bhLs

Eh

Ls

L

bhsU BBB

s

222

12

1

62

1

lbinlbftU s −=−= 6000500

SECTION 4 - SPRINGS

Page 65 of 70

( )

6

2

1030

2

1000,52

12

16000

×

=

Lb

21598 inbL =

L

inb

21598=

inL 36= , inin

inb 4.44

36

1598 2

==

inL 48= , inin

inb 3.33

48

1598 2

==

inL 60= , inin

inb 6.26

60

1598 2

==

inL 72= , inin

inb 2.22

72

1598 2

==

(b) Eh

LsB

2

=δ

inL 36= , ( )( )

( )in4928.4

2

11030

36000,52

6

2

=

×

=δ

inL 48= , ( )( )

( )in9872.7

2

11030

48000,52

6

2

=

×

=δ

inL 60= , ( )( )

( )in48.12

2

11030

60000,52

6

2

=

×

=δ

inL 72= , ( )( )

( )in9712.17

2

11030

72000,52

6

2

=

×

=δ

317. One of the carbon contacts on a circuit breaker is mounted on the free end of

a phosphor-bronze beam ( 35.0=µ ). This beam has the shape of the beam

shown in Fig. 6.24, Text, with .1 inb = , .16

9inb =′ , .

2

14 inL = , and .

16

1inh =

When the contacts are closed, the beam deflects ¾ in. Compute (a) the force

on the contacts, (b) the maximum stress.

Solution:

SECTION 4 - SPRINGS

Page 66 of 70

Figure 6.24

22

36

bh

WL

bh

FLs ==

( ) ( )EI

WLK

EI

FLK

6

1

3

1 23

1

23

1 µµδ

−=

−=

5625.01

16

9

=

=′

b

b

Figure 6.25, 14.11 =K

(a) Force on contacts = F

( )EI

FLK

3

1 23

1 µδ

−=

psiE 61016×= (phosphor bronze)

as a beam, 12

3bh

I =

( )3

23

1 14

Ebh

FLK µδ

−=

( ) ( ) ( )[ ]( )( )

3

6

23

16

111016

35.015.414.14

4

3

×

−=

F

lbF 8=

(b) ( )( )

( )psi

bh

FLs 296,55

16

11

5.486622

=

==

318. A cantilever leaf spring 26 in. long is to support a load of 175 lb. The

construction is similar to that shown in Fig. 6.22 (a), Text. The leaves are to

be 2 in. wide, 3/16 in. thick; SAE 9255 steel, OQT 1000 oF; 107 cycles (§

6.26). (a) How many leaves should be used if the surfaces are left as rolled?

(b) The same as (a) except that the leaves are machined and the surfaces are

not decarburized. (c) The same as (b), except that the surface is peened all

over. (d) Which of these springs absorbs the most energy? Compute for each:

(e) What are the load and deflection of the spring in (b) when the maximum

stress is the standard-test yields strength?

Solution:

Figure 6.22 (a)

2

6

bh

FLsA =

SECTION 4 - SPRINGS

Page 67 of 70

3

36

Ebh

FLA =δ

bNb ′= 1

lbF 175=

inb 2=′

inh16

3=

inL 26=

§ 6.26, SAE 9255, OQT 1000 oF

ksisu 180=

ksisy 160=

inint 1875.016

3==

ksisd 75.83=

(a) As rolled, Figure AF 5


( ) psiksisd 000,232375.83275.0 ===

2

6

bh

FLsA =

( )( )

( )2

116

32

261756000,23

=

N

88.161 =N

say 171 =N

(b) Machined, Figure AF 5


( ) psiksisd 800,628.6275.8375.0 ===

2

6

bh

FLsA =

( )( )

( )2

116

32

261756800,62

=

N

2.61 =N

say 71 =N

(c) Peened surface, (b)

SECTION 4 - SPRINGS

Page 68 of 70

( ) psiksisd 500,785.788.6225.1 ===

2

6

bh

FLsA =

( )( )

( )2

116

32

261756500,78

=

N

95.41 =N

say 51 =N

(d) δFU s2

1=

lbF 175=

3

1

36

hbEN

FL

′=δ

For (a) 171 =N

( )( )

( )( )( )in745.2

16

32171030

2617563

6

3

=

×

=δ

( )( ) lbinU s −== 240745.21752

1

For (b) 71 =N

( )( )

( )( )( )in666.6

16

3271030

2617563

6

3

=

×

=δ

( )( ) lbinU s −== 583666.61752

1

For (c) 51 =N

( )( )

( )( )( )in332.9

16

3251030

2617563

6

3

=

×

=δ

( )( ) lbinU s −== 817332.91752

1

answer – spring (c)

(e) ksiss yd 160== , 71 =N (b)

SECTION 4 - SPRINGS

Page 69 of 70

2

1

2

66

hbN

FL

bh

FLsd ′

==

( )

( )( )2

16

327

266000,160

=

F

load lbF 505=

( )( )

( )( )( )in24.19

16

3271030

2650563

6

3

=

×

=δ

319. The rear spring of an automobile has 9 leaves, each with an average thickness

of 0.242 in. and a width of 2 in.; material is SAE 9261, OQT 1000 oF. The

length of the spring is 56 in. and the total weight on the spring is 1300 lb.

Assume the spring to have the form shown in Fig. 6.22 (b), Text. Determine

(a) the rate of the spring, (b) the maximum stress caused by the dead weight.

(c) What approximate repeated maximum force (0 to maxF ) would cause

impending fatigue in 105 cycles, the number of applications of the maximum

load expected during the ordinary life of a car? (If the leaves are cold rolled

to induce a residual compressive stress on the surfaces, the endurance limit as

2us should be conservative.)

Solution:

Figure 6.22 (b)

22

3

bh

FLsA =

3

3

8

3

Ebh

FLA =δ

lbF 1300=

inh 242.0=

91 =N

inb 2=′

inL 56=

(a) Rate , 3

3

3

8

L

EbhFk

A

==δ

( )( )( )( )( )

inlbL

hbENk 21.116

563

242.02910308

3

83

36

3

3

1 =×

=′

=

(b) ( )( )

( )( )( )psi

hbN

FLsA 590,103

242.0292

5613003

2

322

1

==′

=

SECTION 4 - SPRINGS

Page 70 of 70

(c) SAE 9261. OQT 1000 oF

ksisu 192=

ksisn 962

192==

2

12

3

hbN

FLsA ′

=

( )( )( )( )2

242.0292

563000,96

F=

lbF 1200=

321. The front spring of an automobile is made similar to Fig. 6.23, Text. The

average thickness for each of the 6 leaves, 0.213 in.; material is SAE 9255,

OQT 1000 oF. The load caused by the weight of the car is 775 lb. (a) What

stress is caused by a force of twice the dead weight? (b) What load would

stress the spring to the yield strength?

Solution:

Figure 6.23

2

1

22

336

hbN

WL

bh

WL

bh

FLs

′===

lbW 775= , 61 =N , inb 2=′ , inh 213.0=

inin

L 182

36==

(a) ( ) lbW 15507752 ==

( )( )( )( )( )

psis 740,153213.026

18155032

==

(b) SAE 9255, OQT 1000 oF

ksisy 160=

2

1

22

336

hbN

WL

bh

WL

bh

FLs

′===

( )( )( )( )2

213.026

183000,160

W=

lbW 1613=

- end -

SECTION 5 – COLUMNS

Page 1 of 18

DESIGN PROBLEMS

334. A round steel rod made of structural steel, AISI C1020, as rolled, is to be used as

a column, centrally loaded with 10 kips; 3=N . Determine the diameter for (a)

.25 inL = , (b) .50 inL = (c) The same as (a) and (b) except that the material is

AISI 8640, OQT 1000 F. Is there any advantage in using this material rather than

structural steel?

Solution:

For AISI C1020,as rolled

ksis y 48=

kipsF 10=

3=N

(a) .25 inLLe ==

Consider first J.B. Johnson

−==E

k

Ls

AsNFF

ey

yc 2

2

41

π

4

2D

Aπ

=

4

Dk =

ksiE 31030×=

( )( ) ( )

( )

( )

×

−

=

32

2

2

10304

4

2548

14

48103π

πD

D

−=

22

2 411230

DD

ππ

ππ

481230 2 −= D

inD 096.1=

say ininD 0625.116

11 ==


Page 2 of 18

∴<=

= 12094

4

0625.1

25

k

Le o.k.

(b) .50 inLLe ==

Consider Euler’s Equation

2

2

==

k

L

EANFF

e

c

π

( )( )( )

2

232

4

50

41030

103

×

=

D

Dππ

431875.030 Dπ=

inD 507.1=

say ininD 5.12

11 ==

∴>=

= 120133

4

5.1

50

k

Le o.k.

(c) For AISI 8640, OQT 1000 F

ksisy 150=

2

1

22

=

y

e

s

E

k

L π

( )83.62

150

10302 2

132

=

×=

π

k

Le

For (a) .25 inLLe ==

Consider first J.B. Johnson

−==E

k

Ls

AsNFF

ey

yc 2

2

41

π


Page 3 of 18

( )( ) ( )

( )

( )

×

−

=

32

2

2

10304

4

25150

14

150103π

πD

D

−=

22

2 5.1215.3730

DD

ππ

ππ

75.4685.3730 2 −= D

inD 23.1=

say inD 25.1=

∴>=

= 83.6280

4

25.1

25

k

Le use Euler’s equation

2

2

==

k

L

EANFF

e

c

π

( )( )( )

2

232

4

25

41030

103

×

=

D

Dππ

4375.030 Dπ=

inD 0657.1=

say ininD 0625.116

11 ==

∴>=

= 83.6294

4

0625.1

25

k

Le ok

For (b) .50 inLLe ==

Consider Euler’s Equation

2

2

==

k

L

EANFF

e

c

π


Page 4 of 18

( )( )( )

2

232

4

50

41030

103

×

=

D

Dππ

431875.030 Dπ=

inD 507.1=

say ininD 5.12

11 ==

∴>=

= 83.62133

4

5.1

50

k

Le o.k.

There is no advantage.

335. A hollow circular column, made of AISI C1020, structural steel, as rolled, is to

support a load of 10,000 lb. Let inL 40= , oi DD 75.0= , and 3=N . Determine

oD by (a) using either Euler’s or the parabolic equation; (b) using the straight-

line equation. (c) What factor of safety is given by the secant formula for the

dimensions found in (a)?

Solution:


ksisy 48=

inLLe 40==

kipslbF 10000,10 ==

3=N

oi DD 75.0=

A

Ik =

( ) ( )[ ] 44444

033556.075.064

oooio DDD

DDI =−=

−= π

π

( ) ( )[ ] 2

2222

343612.04

75.0

4o

ooio DDDDD

A =−

=−

=ππ

o

o

o DD

Dk 3125.0

343612.0

033556.02

4

==


Page 5 of 18

(a) Consider parabolic equation

−==E

k

Ls

AsNFF

ey

yc 2

2

41

π

( )( ) ( )( )

( )

( )

×

−=32

2

2

10304

4

3125.0

2548

1343612.048103π

o

o

D

D

9519.10493376.1630 2 −= oD

inDo 576.1=

say ininDo 5625.116

91 ==

( )∴<== 12082

5625.13125.0

40

k

Le o.k.

(b) Straight-line equation

k

L

A

F70000,16 −=

−=

oo DD 3125.0

4070000,16

343612.0

000,102

oo DD 30785498000,10 2 −=

inDo 6574.1=

say ininDo 625.18

51 ==

( )∴<== 1208.78

625.13125.0

40

k

Le o.k.

(c) Secant formula

+=

EA

NF

k

L

k

ec

A

NFs e

y2

sec12

inDo 5625.1=

inDk o 4883.03125.0 == 22 8389.0343612.0 inDA o ==


Page 6 of 18

25.02

=k

ec, (i7.8)

( )( ) ( )( )

×+=

8389.01030

10

4883.02

40sec25.01

8389.0

1048

3

NN

( )[ ]NN 81645.0sec25.0192.1148 +=

289.2=N

336. A column is to be built up of ½-in., AISI C1020, rolled-steel plates, into a square

box-section. It is 6 ft long and centrally loaded to 80,000 lb. (a) Determine the

size of section for 74.2=N . (b) Compute N from the secant formula for the

size found and compare with 2.74.

Solution:

For AISI C1020, rolled-steel plate

ksisy 48=

( ) ( )12

1

12

1

12

4444 −−=

−−=

bbbbI

( )22 1−−= bbA

( )( )[ ]22

44

112

1

−−

−−==

bb

bb

A

Ik

inftLLe 726 ===

kipslbF 80000,80 ==

(a) 74.2=N

Consider J.B. Johnson

−=E

k

Ls

AsNF

ey

y 2

2

41

π

( )( ) ( )( )

( )

×

−=32

2

10304

7248

1488074.2π

kA

2

085.10482.219

k

AA −=


Page 7 of 18

try inb 23.3=

( ) ( )( ) ( )[ ] ink 1331.1

123.323.312

123.323.322

44

=−−

−−=

( ) ( ) ( ) 22222 46.5123.323.31 inbbA =−−=−−=

( ) ( )( )

2.2191331.1

46.5085.1046.5482.219

2=−= ok

Therefore use inb 23.3=

∴<== 12054.631331.1

72

k

Le o.k.

inb 23.3= or inb4

13=

(b)

+=

EA

NF

k

L

k

ec

A

NFs e

y2

sec12

25.02

=k

ec, (i7.8)

( )( ) ( )( )

×+=

46.51030

80

1331.12

72sec25.01

46.5

8048

3

NN

( )[ ]NN 70214.0sec25.01652.1448 +=

74.22.2 <=N

337. A column is to be made of ½-in structural steel plates (AISI 1020, as rolled),

welded into an I-section as shown in Table AT 1 with HG = . The column, 15 ft

long, is to support a load of 125 kips. (a) Determine the cross-sectional

dimensions from the straight-line equation. (Using either Johnson’s or Euler’s

equation, compute the equivalent stress and the factor of safety. (c) Compute N

from the secant formula.

Solution:


ksisy 48=

inftLLe 18015 ===

kipsF 125=


Page 8 of 18

Table AT 1.

HG =

( )( ) ( ) ( )135.05.05.15.05.115.0 222 −=−=+−−=−−−=−= HHHHHHHHghGHA

( )( )( )

( )( )( )136

15.0

135.0

15.0

12

1

12

1343433

−

−−−=

−

−−−=

−

−=

H

HHH

H

HHH

ghGH

ghGHk

(a) Straight-line equation

−=

k

LAF 0044.01000,16

( )

−=

kA

1800044.01000,16000,125

−=

kA

792.018125.7

use inH 37.7=

( ) ( )( )( )( )

ink 04527.3137.736

137.75.037.737.734

=−

−−−=

( )[ ] inA 555.10137.735.0 =−=

81.704527.3

792.01555.108125.7 =

−≈

Therefore use inH 37.7=

Or ininH 375.78

37 ==

(b) Consider J.B. Johnson, 1205904527.3

180<==

k

Le

N

ss

y

e =

( )

( )

ksi

E

k

Ls

A

Fs

ey

e 8.13

10304

04527.3

18048

1555.10

125

41

32

2

2

2=

×

−

=

−

=

ππ


Page 9 of 18

48.38.13

48===

e

y

s

sN

(c) From secant formula

+=

EA

NF

k

L

k

ec

A

NFs e

y2

sec12

25.02

=k

ec, (i7.8)

( )( ) ( )( )

×+=

555.101030

125

04527.32

180sec25.01

555.10

12548

3

NN

( )[ ]NN 5872.0sec25.01843.1148 +=

8.2=N

338. The link shown is to be designed for 5.2=N to support an axial compressive

load that varies from 0 to 15 kips; inL 20= ; Material AISI 1030, as rolled. (a)

Determine the diameter considering buckling only. (b) Determine the diameter

considering varying stresses and using the Soderberg line (perhaps too

conservative). Estimate an appropriate strength-reduction factor (see Fig. AF 6).

(c) Keeping in mind that the stress is always compressive, do you think that the

answer from (a) will do? Discuss.

Problem 338.

Solution:


ksisy 51=

ksisu 80=

( )108

51

103022 2

1322

1

2

=

×=

=

ππ

y

e

s

E

k

L

inL 20=

5.2=N

(a) kipsF 15=

Consider J.B. Johnson


Page 10 of 18

−=E

k

Ls

AsNF

ey

y 2

2

41

π

4

Dk =

4

2D

Aπ

=

inLLe 20==

( )( ) ( )

( )

( )

×

−

=

32

2

2

10304

4

2051

14

51155.2π

πD

D

−=

22

2 72.2175.125.37

DD

ππ

ππ

68.3475.125.37 2 −= D

inD 101.1=

say ininD 1875.116

31 ==

∴<== 10868

4

1875.1

20

k

Le o.k.

(b) Variable stresses

( ) ksiss un 40805.05.0 ===

Size factor = 0.85

( ) ksisn 344085.0 ==

8.2=fK (Figure AF 6)

n

af

y

m

s

sK

s

s

N+=

1

F = 0 to 15 kips

kipsFF am 5.7==

eaem ss =

( )34

8.2

515.2

1 emem ss+=

ksisem 923.3=


Page 11 of 18

−=E

k

Ls

AsF

eem

emm 2

2

41

π

( )

( )

( )

×

−

=

32

2

2

10304

4

2051

14

923.35.7π

πD

D

−=

22

2 72.2198.05.7

DD

ππ

ππ

67.298.05.7 2 −= D

inD 65.1=

say ininD 625.18

51 ==

∴<== 10849

4

625.1

20

k

Le o.k.

(c) The answer in (a) will not do because it is lower than (b)

339. The connecting link for a machine (see figure) is subjected to a load that varies

fro + 450 (tension) to –250 lb. The cross section is to have the proportions

HG 4.0= , Ht 1.0= , fillet radius Hr 05.0≈ ; inL 10= ; material, AISI C1020,

as rolled. (a) Considering buckling only, determine the dimensions for a design

factor of 2.5. (b) For the dimension found compute the factor of safety from the

Soderberg criterion.

Problems 339, 340

Solution:


ksisy 48=

ksisu 65=


Page 12 of 18

Table AT 1

HG 4.0=

Ht 1.0=

Hr 05.0≈

ghGHA −=

HHHtGg 3.01.04.0 =−=−=

( ) HHHh 8.01.02 =−=

( )( ) ( )( ) 216.08.03.04.0 HHHHHA =−=

( )( ) ( )( )( )

HH

HHHH

ghGH

ghGHk 35824.0

16.0

8.03.04.0

12

1

12

12

3333

=

−=

−

−=

(a) Consider J.B. Johnson

−=E

k

Ls

AsNF

ey

y 2

2

41

π

kiplbF 35.0350 ==

inLe 10=

( )( ) ( )( ) ( )( )

×

−=32

2

2

2

10304

35824.0

1048

116.04835.05.2π

HH

2425.068.7875.0 2 −= H

inH 3815.0=

( )( )∴<== 12073

3815.035824.0

10

k

Le ok

say ininH 46875.032

15==

ininHG16

31875.04.0 ===

ininHt64

3046875.01.0 ===


Page 13 of 18

(b) with inH 46875.0=

( ) 220352.046875.016.0 inA ==

( ) ink 1679.046875.035824.0 ==

( )

ksipsi

E

k

Ls

A

F

ss

ey

e 6.11600,11

10304

1679.0

1048

1

0352.0

350

41

32

2

2

2

min

min −=−=

×

−

−

=

−

==

ππ

psipsiA

Fs 8.12800,12

0352.0

450maxmax +=+=

+==

( ) ksism 6.06.118.122

1=−=

( ) ksisa 1.126.118.122

1=+=

( ) ksiss nu 5.32655.05.0 ===

Size factor = 0.85

( ) ksiksisu 62.275.3285.0 ==

Figure AF 9, ( ) 023.046875.005.005.0 === Hr

( ) inHh 7031.046875.0.155.1 ===

inHd 4688.0==

05.005.0

==H

H

d

r

5.15.1

==H

H

d

h

65.2=tK

70.0

023.0

01.01

1

01.01

1=

+

=

+

=

r

q

( ) 2.21165.270.0 =+−=fK

n

qf

y

m

s

sK

s

s

N+=

1

( )( )62.27

1.122.2

48

6.01+=

N

024.1=N

CHECK PROBLEMS

341. The link shown is subjected to an axial compressive load of 15 kips. Made of

AISI C1030, as rolled, it has sectional length of 20 in. Assume a loose fit with the


Page 14 of 18

pins. What is (a) the critical load for this column, (b) the design factor, (c) the

equivalent stress under a load of 15 kips? What material does the secant formula

indicate as satisfactory for the foregoing load, when (e) 25.02 =kec , (f)

400

eLe = .

Problem 341, 342

Solution:


ksis y 51=

inb 75.0=

inh 75.1=

( )( ) 23125.175.175.0 inbhA ===

For loose fit

12

3bh

I =

inh

bh

bh

A

Ik 5052.0

12

75.1

1212

3

=====

1086.395052.0

20<==

k

Le for AISI C1030, as rolled

use J.B. Johnson equation

(a) ( )( )( )

kipsE

k

Ls

AsF

ey

yc 42.62000,304

5052.0

2051

13125.1514

12

2

2

2

=

−=

−=ππ

(b) NFFc =

16.415

42.62===

F

FN c

(c) ksiN

ss

y

e 26.1216.4

51===

(d) Actual ksiA

Fs 43.11

3125.1

15===


Page 15 of 18

Secant Formula

+=

EA

NF

k

L

k

ec

A

NFs e

y2

sec12

(e) 25.02

=k

ec

( ) ( )( )ksisy 4.64

3125.11030

42.62

5052.02

20sec25.01

3125.1

42.623

=

×+=

use AISI C1020, cold drawn, ksisy 66=

(f) inL

e e 05.0400

20

400===

inh

c 875.02

75.1

2===

( )( )( )

1714.05052.0

875.005.022

==k

ec

( ) ( )( )ksisy 12.59

3125.11030

42.62

5052.02

20sec1714.01

3125.1

42.623

=

×+=

use AISI C1045, cold drawn, ksisy 59=

343. A schedule-40, 4-in. pipe is used as a column. Some of its properties are:

inDo 5.4= , inDi 026.4= , ..174.3 insqI = , ftL 15= ; material equivalent to

AISI C1015, as rolled. The total load to be carried is 200 kips. (a) What

minimum number of these columns should be used if a design factor of 2.5 is

desired and the load evenly distributed among them? For the approximately fixed

ends, use LLe 65.0= as recommended by AISC. (b) What is the equivalent stress

in the column?

Solution:


ksis y 5.45=

( )114

5.45

103022 2

1322

1

2

=

×=

=

ππ

y

e

s

E

k

L

inftL 18015 ==

( ) inLLe 11718065.065.0 ===

1145.77509.1

117<==

k

Le


Page 16 of 18

Use J.B. Johnson equation

(a) ( )( ) ( )

( )kips

E

k

Ls

N

AsF

ey

y4.44

000,304

5.775.451

5.2

174.35.45

41

2

2

2

2

=

−=

−=ππ

No. of columns

5.44.44

200== say 5 columns

(b)

−

=

E

k

Ls

A

F

s

ey

e

2

2

41

π

kipsF 405

200==

( )

ksi

k

se 4.16

000,304

5.775.45

1

174.3

40

2

2=

−

=

π

344. A generally loaded column is a 10-in. x 49 lb., wide-flange I-beam whose

properties are (see figure); inkx 35.4= , ink y 54.2= , area ..4.14 insqA = ,

49.272 inI x = , 40.93 inI y = ; length ftL 30= , material AISI 1022, as rolled. Let

the ends be a “little” fixed with LLe 8.0= and determine the critical load (a)

according to the Johndon or the Euler equation; (b) according to the secant

formula if 2kec is assumed to be 0.25.


Page 17 of 18

Solution:


ksis y 52=

( )107

52

103022 2

1322

1

2

=

×=

=

ππ

y

e

s

E

k

L

(a) ink 54.2= 40.93 inI =

( )( ) inLe 28812308.0 ==

1074.11354.2

288>==

k

Le

Use Euler’s Equation

( )( )( )

kips

k

L

EAF

e

c 3324.113

4.14000,302

2

2

2

==

=

ππ

(b) Secant formula

+=

EA

NF

k

L

k

ec

A

NFs e

y2

sec12

( )( )

×+=

4.1410302

4.113sec25.01

4.1452

3

cc FF

[ ]{ }cc F

F0863.0sec25.01

4.1452 +=

kipsFc 273=

348. A 4 x 3 x ½-in. angle is used as a flat-ended column, 5 ft. long, with the resultant

load passing through the centroid G (see figure); inkx 25.1= , ink y 86.0= ,

inku 37.1= , inkv 64.0= , ..25.3 insqA = Find the safe load if 8.2=N and the

material is (a) structural steel, (b) magnesium alloy AZ 91C (i7.12.\, Text), (c)

magnesium alloy AZ 80A, (d) magnesium alloy AZ 80A as before, but use the

Johnson formula and compare.


Page 18 of 18

Solution:

( )( )in

LLe 30

2

125

2===

inkk 64.0min ==

875.4664.0

30==

k

Le

(a) Structural steel, ksisy 48=

120875.46 <=k

Le

use J.B. Johnson

( )( ) ( )( )

kipsE

k

Ls

N

AsF

ey

y75.50

000,304

875.46481

8.2

25.348

41

2

2

2

2

=

−=

−=ππ

(b) magnesium alloy AZ 91C

6

2

104.641

×

+

=

k

LC

C

A

NF

e

000,57=C

( )( )( )

psiF

6

2

104.64

875.46000,571

000,57

25.3

8.2

×+

=

kipslbF 467.22467,22 ==

(c) magnesium alloy AZ 80A

900,82=C

( )( )( )

psiF

6

2

104.64

875.46900,821

900,82

25.3

8.2

×+

=

kipslbF 134.25134,25 ==

(d) By J.B. Johnson

For magnesium alloy AZ 80A, ksisy 36=

( )( ) ( )( )

kipskipsE

k

Ls

N

AsF

ey

y134.2539

000,304

875.46361

8.2

25.336

41

2

2

2

2

>=

−=

−=ππ

- end -

SECTION 7 – SHAFT DESIGN

Page 1 of 76

471. A short stub shaft, made of SAE 1035, as rolled, receivers 30 hp at 300 rpm via a

12-in. spur gear, the power being delivered to another shaft through a flexible

coupling. The gear is keyed (profile keyway) midway between the bearings. The

pressure angle of the gear teeth o20=φ ; 5.1=N based on the octahedral shear

stress theory with varying stresses. (a) Neglecting the radial component R of the

tooth load W , determine the shaft diameter. (b) Considering both the tangential

and the radial components, compute the shaft diameters. (c) Is the difference in

the results of the parts (a) and (b) enough to change your choice of the shaft size?

Problem 471.

Solution:

For SAE 1035, as rolled

ksisy 55=

ksisu 85=

( ) ksiss un 5.42855.05.0 ===

φcosWA =

( )lbin

n

hpT −=== 6300

300

30000,63000,63

2

ADT =

( )2

126300

A=

lbA 1050=

φcosWA =

20cos1050 W=

lbW 1118=

Shear stress

( )33

63001616

dd

Tss

ππ==

3

800,100

dss mss

π==

0=ass


Page 2 of 76

bending stress

From Table AT 2

4

FLM =

(a) Negligible R :

( )( )lbin

ALM −=== 4200

4

161050

4

( )333

400,13442003232

ddd

Ms

πππ===

0=ms

3

400,134

dssa

π==

SF

sKs

s

ss

af

m

y

ne +=

For profile keyway

0.2=fK

6.1=fsK

85.0=SF

( )( )( )( ) 33

661,100

85.0

400,1340.2

ddSF

sKs

af

e ===π

SF

sKs

s

ss

asfs

ms

ys

nses +=

294.1

1

55

5.42===

y

n

ys

ns

s

s

s

s

33

796,24800,100

294.1

1

dds

s

ss ms

ys

nses =

==

π

Octahedral-shear theory

2

122

577.0

1

+

=

n

es

n

e

s

s

s

s

N

( )

2

12

3

2

3 500,42577.0

796,24

500,42

661,100

5.1

1

+

=

dd

ind 569.1=


Page 3 of 76

use ind16

111=

(b) Considering both radial and tangential component.

( )( )lbin

WLM −=== 4472

4

161118

4

( )333

104,14344723232

ddd

Ms

πππ===

0=ms

3

104,143

dssa

π==

( )( )( )( ) 33

180,107

85.0

104,1430.2

ddSF

sKs

af

e ===π

2

122

577.0

1

+

=

n

es

n

e

s

s

s

s

N

( )

2

12

3

2

3 500,42577.0

796,24

500,42

180,107

5.1

1

+

=

dd

ind 597.1=

use ind16

111=

(c) The difference in the results of the parts (a) and (b) is not enough to change the choice

of the shaft size.

472. A cold-finished shaft, AISI 1141, is to transmit power that varies from 200 to 100

and back to 200 hp in each revolution at a speed of 600 rpm. The power is

received by a 20-in. spur gear A and delivered by a 10-in. spur gear C. The

tangential forces have each been converted into a force ( A and C shown) and a

couple (not shown). The radial component R of the tooth load is to be ignored in

the initial design. Let 2 and, considering varying stresses with the maximum

shear theory, compute the shaft diameter.

Problems 472 – 474


Page 4 of 76

Solution:


ksisy 90=

ksisn 50=

8.1

1=

y

n

s

s

85.0=SF

n

hpT

000,63=

( )lbinT −== 000,21

600

200000,63max

( )lbinT −== 500,10

600

100000,63min

( ) ( ) lbinTTTm −=+=+= 750,15500,10000,212

1

2

1minmax

( ) ( ) lbinTTTa −=−=−= 250,5500,10000,212

1

2

1minmax

3

16

d

Tss

π=

( )33

000,252750,1516

ddsms

ππ==

( )33

000,24525016

ddsas

ππ==

SF

sKs

s

ss

asfs

ms

ys

nses +=

For profile keyway

0.2=fK

6.1=fsK

8.1

1==

y

n

ys

ns

s

s

s

s

( )( )333

894,94

85.0

000,846.1000,252

8.1

1

dddses =+

=

ππ

Bending stress, negligible radial load

lbinT −= 000,21 at 200 hp

For A:

TA =

2

20

( ) 000,2110 =A


Page 5 of 76

lbA 2100= at 200 hp

For C:

TC =

2

10

( ) 000,215 =C

lbC 4200= at 200 hp

[ ]∑ = 0BM ( ) ( ) ( )152510 CDA =+

at 200 hp

( )( ) ( ) ( )( )15420025102100 =+ D

lbD 1680=

[ ]∑ = 0VF

DBCA +=+

at 200 hp

168042002100 +=+ B

lbB 4620=

At 200 hp: lbA 2100= , lbB 4620= , lbC 4200= , lbD 1680=

Shear Diagram

Maximum moment at B

( )( ) lbinM −== 000,21102100

( )333

000,672000,213232

ddd

Ms

πππ===

0=ms

3

000,672

dssa

π==

( )( )33

304,503

85.0

000,6720.20

ddSF

sKs

s

ss

af

m

y

ne =+=+=

π


Page 6 of 76

3

894,94

dses =

Maximum Shear Theory

2

122

5.0

1

+

=

n

es

n

e

s

s

s

s

N

( )

2

12

3

2

3 000,505.0

894,94

000,50

304,503

2

1

+

=

dd

ind 78.2=

use ind4

32=

475. A shaft S, of cold-drawn AISI 1137, is to transmit power received from shaft W,

which turns at 2000 rpm, through the 5-in. gear E and 15-in. gear A. The power

is delivered by the 10-in. gear C to gear G, and it varies from 10 hp to 100 hp and

back to 10 hp during each revolution of S. The design is to account for the

varying stresses, with calculations based on the octahedral shear stress theory.

Let 8.1=N and compute the shaft diameter, using only the tangential driving

loads for the first design.

Problem 475 – 477

Solution.


ksisy 93=

ksisu 103=

( ) ksiss un 5.511035.05.0 ===

806.1

1

93

5.51===

ys

ns

y

n

s

s

s

s

n

hpT

000,63=

( ) rpmrpmAin

Einn 6672000

.15

.5==


Page 7 of 76

( )lbinT −== 9450

667

100000,63max

( )lbinT −== 945

667

10000,63min

( ) ( ) lbinTTTm −=+=+= 5.519794594502

1

2

1minmax

( ) ( ) lbinTTTa −=−=−= 5.425294594502

1

2

1minmax

3

16

d

Tss

π=

( )33

160,835.519716

ddsms

ππ==

( )33

040,685.425216

ddsas

ππ==

SF

sKs

s

ss

asfs

ms

ys

nses +=

For profile keyway

0.2=fK

6.1=fsK

85.0=SF

( )( )333

425,55

85.0

040,686.1160,83

806.1

1

dddses =+

=

ππ

Bending stress, using only tangential loads

For 100 hp:

lbinT −= 9450

TA =

2

15

( ) 94505.7 =A

lbA 1260=

For C:

TC =

2

10


Page 8 of 76

( ) 94505 =C

lbC 1890=

[ ]∑ = 0BM CDA 14206 =+

( ) ( )1890142012606 =+ D

lbD 945=

[ ]∑ = 0VF

DBCA +=+

94518901260 +=+ B

lbB 2205=

Shear diagram

Maximum moment at B

( )( ) lbinM −== 756061260

( )333

920,24175603232

ddd

Ms

πππ===

0=ms

3

920,241

dssa

π==

( )( )33

189,181

85.0

920,2410.2

ddSF

sKs

s

ss

af

m

y

ne ==+=

π

3

425,55

dses =

Octahedral Shear Theory

2

122

577.0

1

+

=

n

es

n

e

s

s

s

s

N

( )

2

12

3

2

3 500,51577.0

425,55

500,51

189,181

2

1

+

=

dd

ind 997.1=

use ind 2=


Page 9 of 76

478. A shaft made of AISI 1137, cold rolled, for a forage harvester is shown.

Power is supplied to the shaft by a vertical flat belt on the pulley A. At B, the

roller chain to the cutter exerts a force vertically upwards, and the V-belt to

the blower at C exerts a force vertically upwards. At maximum operating

conditions, the flat belt supplies 35 hp at 425 rpm, of which 25 hp is delivered

to the cutter and 10 hp to the blower. The two sections of the shaft are joined

by a flexible coupling at D and the various wheels are keyed (sled-runner

keyway) to the shafts. Allowing for the varying stresses on the basis of the

von Mises-Hencky theory of failure, decide upon the diameters of the shafts.

Choose a design factor that would include an allowance for rough loading.

Problem 478.

Solution:

For AISI 1137, cold rolled

ksisy 93=

ksisu 103=

( ) ksiss un 5.511035.05.0 ===

806.1

1

93

5.51===

ys

ns

y

n

s

s

s

s

Pulley,

( )lbin

n

hpTA −=== 5188

425

35000,63000,63

For flat-belt

( ) ( )lb

D

TFFFFF

A

AA 692

30

51884222 1221 ==

=−=+=

Sprocket,

( )lbin

n

hpTB −=== 3706

425

25000,63000,63

For chain,

( )lb

D

TF

B

BB 741

10

370622===

Sheave,

( )lbin

n

hpTC −=== 1482

425

10000,63000,63


Page 10 of 76

For V-belt,

( ) ( )lb

D

TFFFFF

C

CC 445

10

1482325.15.1 1221 ==

=−=+=

Consider shaft ABD.

35 hp

Shaft ABD

[ ]∑ = 0'DM

( ) ( ) BA FAF 4'48486 ++=++

( ) ( )7414'1269218 += A

lbA 791'=

[ ]∑ = 0VF

AFDF BA′+=′+

791741692 +=′+ D

lbD 840=′

Shear Diagram

Maximum M at A’.

( )( ) .41526926 lbinM −==

( )333

864,13241523232

ddd

Ms

πππ===

0=ms

3

864,132

dssa

π==


Page 11 of 76

SF

sKs

s

ss

af

m

y

ne +=

For sled-runner keyway (Table AT 13)

6.1=fK

6.1=fsK

85.0=SF

( )( )33

610,79

85.0

864,13260.10

ddSF

sKs

s

ss

af

m

y

ne =+=+=

π

at A’ lbinTT A −== 5188

( )333

008,8351881616

ddd

Tss

πππ===

sms ss =

0=ass

SF

sKs

s

ss

asfs

ms

ys

nses +=

33

630,14000,83

806.1

1

ddses =

=

π

Choose a design factor of 2.0

0.2=N

von Mises-Hencky theory of failure (Octahedral shear theory)

2

122

577.0

1

+

=

n

es

n

e

s

s

s

s

N

( )

2

12

3

2

3 500,51577.0

630,14

500,51

610,79

2

1

+

=

dd

ind 48.1=

use ind2

11=

Consider shaft D-C

( )lbin

n

hpTC −=== 1482

425

10000,63000,63

For V-belt,

( ) ( )lb

D

TFFFFF

C

CC 445

10

1482325.15.1 1221 ==

=−=+=


Page 12 of 76

[ ]∑ = 0'CM

CFD 38 =′′

( )44538 =′′D

lbD 167=′′

[ ]∑ = 0VF

CFDC +′′=′

445167 +=′C

lbC 612=′

Shear Diagram

( )( ) lbinM −== 13368167

( )333

752,4213363232

ddd

Ms

πππ===

0=ms , ssa =

( )( )33

616,25

85.0

752,4260.10

ddSF

sKs

s

ss

af

m

y

ne =+=+=

π

at C’, lbinTC −=1482

( )333

712,2314821616

ddd

Tss

πππ===

sms ss =

0=ass

SF

sKs

s

ss

asfs

ms

ys

nses +=

33

41800

712,23

806.1

1

ddses =+

=

π

2

122

577.0

1

+

=

n

es

n

e

s

s

s

s

N

( )

2

12

3

2

3 500,51577.0

4180

500,51

616,25

2

1

+

=

dd


Page 13 of 76

ind 011.1=

use ind 1=

479. A shaft for a punch press is supported by bearings D and E (with L = 24 in.)

and receives 25 hp while rotating at 250 rpm, from a flat-belt drive on a 44-

in. pulley at B, the belt being at 45o with the vertical. An 8-in. gear at A

delivers the power horizontally to the right for punching operation. A 1500-lb

flywheel at C has a radius of gyration of 18 in. During punching, the shaft

slows and energy for punching comes from the loss of kinetic energy of the

flywheel in addition to the 25 hp constantly received via the belt. A

reasonable assumption for design purposes would be that the power to A

doubles during punching, 25 hp from the belt, 25 hp from the flywheel. The

phase relations are such that a particular point in the section where the

maximum moment occurs is subjected to alternating tension and

compression. Sled-runner keyways are used for A, B, and C; material is cold-

drawn AISI 1137, use a design factor of 5.2=N with the octahedral shear

theory and account for the varying stresses. Determine the shaft diameters.

Problems 479-480

Solution:

Flat-Belt Drive (B)

( )lbin

n

hpTB −=== 6300

250

25000,63000,63

( ) ( )lb

D

TFFFFF

B

BB 573

44

63004222 2121 ==

=−=+=

Gear A, Doubled hp

( )lbin

n

hpTA −=

+== 600,12

250

2525000,63000,63

( )lb

D

TF

A

AA 3150

8

600,1222===

Loading:


Page 14 of 76

Vertical:

lbFB BV 40545cos57345cos ===

[ ]∑ = 0DM

( ) VV EB 24815006 =+

( ) ( ) VE24405815006 =+

lbEV 510=

[ ]∑ = 0VF

VVV BDE +=+1500

4055101500 +=+ VD

lbDV 1605=

Shear Diagram


Page 15 of 76

( )( ) lbinM

VD −== 900015006

( )( ) lbinMVB −== 816051016

( )( ) lbinMVA −== 25505105

Horizontal:

lbFB Bh 40545sin57345sin ===

[ ]∑ = 0DM

Ahh FEB 19248 =+

( ) ( )315019244058 =+ hE

lbEh 2359=

[ ]∑ = 0hF

Ahhh FEBD =++

31502359405 =++hD

lbDh 386=

Shear Diagram


Page 16 of 76

lbinM

hD −= 0

( )( ) lbinMhB −== 30883868

( )( ) lbinMhA −== 795,1123595

( ) ( ) ( ) ( ) lbinMMMVh AAA −=+=+= 068,122550795,11

2222

( ) ( ) ( ) ( ) lbinMMMVh BBB −=+=+= 872581603088

2222

lbinM D −= 9000

Therefore

lbinM −= 068,12max

( )333

176,386068,123232

ddd

Ms

πππ===

Maximum moment subjected to alternating tension and compression

0=ms

3

176,386

dssa

π==

SF

sKs

s

ss

af

m

y

ne +=

For AISI 1137, cold-drawn,

ksisy 93=

ksisu 103=

( ) ksiss un 5.511035.05.0 ===

For sled-runner keyway (Table AT 13)

6.1=fK

6.1=fsK

85.0=SF

( )( )33

386,231

85.0

176,38660.10

ddse =+=

π


Page 17 of 76

At A, 50 hp max. and 25 hp min.

50 hp

( )lbin

n

hpTA −=

+== 600,12

250

2525000,63000,63

( )lb

D

TF

A

AA 3150

8

600,1222===

( )333max

600,201600,121616

ddd

Tss

πππ===

25 hp

( )lbin

n

hpTA −=== 300,6

250

25000,63000,63

( )lb

D

TF

A

AA 1575

8

300,622===

( )333min

800,100300,61616

ddd

Tss

πππ===

( )33minmax

200,151800,100600,201

2

1

2

1

ddsss ssms

ππ=

+=+=

( )33minmax

400,50800,100600,201

2

1

2

1

ddsss ssas

ππ=

−=−=

SF

sKs

s

ss

asfs

ms

ys

nses +=

( )( )333

848,56

85.0

400,506.1200,151

806.1

1

dddses =+

=

ππ

2

122

577.0

1

+

=

n

es

n

e

s

s

s

s

N

( )

2

12

3

2

3 500,51577.0

848,56

500,51

386,231

2

1

+

=

dd

ind 14.2=

say ind16

32=

THRUST LOADS


Page 18 of 76

481. A cold-drawn monel propeller shaft for a launch is to transmit 400 hp at 1500

rpm without being subjected to a significant bending moment; and 40<kLe .

The efficiency of the propeller is 70 % at 30 knots (1.152 mph/knot). Consider

that the number of repetitions of the maximum power at the given speed is 2x

105. Let 2=N based on the maximum shear theory with varying stress.

Compute the shaft diameter.

Solution:

For cold-drawn monel shaft, Table AT 10

ksisy 75=

ksisn 42= at 108

at 2 x 105

ksisn 23.71102

1042

085.0

5

8

=

×≈

053.1

1

75

23.71===

ys

ns

y

n

s

s

s

s

( )lbin

n

hpT −=== 800,16

1500

400000,63000,63

( )333

800,268800,161616

DDD

Tss

πππ===

sms ss =

0=ass

SF

sKs

s

ss

asfs

ms

ys

nses +=

85.0=SF

assume 0.1== fsf KK

33

255,810

800,268

053.1

1

DDses =+

=

π

hpFvm η=

000,33

( )( )( )( ) fpmhrmiftknotmphknotsvm 3041min6015280152.130 ==

( ) ( )( )40070.0000,33

3041=

F

lbF 3040=

( )222

160,12304044

DDD

Fs

πππ===

ssm=

0=as


Page 19 of 76

SF

sKs

s

ss

af

m

y

ne +=

22

36760

160,12

053.1

1

DDse =+

=

π

Maximum Shear Theory

2

122

5.0

1

+

=

n

es

n

e

s

s

s

s

N

( )

2

12

3

2

2 230,715.0

255,81

230,71

3676

2

1

+

=

DD

2

12

3

2

2

2815.2

377.19

1

2

1

+

=

DD

By trial and error

ininD16

11166.1 ==

482. A shaft receives 300 hp while rotating at 600 rpm, through a pair of bevel gears,

and it delivers this power via a flexible coupling at the other end. The shaft is

designed with the average forces ( at the midpoint of the bevel-gear face); the

tangential driving force is F , lbG 580= , lbQ 926= ; which are the rectangular

components of the total reaction between the teeth; inDm 24= , inL 36= ,

ina 10= . Let the material be AISI C1045, cold drawn; 2=N . Considering

varying stresses and using the octahedral shear theory, determine the shaft

diameter.

Problems 482, 485, 486.

Solution:

For AISI C1045, cold drawn

ksisy 85=

ksisu 100=


Page 20 of 76

( ) ksiss un 501005.05.0 ===

85.0=SF

7.1

1

85

50===

ys

ns

y

n

s

s

s

s

( )lbin

hpT −=== 500,31

600

300000,63

600

000,63

( )333

000,504500,311616

DDD

Tss

πππ===

sms ss =

0=ass

SF

sKs

s

ss

asfs

ms

ys

nses +=

33

370,940

000,504

7.1

1

DDses =+

=

π

TD

F m =

2

500,312

24=

F

lbF 2625=

Vertical:

lbinD

Q m −=

=

112,11

2

24926

2

lbG 580=

[ ]0∑ =BM �

( ) ( ) 0102

36 =−+− GQD

A mv

( ) ( )36102

vm AG

QD+=

( ) ( )3610580112,11 vA+=

lbAv 148=


Page 21 of 76

[ ]0∑ =vF

lbBA vv 580=+

lbBv 580148 =+

lbBv 432=

Shear Diagram

Moment Diagram

lbinM

vC −= 112,11

lbinMvB −= 5328

Horizontal:

[ ]0∑ =BM

( ) ( )( )10262536 =hA

lbAh 729=

[ ]0∑ =hF

FAB hh +=

2625725 +=hB


Page 22 of 76

lbBh 3354=

Shear Diagram

0=

hCM

( )( ) lbinMhB −== 244,2672936

Maximum M

( ) ( ) ( ) ( ) lbinMMMMVh BBB −=+=+== 780,265328244,26

2222

( ) ( )232323max

704,3960,8569264780,2632432

DDDDD

Q

D

Ms

ππππππ+=+=+=

3232min

960,8563704324

DDD

M

D

Qs

ππππ−=−=

( )minmax2

1sssm +=

23223

3704960,85637043704960,856

2

1

DDDDDsm

πππππ=

−++=

( )minmax2

1sssa −=

3

960,856

Dsa

π=

SF

sKs

s

ss

af

m

y

ne +=

assume 0.1=fK at B

3232

916,320964960,856

85.0

0.13704

7.1

1

DDDDse +=

+

=

ππ

Octahedral Shear Theory

2

122

577.0

1

+

=

n

es

n

e

s

s

s

s

N


Page 23 of 76

( )

2

12

3

2

32

2

1

2

3

2

32 27.342.6

72

1

000,50577.0

370,94

000,50

916,320694

2

1

+

+=

+

+

=DDDD

DD

By trial and error, use

inD2

12=

483. The worm shown is to deliver 65.5 hp steadily at 1750 rpm. It will be integral

with the shaft if the shaft size needed permits, and its pitch diameter 3 in. The 12-

in. pulley receives the power from a horizontal belt in which the tight tension

21 5.2 FF = . The forces (in kips) on the worm are as shown, with the axial force

taken by bearing B. The strength reduction factor for the thread roots may be

taken as 5.1=fK , shear or bending. The shaft is machined from AISI 1045, as

rolled. (a) For 2.2=N (Soderberg criterion) by the octahedral-shear theory,

compute the required minimum diameter at the root of the worm thread (a first

approximation). (b) What should be the diameter of the shaft 2.5 in. to the left of

the centerline of the worm? (c) Select a shaft size D and check it at the pulley A.

Problem 483.

Solution:


ksisy 59=

ksisu 96=

ksiss un 485.0 ==

229.1

1

59

48===

ys

ns

y

n

s

s

s

s

( )lbin

hpT −=== 2358

1750

5.65000,63

1750

000,63

( ) TFF =

−

2

1221

( )( ) 235865.2 22 =− FF

lbF 2622 =

lbFF 6555.2 21 ==


Page 24 of 76

lbFFFA 91726265521 =+=+=

Horizontal

[ ]0∑ =BM

( )( ) ( )( ) hE135.615706917 =+

lbEh 1208=

[ ]0∑ =hF

1570917 +=+ hh BE

15701208917 +=+ hB

lbBh 555=

Shear Diagram

0=

hAM

( )( ) lbinMhB −== 55026917

( )( ) lbinMhC −== 78525.61208

Vertical:

( ) lbinM −=

=′ 3810

2

32540

[ ]0∑ =EM

( )( ) vBM 135.61170 =+′

( )( ) vB135.611703810 =+

lbBv 878=


Page 25 of 76

[ ]0∑ =vF

1170=+ vv BE

1170878 =+vE

lbEv 292=

Shear Diagram

Moment Diagram

0=

vAM

0=vBM

lbinMvC −= 5707

( ) ( )22

vh MMM +=

( ) ( ) 00022

=+=AM

( ) ( ) lbinM B −=+= 55020550222

( ) ( ) lbinMC −=+= 97075707785222

(a) Minimum diameter at the root of the warm thread.

5.1== fsf KK

lbinMM C −== 9707

lbF 2540=

( ) ( )232323max

160,10624,31025404970732432

rrrrrr DDDDD

F

D

Ms

ππππππ+=+=+=


Page 26 of 76

23min

160,10624,310

rr DDs

ππ+−=

( )minmax2

1sssm +=

2

160,10

r

mD

sπ

=

( )minmax2

1sssa −=

3

624,310

r

aD

sπ

=

SF

sKs

s

ss

af

m

y

ne +=

3232

485,1742632624,310

85.0

5.1160,10

229.1

1

rrrr

eDDDD

s +=

+

=

ππ

( )333

000,1223581616

rrr

sDDD

Ts ===

ππ

sms ss =

0=ass

SF

sKs

s

ss

asfs

ms

ys

nses +=

33

97640

000,12

229.1

1

rr

esDD

s =+

=

2.2=N , Octahedral shear theory

2

122

577.0

1

+

=

n

es

n

e

s

s

s

s

N

( )

2

12

3

2

32

2

1

2

3

2

32

84.2

1635.3

24.18

1

000,48577.0

9764

000,48

485,1742632

2.2

1

+

+=

+

+

=rrrr

rr

DDDD

DD

By trial and error

inDr 023.2=

say inDr16

12=

(b) D – shaft diameter 2.5 in. to the left of the center line of worm


Page 27 of 76

inr16

3=

Figure AF 12

1.0

16

322023.2

16

3

≈

−

=d

r

2.1

16

322023.2

023.2=

−

=d

D

65.1== tf KK

34.1== tsfs KK

at 2.5 in to the shaft

( )( ) ( )( ) lbinM h −=−+= 69505.25.63626917

( )( ) lbinM v −=−= 35125.25.6878

( ) ( ) lbinM −=+= 77873512695022

2

160,10

Dsm

π=

( )333

184,24977873232

DDD

Msa

πππ===

SF

sKs

s

ss

af

m

y

ne +=

3232

970,1532632184,249

85.0

65.1160,10

229.1

1

DDDDse +=

+

=

ππ

3

9764

Dses =

2

122

577.0

1

+

=

n

es

n

e

s

s

s

s

N

( )

2

12

3

2

32

2

1

2

3

2

32

84.2

121.3

24.18

1

000,48577.0

9764

000,48

970,1532632

2.2

1

+

+=

+

+

=DDDD

DD

By trial and error

inD 9432.1=

say inD16

151=


Page 28 of 76

(c) Selecting ininD 9375.116

151 ==

At the pulley A, or 3 in. right of centerline

( )( ) lbinM h −== 27513917

0=vM

lbinM −= 2751

For sled runner keyway

6.1=fK

6.1=fsK

0=ms

( )( )

psiD

Msa 3853

9375.1

2751323233

===ππ

SF

sKs

s

ss

af

m

y

ne +=

( ) psise 7253385385.0

6.10 =

+=

( )psises 1343

9375.1

97643

==

2

122

577.0

1

+

=

n

es

n

e

s

s

s

s

N

( )

2

122

000,48577.0

1343

000,48

72531

+

=

N

2.230.6 >=N , therefore o.k.

484. A propeller shaft as shown is to receive 300 hp at 315 rpm from the right through

a flexible coupling. A 16-in. pulley is used to drive an auxiliary, taking 25 hp.

The belt pull BF is vertically upward. The remainder of the power is delivered to

a propeller that is expected to convert 60% of it into work driving the boat, at

which time the boat speed is 1500 fpm. The thrust is to be taken by the right-hand

bearing. Let 2=N ; material cold-worked stainless 410. Use the octahedral shear

theory with varying stresses. (a) Determine the shaft size needed assuming no

buckling. (b) Compute the equivalent column stress. Is this different enough to

call for another shaft size? Compute N by the maximum shear stress theory,

from both equations (8.4) and (8.11).


Page 29 of 76

Problem 484.

Solution:

For stainless 410, cold-worked

ksisy 85=

ksisn 53=

85.0=SF

Belt drive

( )lbin

n

hpTB −=== 5000

315

25000,63000,63

( ) ( )lb

D

TFFFFF

B

BB 1250

16

50004222 2121 ==

=−=+=

Propeller

( )lbin

n

hpTP −=

−== 000,55

315

25300000,63000,63

Thrust

( )000,33hpFvm η=

( ) ( )( )( )000,332530060.01500 −=F

lbF 3630=

Vertical loading

[ ]0∑ =EM

( )( ) C60125020 =

lbC 417=

[ ]0∑ =vF

BFCA =+

1250417 =+A


Page 30 of 76

lbA 833=

Shear Diagram

( )( ) lbinM B −== 660,1683320

Maximum T at B

lbinTTT PB −=+= 000,60

(a) Shaft size assuming no buckling

lbinM −= 660,16

lbF 3630=

( )222

520,14363044

DDD

Fsm

πππ===

( )333

120,533660,163232

DDD

Msa

πππ===

For sled-runner keyway

6.1=fK

6.1=fsK

604.1

1

85

53===

ys

ns

y

n

s

s

s

s

SF

sKs

s

ss

af

m

y

ne +=

3232

430,3192882120,533

85.0

6.1520,14

604.1

1

DDDDse +=

+

=

ππ

( )333

000,960000,601616

DDDss mss

πππ====

0=ass

SF

sKs

s

ss

asfs

ms

ys

nses +=

32

510,1900

000,960

604.1

1

DDses =+

=

π


Page 31 of 76

2

122

1

+

=

ns

es

n

e

s

s

s

s

N

2=N , Octahedral Shear Theory, nns ss 577.0=

2

122

577.0

1

+

=

n

es

n

e

s

s

s

s

N

( )

2

12

3

2

32

2

1

2

3

2

32 230.6027.6

39.18

1

000,53577.0

510,190

000,53

430,3192882

2

1

+

+=

+

+

=DDDD

DD

By trial and error

inD 6.2=

say ininD 625.28

52 ==

(b) Equivalent Column Stress

απ 2

4

D

Fs =

inLe 82106012 =++=

( ) inDk 65625.0625.24

1

4

1===

12012565625.0

82>==

k

Le

Use Euler’s equation

( )( )

486.41030

1258532

2

2

2

=×

=

=ππ

αE

k

Ls e

y

( )( )

( ) psiD

Fs 3000486.4

625.2

36304422

===π

απ

Since 1>α , it is different enough to call for another shaft size.

Solving for N by maximum shear theory.

( ) ( )psi

DDse 078,18

625.2

430,319

625.2

2882430,31928823232

=+=+=

( )psises 533,10

625.2

510,1903

==

Equation (8.4)


Page 32 of 76

( ) psis

ss 880,132

078,18533,10

2

2

12

22

12

2 =

+=

+=τ

( )91.1

880,13

000,535.05.0===

τns

N

Equation (8.11) nns ss 5.0=

( )

2

1222

122

000,535.0

533,10

000,53

078,181

+

=

+

=

ns

s

n s

s

s

s

N

91.1=N

CHECK PROBLEMS

485. A 3-in. rotating shaft somewhat as shown (482) carries a bevel gear whose mean

diameter is inDm 10= and which is keyed (profile) to the left end. Acting on the

gear are a radial force lbG 8.1570= , a driving force lbQ 6.3141= . The thrust

force is taken by the right-hand bearing. Let ina 5= and inL 15= ; material,

AISI C1040, annealed. Base calculations on the maximum shearing stress theory

with variable stress. Compute the indicated design factor N . With the use of a

sketch, indicate the exact point of which maximum normal stress occurs.

Solution:

For AISI C1040, annealed, Figure AF 1

ksisy 48=

ksisu 80=

ksiss un 405.0 ==

2.1

1

48

40===

ys

ns

y

n

s

s

s

s

( )( )lbin

FDT m −=== 416,31

2

102.6283

2

( )( )

psiD

Tss 5926

3

416,31161633

===ππ

sms ss =

0=ass

SF

sKs

s

ss

asfs

ms

ys

nses +=

( ) psises 4940059262.1

1=+

=

Vertical


Page 33 of 76

( )( )

lbinQDm −== 708,15

2

106.3141

2

[ ]0∑ =EM

Vm AG

QD155

2+=

( ) VA158.15705708,15 +=

lbAV 6.523=

[ ]0∑ =vF

GBA VV =+

8.15706.523 =+ VB

lbBV 2.1047=

Shear Diagram

Moment Diagram

lbinMVC −= 708,15

lbinMVB −= 7854


Page 34 of 76

Horizontal

[ ]0∑ =BM

( )2.6283515 =hA

lbAh 4.2094=

[ ]0∑ =hF

FAB hh +=

2.62834.2094 +=hB

lbBh 6.8377=

Shear Diagram

0=

hCM

( )( ) lbinMhB −== 416,314.209415

Maximum Moment

( ) ( ) lbinMMM BvBh−=+=+= 383,327854416,31

2222

Since thrust force is taken by the right-hand bearing

0=mss

( )( )

psiD

Msas 217,12

3

383,32323233

===ππ

SF

sKs

s

ss

af

m

y

ne +=

Assume 0.1=fK at the bearing B

( ) psise 373,14217,1285.0

0.10 =

+=


Page 35 of 76

Maximum shear theory nns ss 5.0=

2

122

5.0

1

+

=

n

es

n

e

s

s

s

s

N

( )

2

122

000,405.0

4940

000,40

373,141

+

=

N

3.2=N

Location of maximum normal stress

487. A 2 7/16-in. countershaft in a machine shop transmits 52 hp at 315 rpm. It is

made of AISI 1117, as rolled, and supported upon bearing A and B, 59-in. apart.

Pulley C receives the power via a horizontal belt, and pulley D delivers it

vertically downward, as shown. Calculate N based on the octahedral-shear-

stress theory considering varying stresses.

Problem 487, 488

Solution:


ksisy 3.44=


Page 36 of 76

ksisu 6.70=

ksiss un 3.355.0 ==

255.1

1

3.44

3.35===

ys

ns

y

n

s

s

s

s

85.0=SF

( )lbinT −== 400,10

315

52000,63

Pulley C

( ) ( )lb

D

TFFFFF

C

C 231118

400,104222 1221 ==

=−=+=

Pulley D

( ) ( )lb

D

TFFFFF

D

D 166425

400,104222 1221 ==

=−=+=

Horizontal

[ ]0∑ =AM

( ) hB59231115 =

lbBh 588=

[ ]0∑ =hF

2311=+ hh BA

2311588 =+hA

lbAh 1723=

Shear Diagram

( )( ) lbinM

hC −== 845,25151723


Page 37 of 76

( )( ) ( )( ) lbinMhD −=−= 557,1026588151723

Vertical

[ ]0∑ =BM

( ) vA59166418 =

lbAv 508=

[ ]0∑ =vF

1664=+ vv BA

1664508 =+ vB

lbBv 1156=

Shear Diagram

( )( ) lbinMvC −== 762015508

( )( ) lbinMvD −== 808,20181156

( ) ( ) lbinMMMvh CCC −=+=+= 945,267620845,25

2222

( ) ( ) lbinMMMvh DDD −=+=+= 333,23808,20557,10

2222

Maximum M at C

lbinMM C −== 945,26

0=ms

3

32

D

Msa

π=

ininD 4375.216

72 ==


Page 38 of 76

( )( )

psisa 952,184375.2

945,26323

==π

assume 0.1== fsf KK

SF

sKs

s

ss

af

m

y

ne +=

( ) ( )( )psise 300,22

85.0

952,180.10

255.1

1=+

=

( )( )

psiD

Tss 3658

4375.2

400,10161633

===ππ

psiss sms 3658==

0=ass

SF

sKs

s

ss

asfs

ms

ys

nses +=

( ) psises 291503658255.1

1=+

=

Octahedral shear theory nns ss 577.0=

2

122

577.0

1

+

=

n

es

n

e

s

s

s

s

N

( )

2

122

300,35577.0

2915

300,35

300,221

+

=

N

544.1=N

489. A shaft for a general-purpose gear-reduction unit supports two gears as shown.

The 5.75-in. gear B receives 7 hp at 250 rpm. The 2.25-in. gear A delivers the

power, with the forces on the shaft acting as shown; the gear teeth have a

pressure angle of

o

2

114=φ (

v

h

v

h

B

B

A

A==φtan ). Both gears are keyed (profile) to

the shaft of AISI 1141, cold rolled. (a) If the fillet radius is 1/8 in. at bearing D,

where the diameter is 1 3/8 in., compute N based on the octahedral-shear-stress

theory (Soderberg line). The shaft diameter at A is 1 11/16 in. What is N here?


Page 39 of 76

Problem 489, 490

Solution:

For AISI 1141, cold rolled

ksisy 90=

ksisn 50=

8.1

1

90

50===

ys

ns

y

n

s

s

s

s

85.0=SF

( )lbinT −== 1764

250

7000,63

3

16

D

Tsms

π=

0=ass

Gear B:

lbinTBv −==

1764

2

75.5

lbBv 614=

lbBB vh 1595.14tan614tan === φ

Gear A:

lbinTAv −==

1764

2

25.2

lbAv 1568=

lbAA vh 4065.14tan1568tan === φ

Vertical


Page 40 of 76

[ ]0∑ =DM

( ) ( )6143156848 −=vC

lbCv 554=

[ ]0∑ =vF

vvvv BADC +=+

6141568554 +=+ vD

lbDv 1628=

Shear Diagram

( )( ) lbinMvA −== 22164554

( )( ) lbinMvD −== 18423614

Horizontal

[ ]0∑ =CM

( ) ( )1591184064 =+ hD

lbDh 16=

[ ]0∑ =hF

hhhh DABC +=+

16406159 +=+hC

lbCh 263=

Shear Diagram


Page 41 of 76

( )( ) lbinMhA −== 10524263

( )( ) lbinMhD −== 4773159

( ) ( ) lbinMMMvh AAA −=+=+= 245322161052

2222

( ) ( ) lbinMMMvh DDD −=+=+= 19031842477

2222

(a) At bearing D

inr8

1=

ind8

31=

10.0375.1

125.0≈=

d

r

2.1375.1

25.0375.1≈

+=

d

D

6.1=≈ ft KK

34.1=≈ fsts KK

DMM =

0=ms

( )( )

psid

Msa 7456

375.1

1903323233

===ππ

SF

sKs

s

ss

af

m

y

ne +=

( )( )psise 035,14

85.0

74566.10 =+=

( )( )

psiD

Tsms 3456

375.1

1764161633

===ππ

0=ass

SF

sKs

s

ss

asfs

ms

ys

nses +=

( ) psises 1920034568.1

1=+

=

Octahedral shear theory

2

122

577.0

1

+

=

n

es

n

e

s

s

s

s

N


Page 42 of 76

( )

2

122

000,50577.0

3456

000,50

035,141

+

=

N

28.3=N

(b) At A

For profile keyway

0.2=fK , 6.1=fsK

inind 6875.116

111 ==

lbinMM A −== 2453

0=ms

( )( )

psid

Msa 5200

6875.1

2453323233

===ππ

SF

sKs

s

ss

af

m

y

ne +=

( )( )psise 235,12

85.0

52000.20 =+=

( )( )

psiD

Tsms 1870

6875.1

1764161633

===ππ

0=ass

SF

sKs

s

ss

asfs

ms

ys

nses +=

( ) psises 1040018708.1

1=+

=


2

122

577.0

1

+

=

n

es

n

e

s

s

s

s

N

( )

2

122

000,50577.0

1040

000,50

235,121

+

=

N

043.4=N

THRUST LOADS

491. The high-speed shaft of a worm-gear speed reducer, made of carburized AISI

8620, SOQT 450 F, is subjected to a torque of 21,400 in-lb. Applied to the right


Page 43 of 76

end with no bending. The force on the worm has three components: a horizontal

force opposing rotation of lbW 6180= , a vertical radial force lbS 1940= , and a

rightward thrust of lbF 6580= . The shaft has the following dimensions: 6=a ,

8

74=b , 10=c ,

16

94=d ,

4

32=e ,

16

913=f , 646.11=g , 370.10=h ,

740.31 =D , 16

1342 =D , 43 =D , 3469.34 =D , 253.35 =D , 098.01 =r ,

4

332 == rr , 098.04 =r ,

16

15 =r , all in inches. The pitch diameter of the worm,

6.923 in., is the effective diameter for the point of application of the forces. The

root diameter, 5.701 in. is used for stress calculations. The left-hand bearing

takes the thrust load. Calculate N based on the octahedral-shear-stress theory

with varying stresses. (Data courtesy of Cleveland Worm and Gear Company.)

Problem 491

Solution:

Table AT 11n For AISI 8620, SOQT 450 F

ksisy 120=

ksisu 167=

ksiss un 5.835.0 ==

437.1

1

120

5.83===

ys

ns

y

n

s

s

s

s

85.0=SF

lbinT −= 400,21

Vertical


Page 44 of 76

lbinFM −=

=

=′ 777,22

2

923.66580

2

923.6

[ ]0∑ =AM

( )( ) ( ) vG370.10646.111940646.11777,22 +=+

lbGv 2061=

[ ]0∑ =vF

vv GAS =+

20611940 =+ vA

lbAv 121=

Shear Diagram

Moment Diagram

0=

vAM

( )( ) lbinMvB −−=−= 1462035.1121

( )( ) lbinMvC −−=+−= 736875.42035.1121

( )( ) lbinMvD −−=++−= 14095675.5875.42035.1121 at left side

lbinMMvD −=+−=′+−= 368,21777,2214091409 at right side

( )( ) lbinMvE −=−= 233,124325.42061368,21

( )( ) lbinMvF −=−= 28305625.42061233,12

( )( ) 0375.120612830 =−=vGM

Horizontal


Page 45 of 76

[ ]0∑ =AM

( )( ) ( ) hG370.10646.116180646.11 +=

lbGh 3269=

[ ]0∑ =hF

WGA vh =+

61803269 =+hA

lbAh 2911=

Shear Diagram

Moment Diagram

0=

hAM

( )( ) lbinMhB −== 35002035.12911

( )( ) lbinMhC −=+= 695,17875.42035.12911

lbinMhD −= 900,33

( )( ) lbinMhE −=−= 410,194325.43269900,33

( )( ) lbinMhF −=−= 44955625.43269410,19

( )( ) 0375.132694495 =−=hFM

Combined

22

vh MMM +=

( ) ( ) lbinM A −=+= 00022

( ) ( ) lbinM B −=+= 3503146350022

( ) ( ) lbinMC −=+= 710,17736695,1722

( ) ( ) lbinM D −=+= 930,331409900,3322

(left)

( ) ( ) lbinM D −=+= 073,40368,21900,3322

(right)

( ) ( ) lbinM E −=+= 944,22233,12410,1922


Page 46 of 76

( ) ( ) lbinM F −=+= 53124495283022

( ) ( ) lbinMG −=+= 00022

Bending stresses (Maximum)

At A, 0=As

At B, ( )

( )psi

D

Ms B

B 682740.3

3503323233

1

===ππ

At C, ( )

( )psi

D

Ms C

C 16188125.4

710,17323233

2

===ππ

At D, ( )( )

psiD

Ms

r

DD 2203

701.5

073,40323233

===ππ

At E, ( )

( )psi

D

Ms E

E 36524

944,22323233

3

===ππ

At F, ( )

( )psi

D

Ms F

F 14433469.3

5312323233

4

===ππ

At G, 0=Gs

Shear Stresses:

( )( )

psiD

Tss sBsA 2083

740.3

400,21161633

1

====ππ

( )( )

psiD

TssC 978

8125.4

400,21161633

2

===ππ

( )( )

psiD

Ts

r

sD 588701.5

400,21161633

===ππ

( )( )

psiD

TssE 1703

4

400,21161633

3

===ππ

( )( )

psiD

Tss sGsF 2907

3469.3

400,21161633

4

====ππ

Tensile stresses: lbF 6580=

( )( )

psiD

Fss BA 599

740.3

65804422

1

===′=′ππ

( )( )

psiD

FsC 362

8125.4

65804422

2

===′ππ

( )( )

psiD

Fs

r

D 258701.5

65804422

===′ππ

( )( )

psiD

FsE 524

4

65804422

3

===′ππ


Page 47 of 76

( )( )

psiD

Fss FE 748

3469.3

65804422

4

===′=′ππ

At B: 03.0740.3

098.0

1

1 ==D

r

3.1740.3

8125.4

1

2 ==D

D

Figure AF 12

3.2=≈ tf KK

7.1=≈ tsfs KK

SF

sKs

s

ss

af

m

y

ne +=

psiss Bm 599=′=

psiss Ba 682==

( ) ( )( )psise 2262

85.0

6823.2599

437.1

1=+

=

SF

sKs

s

ss

asfs

ms

ys

nses +=

psiss sBms 2083==

0=ass

( ) psises 145002083437.1

1=+

=


2

122

577.0

1

+

=

n

es

n

e

s

s

s

s

N

( )

2

122

500,83577.0

1450

500,83

22621

+

=

N

7.24=N

At C: 16.08125.4

75.0

2

2 ==D

r

2.18125.4

701.5

2

==D

Dr

Figure AF 12

5.1=≈ tf KK

2.1=≈ tsfs KK


Page 48 of 76

SF

sKs

s

ss

af

m

y

ne +=

psism 362=

psisa 1618=

( ) ( )( )psise 3107

85.0

16185.1362

437.1

1=+

=

SF

sKs

s

ss

asfs

ms

ys

nses +=

psiss sCms 978==

0=ass

( ) psises 6810978437.1

1=+

=


2

122

577.0

1

+

=

n

es

n

e

s

s

s

s

N

( )

2

122

500,83577.0

681

500,83

31071

+

=

N

1.25=N

At D:

Assume 5.1=fK as in Prob. 483

SF

sKs

s

ss

af

m

y

ne +=

psism 258=

psisa 2203=

( ) ( )( )psise 4067

85.0

22035.1258

437.1

1=+

=

SF

sKs

s

ss

asfs

ms

ys

nses +=

psiss sDms 588==

0=ass

( ) psises 4090588437.1

1=+

=



Page 49 of 76

2

122

577.0

1

+

=

n

es

n

e

s

s

s

s

N

( )

2

122

500,83577.0

409

500,83

40671

+

=

N

2.20=N

At E: 19.04

75.0

3

3 ==D

r

43.14

701.5

3

==D

Dr

Figure AF 12

45.1=≈ tf KK

25.1=≈ tsfs KK

SF

sKs

s

ss

af

m

y

ne +=

psiss Em 524=′=

psiss Ea 3652==

( ) ( )( )psise 6595

85.0

365245.1524

437.1

1=+

=

SF

sKs

s

ss

asfs

ms

ys

nses +=

psiss sEms 1703==

0=ass

( ) psises 118501703437.1

1=+

=


2

122

577.0

1

+

=

n

es

n

e

s

s

s

s

N

( )

2

122

500,83577.0

1185

500,83

65951

+

=

N

12=N

At F: 03.03469.3

098.0

4

4 ==D

r


Page 50 of 76

2.13469.3

4

4

3 ==D

D

Figure AF 12

3.2=≈ tf KK

7.1=≈ tsfs KK

SF

sKs

s

ss

af

m

y

ne +=

psiss Fm 748=′=

psiss Fa 1443==

( ) ( )( )psise 4425

85.0

14433.2748

437.1

1=+

=

SF

sKs

s

ss

asfs

ms

ys

nses +=

psiss sFms 2907==

0=ass

( ) psises 202302907437.1

1=+

=


2

122

577.0

1

+

=

n

es

n

e

s

s

s

s

N

( )

2

122

500,83577.0

2023

500,83

44251

+

=

N

8.14=N

Then 12=N at inr4

33 = (E)

492. The slow-speed shaft of a speed reducer shown, made of AISI 4140, OQT 1200

F, transmits 100 hp at a speed of 388 rpm. It receives power through a 13.6 in.

gear B. The force on this gear has three components: a horizontal tangential

driving force lbFt 2390= , a vertical radial force lbS 870= , and a thrust force

lbQ 598= taken by the right-hand bearing. The power is delivered to a belt at

F that exerts a downward vertical force of 1620 lb.; sled runner keyways. Use

the octahedral shear theory with the Soderberg line and compute N at sections C

and D. (Data courtesy of Twin Disc Clutch Company.)


Page 51 of 76

Problem 492, 493

Solution:


ksisy 83=

ksisu 112=

ksiss un 565.0 ==

482.1

1

83

56===

ys

ns

y

n

s

s

s

s

85.0=SF

( )lbinT −== 237,16

388

100000,63

Vertical

( ) lbinQM −=

=

=′ 4.4066

2

6.13598

2

6.13

[ ]0∑ =AM


Page 52 of 76

( ) ( )

vG

+++=

+

+++++++

+

32

71

8

33

8

51

16

31

4.406616204

32

16

13

32

111

32

71

8

33

8

51

16

31870

8

51

16

31

lbGv 3573=

[ ]0∑ =vF

vv GFSA =++

35731620870 =++vA

lbAv 1083=

Shear Diagram

Moment Diagram

0=

vAM

( ) lbinMvP −−=

−= 1286

16

311083

( ) lbinMvB −−=

−+−= 3046

8

5110831286 at the left

lbinMvB −=+−= 10214.40663046 at the right

( ) lbinMvC −−=

−= 5570

8

3319531021

( ) lbinMvG −−=

−−= 7950

32

7119535570


Page 53 of 76

( ) lbinMvD −−=

+−= 5773

32

11116207950

( ) lbinMvE −−=

+−= 4457

16

1316205773

( ) lbinMvF −=

+−= 0

4

3216204457

Horizontal

[ ]0∑ =AM

( ) hG

++

32

194

16

1322390

16

132

lbGh 908=

[ ]0∑ =hF

thh FGA =+

2390908 =+hA

lbAh 1482=

Shear Diagram

0=

hAM

( ) lbinMhP −=

= 1760

16

311482

( ) lbinMhB −=

+= 4168

8

5114821760


Page 54 of 76

( ) lbinMhC −=

−= 1104

8

339084168

( ) lbinMhC −=

−= 0

32

719081104

lbinMhD −= 0

lbinMhE −= 0

lbinMhF −= 0

Combined

22

vh MMM +=

lbinM A −= 0

( ) ( ) lbinM P −=+= 21801286176022

( ) ( ) lbinM B −=+= 51633046416822

( ) ( ) lbinMC −=+= 56785570110422

( ) ( ) lbinM D −=+= 57735773022

( ) ( ) lbinM E −=+= 44574457022

( ) ( ) lbinM F −=+= 00022

at C: ininr 125.08

1==

ind 750.2=

inD 953.2=

05.0750.2

125.0==

d

r

10.1750.2

953.2==

d

D

Figure AF 12

9.11 =≈ tf KK

3.11 =≈ tsfs KK

For sled runner keyway

6.12 =fK

6.12 =fsK

( )( ) 4.26.19.18.08.0 21 === fff KKK

( )( ) 7.16.13.18.08.0 21 === fsfsfs KKK


Page 55 of 76

SF

sKs

s

ss

af

m

y

ne +=

( )( )

psid

Qsm 101

750.2

5984422

===ππ

( )( )

psid

Ms C

a 2781750.2

5678323233

===ππ

( ) ( )( )psise 7920

85.0

27814.2101

482.1

1=+

=

SF

sKs

s

ss

asfs

ms

ys

nses +=

( )( )

psid

Tsms 3976

750.2

237,16161633

===ππ

0=ass

( ) psises 268303976482.1

1=+

=


2

122

577.0

1

+

=

n

es

n

e

s

s

s

s

N

( )

2

122

000,56577.0

2683

000,56

79201

+

=

N

6=N

at D: ininr 0625.016

1==

ind 953.2=

ininD 375.38

33 ==

02.0953.2

0625.0==

d

r

14.1953.2

375.3==

d

D

Figure AF 12

4.2=≈ tf KK

6.1=≈ tsfs KK

SF

sKs

s

ss

af

m

y

ne +=


Page 56 of 76

( )( )

psid

Qsm 3.87

953.2

5984422

===ππ

( )( )

psid

Ms C

a 2284953.2

5773323233

===ππ

( ) ( )( )psise 6508

85.0

22844.23.87

482.1

1=+

=

SF

sKs

s

ss

asfs

ms

ys

nses +=

( )( )

psid

Tsms 3211

953.2

237,16161633

===ππ

0=ass

( ) psises 216703211482.1

1=+

=


2

122

577.0

1

+

=

n

es

n

e

s

s

s

s

N

( )

2

122

000,56577.0

2167

000,56

65081

+

=

N

5.7=N

TRANSVERSE DEFLECTIONS

494. The forces on the 2-in. steel shaft shown are kipsA 2= , kipsC 4= . Determine

the maximum deflection and the shaft’s slope at D.

Problems 494-496

Solution:


Page 57 of 76

[ ]0=BM

( ) ( )15425102 =+ D

kipsD 6.1=

[ ]0=vF

DBCA +=+

6.142 +=+ B

kipsB 4.4=

Shear Diagram

Moment Diagram

4

64

DE

M

EI

M

π=

A B C D

( )kipinM − 0 -20 16 0

44 10DEI

M

0 -135.8 108.6 0


Page 58 of 76

Scale ininss 10=

EI

M, Scale inper

Ds

EIM 4

410200 −×=

Slope θ , Scale inradDs

42.0=θ

y deflection, Scale ininDsy

40.2=

Deflection:

At A: inD

yA 4

625.0=


Page 59 of 76

At C: inD

yC 4

375.0=

Slope:

At A: radD

4

075.0=θ

At B: radD

4

0125.0=θ

At D: radD

4

05625.0=θ

Maximum deflection:

( )inyy A 04.0

2

625.04

===

Shaft’s slope at D

( )rad0035.0

2

05625.04

==θ

495. The forces on the steel shaft shown are kipsA 2= , kipsC 4= . Determine the

constant shaft diameter that corresponds to a maximum deflection of 0.006 in. at

section C.

Solution:

(see Problem 494)

006.0375.0

4==

DyC

inD 812.2=

say inD8

72=

496. The forces on the steel shaft shown are kipsA 2= , kipsC 4= . Determine a

constant shaft diameter that would limit the maximum deflection at section A to

0.003 in.

Solution:

(see Problem 494)

003.0625.0

4==

DyA

inD 80.3=

say inD8

73=

497. A steel shaft is loaded as shown and supported in bearings at 1R and 2R .

Determine (a) the slopes at the bearings and (b) the maximum deflection.


Page 60 of 76

Problem 497

Solution:

[ ]∑ = 01RM

( ) ( ) 28

72

4

12

8

71

4

12

8

72100

8

11

8

73000 R

+++=

++−

+

lbR 4442 −=

[ ]∑ = 0F

3000210021 =++ RR

300021004441 =+−R

lbR 13441 =

Loading

Shear Diagram

Moment Diagram


Page 61 of 76

lbinM A −= 0

( ) lbinM B −=

= 1176

8

71344

( ) lbinMC −=

+= 2688

8

11

8

71134

( ) lbinM D −=

−= 825

8

1116562688

( )( ) lbinM E −−=−= 83111656825

( )( ) lbinM F −−=+−= 3871444831

( ) lbinMG −=

+−= 0

8

7444387

A B1 B2 C D1 D2 E F1 F2 G

( )kipsinM − 0 1.18 1.18 2.69 0.83 0.83 -0.83 -0.39 -0.39 0

( )inD 1 ½ 1 ½ 2 2 2 1 ¾ 1 ¾ 1 ¾ 1 ½ 1 ½

( )( )410EI

M 0 1.58 0.50 1.14 0.35 0.60 -0.60 -0.28 -0.52 0

Scale ininss 2=


Page 62 of 76

EI

M, Scale inper

Ds

EIM 4

4102 −×=


44104 −×=θ


44108 −×=

(a) Slopes at the bearings

at 1R , ( ) radA

44 105.1104375.0 −− ×=×=θ

at 2R , radG 0=θ

(b) Maximum deflection

at C, ( ) inyC

44 105.11081875.0 −− ×=×=


Page 63 of 76

498. (a) Determine the diameter of the steel shaft shown if the maximum deflection is

to be 0.01 in.; kipsC 5.1= , kipsA 58.1= , inL 24= . (b) What is the slope of the

shaft at bearing D? See 479.

Problems 498, 505, 506.

Solution:

Vertical

[ ]∑ = 0DM

( ) ( ) vE24424.085.16 =+

kipEv 516.0=

[ ]∑ = 0vF

vv ED +=+ 5.1424.0

516.05.1424.0 +=+vD

kipDv 592.1=

Shear Diagram

0=CM ; ( ) kipsinM D −−=−= 95.16

( ) kipsinM B −−=+−= 264.8092.089


Page 64 of 76

( ) kipsinM A −−=+−= 588.2516.011264.8

( ) 0516.05588.2 =+−=EM

C D B A E

( )kipsinM − 0 -9 -8.264 -2.588 0

( ) 44 10×DEI

M 0 -61.1 -56.1 -17.6 0

Scale ininss 8=

EI

M, Scale inper

Ds

EIM 4

410120 −×=

Slope θ , Scale inradDs4096.0=θ


Page 65 of 76


4768.0=

Deflections.

inD

yvC 4

384.0=

inD

yvB 4

288.0=

inD

yvA 4

168.0=

Slope

radDvD 4

057.0=θ

Horizontal

[ ]∑ = 0DM

( ) ( )58.11924424.08 =+ hE

kipEh 1095.1=

[ ]∑ = 0hF

58.1424.0 =++ hh ED

58.1424.01095.1 =++hD


Page 66 of 76

kipDh 0465.0=

Shear Diagram

Moments

0=CM

0=DM

( ) kipinM B −−=−= 372.00465.08

( ) kipsinM A −−=−+−= 5475.54705.011372.0

( ) 01095.155475.5 =+−=EM

C D B A E

( )kipsinM − 0 0 -0.372 -5.5475 0

( ) 44 10×DEI

M 0 0 -2.53 -37.7 0

Scale ininss 8=

EI

M, Scale inper

Ds

EIM 4

4104 −×=


4032.0=θ


Page 67 of 76


4256.0=

Deflections.

inD

yhC 4

064.0=

inD

yhB 4

072.0=

inD

yhA 4

096.0=

Slope

radDhD 4

012.0=θ

Resultant deflection:

( )2

122

vh yyy +=

( ) ( )[ ]44

2

122

390.0384.0064.0

DDyC =

+=

( ) ( )[ ]44

2

122

297.0288.0072.0

DDyB =

+=

( ) ( )[ ]44

2

122

194.0168.0096.0

DDyA =

+=

Slope:


Page 68 of 76

( )2

122

vh θθθ +=

( ) ( )[ ]rad

DDD 44

2

122

05823.0057.0012.0=

+=θ

(a) Diameter D.

Maximum deflection = inD

yC 01.0390.0

4==

inD 50.2=

(b) slope of the shaft at bearing D

( )rad

DD 0015.0

5.2

05823.005823.044

===θ

CRITICAL SPEED

499. A small, high-speed turbine has a single disk, weighing 0.85 lb., mounted at the

midpoint of a 0.178-in. shaft, whose length between bearings is 6 ½ in. What is

the critical speed if the shaft is considered as simply supported?

Solution:

Table AT 2

( )( )

( ) ( )in

EI

WLy 052634.0

64

178.010303

5.685.0

3 4

6

33

=

×

==π

( )rpm

y

g

Wy

Wygn oo

c 818052634.0

386303030 2

1

2

12

1

2=

=

=

=

∑∑

πππ

500. The bearings on a 1 ½-in. shaft are 30 in. apart. On the shaft are three 300-lb

disks, symmetrically placed 7.5 in. apart. What is the critical speed of the shaft?

Solution:


Page 69 of 76

Table AT 2

Deflection of B.

321 BBBB yyyy ++=

( )( )( ) ( ) ( ) ( )[ ]( ) ( ) ( )

inyB 01273.0

3064

5.110306

5.75.22305.75.223004

6

222

1=

×

−−=

π

( )( )( ) ( ) ( ) ( )[ ]( ) ( ) ( )

inyB 01556.0

3064

5.110306

5.715305.7153004

6

222

2=

×

−−=

π

( )( )( ) ( ) ( ) ( )[ ]( ) ( ) ( )

inyB 00990.0

3064

5.110306

5.75.7305.75.73004

6

222

3=

×

−−=

π

inyB 03819.001556.000990.001273.0 =++=

Deflection of C.

321 CCCC yyyy ++=

( )( )( ) ( ) ( ) ( )[ ]( ) ( ) ( )

inyC 01556.0

3064

5.110306

15305.73015305.73004

6

222

1=

×

−−−−=

π

( )( )( ) ( ) ( ) ( )[ ]( ) ( ) ( )

inyC 02264.0

3064

5.110306

153015301530153004

6

222

2=

×

−−−−=

π

( )( )( ) ( ) ( ) ( )[ ]( ) ( ) ( )

inyC 01556.0

3064

5.110306

155.730155.73004

6

222

3=

×

−−=

π

inyC 05376.001556.002264.001556.0 =++=

Deflection of D.

321 DDDD yyyy ++=


Page 70 of 76

( )( )( ) ( ) ( ) ( )[ ]( ) ( ) ( )

inyD 00990.0

3064

5.110306

5.22305.7305.22305.73004

6

222

1=

×

−−−−=

π

( )( )( ) ( ) ( ) ( )[ ]( ) ( ) ( )

inyD 01556.0

3064

5.110306

5.223015305.2230153004

6

222

2=

×

−−−−=

π

( )( )( ) ( ) ( ) ( )[ ]( ) ( ) ( )

inyD 01273.0

3064

5.110306

5.225.7305.225.73004

6

222

3=

×

−−=

π

inyD 03819.001273.001556.000990.0 =++=

( ) ( ) 2

1

222

2

1

2

3030

++

++=

=

∑∑

DCB

DCBoo

cyyy

yyyg

Wy

Wygn

ππ

( )( ) ( ) ( )

rpmnc 88803819.005376.003819.0

03819.005376.003819.038630 2

1

222=

++

++=

π

501. A fan for an air-conditioning unit has two 50-lb. rotors mounted on a 3-in. steel

shaft, each being 22 in. from an end of the shaft which is 80 in. long and simply

supported at the ends. Determine (a) the deflection curve of the shaft considering

its weight as well as the weight of the rotors, (b) its critical speed.

Solution:

lbW 501 =

lbW 503 =

( ) ( ) ( ) lbW 1608034

284.02

2 =

=

π weight of shaft

inlbw 280

1602 ==

Deflection of B.

321 BBBB yyyy ++=


Page 71 of 76

( )( )( ) ( ) ( ) ( )[ ]( ) ( ) ( )

inyB 002844.0

8064

3510306

2258802250504

6

222

1=

×

−−=

π

( )( )( ) ( ) ( ) ( )[ ]( ) ( ) ( )

inyB 002296.0

8064

310306

2222802222504

6

222

3=

×

−−=

π

( )( ) ( ) ( )( ) ( )[ ]( ) ( )

inyB 006843.0

64

310306

2222802802224

6

323

2=

×

−−=

π

inyB 011983.0002296.0006843.0002844.0 =++=

Deflection of C.

321 CCCC yyyy ++=

( )( )( ) ( ) ( ) ( )[ ]( ) ( ) ( )

inyC 003317.0

8064

3510306

40802280408022504

6

222

1=

×

−−−−=

π

( )( )( ) ( ) ( ) ( )[ ]( ) ( ) ( )

inyC 003317.0

8064

310306

4022804022504

6

222

3=

×

−−=

π

( )( ) ( ) ( )( ) ( )[ ]( ) ( )

inyC 008942.0

64

310306

4040802804024

6

323

2=

×

−−=

π

inyC 015576.0003317.0008942.0003317.0 =++=

By symmetry

inyy BD 011983.0==

(a) Deflection curve


Page 72 of 76

(b) Critical speed

( ) 2

1

2

30

=

∑∑Wy

Wygn

o

cπ

( )( ) ( )( ) ( )( ) 69046.3011983.050015576.0160011983.050 =++=∑Wy

( )( ) ( )( ) ( )( ) 053177.0011983.050015576.0160011983.0502222

=++=∑Wy

( )rpmnc 1563

053177.0

69046.338630 2

1

=

=

π

ASME CODE

502. A cold-rolled transmission shaft, made of annealed AISI C1050, is to transmit a

torque of 27 in-kips with a maximum bending moment of 43 in-kips. What

should be the diameter according to the Code for a mild shock load?

Solution:

For AISI C1050, annealed

ksisy 53=

ksisu 92=

ksisy 9.153.0 =

ksisu 56.1618.0 =

use ksisyd 9.153.0 ==τ

kipsinM −= 43

kipsinT −= 27

( )( ) ( ) 2

12

22

4

3

8

1

1

16

+++

−=

BFDMKTK

BD ms

d

α

πτ

Reduce to

( )( ) ( )[ ]2

122

4

3

1

16MKTK

BD ms

d

+−

=πτ

For mild shock load, rotating shafts

75.1=mK

25.1=sK

0=B

( )( )( )[ ] ( )( )[ ]{ }2

1223 000,4375.1000,2725.1

900,15

16+=

πD

inD 98.2=

say inD 3=


Page 73 of 76

503. A machinery shaft is to transmit 82 hp at a speed of 1150 rpm with mild shock.

The shaft is subjected to a maximum bending moment of 7500 in-lb. and an axial

thrust load of 15,000 lb. The material is AISI 3150, OQT 1000 F. (a) What

should be the diameter when designed according to the Code? (b) Determine the

corresponding conventional factor of safety (static-approach and maximum-shear

theory).

Solution:


ksisy 130=

ksisu 151=

ksisy 393.0 =

ksisu 18.2718.0 =

use ksisud 18.2718.0 ==τ

( )lbinT −== 4492

1150

8263000

lbinM −= 7500

lbF 000,15=

(a) ( )

( ) ( ) 2

12

22

4

3

8

1

1

16

+++

−=

BFDMKTK

BD ms

d

α

πτ

For mild shock load

75.1=mK

25.1=sK

0=B

1=α

( )( )( )[ ] ( )( )

( )( ) 2

12

23

8

000,151750075.1449225.1

27180

16

++=

DD

π

[ ]{ }2

123 875.1125.1353.311874.0 DD ++=

inD 4668.1=

say inD 5.1=

(b) ( )( )

( )( )

ksipsiD

F

D

Ms 124.31124,31

5.1

000,154

5.1

7500324322323

==+=+=ππππ


Page 74 of 76

( )( )

ksipsiD

Tss 7785.65.6778

5.1

4492161633

====ππ

+

=

22

1

ys

s

y s

s

s

s

N

Maximum shear theory

yys ss 5.0=

( )

+

=

22

1305.0

7785.6

130

124.311

N

83.3=N

504. short stub shaft, made of SAE 1035, as rolled, receives 30 hp at 300 rpm via a

12-in. spur gear, the power being delivered to another shaft through a flexible

coupling. The gear is keyed midway between the bearings and its pressure angle o20=φ . See the figure for 471. (a) Neglecting the radial component of the tooth

load, determine the shaft diameter for a mild shock load. (b) Considering both

tangential and radial components, compute the shaft diameter. (c) Is the

difference in the foregoing results enough to change your choice of the shaft

size?

Solution:

Figure for 471.

For SAE 1035, as rolled

ksisy 55=

ksisu 85=

ksisy 5.163.0 =

ksisu 3.1518.0 =



Page 75 of 76

Data are the same as 471.

From Problem 471.

(a) kipsinlbinM −=−= 2.44200


( )( ) ( ) 2

12

22

4

3

8

1

1

16

+++

−=

BFDMKTK

BD ms

d

α

πτ

Reduce to

( )( ) ( )[ ]2

122

4

3

1

16MKTK

BD ms

d

+−

=πτ

For mild shock load, rotating shafts

75.1=mK

25.1=sK

0=B

( )( )( )[ ] ( )( )[ ]{ }2

1223 2.475.13.625.1

3.15

16+=

πD

inD 5306.1=

say inD16

91=

(b) kipsinlbinM −=−= 472.44472


( )( )( )[ ] ( )( )[ ]{ }2

1223 472.475.13.625.1

3.15

16+=

πD

inD 5461.1=

say inD16

91=

(c) Not enough to change the shaft size.

505. Two bearings D and E, a distance inD 24= . Apart, support a shaft for a punch

press on which are an 8-in. gear A, a 44-in. pulley B, and a flywheel C, as

indicated (498). Weight of flywheel is 1500 lb.; pulley B receives the power at an

angle of 45o to the right of the vertical; gear A delivers it horizontally to the right.

The maximum power is 25 hp at 250 rpm is delivered, with heavy shock. For

cold-finish AISI 1137, find the diameter by the ASME Code.

Solution:


Page 76 of 76

Data and figure is the same as in Problem 479. Also figure is the same as in Problem 498.


ksisy 93=

ksisu 103=

ksisy 9.273.0 =

ksisu 54.1818.0 =


From Problem 479

kipsinlbinMM B −=−== 343.14343,14

kipsinlbinTT A −=−== 6.12600,12

For heavy shock load

5.2=mK

75.1=sK

0=B

( )( ) ( ) 2

12

22

4

3

8

1

1

16

+++

−=

BFDMKTK

BD ms

d

α

πτ

( )( ) ( )[ ]2

122

4

3

1

16MKTK

BD ms

d

+−

=πτ

( )( )( )[ ] ( )( )[ ]{ }2

1223 343.145.26.1275.1

54.18

16+=

πD

inD 2613.2=

say inD16

52=

- end -

SECTION 9 – JOURNAL AND PLANE-SURFACE BEARINGS

Page 1 of 63

LIGHTLY LOADED BEARINGS

551. (a) A 3 x 3 – in. full bearing supports a load of 900 lb., 0015.0=Dcd ,

rpmn 400= . The temperature of the SAE 40 oil is maintained at 140 oF.

Considering the bearing lightly loaded (Petroff), compute the frictional torque,

fhp, and the coefficient of friction. (b) The same as (a) except that the oil is

SAE 10W.

Solution.

(a) ( )

=

22

D

c

DLvT

d

ips

f

µπ

inL 3=

inD 3=

( )( )ips

Dnvips π

ππ20

60

4003

60===

0015.0=Dcd

SAE 40 oil, 140 oF, Figure A16.

reynsµµ 25.7=

( )( )( )( )( )( )

( )lb

c

DLvF

d

ips173.17

20015.0

20331025.7

2

6

=×

==− ππµπ

( ) lbinD

FT f −=

=

= 76.25

2

3173.17

2

000,33

mFvfhp =

( )( )fpm

Dnvm 16.314

12

4003

12===

ππ

( )( )hp

Fvfhp m 1635.0

000,33

16.314173.17

000,33===

0191.0900

173.17===

W

Ff

(b) SAE 10W oil, 140 oF, Figure A16.

reynreyns 6102.22.2 −×== µµ

( )( )( )( )( )( )

( )lb

c

DLvF

d

ips211.5

20015.0

2033102.2

2

6

=×

==− ππµπ

( ) lbinD

FT f −=

=

= 817.7

2

3211.5

2

000,33

mFvfhp =


Page 2 of 63

( )( )fpm

Dnvm 16.314

12

4003

12===

ππ

( )( )hp

Fvfhp m 0496.0

000,33

16.314211.5

000,33===

00579.0900

211.5===

W

Ff

553. The average pressure on a 6-in. full bearing is 50 psi, .003.0 incd = , 1=DL .

While the average oil temperature is maintained at 160 oF with rpmn 300= ,

the frictional force is found to be 13 lb. Compute the coefficient of friction

and the average viscosity of the oil. To what grade of oil does this

correspond?

Solution:

LD

Wp =

.6 inD =

1=DL

.6 inL =

( )( )( ) lbpLDW 18006650 ===

lbF 13=

Coefficient of Friction

0072.01800

13===

W

Ff

( )2d

ips

c

DLvF

µπ=

( )( )ips

Dnvips π

ππ30

60

3006

60===

( )( )( )( )( )

( )lb

c

DLvF

d

ips13

2003.0

3066

2===

ππµµπ

reynsreyn µµ 8.1108.1 6 =×= −

Figure AF 16, 160 oF use SAE 10W or SAE 20W

FULL BEARINGS

554. The load on a 4-in. full bearing is 2000 lb.,

rpmn 320= ; 1=DL ; 0011.0=Dcd ; operating temperature = 150 oF;

inho 00088.0= . (a) Select an oil that will closely accord with the started

conditions. For the selected oil determine (b) the frictional loss (ft-lb/min), (c)

the hydrodynamic oil flow through the bearing, (d) the amount of end leakage,


Page 3 of 63

(e) the temperature rise as the oil passes through the bearing, (f) the maximum

pressure.

Solution:

(a) inD 4=

1=DL

inL 4=

( ) inDcd 0044.040011.00011.0 ===

inho 00088.0=

( )6.0

0044.0

00088.021

21 =−=−=

d

o

c

hε

Table AT 20

6.0=ε , 1=DL

Sommerfield Number 2

=

d

s

c

D

p

nS

µ

rpsns 333.560

320==

( )( )psi

LD

Wp 125

44

2000===

0011.0=Dcd

( ) 2

0011.0

1

125

333.5121.0

=

µ

reynsreyn µµ 4.3104.3 6 =×= −

Figure AF-16, 150 oF, use SAE 30 or SAE 20 W

Select SAE 30, the nearest

reyn6109.3 −×=µ

(b) Table AT 20, 1=DL , 6.0=ε

22.3=fc

r

r

0011.0

1==

dr c

D

c

r

22.30011.0

1=

f

003542.0=f

( )( ) lbWfF 084.72000003542.0 ===


Page 4 of 63

( )( )fpm

Dnvm 1.335

12

3204

12===

ππ

Frictional loss = ( )( ) minlbftFvm −= 23741.335084.7

(c) Table AT 20, 1=DL , 6.0=ε

33.4=Lnrc

q

sr

inD

r 0.22

==

inc

c dr 0022.0

2

0044.0

2===

rpsns 333.5=

inL 4=

( )( )( )( ) secinLnrcq sr

34064.04333.50022.00.233.433.4 ===

(d) Table AT 20, 1=DL , 6.0=ε

680.0=q

qs

( ) secinqqs

32764.04064.0680.0680.0 ===

(e) Table AT 20, 1=DL , 6.0=ε

2.14=∆

p

tc oρ

112=cρ , psip 125=

( )F

c

pto

o85.15112

1252.142.14===∆

ρ

(f) Table AT 20, 1=DL , 6.0=ε

415.0max

=p

p

psip 2.301415.0

125max ==

555. A 4-in., 360o bearing, with 1.1=DL (use table and chart values for 1), is to

support 5 kips with a minimum film thickness 0.0008 in.; .004.0 incd = ,

rpmn 600= . Determine (a) the needed absolute viscosity of the oil .(b)

Suitable oil if the average film temperature is 160 F, (c) the frictional loss in

hp. (d) Adjusting only oh to the optimum value for minimum friction,

determine the fhp and compare. (e) This load varies. What could be the

magnitude of the maximum impulsive load if the eccentricity ration ε

becomes 0.8? Ignore “squeeze” effect.


Page 5 of 63

Solution:

inD 4=

( ) inDL 4.441.11.1 ===

( )( )psi

LD

Wp 284

44.4

5000===

inho 0008.0=

.004.0 incd =

( )6.0

004.0

0008.021

21 =−=−=

d

o

c

hε

rps1060

600==η

(a) Table AT20, 1=DL , 6.0=ε

121.0=S 22

=

=

d

ss

r c

D

p

n

p

n

c

rS

µµ

( ) 2

004.0

4

284

10121.0

=

µ

reyn6104.3 −×=µ

(b) Figure AF16, 160 F

Use SAE 30, reyn6102.3 −×=µ

(c) Table AT 20, 1=DL , 6.0=ε

22.3=fc

r

r

22.3=fc

D

d

22.3004.0

4=

f

00322.0=f

( )( ) lblbWfF 1.16500000322.0 ===

( )( )fpm

Dnvm 3.628

12

6004

12===

ππ


Page 6 of 63

( )( )hp

Fvfhp m 3065.0

000,33

3.6281.16

000,33===

(d) adjusting oh , .004.0 incd =

Table AT 20, 1=DL

30.0=ro ch optimum value for minimum friction

46.2=fc

r

r

46.2=fc

D

d

46.2004.0

4=

f

00246.0=f

( )( ) lblbWfF 3.12500000246.0 ===

( )( )fpm

Dnvm 3.628

12

6004

12===

ππ

( )( ) ( )cfhphpFv

fhp m <=== 234.0000,33

3.6283.12

000,33

(e) 8.0=ε , Table AT 20, 1=DL

0446.0=S 22

=

=

d

ss

r c

D

p

n

p

n

c

rS

µµ

( )( ) 26

004.0

410102.30446.0

×=

−

p

psip 5.717=

( )( )( ) lbpDLW 628,124.445.717 ===

556. For an 8 x 4 – in. full bearing, .0075.0 incr = , rpmn 2700= , average

reyn6104 −×=µ . (a) What load may this bearing safely carry if the minimum

film thickness is not to be less than that given by Norton, i11.14, Text? (b)

Compute the corresponding frictional loss (fhp). (c) Complete calculations for

the other quantities in Table AT 20, φ , q , sq , ot∆ , maxp . Compute the

maximum load for an optimum (load) bearing (d) if rc remains the same, (e)

if oh remains the same.


Page 7 of 63

Solution:

48×=× LD

21=DL

incr 0075.0=

inDr 42 ==

reyn6104 −×=µ

(a) by Norton, ( ) inDho 002.0800025.000025.0 ===

27.00075.0

002.0==

r

o

c

h

Table AT 20, 21=DL , 27.0=r

o

c

h

172.0=S

p

n

c

rS s

r

µ2

=

rpsns 4560

2700==

( )( )p

S45104

0075.0

4172.0

62 −×

==

psip 298=

( )( )( ) lbpDLW 953648298 ===

(b) Table AT 20, 21=DL , 27.0=r

o

c

h

o5.38=φ

954.4=fc

r

r

954.4=fc

D

d

954.4004.0

4=

f

0093.0=f

( )( ) lblbWfF 7.8895360093.0 ===

( )( )fpm

Dnvm 5655

12

27008

12===

ππ

( )( )hp

Fvfhp m 2.15

000,33

56557.88

000,33===


Page 8 of 63

(c) Table AT 20, 21=DL , 27.0=r

o

c

h

o5.38=φ

214.5=Lnrc

q

sr

( )( )( )( ) sec2.284450075.04214.5214.5 3inLnrcq sr ===

824.0=q

qs

( ) sec2.232.28824.0 3inqs ==

26.20=∆

p

tcρ

( )Ft

o54112

29826.20==∆

3013.0max

=p

p

psip 9893013.0

298max ==

To solve for maximum load, Table AT 20, 21=DL , 43.0=r

o

c

h

388.0

2

=

=

p

n

c

rS s

r

µ

(d) incr 0075.0=

( )( )p

S45104

0075.0

4388.0

62 −×

==

psip 132=

( )( )( ) lbpDLW 422448132 ===

(e) inho 002.0=

43.0=r

o

c

h

incr 00465.043.0

002.0==

( )( )p

S45104

00465.0

4388.0

62 −×

==

psip 3.343=

( )( )( ) lbpDLW 986,10483.343 ===


Page 9 of 63

557. A 6 x 6 – in full bearing has a frictional loss of 11=fhp when the load is

68,500 lb. and rpmn 1600= ; 001.0=rcr . (a) Compute the minimum film

thickness. Is this in the vicinity of that for an optimum bearing? (b) What is

the viscosity of the oil and a proper grade for an operating temperature of 160

F? (c) For the same oh , but for the maximum-load optimum, determine the

permissible load and the fhp.

Solution:

inL 6=

inD 6=

1=DL

inDr 32 ==

001.0=rcr

rpmn 1600=

( )( )fpm

Dnvm 2513

12

16003

12===

ππ

000,33

mFvfhp =

( )lbF 45.144

2513

11000,33==

00211.0500,68

45.144===

W

Ff

(a) ( ) 11.200211.0001.0

1=

=f

c

r

r

Table AT 20, 1=DL , 11.2=fc

r

r

Near the vicinity of optimum bearing

( ) inrcr 003.03001.0001.0 ===

( ) inch ro 0008.0003.0254.0254.0 ===

(b) Table AT 20, 1=DL , 11.2=fc

r

r

0652.0=S

388.0

2

=

=

p

n

c

rS s

r

µ

rpsns 67.2660

1600==

( )( )psi

LD

Wp 8.1902

66

500,68===


Page 10 of 63

( )( )8.1902

67.26

001.0

10652.0

2µ

==S

reyn6107.4 −×=µ

Figure AF 16, 160 F, use SAE 40.

(c) Table AT 20, 1=DL

optimum bearing, maximum load, 53.0=r

o

c

h

oh the same, inh

c or 0015.0

53.0

0008.0

53.0===

53.0=r

o

c

h, 214.0=S , 89.4=f

c

r

r

p

n

c

rS s

r

µ2

=

( )( )p

S67.26107.4

0015.0

3214.0

62 −×

==

psip 2343=

( )( )( ) lbpDLW 348,84662343 ===

89.4=fc

r

r

89.40015.0

3=

f

00245.0=f

( ) lbWfF 65.206348,8400245.0 ===

fpmvm 2513=

( )( )hp

Fvfhp m 74.15

000,33

251365.206

000,33===

558. The maximum load on a 2.25 x 1.6875 in. main bearing of an automobile is

3140 lb. with wide-open throttle at 1000 rpm. If the oil is SAE 20W at 210 F,

compute the minimum film thickness for a bearing clearance of (a) 0.0008 in.

and (b) 0.0005 in. Which clearance results in the safer operating conditions?

Note: Since a load of this order exists for only 20-25o of rotation, the actual

oh does not reach this computed minimum (squeeze effect).

Solution:

inLD 6875.125.2 ×=×

75.025.2

6875.1==

D

L


Page 11 of 63

SAE 20 W at 210 oF

reyn61096.0 −×=µ

lbW 3140=

rpmn 1000=

( )( )psi

DL

Wp 827

6875.125.2

3140===

rpsns 67.1660

1000==

inD

r 125.12

==

2

=

r

s

c

r

p

nS

µ

(a) incr 0008.0=

( )( )038.0

0008.0

125.1

827

67.161096.026

=

×=

−

S

Table AT 20, 43=DL , 038.0=S

DL ro ch S

1 0.2 0.0446

½ 0.2 0.0923

¾ 0.2 0.0685

DL ro ch S

1 0.1 0.0188

½ 0.1 0.0313

¾ 0.1 0.0251

At 43=DL

( ) 13.01.01.02.00251.00685.0

0251.0038.0=+−

−

−=

r

o

c

h

( ) inch ro 0001.00008.013.013.0 ===

(b) incr 0005.0=

( )( )098.0

0005.0

125.1

827

67.161096.026

=

×=

−

S

Table AT 20, 43=DL , 098.0=S


Page 12 of 63

DL ro ch S

1 0.2 0.0446

½ 0.2 0.0923

¾ 0.2 0.0685

DL ro ch S

1 0.4 0.121

½ 0.4 0.319

¾ 0.4 0.220

At 43=DL

( ) 239.02.02.04.00685.0220.0

0685.0098.0=+−

−

−=

r

o

c

h

( ) inch ro 00012.00005.0239.0239.0 ===

use incr 0005.0= , inho 00012.0=

561. A 360o bearing supports a load of 2500 lb.; .5 inD = , .5.2 inL = ,

.003.0 incr = , rpmn 1800= ; SAE 20 W oil entering at 100 F. (a) Compute

the average temperature avt of the oil through the bearing. (An iteration

procedure. Assume µ ; compute S and the corresponding ot∆ ; then the

average oil temperature 2oiav ttt ∆+= . If this avt and the assumed µ do not

locate a point in Fig. AF 16 on the line for SAE 20 W oil, try again.) Calculate

(b) the minimum film thickness, (c) the fhp, (d) the amount of oil to be

supplied and the end leakage.

Solution:

inD 5=

inL 5.2=

5.05

5.2==

D

L

incr 003.0=

(a) Table AT 20

Parameter, p

tc o∆ρ, 112=cρ

p

n

c

rS s

r

µ2

=


Page 13 of 63

( )( )psi

DL

Wp 200

5.25

2500===

rpsns 3060

1800==

inD

r 5.22

==

incr 003.0=

Fig. AF 16, SAE 20 W, Table AT 20, 5.0=DL , Fti

o100=

Trial µ ( Ft o ), reyns S p

tc o∆ρ Fto

o∆ Fttt oiav

o2∆+=

3.5 x 10-6

(130 F) 0.365 36.56 65 132.5

3.2 x 10-6

(134 F) 0.333 34.08 61 130.5

3.4 x 10-6

(132 F) 0.354 35.71 64 132.0

Therefore, use Ftav

o132= , 354.0=S

(b) Table AT 20, 5.0=DL , 354.0=S

415.0=r

o

c

h

( ) inho 00125.0003.0415.0 ==

(c) Table AT 20, 5.0=DL , 354.0=S

777.8=fc

r

r

777.8003.0

5.2=

f

0105.0=f

( ) lbWfF 25.2625000105.0 ===

( )( )fpm

Dnvm 2356

12

18005

12===

ππ

( )( )hp

Fvfhp m 874.1

000,33

235625.26

000,33===

(d) Table AT 20, 5.0=DL , 354.0=S

807.4=Lnrc

q

sr

( )( )( )( ) sec704.25.230003.05.2807.4807.4 3inLnrcq sr ===

7165.0=q

qs


Page 14 of 63

( ) sec937.1704.27165.0 3inqs ==

PARTIAL BEARINGS

562. A 2 x 2-in. bearing has a clearance incr 001.0= , and .0004.0 inho = ,

rpmn 2400= , and for the oil, reyn6103 −×=µ . Determine the load, frictional

horsepower, the amount of oil to enter, the end leakage of oil, and the

temperature rise of the oil as it passes through for : (a) a full bearing, partial

bearings of (b) 180o, (c) 120

o, (d) 90

o, (e) 60

o.

Solution:

inLD 2==

1=DL

incr 001.0=

inDr 12 ==

rpmn 2400=

rpsns 40=

reyn6103 −×=µ

.004.0 inho =

4.0001.0

0004.0==

r

o

c

h

( )( )fpm

Dnvm 1257

12

24002

12===

ππ

(a) Full bearing

Table AT 20, 1=DL , 4.0=ro ch

121.0=S

22.3=rc

rf

33.4=Lnrc

q

sr

680.0=q

qs

2.14=∆

p

tc oρ

415.0max

=p

p

Load W


Page 15 of 63

p

n

c

rS s

r

µ2

=

( )( )p

40103

001.0

1121.0

62 −×

=

psip 992=

( )( )( ) lbpDLW 396822992 ===

fhp:

WfF =

22.3=rc

rf

22.3001.0

1=

f

00322.0=f

( )( ) lbWfF 78.12396800322.0 ===

( )( )hp

Fvfhp m 4868.0

000,33

125778.12

000,33===

Oil flow, q

33.4=Lnrc

q

sr

( )( )( )( )33.4

240001.01.0=

q

sec3464.0 3inq =

End leakage

680.0=q

qs

( ) sec2356.03464.068.0 3inqs ==

Temperature rise, ot∆

2.14=∆

p

tc oρ

( )2.14

992

112=

∆ ot

Fto

o126=∆

(b) 180o Bearing


128.0=S


Page 16 of 63

28.2=rc

rf

25.3=Lnrc

q

sr

572.0=q

qs

4.12=∆

p

tc oρ

Load W

p

n

c

rS s

r

µ2

=

( )( )p

40103

001.0

1128.0

62 −×

=

psip 5.937=

( )( )( ) lbpDLW 3750225.937 ===

fhp:

WfF =

28.2=rc

rf

28.2001.0

1=

f

00228.0=f

( )( ) lbWfF 55.8375000228.0 ===

( )( )hp

Fvfhp m 3257.0

000,33

125755.8

000,33===

Oil flow, q

25.3=Lnrc

q

sr

( )( )( )( )25.3

240001.01.0=

q

sec26.0 3inq =

End leakage

572.0=q

qs

( ) sec1487.026.0572.0 3inqs ==



Page 17 of 63

4.12=∆

p

tc oρ

( )4.12

5.937

112=

∆ ot

Fto

o104=∆

(c) 12o Bearing


162.0=S

16.2=rc

rf

24.2=Lnrc

q

sr

384.0=q

qs

15=∆

p

tc oρ

Load W

p

n

c

rS s

r

µ2

=

( )( )p

40103

001.0

1162.0

62 −×

=

psip 741=

( )( )( ) lbpDLW 296422741 ===

fhp:

WfF =

16.2=rc

rf

16.2001.0

1=

f

00216.0=f

( )( ) lbWfF 4.6296400216.0 ===

( )( )hp

Fvfhp m 2438.0

000,33

12574.6

000,33===

Oil flow, q


Page 18 of 63

24.2=Lnrc

q

sr

( )( )( )( )24.2

240001.01.0=

q

sec1792.0 3inq =

End leakage

384.0=q

qs

( ) sec0688.01792.0384.0 3inqs ==


15=∆

p

tc oρ

( )15

741

112=

∆ ot

Fto

o99=∆

(d) 60o Bearing

1=DL , 4.0=ro ch

450.0=S

29.3=rc

rf

56.1=Lnrc

q

sr

127.0=q

qs

2.28=∆

p

tc oρ

Load W

p

n

c

rS s

r

µ2

=

( )( )p

40103

001.0

1450.0

62 −×

=

psip 267=

( )( )( ) lbpDLW 106822267 ===

fhp:

WfF =


Page 19 of 63

29.3=rc

rf

29.3001.0

1=

f

00329.0=f

( )( ) lbWfF 514.3106800329.0 ===

( )( )hp

Fvfhp m 1339.0

000,33

1257514.3

000,33===

Oil flow, q

56.1=Lnrc

q

sr

( )( )( )( )56.1

240001.01.0=

q

sec1248.0 3inq =

End leakage

127.0=q

qs

( ) sec0158.01248.0127.0 3inqs ==


2.28=∆

p

tc oρ

( )2.28

267

112=

∆ ot

Fto

o67=∆

563. A 2 x 2 in. bearing sustains a load of .5000 lbW = ; .001.0 incr = ;

rpmn 2400= ; reyn6103 −×=µ . Using Figs. AF 17 and AF 18, determine the

minimum film thickness and the frictional loss (ft-lb/min.) for (a) a full

bearing, and for partial bearings of (b) 180o, (c) 120

o, (d) 90

o, (e) 60

o.

Solution:

inL 2=

inD 2=

lbW 5000=

.001.0 incr =

rpmn 2400=

rpsns 40=

reyn6103 −×=µ

inDr 12 ==


Page 20 of 63

( )( )psi

LD

Wp 1250

22

5000===

( )( )10.0

1250

40103

001.0

1 622

=×

=

=

−

p

n

c

rS s

r

µ

( )( )fpm

Dnvm 1257

12

24002

12===

ππ

Using Fig. AF 17 and AF 18

(a) Full Bearing

346.0=r

o

c

h

8.2=fc

r

r

( ) inho 000346.0001.0346.0 ==

8.2001.0

1=

f

0028.0=f

( )( ) lbWfF 1450000028.0 ===

( )( ) min600,17125714 lbftFvm −==

(b) 180o Bearing

344.0=r

o

c

h

0.2=fc

r

r

( ) inho 000344.0001.0344.0 ==

0.2001.0

1=

f

0020.0=f

( )( ) lbWfF 1050000020.0 ===

( )( ) min570,12125710 lbftFvm −==

(c) 120o Bearing

302.0=r

o

c

h


Page 21 of 63

7.1=fc

r

r

( ) inho 000302.0001.0302.0 ==

7.1001.0

1=

f

0017.0=f

( )( ) lbWfF 5.850000017.0 ===

( )( ) min685,1012575.8 lbftFvm −==

(d) 60o Bearing

20.0=r

o

c

h

4.1=fc

r

r

( ) inho 0002.0001.020.0 ==

4.1001.0

1=

f

0014.0=f

( )( ) lbWfF 750000014.0 ===

( )( ) min800,812577 lbftFvm −==

564. A 120o partial bearing is to support 4500 lb. with .002.0 inho = ; 1=DL ;

.4 inD = ; .010.0 incd = ; rpmn 3600= . Determine (a) the oil’s viscosity,(b)

the frictional loss (ft-lb/min), (c) the eccentricity angle, (d) the needed oil

flow, (e) the end leakage, (f) the temperature rise of the oil as it passes

through, (g) the maximum pressure. (h) If the clearance given is the average,

what approximate class of fit (Table 3.1) is it? (i) What maximum impulsive

load would be on the bearing if the eccentricity ratio suddenly went to 0.8?

Ignore “squeeze” effect.

Solution:

lbW 4500=

inho 002.0=

1=DL

inD 4=

inL 4=

inDr 22 ==

.010.0 incd =

rpmn 3600=


Page 22 of 63

rpsns 6060

3600==

( )( )fpm

Dnvm 3770

12

36002

12===

ππ

( )( )psi

LD

Wp 25.281

44

4500===

( )4.0

010.0

002.022===

r

o

r

o

c

h

c

h


162.0=S o65.35=φ

16.2=fc

r

r

24.2=Lnrc

q

sr

384.0=q

qs

0.15=∆

p

tc oρ

356.0max

=p

p

(a) p

n

c

rS s

r

µ2

=

p

n

c

DS s

d

µ2

=

( )25.281

60

010.0

4162.0

2µ

=

reyn61075.4 −×=µ

(b) 16.2=fc

r

r

16.2=fc

D

d

16.2010.0

4=

f

0054.0=f

( ) lbWfF 30.2445000054.0 ===


Page 23 of 63

( )( ) min611,91377030.24 lbftFvm −==

(c) o65.35=φ

(d) 24.24

==LnDc

q

Lnrc

q

sdsr

( )( )( )( )24.2

460010.04

4=

q

sec4.5 3inq =

(e) 384.0=q

qs

( ) sec07.24.5384.0 3inqs ==

(f) 0.15=∆

p

tc oρ

( )0.15

25.281

112=

∆ ot

Fto

o38=∆

(g) 356.0max

=p

p

psip 790356.0

25.281max ==

(h) incd 010.0= , inD 4=

Table 3.1

RC 8, Hole, average = + 0.0025

Shaft, average = - 0.00875

incd 010.001125.000875.00025.0 ≈=+=

Class of fit = RC 9

(i) 80.0=ε

Table AT 22, , 1=DL

162.0=S

p

n

c

DS s

d

µ2

=

( )( )p

60103

010.0

40531.0

62 −×

=

psip 542=


Page 24 of 63

( )( )( ) lbpDLW 867244542 ===

565. A 120o partial bearing is to support 4500 lb., .3 inD = , .003.0 incd = ;

rpmn 3600= ; SAE 20W entering at 110 F. Calculate (a) the average

temperature of the oil as it passes through,(b) the minimum film thickness, (c)

the fhp, (d) the quantity of oil to be supplied. HINT: In (a) assume µ and

determine the corresponding values of S and ot∆ ; then 2oiav ttt ∆+= . If

assumed µ and avt do not locate a point in Fig. AF 16 that falls on line for

SAE 20W, iterate.

Solution:

lbW 4500=

inD 3=

inL 3=

1=DL

.003.0 incd =

p

n

c

DS s

d

µ2

=

rpsns 6060

3600==

( )( )psi

DL

Wp 500

33

4500===

p

tc o∆ρ, (SAE 20W)

(a) Using Table AT22, 1=DL , 112=cρ , Fti

o110=

Trial µ t , oF S

p

tc o∆ρ

ot∆ 2oiav ttt ∆+=

3.5 x 10-6

130 0.42 19.8 88 154

2.0 x 10-6

160 0.24 15.4 68 144

2.6 x 10-6

145 0.312 17.7 79 149.5

2.35 x 10-6

150 0.282 17.2 76 148

2.4 x 10-6

149 0.288 17.3 78 149

∴ Use Ftav

o149=

(b) Table AT 22, 1=DL , 288.0=S


Page 25 of 63

513.0=r

o

c

h

513.02

=d

o

c

h

( )003.0513.02 =oh

inho 00077.0=

(c) Table At 22, 1=DL , 288.0=S

974.2=fc

r

r

974.2003.0

3== ff

c

D

r

002974.0=f

( )( ) lbWfF 383.134500002974.0 ===

000,33

mFvfhp =

( )( )fpm

Dnvm 2827

12

36003

12===

ππ

( )( )hp

Fvfhp m 15.1

000,33

2827383.13

000,33===

(d) Table At 22, 1=DL , 288.0=S

528.2=Lnrc

q

sr

528.24

=LnDc

q

sd

( )( )( )( )528.2

360003.03

4=

q

sec024.1 3inq =

566. The 6000-lb. reaction on an 8 x 4 –in., 180o partial bearing is centrally

applied; rpmn 1000= ; inho 002.0= . For an optimum bearing with minimum

friction determine (a) the clearance, (b) the oil’s viscosity, (c) the frictional

horsepower. (d) Choose a Dcd ratio either smaller or larger than that

obtained in (a) and show that the friction loss is greater than that in the

optimum bearing. Other data remain the same.

Solution:

lbW 6000=


Page 26 of 63

inD 8=

inL 4=

rpmn 1000=

rpsns 67.1660

1000==

21=DL

inho 002.0=

(a) Table AT 21, 21=DL

Optimum value (minimum friction)

23.0=ro ch

incr 0087.023.0

002.0==

(b) Table AT 21, 21=DL , 23.0=ro ch

126.0=S

p

n

c

rS s

r

µ2

=

( )( )psi

DL

Wp 5.187

84

6000===

inD

r 42

==

( )5.187

67.16

0087.0

4126.0

2µ

==S

reyn61070.6 −×=µ

(c) Table AT 21, 21=DL , 23.0=ro ch

97.2=fc

r

r

97.20087.0

4=

f

00646.0=f

( )( ) lbWfF 76.38600000646.0 ===

( )( )fpm

Dnvm 2094

12

10008

12===

ππ


Page 27 of 63

( )( )hp

Fvfhp m 46.2

000,33

209476.38

000,33===

For (a) ( )

0022.08

0087.022===

D

c

D

c rd

0022.0>D

cd

0030.0=D

cd

( ) incd 0240.080030.0 ==

incr 0120.0=

1667.0012.0

002.0==

r

o

c

h

Table AT 21, 21=DL

67.1=fc

r

r

67.10016.0

4=

f

00668.0=f

( )( ) lbWfF 08.40600000668.0 ===

( )( )fpm

Dnvm 2094

12

10008

12===

ππ

( )( )hphp

Fvfhp m 46.254.2

000,33

209408.40

000,33>===

0022.0<D

cd

0020.0=D

cd

( ) incd 0160.080020.0 ==

incr 0080.0=

25.0008.0

002.0==

r

o

c

h

Table AT 21, 21=DL

26.3=fc

r

r

26.30016.0

4=

f

00652.0=f


Page 28 of 63

( )( ) lbWfF 12.39600000652.0 ===

( )( )fpm

Dnvm 2094

12

10008

12===

ππ

( )( )hphp

Fvfhp m 46.248.2

000,33

209412.39

000,33>===

567. A 120o partial bearing supports 3500 lb. when rpmn 250= ; .5 inD = ,

.5 inL = ; reyn6103 −×=µ . What are the clearance and minimum film

thickness for an optimum bearing (a) for maximum load, (b) for minimum

friction? (c) On the basis of the average clearance in Table 3.1, about what

class fit is involved? Would this fit be on the expensive or inexpensive side?

(d) Find the fhp for each optimum bearing.

Solution:

.5 inD =

.5 inL =

1=D

L

rpmn 250=

rpsns 17.460

250==

reyn6103 −×=µ

lbW 3500=

( )( )psi

DL

Wp 140

55

3500===

(a) Table AT 22, 1=D

L, max. load 46.0=

r

o

c

h

229.0=S

p

n

c

rS s

r

µ2

=

inD

r 5.22

==

( )( )140

17.4100.35.2229.0

62

−×

==

rcS

incr 00156.0=

( ) inch ro 00072.000156.046.046.0 ===


Page 29 of 63

(b) Table AT 22, 1=D

L, min. friction 40.0=

r

o

c

h

162.0=S

p

n

c

rS s

r

µ2

=

inD

r 5.22

==

( )( )140

17.4100.35.2162.0

62

−×

==

rcS

incr 00186.0=

( ) inch ro 00074.000186.040.046.0 ===

(c) ( ) incd 00312.000156.021 ==

( ) incd 00372.000186.022 ==

Use Class RC4, ave. incd 00320.0= , expensive side

(d) Table AT 22, 1=D

L, max. load 46.0=

r

o

c

h

592.2=fc

r

r

592.200156.0

5.2=

f

00162.0=f

( )( ) lbWfF 67.5350000162.0 ===

( )( )fpm

Dnvm 25.327

12

2505

12===

ππ

( )( )hp

Fvfhp m 0562.0

000,33

25.32767.5

000,33===

For minimum friction, 40.0=r

o

c

h

16.2=fc

r

r

16.200186.0

5.2=

f

00161.0=f

( )( ) lbWfF 635.5350000161.0 ===


Page 30 of 63

( )( )fpm

Dnvm 25.327

12

2505

12===

ππ

( )( )hp

Fvfhp m 0559.0

000,33

25.327635.5

000,33===

570. A 180o partial bearing is to support 17,000 lb. with psip 200= ,

rpmn 1500= , inho 003.0= , 1=DL . (a) Determine the clearance for an

optimum bearing with minimum friction. (b) Taking this clearance as the

average, choose a fit (Table 3.1) that is approximately suitable. (c) Select an

oil for an average temperature of 150 F. (d) Compute fhp.

Solution:

lbW 000,17=

psip 200=

rpmn 1500=

rpsns 2560

1500==

1=DL

DL =

DL

Wp =

2

000,17200

D=

inLD 22.9==

inD

r 61.42

22.9

2===

(a) For optimum bearing with minimum friction


44.0=ro ch

44.0003.0

=rc

incr 00682.0=

(b) Table 3.1, inD 22.9=

( ) incc rd 01364.000682.022 ===

Use Class RC7, average incd 01065.0=

Or use Class RC8, average incd 01575.0=

(c) Table AT 21, 1=DL , 44.0=ro ch

158.0=S


Page 31 of 63

p

n

c

rS s

r

µ2

=

( )200

25

00682.0

61.4158.0

2µ

=

reyn6108.2 −×=µ

Fig. AF 16, at 150 F

Use Either SAE 20W or SAE 30.

(d) Table AT 21, 1=DL , 44.0=ro ch

546.2=fc

r

r

546.200682.0

61.4=

f

00377.0=f

( )( )fpm

Dnvm 3621

12

150022.9

12===

ππ

( )( ) lbWfF 09.64000,1700377.0 ===

( )( )hp

Fvfhp m 0.7

000,33

362109.64

000,33===

571. The reaction on a 120o partial bearing is 2000 lb. The 3-in journal turns at

1140 rpm; .003.0 incd = ; the oil is SAE 20W at an average operating

temperature of 150 F. Plot curves for the minimum film thickness and the

frictional loss in the bearing against the ratio DL , using =DL 0.25, 0.5, 1,

and 2. (Note: This problem may be worked as a class problem with each

student being responsible for a particular DL ratio.)

Solution:

lbW 2000=

.3 inD =

rpmn 1140=

rpsns 1960

1140==

incd 003.0=

incr 0015.0=

For SAE 20W, 150 F

reyn61075.2 −×=µ

(a) 25.0=D

L


Page 32 of 63

( ) inDL 75.0325.025.0 ===

( )( )psi

DL

Wp 889

75.03

2000===

Table AT 22, 25.0=D

L

inD

r 5.12

==

( )( )0588.0

889

191075.2

0015.0

5.1 622

=×

=

=

−

p

n

c

rS s

r

µ

083.0=r

o

c

h

( ) inho 000125.00015.0083.0 ==

193.2=fc

r

r

193.20015.0

5.1=

f

002193.0=f

( )( ) lbWfF 386.42000002193.0 ===

( )( )fpm

Dnvm 895

12

11403

12===

ππ

( )( )hp

Fvfhp m 119.0

000,33

895386.4

000,33===

(b) 5.0=D

L

( ) inDL 5.135.05.0 ===

( )( )psi

DL

Wp 444

5.13

2000===

Table AT 22, 5.0=D

L

inD

r 5.12

==

( )( )1177.0

444

191075.2

0015.0

5.1 622

=×

=

=

−

p

n

c

rS s

r

µ

2159.0=r

o

c

h

( ) inho 000324.00015.02159.0 ==

35.2=fc

r

r


Page 33 of 63

35.20015.0

5.1=

f

00235.0=f

( )( ) lbWfF 7.4200000235.0 ===

( )( )fpm

Dnvm 895

12

11403

12===

ππ

( )( )hp

Fvfhp m 1275.0

000,33

8957.4

000,33===

(c) 1=D

L

inDL 3==

( )( )psi

DL

Wp 222

33

2000===

Table AT 22, 1=D

L

inD

r 5.12

==

( )( )2354.0

222

191075.2

0015.0

5.1 622

=×

=

=

−

p

n

c

rS s

r

µ

4658.0=r

o

c

h

( ) inho 000699.00015.04658.0 ==

634.2=fc

r

r

634.20015.0

5.1=

f

002634.0=f

( )( ) lbWfF 268.52000002634.0 ===

( )( )fpm

Dnvm 895

12

11403

12===

ππ

( )( )hp

Fvfhp m 1429.0

000,33

895268.5

000,33===

(d) 2=D

L

( ) inDL 6322 ===

( )( )psi

DL

Wp 111

63

2000===

Table AT 22, 2=D

L


Page 34 of 63

inD

r 5.12

==

( )( )47.0

111

191075.2

0015.0

5.1 622

=×

=

=

−

p

n

c

rS s

r

µ

718.0=r

o

c

h

( ) inho 00108.00015.0718.0 ==

8118.3=fc

r

r

8118.30015.0

5.1=

f

003812.0=f

( )( ) lbWfF 624.72000003812.0 ===

( )( )fpm

Dnvm 895

12

11403

12===

ππ

( )( )hp

Fvfhp m 2068.0

000,33

895624.7

000,33===

D

L inho , fhp

0.25 0.000125 0.119

0.5 0.000324 0.128

1.0 0.000699 0.143

2.0 0.001080 0.207


Page 35 of 63

STEADY-STATE TEMPERATURE

572. A 180o partial bearing is subjected to a load of 12,000 lb.; .88 inLD ×=× ,

0015.0=rcr , .0024.0 inho ≈ , rpmn 500= . The air speed about the bearing

is expected to be in excess of 1000 fpm (on moving vehicle) and the effective

radiating area is DL20 . Determine: (a) the eccentricity factor, (b) µreyns, (c)

the frictional loss (ft-lb/min), (d) the estimated temperature of oil and bearing

( a self-contained oil-bath unit) for steady-state operation, and a suitable

oil.(e) Compute ot∆ of the oil passing through the load-carrying area, remark

on its reasonableness, and decide upon whether some redesign is desirable.

Solution:

.8 inD =

.8 inL =

1=DL

lbW 000,12=

inD

r 42

==

( ) inrcr 0060.040015.00015.0 ===

4.00060.0

0024.0==

r

o

c

h

rpmn 500=

rpsns 33.860

500==

Table AT 21, 4.0=ro ch , 1=DL

128.0=S

28.2=fc

r

r

4.12=∆

p

tc oρ

( )( )psi

DL

Wp 5.187

88

000,12===

(a) 6.04.011 =−=−=r

o

c

hε

(b) p

n

c

rS s

r

µ2

=

( )5.187

33.8

0060.0

4128.0

2µ

==S

reyn6105.6 −×=µ


Page 36 of 63

(c) 28.2=fc

r

r

28.20060.0

4=

f

00342.0=f

( )( ) lbWfF 04.41000,1200342.0 ===

( )( )fpm

Dnvm 1047

12

5008

12===

ππ

( )( )hp

Fvfhp m 302.1

000,33

104704.41

000,33===

Frictional loss = 43,000 ft-lb/min

(d) bbcr tAhQ ∆= ft-lb/min

min000,43 lbftQ −=

rccr hhh +=

Finsqlbfthr −−−= ..min108.0

4.0

6.0

017.0D

vh a

c = , fpmva 1000≥

( )( )

Finsqlbfthc −−−== ..min467.08

1000017.0

4.0

6.0

Finsqlbfthcr −−−=+= ..min575.0108.0467.0

( )( ) ..1280882020 insqDLAb ===

bbcr tAhQ ∆=

( )( )( )bt∆= 1280575.0000,43

Ftb 42.58=∆

Oil-bath, 1000 fpm

( )( )( )boa tt ∆≈∆ 3.12.1

( )( )( ) Ftoa 1.9142.583.12.1 ==∆

assume 100 F ambient temperature

FFtb 42.15842.58100 =+=

FFtb 1.1911.91100 =+=

(c) 4.12=∆

p

tc oρ

( )4.12

5.187

112=

∆ ot

Fto 8.20=∆

Solve for 2ot


Page 37 of 63

( ) Ftt oo 2.3821.191221 ==+

21 2.382 oo tt −=

Ftt oo 8.2012 =−

8.202.362 22 =+− oo tt

FFto 2005.2012 ≈=

∴ not reasonable since the oil oxidizes more rapidly above 200 F, a redesign is

desireable.

573. A 2 x 2-in. full bearing (ring-oiled) has a clearance ratio 001.0=Dcd . The

journal speed is 500 rpm, reyn6104.3 −×=µ , and .0005.0 inho = The ambient

temperature is 100 F; DLAb 25= , and the transmittance is taken as

FftsqhrBtuhcr −−= ..2 . Calculate (a) the total load for this condition; (b)

the frictional loss, (c) the average temperature of the oil for steady-state

operation. Is this temperature satisfactory? (d) For the temperature found,

what oil do you recommend? For this oil will oh be less or greater than the

specified value? (e) Compute the temperature rise of the oil as it passes

through the bearing. Is this compatible with other temperatures found? (f)

What minimum quantity of oil should the ring deliver to the bearing?

Solution:

.2 inL =

.2 inD =

001.0=Dcd

( )( ) incd 0020.02001.0 ==

reyn6104.3 −×=µ

.0005.0 inho =

incr 0010.0=

5.00010.00005.0 ==ro ch

Table AT 20, 1=DL , 5.0=ro ch , Full Bearing

1925.0=S

505.4=fc

r

r

16.4=Lnrc

q

sr

25.19=∆

p

tc oρ

(a) 1925.0=S


Page 38 of 63

p

n

c

rS s

r

µ2

=

inD

r 12

==

rpsns 33.860

500==

( )( )p

S330.8104.3

0010.0

11925.0

62 −×

==

psip 147=

( )( )( ) lbpDLW 58822147 ===

(b) 505.4=fc

r

r

505.4001.0

1=

f

004505.0=f

( )( ) lbWfF 649.2588004505.0 ===

( )( )fpm

Dnvm 8.261

12

5002

12===

ππ

( )( ) min5.6938.261649.2 lbftFvU mf −===

(c) bbcr tAhQ ∆=

FinsqlbftFftsqhrBtuhcr −−−=−−= ..min18.0..2

( )( ) ..100222525 insqDLAb ===

fUQ =

( )( )( ) 5.69310018.0 =∆ bt

Ftb 53.38=∆

( ) Ftt boa 7753.3822 ==∆=∆

Fto 17710077 =+= , near 160 F

∴ satisfactory.

(d) Fto 177= , reyn6104.3 −×=µ

Figure AF 16

Use SAE 40 oil, reyn6103.3 −×=µ

p

n

c

rS s

r

µ2

=


Page 39 of 63

( )( )187.0

147

33.8103.3

0010.0

1 62

=×

=

−

S

Table AT 20, 1=DL , 187.0=S

4923.0=ro ch

( ) ( )inhinh oo 0005.000049.00010.04923.0 =<==

(e) 25.19=∆

p

tc oρ

( )25.19

147

112=

∆ ot

Fto 3.25=∆

( ) Ftt oo 354177221 ==∆+∆

Ftt oo 3.2512 =∆−∆

3.253542 2 +=∆ ot

FFto 2001902 <=∆

∴ compatible.

(f) 16.4=Lnrc

q

sr

( )( )( )( )16.4

233.8001.01=

q

sec0693.0 3inq =

574. An 8 x 9-in. full bearing (consider 1=DL for table and chart use only)

supports 15 kips with rpmn 1200= ; 0012.0=rcr ; construction is medium

heavy with a radiating-and-convecting area of about DL18 ; air flow about the

bearing of 80 fpm may be counted on (nearby) pulley; ambient temperature is

90 F. Decide upon a suitable minimum film thickness. (a) Compute the

frictional loss and the steady state temperature. Is additional cooling needed

for a reasonable temperature? Determine (b) the temperature rise of the oil as

it passes through the load-carrying area and the grade of oil to be used if it

enters the bearing at 130 F, (c) the quantity of oil needed.

Solution:

.8 inD =

.9 inL =

.000,15 lbW =

.1200 rpmn =

rpsns 2060

1200==

0012.0=rcr


Page 40 of 63

inDr 42 ==

( ) incr 0048.040012.0 ==

By Norton: ( ) inDho 002.0800025.000025.0 ===

4.00048.0

002.0==

r

o

c

h


121.0=S

22.3=fc

r

r

33.4=Lnrc

q

sr

2.14=∆

p

tc oρ

(a) 22.3=fc

r

r

22.30048.0

4=

f

003864.0=f

( )( ) lbWfF 96.57000,15003864.0 ===

( )( )fpm

Dnvm 2513

12

12008

12===

ππ

( )( ) min654,145251396.57 lbftFvU mf −===

bbcr tAhQ ∆=


FinsqlbftD

vh a

c −−−= ..min017.04.0

6.0

( )( )

Finsqlbfthc −−−== ..min103.08

80017.0

4.0

6.0

Finsqlbfthhh rccr −−−=+=+= ..min211.0108.0103.0

( )( ) ..1296981818 insqDLAb ===

QU f =

( )( ) bt∆= 1296211.0654,145

Ftb 533=∆ , very high, additional cooling is necessary.

(b) 2.14=∆

p

tc oρ


Page 41 of 63

( )( )psi

DL

Wp 208

98

000,15===

( )2.14

208

112=

∆ ot

Fto 26=∆

Fti 130=

Fto 156=

( ) Ftave 14315613021 =+=

p

n

c

rS s

r

µ2

=

( )208

20

0048.0

4121.0

2µ

==S

reyn6108.1 −×=µ

Figure AF 16, reynsµµ 8.1= , 143 F

Use SAE 10W

(c) 33.4=Lnrc

q

sr

( )( )( )( )33.4

9200048.04=

q

sec96.14 3inq =

575. A 3.5 x 3.5-in., 360o bearing has 0012.0=rcr ; rpmn 300= ; desired

minimum inho 0007.0≈ . It is desired that the bearing be self-contained (oil-

ring); air-circulation of 80 fpm is expected; heavy construction, so that

DLAb 25≈ . For the first look at the bearing, assume reyn6108.2 −×=µ and

compute (a) the frictional loss (ft-lb/min), (b) the average temperature of the

bearing and oil as obtained for steady-state operation, (c) ot∆ as the oil passes

through the load-carrying area (noting whether comparative values are

reasonable). (d) Select an oil for the steady-state temperature and decide

whether there will be any overheating troubles.

Solution:

.5.3 inD =

.5.3 inL =

0012.0=rcr

.75.12 inDr ==

( )( ) incr 0021.075.10012.0 ==

inho 0007.0≈


Page 42 of 63

333.00021.00007.0 ==ro ch

Table AT 20, 360o Bearing, 1=DL , 333.0=ro ch

0954.0=S

71.2=fc

r

r

12.12=∆

p

tc oρ

(a) p

n

c

rS s

r

µ2

=

rpsns 560

300==

reyn6108.2 −×=µ

( )( )p

S5108.2

0021.0

75.10954.0

62 −×

==

psip 102=

( )( )( ) lbpDLW 12505.35.3102 ===

71.2=fc

r

r

71.20021.0

75.1=

f

00325.0=f

( )( ) lbWfF 0625.4125000325.0 ===

( )( )fpm

Dnvm 275

12

3005.3

12===

ππ

( )( ) min11172750625.4 lbftFvU mf −===

(b) bbcr tAhQ ∆=


FinsqlbftD

vh a

c −−−= ..min017.04.0

6.0

( )( )

Finsqlbfthc −−−== ..min143.05.3

80017.0

4.0

6.0


( )( ) ..25.3065.35.32525 insqDLAb ===

QU f =

( )( ) bt∆= 25.306251.01117


Page 43 of 63

Ftb 5.14=∆

( ) Ftt boa 295.1422 ==∆=∆

assume ambient temperature of 100 F

Ftb 5.114=

Fto 129=

(c) 12.12=∆

p

tc oρ

( )12.12

102

112=

∆ ot

Fto 11=∆

( ) Ftt oo 258129221 ==+

Ftt oo 1112 =−

Fto 2692 2 =

FFto 1401352 <=

∴ reasonable

(d) Fto 129= , reyn6108.2 −×=µ

use SAE 10W

Figure AF 16, Fto 126=

Ftoa 26100126 =−=∆

boa tt ∆=∆ 2

Ftb 132

26==∆

( )( )( )fbbcr UlbfttAhQ <−==∆= min9991325.306251.0

∴ there is an overheating problem.

576. A 10-in. full journal for a steam-turbine rotor that turns 3600 rpm supports a

20-kip load with psip 200= ; 00133.0=rcr . The oil is to have

reyn61006.2 −×=µ at an average oil temperature of 130 F. Compute (a) the

minimum film thickness (comment on its adequacy), (b) the fhp, (c) the

altitude angle, the maximum pressure, and the quantity of oil that passes

through the load-carrying area (gpm).(d) At what temperature must the oil be

introduced in order to have 130 F average? (e) Estimate the amount of heat

lost by natural means from the bearing (considered oil bath) with air speed of

300 fpm. If the amount of oil flow computed above is cooled back to the

entering temperature, how much heat is removed? Is this total amount of heat

enough to care for frictional loss? If not, what can be done (i11.21)?


Page 44 of 63

Solution:

.10 inD =

rpmn 3600=

rpsns 6060

3600==

lbW 000,20=

psip 200=

DL

Wp =

L10

000,20200 =

inL 10=

1=DL

inD

r 52

==

00133.0=rcr

( ) incr 00665.0500133.0 ==

reyn61006.2 −×=µ

Ftave 130=

p

n

c

rS s

r

µ2

=

( )( )35.0

200

601006.2

00665.0

5 62

=×

=

−

S

Table AT 20, 1=DL , 35.0=S

647.0=ro ch

o66.65=φ

433.7=fc

r

r

90.3=Lnrc

q

sr

495.0max

=p

p

8.30=∆

p

tc oρ

446.0=q

qs

(a) ( ) inch ro 00430.000665.0647.0647.0 ===


Page 45 of 63

Norton’s recommendation = ( ) ininD 00430.000250.01000025.000025.0 <==

∴ adequate

(b) 433.7=fc

r

r

433.700665.0

5=

f

0099.0=f

( )( ) lbWfF 198000,200099.0 ===

( )( )fpm

Dnvm 9425

12

360010

12===

ππ

( )( )hp

Fvfhp m 55.56

000,33

9425198

000,33===

(c) o66.65=φ

psip

p 404495.0

200

495.0max ===

Lnrcq sr90.3=

( )( )( )( ) sec805.77106000665.0590.3 3inq ==

( )( )( ) gpmingpminq 21.0minsec602311sec805.77 33 ==

(d) 8.30=∆

p

tc oρ

( )8.30

200

112=

∆ ot

Fto 55=∆

2

oiave

ttt

∆+=

2

55130 += it

Fti 5.102=

(e) bbcr tAhQ ∆=


FinsqlbftD

vh a

c −−−= ..min017.04.0

6.0

( )( )

Finsqlbfthc −−−== ..min207.05.3

300017.0

4.0

6.0


Assume


Page 46 of 63

( )( ) ..250010102525 insqDLAb ===

Ftoa 30100130 =−=∆

boa tt ∆=∆ 3.1

Ftb 233.1

30==∆

( )( )( ) min113,18232500315.0 lbftQ −==

( ) seclbintqqcQ osr −∆−= ρ

( )( )( )( )( )( ) min602,327,16012155805.77446.01112 lbftQr −=−=

min735,345,1602,327,1113,18 lbftQQQ rT −=+=+=

( )( )Tmf QlbftFvU >−=== min150,866,19425198

not enough to care for frictional loss, use pressure feed (i11.21).

DESIGN PROBLEMS

578. A 3.5-in. full bearing on an air compressor is to be designed for a load of 1500

lb.; rpmn 300= ; let 1=DL . Probably a medium running for would be

satisfactory. Design for an average clearance that is decided by considering both

Table 3.1 and 11.1. Choose a reasonable oh , say one that gives 5.0≈ro ch .

Compute all parameters that are available via the Text after you have decided on

details. It is desired that the bearing operate at a reasonable steady-state

temperature (perhaps ring-oiled medium construction), without special cooling.

Specify the oil to be used and show all calculations to support your conclusions.

What could be the magnitude of the maximum impulsive load if the eccentricity

ration ε becomes 0.8, “squeeze” effect ignored?

Solution:

1=DL

inD 5.3=

inL 5.3=

lbW 1500=

rpmn 300=

rpsns 560

300==

( )( )psi

DL

Wp 45.122

5.35.3

1500===

Table 3.1, medium running fit,

inD 5.3=

RC 5 or RC 6

Use RC 6

Average incd 0052.0=


Page 47 of 63

Table 11.1, air-compressor

General Machine Practice

Average incd 0055.0=

Using incd 0055.0=

incr 00275.0=

( ) inch ro 001375.000275.05.05.0 ===


5.0=ε

1925.0=S o84.56=φ

505.4=fc

r

r

16.4=Lnrc

q

sr

25.19=∆

p

tc oρ

4995.0max

=p

p

Specifying oil:

bbcr tAhQ ∆=

mf FvU =

505.4=fc

r

r

505.400275.0

75.1=

f

00708.0=f

( )( ) lbWfF 62.10150000708.0 ===

( )( )fpm

Dnvm 275

12

3005.3

12===

ππ

( )( ) min292127562.10 lbftFvU mf −===

bbcr tAhQ ∆=

Assume Finsqlbfthcr −−−= ..min516.0

Medium construction

( )( ) ..875.1895.35.35.155.15 insqDLAb ===

Oil-ring bearing

boa tt ∆=∆ 2

fUQ =

( )( )( ) 2921875.189516.0 =∆ bt


Page 48 of 63

Ftb 30=∆

( ) Ftt boa 603022 ==∆=∆

assume ambient temperature = 90 F

Fto 150=

p

n

c

rS s

r

µ2

=

( )45.122

5

00275.0

75.11925.0

2µ

==S

reyn6106.11 −×=µ

Figure AF 16, 150 F, reyn6106.11 −×≈µ

Use SAE 70 oil

Maximum load, W with 8.0=ε

Table AT 20, 1=DL

0446.0=S

p

n

c

rS s

r

µ2

=

( )( )p

S5106.11

00275.0

75.10446.0

62 −×

==

psip 527=

( )( )( ) 64565.35.3527 === pDLW

580. A 2500-kva generator, driven by a water wheel, operates at 900 rpm. The weight

of the rotor and shaft is 15,100 lb. The left-hand, 5 –in, full bearing supports the

larger load, lbR 8920= . The bearing should be above medium-heavy

construction (for estimating bA ). (a) Decide upon an average clearance

considering both Table 3.1 and 11.1, and upon a minimum film thickness

( 5.0≈ro ch is on the safer side). (b) Investigate first the possibility of the

bearing being a self-contained unit without need of special cooling. Not much air

movement about the bearing is expected. Then make final decisions concerning

oil-clearance, and film thickness and compute all the parameters given in the text,

being sure that everything is reasonable.

Solution:

rpmn 900=

rpsns 1560

900==

inD 5=

lbRW 8920==


Page 49 of 63

(a) Table 3.1, inD 5=

RC 5, average incd 0051.0=

incr 00255.0=

( ) inch ro 00128.000255.05.05.0 ===

(b) Use 1=DL

inL 5=

inD

r 5.22

==

( )( )psi

DL

Wp 8.356

55

8920===


1925.0=S

505.4=fc

r

r

16.4=Lnrc

q

sr

25.19=∆

p

tc oρ

p

n

c

rS s

r

µ2

=

( )8.356

15

00255.0

5.21925.0

2µ

==S

reyn6108.4 −×=µ

505.4=fc

r

r

505.400255.0

5.2=

f

00460.0=f

( )( ) lbWfF 032.41892000460.0 ===

( )( )fpm

Dnvm 1178

12

9005

12===

ππ

( )( ) min336,481178032.41 lbftFvU mf −===

bbcr tAhQ ∆=

Medium-Heavy

( )( ) ..25.5065525.2025.20 insqDLAb ===

Assume Finsqlbfthcr −−−= ..min516.0

fUQ =


Page 50 of 63

( )( )( ) 336,4825.506516.0 =∆ bt

Ftb 185=∆ , very high

Therefore, special cooling is needed.

25.19=∆

p

tc oρ

( )25.19

8.356

112=

∆ ot

Fto 61=∆

Assume Fti 100=

Ftave 1302

61100 ≈+=

Figure AF 16, reynsµµ 8.4= , 130 F

Select SAE 30 oil. reynsµµ 0.6=

p

n

c

rS s

r

µ2

=

( )( )242.0

8.356

15100.6

00255.0

5.2 62

=×

=

−

S

Table AT 20, 1=DL , 242.0=S

SAE 30 oil at 130 F

569.0=r

o

c

h

o17.61=φ

395.5=fc

r

r

04.4=Lnrc

q

sr

75.22=∆

p

tc oρ

4734.0max

=p

p

Oil, SAE 30

incr 00255.0=

( ) inho 00145.000255.0569.0 ==

PRESSURE FEED

581. An 8 x 8-in. full bearing supports 5 kips at 600 rpm of the journal; .006.0 incr = ;

let the average reyn6105.2 −×=µ . (a) Compute the frictional loss fU . (b) The


Page 51 of 63

oil is supplied under a 40-psi gage pressure with a longitudinal groove at the

point of entry. Assuming that other factors, including fU , remain the same and

that the heat loss to the surrounding is negligible, determine the average

temperature rise of the circulating oil.

Solution:

inL 5=

inD 5=

lbW 5000=

rpmn 600=

rpsns 1060

600==

incr 006.0=

reyn6105.2 −×=µ

1=DL

( )( )psi

DL

Wp 125.78

88

5000===

p

n

c

rS s

r

µ2

=

( )( )1422.0

125.78

10105.2

006.0

4 62

=×

=

−

S

(a) Table AT 20, 1=DL , 1422.0=S

6.3=fc

r

r

, 57.0=ε

6.3006.0

4=

f

0054.0=f

( )( ) lbWfF 2750000054.0 ===

( )( )fpm

Dnvm 1257

12

6008

12===

ππ

( )( ) min940,33125727 lbftFvU mf −===

(b) Longitudinal Groove.

( ) sec5.112

tan3

5.2 3213

inL

rpcq ir ε

π

µ+

= −

psipi 40=

( ) ( )( )

( ) ( )[ ] sec57.05.118

42tan

105.23

40006.05.2 321

6

3

inq +

×= −

−

π


Page 52 of 63

sec41.5 3inq =

of tcqU ∆= ρ

( )( )( ) ( )( ) otftinlbft ∆=− 41.512sec60min112min940,33

Fto 2.11=∆

583. A 4-in. 360o bearing, with 1=DL , supports 2.5 kips with a minimum film of

.0008.0 inho = , .01.0 incd = , .600 rpmn = The average temperature rise of the oil

is to be about 25 F. Compute the pressure at which oil should be pumped into the

bearing if (a) all bearing surfaces are smooth, (b) there is a longitudinal groove at

the oil-hole inlet. (c) same as (a) except that there is a 360o circumferential

groove dividing the bearing into 2-in. lengths.

Solution:

inD 4=

inL 4=

inr 2=

lbW 2500=

incd 010.0=

incr 005.0=

rpmn 600=

rpsns 1060

600==

Fto 25=∆

( )( )psi

DL

Wp 25.156

44

2500===

inho 00080.0=

16.0005.0

0008.0==

r

o

c

h


44.1=fc

r

r

, 84.0=ε

44.1005.0

2=

f

0036.0=f

( )( ) lbWfF 925000036.0 ===

( )( )fpm

Dnvm 628

12

6004

12===

ππ

( )( ) sec1130min56526289 lbinlbftFvU mf −=−===

0343.0=S


Page 53 of 63

p

n

c

rS s

r

µ2

=

( )25.156

10

005.0

20343.0

2µ

==S

reyn61035.3 −×=µ

of tcqU ∆= ρ

( )( )( )251121130 q=

sec404.0 3inq =

(a) Smooth

( ) sec5.112

tan3

3213

inL

rpcq ir ε

π

µ+

= −

( ) ( )( )

( ) ( )[ ] sec84.05.114

22tan

1035.33

005.0404.0 321

6

3

inpi +

×= −

−

π

psipi 5.12=

(b) Longitudinal groove

( ) sec5.112

tan3

5.2 3213

inL

rpcq ir ε

π

µ+

= −

( ) ( )( )

( ) ( )[ ] sec84.05.114

22tan

1035.33

005.05.2404.0 321

6

3

inpi +

×= −

−

π

psipi 5=

(c) Circumferential groove

( ) sec5.113

2 323

inL

prcq ir ε

µ

π+=

( )( ) ( )( )( )

( )[ ] sec84.05.1141035.33

005.022404.0 32

6

3

inpi +

×=

−

π

psipi 5=

BEARING CAPS

584. An 8-in. journal, supported on a 150o partial bearing, is turning at 500 rpm;

bearing length = 10.5 in., incd 0035.0= ., inho 00106.0= . The average

temperature of the SAE 20 oil is 170 F. Estimate the frictional loss in a 160o

cap

for this bearing.

Solution:


Page 54 of 63

inho 00106.0=

incd 0035.0=

incr 00175.0=

inc

hch

r

orav

−+=

2

174.01

( ) inhav 00195.000175.0

00106.0174.0100175.0

2

=

−+=

For SAE 20, 170 F

reyn6107.1 −×=µ

av

ips

h

AvF

µ=

DLA θ2

1=

inD 8=

inL 5.10=

9

8

180

160160

ππθ === o

( )( ) ..3.1175.1089

8

2

1insqA =

=

π

( ) ipsDnv sips 5.20960

5008 =

== ππ

( )( )( )lbF 424.21

00195.0

5.2093.117107.1 6

=×

=−

( )( )fpm

Dnvm 1047

12

5008

12===

ππ

( )( ) sec1130min430,221047424.21 lbinlbftFvU mf −=−===

585. A partial 160o bearing has a 160

o cap; inD 2= .,

inL 2= ., incd 002.0= ., inho 0007.0= ., rpmn 500= , and reyn6105.2 −×=µ .

For the cap only, what is the frictional loss?

Solution:

incd 002.0=

incr 001.0=

inho 0007.0=

7.0001.0

0007.0==

r

o

c

h


Page 55 of 63

3.07.011 =−=−=r

o

c

hε

( ) ( ) ( )[ ] inch rav 001067.03.074.01001.074.0122 =+=+= ε

av

ips

h

AvF

µ=

( )( )fpm

Dnvm 8.261

12

5002

12===

ππ

( ) ipsvips 36.5260

128.261 =

=

( )( )( ) ..585.522180

160

2

1

180

160

2

1insqDLA =

=

= ππ

( )( )( )lbF 685.0

001067.0

36.52585.5105.2 6

=×

=−

( )( ) min3.1798.261685.0 lbftFvU mf −===

586. The central reaction on a 120o partial bearing is 10 kips; inD 8= .,

1=DL ., 001.0=rcr . Let rpmn 400= and reyn6104.3 −×=µ . The bearing has

a 150o cap. (a) For the bearing and the cap, compute the total frictional loss by

adding the loss in the cap to that in the bearing. (b) If the bearing were 360o,

instead of partial, calculate the frictional loss and compare.

Solution:

p

n

c

rS s

r

µ2

=

rpsns 67.660

400==

( )( )psi

DL

Wp 25.156

88

000,10===

( )( )145.0

25.156

67.6104.3

001.0

1 62

=×

=

−

S

(a) Table AT 22, 1=DL , 145.0=S

021.2=fc

r

r

6367.0=ε

021.2=fc

r

r

021.2001.0

1=

f

002021.0=f


Page 56 of 63

( )( ) lbWfF 21.20000,10002021.0 ===

( )( )fpm

Dnvm 838

12

4008

12===

ππ

( )( ) min936,1683821.201 lbftFvU mf −===

CAP:

( )274.01 ε+= rav ch

rcr 001.0=

inD

r 42

==

( ) incr 004.04001.0 ==

( ) ( ) ( )[ ] inch rav 0052.06367.074.01004.074.0122 =+=+= ε

av

ips

h

AvF

µ=

( ) ipsvips 6.16760

12838 =

=

( )( )( ) ..78.8388180

150

2

1

180

150

2

1insqDLA =

=

= ππ

( )( )( )lbF 18.9

0052.0

6.16778.83104.3 6

=×

=−

( )( ) min769383818.92 lbftFvU mf −===

Total Frictional Loss

= min629,247693936,1621 lbftUU ff −=+=+

(b) 360o Bearing, 1=DL , 145.0=S

65.3=fc

r

r

5664.0=ε

BEARING:

65.3001.0

1=

f

00365.0=f

( )( ) lbWfF 5.36000,1000365.0 ===

( )( )fpm

Dnvm 838

12

4008

12===

ππ

( )( ) min587,308385.361 lbftFvU mf −===

CAP:

( )274.01 ε+= rav ch

( ) ( ) ( )[ ] inch rav 00495.05664.074.01004.074.0122 =+=+= ε


Page 57 of 63

av

ips

h

AvF

µ=

( )( )( )lbF 645.9

00495.0

6.16778.83104.3 6

=×

=−

( )( ) min8083838645.92 lbftFvU mf −===


= min670,388083587,3021 lbftUU ff −=+=+

587. The central reaction on a 120o partial bearing is a 10 kips; .8 inD = , 1=DL ,

001.0=rcr ; rpmn 1200= . Let reyn6105.2 −×=µ . The bearing has a 160o cap.

(a) Compute oh and fhp for the bearing and for the cap to get the total fhp. (b)

Calculate the fhp for a full bearing of the same dimensions and compare.

Determine (c) the needed rate of flow into the bearing, (d) the side leakage sq .

(e) the temperature rise of the oil in the bearing both by equation (o), i11.13,

Text, and by Table AT 22. (f) What is the heat loss from the bearing if the oil

temperature is 180 F? Is the natural heat loss enough to cool the bearing? (g) It is

desired to pump oil through the bearing with a temperature rise of 12 F. How

much oil is required? (h) For the oil temperature in (f), what is a suitable oil to

use?

Solution:

p

n

c

rS s

r

µ2

=

rpsns 2060

1200==

( )( )psi

DL

Wp 25.156

88

000,10===

( )( )32.0

25.156

20105.2

001.0

1 62

=×

=

−

S

(a) Table AT 22, 1=DL , 32.0=S

5417.0=ε

4583.0=r

o

c

h

18.3=fc

r

r

60.2=Lnrc

q

sr


Page 58 of 63

305.0=q

qs

834.17=∆

p

tc oρ

38434.0max

=p

p

( )( ) inch ro 00183.04001.04583.04583.0 ===

BEARING:

18.3=fc

r

r

18.3001.0

1=

f

00318.0=f

( )( ) lbWfF 8.31000,1000318.0 ===

( )( )fpm

Dnvm 2513

12

12008

12===

ππ

( )( ) min913,7925138.311 lbftFvU mf −=== , hp42.2

CAP:

( )274.01 ε+= rav ch

rcr 001.0=

inD

r 42

==

( ) incr 004.04001.0 ==

( ) ( ) ( )[ ] inch rav 00487.05417.074.01004.074.0122 =+=+= ε

av

ips

h

AvF

µ=

( ) ipsvips 50360

122513 =

=

( )( )( ) ..36.8988180

160

2

1

180

160

2

1insqDLA =

=

= ππ

( )( )( )lbF 1.23

00487.0

503636.89105.2 6

=×

=−

( )( ) min050,5825131.232 lbftFvU mf −=== , hp76.1


= min963,137050,58913,7921 lbftUU ff −=+=+

hpU

fhpf

18.4000,33

963,137

000,33===

(b) Full Bearing, 1=DL , 32.0=S


Page 59 of 63

Table AT 20

6305.0=r

o

c

h

86.6=fc

r

r

3695.0=ε

( ) inho 002522.0004.06305.0 ==

BEARING:

86.6=fc

r

r

86.6001.0

1=

f

00686.0=f

( )( ) lbWfF 6.68000,1000686.0 ===

( )( ) min392,17225136.681 lbftFvU mf −=== , hp224.5

CAP:

( )274.01 ε+= rav ch

( ) ( ) ( )[ ] inch rav 00440.03695.074.01004.074.0122 =+=+= ε

av

ips

h

AvF

µ=

( )( )( )lbF 54.25

00440.0

50336.89105.2 6

=×

=−

( )( ) min182,64251354.252 lbftFvU mf −=== , hp946.1


= min574,236182,64392,17221 lbftUU ff −=+=+

hpU

fhpf

17.7000,33

574,236

000,33===

(c) 120o Bearing

60.2=Lnrc

q

sr

( )( )( )( )60.2

820004.04=

q

sec656.6 3inq =

(d) 305.0=q

qs

305.0656.6

=sq


Page 60 of 63

sec03.2 3inqs =

(e) Equation(o)

of tcqU ∆= ρ1

sec983,15sec60

12913,79min913,791 lbinlbinlbftU f −=−

=−=

( )( ) of tU ∆== 656.6112983,151

Fto 4.21=∆

Table 22.

834.17=∆

p

tc oρ

834.1725.156

112=

∆ ot

Fto 9.24=∆

(f) bbcr tAhQ ∆=

assume Finsqlbfthcr −−−= ..min516.0

( )( ) ..1600882525 insqDLAb ===

2

oab

tt

∆=∆

assume ambient = 100 F

Ftb 402

100180=

−=∆

( )( )( ) 1min024,33401600516.0 fUlbftQ <−==

Therefore not enough to cool the bearing.

(g) 21 ffr UUQQ +=+

963,137024,33 =+rQ

min939,104 lbftQr −=

sec988,20 lbinQr −=

or tcqQ ∆= ρ

( ) ( )12112988,20 q=

sec62.15 3inq =

(h) Fig. AF 16, 180 F, reyn6105.2 −×=µ

use SAE 30 oil


Page 61 of 63

IMPERFECT LUBRICATION:

588. A 0.5 x 0.75-in. journal turns at 1140 rpm. What maximum load may be

supported and what is the frictional loss if the bearing is (a) SAE Type I, bronze

base, sintered bearing, (b) nylon (Zytel) water lubricated, (c) Teflon, with

intermittent use, (d) one with carbon graphite inserts.

Solution:

(a) 12.0=f

( )( )fpm

Dnvm 23.149

12

11405.0

12===

ππ

000,50=mpv

( ) 000,5023.149 =p

psip 335=

( )( )( ) lbpDLW 12675.05.0335 ===

( )( ) lbWfF 12.1512612.0 ===

( )( ) min225623.14912.15 lbftFvU mf −===

(b) 18.0~14.0=f , use 16.0=f

2500=mpv , water

( ) 250023.149 =p

psip 75.16=

( )( )( ) lbpDLW 28.675.05.075.16 ===

( )( ) lbWfF 005.128.616.0 ===

( )( ) min15023.149005.1 lbftFvU mf −===

(c) fpmvm 100>

25.0=f

000,20=mpv , intermittent

( ) 000,2023.149 =p

psip 134=

( )( )( ) lbpDLW 25.5075.05.0134 ===

( )( ) lbWfF 5625.1225.5025.0 ===

( )( ) min187523.1495625.12 lbftFvU mf −===

(d) 000,15=mpv

( ) 000,1523.149 =p

psip 5.100=

( )( )( ) lbpDLW 69.3775.05.05.100 ===

assume 20.0=f


Page 62 of 63

( )( ) lbWfF 54.769.3720.0 ===

( )( ) min112523.14954.7 lbftFvU mf −===

590. A bearing to support a load of 150 lb at 800 rpm is needed; .1 inD = ; semi-

lubricated. Decide upon a material and length of bearing, considering sintered

metals, Zytel, Teflon, and graphite inserts.

Solution:

( )( )fpm

Dnvm 44.209

12

8001

12===

ππ

assume, inDL 1==

( )( )psi

DL

Wp 150

11

150===

( )( ) 416,3144.209150 ==mpv

Use sintered metal, limit 000,50=mpv

THRUST BEARINGS

592. A 4-in. shaft has on it an axial load of 8000 lb., taken by a collar thrust

bearing made up of five collars, each with an outside diameter of 6 in. The

shaft turns 150 rpm. Compute (a) the average bearing pressure, (b) the

approximate work of friction.

Solution:

(a) ( )

( )psi

D

Wp

o

2836

80004422

===ππ

(b) assume 065.0=f , average

( )( ) lbWfF 5208000065.0 ===

( )( )fpm

Dnvm 81.117

12

1503

12===

ππ

( )( )( ) min306,30681.1175205 lbftnFvU mf −===

593. A 4-in. shaft, turning at 175 rpm, is supported on a step bearing. The bearing

area is annular, with a 4-in. outside diameter and a 3/4 –in. inside diameter.

Take the allowable average bearing pressure as 180 psi. (a) What axial load

may be supported? (b) What is the approximate work of friction?

Solution:

12

Dnvm

π=


Page 63 of 63

( ) inD 375.275.042

1=+=

( )( )fpm

Dnvm 81.108

12

175375.2

12===

ππ

assume 065.0=f , average

(a) ( )22

4

io DD

Wp

−=

π

( ) ( ) lbW 21821804

34

4

2

2=

−=

π

(b) ( )( )( ) min433,1581.1082182065.0 lbftWvfU mf −===

- end -

SECTION 10 - BALL AND ROLLER BEARINGS

Page 1 of 17

601. The radial reaction on a bearing is 1500 lb.; it also carries a thrust of 1000 lb.;

shaft rotates 1500 rpm; outer ring stationary; smooth load, 8-hr./day service, say

15,000 hr. (a) Select a deep-groove ball bearing. (b) What is the rated 90 % life

of the selected bearing? (c) For 34.1=b , compute the probability of the selected

bearing surviving 15,000 hr.

Solution:

lbFx 1500=

lbFy 1000=

( )( )( )( ) mrB 135010150060000,15 6

10 == −

ztxre FCFCF += 56.0

1=rC , outer ring stationary

assume 8.1=tC

( )( ) ( )( ) lbF e 264010008.11500156.0 =+=

( ) ( ) lbFB

BF e

r

r 178,2926401350 3

13

1

10 ==

=

(a) Table 12.3

use 320, lbFr 900,29=

lbFs 900,29=

To check:

0340.0400,29

1000==

s

z

F

F

Table 12.2, 93.1=tC , 2286.0=Q

( )( )Q

FC

F

xr

z >== 667.015000.1

1000

ztxre FCFCF += 56.0

( )( ) ( )( ) lbF e 2770100093.11500156.0 =+=

( ) ( ) lbFB

BF e

r

r 614,3027701350 3

13

1

10 ==

=

2.4 % higher than 29,900 lb. Safe.

Therefore use Bearing 320, Deep-Groove Ball Bearing.

(b) lbFr 900,29=

lbF e 2770=

( ) lbmr

B614,302770

1900,29

3

1

10 =

=

mrB 125810 =


Page 2 of 17

( )( )( )( ) 125810150060 6

10 == −HRB

hrHR 000,14≈

(c)

b

P

P

B

B

1

10

10 1ln

1ln

=

=

9.0

1ln

1ln

10P

mrB 125810 =

mrB 1350=

34.1

1

9.0

1ln

1ln

1258

1350

=P

891.0=P

602. A certain bearing is to carry a radial load of 500 lb. and a thrust of 300 lb. The

load imposes light shock; the desired 90 % life is 10 hr./day for 5 years at

rpmn 3000= . (a) Select a deep-groove ball bearing. What is its bore? Consider

all bearings that may serve. (b) What is the computed rated 90 % life of the

selected bearing? (c) What is the computed probability of the bearing surviving

the specified life? (d) If the loads were changed to 400 and 240 lb., respectively,

determine the probability of the bearing surviving the specified life, and the 90 %

life under the new load.

Solution:

lbFx 500=

lbFz 300=

Assume 1=rC

( )( )6.0

5000.1

300==

xr

z

FC

F

Table 12.2, QFC

F

xr

z >

(a) ztxre FCFCF += 56.0

1=rC

Assume 8.1=tC

( )( ) ( )( ) lbF e 82030093.1500156.0 =+=


Page 3 of 17

For light shock, service factor ~ 1.1

( )( ) lbF e 9028201.1 ==

( ) ( ) lbFB

BF e

r

r 614,3027701350 3

13

1

10 ==

=

( )( )( )( )( )( ) mrB 328510300060103655 6

10 == −

( ) ( ) lbFB

BF e

r

r 409,139023285 3

13

1

10 ==

=

Table 12.3,

Bearing No. rF , lb sF , lb Bore

217 14,400 12,000 85 mm

312 14,100 10,900 60 mm

Select, Bearing No. 312

lbFr 100,14=

lbFs 900,10=

(b) Table 12.2

0285.0900,10

300==

s

z

F

F

99.1=tC

22.0=Q

ztxre FCFCF += 56.0

( )( ) ( )( ) lbF e 87730099.1500156.0 =+=

( )( ) lbF e 9658771.1 ==

e

r

r FB

BF

3

1

10

=

( )9651

100,143

1

10

=

B

mrB 311910 =

( )( )( )( )( )( ) 31191030006010365 6

10 == −YRB

yearsYR 75.4=

(c)

b

P

P

B

B

1

10

10 1ln

1ln

=


Page 4 of 17

use 125.1=b

mrB 311910 =

mrB 3285=

125.1

1

9.0

1ln

1ln

3119

3285

=P

8943.0=P

(d) lbFx 400=

lbFz 240=

1=rC

( )( )6.0

4000.1

240==

xr

z

FC

F

Table 12.2

15.2=tC

6.021.0 <=Q

ztxre FCFCF += 56.0

( )( ) ( )( ) lbF e 74024015.2400156.0 =+=

( )( ) lbF e 8147401.1 ==

e

r

r FB

BF

3

1

10

=

( )8141

100,143

1

10

=

B

mrB 519710 =

b

P

P

B

B

1

10

10 1ln

1ln

=

125.1

1

9.0

1ln

1ln

5197

3285

=P

939.0=P

Life:

( )( )( )( )( )( ) 51971030006010365 6

10 == −YRB


Page 5 of 17

yearsYR 8=

603. The smooth loading on a bearing is 500-lb radial, 100 lb. thrust; rpmn 300= . An

electric motor drives through gears; 8 hr./day, fully utilized. (a) Considering

deep-groove ball bearings that may serve, choose one end specify its bore. For

the bearing chosen, determine (b) the rated 90 % life and (c) the probability of

survival for the design lufe.

Solution:

lbFx 500=

lbFz 100=

Table 12.1, 8 hr./day fully utilized, assume 25,000 hr

( )( )( )( ) mrB 4501030060000,25 6

10 == −

(a) assume 1=rC

( )( )2.0

5000.1

100==

xr

z

FC

F

consider xr

z

FC

FQ >

( )( ) lbFCF xre 5005000.1 ===

( ) ( ) lbFB

BF e

r

r 3832500450 3

13

1

10 ==

=

Table 12.3

Bearing No. rF , lb sF , lb

207 4440 3070

306 4850 3340

305 3660 2390

Select 305, lbFr 3660= , lbFs 2390=

Bore (Table 12.4) = 25 mm

(a) 0418.02390

100==

s

z

F

F

Table 12.2, 26.022.0 Q<

xr

z

FC

FQ >

( )( ) lbFCF xre 5005000.1 ===

( )5001

36603

1

10

=

B

mrB 39210 =

Rated Life:


Page 6 of 17

( )( )( )( ) 3921030060 6

10 == −HRB

hrHR 000,22≈

(c)

b

P

P

B

B

1

10

10 1ln

1ln

=

125.1=b

125.1

1

9.0

1ln

1ln

392

450

=P

884.0=P

605. A No. 311, single-row, deep-groove ball bearing is used to carry a radial load of

1500 lb. at a speed of 500 rpm. (a) What is the 90 % life of the bearing in hours?

What is the approximate median life? What is the probability of survival if the

actual life is desired to be (b) 105 hr., (c) 10

4 hr.?

Solution:

Table 12.3, No. 311

lbFs 9400=

lbFr 12400=

lbFx 1500=

assume 1=rC

( )( ) lbFCF xre 150015001 ===

(a) e

r

r FB

BF

3

1

10

=

( )15001

124003

1

10

=

B

mrB 56510 =

( )( )( )( ) 5651050060 6

10 == −HRB

hrHR 800,18≈

For median life = 5( 90 % life) = ( ) hr000,94800,185 =


Page 7 of 17

(b) ( )( )( )( ) mrB 3000105006010 65 == −

b

P

P

B

B

1

10

10 1ln

1ln

=

125.1=b

125.1

1

9.0

1ln

1ln

565

3000

=P

502.0=P

(c) 104 hr

( )( )( )( ) mrB 300105006010 64 == −

b

P

P

B

B

1

10

10 1ln

1ln

=

125.1=b

125.1

1

9.0

1ln

1ln

565

300

=P

950.0=P

606. The load on an electric-motor bearing is 350 lb., radial; 24 hr. service,

rpmn 1200= ; compressor drive; outer race stationary. (a) Decide upon a deep-

groove ball bearing, giving its significant dimensions. Then compute the selected

bearing’s 90 % life, and the probable percentage of failures that would occur

during the design life. What is the approximate median life of this bearing? (b)

The same as (a), except that a 200 series roller bearing is to be selected.

Solution:

lbFx 350=

xre FCF =

outer race stationary, 1=rC

( )( ) lbFe 3503501 ==


Page 8 of 17

Table 12.1

90 % Life, hrs = 50,000 hrs

( )( )( )( ) mrB 360010120060000,50 6 == −

(a) ( ) ( ) lbFB

BF e

r

r 53643503600 3

13

1

10 ==

=

Table AT 12.3

earing No. rF , lb sF , lb

208 5040 3520

209 5660 4010

306 4850 3340

307 5750 4020

Use No. 209 lbFr 5660=

Table 12.4, Dimension

Bore = 45 mm

O.D. = 85 mm

Width of Races = 19 mm

Max. Fillet r = 0.039 mm

90 % Life:

e

r

r FB

BF

3

1

10

=

( )3501

56603

1

10

=

B

mrB 422910 =

( )( )( )( ) 422910120060 6

10 == −HRB

hrHR 740,58≈

Probability.

b

P

P

B

B

1

10

10 1ln

1ln

=

125.1=b

125.1

1

9.0

1ln

1ln

4229

3600

=P

916.0=P

% failures = 1 – 0.916 = 0.084 = 8.4 %


Page 9 of 17

Median Life = 5(58,740) = 293,700 hrs

(b) Table 12.3, lbFr 5364=

use No. 207, lbFr 5900=

Bore = 35 mm

O.D. = 72 mm

Width of Races = 17 mm

90 % life:

e

r

r FB

BF

3

1

10

=

( )3501

59003

1

10

=

B

mrB 479010 =

( )( )( )( ) 479010120060 6

10 == −HRB

hrHR 530,66≈

Probability.

b

P

P

B

B

1

10

10 1ln

1ln

=

125.1=b

125.1

1

9.0

1ln

1ln

4790

3600

=P

926.0=P

% failures = 1 – 0.926 = 0.074 = 7.4 %

Median Life = 5(66,530) = 332,650 hrs

608. A deep-groove ball bearing on a missile, supporting a radial load of 200 lb., is to

have a design life of 20 hr.; with only a 0.5 % probability of failure while

rpmn 4000= . Using a service factor of 2.1 , choose a bearing. ( A 5- or 6- place

log table is desirable.)

Solution: No need to use log table.

lbFx 200=

assume 1=rC


Page 10 of 17

( )( ) lbFCF xre 2002000.1 ===

( )( ) lbFe 2402002.1 ==

( )( )( )( ) mrB 8.41040006020 6

10 == −

995.0005.01 =−=P

b

P

P

B

B

1

10

10 1ln

1ln

=

125.1=b

125.1

1

10

9.0

1ln

995.0

1ln

8.4

=B

mrB 7210 =

( ) ( ) lbFB

BF e

r

r 4.99824072 3

13

1

10 ==

=

Table 12.3

Select No. 201, lbFr 1180=

VARIABLE LOADS

610. A certain bearing is to carry a radial load of 10 kip at a speed of 10 rpm for 20 %

of the time, a load of 8 kips at a speed of 50 rpm for 50 % of the time, and a load

of 5 kips at 100 rpm during 30 % of the time, with a desired life of 3000 hr.; no

thrust. (a) What is the cubic mean load? (b) What ball bearings may be used?

What roller bearings?

Solution:

(a) 3

1

3

3

32

3

21

3

1

+++=

∑n

nFnFnFFm

L

321 nnnn ++=∑

For 1 min.

( )( ) revn 2102.01 ==

( )( ) revn 25505.02 ==

( )( ) revn 301003.03 ==

revn 5730252 =++=∑

kipsF 101 =


Page 11 of 17

kipsF 82 =

kipsF 53 =

( ) ( ) ( ) ( ) ( ) ( )kipsFm 88.6

57

305258210 3

1333

=

++=

(b) lbkipsFx 688088.6 ==

assume 1=rC

( )( ) lbFe 688068800.1 ==

1 min = 57 rev

( )( )( )( ) mrB 26.101057603000 6

10 == −

( ) ( ) lbFB

BF e

r

r 950,14688026.10 3

13

1

10 ==

=

Table 12.3, Ball Bearing

Use Bearing No. 217, lbFr 400,14=

(c) Table 12.3 (Roller Bearing)

Use Bearing No. 213, lbFr 900,14=

612. A deep-groove ball bearing No. 215 is to operate 30 % of the time at 500 rpm

with lbFx 1200= and lbFz 600= , 55 % of the time at 800 rpm with

lbFx 1000= and lbFz 500= , and 15 % of the time at 1200 rpm with

lbFx 800= and lbFz 400= . Determine (a) the cubic mean load; (b) the 90 % life

of this bearing in hours, (c) the average life in hours.

Solution:

Bearing No. 215, lbFr 400,11= , lbFs 250,9=

Table 12.2, sz FF

At 30 % of the time, 500 rpm

065.09250

600==

s

z

F

F

66.1=tC

266.0=Q

( )( )Q

FC

F

xr

z >== 5.012001

600

( )( ) ( )( ) lbFCFCF ztcre 166860066.11200156.056.01 =+=+=


054.09250

500==

s

z

F

F


Page 12 of 17

73.1=tC

257.0=Q

( )( )Q

FC

F

xr

z >== 5.010001

500

( )( ) ( )( ) lbFCFCF ztcre 142550073.11000156.056.02 =+=+=


043.09250

400==

s

z

F

F

84.1=tC

242.0=Q

( )( )Q

FC

F

xr

z >== 5.08001

400

( )( ) ( )( ) lbFCFCF ztcre 118440084.1800156.056.01 =+=+=

(a) 3

1

3

3

32

3

21

3

1

+++=

∑n

nFnFnFFm

L

321 nnnn ++=∑

lbF 16681 =

lbF 14252 =

lbF 11843 =

For 1 min.

( )( ) revn 1505003.01 ==

( )( ) revn 44080055.02 ==

( )( ) revn 180120015.03 ==

revn 770180440150 =++=∑

( ) ( ) ( ) ( ) ( ) ( )kipsFm 1434

770

180118444014251501668 3

1333

=

++=

(b) lbFF me 1434==

e

r

r FB

BF

3

1

10

=

( )14341

400,113

1

10

=

B

mrB 50310 =

For 1 min = 770 rev


Page 13 of 17

( )( )( )( ) 5031077060 6

10 == −HRB

hrHR 000,11≈

(c) Average life = 5(11,000) = 55,000 hrs

MANUFACTURER’S CATALOG NEEDED

614. A shaft for the general-purpose gear-reduction unit described in 489 has radial

bearing reactions of lbRC 613= and lbRD 1629= ; rpmn 250= . Assume that

the unit will be fully utilized for at least 8 hr./day, with the likelihood of the same

uses involving minor shock. (a) Select ball bearings for this shaft. (b) Select

roller bearings. (c) What is the probability of both bearings C and D surviving for

the design life?

Solution:

Problem 489, ininD 375.18

31 ==

Ref: Design of Machine Members, Doughtie and Vallance

( )rtspolac FKKKKKKF =

at C. lbRF Cr 613==

0.1=tK

0.1=pK

0.1=oK

3

c

ars

N

NKK =

rpmNa 250=

rpmNc 500=

5.1=rK

( )( )90856.0

500

2505.13 ==sK

0.1=aK

3

relc

al

KH

HK =

Table 12.1, 8 hr/day, fully utilized, Text

hrHa 000,25=

hrHc 000,10=

assume 0.1=relK for 90 % reliability

3572.1000,10

000,253 ==lK


Page 14 of 17


( )( )( )( )( )( )( ) lbFc 7566130.190856.00.10.13572.10.1 ==

Table 9-7, Doughtie and Vallance,

Two-row spherical Type, No. 207

Bore = 1.3780 in, lbFc 880=

At D. lbRF Dr 1629==


( )( )( )( )( )( )( ) lbFc 200916290.190856.00.10.13572.10.1 ==

Table 9-7, Doughtie and Vallance,

Two-row spherical Type, No. 407

Bore = 1.3780 in, lbFc 2290=

(b) at C, lbFc 756=

Table 9.8, Doughtie and Vallance

Use No. 207, Bore = 1.3780 in, lbFc 1540=

at C, lbFc 2009=

Table 9.8, Doughtie and Vallance

Use No. 307, Bore = 1.3780 in, lbFc 2660=

(c) For probability:

(c.1) at C, Bearing No. 207, Two-row spherical bearing, lbFc 880=

( ) ( )( )( )( )( )613190856.0111880 lc KlbF ==

58.1=lK

3

relc

al

KH

HK =

3

000,10

000,2558.1

relK=

634.0=relK

Table 9-3, Reference

Probability = 95.8 %

at D, Bearing No. 407, Deep-groove bearing, lbFc 2290=

( ) ( )( )( )( )( )1627190856.01112290 lc KlbF ==

547.1=lK

3

relc

al

KH

HK =


Page 15 of 17

3

000,10

000,25547.1

relK=

675.0=relK



(c.2) at C, Roller Bearing No. 207, lbFc 1540=

( ) ( )( )( )( )( )613190856.01111540 lc KlbF ==

765.2=lK

3

relc

al

KH

HK =

3

000,10

000,25765.2

relK=

118.0=relK



at D, Roller Bearing No. 407, lbFc 2660=

( ) ( )( )( )( )( )1627190856.01112660 lc KlbF ==

80.1=lK

3

relc

al

KH

HK =

3

000,10

000,2580.1

relK=

43.0=relK



615. A shaft similar to that in 478 has the following radial loads on the bearings, left

to right: 803 lb, 988 lb, 84 lb, and 307 lb; no thrust. The minimum shaft diameter

at the bearings are 1.250 in, 1.125 in, 1.000 in, and 1.0625 in. Assume that the

service will not be particularly gentle; intermittently used, with rpmn 425= . (a)

Select ball bearing for this shaft. (b) Select roller bearings.

Solution:

Ref: Design of Machine Members by Doughtie and Vallance


0.1=aK


Page 16 of 17

3

relc

al

KH

HK =

hrHc 000,10=

Table 12.1, Text, hrHa 000,10= (intermittent)

90 % reliability, 0.1=relK

0.1000,10

000,103 ==lK

0.1=oK

0.1=pK

5.1=rK assumed

3

c

ars

N

NKK =

rpmNa 425=

rpmNc 500=

( )( )0844.1

500

4255.13 ==sK

0.1=tK

(a) Ball Bearing

(a.1) 803 lb, inD 250.1=

( )( )( )( )( )( )( ) lbFc 8708030.10844.10.10.10.10.1 ==

Table 9-7, Ref.

Two-row spherical type, 207

lbFc 880=

Bore = 1.3780 in

(a.2) 988 lb, inD 125.1=

( )( )( )( )( )( )( ) lbFc 10719880.10844.10.10.10.10.1 ==

Table 9-7, Ref.

Two-row spherical type, 306

lbFc 1050=

Bore = 1.1811 in

(a.3) 84 lb, inD 000.1=

( )( )( )( )( )( )( ) lbFc 91840.10844.10.10.10.10.1 ==


Page 17 of 17

Table 9-7, Ref.

Deep-groove type, 106

lbFc 544=

Bore = 1.1811 in

(a.4) 307 lb, inD 0625.1=

( )( )( )( )( )( )( ) lbFc 3333070.10844.10.10.10.10.1 ==

Table 9-7, Ref.

Deep-groove type, 106

lbFc 544=

Bore = 1.1811 in

(b) Roller Bearing

(b.1) 803 lb, inD 250.1=

lbFc 870= , Bore = 1.3780 in

use No. 207, lbFc 1540=

(b.2) 988 lb, inD 125.1=

lbFc 1071= , Bore = 1.1811 in

use No. 206, lbFc 1320=

(b.3) 84 lb, inD 000.1=

lbFc 91= , Bore = 1.1811 in

use No. 206, lbFc 1320=

(b.4) 307 lb, inD 0625.1=

lbFc 333= , Bore = 1.1811 in

use No. 206, lbFc 1320=

- end -

SECTION 12 – HELICAL GEARS

Page 1 of 14

DESIGN PROBLEMS

701. For continuous duty in a speed reducer, two helical gears are to be rated at 7.4 hp

at a pinion speed of 1750 rpm; 75.2≈wm ; the helix angle 15o ; 20

o F.D. teeth in

the normal plane; let 21=pN teeth, and keep pDb 2< . Determine the pitch, face,

gN , and the material and heat treatment. Use through-hardened teeth with a

maximum of 250 BHM (teeth may be cut after heat treatment).

Solution: o15=ψ o

n 20=φ

12

pp

m

nDv

π=

dd

p

pPP

ND

21==

rpmnp 1750=

( )

d

d

mP

Pv

9621

12

175021

=

=

π

( )( )d

d

m

t P

P

v

hpF 38.25

9621

4.7000,33000,33=

==

pDb 2≤

dd PPb

42212 =

=

( )

( )lb

CbFv

CbFvFF

tm

tmtd

2

12

2

cos05.0

coscos05.0

ψ

ψψ

++

++=

Table AT 25

Assume 1660=C o15=ψ

lb

PP

P

PP

PPF

d

d

d

d

d

d

dd

2

1

2

2

15cos42

166038.259621

05.0

15cos15cos42

166038.259621

05.0

38.25

++

+

+=


Page 2 of 14

lb

PP

P

PP

PPF

d

d

d

d

d

d

dd

2

1

6505038.25

481

6505038.25

465

38.25

++

+

+=

For continuous service: dw FF ≥

ψ2cos

gp

w

QKbDF =

( )467.1

175.2

75.22

1

2=

+=

+=

g

g

m

mQ

Table At 26, Bhn = 250

Sum of BHN = 500, o

n 20=φ

131=gK

( )( )22

670,181

15cos

131467.12142

ddd

wPPP

F =

=

dw FF ≥

By trial and error method

dP dF wF

7 3967 3708

6 4758 5046

use 6=dP

inP

Dd

p 5.36

2121===

inP

bd

76

4242===

fpmP

vd

m 16046

96219621===

Fig. AF 19, permissible error = 0.0018 in

Fig. AF 20

Use carefully cut gears, 6=dP

Error = 0.001 in is o.k.

For material

Strength

df

sPK

sbYF

ψcos=


Page 3 of 14

2315cos

21

cos 33===

ψp

ep

NN

Table AT 24, Load near middle

23=epN , FDn

o20=φ

565.0=Y

assume 0.2=fK

dsfs FNF =

assume 0.2=sfN

( )( )( )( )

( )( )2475862

15cos565.07=

s

psis 892,29=

use 3

un

ss =

( ) psisu 676,89892,293 ==

Use C1050, OQT 1100 F,

ksisu 122= , 250248 <=BHN

Ans.

6=dP

inb 7=

( )( ) 582175.2 === pwg NmN

Material. C1050, OQT 1100 F

703. A pair of helical gears, subjected to heavy shock loading, is to transmit 50 hp at

1750 rpm of the pinion.; 25.4=gm ; o15=ψ ; minimum .4

34 inDp = ; continuous

service, 24 hr/day; 20o F.D. teeth in the normal plane, carefully cut; through-

hardened to a maximum BHN = 350. Decide upon the pitch, face width, material

and its treatment.

Solution:

( )( )fpmvm 2176

12

175075.4==

π

( )( )( )

lbv

hpF

m

t 7582176

50000,33000,33===

Dynamic load:

( )

( )lb

CbFv

CbFvFF

tm

tmtd

2

12

2

cos05.0

coscos05.0

ψ

ψψ

++

++=

Fig. AF 19, fpmvm 2176=

Permissible error = 0.0014 in


Page 4 of 14

Use carefully cut gears, ine 001.0= , 5=dP as standard

Table AT 25,

Steel and steel, 20o FD

1660=C

( )( )

( ) ( )lb

b

bFd

2

12

2

15cos1660758217605.0

15cos15cos1660758217605.0758

++

++=

( )

( )lb

b

bFd

2

1

8.15487588.108

8.15487581.105758

++

++=

Wear load:

ψ2cos

gp

w

QKbDF =

( )619.1

125.4

25.42

1

2=

+=

+=

g

g

m

mQ

Table At 26, 20o FD,

Sum of BHN =2(350)=700

270=gK

( )( )( )b

bFw 2225

15cos

270619.175.42

==

dw FF ≥ , .69.4tan

22min in

PPb

d

a ===ψ

π


b dF wF

5 5203 11125

6 5811 13350

use inb 5=

Material:

Strength:

dfdnf

sPK

sbY

PK

sbYF

ψcos==

ψ3cos

p

ep

NN =

( )( ) 22375.45 === pdp DPN

2515cos

223

==epN



Page 5 of 14

25=epN , FDn

o20=φ

580.0=Y

assume 7.1=fK

( )( )( )( )

ss

Fs 32955.057.1

15cos580.05==

dsfs FNF =

for 24 hr/day service, heavy shock loading

75.1=sfN

( )( )520375.132955.0 =s

psis 629,27=

use 3

un

ss =

( ) psisu 887,82629,273 ==

Table AT 9

Use 4150, OQT 1200 F,

ksisu 159= , 350331<=BHN

Ans.

5=dP

inb 5=

Material. 4150, OQT 1200 F

705. Design the teeth for two herringbone gears for a single-reduction speed reducer

with 80.3=wm . The capacity is 36 hp at 3000 rpm of the pinion; o30=ψ ; F.D.

teeth with o20=nφ . Since space is at a premium, the initial design is for 15=pN

teeth and carburized teeth of AISI 8620; preferably pDb 2< .

Solution:

dd

p

pPP

ND

15==

pDb 2≈

d

pP

Db30

2 ==

12

pp

m

nDv

π=

( )

d

d

mP

Pv

781,11

12

300015

=

=

π


Page 6 of 14

( )( )d

d

m

t P

P

v

hpF 101

781,11

36000,33000,33=

==

Dynamic load

( )

( )lb

CbFv

CbFvFF

tm

tmtd

2

12

2

cos05.0

coscos05.0

ψ

ψψ

++

++=

o

n 20=φ

o30=ψ

Assume 1660=C , Table AT 25, 20o FD

lb

PP

P

PP

PPF

d

d

d

d

d

d

dd

2

1

2

2

30cos30

1660101781,11

05.0

30cos30cos30

1660101781,11

05.0

101

++

+

+=

lb

PP

P

PP

PPF

d

d

d

d

d

d

dd

2

1

350,37101

589

350,37101

510

101

++

+

+=

Wear load

ψ2cos

gp

w

QKbDF =

( )583.1

180.3

80.32

1

2=

+=

+=

g

g

m

mQ

For AISI 8620, carburized, 20o FD

750=gK for 1010

cycles

( )( )22

350,712

30cos

750583.11530

ddd

wPPP

F =

=

By trial and error, dw FF ≥

dP dF wF

5 4433 28,494

4 5454 44,522

6 3817 19,788

8 3173 11,130

9 3008 8794

For carefully cut gears, 001.0=e

fpmv 1400max = (Fig. AF 9)


Page 7 of 14

dP d

mP

v781,11

=

5 2356.2

4 1963.5

6 1683

8 1473

9 1309 fpm

use 9=dP

lbFd 3008=

dw FlbF >= 5794

inP

bd

3.39

3030===

use inb 0.3=

To check for strength

dfdnf

sPK

sbY

PK

sbYF

ψcos==

ψ3cos

p

ep

NN =

15=pN

2330cos

153

==epN


23=epN , FDn

o20=φ

565.0=Y

assume 7.1=fK

8620, SOQT 450, ksisu 167=

3

un

ss =

5.832

167

2=== u

n

ss

( )( )( )( )( )

( )lbFlbF ds 3008801197.1

30cos565.00.3500,83=>==

Designed Data:

9=dP

inb 0.3=

15=pN

( )( ) 57158.3 === pwg NmN


Page 8 of 14

inP

ND

d

p

p 67.19

15===

inP

ND

d

g

g 33.69

57===

CHECK PROBLEMS

707. The data for a pair of carefully cut gears are: 5=dnP , o20=nφ , o12=ψ ,

.5.3 inb = , 18=pN , 108=gN teeth; pinion turns 1750 rpm. Materials: pinion,

SAE 4150, OQT to BHN = 350; gear, SAE 3150, OQT to BHN = 300. Operation

is with moderate shock for 8 to 10 hr./day. What horsepower may be transmitted

continuously?

Solution:

d

p

pP

ND =

( ) 89.415cos5cos === ψdnd PP

inDp 681.389.4

18==

Wear load

ψ2cos

gp

w

QKbDF =

.5.3 inb =

( )7143.1

10818

10822=

+=

+=

gp

g

NN

NQ

Table AT 26, o20=nφ

Sum of BHN = 350 + 300 = 650

233=gK

( )( )( )( )lbFw 5379

12cos

2337143.1681.35.32

==

Strength of gear

lbPK

sbYF

dnf

s =

For gear: SAE 3150, OQT to BHN = 300

ksisu 151=

( ) ksiss un 5.751515.05.0 ===

11612cos

108

cos 33===

ψg

eg

NN


Page 9 of 14

Table AT 24, Load near middle, o20=nφ

763.0=Y

( )( ) 6.57763.05.75 ==Ysn

For pinion: SAE 4150, OQT to BHN = 350

( ) ksiBHNsu 1753505.05.0 ===

( ) ksiss un 5.871755.05.0 ===

1912cos

18

cos 33===

ψp

ep

NN


534.0=Y

( )( ) 7.46534.05.87 ==Ysn

Therefore use pinion as weak

Assume 7.1=fK

( )( )( )( )( )

lbFs 240,1957.1

534.05.3500,87==

For moderate shock, 8 to 10 hr./day

Use 5.1=sfN

dsfs FNF ≥

dF5.1240,19 =

lbFd 827,12≤

Therefore use lbFF wd 5379==

( )

( )lb

CbFv

CbFvFF

tm

tmtd

2

12

2

cos05.0

coscos05.0

ψ

ψψ

++

++=

Fig. AF 20, carefully cut gears, 5=dnP , ine 001.0=

Table AT 25, steel and steel, 20o FD

1660=C

( )( )fpm

nDv

pp

m 168612

1750681.3

12===

ππ

( ) ( )[ ]( ) ( )[ ]

lb

F

FFF

t

ttd

2

12

2

12cos5.31660168605.0

12cos12cos5.31660168605.0

++

++=

[ ]

[ ]lb

F

FFF

t

ttd 5379

55593.84

555946.82

2

1=

++

++=

Trial and error

lbFt 1800=

( )( )hp

vFhp mt 92

000,33

16861800

000,33===


Page 10 of 14

708. Two helical gears are used in a single reduction speed reducer rated at 27.4 hp at

a motor speed of 1750 rpm; continuous duty. The rating allows an occasional 100

% momentary overload. The pinion has 33 teeth. 10=dnP , .2 inb = , o20=nφ ,

o20=ψ , 82.2=wm . For both gears, the teeth are carefully cut from SAE 1045

with BHN = 180. Compute (a) the dynamic load, (b) the endurance strength;

estimate 7.1=fK . Also decide whether or not the 100 % overload is damaging.

(c) Are these teeth suitable for continuous service? If they are not suitable

suggest a cure. (The gears are already cut.)

Solution:

d

p

pP

ND =

( ) 66.915cos10cos === ψdnd PP

inDp 42.366.9

33==

( )( )fpm

nDv

pp

m 156712

175042.3

12===

ππ

( )lb

v

hpF

m

t 5771567

4.27000,33000,33===

(a) Dynamic load

( )

( )lb

CbFv

CbFvFF

tm

tmtd

2

12

2

cos05.0

coscos05.0

ψ

ψψ

++

++=



1660=C

inb 2=

( ) ( )[ ]( ) ( )[ ]

lbFd 2578

15cos21660577156705.0

15cos15cos21660577156705.0577

2

12

2

=

++

++=

(b) Endurance strength

lbPK

sbYF

dnf

s =

For SAE 1045, BHN = 180

( ) ksiBHNsu 901805.05.0 ===

( ) ksiss un 45905.05.0 ===

3715cos

33

cos 33===

ψp

ep

NN



Page 11 of 14

645.0=Y

7.1=fK

( )( )( )( )( )

lbPK

sbYF

dnf

s 3415107.1

645.02000,45===

For 100 % overload

( ) lbFt 11545772 ==

( )

( )lb

CbFv

CbFvFF

tm

tmtd

2

12

2

cos05.0

coscos05.0

ψ

ψψ

++

++=

( ) ( )[ ]( ) ( )[ ]

lbFd 3475

15cos216601154156705.0

15cos15cos216601154156705.01154

2

12

2

=

++

++=

Since ds FF ≈ , 100 % overload is not damaging

(c) ψ2cos

gp

w

QKbDF =

.2 inb =

( )476.1

182.2

82.22

1

2=

+=

+=

w

w

m

mQ


Sum of BHN = 2(180) = 360

5.62=gK

( )( )( )( ) ( )lbFlbF dw 257867615cos

5.62476.142.322

=<==

Therefore not suitable for continuous service.

Cure: Through hardened teeth

For Bhn

( ) 2385.62676

2578==gK

min Bhn = 0.5(650) = 325

709. Two helical gears are used in a speed reducer whose input is 100 hp at 1200 rpm,

from an internal combustion engine. Both gears are made of SAE 4140, with the

pinion heat treated to a BHN 363 – 415, and the gear to 321 – 363; let the teeth

be F.D.; 20o

pressure angle in the normal plane; carefully cut; helix angle o15=ψ ; 22=pN , 68=gN teeth; 5=dP , inb 4= . Calculate the dynamic load,

the endurance strength load, and the limiting wear load for the teeth. Should these

gears have long life if they operate continuously? (Data courtesy of the Twin

Disc Clutch Co.)

Solution:


Page 12 of 14

inP

ND

d

p

p 4.45

22===

( )( )fpm

nDv

pp

m 138212

12004.4

12===

ππ

( )lb

v

hpF

m

t 23881382

100000,33000,33===

Dynamic load

( )

( )lb

CbFv

CbFvFF

tm

tmtd

2

12

2

cos05.0

coscos05.0

ψ

ψψ

++

++=



1660=C

inb 4=

( ) ( )[ ]( ) ( )[ ]

lbFd 5930

15cos416602388138205.0

15cos15cos416602388138205.02388

2

12

2

=

++

++=

Endurance strength load

lbPK

sbYF

df

s

ψcos=

Assume 7.1=fK

Pinion

( ) ksiBHNsn 75.9036325.025.0 ===

2515cos

22

cos 33===

ψp

ep

NN


580.0=Y

( )( )( )( )( )

lbPK

sbYF

df

s 925,2357.1

15cos580.04750,90cos===

ψ

Gear

( ) ksiBHNsn 25.8032125.025.0 ===

7515cos

68

cos 33===

ψp

ep

NN


735.0=Y

( )( )( )( )( )

lbPK

sbYF

df

s 811,2657.1

15cos735.04250,80cos===

ψ

use lbFs 925,23=


Page 13 of 14

Limiting Wear Load

ψ2cos

gp

w

QKbDF =


Sum of BHN = 684 to 778 use 700

270=gK

( )511.1

6822

6822=

+=

+=

gp

g

NN

NQ

( )( )( )( )lbFw 7696

15cos

270511.14.442

==

Since ( ) ( )lbFlbF dw 59307696 =>= these gears have long life if they operate

continuously.

CROSSED HELICAL

710. Helical gears are to connect two shafts that are at right angles

( 201 =N , 402 =N , 10=dnP , o4521 ==ψψ ). Determine the center distance.

Solution:

1111

1 coscos

ψψπ

DPP

DN dn

cn

==

( )( ) 45cos1020 1D=

inD 83.21 =

222 cosψDPN dn=

( )( ) 45cos1040 2D=

inD 66.52 =

( ) ( ) inDDC 25.466.583.221

2121 =+=+=

712. Two shafts that are at right angles are to be connected by helical gears. A

tentative design is to use 201 =N , 602 =N , 10=dnP , and a center distance of 6

in. What must be the helix angles?

Solution: o9021 =+=Σ ψψ

1

11

cosψdnP

ND =

2

22

cosψdnP

ND =

( )2121 DDC +=


Page 14 of 14

2

2

1

1

coscos2

ψψ dndn P

N

P

NC +=

( )21 cos10

60

cos10

2062

ψψ+=

21 cos

6

cos

212

ψψ+=

21 cos

3

cos

16

ψψ+=


11 sin

3

cos

16

ψψ+=

o5.391 =ψ o5.502 =ψ

- end -

SECTION 13 – BEVEL GEARS

Page 1 of 17

DESIGN PROBLEMS

751. Decide upon the pitch, face, gN , material, and heat treatment of a pair of straight

bevel gears to transmit continuously and indefinitely a uniform loading of 5 hp at

900 rpm of the pinion, reasonable operating temperature, high reliability;

75.1≈gm ; inDp 333.3≈ . Pinion overhangs, gear is straddle mounted.

Solution:

( )2

122

gp rrL +=

75.1

11tan ==

g

pm

γ

o75.29=pγ

pp rL =γsin

2

333.375.29sin =L

inL 358.3=

lbv

hpF

m

t

000,33=

( )( )fpm

nDv

pp

m 4.78512

900333.3

12===

ππ

( )lbFt 210

4.785

5000,33==

( )tmsfd FKNVFF =

( )56.1

50

4.78550

50

50 21

21

=+

=+

= mvVF

One gear straddle, one not

2.1=mK

Table 15.2, uniform

0.1=sfN

( )( )( )( ) lbFd 3932102.10.156.1 ==

Wear load 2

2

2

=

rt

lcdpw

CK

C

C

sbIDF

e

inDp 333.3=

( ) inLb 0.1358.33.03.0 ===

Temperature factor

0.1=tK , reasonable operating temperature

Life factor for wear


Page 2 of 17

0.1=lC for indefinite life

Reliability factor for wear

25.1=rC high reliability

Geometry factor for wear, Fig. 15.7

Assume 080.0=I

Elastic coefficient (Table 15.4)

Steel on steel , 2800=eC

dw FF =

( )( )( ) ( )( ) ( )( )

39325.10.1

0.1

280008.00.1333.3

2

2

2

=

cds

psiscd 370,134=

Table 15.3, use Steel, (300)

ksiscd 135=

Strength of bevel gears

rts

l

d

ds

KKK

K

P

bJsF =

Size factor, assume 71.0=sK

Life factor for strength

1=lK for indefinite life

Temperature factor,

1=tK good operating condition

Reliability factor

5.1=rK high reliability

Geometry factor for strength (Fig. 15.5)

Assume 240.0=J

inb 0.1=

ds = design flexural stress

Min. BHN = 300

ksisd 19=

ds FF =

( )( )( )( )( )( )

3935.1171.0

1240.00.1000,19=

dP

11=dP

say 10=dP

so that inP

bd

0.110

1010===

( )( ) inmDD gpg 833.575.1333.3 ===


Page 3 of 17

( )( ) 33.58833.510 === gdg DPN

say 58=gN

Use 10=dP , inb 0.1= , 58=gN

Material = steel, min. Bhn = 300

752. A pair of steel Zerol bevel gears to transmit 25 hp at 600 rpm of the pinion;

3=gm ; let 20≈pN teeth; highest reliability; the pinion is overhung, the gear

straddle mounted. An electric motor drives a multi-cylinder pump. (a) Decide

upon the pitch, face width, diameters, and steel (with treatment) for intermittent

service. (b) The same as (a) except that indefinite life is desired.

Solution:

dd

p

pPP

ND

20==

( )fpm

P

PnDv

d

dpp

m

ππ

π 1000

12

60020

12=

==

Let dP

b10

=

Dynamic load

( ) tmsfd FKNVFF =

lbv

hpF

m

t

000,33=

( )d

d

t P

P

F 6.2621000

25000,33=

=

π

dd

dm

PP

PvVF

121.11

121.11

50

100050

50

502

1

21

21

+=+=

+

=+

=

π

Table 15.2, electric motor drives a multi-cylinder pump

Service factor, 25.1=sfN

One gear straddle, one not, 2.1=mK

( )( )( )

+=

+=

d

dd

d

dP

PPP

F121.1

13946.2622.125.1121.1

1

(a) Strength of Bevel Gears

rts

l

d

ds

KKK

K

P

bJsF =



Page 4 of 17


Intermittent service, use 6.4=lK

Temperature factor, say 0.1=tK

Reliability factor, highest reliability

0.3=rK

Geometry factor for strength

p

g

gN

Nm =

20=pN

( ) 60203 ==pN

Fig. 15.5, 205.0=J

dPb

10=

Design flexural stress, steel

Assume ksisd 15=

ds FF =

( ) ( )( )

( )( )( )

+=

d

d

d

d

PP

P

P 121.11394

30.171.0

6.4205.010

000,15

+=

d

d

d PP

P

121.11394

408,662

814.4=dP

say 5=dP

inP

bd

0.25

1010===

Wear load for bevel gears 2

2

2

=

rt

lcdpw

CK

C

C

sbIDF

e

inP

ND

d

p

p 45

20===

0.1=tK

Life factor for wear, intermittent service

5.1=lC

Reliability factor for wear, highest reliability

25.1=rC



Page 5 of 17

20=pN , 60=gN

083.0=I

Elastic coefficient, steel on steel (Table 15.4)

2800=eC

5=dP

dw FF =

( )( )( )( ) ( )( )

( )

+=

5

121.115394

25.10.1

5.1

2800083.024

2

2

2

cds

psiscd 730,155=

Table 15.3

Use steel, min. BHN = 360, ksiscd 160=

5=dP

inb 2=

inDp 4=

( )( ) inDmD pgg 1243 ===

steel, min. BHN = 360

(b) For indefinite life,

0.1=lK , life factor for strength

0.1=lC , life factor for wear

Strength:

rts

l

d

ds

KKK

K

P

bJsF =

ds FF =

( ) ( )( )

( )( )( )

+=

d

d

d

d

PP

P

P 121.11394

30.171.0

0.1205.010

000,15

+=

d

d

d PP

P

121.11394

437,142

799.2=dP

say 3=dP

inP

bd

33.33

1010===

Wear load


Page 6 of 17

2

2

2

=

rt

lcdpw

CK

C

C

sbIDF

e

inP

ND

d

p

p 67.63

20===

dw FF =

( )( )( )( ) ( )( )

( )

+=

3

121.113394

25.10.1

0.1

2800083.033.367.6

2

2

2

cds

psiscd 744,113=

Table 15.3

Use steel, min. BHN = 240, ksiscd 115=

3=dP

inb 33.3=

inDp 67.6=

( )( ) inDmD pgg 2067.63 ===

steel, min. BHN = 240

753. Decide upon the pitch, face, and number of teeth for two spiral-bevel gears for a

speed reducer. The input to the pinion is 20 hp at 1750 rpm; 9.1≈gm ; pinion

overhung, gear-straddle mounted. It is hoped not to exceed a maximum pD of 4

3/8-in.; steel gears with minimum 245 BHN on pinion and 210 BHN on gear.

The gear is motor-driven, subject to miscellaneous drives involving moderate

shock; indefinite life against breakage and wear with high reliability. If the gears

designed for the foregoing data are to be subjected to intermittent service only,

how much power could they be expected to transmit?

Solution:

(a) ( )( )

fpmnD

vpp

m 200012

1750375.4

12===

ππ

( )lb

v

hpF

m

t 3302000

20000,33000,33===

Dynamic load

( )tmsfd FKNVFF =


2.1=mK

Table 15.2

Motor-driven, moderate shock


Page 7 of 17

25.1=sfN

21

21

70

70

+= mv

VF , spiral

( )254.1

70

2000702

1

21

=

+=VF

( )( )( )( ) lbFd 6213302.125.1254.1 ==

Wear load 2

2

2

=

rt

lcdpw

CK

C

C

sbIDF

e

inDp 375.4=

Temperature Factor, 0.1=tK

Design contact stresses,

245=BHN , pinion

ksiscd 116=

Life factor for wear

0.1=lC , indefinite life

Reliability factor for wear

25.1=rC , high reliability


Assume 12.0=I

Elastic coefficient, steel on steel (Table 15.4)

2800=eC

( )( )( ) ( )( ) ( )( )

bbFw 72125.10.1

0.1

2800

000,11612.0375.4

2

2

2

=

=

dw FF =

621721 =b

inb 8613.0=

say ininb 875.08

7==

Strength of gear

rts

l

d

ds

KKK

K

P

bJsF =

ds = design flexural stress

min. BHN = 210

ksisd 4.15=



Page 8 of 17


1=lK for indefinite life

Temperature factor,

1=tK

Reliability factor

5.1=rK high reliability

Geometry factor Fig. 15.6

Assume 28.0=J

( )( )( )( )( )( ) dd

sPP

F3543

5.1171.0

128.0875.0400,15=

=

ds FF =

6213543

=dP

7.5=dP

say 6=dP

Then, 6=dP , inb8

7= , ( )( ) 266375.4 === dpp PDN

( )( ) 50269.1 === pwg NmN

(b) Intermittent service only

Strength

rts

l

d

ds

KKK

K

P

bJsF =

psisd 400,15= (Gear)

For 6=dP , 64.0=sK

For indefinite service, 6.4=lK

0.1=tK , 5.1=rK

Geometry factor, Fig. 15.6, 26=pN , 50=gN

292.0=J

( )( )( )( )( )( )

lbFs 31425.1171.0

6.4

6

292.0875.0400,15=

=

Wear load 2

2

2

=

rt

lcdpw

CK

C

C

sbIDF

e

inDp 375.4=

0.1=tK

ksiscd 116=

2800=eC


Page 9 of 17

5.1=lC intermittent service

25.1=rC


26=pN , 50=gN

116.0=I

( )( )( ) ( )( ) ( )( )

lbFw 109825.10.1

5.1

2800

000,116116.0875.0375.4

2

2

2

=

=

use dw FF =

( )tmsfd FKNVFF =

( )( )( ) tF2.1125254.11098 =

lbFt 584=

( )( )hp

vFhp mt 35

000,33

2000584

000,33===

CHECK PROBLEMS

755. A pair of straight-bevel gears are to transmit a smooth load of 45 hp at 500 rpm

of the pinion; 3=gm . A proposed design is .15 inDg = , .8

32 inb = , 4=dP .

Teeth are carburized AISI 8620, SOQT 450 F. The pinion overhangs, the gear is

straddle-mounted. Would these gears be expected to perform with high reliability

in continuous service? If not would you expect more than 1 failure in 100?

Solution:

inm

DD

g

g

p 53

15===

( )( )fpm

nDv

pp

m 65512

5005

12===

ππ

( )lb

v

hpF

m

t 2267655

45000,33000,33===

Dynamic load

( )tmsfd FKNVFF =

( )512.1

50

65550

50

50 21

21

=+

=+

= mvVF


2.1=mK

Smooth load, 0.1=sfN

( )( )( )( ) lbFd 411322672.10.1512.1 ==


Page 10 of 17

Strength of bevel gears

rts

l

d

ds

KKK

K

P

bJsF =

Size factor, for 4=dP ,

71.0=sK


1=lK

Temperature factor,

1=tK

Geometry factor for strength (Fig. 15.5)

( )( ) 2054 === pdp DPN

( )( ) 60154 === gdg DPN

205.0=J

ksisd 30= (55 – 63 Rc) for carburized teeth

( )( )( )( )( )( )

rr

sKK

F5143

171.0

1

4

205.0375.2000,30=

=

ds FF =

41135143

=rK

5.125.1 <=rK will not perform high reliability.

Wear load 2

2

2

=

rt

lcdpw

CK

C

C

sbIDF

e

inDp 5=

inb 375.2=

Table 15.3, ksiscd 225=

Table 15.4, 2800=eC

Geometry factor for wear, Figure 15.7

20=pN , 60=gN

083.0=I

1=tK

life factor for wear 1=lC

( )( )( ) ( )( ) ( )( ) 2

2

2

26364

1

1

2800

000,225083.0375.25

rr

wCC

F =

=

dw FF =


Page 11 of 17

41136364

2=

rC

25.1244.1 ≈=rC , high reliability

Since 5.1<rK ,this will not perform high reliability but 1 in 100, 25.112.1 <≈rK

756. A gear catalog rates a pair of cast-iron, straight-bevel gears at 15.26 hp at 800

rpm of the 16-tooth pinion; 5.3=gm , .3 inb = , 3=dP ; pinion overhangs,

straddle-mounted gear. Assume the cast iron to be class 30. If the load is smooth

is this rating satisfactory, judging by the design approach of the Text for good

reliability (a) when strength alone is considered, (b) when long continuous

service is desired?

Solution:

inP

ND

d

p

p 333.53

16===

( )( )fpm

nDv

pp

m 111712

800333.5

12===

ππ

( )lb

v

hpF

m

t 4511117

26.15000,33000,33===

Dynamic load

( )tmsfd FKNVFF =

( )668.1

50

111750

50

50 21

21

=+

=+

= mvVF


2.1=mK

Smooth load, 0.1=sfN

( )( )( )( ) lbFd 9034512.10.1668.1 ==

(a) Strength

rts

l

d

ds

KKK

K

P

bJsF =

3=dP ,

76.0=sK

1=lK

1=tK

5.1=rK

ksisd 6.4= , cast-iron class 30

16=pN


Page 12 of 17

( )( ) 56165.3 === pwg NmN

184.0=J

( )( )( )( )( )( )

( )ds FlblbF =<=

= 903742

5.1176.0

1

3

184.03600,4

with

4.1=lK for 106 cycles

( )( ) ( )ds FlblbF =>== 90310407424.1

Therefore satisfactory for 106 cycles.

(b) Continuous service

Wear load 2

2

2

=

rt

lcdpw

CK

C

C

sbIDF

e

inDp 333.5=

inb 3=

Table 15.3, ksiscd 50= , cast-iron class 30

Table 15.4, cast-iron and cast-iron 2250=eC

1=lC

1=tK

25.1=rC


16=pN , 56=gN

077.0=I

( )( )( ) ( )( ) ( )( )

( )lbFlbF dw 90338925.11

1

2250

000,50077.03333.5

2

2

2

=<=

=

Therefore, not satisfactory for long continuous service.

757. An 870-rpm motor drives a belt conveyor through bevel gears having 18 and 72

teeth; 6=dP , inb4

31= . Both gears are straddle-mounted. What horsepower may

these gears transmit for an indefinite life with high reliability if both gears are (a)

cast-iron, class 40; (b) AISI 5140, OQT 1000 F; (c) AISI 5140, OQT 1000 F,

flame hardened (d) AISI 8620, SOQT 450 F?

Solution:

inP

ND

d

p

p 36

18===

( )( )fpm

nDv

pp

m 68312

8703

12===

ππ

Dynamic load


Page 13 of 17

( )tmsfd FKNVFF =

Both gears straddle mounted

0.1=mK

Table 15.2, 0.1=sfN

( )523.1

50

68350

50

50 21

21

=+

=+

= mvVF

( )( )( ) ttd FFF 523.10.10.1523.1 ==

(a) Cast-iron, class 40

Strength

rts

l

d

ds

KKK

K

P

bJsF =

ksisd 7= , cast-iron class 40

inb4

31=

1=lK , indefinite life

6=dP

64.0=sK

1=tK

5.1=rK , high reliability

Figure 15.5, 18=pN , 72=gN

204.0=J

( )( )( )( )( )( )

lbFs 4345.1164.0

1

6

204.075.17000=

=

Wear: 2

2

2

=

rt

lcdpw

CK

C

C

sbIDF

e

inDp 3=

inb4

31=

Table 15.3, ksiscd 65= , cast-iron class 40

Table 15.4, cast-iron and cast-iron 2250=eC

1=lC , indefinite life

1=tK

25.1=rC , high reliability


18=pN , 72=gN

082.0=I


Page 14 of 17

( )( )( ) ( )( ) ( )( )

lbFw 23025.11

1

2250

000,65082.075.13

2

2

2

=

=

wd FF =

230523.1 =tF

lbFt 151=

( )( )hp

vFhp mt 3

000,33

683151

000,33===

(b) AISI 5140, OQT 1000 F, BHN = 300

Strength

psisd 000,19=

( )( )( )( )( )( )

lbFs 11785.1164.0

1

6

204.075.1000,19=

=

Wear: 2

2

2

=

rt

lcdpw

CK

C

C

sbIDF

e


Table 15.4, steel and steel, 2800=eC

( )( )( ) ( )( ) ( )( )

lbFw 64025.11

1

2800

000,135082.075.13

2

2

2

=

=

wd FF =

640523.1 =tF

lbFt 420=

( )( )hp

vFhp mt 7.8

000,33

683420

000,33===

(c) AISI 5140, OQT 1000 F, Flame Hardened

Strength

ksisd 5.13=

( )( )( )( )( )( )

lbFs 8375.1164.0

1

6

204.075.1500,13=

=

Wear: 2

2

2

=

rt

lcdpw

CK

C

C

sbIDF

e




Page 15 of 17

( )( )( ) ( )( ) ( )( )

lbFw 126925.11

1

2800

000,190082.075.13

2

2

2

=

=

sd FF =

837523.1 =tF

lbFt 550=

( )( )hp

vFhp mt 4.11

000,33

683550

000,33===

(d) AISI 86200, SOQT 450 F, carburized

Strength

ksisd 30= (55 – 63 Rc)

( )( )( )( )( )( )

lbFs 18595.1164.0

1

6

204.075.1000,30=

=

Wear: 2

2

2

=

rt

lcdpw

CK

C

C

sbIDF

e



( )( )( ) ( )( ) ( )( )

lbFw 177925.11

1

2800

000,225082.075.13

2

2

2

=

=

wd FF =

1779523.1 =tF

lbFt 1168=

( )( )hp

vFhp mt 2.24

000,33

6831168

000,33===

758. A pair of straight-bevel gears transmits 15 hp at a pinion speed of 800 rpm;

5=dP , 20=pN , 60=pN , inb 2= . Both gears are made of AISI 4140 steel,

OQT 800 F. What reliability factor is indicated for these gears for strength and

for wear (a) for smooth loads, (b) for light shock load from the power source and

heavy shock on the driven machine?

Solution:

inP

ND

d

p

p 45

20===

( )( )fpm

nDv

pp

m 83812

8004

12===

ππ


Page 16 of 17

( )lb

v

hpF

m

t 591838

15000,33000,33===

( )tmsfd FKNVFF =

( )579.1

50

83850

50

50 21

21

=+

=+

= mvVF

assume 0.1=mK

( )( )( )( )sfsfd NNF 9335910.1579.1 ==

Strength of bevel gear

rts

l

d

ds

KKK

K

P

bJsF =

For AISI 4140, OQT 800 F, BHN = 429

ksisd 24=

assume 1=lK

1=tK

5=dP

675.0=sK

Figure 15.5, 20=pN , 60=gN

205.0=J

( )( )( )( )( )( )

rr

sKK

F2916

1675.0

1

5

205.02000,24=

=

ds FF =

sf

r

NK

9332916

=

sf

rN

K1254.3

=

Wear load: 2

2

2

=

rt

lcdpw

CK

C

C

sbIDF

e

BHN = 429



inDp 4=

inb 2=

Assume 0.1=lC , 0.1=tK

Fig. 15.7, 20=pN , 60=gN

083.0=I


Page 17 of 17

( )( )( ) ( )( ) ( )( ) 2

2

2

23058

1

1

2800

000,190083.024

rr

wCC

F =

=

wd FF =

2

3058933

r

sfC

N =

sf

rN

C810.1

=

(a) Table 15.2, smooth load

0.1=sfN

For strength, 1254.31

1254.31254.3===

sf

rN

K

For wear, 810.11

810.1810.1===

sf

rN

C

(b) Table 15.2, light shock source, heavy shock driven

0.2=sfN

For strength, 5627.12

1254.31254.3===

sf

rN

K

For wear, 2799.12

810.1810.1===

sf

rN

C

- end -

SECTION 14 – WORM GEARS

Page 1 of 19

DESIGN PROBLEMS

791. (a) Determine a standard circular pitch and face width for a worm gear drive with

an input of 2 hp at 1200 rpm of the triple-threaded worm; the 1.58-in. ( wD ) is

steel with a minimum BHN = 250; gear is manganese bronze (Table AT 3);

12=wm . Consider wear and strength only. Use a nφ to match the lead angle λ .

(See i16.13, Text.) (b) compute the efficiency.

Solution:

a) lbFv

F t

mg

d

+=

1200

1200

mg

tv

hpF

000,33=

12

gg

mg

nDv

π=

rpmm

nn

w

wg 100

12

1200===

λtanwwg DmD =

w

ct

D

PN

πλ =tan

( )( )c

ccwt

w

ctwwg P

PPmN

D

PNDmD 46.11

123===

=

πππ

( )( )c

cmg P

Pv 300

12

10045.11==

π

( )

cc

tPP

F220

300

2000,33==

( )lb

P

P

P

PF

c

c

c

cd

+=

+=

455220

1200

3001200

Wear load

wgw bKDF =

say cPb 2= ,

cg PD 46.11=

dw FF =

( )( )( ) ( )

c

cwcc

P

PKPP

+=

455246.11

( )

c

cwc

P

PKP

+=

45592.22 2


Page 2 of 19

( )( ) cc

ww

gP

P

Dm

D60443.0

58.112

46.11tan ===λ

By trial and error and using Table AT 27 ( λφ ≈n )

wK cP cP (std) λ maxλ i16.11 nφ

36 0.678 ¾ 24.4 16 14 ½

50 0.605 5/8 20.7 25 20

Use o20=nφ , o7.20=λ , inPc8

5=

dw FF =

( )( )( ) ( )

c

cwc

P

PKbP

+=

45546.11

( ) ( )( )

8

5

8

5455

508

546.11

+

=

b

inb 1365.1=

say inb32

51=

To check for strength.

π

λ

π

cosccns

sYbPsYbPF ==

For manganese bronze,

psis 000,30=

o20=nφ

392.0=Y o7.20=λ

inPc8

5=

inb32

51=

( )( )ds FlbF >=

= 2530

7.20cos8

5

32

51392.0000,30

π

use inPc8

5=

inb32

51=


Page 3 of 19

(b)

+

−=

f

fe

n

n

λφ

λφλ

tancos

tancostan

o20=nφ o7.20=λ

( )( )fpmfpm

nDv ww

r 705317.20cos12

120058.1

cos12>===

π

λ

π

( )0334.0

531

32.032.036.036.0

===rv

f

%2.90902.00334.07.20tan20cos

7.20tan0334.020cos7.20tan ==

+

−=e

792. A high-efficiency worm-gear speed reducer is desired, to accept 20 hp from a

1750-rpm motor. The diameter wD of the integral worm has been estimated to be

.8

71 in ; the next computations are to be for a steel worm with a minimum BHN =

250; phosphor-bronze gear (Table AT 3); 11=wm . Probably, the worm should

not have less than 4 threads. (a) Considering wear and strength only (i16.13),

decide upon a pitch and face width that satisfies these requirements (i16.11,

Text); specifying the pressure angle, diameters, and center distance. How does

wD used compare with that from equation (m), i16.11, Text? What addendum

and dedendum are recommended by Dudley? Compute a face length for the

worm. (b) Compute the efficiency. What do you recommend as the next trial for

a “better” reducer?

Solution:

mg

tv

hpF

000,33=

12

gg

mg

nDv

π=

rpmm

nn

w

wg 1.159

11

1750===

( )( )c

ctwcgc

g PPNmPNP

D 14411

====πππ

( )( )c

cmg P

Pv 583

12

1.15914==

π

( )

cc

tPP

F1132

583

20000,33==


Page 4 of 19

lbFv

F t

mg

d

+=

1200

1200

( )lb

P

P

P

PF

c

c

c

cd

4858.0111321132

1200

5831200 +=

+=

(a) Wear

wgw bKDF =

cPb 2= ,

cg PD 14=

dw FF =

( )( )( ) ( )

c

cwcc

P

PKPP

4858.011132214

+=

( )

c

cwc

P

PKP

4858.01113228 2 +

=

Table AT 27, steel, min. BHN = 250, and bronze

And by trial and error ethod

( )( )( ) c

c

w

ct PP

D

PN6791.0

875.1

4tan ===

ππλ

By trial and error and using Table AT 27


36 1.213 1 ¼ 40.33 16 14 ½

50 1.071 1 ¼ 40.33 25 20

60 1.000 1.0 34.18 35 25

Use o25=nφ , o18.34=λ , inPc 1=

dw FF =

( )( )( ) ( )

c

cwc

P

PKbP

4858.01113214

+=

( )( )( )( ) ( )1

4858.01113260114

+=b

inb 2=

To check for strength

π

λ

π

cosccns

sYbPsYbPF ==

For phosphor-bronze,

psiss n 000,31==


Page 5 of 19

For o25=nφ , 470.0=Y

( )( )( )( )ds FlbF >== 7674

18.34cos0.10.2470.0000,31

π, ok

use inPc 0.1=

inb 0.2= o25=nφ

inDw8

71=

( ) inDmD wwg 0.1418.34tan8

7111tan =

== λ

( ) inDDC gw 9375.7148

71

2

1

2

1=

+=+=

Equation (m)

( )ininin

CDw 875.1785.2

2.2

9375.7

2.2

875.0875.0

>=== , ok

Addendum and dedendum (by Dudley)

Addendum = ( ) inPPa ccn 2633.018.34cos0.13183.0cos3183.03183.0 ==== λ

Whole depth = ( ) inPP ccn 5791.018.34cos0.17.0cos7.07.0 === λ

Dedendum = whole depth – addendum = 0.5791 in – 0.2633 in = 0.3158 in

Face length =

+

505.4

g

c

NP

( )( ) 44411 === pwg NmN

Face length = in38.550

445.40.1 =

+

Or

Face length = ( )[ ] 21

222 aDa g −

inDg 14=

ina 2633.0=

Face length = ( ) ( )[ ]{ } in33.52633.02142633.022 21

=−

Use Face length = 5.38 in

(b)

+

−=

f

fe

n

n

λφ

λφλ

tancos

tancostan

( )( )fpmfpm

nDv ww

r 70103818.34cos12

1750875.1

cos12>===

π

λ

π


Page 6 of 19

( )0263.0

1038

32.032.036.036.0

===rv

f ( fpmvr 300070 << )

o25=nφ , o18.34=λ ,

%9494.00263.018.34tan25cos

18.34tan0263.025cos18.34tan ==

+

−=e

recommendation for next trial o30=nφ

o45max =λ

793. The input to a worm-gear set is to be 25 hp at 600 rpm of the worm with

20=wm . The hardened-steel worm is to be the shell type with a diameter

approximately as given in i16.11, Text, and a minimum of 4 threads; the gear is

to be chilled phosphor bronze (Table AT 3). (a) Considering wear and strength

only determine suitable values of the pitch and face width. Let nφ be appropriate

to the value of λ . (b) Compute the efficiency. (c) Estimate the radiating area of

the case and compute the temperature rise of lubricant. Is special cooling needed?

Solution:

mg

tv

hpF

000,33=

12

gg

mg

nDv

π=

rpmm

nn

w

wg 30

20

600===

( )( )ππππ

cctwcgc

g

PPNmPNPD

80420====

( )c

c

mg P

P

v 20012

3080

=

=π

π

( )

cc

tPP

F4125

200

25000,33==

lbFv

F t

mg

d

+=

1200

1200

( )lb

P

P

P

PF

c

c

c

cd

+=

+=

65.6874125

1200

2001200

shell type: inPD cw 1.14.2 +=

( )( )( ) ( )1.14.2

4

1.14.2

4tan

+=

+==

c

c

c

c

w

ct

P

P

P

P

D

PN

πππλ


Page 7 of 19

(a) Wear load

wgw bKDF =

cPb 2= ,

πc

g

PD

80=

dw FF =

( )( ) ( )

c

cwc

c

P

PKP

P +=

65.6872

80

π

( )

c

cwc

P

PKP

+=

65.68793.50 2

Table AT 27, Hardened steel and chilled bronze




90 1.017 1.0 20 16 14 ½

125 0.907 1.0 20 25 20

Use o20=nφ , o20=λ , inPc 1=

dw FF =

( )( ) ( )1

165.1687125

80 +=

b

π

inb 512.1=

say inb8

51=

(b)

+

−=

f

fe

n

n

λφ

λφλ

tancos

tancostan

o20=nφ

o20=λ

inPD cw 5.31.14.21.14.2 =+=+=

( )( )fpm

nDv ww

r 58520cos12

6005.3

cos12===

π

λ

π

( )0323.0

585

32.032.036.036.0

===rv

f ( fpmvr 300070 << )

%23.909023.00323.020tan20cos

20tan0323.020cos20tan ==

+

−=e


Page 8 of 19

(c) Radiating area 7.1

min 2.43 CA =≈ sq. in.

( )gw DDC +=2

1

inDw 5.3=

( )in

PD c

g 5.2518080

===ππ

( ) inC 5.145.255.32

1=+=

( ) ..40725.142.437.1

min insqA ==

Temperature rise = t∆

minlbfttAhQ crc −∆=

( )( ) ( )( ) ( ) min600,80min000,334425.2259023.011 lbfthplbfthphpeQ i −=−−=−=−=

Figure AF 21, ..3.28..4072 ftsqinsqA ==

Finsqlbfthcr −−−= ..min42.0

cQQ =

( )( )( )t∆= 407242.0600,80

Ft 47=∆

with Ft 1001 =

FFt 1501472 <=

Therefore, no special cooling needed.

794. A 50-hp motor turning at 1750 rpm is to deliver its power to a worm-gear

reducer, whose velocity ratio is to be 20. The shell-type worm is to be made of

high-test cast iron; since a reasonably good efficiency is desired, use at least 4

threads; manganese –bronze gear (Table AT 3). (a) Decide upon wD and nφ , and

determine suitable values of the pitch and face width. Compute (b) the efficiency,

(c) the temperature rise of the lubricant. Estimate the radiating area of the case. Is

special cooling needed?

Solution:

mg

tv

hpF

000,33=

12

gg

mg

nDv

π=

rpmm

nn

w

wg 5.87

20

1750===

( )( )ππππ

cctwcgc

g

PPNmPNPD

80420====


Page 9 of 19

( )c

c

mg P

P

v 58312

5.8780

=

=π

π

( )

cc

tPP

F2830

583

50000,33==

(a) lbFv

F t

mg

d

+=

1200

1200

( )lb

P

P

P

PF

c

c

c

cd

+=

+=

06.213752830

1200

5831200

Wear load

wgw bKDF =

cPb 2= ,

πc

g

PD

80=

dw FF =

( )( ) ( )

c

cwc

c

P

PKP

P +=

06.213752

80

π

( )

c

cwc

P

PKP

+=

06.2137593.50 2

w

ct

D

PN

πλ =tan

Shell-type

inPD cw 1.14.2 +=

( )1.14.2

4tan

+=

c

c

P

P

πλ

Table AT 27, high-test cast-iron and manganese bronze



80 1.012 1.0 20 16 14 ½

115 0.885 7/8 19.2 25 20

Use o2.19=λ , o20=nφ , inPc8

7=

inPD cw 2.31.18

74.21.14.2 =+

=+=

dw FF =


Page 10 of 19

( )( )

8

7

8

706.21375

1158

780

+

=

b

π

inb 80.1=

say inb8

71=

(b)

+

−=

f

fe

n

n

λφ

λφλ

tancos

tancostan

o2.19=λ o20=nφ

λ

π

cos12

wwr

nDv =

rpmnw 1750=

inDw 2.3=

( )( )fpm

nDv ww

r 15522.19cos12

17502.3

cos12===

π

λ

π

( )0227.0

1552

32.032.036.036.0

===rv

f ( fpmvr 300070 << )

%73.929273.00227.02.19tan20cos

2.19tan0227.020cos2.19tan ==

+

−=e

(c) ( )( ) ( )( ) min955,119635.3509273.011 lbfthphpeQ i −==−=−=


..2.43 7.1

min insqCAA ==

( )gw DDC +=2

1

inDw 2.3=

inP

D cg 3.22

8

780

80=

==ππ

( ) inC 75.1235.222.32

1=+=

( ) ..327275.122.437.1

insqA ==

Figure AF 1

27.22144

3272ftA ==



Page 11 of 19

cQQ =

( )( )( )t∆= 327243.0955,119

Ft 85=∆

with Ft 1001 =

FFt 1501852 >=

Therefore, special cooling is needed.

CHECK PROBLEMS

795. A worm-gear speed reducer has a hardened-steel worm and a manganese-bronze

gear (Table AT 3); triple-threaded worm with .15278.1 inPc = , .136.3 inDw = ,

o25=nφ , .4

12 inb = , 12=wm , rpmnw 580= . The output is 16 hp. Compute (a)

the dynamic load, (b) the endurance strength of the teeth and the indicated

service factor on strength, (c) the limiting wear load (is it good for indefinitely

continuous service?), (d) the efficiency and input hp, (e) the temperature rise of

the oil (estimate case area as minA , i16.6). (f) Determine the tangential and radial

components of the tooth load. (g) Is this drive self-locking?

Solution:

mg

tv

hpF

000,33=

12

gg

mg

nDv

π=

rpmm

nn

w

wg 3.48

20

580===

( )( )( )in

NmPNPD twcgc

g 21.1331215278.1

====πππ

( )( )fpmvmg 167

12

3.4821.13==

π

(a) t

mg

d Fv

F

+=

1200

1200

td FF

+=

1200

1671200

( )lbFt 3162

167

16000,33==

( ) lbFd 360231621200

1671200=

+=


Page 12 of 19

(b) π

λ

π

cosccns

sYbPsYbPF ==

( )( )( )136.3

15278.13tan

ππλ ==

w

ct

D

PN

o34.19=λ

For manganese-bronze, psiss n 000,30==

For o25=nφ , 470.0=Y

( )( ) ( )lbFs 984,10

34.19cos15278.14

12470.0000,30

=

=π

Service factor

05.33602

984,10===

d

ssf

F

FN

(c) wgw bKDF =

inDg 21.13=

inb 25.2=

Table AT 27, hardened-steel worn and manganese bronze gear o25=nφ

100=wK

( )( )( ) ( )lbFlbF dw 3602297210025.221.13 =<==

Therefore, not good for indefinitely continuous service

(d)

+

−=

f

fe

n

n

λφ

λφλ

tancos

tancostan

( )( )fpm

nDv ww

r 5.18534.19cos12

58015278.1

cos12===

π

λ

π

( )0488.0

5.185

32.032.036.036.0

===rv

f ( fpmvr 300070 << )

%8585.00488.034.19tan25cos

34.19tan0488.025cos34.19tan ==

+

−=e

hphp

e

hphp o

i 82.1885.0

16===

(e) Temperature rise, t∆

( )( ) ( )( )( ) min159,93000,3382.1885.011 lbfthpeQ i −=−=−=


..2.43 7.1

min insqCAA ==


Page 13 of 19

( )gw DDC +=2

1

( ) inC 18.721.1315278.12

1=+=

( ) ..123318.72.437.1

insqA ==

Figure AF 1

26.8144

1233ftA ==


cQQ =

( )( )( )t∆= 123347.0159,93

Ft 161=∆

(f) Tangential components on the worm

lbf

fFW

n

ntt 1305

34.19sin0488.034.19cos25cos

34.19cos0488.034.19sin25cos3162

sincoscos

cossincos=

−

+=

−

+=

λλφ

λλφ

on the gear

lbFt 3162=

radial components

lbf

FS

n

nt 159334.19sin0488.034.19cos25cos

25sin3162

sincoscos

sin=

−=

−=

λλφ

φ

(g) oo 534.19 >=λ , not self-locking

797. A worm-gear speed reducer has a hardened-steel worm and a phosphor-bronze

gear. The lead angle of the 5-threaded worm '5728o=λ , .2812.1 inPc = ,

o25=nφ , .2

12 inb = , 8=wm ; worm speed = 1750 rpm. The gear case is 35 3/8

in. high, 22 in. wide, 14 in. deep. Compute (a) the efficiency, (b) the limiting

wear load, the strength load, and the corresponding safe input and output

horsepowers. (c) The manufacturer rates this reducer at 53-hp input. Is this rating

conservative or risky? (d) What is the calculated temperature rise of the oil with

no special cooling? (e) The manufacturer specifies that for continuous service

power should not exceed 36.5 hp if there is to be no artificial cooling and if t∆ is

to be less than 90 F. Make calculations and decide whether the vendor is on the

safe side. (Data courtesy of the Cleveland Worm Gear Co.)

Solution:


Page 14 of 19

(a)

+

−=

f

fe

n

n

λφ

λφλ

tancos

tancostan

oo 95.28'5728 ==λ o25=nφ

λ

π

cos12

wwr

nDv =

rpmnw 1750=

πcg

g

PND =

( )( ) 4058 === twg NmN

( )( )inDg 31.16

2812.140==

π

w

ct

D

PN

πλ =tan

( )( )

wDπ

2812.1595.28tan =

inDw 686.3=

( )( )fpmvr 1923

'5728cos12

150686.3==

o

π

( )0210.0

1923

32.032.036.036.0

===rv

f ( fpmvr 300070 << )

%75.949475.00210.095.28tan25cos

95.28tan0210.025cos95.28tan ==

+

−=e

(b) wgw bKDF =

inDg 31.16=

inb 5.2=

Table AT 27, hardened-steel worn and phosphor bronze gear o25=nφ

100=wK

( )( )( ) lbFw 40781005.231.16 ==

π

λ

π

cosccns

sYbPsYbPF ==

For phosphor-bronze, psiss n 000,31==

For o25=nφ , 470.0=Y

( )( )( )( )lbFs 000,13

95.28cos2812.15.2470.0000,31==

π


Page 15 of 19

For safe input and output

t

mg

d Fv

F

+=

1200

1200

12

gg

mg

nDv

π=

rpmm

nn

w

wg 75.218

8

1750===

( )( )fpmvmg 934

12

75.21831.16==

π

dw FF =

tF

+=

1200

93412004078

lbFt 2293=

safe output = ( )( )

hpvF

hpmgt

o 9.64000,33

9342293

000,33===

safe input = hpe

hphp o

i 5.689475.0

9.64===

(c) 53-hp input < 68.5 hp. ∴ conservative.

(d) ( )( ) ( )( )( ) min676,118000,335.689475.011 lbfthpeQ i −=−=−=


( )( ) ( )( )[ ] ..5.2172375.352214222 insqA =+=

Figure AF 1

215144

5.2172ftA ==


cQQ =

( )( )( )t∆= 5.217245.0676,118

Ft 4.121=∆

(e) Ft 90=′∆

t

t

hp

ph

i

i

∆

′∆=

′

124

90

5.68=

′iph

hpph i 8.50=′

Since 36.5 hp < 50.8 hp, therefore on the safe side.


Page 16 of 19

HEATING

799. The input to a worm-gear reducer is 50.5 hp at 580 rpm of the 4-threaded worm.

The gear case is 22 x 31 x 45 in. in size; o

n 25=φ , inPc 5.1= , inDw 432.4= ,

035.0=f , room temperature = 80 F. Compute the steady-state temperature for

average cooling.

Solution:

( )( )432.4

5.14tan

ππλ ==

w

ct

D

PN

o3.23=λ

+

−=

f

fe

n

n

λφ

λφλ

tancos

tancostan

9025.0035.03.23tan25cos

3.23tan035.025cos3.23tan =

+

−=e

( )( ) ( )( )( ) min484,162000,335.509025.011 lbfthpeQ i −=−=−=


( )( ) ( )( )[ ] ..4154453131222 insqA =+=

Figure AF 1

285.28144

4154ftA ==


cQQ =

( )( )( )t∆= 415442.0484,162

Ft 93=∆

Ft 801 =

Ft 1732 =

801. A hardened-steel, 4-threaded worm drives a bronze gear; inDw 875.1= ,

inDg 14≈ , inPc 0.1= , o

n 25=φ , area of case ..1500 insq≈ , fpmvr 1037≈ ;

input = 20 hp at 1750 rpm of the worm; room temperature = 80 F. Compute the

steady-state temperature of the lubricant for average ventilation.

Solution:

( )( )875.1

0.14tan

ππλ ==

w

ct

D

PN

o2.34=λ

( )0263.0

1037

32.032.036.036.0

===rv

f ( fpmvr 300070 << )


Page 17 of 19

+

−=

f

fe

n

n

λφ

λφλ

tancos

tancostan

94.00263.02.34tan25cos

2.34tan0263.025cos2.34tan =

+

−=e

( )( ) ( )( )( ) min600,39000,332094.011 lbfthpeQ i −=−=−=


..1500 insqA =

Figure AF 1

24.10144

1500ftA ==


cQQ =

( )( )( )t∆= 150046.0600,39

Ft 57=∆

Ft 801 =

Ft 1372 =

802. The input to a 4-threaded worm is measured to be 20.8 hp; inPc 0.1= , inDw 2= , o

n 25=φ . The area of the case is closely 1800 sq. in.; ambient temperature = 100

F; oil temperature = 180 F. Operation is at a steady thermal state. Compute the

indicated coefficient of friction.

Solution:


Figure AF 1

25.12144

1800ftA ==


..1800 insqA =

Ft 80100180 =−=∆

( )( )( ) min240,6680180046.0 lbfttAhQ crc −==∆=

( )( )( ) min000,331 lbfthpeQ i −−=

cQQ =

( )( )( ) 240,66000,331 =− ihpe

9035.0=e

+

−=

f

fe

n

n

λφ

λφλ

tancos

tancostan


Page 18 of 19

( )( )2

0.14tan

ππλ ==

w

ct

D

PN

o5.32=λ

+

−=

f

f

5.32tan25cos

5.32tan25cos5.32tan9035.0

ff 4059.05774.09035.05217.0 −=+

0425.0=f

FORCE ANALYSIS

804. The input to a 4-threaded worm is 21 hp at 1750 rpm; %90=e , inDw4

12= ,

inDg 14= , 44=gN , o

n 25=φ . (a) From the horsepowers in and out, compute

the tangential forces on the worm tW and the gear tF . (b) Using this value of tF ,

compute tW from equation (k), i16.8, Text. (Check?) (c) Compute the separating

force. (d) What is the end thrust on the worm shaft? On the gear shaft?

Solution:

hphpi 21=

( )( ) ( )( ) hpehphp io 9.1890.021 ===

λtanw

g

t

g

wD

D

N

Nm ==

λtan4

12

14

4

44

=

o5.29=λ

+

−=

f

fe

n

n

λφ

λφλ

tancos

tancostan

+

−=

f

f

5.29tan25cos

5.29tan25cos5.29tan90.0

ff 32.05128.090.04615.0 −=+

0420.0=f

12

gg

mg

nDv

π=

g

t

w

g

N

N

n

n=


Page 19 of 19

44

4

1750=

gn

rpmng 159=

( )( )fpmvmg 583

12

15914==

π

(a) ( )

lbv

hpF

mg

ot 1070

583

9.18000,33000,33===

w

it

v

hpW

000,33=

( )( )fpm

nDv ww

w 103112

175025.2

12===

ππ

( )lbWt 672

1031

21000,33==

(b) lbf

fFW

n

ntt 672

5.29sin0420.05.29cos25cos

5.29cos0420.05.29sin25cos1070

sincoscos

cossincos=

−

+=

−

+=

λλφ

λλφ

(c) lbf

FS

n

nt 5895.29sin0420.05.29cos25cos

25sin1070

sincoscos

sin=

−=

−=

λλφ

φ

(d) End thrust

Worm shaft = lbFt 1070=

Gear shaft = lbWt 672=

- end -

SECTION 15 – FLEXIBLE POWER-TRANSMITTING ELEMENTS

Page 1 of 56

LEATHER BELTS

DESIGN PROBLEMS

841. A belt drive is to be designed for 321 =FF , while transmitting 60 hp at 2700

rpm of the driver 1D ; 85.1≈wm ; use a medium double belt, cemented joint, a

squirrel-cage, compensator-motor drive with mildly jerking loads; center distance

is expected to be about twice the diameter of larger pulley. (a) Choose suitable

iron-pulley sizes and determine the belt width for a maximum permissible

psis 300= . (b) How does this width compare with that obtained by the ALBA

procedure? (c) Compute the maximum stress in the straight port of the ALBA

belt. (d) If the belt in (a) stretches until the tight tension lbF 5251 = ., what is

21 FF ?

Solution:

(a) Table 17.1, Medium Double Ply,

Select inD 71 = . min.

int64

20=

( )( )fpm

nDvm 4948

12

27007

12

11 ===ππ

fpmfpmfpm 600049484000 <<

( )000,33

21 mvFFhp

−=

( )( )000,33

494860 21 FF −

=

lbFF 40021 =−

21 3FF =

lbFF 4003 22 =−

lbF 2002 =

( ) lbFF 60020033 21 ===

sbtF =1

η300=ds

For cemented joint, 0.1=η

psisd 300=

( )( )

==

64

203006001 bF

inb 4.6=

say inb 5.6=


Page 2 of 56

(b) ALBA Procedure

( )( )( )L21

1.17., ffpm CCCbCTableinhphp =

Table 17.1, fpmvm 4948=

Medium Double Ply

448.12=inhp

Table 17.2

Squirrel cage, compensator, starting

67.0=mC

Pulley Size, inD 71 =

6.0=pC

Jerky loads, 83.0=fC

( )( )( )( )( )83.06.067.0448.1260 bhp ==

inb 5.14=

say inb 15=

(c)

( )( )psi

bt

Fs 128

64

20151

6001 =

==

η

(d) ( ) ( )2

1

2

12

1

22

1

12

1

2006002 +=+= FFFo

lbFo 2.373=

lbF 5251 =

( ) ( ) 2

1

22

1

2

1

5252.3732 F+=

lbF 2472 =

1255.2247

525

2

1 ==F

F

842. A 20-hp, 1750 rpm, slip-ring motor is to drive a ventilating fan at 330 rpm. The

horizontal center distance must be about 8 to 9 ft. for clearance, and operation is

continuous, 24 hr./day. (a) What driving-pulley size is needed for a speed

recommended as about optimum in the Text? (b) Decide upon a pulley size (iron

or steel) and belt thickness, and determine the belt width by the ALBA tables. (c)

Compute the stress from the general belt equation assuming that the applicable

coefficient of friction is that suggested by the Text. (d) Suppose the belt is

installed with an initial tension inlbFo 70= . (§17.10), compute 21 FF and the

stress on the tight side if the approximate relationship of the operating tensions

and the initial tensions is 2

1

2

1

22

1

1 2 oFFF =+ .


Page 3 of 56

Solution:

fpmtovm 45004000=

assume fpmvm 4250=

12

11nDvm

π=

( )12

17504250 1Dπ

=

inD 26.91 =

say inD 101 =

(b) Using Heavy Double Ply Belt, int64

23=

Minimum pulley diameter for fpmvm 4250≈ , inD 101 =

Use inD 101 =

( )( )fpm

nDvm 4581

12

175010

12

11 ===ππ

ALBA Tables

( )( )( )L21


8.13=inhp

Slip ring motor, 4.0=mC

Pulley Size, inD 101 =

7.0=pC

Table 17.7, 24 hr/day, continuous

8.1=sfN

Assume 74.0=fC

( )( ) ( )( )( )( )( )74.07.04.08.13208.1 bhp ==

inb 59.12=

use inb 13=

(c) General belt equation

−

−=−

θ

θρf

f

s

e

evsbtFF

1

2.32

12 2

21

fpsvs 35.7660

4581==

..035.0 inculb=ρ for leather

int64

23=

inb 13=


Page 4 of 56

( )( )lbFF 260

4581

208.1000,3321 ==−

3.0=f on iron or steel

C

DD 12 −±≈ πθ

ftC 9~8= use 8.5 ft

( ) inD 5310330

17502 =

=

( )rad72.2

125.8

1053=

−−= πθ

( )( ) 816.072.23.0 ==θf

5578.011

816.0

816.0

=−

=−

e

e

e

ef

f

θ

θ

( ) ( )( ) ( )5578.02.32

35.76035.012

64

2313260

2

21

−

==− sFF

psis 176=

(d) 2

1

2

1

22

1

1 2 oFFF =+

( )( ) lbininlbFo 9101370 ==

lbFF 26021 =−

lbFF 26012 −=

( ) ( ) 33.609102260 2

1

2

1

12

1

1 ==−+ FF

lbF 10451 =

lbF 78526010452 =−=

( )psi

bt

Fs 224

64

2313

10451 =

==

331.1785

1045

2

1 ==F

F

843. A 100-hp squirrel-cage, line-starting electric motor is used to drive a Freon

reciprocating compressor and turns at 1140 rpm; for the cast-iron motor pulley,

inD 161 = ; inD 532 = , a flywheel; cemented joints;l ftC 8= . (a) Choose an

appropriate belt thickness and determine the belt width by the ALBA tables. (b)

Using the design stress of §17.6, compute the coefficient of friction that would be

needed. Is this value satisfactory? (c) Suppose that in the beginning, the initial

tension was set so that the operating 221 =FF . Compute the maximum stress in

a straight part. (d) The approximate relation of the operating tensions and the


Page 5 of 56

initial tension oF is 2

1

2

1

22

1

1 2 oFFF =+ . For the condition in (c), compute oF . Is it

reasonable compared to Taylor’s recommendation?

Solution:

(a) Table 17.1

( )( )fpm

nDvm 4775

12

114016

12

11 ===ππ

Use heavy double-ply belt

int64

23=

1.14=inhp

( )( )( )L21


line starting electric motor , 5.0=mC

Table 17.7, squirrel-cage, electric motor, line starting, reciprocating compressor

4.1=sfN

inD 161 = , 8.0=pC

assume, 74.0=fC

( )( ) hphp 1401004.1 ==

( )( )( )( )( )74.08.05.01.14140 bhp ==

inb 5.33=

use inb 34=

(b) §17.6, η400=ds

00.1=η for cemented joint.

psisd 400=

−

−=−

θ

θρf

f

s

e

evsbtFF

1

2.32

12 2

21

fpsvs 6.7960

4775==


int64

23=

inb 34=

( )( )lbFF 968

4775

1004.1000,3321 ==−

( ) ( )( )

−

−

==−

θ

θ

f

f

e

eFF

1

2.32

6.79035.012400

64

2334968

2

21


Page 6 of 56

2496.01

=−θ

θ

f

f

e

e

28715.0=θf

C

DD 12 −±≈ πθ

ftC 8=

( )rad7562.2

128

1653=

−−= πθ

( ) 28715.07562.2 =f

3.01042.0 <=f

Therefore satisfactory.

(c) lbFF 96821 =−

21 2FF =

lbFF 9682 22 =−

( ) lbFF 193696822 21 ===

( )psi

bt

Fs 159

64

2334

19361 =

==

(d) lbF 19361 = , lbF 9682 =

2

1

22

1

12

1

2 FFFo +=

( ) ( )2

1

2

12

1

96819362 +=oF

lbFo 1411=

inlbFo 5.4134

1411== of width is less than Taylor’s recommendation and is reasonable.

844. A 50-hp compensator-started motor running at 585 rpm drives a reciprocating

compressor for a 40-ton refrigerating plant, flat leather belt, cemented joints. The

diameter of the fiber driving pulley is 13 in., inD 702 = ., a cast-iron flywheel;

.11.6 inftC = Because of space limitations, the belt is nearly vertical; the

surroundings are quite moist. (a) Choose a belt thickness and determine the width

by the ALBA tables. (b) Using recommendations in the Text, compute s from

the general belt equation. (c) With this value of s , compute 1F and 21 FF . (d)

Approximately, 2

1

2

1

22

1

1 2 oFFF =+ , where oF is the initial tension. For the

condition in (c), what should be the initial tension? Compare with Taylor, §17.10.

(e) Compute the belt length. (f) The data are from an actual drive. Do you have

any recommendations for redesign on a more economical basis?


Page 7 of 56

Solution:

(a) ( )( )

fpmnD

vm 294412

86513

12

11 ===ππ

Table 17.1, use Heavy Double Ply,

inD 9min = for fpmvm 2944=

belts less than 8 in wide

int64

23=

( )( )( )L21


86.9=inhp

Table 17.2

67.0=mC

8.0=pC

( )( ) 592.080.074.0 ==fC

Table 17.7, electric motor, compensator-started (squirrel cage) and reciprocating

compressor

4.1=sfN

( )( ) hphp 70504.1 ==

( )( )( )( )( )592.08.067.086.970 bhp ==

inb 4.22=

use inb 25=

(b) General Belt Equation

−

−=−

θ

θρf

f

s

e

evsbtFF

1

2.32

12 2

21

inb 25=

int64

23=


fpsvs 1.4960

2944==

Leather on iron, 3.0=f

C

DD 12 −−= πθ

( )rad35.2

126

1370=

−−= πθ

( )( ) 705.035.23.0 ==θf


Page 8 of 56

5059.011

705.0

705.0

=−

=−

e

e

e

ef

f

θ

θ

( )( )lbFF 785

2944

504.1000,3321 ==−

( ) ( )( ) ( )5059.02.32

1.49035.012

64

2325785

2

21

−

==− sFF

psis 204=

Cemented joint, 0.1=η

psis 204=

(c) ( )( ) lbsbtF 183364

23252041 =

==

lbF 104878518332 =−=

749.11048

1833

2

1 ==F

F

(d) 2

1

22

1

12

1

2 FFFo +=

( ) ( )2

1

2

12

1

104818332 +=oF

lbFo 1413=

inlbFo 5.5625

1413==

Approximately less than Taylor’s recommendation ( = 70 lb/in.)

(e) ( ) ( )C

DDDDCL

457.12

2

1212

−+++≈

( )( ) ( ) ( )( )( )

inL 2861264

1370137057.11262

2

=−

+++=

(f) More economical basis

12

11nDvm

π=

( )12

8654500 1Dπ

=

inD 87.191 =

use inD 201 =


Page 9 of 56

CHECK PROBLEMS

846. An exhaust fan in a wood shop is driven by a belt from a squirrel-cage motor that

runs at 880 rpm, compensator started. A medium double leather belt, 10 in. wide

is used; inC 54= .; inD 141 = . (motor), inD 542 = ., both iron. (a) What

horsepower, by ALBA tables, may this belt transmit? (b) For this power,

compute the stress from the general belt equation. (c) For this stress, what is

21 FF ? (d) If the belt has stretched until psis 200= on the tight side, what is

21 FF ? (e) Compute the belt length.

Solution:

(a) For medium double leather belt

int64

20=

( )( )fpm CCCbinhphp =

Table 17.1 and 17.2

67.0=mC

8.0=pC

74.0=fC

inb 10=

( )( )fpm

nDvm 3225

12

88014

12

11 ===ππ

6625.6=inhp

( )( )( )( )( ) hphp 43.2674.08.067.0106625.6 ==

(b)

−

−=−

θ

θρf

f

s

e

evsbtFF

1

2.32

12 2

21

inb 10=

int64

20=

..035.0 inculb=ρ

fpsvs 75.5360

3225==

C

DD 12 −−= πθ

rad4.254

1454=

−−= πθ

Leather on iron 3.0=f

( )( ) 72.04.23.0 ==θf


Page 10 of 56

51325.011

72.0

72.0

=−

=−

e

e

e

ef

f

θ

θ

( )lbFF 270

3225

43.26000,3321 ==−

( ) ( )( ) ( )51325.02.32

75.53035.012

64

2010270

2

21

−

==− sFF

psis 206=

(c) ( )( ) lbsbtF 64464

20102061 =

==

lbF 3742706442 =−=

72.1374

644

2

1 ==F

F

(d) psis 200=

( )( ) lbsbtF 62564

20102001 =

==

lbF 3552706252 =−=

76.1355

625

2

1 ==F

F

(e) ( ) ( )C

DDDDCL

457.12

2

1212

−+++≈

( ) ( ) ( )( )

inL 222544

1454145457.1542

2

=−

+++=

847. A motor is driving a centrifugal compressor through a 6-in. heavy, single-ply

leather belt in a dusty location. The 8-in motor pulley turns 1750 rpm;

inD 122 = . (compressor shaft); ftC 5= . The belt has been designed for a net

belt pull of inlbFF 4021 =− of width and 321 =FF . Compute (a) the

horsepower, (b) the stress in tight side. (c) For this stress, what needed value of

f is indicated by the general belt equation? (d) Considering the original

data,what horsepower is obtained from the ALBA tables? Any remarks?

Solution:

(a) ( )( )

fpmnD

vm 366512

17508

12

11 ===ππ

inb 6=

( )( ) lbFF 24064021 ==−


Page 11 of 56

( ) ( )( )hp

vFFhp m 65.26

000,33

3665240

000,33

21 ==−

=

(b) 21 3FF =

lbFF 2403 22 =−

lbF 1202 =

lbF 3601 =

bt

Fs 1=

For heavy single-ply leather belt

int64

13=

( )psis 295

64

136

360=

=

(c)

−

−=−

θ

θρf

f

s

e

evsbtFF

1

2.32

12 2

21

..035.0 inculb=ρ

fpsvs 1.6160

3665==

lbFF 24021 =−

( ) ( )( )

−

−

==−

θ

θ

f

f

e

eFF

1

2.32

1.61035.012295

64

136240

2

21

7995.01

=−θ

θ

f

f

e

e

C

DD 12 −−= πθ

( )rad075.3

125

812=

−−= πθ

9875.4=θfe

607.1=θf

( ) 607.1075.3 =f

5226.0=f

(d) ALBA Tables (Table 17.1 and 17.2)

( )( )fpm CCCbinhphp =

fpmvm 3665=

965.6=inhp


Page 12 of 56

inb 10=

0.1=mC (assumed)

6.0=pC

74.0=fC

( )( )( )( )( ) hphphp 65.266.1874.06.00.16965.6 <==

848. A 10-in. medium double leather belt, cemented joints, transmits 60 hp from a 9-

in. paper pulley to a 15-in. pulley on a mine fab; dusty conditions. The

compensator-started motor turns 1750 rpm; inC 42= . This is an actual

installation. (a) Determine the horsepower from the ALBA tables. (b) Using the

general equation, determine the horsepower for this belt. (c) Estimate the service

factor from Table 17.7 and apply it to the answer in (b). Does this result in better

or worse agreement of (a) and (b)? What is your opinion as to the life of the belt?

Solution:

( )( )fpm

nDvm 4123

12

17509

12

11 ===ππ

(a) ( )( )fpm CCCbinhphp =

Table 17.1 and 17.2

Medium double leather belt

int64

20=

fpmvm 4123=

15.11=inhp

67.0=mC

7.0=pC

74.0=fC

inb 10=

( )( )( )( )( ) hphp 7.3874.07.067.01015.11 ==

(b)

−

−=−

θ

θρf

f

s

e

evsbtFF

1

2.32

12 2

21

inb 10=

..035.0 inculb=ρ

η400=s

0.1=η cemented joint

psis 400=

C

DD 12 −−= πθ


Page 13 of 56

rad9987.242

915=

−−= πθ

Leather on paper pulleys, 5.0=f

( )( ) 5.19987.25.0 ==θf

77687.01

=−θ

θ

f

f

e

e

fpsvs 72.6860

4123==

( ) ( )( ) ( ) lbFF 82277687.02.32

72.68035.012400

64

2010

2

21 =

−

=−

( ) ( )( )hp

vFFhp m 7.102

000,33

4123822

000,33

21 ==−

=

(c) Table 17.7

6.1=sfN

hphphp 7.1022.646.1

7.102<==

Therefore, better agreement

Life of belt, not continuous, hphp 7.3860 > .

MISCELLANEOUS

849. Let the coefficient of friction be constant. Find the speed at which a leather belt

may transmit maximum power if the stress in the belt is (a) 400 psi, (b) 320 psi.

(c) How do these speeds compare with those mentioned in §17.9, Text? (d)

Would the corresponding speeds for a rubber belt be larger or smaller? (HINT:

Try the first derivative of the power with respect to velocity.)

Solution:

−

−=−

θ

θρf

f

s

e

evsbtFF

1

2.32

12 2

21

( )000,33

21 mvFFhp

−=

( )000,33

60 21 svFFhp

−=

−

−=

θ

θρf

f

ss

e

evs

btvhp

1

2.32

12

000,33

60 2

ss

f

f

vv

se

ebthp

−

−=

2.32

121

000,33

60 2ρθ

θ


Page 14 of 56

( )( )

02.32

24

2.32

121

000,33

60 22

=

−

−

−= ss

f

f

s

vvs

e

ebt

vd

hpd ρρθ

θ

2.32

36 2

svs

ρ=

..035.0 inculb=ρ

(a) psis 400=

( )2.32

035.036400

2

sv=

fpsvs 105.101=

fpmvm 6066=

(b) psis 320=

( )2.32

035.036320

2

sv=

fpsvs 431.90=

fpmvm 5426=

(c) Larger than those mentioned in §17.9 (4000 – 4500 fpm)

(d) Rubber belt, ..045.0 inculb=ρ

(a) psis 400=

( )2.32

045.036400

2

sv=

fpsvs 166.89=

fpmfpmvm 60665350 <=

Therefore, speeds for a rubber belt is smaller.

850. A 40-in. pulley transmits power to a 20-in. pulley by means of a medium double

leather belt, 20 in. wide; ftC 14= , let 3.0=f . (a) What is the speed of the 40-in

pulley in order to stress the belt to 300 psi at zero power? (b) What maximum

horsepower can be transmitted if the indicated stress in the belt is 300 psi? What

is the speed of the belt when this power is transmitted? (See HINT in 849).

Solution:

−

−=−

θ

θρf

f

s

e

evsbtFF

1

2.32

12 2

21


Page 15 of 56

( )000,33

60 21 svFFhp

−=

ss

f

f

vv

se

ebthp

−

−=

2.32

121

000,33

60 2ρθ

θ

( )( )

02.32

24

2.32

121

000,33

60 22

=

−

−

−= ss

f

f

s

vvs

e

ebt

vd

hpd ρρθ

θ

2.32

36 2

svs

ρ= for maximum power

(a) At zero power:

2.32

12 2

svs

ρ=

psis 300=

..035.0 inculb=ρ

( )2.32

035.012300

2

sv=

fpsvs 6575.151=

fpmvm 9100=

Speed, 40 in pulley, ( )( )

rpmD

vn m 869

40

91001212

2

2 ===ππ

(b) Maximum power

2.32

36 2

svs

ρ=

( )2.32

035.036300

2

sv=

fpsvs 5595.87=

fpmvm 5254=

ss

f

f

vv

se

ebthp

−

−=

2.32

121

000,33

60 2ρθ

θ

int64

20=

inb 20=

C

DD 12 −−= πθ

( )rad0225.3

1214

2040=

−−= πθ

3.0=f


Page 16 of 56

( )( ) 90675.00225.33.0 ==θf

5962.01

=−θ

θ

f

f

e

e

( )( ) ( )( ) ( ) 64.1185595.87

2.32

5595.87035.0123005962.0

000,33

64

202060 2

=

−

=hp

fpmvm 5254=

AUTOMATIC TENSION DEVICES

851. An ammonia compressor is driven by a 100-hp synchronous motor that turns

1200 rpm; 12-in. paper motor pulley; 78-in. compressor pulley, cast-iron;

inC 84= . A tension pulley is placed so that the angle of contact on the motor

pulley is 193o and on the compressor pulley, 240

o. A 12-in. medium double

leather belt with a cemented joint is used. (a) What will be the tension in the

tight side of the belt if the stress is 375 psi? (b) What will be the tension in the

slack side? (c) What coefficient of friction is required on each pulley as indicated

by the general equation? (d) What force must be exerted on the tension pulley to

hold the belt tight, and what size do you recommend?

Solution:

(a) sbtF =1

inb 12=

int64

20=

( )( )

=

64

20123751F

(b) mv

hpFF

000,3321 =−

( )( )fpm

nDvm 3770

12

120012

12

11 ===ππ

Table 17.7, 2.1=sfN

( )( )lbFF 1050

3770

1002.1000,3321 ==−

lbFF 35610501406105012 =−=−=

(c)

−

−=−

θ

θρf

f

s

e

evsbtFF

1

2.32

12 2

21

fpsvs 83.6260

3770==


Page 17 of 56

..035.0 inculb=ρ

( ) ( )( )

−

−

=

θ

θ

f

f

e

e 1

2.32

83.62035.012375

64

20121050

8655.01

=−θ

θ

f

f

e

e

006.2=θf

Motor pulley

rad3685.3180

193193 =

==

πθ o

( ) 006.23685.3 =f

5955.0=f

Compressor Pulley

rad1888.4180

2402403 =

==

πθ o

( ) 006.21888.4 =f

4789.0=f

(d) Force:

Without tension pulley

radC

DD356.2

84

1278121 =

−−=

−−= ππθ

radC

DD9273.3

84

1278122 =

−+=

−+= ππθ


Page 18 of 56

o5.356197.0

2

356.2356.23685.3

2

1111 ==

−−−=

−−−′= rad

πθπθθα

o5.376544.09273.31888.42

9273.3

222

22 ==−+

−=−′+

−= rad

πθθ

πθα

( ) ( ) lbFQ 16725.37sin5.35sin1406sinsin 211 =+=+= αα of force exerted

Size of pulley; For medium double leather belt,

fpmvm 3770= , width = inin 812 >

inD 826 =+=


Page 19 of 56

852. A 40-hp motor, weighing 1915 lb., runs at 685 rpm and is mounted on a pivoted

base. In Fig. 17.11, Text, ine 10= ., inh16

319= . The center of the 11 ½-in.

motor pulley is 11 ½ in. lower than the center of the 60-in. driven pulley;

inC 48= . (a) With the aid of a graphical layout, find the tensions in the belt for

maximum output of the motor if it is compensator started. What should be the

width of the medium double leather belt if psis 300= ? (c) What coefficient of

friction is indicated by the general belt equation? (Data courtesy of Rockwood

Mfg. Co.)

Solution:

(a)

lbR 1915=

Graphically

inb 26≈

ina 9≈

[ ]∑ = 0BM

bFaFeR 21 +=

( )( ) ( )( ) ( )( )269191510 21 FF +=

150,19269 21 =+ FF

For compensator started

( ) ( ) hphpratedhp 56404.14.1 ===

mv

hpFF

000,3321 =−


Page 20 of 56

( )( )fpm

nDvm 2062

12

6855.11

12

11 ===ππ

( )lbFF 896

2062

56000,3321 ==−

89612 −= FF

Substituting

( ) 150,19896269 11 =−+ FF

lbF 12131 =

lbF 31789612132 =−=

For medium leather belt, int64

20=

sbtF =1

( )( )

=

64

203001213 b

inb 13=

(c)

−

−=−

θ

θρf

f

s

e

evsbtFF

1

2.32

12 2

21

fpsvs 37.3460

2062==

..035.0 inculb=ρ

( ) ( )( )

−

−

=

θ

θ

f

f

e

e 1

2.32

37.34035.012300

64

2013896

775.01

=−θ

θ

f

f

e

e

492.1=θf

radC

DD1312.2

48

5.116012 =−

−=−

−= ππθ

( ) 492.11312.2 =f

70.0=f

853. A 50-hp motor, weighing 1900 lb., is mounted on a pivoted base, turns 1140 rpm,

and drives a reciprocating compressor; in Fig. 17.11, Text, ine4

38= .,

inh16

517= . The center of the 12-in. motor pulley is on the same level as the

center of the 54-in. compressor pulley; inC 40= . (a) With the aid of a graphical

layout, find the tensions in the belt for maximum output of the motor if it is

compensator started. (b) What will be the stress in the belt if it is a heavy double


Page 21 of 56

leather belt, 11 in. wide? (c) What coefficient of friction is indicated by the

general belt equation? (Data courtesy of Rockwood Mfg. Co.)

Solution:

(a) For compensator-started

( ) hphp 70504.1 ==

mv

hpFF

000,3321 =−

( )( )fpm

nDvm 3581

12

114012

12

11 ===ππ

( )lbFF 645

2062

70000,3321 ==−

inb 25≈

ina 5≈

lbR 1900=

bFaFeR 21 +=

( )( ) ( ) ( )255190075.8 21 FF +=

lbFF 33255 21 =+

lbFF 33255645 22 =++

lbF 4472 =

lbFF 1092447645645 21 =+=+=

(b) For heavy double leather belt

int64

23=

inb 11=


Page 22 of 56

( )psi

bt

Fs 276

64

2011

10921 =

==

(c)

−

−=−

θ

θρf

f

s

e

evsbtFF

1

2.32

12 2

21

fpsvs 68.5960

3581==

..035.0 inculb=ρ

( ) ( )( )

−

−

=

θ

θ

f

f

e

e 1

2.32

68.59035.012276

64

2311645

241.1=θf

radC

DD092.2

40

125412 =−

−=−

−= ππθ

( ) 492.1092.2 =f

60.0=f

RUBBER BELTS

854. A 5-ply rubber belt transmits 20 horsepower to drive a mine fan. An 8-in., motor

pulley turns 1150 rpm; inD 362 = ., fan pulley; ftC 23= . (a) Design a rubber

belt to suit these conditions, using a net belt pull as recommended in §17.15,

Text. (b) Actually, a 9-in., 5-ply Goodrich high-flex rubber belt was used. What

are the indications for a good life?

Solution:

(a) ( )

o174040.31223

83612 ==−

−=−

−= radC

DDππθ

976.0=θK

2400

θKNbvhp

pm=

976.0=θK

( )( )fpm

nDvm 2409

12

11508

12

11 ===ππ

5=pN

( )( )( )2400

976.05240920

bhp ==

inb 1.4=

min. inb 5=


Page 23 of 56

(b) With inb 9= is safe for good life.

855. A 20-in., 10-ply rubber belt transmits power from a 300-hp motor, running at 650

rpm, to an ore crusher. The center distance between the 33-in. motor pulley and

the 108-in. driven pulley is 18 ft. The motor and crusher are so located that the

belt must operate at an angle 75o with the horizontal. What is the overload

capacity of this belt if the rated capacity is as defined in §17.15, Text?

Solution:

2400

pmNbvhp =

inb 20=

( )( )fpm

nDvm 5616

12

65033

12

11 ===ππ

10=pN

( )( )( )hphp 468

2400

10561620==

Overlaod Capacity = ( ) %56%100300

300468=

−

V-BELTS

NOTE: If manufacturer’s catalogs are available, solve these problems from catalogs as

well as from data in the Text.

856. A centrifugal pump, running at 340 rpm, consuming 105 hp in 24-hr service, is to

be driven by a 125-hp, 1180-rpm, compensator-started motor; intoC 4943= .

Determine the details of a multiple V-belt drive for this installation. The B.F.

Goodrich Company recommended six C195 V-belts with 14.4-in. and 50-in.

sheaves; inC 2.45≈ .

Solution:

Table 17.7

4.12.02.1 =+=sfN (24 hr/day)

Design hp = sfN (transmitted hp) = ( )( ) hp1751254.1 =

Fig. 17.4, 175 hp, 1180 rpm

inD 13min = , D-section

4.14

50

340

1180

1

2 ==D

D

use ininD 134.141 >=

inD 502 =


Page 24 of 56

( )( )fpm

nDvm 4449

12

11804.14

12

11 ===ππ

36

2

1

09.03

1010

10 mm

dm

vve

DK

c

vahpRated

−−

=

Table 17.3, D-section

788.18=a , 7.137=c , 0848.0=e

Table 17.4, 47.31

2 =D

D

14.1=dK

( )( )( ) ( )

hphpRated 294.2810

4449

10

44490848.0

4.1414.1

7.137

4449

10788.18

36

209.03

=

−−

=

Back to Fig. 17.14, C-section must be used.

792.8=a , 819.38=c , 0416.0=e

36

2

1

09.03

1010

10 mm

dm

vve

DK

c

vahpRated

−−

=

( )( )( ) ( )

hphpRated 0.2010

4449

10

44490416.0

4.1414.1

819.38

4449

10792.8

36

209.03

=

−−

=

Adjusted rated hp = ( )hpratedKK Lθ

Table 17.5,

77.046

4.145012 =−

=−

C

DD

88.0=θK

Table 17.6

( ) ( )C

DDDDCL

457.12

2

1212

−+++≈

( ) ( ) ( )( )

inL 200464

4.14504.145057.1462

2

=−

+++=

use C195, inL 9.197=

07.1=LK

Adjusted rated hp = ( )( )( ) hp83.182007.188.0 =

beltshpratedAdjusted

hpDesignbeltsofNo 3.9

83.18

175. === use 9 belts

Use 9 , C195 V-belts with 14.4 in and 50 in sheaves


Page 25 of 56

( )16

322

12

2 DDBBC

−−+=

( ) ( ) ( ) inDDLB 2.3874.145028.69.197428.64 12 =+−=+−=

( ) ( )inC 9.44

16

4.1450322.3872.38722

=−−+

=

857. A 50-hp, 1160-rpm, AC split-phase motor is to be used to drive a reciprocating

pump at a speed of 330 rpm. The pump is for 12-hr. service and normally

requires 44 hp, but it is subjected to peak loads of 175 % of full load; inC 50≈ .

Determine the details of a multiple V-belt drive for this application. The Dodge

Manufacturing Corporation recommended a Dyna-V Drive consisting of six

5V1800 belts with 10.9-in. and 37.5-in. sheaves; inC 2.50≈ .

Solution:

Table 17.7, (12 hr/day)

2.12.04.1 =−=sfN

Design hp = ( )( )( ) hp1055075.12.1 =

Fig. 17.4, 105 hp, 1160 rpm


2.13

4.46

330

1160

1

2 ≈=D

D

use ininD 132.131 >=

inD 4.462 =

( )( )fpm

nDvm 4009

12

11602.13

12

11 ===ππ

36

2

1

09.03

1010

10 mm

dm

vve

DK

c

vahpRated

−−

=


788.18=a , 7.137=c , 0848.0=e

Table 17.4, 5.32.13

4.46

1

2 ==D

D

14.1=dK

( )( )( ) ( )

hphpRated 32.2410

4009

10

40090848.0

2.1314.1

7.137

4009

10788.18

36

209.03

=

−−

=

Back to Fig. 17.14, C-section must be used.

792.8=a , 819.38=c , 0416.0=e

inD 9min =


Page 26 of 56

1.9

32

330

1160

1

2 ≈=D

D

use inD 1.91 =

( )( )fpm

nDvm 2764

12

11601.9

12

11 ===ππ

( )( )( ) ( )

hphpRated 96.1010

2764

10

27640416.0

1.914.1

819.38

2764

10792.8

36

209.03

=

−−

=


Table 17.5,

458.050

1.93212 =−

=−

C

DD

935.0=θK

Table 17.6

( ) ( )C

DDDDCL

457.12

2

1212

−+++≈

( ) ( ) ( )( )

inL 167504

1.9321.93257.1502

2

=−

+++=

use C158, inL 9.160=

02.1=LK

Adjusted rated hp = ( )( )( ) hp45.1096.1002.1935.0 =


hpDesignbeltsofNo 10

43.10

105. ===

( )16

322

12

2 DDBBC

−−+=

( ) ( ) ( ) inDDLB 5.3851.93228.69.160428.64 12 =+−=+−=

( ) ( )inC 8.46

16

1.932325.3855.38522

=−−+

=

Use 10-C158 belts, inD 1.91 =

inD 322 = , inC 8.46=

858. A 200-hp, 600-rpm induction motor is to drive a jaw crusher at 125 rpm; starting

load is heavy; operating with shock; intermittent service; intoC 123113= .

Recommend a multiple V-flat drive for this installation. The B.F. Goodrich

Company recommended eight D480 V-belts with a 26-in. sheave and a 120.175-

in. pulley; inC 3.116≈ .

Solution:


Page 27 of 56

Table 17.7

4.12.06.1 =−=sfN

( )( ) hphp 2802004.1 ==

Fig. 17.14, 280 hp, 600 rpm

Use Section E

But in Table 17.3, section E is not available, use section D

13min =D

8.4125

600

1

2 ==D

D

For max1D :

121

2min D

DDC +

+=

111

2

8.4113 D

DD+

+=

inD 281 =

2min DC =

inD 1132 =

inD 5.238.4

1131 ==

use ( ) inD 185.23132

11 =+≈

( )( ) inD 4.86188.42 ==

( ) ( )C

DDDDCL

457.12

2

1212

−+++≈

( ) ( ) ( )( )

inL 4101184

184.86184.8657.11182

2

=−

+++=

using inD 191 = , inD 2.912 = , inC 118=

( ) ( ) ( )( )

inL 4201184

192.91192.9157.11182

2

=−

+++=

Therefore use D420 sections

inD 191 = , inD 2.912 =

( )( )fpm

nDvm 2985

12

60019

12

11 ===ππ

36

2

1

09.03

1010

10 mm

dm

vve

DK

c

vahpRated

−−

=


788.18=a , 7.137=c , 0848.0=e

Table 17.4, 8.41

2 =D

D


Page 28 of 56

14.1=dK

( )( )( ) ( )

hphpRated 6.2910

2985

10

29850848.0

1914.1

7.137

2985

10788.18

36

209.03

=

−−

=

Therefore, Fig. 17.14, section D is used.


Table 17.5,

612.0118

192.9112 =−

=−

C

DD

83.0=θK (V-flat)

Table 17.6, D420

inL 8.420=

12.1=LK



hpDesignbeltsofNo 10

52.27

280. ===

Use10 , D420, inD 191 = , inD 2.912 = , inC 118=

859. A 150-hp, 700-rpm, slip-ring induction motor is to drive a ball mill at 195 rpm;

heavy starting load; intermittent seasonal service; outdoors. Determine all details

for a V-flat drive. The B.F. Goodrich Company recommended eight D270 V-

belts, 17.24-in sheave, 61-in. pully, inC 7.69≈ .

Solution:

Table 17.7,

4.12.06.1 =−=sfN

Design hp = ( )( ) hp2101504.1 =

Fig. 17.4, 210 hp, 700 rpm


36

2

1

09.03

1010

10 mm

dm

vve

DK

c

vahpRated

−−

=

For Max. Rated hp, ( )

0

103

=

mvd

hpd

3

33

1

91.0

3 101010

−

−

= mm

d

m ve

v

DK

cvahpRated

Let 310

mvX =


Page 29 of 56

3

1

91.0eXX

DK

caXhp

d

−−=

( )3

1

3

11

3 1012

700

101210 ×=

×==

DnDvX m ππ

π700

1012 3

1

XD

×=

3

3

91.0

1012

700eX

K

caXhp

d

−×

−=π

( )( )

0391.0 209.0 =−= − eXaXXd

hpd

e

aX

3

91.009.2 =


788.18=a , 7.137=c , 0848.0=e

( )( )0848.03

788.1891.0

10

09.2

3

09.2 =

= mv

X

fpmvm 7488=

748812

11 ==nD

vm

π

( )7488

12

7001 ==D

vm

π

inD 86.401 =

max inD 86.401 =

ave. ( ) inD 93.2686.40132

11 =+=

use inD 221 =

22

79

195

700

1

2 ≈=D

D

inD 221 = , inD 792 =

Min. inDDD

C 5.72222

7922

21

21 =++

=++

=

Or Min. inDC 792 ==

( ) ( )C

DDDDCL

457.12

2

1212

−+++≈

( ) ( ) ( )( )

inL 327794

2279227957.1792

2

=−

+++=

use D330, inL 8.330=

( )16

322

12

2 DDBBC

−−+=


Page 30 of 56

( ) ( ) ( ) inDDLB 689227928.68.330428.64 12 =+−=+−=

( ) ( )inC 12.81

16

22793268968922

=−−+

=

( )( )fpm

nDvm 4032

12

70022

12

11 ===ππ

14.1=dK

( )( )( ) ( )

hphpRated 124.3910

4032

10

40320848.0

2214.1

7.137

4032

10788.18

36

209.03

=

−−

=


Table 17.5,

70.012.81

227912 =−

=−

C

DD

84.0=θK (V-flat)

Table 17.6

D330

07.1=LK




165.35

210. === use 6 belts

Use 6 , D330 V-belts , inD 221 = , inD 792 = , inC 1.81≈

860. A 30-hp, 1160-rpm, squirrel-cage motor is to be used to drive a fan. During the

summer, the load is 29.3 hp at a fan speed of 280 rpm; during the winter, it is 24

hp at 238 rpm; inC 5044 << .; 20 hr./day operation with no overload. Decide

upon the size and number of V-belts, sheave sizes, and belt length. (Data

courtesy of The Worthington Corporation.)

Solution:

Table 17.7

8.12.06.1 =+=sfN

Design hp = ( )( ) hp54308.1 =

Speed of fan at 30 hp

( ) rpmn 286238238280243.29

24302 =+−

−

−=

at 54 hp, 1160 rpm. Fig. 17.4

use either section C or section D

Minimum center distance:

2DC =


Page 31 of 56

or 121

2D

DDC +

+=

056.4286

1160

1

2 ==D

D

use 1056.4 DC =

inCin 5044 << , use inC 47=

inD 6.11056.4

47max1 ==

use C-section, inD 9min =

Let inD 101 = , inD 411 =

( ) ( )C

DDDDCL

457.12

2

1212

−+++≈

( ) ( ) ( )( )

inL 3.179474

1.10411.104157.1472

2

=−

+++=

use C137, inL 9.175=

( )16

322

12

2 DDBBC

−−+=

( ) ( ) ( ) inDDLB 7.3281.104128.69.175428.64 12 =+−=+−=

( ) ( )ininC 442.45

16

1.1041327.3827.38222

≈=−−+

=

C173, satisfies inCin 5044 << 3

33

1

91.0

3 101010

−

−

= mm

d

m ve

v

DK

cvahpRated

( )( )fpm

nDvm 3067

12

11601.10

12

11 ===ππ

Table 17.4

056.41

2 =D

D, 14.1=dK


792.8=a , 819.38=c , 0416.0=e

( )( )( ) ( )

hphpRated 838.1210

3067

10

30670416.0

1.1014.1

819.38

3067

10792.8

36

209.03

=

−−

=


Table 17.5,

68.02.45

1.104112 =−

=−

C

DD

90.0=θK


Page 32 of 56

Table 17.6

9.175=L , C173

04.1=LK




02.12

54. === use 5 belts

Use 5 , C173 V-belts , inD 1.101 = , inD 412 =

POWER CHAINS

NOTE: If manufacturer’s catalogs are available, solve these problems from catalogs as

well as from data in the Text.

861. A roller chain is to be used on a paving machine to transmit 30 hp from the 4-

cylinder Diesel engine to a counter-shaft; engine speed 1000 rpm, counter-shaft

speed 500 rpm. The center distance is fixed at 24 in. The cain will be subjected to

intermittent overloads of 100 %. (a) Determine the pitch and the number of

chains required to transmit this power. (b) What is the length of the chain

required? How much slack must be allowed in order to have a whole number of

pitches? A chain drive with significant slack and subjected to impulsive loading

should have an idler sprocket against the slack strand. If it were possible to

change the speed ratio slightly, it might be possible to have a chain with no

appreciable slack. (c) How much is the bearing pressure between the roller and

pin?

Solution:

(a) ( ) hphpdesign 60302 == intermittent

2500

1000

2

1

1

2 ==≈n

n

D

D

12 2DD =

inD

DC 242

12 =+=

242

2 11 =+

DD

inD 6.91 =

( ) inDD 2.196.922 12 ===

( )( )fpm

nDvm 2513

12

10006.9

12

11 ===ππ

Table 17.8, use Chain No. 35,

Limiting Speed = 2800 fpm


Page 33 of 56

Minimum number of teeth

Assume 211 =N

422 12 == NN

[Roller-Bushing Impact]

8.0

5.1100

Pn

NKhp ts

r

=

Chain No. 35

inP8

3=

21=tsN

rpmn 1000=

29=rK

( )hphp 3.40

8

3

1000

2110029

8.05.1

=

=

[Link Plate Fatigue] P

ts PnNhp07.039.008.1004.0 −=

( ) ( ) hphp 91.28

3100021004.0

8

307.03

9.008.1=

=

−

No. of strands = 2191.2

60==

hprated

hpdesign

Use Chain No. 35, inP8

3= , 21 strands

(b) ( )

C

NNNNCL

4022

2

1221 −+

++≈ pitches

64

8

3

24=

=C

211 =N

422 =N

( ) ( )( )

pitchespitchesL 16067.1596440

2142

2

4221642

2

≈=−

++

+=

Amount of slack

( )2

122433.0 LSh −=

inCL 24==


Page 34 of 56

( )in

in

inS 062.242

8

367.159160

24 =

−

+=

( ) ( )[ ] ininh4

375.024062.24433.0 2

122

==−=

(c) bp = bearing pressure

Table 17.8, Chain No. 25

inC 141.0=

inE16

3=

inJ 05.0=

( ) ( ) 204054.005.0216

3141.02 inJECA =

+=+=

hpFV

60000,33

=

( )hp

F60

000,33

2513=

lbF 9.787=

strandlbF 5.3721

9.787==

psipb 92504054.0

5.37==

862. A conveyor is driven by a 2-hp high-starting-torque electric motor through a

flexible coupling to a worm-gear speed reducer, whose 35≈wm , and then via a

roller chain to the conveyor shaft that is to turn about 12 rpm; motor rpm is 1750.

Operation is smooth, 8 hr./day. (a) Decide upon suitable sprocket sizes, center

distance, and chain pitch. Compute (b) the length of chain, (c) the bearing

pressure between the roller and pin. The Morse Chain Company recommended

15- and 60-tooth sprockets, 1-in. pitch, inC 24= ., pitchesL 88= .

Solution:

Table 17.7

0.12.02.1 =−=sfN (8 hr/day)

( ) hphpdesign 0.220.1 ==

rpmn 5035

17501 ==

rpmn 122 =

Minimum number of teeth = 12


Page 35 of 56

Use 121 =N

[Link Plate Fatigue] P

ts PnNhp07.039.008.1004.0 −=

( ) ( )0.1

5012004.0

0.2

004.09.008.19.008.1

307.03 ===≈−

nN

hpPP

ts

P

Use Chain No. 80, inP 0.1=

To check for roller-bushing fatigue

8.0

5.1100

Pn

NKhp ts

r

=

29=rK

( ) ( ) hphphp 2274711000

1210017

8.0

5.1

>=

=

(a) 121 =N

( ) teethNn

nN 5012

12

501

2

12 =

=

=

2

12

DDC +=

( )( )in

PND 82.3

120.111 ==≈

ππ

( )( )in

PND 92.15

500.112 ==≈

ππ

inC 83.172

82.392.15 =+≈

use inC 18=

pitchesC 18=

chain pitch = 1.0 in, Chain No. 80

(b) ( )

C

NNNNCL

4022

2

1221 −+

++≈

( ) ( )( )

691840

1250

2

5912182

2

=−

++

+≈L pitches

use pitchesL 70=


Table 17.8, Chain No. 80

inC 312.0=

inE8

5=


Page 36 of 56

inJ 125.0=

( )( )( )fpm

nPNv ts

m 5012

50121

12

1 ===

( ) ( ) 204054.005.0216

3141.02 inJECA =

+=+=

hpFV

60000,33

=

( )lbF 1320

50

2000,33==

( ) ( )psi

JEC

Fpb 4835

125.028

5312.0

1320

2=

+

=+

=

863. A roller chain is to transmit 5 hp from a gearmotor to a wood-working machine,

with moderate shock. The 1-in output shaft of the gearmotor turns rpmn 500= .

The 1 ¼-in. driven shaft turns 250 rpm; inC 16≈ . (a) Determine the size of

sprockets and pitch of chain that may be used. If a catalog is available, be sure

maximum bore of sprocket is sufficient to fit the shafts. (b) Compute the center

distance and length of chain. (c) What method should be used to supply oil to the

chain? (d) If a catalog is available, design also for an inverted tooth chain.

Solution:

Table 17.7

2.1=sfN

( ) hphpdesign 652.1 ==

2250

500

1

2 ==D

D

2

12

DDC +=

2216 1

1

DD +=

inD 4.61 =

( ) inDD 8.124.622 12 ===

( )( )fpm

nDvm 838

12

5004.6

12

11 ===ππ

(a) Link Plate Fatigue P

ts PnNhp07.039.008.1004.0 −=

( )PPP

DNNts

11.204.611 ==≈=

ππ


Page 37 of 56

( ) PPP

hp 07.039.0

08.1

50011.20

004.0 −

=

PP

07.092.147.276 −=

inP 45.0=

use inP2

1= , Chain No. 40

( )40

2

1

4.611 =

=≈

ππ

P

DN

802 12 == NN

Size of sprocket, 401 =N , 802 =N , inP2

1= .

(b) inC 16=

pitches

in

inC 32

2

1

16==

( )C

NNNNCL

4022

2

1221 −+

++≈

( ) ( )( )

25.1253240

4080

2

8040322

2

=−

++

+≈L pitches

use pitchesL 126=

(c) Method: fpmvm 838= .

Use Type II Lubrication ( fpmv 1300max = ) – oil is supplied from a drip lubricator to link

plate edges.

864. A roller chain is to transmit 20 hp from a split-phase motor, turning 570 rpm, to a

reciprocating pump, turning at 200 rpm; 24 hr./day service. (a) Decide upon the

tooth numbers for the sprockets, the pitch and width of chain, and center

distance. Consider both single and multiple strands. Compute (b) the chain

length, (c) the bearing pressure between the roller and pin, (d) the factor of safety

against fatigue failure (Table 17.8), with the chain pull as the force on the chain.

(e) If a catalog is available, design also an inverted-tooth chain drive.

Solution:

Table 17.7

2.04.1 +=sfN (24 hr/day)

( ) hphpdesign 32206.1 ==

(a) 85.2200

570

2

1 ==n

n


Page 38 of 56

85.22

1

1

2 =≈n

n

D

D

Considering single strand P

ts PnNhp07.039.008.1004.0 −=

min 17=tsN

( ) ( ) PPhp 07.039.008.157017004.032 −==

24.107.03 =− PP

inP 07.1=

use inP 0.1=

( ) ( ) ( ) ( )107.039.008.1

1 1570004.032−

== Nhp

211 =N

( ) 6021200

5702 =

=N

Roller width = in8

5

2

12

DDC +=

( )( )in

PND 685.6

21111 ==≈

ππ

( )( )in

PND 10.19

60122 ==≈

ππ

inC 44.222

685.610.19 =+=

Use inC 23=

pitchesC1

23=

Considering multiple strands

Assume, inP2

1=

P

ts PnNhp07.039.008.1004.0 −=

( ) ( ) ( ) ( )hphp 148.45.057021004.0

5.007.039.008.1==

−

No. of strands = 7.7148.4

32=

hp

hp

Use 8 strands

(b) Chain Length

( )C

NNNNCL

4022

2

1221 −+

++≈


Page 39 of 56

( ) ( )( )

15.882340

2160

2

6021232

2

=−

++

+≈L pitches

use pitchesL 88=


Table 17.8, inP 1=

inE8

5=

inJ 125.0=

inC 312.0=

mv

hpF

000,33=

( )( )fpm

nDvm 998

12

570685.6

12

11 ===ππ

( )lbF 1058

998

32000,33==

( ) ( )psi

JEC

Fpb 3876

125.028

5312.0

1058

2=

+

=+

=

(d) Factor of Safety = F

Fu

4, based on fatigue

lbFu 500,14= , Table 17.8

Factor of Safety = ( )

43.310584

500,14

4==

F

Fu

865. A 5/8-in. roller chain is used on a hoist to lift a 500-lb. load through 14 ft. in 24

sec. at constant velocity. If the load on the chain is doubled during the speed-up

period, compute the factor of safety (a) based on the chain’s ultimate strength, (b)

based on its fatigue strength. (c) At the given speed, what is the chain’s rated

capacity ( teethN s 20= ) in hp? Compare with the power needed at the constant

speed. Does it look as though the drive will have a “long” life?

Solution:

Table 17.8

inP8

5=

lbFu 6100=


Page 40 of 56

(a) Factor of Safety =F

Fu

( )( ) lbF 10002500 ==

Factor of Safety = 1.61000

6100=

(b) Factor of Safety = F

Fu

4 (fatigue)

Factor of Safety = ( )

5.110004

6100=

(c) fpmft

vm 35min1

sec60

sec24

14=

=

20=sN

inP8

5=

Rated P

ts PnNhp07.039.008.1004.0 −= [Link Plate Fatigue]

( )fpm

nnPN

v sm 35

12

208

5

12=

==

rpmn 6.33=

Rated ( ) ( ) hphp 6.08

56.3320004.0

8

507.03

9.008.1=

=

−

Hp needed at constant speed

( )( )hphp

Fvhp m 6.053.0

000,33

35500

000,33<===

Therefore safe for “long” life.

WIRE ROPES

866. In a coal-mine hoist, the weight of the cage and load is 20 kips; the shaft is 400

ft. deep. The cage is accelerated from rest to 1600 fpm in 6 sec. A single 6 x 19,

IPS, 1 ¾-in. rope is used, wound on an 8-ft. drum. (a) Include the inertia force

but take the static view and compute the factor of safety with and without

allowances for the bending load. (b) If 35.1=N , based on fatigue, what is the

expected life? (c) Let the cage be at the bottom of the shaft and ignore the effect

of the rope’s weight. A load of 14 kips is gradually applied on the 6-kip cage.

How much is the deflection of the cable due to the load and the additional energy

absorbed? (d) For educational purposes and for a load of uF2.0 , compute the

energy that this 400-ft rope can absorb and compare it with that for a 400-ft., 1

¾-in., as-rolled-1045 steel rod. Omit the weights of the rope and rod. What is the

energy per pound of material in each case?


Page 41 of 56

Solution:

(a)

( )212 445.4

sec6

sec60

min11600

fps

fpm

t

vva =

=−

=

kipsWh 20=

For 6 x 19 IPS,

ftlbDw r

26.1≈

kipsDkipsDwL rr

22 64.01000

4006.1 =

=

maWwLF ht =−−

2.32

64.020 2

rDm

+=

( )445.42.32

64.0202064.0

22

+=−− r

rt

DDF

273.076.22 rt DF +=

inDr4

31=

kipsFt 254

3173.076.22

2

=

+=

t

bu

F

FFN

−=

Table AT 28, IPS

tonsDF ru

242≈

( ) kipstonsFu 25812975.1422

===

with bending load

mbb AsF =

s

wb

D

EDs =


Page 42 of 56

s

wmb

D

DEAF =

Table At 28, 6 x 19 Wire Rope

( ) inDD rw 11725.075.1067.0067.0 ===

inftDs 968 ==

ksiE 000,30= 24.0 rm DA ≈

( ) insqAm 225.175.14.02

==

( )( )( )( )

kipsFb 4596

11725.0225.1000,30==

52.825

45258=

−=

−=

t

bu

F

FFN

without bending load

32.1025

258===

t

u

F

FN

(b) 35.1=N on fatigue

IPS, ksisu 260≈

( ) uu

tsr

ssp

NFDD

2=

( )( ) ( )( )( )( )260

2535.129675.1

usp=

0015.0=usp

Fig. 17.30, 6 x 19 IPS

Number of bends to failure = 7 x 105

(c) rmEA

FL=δ

insqAm 225.1=

ksiEr 000,12≈ (6 x 19 IPS)

kipsF 14=

inftL 4800400 ==

( )( )( )( )

in57.4000,12225.1

480014==δ

( )( ) kipsinFU −=== 3257.4142

1

2

1δ

(d) ( ) kipsFF u 6.512582.02.0 ===


Page 43 of 56

rmEA

FL=δ

( )( )( )( )

in85.16000,12225.1

48006.51==δ

( )( ) kipsinFU −=== 43485.166.512

1

2

1δ

For 1 ¾ in, as-rolled 1045 steel rod

ksisu 96=

( ) ( ) kipsAsF uu 9.23075.14

962

=

==

π

( ) kipsFF u 2.469.2302.02.0 ===

AE

FL=δ

( )( )

( ) ( )in073.3

000,3075.14

48002.46

2

=

=

πδ

( )( ) UkipsinFU <−=== 71073.32.462

1

2

1δ of wire rope.

868. A hoist in a copper mine lifts ore a maximum of 2000 ft. The weight of car, cage,

and ore per trip is 10 kips, accelerated in 6 sec. to 2000 fpm; drum diameter is 6

ft. Use a 6 x 19 plow-steel rope. Determine the size (a) for a life of 200,000

cycles and 3.1=N on the basis of fatigue, (b) for 5=N by equation (v), §17.25,

Text. (c) What is the expected life of the rope found in (b) for 3.1=N on the

basis of fatigue? (d) If a loaded car weighing 7 kips can be moved gradually onto

the freely hanging cage, how much would the rope stretch? (e) What total energy

is stored in the rope with full load at the bottom of te shaft? Neglect the rope’s

weight for this calculation. (f) Compute the pressure of the rope on the cast-iron

drum. Is it reasonable?

Solution:

( )212 56.5

sec6

sec60

min12000

fps

fpm

t

vva =

=−

=


Page 44 of 56

For 6 x 19 IPS,

ftlbDw r

26.1≈

kipsDkipsDwL rr

22 2.31000

20006.1 =

=

kipsWh 10=

aWwL

WwLF hht

+=−−

2.32

( ) ( ) ( )102.317267.1102.312.32

56.51

2.32

22 +=+

+=+

+= rrht DDWwL

aF

(a) ( ) uu

tsr

ssp

NFDD

2=

Fig. 17.30, 200,000 cycles, 6 x 19

0028.0=usp

PS: ksisu 225≈

inftDs 726 ==

3.1=N

( ) ( )( )( )( )( )2250028.0

102.317267.13.1272

2 += r

r

DD

49.307566.936.45 2 += rr DD

01251.364916.42 =+− rr DD

inDr 815.0=

say inDr8

7=

(b) by 5=N , Equation (v)

t

bu

F

FFN

−=

s

wb

D

EDs =

rw DD 067.0=

( )( )r

rb D

Ds 92.27

72

067.0000,30==

mbb AsF = 24.0 rm DA =

( )( ) 32 17.114.092.27 rrrb DDDF ==

tonsDF ru

236= for PS

kipsDF ru

272=

tbu NFFF =−


Page 45 of 56

( )( )( )102.317267.1517.1172 232 +=− rrr DDD

( )( )102.38634.517.1172 232 +=− rrr DDD

inDr 216.1=

use inDr4

11=

(c) ( ) uu

tsr

ssp

NFDD

2=

( )( ) ( )( ) ( )[ ]( )( )225

1025.12.317267.13.127225.1

2

usp

+=

00226.0=usp

Fig. 17.20

Expected Life = 3 x 105 cycles

(d) kipsF 7=

ksiEr 000,12=

inftL 000,242000 ==

For (a) inDr8

7=

rmEA

FL=δ

insqDA rm 30625.08

74.04.0

2

3 =

=≈

( )( )( )( )

in7.45000,1230625.0

000,247==δ

For (b) inDr4

11=

rmEA

FL=δ

insqDA rm 625.04

114.04.0

2

3 =

=≈

( )( )( )( )

in4.22000,12625.0

000,247==δ

(e) For (a) ( )( ) kipsinFU −=== 1607.4572

1

2

1δ

For (b) ( )( ) kipsinFU −=== 4.784.2272

1

2

1δ

(f) Limiting pressure, cast-iron sheaves, 6 x19, psip 500= .


Page 46 of 56

For (a) 0028.0=usp

( ) psipsikipsp 500630630.02250028.0 >=== , not reasonable.

For (b) 00226.0=usp

( ) psipsikipsp 5005.5085085.022500226.0 ≈=== , reasonable.

869. For a mine hoist, the cage weighs 5900 lb., the cars 2100 lb., and the load of coal

in the car 2800 lb.; one car loaded loaded at a time on the hoist. The drum

diameter is 5 ft., the maximum depth is 1500 ft. It takes 6 sec. to accelerate the

loaded cage to 3285 fpm. Decide on a grade of wire and the kind and size of rope

on the basis of (a) a life of 5102× cycles and 3.1=N against fatigue failure, (b)

static consideration (but not omitting inertia effect) and 5=N . (c) Make a final

recommendation. (d) If the loaded car can be moved gradually onto the freely

hanging cage, how much would the rope stretch? (e) What total energy has the

rope absorbed, fully loaded at the bottom of the shaft? Neglect the rope’s weight

for this calculation. (f) Compute the pressure of the rope on the cast-iron drum. Is

it all right?

Solution:

kipslbWh 8.10800,10280021005900 ==++=

( )212 125.9

sec6

sec60

min13285

fps

fpm

t

vva =

=−

=

aWwL

WwLF hht

+=−−

2.32

Assume 6 x 19 IPS,

ftlbDw r

26.1≈

kipsDkipsDwL rr

22 4.21000

15006.1 =

=

( ) ( ) 86.1308.3104.212.32

125.91

2.32

22 +=+

+=+

+= rrht DDWwL

aF

(a) Fig. 17.30, 2 x 105 cycles

0028.0=usp


Page 47 of 56

( ) uu

tsr

ssp

NFDD

2=

inftDs 605 ==

rs DD 45≈

inDr 33.145

60max ==

use inDr4

11=

kipsFt 67.1886.134

1108.3

2

=+

=

( )( )

( ) ( )ksisu 231

604

110028.0

67.183.12=

=

Use Plow Steel, 6 x 19 Wire Rope, inDr4

11= .

(b) t

bu

F

FFN

−=

s

wb

D

EDs =

inDD rw 08375.04

11067.0067.0 =

==

inDs 60=

ksiE 000,30=

( )( )ksisb 875.41

60

08375.0000,30==

2

2

2 625.04

114.04.0 inDA rm =

==

( )( ) kipsAsF mbb 17.26625.0875.41 ===

5=N

( )( ) tonskipsFNFF btu 76.5952.11917.2667.185 ==+=+=

25.38

4

11

76.5922

=

=

r

u

D

F

Table AT 28,

Use IPS, 6 x 19, 25.38422

>=r

u

D

F


Page 48 of 56

(c) Recommendation:

6 x 19, improved plow steel, inDr4

11=

(d) rmEA

FL=δ

lbF 490028002100 =+=

psiEr

61012×≈

inftL 000,181500 ==

( )( )( )( )

in76.111012625.0

000,1849006

=×

=δ

(e) ( )( ) lbinFU −=== 800,2876.1149002

1

2

1δ

(f) 0028.0=usp

ksisu 231=

( ) psip 8.646000,2310028.0 ==

For cast-iron sheave, limiting pressure is 500 psi

psipsip 5008.646 >= , not al right.

870. The wire rope of a hoist with a short lift handles a total maximum load of 14 kips

each trip. It is estimated that the maximum number of trips per week will be

1000. The rope is 6 x 37, IPS, 1 3/8 in. in diameter, with steel core. (a) On the

basis of 1=N for fatigue, what size drum should be used for a 6-yr. life? (n)

Because of space limitations, the actual size used was a 2.5-ft. drum. What is the

factor of safety on a static basis? What life can be expected ( 1=N )?

Solution:

(a)

No. of cycles = ( ) cyclescycleswk

trips

days

wk

yr

daysyr

5103857,3121

1000

7

1

1

3656 ×≈=

Figure 17.30, 6 x 37, IPS

00225.0=usp

( ) uu

tsr

ssp

NFDD

2=

For IPS, ksisu 260≈

kipsFt 14=

0.1=N

inDr 375.1=

( ) uu

tsr

ssp

NFDD

2=


Page 49 of 56

( ) ( )( )( )( )26000225.0

140.12375.1 =sD

inDs 8.34=

(b) inftDs 302 ==

Static Basis

t

bu

F

FFN

−=

Table AT 28, 6 x 37

( ) inDD rw 066.0375.1048.0048.0 ==≈

( ) 222 75625.0375.14.04.0 inDA rm ==≈

( )( ) kipsAsF muu 6.19675625.0260 ===

( )( )( )kips

D

AEDAsF

s

mwmbb 9.49

30

75625.0066.0000,30====

5.105.10

9.496.196=

−=

−=

t

bu

F

FFN

Life: 0.1=N (fatigue)

( ) uu

tsr

ssp

NFDD

2=

( )( ) ( )( )( )( )260

140.1230375.1

usp=

0026.0=usp

Figure 17.30, Life cycles5105.2 ×≈ , 6 x 37.

871. A wire rope passes about a driving sheave making an angle of contact of 540o, as

shown. A counterweight of 3220 lb. is suspended from one side and the

acceleration is 4 fps2. (a) If 1.0=f , what load may be noised without slipping on

the rope? (b) If the sheave is rubber lined and the rope is dry, what load may be

raised without slipping? (c) Neglecting the stress caused by bending about the

sheave, find the size of 6 x 19 MPS rope required for 6=N and for the load

found in (a). (d) Compute the diameter of the sheave for indefinite life with say

1.1=N on fatigue. What changes could be made in the solution to allow the use

of a smaller sheave?


Page 50 of 56


Solution:

( ) lbfps

fpslbF 2820

2.32

413220

2

2

2 =

−=

(a) θfeFF 21 =

πθ 3540 == o

10.0=f

( ) ( )( ) lbeF 72372820 310.0

1 == π

(b) For rubber lined, dry rope

495.0=f

( ) ( )( ) lbeF 466,2492820 3495.0

1 == π

(c) lbFFt 72371 ==

( )

t

u

t

bu

F

F

F

FFN =

≈−=

0

tonsDF ru

232≈ for MPS

kipsDF ru

264≈

lbDF ru

2000,64=

tu NFF =

( )( )72376000,64 2 =rD

inDr 824.0=

use inDr 875.0=


Page 51 of 56

(d) ( ) uu

tsr

ssp

NFDD

2=

Indefinite life, 0015.0=usp

MPS: psiksisu 000,195195 =≈

( ) ( )( )( )( )000,1950015.0

72371.12875.0 =sD

inDs 2.62=

To reduce the size of sheave, increase the size of rope.

872. A traction elevator with a total weight of 8 kips has an acceleration of 3 fps2; the

6 cables pass over the upper sheave twice, the lower one once, as shown..

Compute the minimum weight of counterweight to prevent slipping on the

driving sheave if it is (a) iron with a greasy rope, (b) iron with a dry rope, (c)

rubber lined with a greasy rope. (d) Using MPS and the combination in (a),

decide upon a rope and sheave size that will have indefinite life ( 1=N will do).

(e) Compute the factor of safety defined in the Text. (f) If it were decided that 5105× bending cycles would be enough life, would there be a significant

difference in the results?

Solution:

( ) kipsfps

fpskipsF 745.8

2.32

318

2

2

1 =

+=

( ) πθ 31803 == o

θfe

FF 1

2 =

cW = weight of counterweight

22 10274.1

2.32

31

FF

Wc =

−

=

θfce

FW 110274.1

=

(a) Iron sheave, greasy rope, 07.0=f

( )( )( ) kips

eWc 986.4

745.810274.1307.0

==π

(b) Iron sheave, dry rope, 12.0=f

( )( )( ) kips

eWc 112.3

745.810274.1312.0

==π

(c) Rubber lined with a greasy rope, 205.0=f

( )( )( ) kips

eWc 397.1

745.810274.13205.0

==π


Page 52 of 56

(d) ( ) uu

tsr

ssp

NFDD

2=


kipsFFt 745.81 == total

kipsFt 458.16

745.8== each rope

lbsFt 1458=

1=N

Table AT 28, 6 x 19

rs DD 45≈

( ) ( )( )( )( )000,1950015.0

14581245 =rr DD

inDr 47.0=

Use ininDr 5.02

1==

(e) t

bu

F

FFN

−=

Table AT 28, MPS

( ) lblblbDtonsDF rru 000,165.0000,64000,6432222 ====

s

mwb

D

AEDF =

psiE 61030×=

6 x 19, rw DD 067.0=

rs DD 45≈

( ) ..1.05.04.04.022

insqDA rm ===

( )( )( )lbFb 4467

45

1.0067.01030 6

=×

=

91.71458

4467000,16=

−=N

(f) 5 x 105 cycles

Fig. 17.30, 6 x 19.

0017.0=usp

( ) uu

tsr

ssp

NFDD

2=


Page 53 of 56

( ) ( )( )( )( )000,1950017.0

14581245 =rr DD

inDr 44.0=

since ininDr 47.044.0 ≈= as in (d), therefore, no significant difference will result.

873. A 5000-lb. elevator with a traction drive is supported by a 6 wire ropes, each

passing over the driving sheave twice, the idler once, as shown. Maximum values

are 4500-lb load, 4 fps2 acceleration during stopping. The brake is applied to a

drum on the motor shaft, so that the entire decelerating force comes on the

cables, whose maximum length will be 120 ft. (a) Using the desirable sD in

terms of rD , decide on the diameter and type of wire rope. (b) For this rope and

05.1=N , compute the sheave diameter that would be needed for indefinite life.

(c) Compute the factor of safety defined in the Text for the result in (b). (d)

Determine the minimum counterweight to prevent slipping with a dry rope on an

iron sheave. (e) Compute the probable life of the rope on the sheave found in (a)

and recommend a final choice.

Solution:

(a)

lbFt 4500=

lbWh 5000=

awLW

FwLW hth

+=−+

2.32

assume 6 x 19

ftlbDw r

26.1=

( )( ) 22 1921206.1 rr DDwL == per rope

( ) 22 11521926 rr DDwL ==

( )42.32

11525000450011525000

22

+=−+ r

r

DD

22 11.14312.6215001152 rr DD +=+

inDr 3465.0=

say ininDr8

3375.0 ==


Page 54 of 56

inDD rs8

716

8

34545 =

=≈

Six – 6 x 19 rope, inDr8

3=

(a) ininDr8

3375.0 ==

lbFt 7506

4500==

05.1=N

( ) uu

tsr

ssp

NFDD

2=

assume IPS, psiksisu 000,260260 ==


( ) ( )( )( )( )000,2600015.0

75005.12375.0 =sD

inDs 77.10=

(c) t

bu

F

FFN

−=

lbFt 750=

IPS

lblblbDtonsDF rru 813,118

3000,84000,8442

2

22 =

==≈

s

mwb

D

AEDF =

6 x 19,

inDs 77.10= as in (b)

inDD rw 025.08

3067.0067.0 =

==

..05625.08

34.04.0

2

2 insqDA rm =

==

psiE 61030×=

( )( )( )lbFb 3917

77.10

05625.0025.01030 6

=×

=


Page 55 of 56

53.10750

3917813,11=

−=N

(c) lbFF t 45001 ==

θfeFF 21 =

For iron sheave, dry rope, 12.0=f

πθ 3540 == o

( )( ) lbee

FF

f1452

4500312.0

12 ===

πθ

22.32

1 Fa

CW =

+

14522.32

41 =

+CW

lbCW 1291=

874. A traction elevator has a maximum deceleration of 5 fps2 when being braked on

the downward motion with a total load of 10 kips. There are 5 cables that pass

twice over the driving sheave. The counterweight weighs 8 kips. (a) Compute the

minimum coefficient of friction needed between ropes and sheaves for no

slipping. Is a special sheave surface needed? (b) What size 6 x 19 mild-plow-

steel rope should be used for 4=N , including the bending effect? (Static

approach.) (c) What is the estimated life of these ropes ( 1=N )?

Solution:

205.8 fpsa =

(a) kipsF 101 =

( ) kipskipsF 62.32

05.8182 =

−=

πθ 3=

θfe

F

F=

2

1


Page 56 of 56

( )π3

6

10 fe=

0542.0=f

Special sheave surface is needed for this coefficient of friction, §17.21.

(b) t

bu

F

FFN

−=

kipsFt 25

10==

s

mwb

D

AEDF =

Table AT 28, 6 x 19, MPS

rw DD 067.0=

rs DD 45≈ 24.0 rm DA ≈

psiE 61030×=

( )( )( )kipsD

D

DDF r

r

rrb

226

87.1745

4.0067.01030=

×=

kipsDtonsDF rru

22 6432 =≈

2

87.17644

22

rr DDN

−==

inDr 4164.0=

use inDr16

7=

(c) inDD rs 2016

74545 =

=≈

( ) uu

tsr

ssp

NFDD

2=

kipsFt 2= each rope

MPS, ksisu 195=

0.1=N

( ) ( )( )( )( )195

20.1220

16

7

usp=

0023.0=usp

Expected life, Figure 17.30, 3 x 105 bending cycles.

- end -

SECTION 16 – BRAKES AND CLUTCHES

Page 1 of 97

ENERGY TO BRAKES

881. A motor operates a hoist through a pair of spur gears, with a velocity ratio of 4.

The drum on which the cable wraps is on the same shaft as the gear, and the

torque cause by the weight of the load and hoist is 12,000 ft-lb. The pinion is on

the motor shaft. Consider first on which shaft to mount the brake drum; in the

process make trial calculations, and try to think of pros and cons. Make a

decision and determine the size of a drum that will not have a temperature rise

greater than Fto150=∆ when a 4000-lb. load moves down 200 ft. at a constant

speed. Include a calculation for the frp/sq. in. of the drum’s surface.

Solution:

Consider that brake drum is mounted on motor shaft that has lesser torque.

lbinlbftlbft

T f −=−=−

= 000,3630004

000,12

From Table AT 29,

Assume 35.0=f , psip 75= , max. fpmvm 5000=

2

FDT f =

D

TfNF

f2==

Df

TN

f2=

A

Np =

DbA π=

( )( )

7535.0

000,362222

====bDbfD

T

Db

Np

f

πππ

8732 =bD

use 8732 =bD

2

873

Db =

Then,

cW

lbftUFt

m

f −=∆ o

Assume a cast-iron, 3253.0 inlb=ρ

101=c

VWm ρ=


Page 2 of 97

+=+=

44

22 D

DbttDDbtV ππ

π

( )( ) lbftU f −== 000,8002004000

Fto150=∆

tc

UVW

f

m∆

== ρ

( )( )101150

000,800253.0 =V

37.208 inV =

But

+=

4

2DDbtV π

2

873

Db =

+=

4

873 2D

DtV π

For minimum V :

02

8732

=

+

−=

D

Dt

dD

dVπ

( )87323 =D

inD 12=

For t :

( )

+==

4

12

12

8737.208

2

tV π

int 611.0=

say int8

5=

( )ininb

16

160625.6

12

8732

===

Therefore use inD 12= , int8

5= , inb

16

16=

For A

fhpinsqfhp =..

000,33

mFvfhp =


Page 3 of 97

( )lb

D

TF

f6000

12

000,3622===

fpmvm 5000= (max.)

( )( )hpfhp 909

000,33

50006000==

( ) 2

16

1612 inDbA

== ππ

98.355.228

909.. ===

A

fhpinsqfhp (peak value)

882. A 3500-lb. automobile moving on level ground at 60 mph, is to be stopped in a

distance of 260 ft. Tire diameter is 30 in.; all frictional energy except for the

brake is to be neglected. (a) What total averaging braking torque must be

applied? (b) What must be the minimum coefficient of friction between the tires

and the road in order for the wheels not to skid if it is assumed that weight is

equally distributed among the four wheels (not true)? (c) If the frictional energy

is momentarily stored in 50 lb. of cast iron brake drums, what is the average

temperature rise of the drums?

Solution:

(a) Solving for the total braking torque.

( )22

212ssf vv

g

WKEU −=∆−=

lbW 3500=

fpsmphvs 88601

==

fpsmphvs 002

==

22.32 fpsg =

( )( ) lbftU f −=−= 000,421088

2.322

3500 22

( ) ( )000,63000,33

nlbinTlbftTfhp

fmf −=

−=

ω

( )( )

2

222

892.142602

880

2

12 fpss

vva

ss−=

−=

−=

sec91.5892.14

88012 =

−

−=

−=

a

vvt

ss

( )( ) ( )hp

t

U

t

KEfhp

f130

91.5550

000,421

550550===

∆−=


Page 4 of 97

( )( )rpm

ft

fps

D

vn m 336

12

30

minsec60882

1

=

==

ππ

000,63

nTfhp

f=

( )lbinT f −== 375,24

336

130000,63

(b) N

Ff =

for each wheel, lbN 8754

3500==

lbinT f −== 60944

375,24

( )lbin

D

TF

f−=== 406

30

609422

464.0875

406===

N

Ff

(c) cW

Ut

m

f=∆

lbftU f −= 000,421

lbWm 50=

Flblbftc −−=101 for cast-iron

( )( )Ft o4.83

10150

000,421==∆

884. An overhead traveling crane weighs 160,000 lb. with its load and runs 253 fpm.

It is driven by a 25-hp motor operating at 1750 rpm.The speed reduction from the

motor to the 18-in. wheels is 32 to 1. Frictional energy other than at the brake is

negligible. (a) How much energy must be absorbed by the brake to stop this crane

in a distance of 18 ft.? (b) Determine the constant average braking torque that

must be exerted on the motor shaft. (c) If all the energy is absorbed by the rim of

the cast-iron brake drum, which is 8 in. in diameter, 1 ½ in. thick, with a 3 ¼-in.

face, what will be its temperature rise? (d) Compute the average rate at which the

energy is absorbed during the first second (fhp). Is it reasonable?

Solution:


Page 5 of 97

( )22

212ssf vv

g

WKEU −=∆−=

lbW 000,160= 22.32 fpsg =

fpsfpmvs 22.42531

==

fpsvs 02

=

( )( )[ ] lbftU f −=−= 245,44022.4

2.322

000,160 22

(b) ( )

n

fhpT f

000,63=

( )( )

2

222

495.0182

22.40

2

12 fpss

vva

ss−=

−=

−=

sec53.8495.0

22.4012 =

−

−=

−=

a

vvt

ss

( )hp

t

Ufhp

f43.9

53.8550

245,44

550===

( ) ( )( )

( )lbin

n

fhpT f −=== 68

17502

1

000,6343.9000,63 on the motor shaft.

(c) cW

Ut

m

f=∆

DbtV π= (rim only) on the motor shaft

inD 8=

inb 25.3=

int 5.0=

( )( )( ) 384.405.025.38 inV == π

VWm ρ=

3253.0 inlb=ρ for cast iron

Flblbftc −−=101 for cast-iron

( )( ) lbWm 33.1084.40253.0 ==

( )( )Ft o4.42

10133.10

245,44==∆

(d) First second:


Page 6 of 97

fpsvs 22.41

=

2495.0 fpsa −=

( ) fpsatvv ss 73.31495.022.412

=−=+=

( )( ) ( )[ ] lbftKEU f −=−=∆−= 968073.322.4

2.322

000,160 22

( )hphp

t

Ufhp

f256.17

1550

9680

550<=== , therefore reasonable.

885. The diagrammatic hoist shown with its load weighs 6000 lb. The drum weighs

8000 lb., has a radius of gyration ftk 8.1= ; ftD 4= . A brake on the drum shaft

brings the hoist to rest in 10 ft. from fpsvs 8= (down). Only the brake frictional

energy is significant, and it can be reasonably assumed that the acceleration is

constant. (a) From the frictional energy, compute the average braking torque. (b)

If the average fhp/sq. in. is limited to 0.15 during the first second, what brake

contact area is needed?

Problems 885, 886

Solution:

n

fhpT f

000,63=

( ) ( )2222

2

2

11

21 1122ssf vv

g

WIKEKEU −+−=∆−∆−= ωω

fpsvs 81

= , fpsvs 02

=

( )srad

D

vs4

4

8221

1 ===ω , srad02 =ω

g

kWI

2

11 =

lbW 80001 =

ftk 8.1=

lbW 60002 =


Page 7 of 97

22.32 fpsg =

( ) ( ) ( )( )

( )( )

( ) lbftvvg

WIU ssf −=+=−+−= 400,128

2.322

600004

2.322

8.180000

22

222

2222

2

2

11

11ωω

s

vva

ss

2

22

12−

=

fts 10=

( )2

22

2.3102

80fpsa −=

−=

sec5.22.3

8012 =

−

−=

−=

a

vvt

ss

( )hp

t

Ufhp

f9

5.2550

400,12

550===

rpmnπ

ω

2

60=

( ) 02042

1−=−= sradsradω

( )rpmn 1.19

2

260==

π

( )lbin

n

fhpT f −=== 700,29

1.19

9000,63000,63

(b) 15.0.. =insqfhp (first second)

( ) fpsatvv ss 8.412.3812

=−=+=

( )sec4.2

4

8.4222

2 radD

vs===ω

( )( )

( ) ( )[ ]( )

( ) ( )[ ] lbftU f −=−+−= 61068.482.322

600004.24

2.322

8.180000 22222

( )hp

t

Ufhp

f10.11

1550

6106

550===

27415.0

10.11

..in

insqfhp

fhpA ===

887. The same as 885, except that a traction drive, arranged as shown, is used; the

counterweight weighs 4000 lb. The ropes pass twice about the driving sheave; the

brake drum is on this same shaft.


Page 8 of 97

Problem 887.

Solution:

(a) ( )22

212ss

Tf vv

g

WKEU −=∆−=

lblblbWT 000,1060004000 =+=

KE∆− of pulley is negligible

fpsvs 81

= , fpsvs 02

=

( )( ) lbftU f −== 940,98

2.322

000,10 2

( )2

2222

2.3102

80

2

12 fpss

vva

ss−=

−=

−=

sec5.22.3

8012 =

−

−=

−=

a

vvt

ss

( )hp

t

Ufhp

f23.7

5.2550

9940

550===

ftD 4=

( )sec4

4

8221

1 radD

vs===ω

( )sec0

4

0222

2 radD

vs===ω

( ) ( ) sec2042

1

2

121 rad=+=+= ωωω

( )rpmn 1.19

2

260

2

60

ππ

ω==


Page 9 of 97

Braking torque, ( )

lbinn

fhpT f −=== 850,23

1.19

23.7000,63000,63

(b) 15.0.. =insqfhp (first second)

fpsvs 81

=

atvv ss =−12

( )12.382

−=−sv

fpsvs 8.42

=

( )( ) ( )[ ] lbftU f −=−= 63608.48

2.322

000,10 22

( )hp

t

Ufhp

f56.11

1550

6360

550===

Contact area = 21.7715.0

56.11

..in

insqfhp

fhpA ===

SINGLE-SHOE BRAKES

888. For the single-shoe, short-block brake shown (solid lines) derive the expressions

for brake torque for (a) clockwise rotation, (b) counterclockwise rotation. (c) In

which direction of rotation does the brake have self-actuating properties? If

25.0=f , for what proportions of e and c would the brake be self-actuating?

Problems 888 – 891, 893.

Solution:

(a) Clockwise rotation (as shown)


Page 10 of 97

2

FDT f =

fNF =

[ ]∑ = 0HM

cNWaefN =+

WaefNcN =−

fec

WaN

−=

fec

fWaF

−=

( )fec

fWaDT f

−=

2

(b) Counter Clockwise Rotation

2

FDT f =

fNF =

[ ]∑ = 0HM


Page 11 of 97

cNefNWa +=

fec

WaN

+=

fec

fWaF

+=

( )fec

fWaDT f

+=

2

(c) Clockwise rotation is self-actuating

fec >

with 25.0=f

ec 25.0>

889. The same as 888, except that the wheel and brake shoe are grooved, θ2 degrees

between the sides of the grooves (as in a sheave, Fig. 17.38, Text).

Solution:

[ ]∑ = 0VF

NN =θsin2 1

12 NfF =


Page 12 of 97

θθ sinsin22

NfNfF =

=

(a) Clockwise rotation

fec

WaN

−=

( ) θsinfec

fWaF

−=

( ) θsin2 fec

fWaDT f

−=

(b) Counter clockwise rotation

fec

WaN

+=

( ) θsinfec

fWaF

+=

( ) θsin2 fec

fWaDT f

+=

(c) Clockwise rotation is self-actuating

fec >

with 25.0=f

ec 25.0>

890. Consider the single-shoe, short-block brake shown (solid lines) with the drum

rotating clockwise; let e be positive measured downward and cD 6.1= . (a) Plot

the mechanical advantage MA (ordinate) against f values of 0.1, 0.2, 0.3, 0.4,

0.5 (abscissa) when ce has values 2, 0.5, 0, -0.5, -1. (b) If f may vary from 0.3

to 0.4, which proportions give the more nearly constant brake response? Are

proportions good? (c) What proportions are best if braking is needed for both

directions of rotation?

Solution:


Page 13 of 97

(a) Wa

TMA

f= , Clockwise rotation

( )fec

DfMA

−=

2

cD 6.1=

( )fec

fcMA

−=

2

6.1

−

=

c

fe

fMA

1

8.0

Tabulation:

Values of MA

ce

f 2 0.5 0 -0.5 -1

0.1 0.100 0.084 0.08 0.076 0.073

0.2 0.267 0.178 0.16 0.145 0.133

0.3 0.600 0.284 0.24 0.209 0.185

0.4 1.600 0.400 0.32 0.267 0.229

0.5 ∞ 0.533 0.40 0.320 0.267


Page 14 of 97

Plot:

(b) 4.03.0 tof = , 1−=ce , with ttanconsMA ≈ .

They are good because c

fe>1 except 2=ce .

(c) 0=ce is the best if braking is needed for both directions of rotation with MA the

same.

891. A single-block brake has the dimensions: cast-iron wheel of inD 15= .,

ina2

132= ., inc

8

39= ., ine

16

114= ., width of contact surface = 2 in. The brake

block lined with molded asbestos, subtends 80o, symmetrical about the center

line; it is permitted to absorb energy at the rate of 0.4 hp/in.2; rpmn 200= .

Assume that p is constant, that F and N act at K , and compute (a) mpv and

the approximate braking torque, (b) the force W to produce this torque, (c) the

mechanical advantage, (d) the temperature rise of the 3/8-in.-thick rim, if it

absorbs all the energy with operation as specified, in 1 min. (e) How long could

this brake be so applied for Fto400=∆ ? See 893.

Solution:


Page 15 of 97

inD 15=

ina 5.32=

inc 375.9=

ine 6875.4=

inb 2=

(a) Solving for mpv

minlbftfpAvFv mm −=

24.0 inhpA

Fvm =

( )( )22

min200,13min000,334.0

in

lbft

in

hplbfthp

A

Fvm −=

−−=

mm fpv

A

Fv=

35.0=f from Table AT 29, molded asbestos on cast iron

mm pv

A

Fv35.0200,13 ==

min700,37 −−= insqlbftpvm

Solving for braking torque

min..200,13 −−= insqlbftA

Fvm

( ) fpmDnvm 78520012

15=

== ππ

2

DbA

θ=

( ) rad3963.1180

80 =

=

πθ


Page 16 of 97

( )( )( )..21

2

2153963.1

2insq

DbA ===

θ

( )200,13

21

785=

F

lbF 353=

( )( )lbin

FDT f −=== 2650

2

15353

2

(b) Solving for W

fec

WafF

−=

lbF 353=

35.0=f

ina 5.32=

ine 6875.4=

inc 375.9=

( ) ( ) ( )( )[ ]( )( )

lbfa

fecFW 240

5.3235.0

6875.435.0375.9353=

−=

−=

(c) Solving for MA

( )( )( )

( )( )[ ]34.0

6875.435.0375.92

1535.0

2=

−=

−=

fec

DfMA

(d) Solving for t∆

cW

lbftUFt

m

f −=∆

,o

DbtWm ρπ=

inD 15=

inb 2=

inint 375.08

3==

3253.0 inlb=ρ for cast iron

( )( )( )( )( ) lbWm 942.8375.0215253.0 == π

Flblbftc −−= 101 for cast iron


Page 17 of 97

( )fhptU f′= 550

sec60min1 ==′t

( )( ) fhpfhpU f 000,3360550 ==

( )( )hp

nTfhp

f4127.8

000,63

2002650

000,63===

( ) lbftU f −== 619,2774127.8000,33

( )( )F

cW

Ut

m

f o310101942.8

619,277===∆

(e) Solving for t′ , (time) with Fto400=∆

tcWU mf ∆=

( )( )( ) lbftU f −== 260,361400101942.8

( )( )fUtfhp =′550

( )( ) 260,3614127.8550 =′t

min3.1sec78 ==′t

892. For a single-block brake, as shown, ina 26= ., inc2

17= ., ine 75.3= .,

inD 15= ., drum contact width inb2

13= . The molded asbestos lining subtends

o60=θ , symmetrical about the vertical axis; force lbW 300= .; rpmn 600= .

Assume that p is constant, that F and N act at K , and compute (a) mpv and

the braking torque, (b) the energy rate in fhp/in.2 of contact surface. (c) the

mechanical advantage, (d) the temperature of the 3/8-in.-thick rim, if it absorbs

all the energy with the operation as specified in 1 min. (e) How long could this

brake be so applied for Ftrim

o400=∆ ? See 894.


Page 18 of 97

Problems 892, 894.

Solution:

For greater braking torque, fT , use counterclockwise rotation

[ ]∑ = 0AM

cNefNaW =+

efc

WaN

−=

efc

WafF

−=

From Table AT 29, 35.0=f for molded asbestos

lbW 300=

ina 26=

inc 5.7=

75.3=e

( )( )( )( )( )

lbF 44235.075.35.7

2630035.0=

−=

(a) Solving for mpv

mm fpAvFv =

( )( )fpm

Dnvm 2536

12

60015

12===

ππ

2

DbA

θ=

rad047.1180

60 =

=

πθ

( )( )( ) 25.272

5.315047.1inA ==

( )( ) ( )( ) mm pvFv 5.2735.02536442 ==

min..500,116 −−= insqlbftpvm


Page 19 of 97

Solving for the braking torque,

( )( )lbin

FDT f −=== 3315

2

15442

2

(b) Energy rate, fhp.in2.

( )( )hp

nTfhp

f6.31

000,63

6003315

000,63===

25.27 inA =

2

2

2 15.15.27

6.31inhp

in

hpinfhp ==

(c) ( )( )

425.026300

3315===

Wa

TMA

f

(d) cW

lbftUFt

m

f −=∆

,o

DbtWm ρπ=

inint 375.08

3==

inD 15=

inb 5.3= 3253.0 inlb=ρ for cast iron

Flblbftc −−= 101 for cast iron

( )( )( )( )( ) lbWm 648.15375.05.315253.0 == π

For 1 min

( )( ) ( )( ) lbftfhpU f −=== 800,042,16.311000,331000,33

( )( )Ft o660

101648.15

800,042,1==∆

(e) Ftrim

o400=∆

( )( )( ) lbftU f −== 179,632101648.15400

( )min61.0

6.31000,33

179,632

000,33min ===′

fhp

Ut

f


Page 20 of 97

LONG-SHOE BRAKES

FIXED SHOES

893. The brake is as described in 891 and is to absorb energy at the same rate but the

pressure varies as θsinPp = . Derive the equations needed and compute (a) the

maximum pressure, (b) the moment HFM of F about H , (c) the moment HNM

of N about H , (d) the force W , (e) the braking torque, (f) the x and y

components of the force at H .

Solution:

φθ sinsin PPp ==

2

Dr =

φpbrddN =

φfpbrddF =


Page 21 of 97

∫= rdFT f

∫= φdfpbrT f

2

∫= φφdPfbrT f sin2

( )21

2 coscos φφ −= PfbrT f

(a) Solving for P

( )21

2 coscos φφ −=

fbr

TP

f

2

Dr =

er

c

−=αtan

inc 375.9=

inr 5.72

15==

ine 6875.4=

6875.45.7

375.9tan

−=α

o3.73=α o80=θ

o3.332

803.73

21 =−=−=

θαφ

o3.1132

803.73

22 =+=+=

θαφ

35.0=f

inb 2=

inr 5.7=

( )21

2 coscos φφ −=

fbr

TP

f

( )( )( ) ( )psi

TTP

ff

5.483.113cos3.33cos5.7235.02

=−

=

n

fhpT f

000,63=

( )( )Ainfhpfhp 2=

2

DbA

θ=

rad396.1180

80 =

=

πθ


Page 22 of 97

( )( )( ) 2212

215396.1inA ==

4.02 =infhp

( )( ) hpfhp 4.8214.0 ==

rpmn 200=

( )lbinT f −== 2646

200

4.8000,63

( )o90.max555.48

2646

5.482 >==== φPpsi

TP

f

(b) ( )∫ −= dFRrM HF φcos

( )∫ −=2

1

sincosφ

φφφφ dfbrPRrM HF

( )∫ −=2

1

cossinsinφ

φφφφφ dRrfbrPM HF

2

1

2sin2

cos

φ

φ

φφ

−−=

RrfbrPM HF

( ) ( )

−−−= 1

2

2

2

21 sinsin2

coscos φφφφR

rfbrPM HF

( ) ( ) ( ) inercR 788.96875.45.7375.92222 =−+=−+=

( )( )( )( ) ( ) ( )

−−−= 3.33sin3.113sin

2

788.93.113cos3.33cos5.7555.7235.0 22

HFM

lbinM HF −=1900

(c) ∫= dNRM HN φsin

∫=2

1

2sinφ

φφφbrdRPM HN

∫=2

1

2sinφ

φφφdbrRPM HN

( )∫ −=2

1

2cos12

φ

φφφ d

brRPM HN

2

1

2sin2

1

2

φ

φ

φφ

−=

brRPM HN

( ) ( )[ ]1212 2sin2sin24

φφφφ −−−=brRP

M HN

rad396.112 ==− θφφ

( ) o6.2263.11322 2 ==φ

( ) o6.663.3322 1 ==φ


Page 23 of 97

( )( )( )( ) ( ) ( )[ ]6.66sin6.226sin396.124

55788.95.72−−=HNM

lbinM HN −= 8956

(d) ∑ = 0HM

0=−+ HNHF MMWa

ina 5.32=

( ) 0895619005.32 =−+W

lbW 217=

(e) lbinT f −= 2646

(f) ∑ = 0xF

∫ ∫ =++−− 0cossincos φφα dFdNWH x

∫∫ −−=−2

1

2

1

cossinsincos 2φ

φ

φ

φφφφφφα dfPbrdPbrWH x

( ) ( )[ ] ( )1

2

2

2

1212 sinsin2

2sin2sin24

cos φφφφφφα −−−−−−=−fbrPbrP

WH x

( )( )( ) ( ) ( )[ ]

( )( )( )( ) ( )3.113sin3.113sin2

555.7235.0

6.66sin6.226sin396.124

555.723.73cos217

22 −−

−−−=− xH

lbH x 931−=−

lbH x 931=

∑ = 0yF

∫ ∫ =+−+− 0sincossin φφα dFdNWH y

αφφφφφφ

φ

φ

φsinsincossin

2

1

2

1

2 WdfbrPdbrPH y −−=− ∫∫

( ) ( ) ( )[ ] αφφφφφφ sin2sin2sin24

sinsin2

12121

2

2

2W

fbrPbrPH y −−−−−−=−

( )( )( ) ( )

( )( )( )( ) ( ) ( )[ ] 3.73sin2176.66sin6.226sin396.124

555.7235.0

3.33sin3.113sin2

555.72 22

−−−−

−=− yH

lbH y 305−=−

lbH y 305=


Page 24 of 97

894. The brake is as described in 892, but the pressure varies as φsinPp = . Assume

the direction of rotation for which a given W produces the greater fT , derive the

equations needed, and compute (a) the maximum pressure, (b) the moment of F

about A , (c) The moment of N about A , (d) the braking torque, (e) the x and y

components of the force at A .

Solution:

φsinPp =

φpbrddN =

φφdPbrdN sin=

φφdfPbrfdNdF sin==

Solving for 1φ and 2φ


Page 25 of 97

er

c

+=αtan

inD

r 5.72

==

75.35.7

5.7tan

+=α

o69.33=α

o69.32

6069.33

21 =−=−=

θαφ

o69.632

6069.33

21 =+=+=

θαφ

( )∫ −= dFrRM AF φcos

( ) φφφφ

φdfPbrrRM AF sincos

2

1∫ −=

( ) φφφφφ

φdrRfPbrM AF ∫ −=

2

1

sincossin

( ) ( )

−+−= 121

2

2

2 coscossinsin2

φφφφ rR

fPbrM AF

( ) ( ) ( ) inrecR 52.135.775.35.72222 =++=++=

( ) ( )( ) ( ) ( )

−+−

= 69.3cos69.63cos5.769.3sin69.63sin

2

52.135.75.335.0 22

PM AF

PM AF 43.11=

∫= dNRM AN φsin

∫=2

1

2sinφ

φφφdRPbrM AN

( )∫ −=2

1

2cos12

φ

φφφ d

brPRM AN

( ) ( )[ ]1212 2sin2sin24

φφφφ −−−=brPR

M AN

rad047.112 ==− θφφ

( ) o38.12769.6322 2 ==φ

( ) o38.769.322 1 ==φ

( )( ) ( ) ( ) ( )[ ]38.7sin38.127sin047.124

52.135.75.3−−=

PM AN

PM AN 68.126=

(a) ∑ = 0AM

0=−+ ANAF MMWa

lbW 300=


Page 26 of 97

ina 26=

( )( ) 068.12643.1126300 =−+ PP

psiP 68.67=

max. psiPp 67.6069.63sin68.67sin 2 === φ

(b) ( ) lbinM AF −== 77468.6743.11

(c) ( ) lbinM AN −== 857568.6768.126

(d) ∫= rdFT f

∫=2

1

sin2φ

φφφdfPbrT f

( )21

2 coscos φφ −= fPbrT f

( )( )( )( ) ( )69.63cos69.3cos5.75.368.6035.02

−=fT

lbinT f −= 2587

(e) [ ]∑ = 0xF

∫ ∫ =−+−− 0cossincos φφα dFdNWH x

∫∫ +−=−2

1

2

1

cossinsincos 2φ

φ

φ

φφφφφφα dfPbrdPbrWH x

( ) ( )[ ] ( )1

2

2

2

1212 sinsin2

2sin2sin24

cos φφφφφφα −+−−−−=−fPbrPbr

WH x

( )( )( ) ( ) ( )[ ]

( )( )( )( ) ( )69.3sin69.63sin2

5.75.368.6735.0

38.7sin38.127sin047.124

5.75.368.6769.33cos300

22 −+

−−−=− xH

lbH x 136−=−

lbH x 136=

[ ]∑ = 0yF

∫ ∫ =−−+ 0sincossin φφα dFdNWH y

αφφφφφφ

φ

φ

φsinsincossin

2

1

2

1

2 WdfPbrdPbrH y −+= ∫∫

( ) ( ) ( )[ ] αφφφφφφ sin2sin2sin24

sinsin2

12121

2

2

2W

fPbrPbrH y −−−−+−=

( )( )( ) ( )

( )( )( )( ) ( ) ( )[ ] 69.33sin30038.7sin38.127sin047.124

5.75.368.6735.0

69.3sin69.63sin2

5.75.368.67 22

−−−+

−=yH


Page 27 of 97

lbH y 766=

895. (a) For the brake shown, assume αcosPp = and the direction of rotation for

which a given force W results in the greater braking torque, and derive equations

for fT in terms of W , f , and the dimensions of the brake. (b) Under what

circumstances will the brake be self-acting? (c) Determine the magnitude and

location of the resultant forces N and F .

Solution:

(a) Clockwise rotation has greatest braking torque.

αcosPp =

ααα dPbrpbrddN cos==

ααα dfPbrfpbrdfdNdF cos===

( )∫− +=2

1

sinθ

θα dFcrM HF

( )∫− +=2

1

cossinθ

θααα dfPbrcrM HF


Page 28 of 97

( )∫− +=2

1

cossincosθ

θαααα dcrfPbrM HF

2

1

2sin2

1sin

θ

θ

αα−

+= crfPbrM HF

( ) ( )[ ] ( ) ( )[ ]

−−+−−= 1

2

2

2

12 sinsin2

1sinsin θθθθ crfPbrM HF

( ) ( )

−++= 1

2

2

2

12 sinsin2


∫−=2

1

cosθ

θαdNM HN

∫−=2

1

2cosθ

θααdcPbrM HN

( )∫− +=2

1

2cos12

θ

θαα d

cPbrM HN

[ ] 2

12sin2

4

θ

θα −+=cPbr

M HN

( ) ( )[ ]1212 2sin2sin24

θθθθ +++=cPbr

M HN

[ ]∑ = 0HM

0=−+ HNHF MMWa

( ) ( ) ( ) ( )[ ]12121

2

2

2

12 2sin2sin24

sinsin2

1sinsin θθθθθθθθ +++=

−+++

cPbrcrfPbrWa

( ) ( )[ ] ( ) ( )[ ]1

2

2

2

121212 sinsinsinsin22

2sin2sin24

θθθθθθθθ −++−+++

=

crfbrcbr

WaP

( ) ( )[ ] ( ) ( )[ ]{ }1

2

2

2

121212 sinsinsinsin222sin2sin2

4

θθθθθθθθ −++−+++=

crfcbr

WaP

∫= rdFT f

∫−=2

1

cos2θ

θααdfPbrT f

[ ] 2

1sin2 θ

θα −= fPbrT f

( )12

2 sinsin θθ += fPbrT f

( )( ) ( )[ ] ( ) ( )[ ]{ }1

2

2

2

121212

12

2

sinsinsinsin222sin2sin2

sinsin4

θθθθθθθθ

θθ

−++−+++

+=

crfcbr

fWabrT f

( )( ) ( )[ ] ( ) ( )[ ]1

2

2

2

121212

12

sinsinsinsin222sin2sin2

sinsin4

θθθθθθθθ

θθ

−++−+++

+=

crfc

fWarT f

where 2

Der ==


Page 29 of 97

(b) ( ) ( )[ ] ( ) ( )[ ]1

2

2

2

121212 sinsinsinsin222sin2sin2 θθθθθθθθ −++>+++ crfc

( )( ) ( ) ( )1

2

2

2

1212

12

sinsin22sin2sin2

sinsin4

θθθθθθ

θθ

−−+++

+>

f

frc

(c) ∫= dNN

∫−=2

1

cosθ

θααdPbrN

[ ] 2

1sin

θ

θα −= PbrN

( )12 sinsin θθ += PbrN

fNF =

( )12 sinsin θθ += fPbrF

Solving for the location of F and N .

Let A = vertical distance from O .

( )∫∑−

−=2

1

cos.

θ

θα dFrAM LocF

( )∫∑−

−=2

1

2

. coscosθ

θααα fbrdrAPM LocF

( )∫∑−

−=2

1

2

. coscosθ

θααα drAPfbrM LocF

( )∫∑−

+−=

2

1

2cos12

1cos.

θ

θααα drAPfbrM LocF

2

1

2sin2

1

2

1sin.

θ

θ

ααα−

+−=∑ rAPfbrM LocF

( )[ ] ( ) ( )

+++−+=∑ 121212. 2sin2sin

2

1

2

1sinsin θθθθθθ rAPfbrM LocF

Then 0. =∑ LocFM

( )[ ] ( ) ( ) 02sin2sin2

1

2

1sinsin 121212 =

+++−+ θθθθθθ rA

( ) ( ) ( )

+++=+ 121212 2sin2sin

2

1

2

1sinsin θθθθθθ rA

( ) ( )

( )12

1212

sinsin

2sin2sin2

1

2

1

θθ

θθθθ

+

+++

=

r

A

( ) ( )[ ]( )12

1212

sinsin4

2sin2sin2

θθ

θθθθ

+

+++=

rA


Page 30 of 97

896. For the brake shown with 21 θθ ≠ , assume that the direction of rotation is such

that a given W results in the greater braking torque and that φsinPp = . (a)

Derive equations in terms of 1θ and 2θ for the braking torque, for the moment

HFM and for HNM . (b) Reduce the foregoing equations for the condition

21 θθ = . (c) Now suppose that θ , taken as 21 θθθ += , is small enough that

θθ ≈sin , 1cos ≈θ , 2

21

θθθ == . What are the resulting equations?

Solution:

(a) Use clockwise rotation

φsinPp =

φφdPbrdN sin=


11 90 θφ −=

22 90 θφ +=

∫= rdFT f


Page 31 of 97

∫=2

1

sin2φ

φφφdfPbrT f

( )21

2 coscos φφ −= fPbrT f

( ) ( )[ ]21

2 90cos90cos θθ +−−= fPbrT f

( )21

2 sinsin θθ += fPbrT f

( )dFcrM HF ∫ −= φcos

( ) φφφφ

φdcrPrfbM HF sincos

2

1∫ −=

( ) φφφφφ

φdcrfPbrM HF ∫ −=

2

1

cossinsin

2

1

2sin2

1cos

φ

φ

φφ

−−= crfPbrM HF

( ) ( )

−−−= 1

2

2

2

21 sinsin2

1coscos φφφφ crfPbrM HF

( ) ( )[ ] ( ) ( )[ ]

−−+−+−−= 1

2

2

2

21 90sin90sin2

190cos90cos θθθθ crfPbrM HF

( ) ( )

−−+= 1

2

2

2

21 coscos2


( ) ( ) ( )[ ]

−−−−+= 1

2

2

2

21 sin1sin12


( ) ( )

−++= 1

2

2

2

21 sinsin2


dNrM HN φ∫= sin

φφφ

φdPrbM HN ∫=

2

1

22 sin

( ) φφφ

φd

PbrM HN ∫ −=

2

1

2cos12

2

[ ] 2

12sin2

4

2φ

φφ−=Pbr

M HN

( ) ( )[ ]1212

2

2sin2sin24

φφφφ −−−=Pbr

M HN

( ) ( )[ ] ( ) ( )[ ]{ }1212

2

902sin902sin909024

θθθθ −−+−−−+=Pbr

M HN

( ) ( )[ ]1212

2

2sin2sin24

θθθθ −−−+=Pbr

M HN


Page 32 of 97

( ) ( )[ ]1212

2

2sin2sin24

θθθθ +++=Pbr

M HN

(b) 21 θθ =

( )21

2 sinsin θθ += PrfbT f

1

2 sin2 θPrfbT f =

( ) ( )

−++= 1

2

2

2

21 sinsin2


1

2 sin2 θfPbrM HF =

( ) ( )[ ]1212

2

2sin2sin24

θθθθ −−−+=Pbr

M HN

( )11

2

2sin244

θθ +=bPr

M HN

( )111

2

cossin444

θθθ +=bPr

M HN

( )111

2 cossin θθθ += bPrM HN

(c) 21 θθθ +=

θθ ≈sin

1cos ≈θ

221

θθθ ==

1

2 sin2 θPrfbT f =

θθθ 222

22

2sin2 PrfbPrfbPrfbT f =

=

=

1

2 sin2 θfPbrM HF =

θθθ 222

22

2sin2 PrfbPrfbPrfbM HF =

=

=

( )111

2 cossin θθθ += bPrM HN

( ) θθθ 2

HN bPrbPrM =

+= 1

22

2


Page 33 of 97

897. The brake shown is lined with woven asbestos; the cast-iron wheel is turning at

60 rpm CC; width of contact surface is 4 in. A force lbW 1300= . is applied via

linkage systemnot shown; o90=θ . Let φsinPp = . (a) With the brake lever as a

free body, take moments about the pivot J and determine the maximum pressure

and compare with permissible values. Compute (b) the braking torque, (c) the

frictional energy in fhp. (d) Compute the normal force N , the average pressure

on the projected area, and decide if the brake application can safely be

continuous.

Solution:

(a)

fdNdF =

φsinPp =

φφφ dPbrpbrddN sin==

φφdfPbrdF sin=

( )dFrRM JF ∫ −= φcos

( )∫ −=2

1

sincosφ

φφφφ drRfPbrM JF


Page 34 of 97

( )∫ −=2

1

sincossinφ

φφφφφ drRfPbrM JF

2

1

cossin2

1 2

φ

φ

φφ

+= rRfPbrM JF

( ) ( )

−+−= 121

2

2

2 coscossinsin2

1φφφφ rRfPbrM JF

10

5.12tan =β

o34.51=β

21

θβφ −=

o90=θ

o34.62

9034.511 =−=φ

o34.962

9034.51

21 =+=+=

θβφ

inb 4=

inr 10=

for woven asbestos 4.0=f (Table At 29)

( ) ( ) inR 16105.1222

=+=

( ) ( )

−+−= 121

2

2

2 coscossinsin2

1φφφφ rRfPbrM JF

( ) ( )( ) ( ) ( )

−+−= 34.6cos34.96cos1034.6sin34.96sin

2

161044.0 22

PM JF

PM JF 81.51−=

∫= dNRM JN φsin

∫=2

1

2sinφ

φφφdPbrRM JN

[ ] 2

12cos1

2

φ

φφ−=PbrR

M JN

( ) ( )[ ]1212 2sin2sin24

φφφφ −−−=PbrR

M JN

( )( )( )( ) ( ) ( )( )

−−

−= 34.62sin34.962sin

18034.634.962

4

16104 πPM JN

PM JN 9.572=

0=−+=∑ JNJFJ MMWaM


Page 35 of 97

( )( ) ( ) 09.57281.51251300 =−−+ PP

psiP 52=

max. psiPp 52== , 902 >φ

From Table AT 29, permissible psip 50=

Therefore epermissiblpp ≈max

(b) ∫= rdFT f

∫=2

1

sinφ

φφφdfPbrT f

( )21 coscos φφ −= fPbrT f

( )( )( )( )( ) lbinT f −=−= 918834.96cos34.6cos104524.0

(c) 000,63

nTfhp

f= , rpmn 60=

( )( )hpfhp 75.8

000,63

609188==

(d) ∫= dNN

∫=2

1

sinφ

φφφdPbrN

( )21 coscos φφ −= PbrN

( )( )( )( ) lbN 229734.96cos34.6cos10452 =−=

2sin2

.θ

br

Npave =

o90=θ

( )( )psipave 6.40

2

90sin1042

2297. ==

( ) ( )( ) min..755,12602012

6.4012

−−=

=

= insqlbft

Dnppvm

ππ

since min..000,28 −−< insqlbftpvm (§18.4)

Application is continuous.


Page 36 of 97

PIVOTED-SHOE BRAKES

898. In the brake shown, the shoe is lined with flexible woven asbestos, and pivoted at

point K in the lever; face width is 4 in.; o90=θ . The cast-iron wheel turns 60

rpm CL; let the maximum pressure be the value recommended in Table At 29.

On the assumption that K will be closely at the center of pressure, as planned,

compute (a) the brake torque, (b) the magnitude of force W , (c) the rate at which

frictional energy grows, (d) the time of an application if it is assumed that all this

energy is stored in the 1-in. thick rim with Ftrim 350=∆ , (e) the average pressure

on projected area. May this brake be applied for a “long time” without damage?

(f) What would change for CC rotation?

Problem 898.

Solution:

ina 27= , inb 4= , rpmn 60= CL

θθ

θ

sin

2sin2

+=

D

c

inD 20= , inr 10=

rad571.190 == oθ

( )inc 0.11

90sin571.1

2

90sin202

=+

=

(a) 2

sin2 2 θfPbrT f =

For woven asbestos, Table AT 29, 4.0=f

psiP 50=

( )( )( )( ) lbinT f −== 314,112

90sin104504.02

2


Page 37 of 97

(b)

( )( )( ) lbPbrN 25712

90sin571.110450

2

sin=

+=

+=

θθ

[ ]∑ = 0JM

NWa 12=

( ) ( )25711215 =W

lbW 2057=

(c) ( )( )

hpnT

fhpf

78.10000,63

60314,11

000,63===

rate of frictional energy ( ) min740,35578.10000,33000,33 lbftfhp −===

(d) Time (min) fhp

U f

000,33=

cW

lbftUFt

m

f −=∆ o

DbtWm ρπ=

For cast iron 3253.0 inlb=ρ

Flblbftc −−= 101

int 1=

( ) ( )( )( ) lbWm 6.631420253.0 == π

( )( )Flblbftlb

lbftUFt

f

−−

−==∆

1016.63350o

lbftU f −= 260,248,2

Time (min) ( )

min32.678.10000,33

260,248,2==

(e) Ave.

( )( )psi

br

Np 45.45

2

90sin1042

2571

2sin2

===θ


Page 38 of 97

( )( )( )( )Finsqlbft

Dnppvm −−=== ..280,14

12

602045.45

12

ππ

since 000,28<mpv , this brake may be applied for a long time.

(f) Since the moment arn of F is zero, no change or CC rotation.

899. The pivoted-shoe brake shown is rated at 450 ft-lb. of torque; o90=θ ; contact

width is 6.25 in.; cast-iron wheel turns at 600 rpm; assume a symmetric

sinusoidal distribution of pressure. (a) Locate the center of pressure and compute

with the location of K. Compute (b) the maximum pressure and compare with

allowable value, (c) the value of force W , (d) the reaction at the pin H , (e) the

average pressure and mpv , and decide whether or not the application could be

continuous at the rated torque. (f) Compute the frictional work from ωT and

estimate the time it will take for the rim temperature to reach 450 F (ambient, 100

F).

Problem 899.

Solution:

(a) θθ

θ

sin

2sin2

+=

D

c

inD 18=

rad571.190 == oθ

( )inc 9011.9

90sin571.1

2

90sin182

=+

=

but location of K = 9.8125 in

then, Klocationc ≈

(b) 2

sin2 2 θfPbrT f =

lbinlbftT f −=−= 5400450


Page 39 of 97

inb 25.6=

inr 9=

use 4.0=f (on cast-iron)

2sin2 2 θ

fPbrT f =

( ) ( )( )2

90sin925.64.025400

2P=

psiP 86.18= < allowable (Table AT 9)

(c) ( ) ( )375.10375.20 NW =

( )( )( ) lbPbrN 13642

90sin571.1925.686.18

2

sin=

+=

+=

θθ

( )( )lbW 695

375.20

375.101364==

(d) ↓=−=−= lbWNH 6696951364

(e) Ave.

( )( )psi

br

Np 15.17

2

90sin925.62

1364

2sin2

===θ

rpmn 600=

( )( )( )( )Finsqlbft

Dnppvm −−=== ..490,48

12

6001815.17

12

ππ

since 000,28>mpv , not continuous

(f) Frictional work ( ) ( )sec275,28

minsec60

6002450 perlbft

rpmlbftT −=

−==

πω

cW

lbftUFt

m

f −=∆ o

DbtWm ρπ=

For cast iron 3253.0 inlb=ρ

Flblbftc −−= 101

( ) ( )( ) ( )tttWm 154

4

1825.618253.0

2

=

+=

ππ

Ft 350100450 =−=∆

( )( )( ) lbftttctWU mf −==∆= 900,443,5101154350


Page 40 of 97

lbftU f −= 260,248,2

sec5.192275,28

900,443,5t

tTime ==

Assume int2

1=

sec96=Time

TWO-SHOE BRAKES

PIVOTED SHOES

900. The double-block brake shown is to be used on a crane; the force W is applied

by a spring, and the brake is released by a magnet (not shown); o90=θ ; contact

width = 2.5 in. Assume that the shoes are pivoted at the center of pressure. The

maximum pressure is the permissible value of Table AT 29. Compute (a) the

braking torque, (b) the force W , (c) the rate of growth of frictional energy at 870

rpm, (d) the time it would take to raise the temperature of the 0.5-in.-thick rim by

Ft 300=∆ (usual assumption of energy storage), (e) mpv . (f) Where should the

pivot center be for the calculations to apply strictly?

Problem 900.

Solution:

( )in

D

c 5.5

90sin2

2

90sin102

sin

2sin2

=

+

=+

=πθθ

θ


Page 41 of 97

[ ]∑ = 0

1RM

( ) 11 75.675.12875.05.5 NWF =+−

( ) 11 75.675.12625.4 NWfN =+

f

WN

625.425.6

75.121

−=

[ ]∑ = 0

2RM

( ) 22 75.6875.05.575.12 NFW +−=

22 75.6625.475.12 NfNW +=

f

WN

625.425.6

75.122

+=

Assume flexible woven asbestos,

40.0=f , psip 50=

( )W

WN 898.2

40.0625.425.6

75.121 =

−=

( )( ) WWfNF 16.1898.24.011 ===


Page 42 of 97

( )W

WN 574.1

40.0625.425.6

75.122 =

+=

( )( ) WWfNF 63.0574.14.022 ===

1.max ff TT =

cFfPbr 1

2

2sin2 =

θ

( )( )( ) ( )( )5.516.12

90sin

2

105.25040.02

2

W=

lbW 277=

(a) Braking torque = ( ) ( )( )( ) lbincFFTT ff −=+=+=+ 27275.527763.016.12121

(b) lbW 277=

(c) ( )( )

hpnT

fhpf

66.37000,63

8702727

000,63===

(d) Solving for tine:

cW

lbftUFt

m

f −=∆ o

FFtoo 300=∆

101=c , 253.0=ρ for cast iron

VWm ρ=

( )( )( ) ( ) ( ) 3

22

54.784

5.0105.05.210

4in

tDDbtV =+=+=

ππ

ππ

( )( ) lbWm 87.1954.78253.0 ==

( )( )( ) lbftU f −== 061,60210187.19300

( )sec29min4844.0

66.37000,33

061,602

000,33====

fhp

UTime

f

(e) mpv :

( )( )fpm

Dnvm 2278

12

87010

12===

ππ

( )( ) 900,113227850 ==mpv

(f) inc 5.5=

901. A pivoted-shoe brake, rated at 900 ft-lb. torque, is shown. There are 180 sq. in. of

braking surface; woven asbestos lining; 600 rpm of the wheel; 90o arc of brake

contact on each shoe. The effect of spring A is negligible. (a) Is the pin for the


Page 43 of 97

shoe located at the center of pressure? (b) How does the maximum pressure

compare with that in Table AT 29? (c) What load W produces the rated torque?

(d) At what rate is energy absorbed? Express in horsepower. Is it likely that this

brake can operate continuously without overheating? (e) Does the direction of

rotation affect the effectiveness of this brake?

Problem 901.

Solution:

(a)

( )in

D

c 9.9

90sin2

2

90sin182

sin

2sin2

=

+

=+

=πθθ

θ

and in9.92

16

1319

≈ , therefore the pin located at the center of pressure

(b)

16

1319

4tan =α

o4.11=α

[ ]∑ = 0QM

WFA 5.8cos4 =α


Page 44 of 97

WFA 5.84.11cos4 =

WFA 168.2=

[ ]∑ = 0VF and [ ]∑ = 0HF

( ) WWWWFQ Av 429.14.11sin168.2sin =+=+= α

( ) WWFQ Ah 125.24.11cos168.2cos === α

[ ]∑ = 0

1RM

( ) hQN 375.20375.101 =

( ) ( )WN 125.2375.20375.101 =

WN 173.41 =

11 NfF =

For woven asbestos lining, 40.0=f , psip 50=

( )( ) WWF 67.1173.440.01 == (either direction)


Page 45 of 97

[ ]∑ = 02RM

αcos375.20375.10 2 AFN =

( ) WWN 174.44.11cos168.2375.10

375.202 ==

( )( ) WWF 67.1174.440.02 == (either direction)

( )cFFT f 21 +=

( )( ) ( )( )( )9.967.167.112900 W+=

lbW 6.326=

2sin2 2

21

θfPbrFcTT ff ===

but brA θ=

θ

Arbr =2

( )( )( )( )( )( )( )

2

2

90sin91804.02

9.96.32667.1π

P

=

psipsiP 5026.9 <=

(c) lbW 6.326=

(d) ( )( )( )

hpnT

fhpf

103000,63

60012900

000,63===

( )( )fpm

Dnvm 2827

12

60018

12===

ππ

( )( ) Finsqlbftpvm −−== ..178,26282726.9

since 000,28<mpv , it is likely to operate continuously.

(e) Since the value of F is independent of rotation, the direction doesn’t affect the

effectiveness of this brake.

902. Refer to the diagrammatic representation of the brake of Fig. 18.2, Text, and let

the dimensions be: 16

94==== tmba , 14=c , 15=D , inh

16

99= ., and the

contact width is 4 in.; arc of contact = 90o; lining is asbestos in resin binder,

wheel rotation of 100 rpm CC; applied load lbW 2000= . (a) Locate the center of

pressure for a symmetrical sinusoidal pressure distribution and compare with the

actual pin centers. Assume that this relationship is close enough for approximate


Page 46 of 97

results and compute (b) the dimensions k and e if the braking force on each

shoe is to be the same, (c) the normal force and the maximum pressure, (d) the

braking torque, (e) mpv . Would more-or-less continuous application be

reasonable?

Figure 18.2

Solution:

(a)

( )in

D

c 25.8

90sin2

2

90sin152

sin

2sin2

=

+

=+

=πθθ

θ

On Centers:

cinmtK >=+=+ 125.916

94

16

94:

cinbaB >=+=+ 125.916

94

16

94:

[ ]∑ = 0CRM


Page 47 of 97

( )WceeRF +=

We

ceRF

+=

WRR FC −=

e

cWWW

e

ceRC =−

+=

[ ]∑ = 0

HRM

aRbFhN F=− 11

aRbfNhN F=− 11

fbh

aRN F

−=1

fbh

afRF F

−=1

( )( )fbhe

WcefaF

−

+=1

[ ]∑ = 0

ERM

kRtFhN C=+ 22


Page 48 of 97

kRtfNhN C=+ 22

fth

kRN C

+=2

fth

kfRF C

−=2

( )fthe

fkcWF

+=2

(b) 21 ff TT =

cFcF 21 =

21 FF =

( )( ) ( )fthe

fkcW

fbhe

Wcefa

+=

−

+

( )fth

kc

fbh

cea

+=

−

+

For asbestos in resin binder,

35.0=f , Table AT 29

inina 5625.416

94 ==

ininb 5625.416

94 ==

ininm 5625.416

94 ==

inint 5625.416

94 ==

inc 14=

ininh 5625.916

99 ==

( )( )

( )( )5625.435.05625.9

14

5625.435.05625.9

145625.4

+=

−

+ ke

ke 1903.214 =+

but emk =+

or 5625.4+= ke

then kk 1903.2145625.4 =++

ink 6.15=

ine 1625.205625.46.15 =+=

(c) ( )

( )( )( )( ) ( )[ ]

lbfthe

kcWNNN 2720

5625.435.05625.91625.20

2000146.1521 =

−=

+===


Page 49 of 97

(d) ( ) ( )( )( ) lbincNNfTTT fff −==+=+= 708,1525.82720235.02121

(e) ( )( )

fpmDn

vm 39312

10015

12===

ππ

( )( ) Finsqlbftpvm −−== ..195,2539311.64

since 000,28<mpv , continuous application is reasonable.

FIXED SHOES

903. A double-block brake has certain dimensions as shown. Shoes are lined with

woven asbestos; cast-iron wheel turns 60 rpm; applied force lbW 70= . For each

direction of rotation, compute (a) the braking torque, (b) the rate of generating

frictional energy (fhp). (c) If the maximum pressure is to be psiP 50= (Table

AT 29), what contact width should be used? (d) With this width, compute mpv

and decide whether or not the applications must be intermittent.

Problems 903, 904.

Solution:

[ ]0=∑ BM

WQ 264 =


Page 50 of 97

WQ 5.6=

[ ]0=∑ RM

( )WQS 5.66625.2 ==

WS 33.17=

WSRH 33.17==

WQRV 5.6==

ine 10=

inR 5.12=

ina 75.235.12925.2 =++=

011

=−−=∑ HNHFH MMSaM (CC)

011

=−+=∑ HNHFH MMSaM (CL)

( ) ( )

−−−= 1

2

2

2

21 sinsin2

coscos1

φφφφR

rfbrPM HF

( ) ( )[ ]1212 2sin2sin241


M HN

2sin2 2

1

θfPbrT f =

2sin2

1

θfr

TPbr

f=

inr 10=


Page 51 of 97

inr 112

sin2 =θ

( ) in112

sin102 =θ

rad165.143.66 == oθ

4.0=f for woven asbestos

( ) ( )

2sin2

sinsin2

coscos 1

2

2

2

211

1 θ

φφφφ

fr

RrfT

Mf

HF

−−−

=

( ) ( )

2sin2

sinsin2

coscos 1

2

2

2

211

1 θ

φφφφ

r

RrT

Mf

HF

−−−

=

rad9886.064.562

73.6690

2901 ==−=−= oθ

φ

o28.1132 1 =φ

rad1530.236.1232

73.6690

2902 ==+=+= oθ

φ

o72.2462 2 =φ

θφφ =− 12

( ) ( )

( )1

1

1

2

73.66sin102

64.56sin36.123sin2

5.1236.123cos64.56cos10 22

f

f

HF T

T

M =

−−−

=

( ) ( )[ ]

−−−=

2sin24

2sin2sin2 12121

1 θ

φφφφ

fr

RTM

f

HN

( ) ( )[ ]

( )( )1

1

196.2

2

73.66sin104.08

28.113sin72.246sin165.125.12f

f

HN TT

M =−−

=

CC:

011

=−−=∑ HNHFH MMSaM

( )( )( ) 0960.275.237033.1711

=−− ff TT

lbinT f −= 72761

CL:

011

=−+=∑ HNHFH MMSaM


Page 52 of 97

( )( )( ) 0960.275.237033.1711

=−+ ff TT

lbinT f −= 700,141

ine 10=

ind 5.12=

CC: [ ]∑ = 0HM

022

=−+−′HNHFVH MMdRaR

CL: [ ]∑ = 0HM

022

=−−−′HNHFVH MMdRaR

2

2

12

1

f

f

f

HFHF TT

TMM =

=

2

2

12960.2

1

f

f

f

HNHN TT

TMM =

=

CC:

022


( )( ) ( )( )[ ]( ) 0960.2705.125.65.2133.1722

=−+− ff TT

lbinT f −= 405,102

CL:

022


( )( ) ( )( )[ ]( ) 0960.2705.125.65.2133.1722

=−−− ff TT

lbinT f −= 51502

(a) Braking Torque 21 ff TT +=


Page 53 of 97

CC:

lbinTTT fff −=+=+= 681,17405,10727621

CL:

lbinTTT fff −=+=+= 850,195150700,1421

(b) Rate of generating frictional energy

000,63

nTfhp

f=

CC: ( )( )

hpfhp 84.16000,63

60681,17==

CL: ( )( )

hpfhp 90.18000,63

60850,19==

(c) psip 50=

2sin2 2

21

θfPbrTorT ff =

CC:

( )( )( )in

Prf

Tb

f73.4

2

73.66sin10504.02

405,10

2sin2

22

1 ===θ

CL:

( )( )( )in

Prf

Tb

f68.6

2

73.66sin10504.02

700,14

2sin2

22

2 ===θ

(d) mpv

( )( )fpm

Dnvm 314

12

6020

12===

ππ

( )( ) 000,55700,1531450 <==mpv

( )( ) 000,28700,1531450 <==mpv

application can be continuous or intermittent.

904. If the brake shown has a torque rating of 7000 lb-in. for counter-clockwise

rotation, what braking torque would it exert for clockwise rotation, force W the

same?

Solution:

CC:

011

=−− HNHF MMSa

11 fHF TM =

11960.2 fHN TM =


Page 54 of 97

WS 33.17=

ina 75.23=

( )( ) 096.275.2333.1711

=−− ff TTW

WT f 9.1031

=

022


WRH 33.17=

WRV 53.6=

ina 5.21=′

22 fHF TM =

22960.2 fHN TM =

( )( ) ( )( ) 0960.25.125.65.2133.1722

=−+− ff TTWW

WT f 65.1482

=

21 fff TTT +=

WW 65.1489.1037000 +=

lbW 7.27=

CL:

011

=−+ HNHF MMSa

( )( )( ) 096.275.237.2733.1711

=−− ff TT

lbinT f −= 58171

022


( )( ) ( )( )[ ]( ) 0960.27.275.125.65.2133.1722

=−+− ff TT

lbinT f −= 20382

lbinTTT fff −=+=+= 78552038581721

(CL)

905. A double-block brake is shown for which o90=θ , inb 5= ., rpmn 300= , rim

thickness = ¾ in., and lbW 400= . The shoes are lined with asbestos in resin

binder. Determine the frictional torque for (a) clockwise rotation, (b)

counterclockwise rotation. (c) How much energy is absorbed by the brake?

Express in horsepower. (d) Will the brake operate continuously without danger of

overheating? How long for a Ftrim 300=∆ ? How does mpv compare with Text

values?

ind 5.12=


Page 55 of 97

Problem 905

Solution:

44

4tan

+=α

o565.26=α

[ ]0=∑ RM

( )( ) WQ 164cos =α

( )( ) ( )400164565.26cos =Q

lbQ 1789=

lbQRH 1600565.26cos1789cos === α

lbWQRV 1200400565.26sin1789sin =+=+= α


Page 56 of 97

( ) ( )

−−−= 1

2

2

2

21 sinsin2

coscos φφφφR

rfbrPM HF

( ) ( )[ ]1212 2sin2sin24


M HN

2sin2 2 θ

fPbrT f =

( ) ( )

2sin2

sinsin2

coscos 1

2

2

2

21

θ

φφφφ

r

RrT

Mf

HF

−−−

=

( ) ( )[ ]

2sin8

2sin2sin2 1212

θ

φφφφ

fr

RTM

f

HN

−−−=

inr 102

20==

12

4tan =β

o435.18=β

rad571.190 == oθ

rad464.0565.26435.182

9090

2901 ==−−=−−= oβ

θφ

( ) o13.53565.2622 1 ==φ

rad034.2565.116435.182

9090

2902 ==−+=−+= oβ

θφ

( ) o13.233565.11622 2 ==φ

inR 65.12124 22 =+=

Asbestos in resin binder 35.0=f


Page 57 of 97

( ) ( )

( )f

f

HF T

T

M 6803.0

2

90sin102

565.26sin565.116sin2

65.125656.116cos565.26cos10 22

=

−−−

=

( ) ( ) ( )[ ]

( )( )f

f

HN TT

M 03.3

2

90sin1035.08

13.53sin13.233sin464.0034.2265.12=

−−−=

(a) Clockwise

[ ]∑ = 0

1HM

( )( ) ( )( ) 024cos5.2sin1111

=−++ HNHF MMQQ αα

( )( ) ( )( ) 003.36803.024565.26cos17895.2565.26sin178911

=−++ ff TT

lbinT f −= 195,171

[ ]∑ = 0

2HM

0245.22222

=++− HFHNHV MMRR

( ) ( ) 06803.003.316002412005.222

=++− ff TT

lbinT f −= 95412


Page 58 of 97

lbinTTT fff −=+=+= 736,269541195,1721

(b) Counterclockwise

[ ]∑ = 0

1HM

0sin5.2cos241111

=−−+ HNHF MMQQ αα

( )( ) ( )( ) 003.36803.0565.26sin17895.2565.26cos17892411

=−−+ ff TT

lbinT f −= 890,101

[ ]∑ = 0

2HM

0245.22222

=+−− HNHFHV MMRR

( ) ( ) 003.36803.016002412005.222

=+−− ff TT

lbinT f −= 066,152

lbinTTT fff −=+=+= 956,25066,15890,1021


Page 59 of 97

(c) CL:

( )( )hp

nTfhp

f3.127

000,63

300736,26

000,63===

CC:

( )( )hp

nTfhp

f6.123

000,63

300956,25

000,63===

(d) ( )( )

fpmDn

vm 157112

30020

12===

ππ

For p :

12sin2 2

ff TfPbrT ==θ

(CL)

( )( )( )( )2

90sin10535.02195,17

2P=

psiP 48.69=

( )( ) 000,28153,109157148.69 >==mpv

the brake operate continuously with danger of overheating.

For time:

cW

lbftUFt

m

f −=∆ o

101=c , 253.0=ρ

VWm ρ=

4

2tD

DbtVπ

π +=

( )( ) ( ) 3

2

24.4714

3

4

20

4

3520 inV =

+

=

ππ

( )( ) lbVWm 22.11924.471253.0 === ρ

( )( )( ) lbfttcWU mf −==∆= 366,612,330010122.119

Time = fhp

U f

000,33

CL: Time = ( )

sec53min886.06.123000,33

366,612,3

000,33===

fhp

U f

CC: Time = ( )

sec52min860.03.127000,33

366,612,3

000,33===

fhp

U f

000,28>mpv , not good for continuous application.


Page 60 of 97

906. The double-block brake for a crane has the dimensions: 3.14=a , 37.2=b ,

10=D , 05.11=e , 1.7=g , 12=h , 6.6=j , 55.10=k , inm 5.3= ., the width of

shoes is 4 in., and the subtended angle is o90=θ ; wocen asbestos lining. Its

rated braking torque is 200 ft-lb. The shoes contact the arms in such a manner

that they are virtually fixed to the arms. What force W must be exerted by a

hydraulic cylinder to develop the rated torque for (a) counterclockwise rotation,

(b) clockwise rotation? Is the torque materially affected by the direction of

rotation? (c) Compute the maximum pressure and compare with that in Table AT

29. (Data courtesy of Wagner Electric Corporation.)

Problem 906.

Solution:

83.005.11

37.2tan

−=

−=

ce

bα

o056.13=α


Page 61 of 97

[ ]∑ = 0RM

eWcQbQ =+ αα sincos

( )( ) ( )( ) WQQ 3.14056.13sin83.0056.13cos37.2 =+

WQ 7286.5=

WWQRH 58.5056.13cos7286.5cos === α

WWWWQRV 294.0056.13sin7286.5sin =−=−= α

6.6

275.5

2tan ==

j

kβ

o63.38=β

rad1112.037.663.382

9090

2901 ==−−=−−= oβ

θφ

o74.122 1 =φ

rad6820.137.9663.382

9090

2902 ==−+=−+= oβ

θφ

o74.1922 2 =φ

( ) injk

R 449.86.62

55.10

2

2

2

2

2

=+

=+

=

inD

r 52

10

2===


Page 62 of 97

( ) ( )

2sin2

sinsin2

coscos 1

2

2

2

21

θ

φφφφ

r

RrT

Mf

HF

−−−

=

( ) ( )[ ]

2sin8

2sin2sin2 1212

θ

φφφφ

fr

RTM

f

HN

−−−=

For woven asbestos lining, 40.0=f

( ) ( )

( )f

f

HF T

T

M 1985.0

2

90sin52

37.6sin37.96sin2

449.837.96cos37.6cos5 22

=

−−−

=

( ) ( )[ ]

( )( )f

f

HN TT

M 6755.2

2

90sin54.08

74.12sin74.192sin1112.0682.12449.8=

−−−=

(a) CC:

[ ]∑ = 01HM

( ) ( ) 025.0121111

=−−− HNHFVH MMRR

( )( ) ( )( ) 06755.21985.025.0294.01258.511

=−−− ff TTWW

WT f 3.231

=


Page 63 of 97

[ ]∑ = 0

2HM

05.3sin25.0cos122222

=−−++ WMMQQ HNHFαα

( ) ( )22

1985.06755.25.3056.13sin3.1425.0056.13cos3.1412 ff TTWWW −=−+

WT f 4.662

=

21 fff TTT +=

lbinlbftT f −=−= 2400200

WW 4.663.232400 +=

lbW 8.26=

(b) CL:


Page 64 of 97

[ ]∑ = 01HM

( ) ( ) 025.0121111

=−+− HNHFVH MMRR

( )( ) ( )( ) 06755.21985.025.0294.01258.511

=−+− ff TTWW

WT f 0.271

=

[ ]∑ = 0

2HM

05.3sin25.0cos122222

=−−−+ WMMQQ HNHFαα

( ) ( )22

1985.06755.25.3056.13sin3.1425.0056.13cos3.1412 ff TTWWW +=−+

WT f 2.572

=

21 fff TTT +=

WW 2.570.272400 +=

lbW 5.28=

Since W has different values, torque is materially affected by the direction of rotation.

(c) 2

sin2 2 θfPbrT f =

For woven asbestos lining, 40.0=f

Use ( ) lbinWT f −=== 17808.264.664.66

inb 4=

inr 5= o90=θ

( ) ( )( )2

90sin544.021780

2PT f ==

psiP 47.31=


Page 65 of 97

From Table AT 29, psip 50max =

psipsi 5047.31 <

INTERNAL-SHOE BRAKES

908. Assuming that the distribution of pressure on the internal shoe shown is given by

φsinPp = , show that the moments BNM , BFM , and OFT of N with respect to

B and of F with respect to B and to O are (b = face width)

( ) ( )[ ]22sin2sin2 12 φφθ −−= PbarM BN ,

( ) ( )[ ]2sinsincoscos 1

2

2

2

21 φφφφ −−−= arfPbrM BF ,

( )21

2 coscos φφ −= fPbrT OF .


Solution:

φsinPp =

( ) kdNMd BN =

( ) φφφφ dPbrbrdPdN sinsin ==

( ) φφ sin90cos aak =−=

( ) ( )( ) φφφφφ dPabrdPbraMd BN

2sinsinsin ==

( )∫∫ −==2

1

2

1

2cos12

sin2φ

φ

φ

φφφφφ d

PabrdPabrM BN

( ) ( )

−−−=

−=

2

2sin2sin

22sin

2

1

2

1212

2

1

φφφφφφ

φ

φ

PabrPabrM BN

but θφφ =− 12


Page 66 of 97

( )

−−=

2

2sin2sin

2

12 φφθ

PabrM BN

( ) edFMd BF =


( ) φφ cos90sin arare −=−+=

( ) ( )( ) ( ) φφφφφφφ darfPbrdfPbrarMd BF cossinsinsincos −=−=

[ ] 2

1

2sincosφ

φφφ arfPbrM BF −−=

( ) ( )

−−−=

2

sinsincoscos 1

2

2

2

21

φφφφ

arfPbrM BF

( ) φφdfPbrrdFTd OF sin2==

[ ] 2

1cos2 φ

φφ−= fPbrT OF

( )21

2 coscos φφ −= fPbrT OF

909. The same as 908, except that a pressure distribution of αcosPp = is assumed.

( ) ( )[ ]2sinsin42sin2sin2 1

2

2

2

12 ααααθ −+++= chPbrM BN ,

( ) ( ) ( )[ ]42sin2sin22sinsinsinsin 121

2

2

2

12 ααθαααα ++−−++= chrfPbrM BF

( )12

2 sinsin αα += fPbrM OF .

Solution:

αα sincos chk +=

αα cossin chre −+=

ααα dPbrpbrddN cos==

ααdfPbrfdNdF cos==

( )( )αααα dPbrchkdNdM BN cossincos +==

( ) αααα dchPbrdM BN cossincos2 +=

( )∫− +=2

1

cossincos2α

ααααα dchPbrM BN

( ) 2

1

2

sin

4

2sin2 2α

α

ααα

−

+

+=

chPbrM BN

but 21 ααθ +=

( ) ( )[ ] ( )[ ]

−−

+−−++

=2

sinsin

4

2sin2sin2 1

2

2

2

1212 αααααα chPbrM BN


Page 67 of 97

( ) ( )

−+

++=

2

sinsin

4

2sin2sin2 1

2

2

2

12 ααααθ chPbrM BN

( )( )αααα dfPbrchredFdM BF coscossin −+==

( ) ααααα dchrfPbrdM BF

2coscossincos −+=

( ) 2

1

4

2sin2

2

sinsin

2α

α

αααα

−

+−+=

chrfPbrM BF

( )[ ] ( )[ ] ( ) ( )[ ]

−−++

−−−

+−−=4

2sin2sin2

2

sinsinsinsin 12121

2

2

2

12

αααααααα

chrfPbrM BF

( ) ( ) ( )

++−

−++=

4

2sin2sin2

2

sinsinsinsin 121

2

2

2

12

ααθαααα

chrfPbrM BF

( ) αααα dfPbrdfPbrrrdFdM OF coscos 2===

[ ] ( )[ ]12

22 sinsinsin 2

1ααα

α

α −−=−= − fPbrfPbrM OF

( )12

2 sinsin αα += fPbrM OF

910. The same as 909, except that the α is to be measured from OG , a perpendicular

to OB ; limits from 1α− to 2α+ .

Solution:

αcosak =

αsinare +=

( ) ( ) ααααααα dPbar

dPbardPbrakdNdM BN 2cos12

coscoscos 2 +====

( ) ( )[ ]1212 2sin2sin242

2sin2

2

2

1

αααααα

α

α

−−++=

+=

−

PbarPbarM BN

( )12 2sin2sin24

ααθ −+=Pbar

M BN

( )( ) ( ) ααααααα darfPbrdfPbraredFdM BF cossincoscossin +=+==

( )[ ] ( )[ ]

−−

+−−=

+=

−2

sinsinsinsin

2

sinsin 1

2

2

2

12

2 2

1

αααα

αα

α

α

arfPbr

arfPbrM BF

( ) ( )

−++=

2

sinsinsinsin 1

2

2

2

12

αααα

arfPbrM BF

( ) αααα dfPbrdfPbrrrdFdM OF coscos 2===


Page 68 of 97

[ ] ( )[ ]12

22 sinsinsin 2

1ααα

α

α −−=−= − fPbrfPbrM OF

( )12

2 sinsin αα += fPbrM OF

911. The following dimensions apply to a two-shoe truck brake somewhat as shown:

face 5=b , 8=r , 1.5=h , 6.2=c , inuw 4.6== ., o110=θ , o151 =φ . Lining is

asbestos in rubber compound. For a maximum pressure on each shoe of 100 psi,

determine the force Q , and the braking torque for (a) clockwise rotation, (b)

counterclockwise rotation. See 908. (Data courtesy of Wagner Electric

Corporation.)

Problems 911, 912.

Solution: See 908.

( )

−−=

2

2sin2sin

2

12 φφθ

PbarM BN

( ) ( )

−−−=

2

sinsincoscos 1

2

2

2

21

φφφφ

arfPbrM BF ,

( )21


o151 =φ , o302 1 =φ , o1251101512 =+=+= θφφ , o2502 1 =φ

psip 100=

inb 5=

inr 8=

( ) ( ) incha 7245.56.21.52222 =+=+=

rad92.1110 == oθ

For asbestos in rubber compound

35.0=f


Page 69 of 97

(a) Both sides (clockwise rotation)

( ) 0=−++ BNBF MMwhQ

( )( )( )( ) ( )lbinM BN −=

−−= 224,30

2

30sin250sin92.1

2

87245.55100

( )( )( )( ) ( ) ( )lbinM BF −=

−−−= 436,14

2

30sin125sin7245.5125cos30cos88510035.0

22

inh 1.5= , inw 4.6=

( ) 0224,30436,144.61.5 =−++Q

lbQ 1373=

( )( )( )( ) ( ) lbinT OF −=−= 242,17125cos15cos8510035.02

( ) lbinTT OFf −=== 484,34242,1722

(b) Counterclockwise rotation

( ) 0=−−+ BFBN MMwhQ


Page 70 of 97

( ) 0436,14224,304.61.5 =−−+Q

lbQ 3883=

( ) lbinTT OFf −=== 484,34242,1722

913. The data are the same as 911, but the shoe arrangement is as shown for this

problem. For a maximum pressure on the shoes of 100 psim determine the force

Q and OFT for (a) Cl rotation, (b) CC rotation, See 908.

Problem 913.

Solution:

( )21

2 coscos φφ −== fPbrTT fOF

( )21 coscos φφ −=

fr

TPbr

f

( )( )

( )

−−

−=

−−=

2

2sin2sin

coscos22

2sin2sin

2

12

21

12 φφθ

φφ

φφθ

fr

aTPbarM

f

BN

( ) ( )

−−−=

2

sinsincoscos 1

2

2

2

21

φφφφ

arfPbrM BF

( )( ) ( )

−−−

−=

2

sinsincoscos

coscos2

1

2

2

2

21

21

φφφφ

φφ

ar

r

TM

f

BF

From 911: o151 =φ , o302 1 =φ , o1251101512 =+=+= θφφ , o2502 1 =φ

rad92.1110 == oθ

( ) ( ) incha 7245.56.21.52222 =+=+=

35.0=f


Page 71 of 97

( )( )( )( )

f

f

BN TT

M 753.12

30sin250sin92.1

125cos15cos835.02

7245.5=

−−

−=

( )( ) ( )

f

f

BF TT

M 43.02

15sin125sin7245.5125cos15cos8

125cos15cos28

22

=

−−−

−=

(a) CL rotation:

Left Side

[ ]∑ = 0BM

( ) 011

=−−+ BFBN MMwhQ

( ) 043.0753.14.61.511

=−−+ ff TTQ

QT f 268.51

=

Right Side:

[ ]∑ = 0BM


Page 72 of 97

( ) 022

=−++ BNBF MMwhQ

( ) 0753.143.04.61.522

=−++ ff TTQ

QT f 6924.82

=

QTT ff 6924.82max

==

( )21

2 coscosmax

φφ −= fPbrT f

( )( )( )( ) ( )125cos15cos8510035.06924.82

−=Q

lbQ 1984=

( ) lbinQT f −=== 452,101984268.5268.51

( ) lbinQT f −=== 246,1719846924.86924.82

Total lbinTTT ffOF −=+=+= 698,27246,17452,1021

(b) CC rotation

Left Side

[ ]∑ = 0BM

( ) 011

=−++ BNBF MMwhQ

( ) 0753.143.04.61.511

=−++ ff TTQ

QT f 6924.81

=

Right Side:


Page 73 of 97

[ ]∑ = 0BM

( ) 022

=−−+ BNBF MMwhQ

( ) 0753.143.04.61.522

=−−+ ff TTQ

QT f 268.52

=

Since values are just interchanged

lbQ 1984=

Total lbinT OF −= 698,27 as in (a)

914. A double-shoe internal brake is actuated by an involute cam as shown, where RQ

is the force on the right shoe at a radius Rw and LQ is the force on the left shoe at

a radius Lw . The pressure of each shoe is proportional to the rotation of the shoe

about B which is inversely proportional to w ; therefore, the ratio of the

maximum pressures is LRRL wwPP = . The dimensions are: face width 4=b ,

6=r , 16

94=h ,

8

11=c ,

16

59=Lw , inwR

16

58= .: for each shoe, o120=θ ,

o301 =φ . The lining is asbestos in rubber compound, Determine the braking

torque and forces RQ and LQ for the maximum permissible pressure for (a)

clockwise rotation, (b) counterclockwise rotation.


Page 74 of 97

Problem 914.

Solution:

( )

−−=

2

2sin2sin

2

12 φφθ

PbarM BN

( ) ( )

−−−=

2

sinsincoscos 1

2

2

2

21

φφφφ

arfPbrM BF ,

( )21


incha 70.48

11

16

94

22

22 =

+

=+=

8926.0

16

59

16

58

===L

R

R

L

w

w

p

p

For asbestos in rubber compound, 35.0=f , psip 75=

psipR 75=

( ) psipL 67758926.0 ==

(a) Clockwise rotation

Left Side:


Page 75 of 97

[ ]∑ = 0

LBM

0=−−LLLL BNBFLL MMwQ

( ) ( )

−−−=

2

sinsincoscos 1

2

2

2

21

φφφφ

arbrfPM LBF LL

o301 =φ o602 1 =φ

o1503012012 =+=+= φθφ o3002 2 =φ

rad094.2120 == oθ

( )( )( )( ) ( ) ( )lbinM

LL BF −=

−−−= 5849

2

30sin150sin70.4150cos30cos6646735.0

22

( )

−−=

2

2sin2sin

2

12 φφθ

barPM L

BNLL

( )( )( )( ) ( )lbinM

LL BN −=

−−= 185,11

2

60sin300sin094.2

2

67.4467

0185,11584916

59 =−−

LQ

lbQL 1829=

( ) ( )21

2 coscos φφ −= brfPT LOFL

( ) ( )( )( )( ) ( ) lbinTL

OF −=−= 5849150cos30cos646735.02

Right side:


Page 76 of 97

[ ]∑ = 0

RBM

0=−+RRRR BNBFRR MMwQ

( )( )lbin

P

PMM

L

RBF

BFLL

RR−=== 6547

67

755849

( )( )lbin

P

PMM

L

RBN

BNLL

RR−=== 520,12

67

75185,11

0520,12654716

58 =−+

RQ

lbQR 719=

( )( ) ( )( )

lbinP

PTT

L

ROF

OFL

R−=== 6547

67

755849

( ) ( ) ( ) lbinTTTLR

OFOFOF −=+=+= 396,1258496547

(b) Counterclockwise rotation

Left side:


Page 77 of 97

[ ]∑ = 0LBM

0=−+LLLL BNBFLL MMwQ

0185,11584916

59 =−+

LQ

lbQL 573=

( ) lbinTL

OF −= 5849

Right Side:

[ ]∑ = 0

RBM

0=−−RRRR BNBFRR MMwQ

0520,12654716

58 =−−

RQ

lbQR 2294=

( ) lbinTR

OF −= 6547


Page 78 of 97

( ) ( ) ( ) lbinTTTLR

OFOFOF −=+=+= 396,1258496547

BAND BRAKES

915. The steel band for the brake shown is lined with flexible asbestos and it is

expected tha the permissible pressure of Table AT 29 is satisfactory; o245=θ ,

ina 20= ., inm2

13= ., inD 18= ., and face width inb 4= .; rotation CL. The

cast-iron wheel turns 200 rpm. Set up suitable equations, use the average f

given and compute (a) the force in each end of the band, (b) the brake torque and

fhp. (c) Determine the mechanical advantage for the limit values of f in Table

AT 29 and its percentage variation fron that for the average f . (d) Investigate

the overheating problem using relevant information given in the Text.

Problem 915.

Solution:

(1) θfe

F

F=

2

1

[ ]∑ = 0intpoFixedM

mFWa 2=

(2) m

WaF =2


Page 79 of 97

(3) fpAFFF =−= 21

(4) 2

FDT f =

From Table AT 29, flexible asbestos

Ave. 40.0=f , psip 50=

(a) For 1F and 2F :

2

DbA

θ=

rad276.4245 == oθ

( )( )( ) 21542

418276.4inA ==

( )( )( ) lbfpAFFF 30801545040.021 ===−=

( )( ) 5312.5276.440.0

2

1 === eeF

F fθ

21 5312.5 FF =

30805312.5 22 =−= FFF

lbF 6802 =

( ) lbF 37606805312.51 ==

(b) fT and fhp

( )( )lbin

FDT f −=== 720,27

2

183080

2

( )( )hp

nTfhp

f88

000,63

200720,27

000,63===

(c) For MA

mF

T

Wa

TMA

ff

2

==

2

FDT f =

12

−=

θfe

FF

( )m

eD

e

Fm

FD

MAf

f

2

1

1

2 −=

−

=θ

θ

inD 18=

inm 5.3=

rad276.4=θ


Page 80 of 97

Limit values (Table AT 29) 35.0=f to 45.0 .

35.0=f ( )( )[ ]

( )914.8

5.32

118 276.435.0

=−

=e

MA

45.0=f ( )( )[ ]

( )042.15

5.32

118 276.445.0

=−

=e

MA

with 40.0=f (average) ( )( )[ ]

( )652.11

5.32

118 276.440.0

=−

=e

MA

Percentage variation from 40.0=f .

35.0=f

( ) %5.23%100652.11

914.8652.11var% =

−=

45.0=f

( ) %1.29%100652.11

652.11042.15var% =

−=

(d) Overheating problem

257.0154

88infhp

A

fhp==

Therefore, a problem of overheating is expected as Rasmussen recommends 0.2 to 0.3

fhp per square inch of brake contact area.

916. (a) For the band brake shown, derive the expressions for the braking torque in

terms of W , etc., for CL rotation and for CC rotation, and specify the ratio bc

for equal effectiveness in both directions of rotation. Are there any proportions of

b and c as shown that would result in the brae being self locking? (b) When o270=θ , ina 16= ., incb 3== ., and inD 12= ., it was found that a

force lbW 50= . Produced a frictional torque of 1000 in-lb. Compute the

coefficient of friction.


Page 81 of 97

Problem 916.

Solution:

(a)

CL:

[ ]∑ = 0OM

cFbFaW 21 += θfeFF 21 =

cFbeFaW f

22 += θ

cbe

aWF

f +=

θ2

cbe

aWeeFF

f

ff

+==

θ

θθ

21

( )cbe

eaW

cbe

aWaWeFFF

f

f

f

f

+

−=

+

−=−=

θ

θ

θ

θ 121


Page 82 of 97

+

−==

cbe

eWaDFDT

f

f

f θ

θ 1

22

CC:

[ ]∑ = 0OM

cFbFaW 12 +=

+

−=

bce

eWaDT

f

f

f θ

θ 1

2

No proportions of b and c as shown that would result in the brake being self-locking.

(b) lbW 50=

lbinT f −=1000

inD 12=

ina 16=

incb 3==

rad7124.4270 == oθ

( )( )( )

+

−==

33

1

2

1216501000

θ

θ

f

f

fe

eT

625.01

1=

+

−θ

θ

f

f

e

e

333.47124.4 == ffee

θ

311.0=f

917. (a) For the brake shown, assume the proper direction of rotation of the cast-iron

wheel for differential acion and derive expressions for the braking torque. (b) Let

inD 14= ., inn4

31= ., inm 4= ., o235=θ , and assume the band to be lined with

woven asbestos. Is there a chance that this brake will be self-acting? If true, will


Page 83 of 97

it always be for the range of values of f given in Table AT 29? (c) The ratio

mn should exceed what value in order for the brake to be self-locking? (d) If the

direction of rotation of the wheel is opposite to that taken in (a), what is the

braking torque with a force lbW 10= . at ina 8= .? (e) Suppose the brake is

used as a stop to prevent reverse motion on a hoist. What is the frictional

horsepower for the forward motion if the wheel turns 63 rpm?

Problems 917, 918.

Solution:

(a) Assume CL

[ ]∑ = 0OM

mFnFWa 21 =+ θfeFF 21 =

( )θθ ff nemFneFmFWa −=−= 222

( )θfnem

WaF

−=2


Page 84 of 97

( )θ

θ

f

f

nem

WaeF

−=1

( )θ

θ

f

f

nem

eWaFFF

−

−=−=

121 , Braking force.

−

−==

θ

θ

f

f

fnem

eWaDFDT

1

22, Braking torque.

(b) inD 14=

inn4

31=

inm 4=

rad10.4235 == oθ

Table AT 29, woven asbestos

35.0=f to 45.0

There is a chance of self-acting if

mnef >θ

inm 4=

use 40.0=f ( )( )

menef >== 0.975.1 10.440.0θ

use 35.0=f ( )( )

menef >== 35.775.1 10.435.0θ

use 45.0=f ( )( )

menef >== 07.1175.1 10.445.0θ

Therefore true for the range of values of f .

(c) mnef >θ , 40.0=f (average)

θfem

n 1>

( )( )10.44.0

1

em

n>

2.0>m

n

(d) For CC:


Page 85 of 97

[ ]∑ = 0OM

mFnFWa 12 =+

nFmFWa 21 −= θfeFF 21 =

( )nmeFnFmeFWa ff −=−= θθ222

nme

WaF

f −=

θ2

nme

WaeF

f

f

−=

θ

θ

1

( )nme

eWaFFF

f

f

−

−=−=

θ

θ 121

−

−==

nme

eWaDFDT

f

f

f θ

θ 1

22

( )( )( ) ( )( )

( )( ) lbine

eT f −=

−

−= 3.123

75.14

1

2

1481010.440.0

10.440.0

(e) ( )( )

hpnT

fhpf

1233.0000,63

633.123

000,63===

918. A differential band brake similar to that shown and lined with woven asbestos,

has the dimensions: inD 18= ., inn 2= ., inm 12= ., o195=θ . (a) Is there a

chance that this brake will be self-acting? (b) If lbW 30= . and ina 26= . ,

compute the maximum braking torque and the corresponding mechanical

advantage. (c) What is the ratio of the braking torque for CL rotation to the

braking torque for CC rotation? (d) A 1/16-in.-thick steel band, SAE 1020 as

rolled, carries the asbestos lining. What should be its width for a factor of safety

of 8, based on the ultimate stress? What should be the face width if the average

pressure is 50 psi?

Solution:


Page 86 of 97

(a) For CL:

mnef >θ

rad4.3195 == oθ

inm 12=

inn 2=

4.0=f

( )

me <= 8.72 4.34.0 , not self-acting

For CC: θf

men >

nmef <θ

( )ne >= 8.4612 4.34.0 , not self-acting

Therefore, there is no change that this brake will be self-acting.

(b) ff TT =max

(CL)

( )( )( ) ( )

( ) lbine

e

nem

eWadT

f

f

f −=

−

−

=

−

−

= 4832

212

1

2

1826301

2 4.34.0

4.34.0

θ

θ

( )( )2.6

2630

4832===

Wa

TMA

f

(c) ( ) lbinCLT f −= 4832

( ) ( )( )( ) ( )

( ) lbine

e

nme

eWadCCT

f

f

f −=

−

−

=

−

−

= 454

212

1

2

1826301

2 4.34.0

4.34.0

θ

θ

( )( )

64.10454

4832===

CCT

CLTRatio

f

f

(d) For SAE 1020, as rolled.

ksisu 65=

psiksiN

ss u 8125125.8

8

65====

bt

Fs 1=

inint 0625.016

1==

max. θ

θ

f

f

nem

WaeF

−=1 (CL)

( )( ) ( )

( ) lbe

eF 3.722

212

26304.34.0

4.34.0

1 =−

=


Page 87 of 97

( )0625.0

3.7228125

bs ==

inb 422.1=

With psip 50=

fpAF =

2

bDA

θ=

max. ( ) ( )( ) ( )( )

( ) lbe

e

nem

eWaF

f

f

9.536212

1263014.34.0

4.34.0

=−

−=

−

−=

θ

θ

2

bDfpF

θ=

( )( )( )( )( )2

184.3504.09.536

b=

inb 88.0=

919. A differential band brake is to be design to absorb 10 fhp at 250 rpm. (a)

Compute the maximum and minimum diameters from both equations (z) and (a),

p. 495, Text. Decide on a size. (b) The band is to be lined with woven asbestos.

The Rasmussen recommendation (§18.4) will help in deciding on the face width.

Also check the permissible pressure in Table AT 29. Choose dimensions of the

lever, its location and shape and the corresponding θ . Be sure the brake is not

self locking. What is the percentage variation of the mechanical advantage from

the minimum value ( minf ) for the f limits in Table AT 29?

Solution:

( )lbin

n

hpT f −=== 2520

250

10000,63000,63

(a) Eq. (z)

inT

Df

96.75

2520

5

3

1

3

1

min =

=

=

inT

Df

57.84

2520

4

3

1

3

1

max =

=

=

Eq. (a)

( ) ( )[ ] infhpD 44.8106060 3

1

3

1

min ===

( ) ( )[ ] infhpD 28.9108080 3

1

3

1

min ===

use inD 5.8=

(b) By Rasmussen

Energy absorption capacity = 0.2 to 0.3 fhp per sq. in. of brake contact area.


Page 88 of 97

Say 0.25 fhp per sq. in.

2

bDA

θ=

hpfhp 10=

inD 5.8=

assume radπθ == o180

A

fhpinfhp =2

( )

=

2

5.8

1025.0

bπ

inb 3=

From Table AT 29, 40.0=f , psipper 50. =

fA

Fp =

( )( ) 2402

5.83inA ==

π

( )lb

D

TF

f593

5.8

252022===

( )psipsip 501.37

404.0

593<== (OK)

For MA :

( )( )θ

θ

f

ff

bec

eD

WA

TMA

−

−==

2

1

Not self-locking θf

bec >

θfe

b

c>

π4.0e

b

c>

5.3>b

c

say 4=b

c or bc 4=

For 40.0=f

( )( )

( )( ) bbeb

e

bec

eD

WA

TMA

f

ff 96.21

42

15.8

2

14.0

4.0

=−

−=

−

−==

π

π

θ

θ

For min35.0 ff ==


Page 89 of 97

( )( )

( )( ) bbeb

e

bec

eD

WA

TMA

f

ff 54.8

42

15.8

2

135.0

35.0

=−

−=

−

−==

π

π

θ

θ

( ) %157%10054.8

54.896.21% =

−=iationvar

DISK CLUTCHES

920. An automobile engine develops its maximum brake torque at 2800 rpm when the

bhp = 200. A design value of 25.0=f is expected to be reasonable for the

asbestos facing and it is desired that the mean diameter not exceed 8.5 in.;

permissible pressure is 35 psi. Designing for a single plate clutch, Fig. 18.10,

Text, determine the outer and inner diameters of the disk.

Solution:

( ) inDDD iom 5.82

1=+=

inrr io 5.8=+

( )lbin

n

hpT f −=== 4500

2800

200000,63000,63

psip 35=

( )2

iof

rrNfT

+=

( )( )( )2

5.825.04500

N=

lbN 4235=

ave. ( )22

io rr

Np

−=

π

( )22

423535

io rr −=

π

5.3822 =− io rr

io rr −= 5.8

( ) 5.385.8 22=−− ii rr

5.381725.72 22 =−+− iii rrr

inri 985.1=

say inri 0.2=

inro 5.60.25.8 =−=

( ) inrD oo 135.622 ===

( ) inrD ii 40.222 ===


Page 90 of 97

921. An automobile engine can develop a maximum brake torque of 2448 in-lb.

Which of the following plate clutches, which make up a manufacturer’s standard

“line,” should be chosen for this car? Facing sizes: (a) 8

78=oD , inDi

8

16= ., (b)

10=oD , inDi8

16= ., (c)

16

111=oD , inDi

8

16= . In each case, assume 3.0=f .

The unit pressures are (a) 34 psi, (b) 30 psi, and (c) 26.2 psi.

Solution:

( )2

iof

rrNfT

+=

( )4

iof

DDNfT

+=

( )22

4ioave DDpN −

=

π

( )( )16

22

ioioavef

DDDDpT

+−=

π

(a) inDo 875.8= ; inDi 125.6= , psip 34= , 3.0=f

( )( )16

22

ioioavef

DDDDpT

+−=

π

( )( ) ( ) ( )[ ]( )lbinT f −=

+−= 1239

16

125.6875.8125.6875.8343.022

π

(b) inDo 10= ; inDi 125.6= , psip 30= , 3.0=f

( )( )16

22

ioioavef

DDDDpT

+−=

π

( )( ) ( ) ( )[ ]( )lbinT f −=

+−= 1780

16

125.610125.610303.022

π

(c) inDo 0625.11= ; inDi 125.6= , psip 2.26= , 3.0=f

( )( )16

22

ioioavef

DDDDpT

+−=

π

( )( ) ( ) ( )[ ]( )lbinT f −=

+−= 2251

16

125.60625.11125.60625.112.263.022

π

use (c)


Page 91 of 97

922. A single-disk clutch for an industrial application, similar to that in Fig. 18.11,

Text, except that there are two disks attached to one shaft and one attached to the

other. The clutch is rated at 50 hp at 500 rpm. The asbestos-in-resin-binder facing

has a inDo2

18= . and inDi

4

34= . What must be the axial force and average

pressure? How does this pressure compare with that recommended by Table AT

29?

Solution:

2=n pairs in contact

35.0=f (Table AT 29)

psip 75=

( )lbin

n

hpT

m

f −=== 6300500

50000,63000,63

inDo 5.8=

inDi 75.4=

( )2

iof

rrNnfT

+=

( )4

iof

DDNnfT

+=

( )( )( )( )4

75.45.835.026300

+=

N

lbN 2717=

ave. ( )

( )( ) ( )[ ] psipsi

DD

Np

io

756.6975.45.8

2717442222

<=−

=−

=ππ

923. A multiple-disk clutch similar to Fig. 18.11, Text, is rated at 22 hp at 100 rpm.

The outside and inside diameters of the disks are 14 and 7 ½ in., respectively. If

25.0=f , find (a) the axial force required to transmit the rated load, and (b) the

unit pressure between the disks.

Solution:

(a) Fig. 18-11, 4=n pairs in contact

( )lbin

n

hpT

m

f −=== 860,13100

22000,63000,63

( )4

iof

DDNnfT

+=

inDo 14=

inDi 5.7=

( )( )( )4

5.71425.04860,13

+=

N


Page 92 of 97

lbN 2579=

(b) ( )

( )( ) ( )[ ] psi

DD

Np

io

5.235.714

2579442222

=−

=−

=ππ

924. A multiple-disk clutch for a machine tool operation has 4 phosphor-bronze

driving disks and 5 hardened-steel driven disks. This clutch is rated at 5.8 hp at

100 rpm when operated dry. The outside and inside diameters of the disks are 5

½ and 4 3/16 in., respectively. (a) If the pressure between the disks is that

recommended for metal on metal in Table AT 29, what coefficient of friction is

required to transmit the rated power? (b) What power may be transmitted for f

and p as recommended in Table AT 29?

Solution:

inDo 5.5=

inDi 1875.4=

( )lbin

n

hpT

m

f −=== 3654100

8.5000,63000,63


(a) Table AT 29, psip 150= , metal to metal

( )4

iof

DDNnfT

+=

( ) ( ) ( )[ ] lbDDpN io 14981875.45.54

1504

2222 =−

=−

=

ππ

( )( )( )( )4

1875.45.5149883654

+==

fT f

126.0=f

(b) from Table AT 29, psip 150= , 2.0=f

( )( )( )( )lbinT f −=

+= 5805

4

1875.45.514982.08

( )( )hp

nThp

mf2.9

000,63

1005805

000,63===

925. A multiple-disk clutch with three disks on one shaft and two on the other, similar

to that in Fig. 18.11, Text, is rated at 53 hp at 500 rpm. (a) What is the largest

value of iD if f and p are given by Table AT 29 for asbestos in resin binder

and inDo 5.10= . (b) For the diameter used of inDi 7= .,what is the required

axial force and the average pressure?

Solution:


Page 93 of 97

Table AT 29, asbestos in resin binder, 3.0=f , psip 75=

( )2

iof

rrNnfT

+=

( )4

iof

DDNnfT

+=

( )io

f

DDnf

TN

+=

4

but

( )22

4io DDpN −

=

π

( )( )ioiof DDDDnfpT +−

= 22

44

π

( )lbin

n

hpT

m

f −=== 6678500

53000,63000,63


( )( )ioiof DDDDnfpT +−

= 22

44

π

( ) ( )( )( ) ( )[ ]( )ii DD +−

= 5.105.10

4753.0466784 22π

inDi 5607.9=

(b) inDi 7=

( )4

iof

DDNnfT

+=

( )( )( )( )4

75.103.046678

+=

N

lbN 1272=

ave. ( )

( )( ) ( )[ ] psi

DD

Np

io

44.2675.10

1272442222

=−

=−

=ππ

MISCELLANEOUS CLUTCHES AND BRAKES

926. For the cone brake shown, find an expression for the braking torque for a given

applied force W on the bell crank. Consider the force F ′ , Fig. 18.12, Text, in

obtaining the expression.


Page 94 of 97

Problems 926-928.

Solution:

927. For the cone brake similar to that shown, certain dimensions are: inDm 15= .,

inc2

12= ., o12=α , inb 9= ., and ina 20= . The contact surfaces are metal and

asbestos. (a) For an applied force lbW 80= ., what braking torque may be

expected of this brake? Consider the resistance F ′ , Fig. 18.12, Text. (b) If the

rotating shaft comes to rest from 300 rpm during 100 revolutions, what frictional

work has been done? (c) What must be the diameter of the steel pin P , SAE

1020 as rolled, for a factor fo safety of 6 against being sheard off? The diameter

of the hub ind2

14= . (d) What is the unit pressure on the face of the brake?

Solution:

(a) ( ) ( )αααα cossin2cossin2 fb

aWDf

f

RDfT mm

f+

=+

=

Table AT 29, asbestos on metal, 40.0=f

( )( )( )( )( )( )

lbinT f −=+

= 89012cos4.012sin92

80201540.0

(b) ( )

sec42.3160

30021 rad==

πω

sec02 rad=ω

( ) rad3.6282100 == πθ

( )t212

1ωωθ +=

( )t042.312

13.628 +=


Page 95 of 97

sec40=t

( ) rpmnm 15003002

1=+=

( )( )hp

nTfhp

mf119.2

000,63

150890

000,63===

( )( ) ( )( ) lbfttfhpU f −=== 618,4640119.2550550

(c) For SAE 1020, as rolled,

ksissu 49=

ksiN

ss su

s 17.86

49===

2

4

d

Rss

π=

( )( )lb

b

aWR 8.177

9

8020===

( )2

8.17748170

dss

π==

ind 1665.0=

say ind16

3=

(d) cD

Np

mπ=

lbf

RN 297

12cos4.012sin

8.177

cossin=

+=

+=

αα

( )( )psip 52.2

5.215

297==

π

928. A cone clutch for industrial use is to transmit 15 hp at 400 rpm. The mean

diameter of the clutch is 10 in. and the face angle o10=α ; let 3.0=f for the

cast-iron cup and the asbestos lined cone; permissible psip 35= . Compute (a)

the needed axial force, (b) the face width, (c) the minimum axial force to achieve

engagement under load.

Solution:

( )lbin

n

hpT f −=== 5.2362

400

15000,63000,63

(a) ( )αα cossin2 f

RDfT m

f+

=


Page 96 of 97

( )( )( )10cos3.010sin2

103.05.2362

+=

R

lbR 739=

(b) cD

Np

mπ=

lbf

RN 1575

10cos3.010sin

739

cossin=

+=

+=

αα

( )c15

157535

π=

inc 44.1=

(c) max. 4.0=f (Table AT 29)

( )αα cossin2 f

RDfT m

f+

=

( )( )( )10cos4.010sin2

104.05.2362

+=

R

lbR 670= , minimum.

929. An “Airflex” clutch, Fig. 18.15, Text, has a 16-in drum with a 5-in. face. This

clutch is rated at 110 hp at 100 rpm with an air pressure of 75 psi. What must be

the coefficient of friction if the effect of centrifugal force is neglected? (Data

courtesy of Federal Fawick Corporation.)

Solution:

inD 16=

inb 5=

hphp 110=

rpmrpm 100=

psip 75=

( )lbin

n

hpT f −=== 300,69

100

110000,63000,63

2

FDT f =

( )2

16300,69

F=

lbF 5.8662=

( ) ( )( )( )( ) lbDbpN 850,1851675 === ππ

46.0850,18

5.8662===

N

Ff


Page 97 of 97

930. The same as 929 except that the diameter is 6 in., the face width is 2 in., and the

rated horsepower is 3.

Solution:

( )lbin

n

hpT f −=== 1890

100

3000,63000,63

2

FDT f =

( )2

6300,69

F=

lbF 630=

( ) ( )( )( )( ) lbDbpN 28272675 === ππ

22.02827

630===

N

Ff

- end -

SECTION 18 – MISCELLANEOUS PROBLEMS

Page 1 of 25

THIN SHELLS, EXTERNAL PRESSURE

981. A closed cylindrical tank is used for a steam heater. The inner shell, 200 in.

outside diameter and 50 ft. long, is subjected to an external pressure of 40 psi.

The material is equivalent to SA 30 (ASME Pressure-Vessel Code: min.

ksisu 55= ); assume an elastic limit of 2uy ss = ; let 5=N . (a) What thickness

of shell is needed from a stress standpoint? (b) For this thickness, what must be

the maximum length of unsupported section to insure against collapse? (c)

Choose a spacing L to give a symmetric arrangement and determine the moment

of inertia of the steel stiffening rings. (d) For a similar problem, the Code

recommends that int 76.0≥ , inL 50= , and 496 inI = . How do these values

check with those obtained above? (e) Without stiffening rings, what thickness

would be needed?

Solution:

(a) Solving for the thickness of shell,

y

c

s

Dpt

2=

( ) psippc 2004055 ===

psis

s uy 500,27

2

000,55

2===

inD 200=

( )( )( )

ins

Dpt

y

c 73.0500,272

200200

2===

say int 75.0=

(b) Solving for the maximum length of unsupported section, use Eq. (20-1)

psi

D

t

D

L

D

tE

pc

2

1

2

5

45.0

60.2

−

=

2

12

5

45.0

60.2

+

=D

t

p

D

tE

D

L

c

psiE 61030×=

psipc 200=

int 75.0=


Page 2 of 25

inD 200=

( )2

12

5

6

200

75.045.0

200

200

75.0103060.2

200

+

×

=L

inL 68.72=

(c) Solving for the moment of inertia of the steel stiffening rings.

Choosing inL 60= for 50 ft long shell

( ) ( )( ) 4

6

33

1121030

20060200035.0035.0in

E

LpDI c =

×==

(d) For int 76.0≥ - above minimum

inL 50= - below maximum 496 inI = - lighter than above.

(e) inftL 60050 ==

Solving for thickness without stiffening rings

By Saunders and Windenburg, Eq. 20-1

psi

D

t

D

L

D

tE

pc

2

1

2

5

45.0

60.2

−

=

E

D

t

D

Lp

D

tc

60.2

45.02

1

2

5

−

=

( )6

2

1

2

5

103060.2

20045.0

200

600200

200 ×

−

=

t

t

2

1

2

5

364.6600886.137 tt −=

600364.6886.137 2

1

2

5

=+ tt

int 791.1=


Page 3 of 25

say int16

131=

982. The same as 981, except that psip 175= , ftD 4= , and the length of the tank is

18 ft.

(a) Solving for the thickness of shell,

y

c

s

Dpt

2=

( ) psippc 87517555 ===

psis

s uy 500,27

2

000,55

2===

inftD 484 ==

( )( )( )

ins

Dpt

y

c 53.1500,272

48875

2===

say int 5625.116

91 ==


psi

D

t

D

L

D

tE

pc

2

1

2

5

45.0

60.2

−

=

2

12

5

45.0

60.2

+

=D

t

p

D

tE

D

L

c

psiE 61030×=

psipc 875=

int 5625.1=

inD 48=

( )2

12

5

6

48

5625.145.0

875

48

5625.1103060.2

48

+

×

=L

inL 822=

(c) Since length of shell = 18 ft = 216 in < 822 in, there is no need for stiffeners.


Page 4 of 25

(d) inL 216=



psi

D

t

D

L

D

tE

pc

2

1

2

5

45.0

60.2

−

=

E

D

t

D

Lp

D

tc

60.2

45.02

1

2

5

−

=

( )6

2

1

2

5

103060.2

4845.0

48

216875

48 ×

−

=

t

t

2

1

2

5

833.565.3937428.4886 tt −=

5.3937833.56428.4886 2

1

2

5

=+ tt

int 9122.0=

say int16

15=

but minimum int16

191=

use int16

191=

Approximate ratio of weight of this shell to the weight of the shell found in (a) =

thickness of shell without stiffening rings / thickness of shell with stiffening rings

= 0.34375 / 0.09375 = 3.6667

STEEL TUBES, EXTERNAL PRESSURE

983. A closed cylindrical tank, 6 ft in diameter, 10 ft long, is subjected to an internal

pressure of 1 psi absolute. The atmospheric pressure on the outside is 14.7 psi.

The material is equivalent to SA 30 (ASME Pressure Vessel Code:

min ksisu 55= ); assume an elastic limit of 2uy ss = ; let 5=ppc . (a) What

thickness of shell is needed for the specified design stress? (b) For this thickness,

what must be the maximum length of unsupported section to insure against

collapse? (c) Choose a symmetric spacing L of stiffening rings, and compute


Page 5 of 25

their moment of inertia and the cross-sectional dimensions h and b if they are

rectangular with bh 2= . (d) Suppose that the tank had no stiffening rings. What

thickness of shell would be needed? What is the approximate ratio of the weight

of the shell found in (a)? Material costs are roughly proportional to the weight.

Solution:

(a) Solving for the thickness of shell

y

c

s

Dpt

2=

( ) psippc 5.6817.1455 =−==

psis

s uy 500,27

2

000,55

2===

inftD 726 ==

( )( )( )

ins

Dpt

y

c 08967.0500,272

725.68

2===

say int 09375.032

3==


psi

D

t

D

L

D

tE

pc

2

1

2

5

45.0

60.2

−

=

2

12

5

45.0

60.2

+

=D

t

p

D

tE

D

L

c

psiE 61030×=

psipc 5.68=

int 09375.0=

inD 72=

( )2

12

5

6

72

09375.045.0

5.68

72

09375.0103060.2

72

+

×

=L

inL 2.6=

(c) For the length of shell = 10 ft = 120 in.

use inL 0.6=


Page 6 of 25

Moment of inertia of stiffening rings.

( ) ( )( ) 4

6

33

1790.01030

5.68672035.0035.0in

E

LpDI c =

×==

Solving for cross-sectional dimension

bh 2=

4443

1790.03

2

12

8

12in

bbbhI ====

inb 72.0=

say inb4

3=

inh2

11

4

32 =

=

(d) inftL 12010 ==



psi

D

t

D

L

D

tE

pc

2

1

2

5

45.0

60.2

−

=

E

D

t

D

Lp

D

tc

60.2

45.02

1

2

5

−

=

( )6

2

1

2

5

103060.2

7245.0

72

1205.68

72 ×

−

=

t

t

2

1

2

5

053.0667.18865.25 tt −=

667.1053.08865.25 2

1

2

5

=+ tt

int 3314.0=

say inint 34375.032

11==


Page 7 of 25

984. A long lap-welded steel tube, 8-in. OD, is to withstand an external pressure of

120 psi. with 5=N . (a) What should be the thickness of the wall of the tube? (b)

What is the ratio tD ? Is it within the range of the Stewart equation? (c)

Assuming the internal pressure to be negligible relative to the external pressure,

calculate the maximum principal stress from equation (8.13), p. 255, text. What

design factor is given by this stress compared with ys for AISI C1015 annealed?

(d) Compute the stress from the thin shell formula.

Solution:

(a) Solving for the thickness of the wall

Stewart’s formula 3

000,200,50

=

D

tpc

( )( ) psiNppc 6001205 ===

3

8000,200,50600

=

t

int 1829.0=

say inint 1875.016

3==

(b) ratio 67.421875.0

8==

t

D

or tD 67.42=

outside the range of the Steward equation ( tD 40< )

(c) Using eq. (8.13) (Lame’s formula)

( )22

22222

io

oioiooiit

rr

rpprrrprp

−

−+−=σ

psipo 120=

0≈ip

inOD

ro 42

8

2===

inri 8125.31875.04 =−=

irr =

( ) ( )( )( ) ( )

psirr

rp

rr

prrp

io

oo

io

oooot 2621

8125.34

412022022

2

22

2

22

22

−=−

−=

−

−=

−

−+−=σ

ys of AISI C1015, annealed = 42 ksi


Page 8 of 25

Design factor, 162621

000,42===

t

y

y

sN

σ

(d) Solving for the stress from the thin shell formula

( )( )( )

psit

pDs 2560

1875.02

8120

2===

985. A long lap-welded steel tube, 3 –in. OD, is to withstand an external pressure of

150 psi with 5=N . Parts (a) – (c) are the same as in 984.

Solution:

(a) Solving for the thickness of the wall

Stewart’s equation 3

000,200,50

=

D

tpc

( )( ) psiNppc 7501505 ===

3

3000,200,50750

=

t

int 074.0=

say inint 078125.064

5==

(b) ratio 4.38078125.0

3==

t

D

or tD 4.38=

within the range of the Steward equation ( tD 40< )

(c) Using eq. (8.13) (Lame’s formula)

psipo 150=

inOD

ro 5.12

3

2===

inri 421875.1078125.05.1 =−=

( )( )( ) ( )

psirr

rp

io

oot 2957

421875.15.1

5.11502222

2

22

2

−=−

−=

−

−=σ

ys of AISI C1015, annealed = 42 ksi

Design factor, 2.142927

000,42===

t

y

y

sN

σ


Page 9 of 25

986. A long lap-welded tube, 3-in. OD, is made of SAE 1015, annealed. Let the shell

thickness 40Dt = and 5=N . (a) What is the corresponding safe external

pressure? (b) Compute the maximum principal stress (p. 255, Text), assuming a

negligible internal pressure. What design factor is given by this stress compared

with ys ? (c) Compare with stress computed from the thin-shell formula.

Solution:

(a) Solving for safe external pressure,

3

000,200,50

=

D

tpc

40

1=

D

t

psipc 78440

1000,200,50

3

=

=

psiN

pp c 157

5

784===

(b) Solving for maximum principal stress, neglecting internal pressure

psipo 157=

inOD

ro 5.12

3

2===

inD

t 075.040

3

40===

inri 425.1075.05.1 =−=

( )( )( ) ( )

psirr

rp

io

oot 3221

425.15.1

5.11572222

2

22

2

−=−

−=

−

−=σ

ys of SAE 1015 annealed = 42 ksi

Design factor, 0.133221

000,42===

t

y

y

sN

σ

(c) Solving for stress from the thin-shell formula

( )( )( )

psit

pDs 3140

075.02

3157

2===

FLAT PLATES

987. A circular plate 24 in. in diameter and supported but not fixed at the edges, is

subjected to a uniformly distributed load of 125 psi. The material is SAE 1020, as


Page 10 of 25

rolled, and 5.2=N based on the yield strength. Determine the thickness of the

plate.

Solution:

Solving for the thickness of the plate

psit

rps

2

=

for SAE 1020,a s rolled, ksisy 48=

psiN

ss

y200,19

5.2

000,48===

psip 125=

inr 122

24==

212

125200,19

==

ts

int 968.0=

say int 1=

988. The cylinder head of a compressor is a circular cast-iron plate (ASTM class 20),

mounted on a 12-in. cylinder in which the pressure is 250 psi. Assuming the head

to be supported but not fixed at the edges, compute its thickness for 6=N based

on ultimate strength.

Solution:

Solving for the thickness of the head

psit

rps

2

=

for cast-iron (ASTM class 20), ksisu 20=

psiN

ss u 3333

6

000,20===

psip 250=

inr 62

12==

26

2503333

==

ts

int 6432.1=

say inint 65625.132

211 ==


Page 11 of 25

989. A 10x15-in. rectangular opening in the head of a pressure vessel, whose internal

pressure is 175 psi, is covered with a flat plate of SAE 1015, annealed. Assuming

the plate to be supported at the edges, compute its thickness for 6=N based on

ultimate strength.

Solution:

Solving for the thickness of the head

( )psi

bat

pbas

222

22

2 +=

for SAE 1015, annealed, ksisu 56=

psiN

ss u 9333

6

000,56===

ina 10=

inb 15=

psip 175=

( ) ( ) ( )( ) ( )[ ] psi

ts

222

22

15102

17515109333

+==

int 8056.0=

say inint 8125.016

13==

CAMS

990. The force between a 5/8-in. hardened steel roller and a cast-iron (140 BHN) cam

is 100 lb.; radius of cam curvature at this point is 1 ¼ in. Compute the contact

width.

Solution:

Solving for the contact width

+

=

21

11

rrN

bKP c

From Table 20-2, hardened steel and a cast-iron, BHN = 400

Use 9002

=cK

lbP 100=

inr 3125.08

5

2

11 =

=

inr 25.14

112 ==

15.1=N


Page 12 of 25

+

=

25.1

1

3125.0

115.1

900100

b

inb 511.0=

991. A radial cam is to lift a roller follower 3 in. with harmonic motion during a 150o

turn of the cam; 1 1.2-in. roller of hardened steel. The reciprocating parts weigh

10 lb., the spring force is 175 lb., the external force during the lift is 250 lb. The

cast-iron (225 BHN) cam turns 175 rpm. The cam curvature at the point of

maximum acceleration is 1 ½-in. radius. Compute the contact width.

Solution:

Neglecting frictional forces

φcosPFFFQ resg =+++

Q = external force during lift = 250 lb

gF = weight of reciprocating parts = 10 lb

sF = spring force = 175 lb

reF = reversed effective force qma−=

maFre 2−= for harmonic motion

== θ

β

π

β

πωcos

2

2

Lxa &&

inL 3=

( )sec326.18

60

1752rad==

πω

rad618.2180

150150 === πβ o

at maximum acceleration

( ) 2

2

sec726618.2

326.18

2

3inx =

=

π&&

xmFFFQP resg&&217510250cos −++=+++=φ

( )( ) in

lb

ftinft

lb

g

Fm

g2

2

sec

4.386

10

12sec2.32

10 −===

assume 1cos ≈φ

( ) lbP 3977264.386

102435 =

−=


Page 13 of 25


+

=

21

11

rrN

bKP c


Use 21002

=cK

ininr 5.12

111 ==

ininr 5.12

112 ==

15.1=N

+

=

5.1

1

5.1

115.1

2100397

b

inb 29.0=

992. The same as 991, except that the motion of the follower is cycloidal.

Solution: maFre 1.1−= for cycloidal motion

As a continuation of 991, but

== θ

β

π

β

ωπ 2sin

22

2L

xa && for cycloidal

inL 3=

( )sec326.18

60

1752rad==

πω

rad618.2180

150150 === πβ o


( )( )( )

2

2

2

2

2

sec924618.2

326.18322in

Lx ===

π

β

ωπ&&

( ) lbP 4099244.386

101.1435 =

−=


+

=

21

11

rrN

bKP c


Use 21002

=cK


Page 14 of 25

ininr 5.12

111 ==

ininr 5.12

112 ==

15.1=N

+

=

5.1

1

5.1

115.1

2100409

b

inb 30.0=

993. The same as 991, except that the motion of the follower is parabolic.

Solution: maFre 3−= for parabolic motion

As a continuation of 991, but 2

2

==

β

ωLxa && for parabolic

inL 3=

( )sec326.18

60

1752rad==

πω

rad618.2180

150150 === πβ o


( ) ( ) 2

2

sec588618.2

326.1823 inx =

=&&

( ) lbP 3895884.386

103435 =

−=


+

=

21

11

rrN

bKP c


Use 21002

=cK

ininr 5.12

111 ==

ininr 5.12

112 ==

15.1=N


Page 15 of 25

+

=

5.1

1

5.1

115.1

2100389

b

inb 284.0=

FLYWHEELS AND DISK

994. A cast-iron flywheel with a mean diameter of 36 in. changes speed from 400 rpm

to 380 rpm while it gives up 8000 ft-lb of energy. What is the coefficient of

fluctuation, the weight, and the approximate sectional area of the rim?

Solution:

Solving for coefficient of fluctuation

n

nnC f

21 −=

2

21 nnn

+=

( )

21

212

nn

nnC f

+

−=

rpmn 4001 =

rpmn 3802 =

( )0513.0

380400

3804002=

+

−=fC

Solving for the weight

2

2.32

sf vC

KEw

∆=

lbftKE −=∆ 8000

0513.0=fC

( )6012

Dnvs

π=

inD 36=

rpmn 3902

380400=

+=

( )( )( )

fpsvs 26.616012

39036==

π

( )( )

lbw 133826.610513.0

80002.322

==


Page 16 of 25

Solving for the approximate sectional area of the rim

Vw ρ=

assume 3254.0 inlb=ρ for cast iron

DAV π=

inD 36=

DAw ρπ=

( )A36254.01338 π= 236 inA =

995. The energy required to shear a 1-in. round bar is approximately 1000 ft-lb. In use,

the shearing machine is expected to make a maximum of 40 cutting strokes a

minute. The frictional losses should not exceed 15 % of the motor output. The

shaft carrying the flywheel is to average 150 rpm. (a) What motor horsepower is

required? (b) Assuming a size of flywheel and choosing appropriate fC , find the

mass and sectional dimensions of the rim of a cast-iron flywheel. The width of

the rim is to equal the depth and is not to exceed 3 ½ in. It would be safe to

assume that all the work of shearing is supplied by the kinetic energy given up by

the flywheel.

Solution:

(a) Solving for the horsepower required

( )( )( )( )min000,331

min

−−−=

hplbftlossesFrictional

perStrokesrequiredEnergyhp

( )( )( )

hphplbft

hp 426.1min000,3315.01

401000=

−−−=

(b) Solving for the mass of the rim and size of section

2

2.32

sf vC

KEw

∆=

Vw ρ=

assume 3254.0 inlb=ρ for cast iron

DAV π=

2

2.32

sf vC

KEDAw

∆== ρπ

assume 06.0=fC (Table 20-3)

lbftKE −=∆ 1000


Page 17 of 25

( )6012

Dnvs

π=

rpmn 150=

2

24

5

2.32

∆==

DC

KEDAw

f

πρπ

AC

KED

f ρππ

2

3

24

5

2.32

∆=

using width = depth = 3 ½ in

( )( ) 225.125.35.3 inA ==

( )

( )( )( )25.12254.024

506.0

10002.322

3

ππ

=D

inD 42.50=

assume inD 51=

( )fpsvs 38.33

24

515==

π

( )( )

lbw 48238.3306.0

10002.322

==

slugsg

wm 15

2.32

482===

( )( )( )

284.1151254.0

482in

D

wwidthdepthA ====

πρπ

ininwidthdepth 44.384.11 2 ===

say inwidthdepth2

13==

996. The same as 995, except that the capacity of the machine is such as to cut 1 ½-in.

round brass rod, for which the energy required is about 400 ft-lb./sq. in. of

section.

Solution:

(a) Solving for the horsepower required

( )( )( )( )min000,331

min

−−−=

hplbftlossesFrictional

perStrokesrequiredEnergyhp


Page 18 of 25

( ) ( ) lbftininlbftrequiredEnergy −=

−= 7075.1

4400

22 π

( )( )( )( )

hphplbft

hp 01.1min000,3315.01

40707=

−−−=

(b) Solving for the mass of the rim and size of section

AC

KED

f ρππ

2

3

24

5

2.32

∆=

lbftKE −=∆ 707

06.0=fC

3254.0 inlb=ρ

( )( ) 225.125.35.3 inA ==

( )

( )( )( )25.12254.024

506.0

7072.322

3

ππ

=D

inD 45=

use inD 45=

2

2.32

sf vC

KEw

∆=

( )fps

Dvs 45.29

24

455

24

5===

ππ

( )( )

lbw 43845.2906.0

7072.322

==

slugsg

wm 6.13

2.32

438===

( )( )( )

220.1245254.0

438in

D

wwidthdepthA ====

πρπ

ininwidthdepth 49.320.12 2 ===

say inwidthdepth2

13==

997. A 75-hp Diesel engine, running at 517 rpm, has a maximum variation of output

of energy of 3730 ft-lb. The engine has three 8 x 10 ½ in. cylinders and is

directly connected to an a-c generator. (a) What should be the weight and

sectional area of the flywheel rim if it has an outside diameter of 48-in.? (b) The

actual flywheel and generator have 22 6787 ftlbWk −= . Compute the

corresponding coefficient of fluctuation and compare.


Page 19 of 25

Solution:

(a) Solving for the weight ad sectional areas

lbvC

KEw

sf

2

2.32 ∆=

assume 0035.0=fC , Table 20-3

lbftKE −=∆ 3730

( )6012

Dnvs

π=

rpmn 570=

assume inD 48=

( )( )( )

rpsvs 28.1086012

51748==

π

( )( )( )

lbw 292728.1080035.0

37302.322

==

DAw ρπ=

D

wA

ρπ=

assume 3254.0 inlb=ρ (cast iron)

( )( )( )242.76

48254.0

2927inA ==

π

(b) Solving for coefficient of fluctuation 22 6787 ftlbWkIg −==

( )lbft

IKE −

−=∆

2

2

2

2

1 ωω

( )lbft

g

WkKE −

−=∆

2

2

2

2

1

2 ωω

( ) ( )lbft

g

WkKE −

+−=∆

2

2121

2 ωωωω

( )ω

ωω=

+

2

21

( )sec14.54

60

5172rad==

πω

lbftKE −=∆ 3730

( ) 14.542.32

67873730 21 ωω −==∆KE

327.021 =−ωω


Page 20 of 25

0035.0006.014.54

327.021 >==−

=ω

ωωfC

998. A 4-ft flywheel, with an rim 4 in. thick and 3 in. wide, rotates at 400 rpm. If there

are 6 arms, what is the approximate stress in the rim? Is this a safe stress? At

what maximum speed should this flywheel rotate if it is made of cast iron, class

30?

Solution:

Solving for the approximate stress,

psig

vs

o

s

144

2ρ=

( )( )fps

Dnvs 78.83

60

4004

60===

ππ

3254.0 inlb=ρ (class 30, cast iron)

( )( ) 333 4391728254.0254.0 ftlbftlbinlb ===ρ 22.32 sftgo =

( ) ( )( )

psis 6652.32144

43978.832

==

since ( )( ) fpmfpmfpmv 600050276078.83 <== (cast iron)

this is a safe stress

Solving for maximum speed, max. fpmv 6000=

( ) fpmnDnv 60004 === ππ

maximum, rpmn 477=

999. A hollow steel shaft with inDo 6= and inDi 3= rotates at 10,000 rpm. (a) What

is the maximum stress in the shaft due to rotation? Will this stress materially

affect the strength of the shaft? (b) The same as (a), except that the shaft is solid.

Solution:

(a) Solving for maximum stress

( ) ( )[ ] psirrg

s io

o

t

222

134

µµρω

−++=

where

inD

r oo 3

2

6

2===

inD

r ii 5.1

2

3

2===


Page 21 of 25

for steel, 3284.0 inlb=ρ , 30.0=µ

sec386 ingo =

sec104760

000,102 rad=

= πω

( ) ( )[ ] psirrg

s io

o

t

222

134

µµρω

−++=

( )( )( )

( )( ) ( )( )[ ] psist 63065.13.0133.033864

1047284.0 222

=−++=

(does not affect the strength of the shaft)

(b) Solving for the maximum stress for solid

( ) ( )( ) ( )( )( )

psig

rs

o

ot 2994

3864

33.031047284.0

4

32222

=+

=+

=µρω

1000. A circular steel disk has an outside diameter inDo 10= and an inside diameter

inDi 2= . Compute the maximum stress for a speed of (a) 10,000 rpm, (b)

20,000 rpm. (c) What will be the maximum speed without danger of permanent

deformation if the material is AISI 3150, OQT at 1000 F?

Solution:

( ) ( )[ ] psirrg

s io

o

t

222

134

µµρω

−++=

where 3284.0 inlb=ρ

30.0=µ

inD

r oo 5

2

10

2===

inD

r ii 1

2

2

2===

sec386 ingo =

(a) Solving for maximum stress for a speed of 10,000 rpm

sec104760

000,102 rad=

= πω

( )( )( )

( )( ) ( )( )[ ] psist 776,1613.0153.033864

1047284.0 222

=−++=


Page 22 of 25

(b) Solving for maximum stress for a speed of 10,000 rpm

sec209460

000,202 rad=

= πω

( )( )( )

( )( ) ( )( )[ ] psist 104,6713.0153.033864

2094284.0 222

=−++=

(c) Solving for maximum speed, ω .

For AISI 3150, OQT at 1000 F, ksisy 130=

( ) ( )[ ] psirrg

ss io

o

yt

222

134

µµρω

−++==

( )( )

( )( ) ( )( )[ ]222

13.0153.033864

284.0000,130 −++=

ω

sec57.2914 rad=ω

( )rpmRPM 832,27

2

57.291460

2

60===

ππ

ω

1001. The same as 1000, except that inDi 1= .

Solution:

inDi 1=

ininD

r ii 5.0

2

1

2===

use other data as in 1000.

(a) Solving for maximum stress for a speed of 10,000 rpm

sec104760

000,102 rad=

= πω

( )( )( )

( )( ) ( )( )[ ] psist 670,165.03.0153.033864

1047284.0 222

=−++=

(b) Solving for maximum stress for a speed of 10,000 rpm

sec209460

000,202 rad=

= πω

( )( )( )

( )( ) ( )( )[ ] psist 680,665.03.0153.033864

2094284.0 222

=−++=

(c) Solving for maximum speed, ω .


Page 23 of 25

For AISI 3150, OQT at 1000 F, ksisy 130=

( ) ( )[ ] psirrg

ss io

o

yt

222

134

µµρω

−++==

( )( )

( )( ) ( )( )[ ]222

5.03.0153.033864

284.0000,130 −++=

ω

sec8.2923 rad=ω

( )rpmRPM 920,27

2

8.292360

2

60===

ππ

ω

1002. A circular steel disk, with inDo 8= and inDi 2= , is shrunk onto a solid steel

shaft with an interference of metal ini 002.0= . (a) At what speed will the

pressure in the fit become zero as a result of the rotation? Assume that the shaft is

unaffected by centrifugal action. (This effect is relatively small.) (b) Compute the

maximum stress in the disk and the pressure at the interface when the speed is

10,000 rpm. Note: The maximum stress in the disk is obtained by adding

equations (8.15) of i8.26, Text, and (n of i20.9. The resulting equation together

with equation (s) of i8.27 can then be used to obtain ip and thσ ; where

its p−=σ for a solid shaft.

Solution:

(a) Solving for speed, 0=ip

From equation 8.15, i8.26, Text.

+−

+=

s

ists

h

ihthi

E

p

E

pDi

µσµσ

inDi 2=

ini 002.0=

psiEE sh

61030×==

30.0== sh µµ

0=−= its pσ

( )( ) ( )( )

×

+−

×

+==

66 1030

03.00

1030

03.02002.0 thi

σ

psith 000,30=σ

ttith s+= σσ

From Equation 8-15

( )22

222 2

io

ooioiti

rr

rprrp

−

−+=σ


Page 24 of 25

but 0=op

( )22

22

io

ioiti

rr

rrp

−

+=σ

inD

r oo 4

2

8

2===

inD

r ii 1

2

2

2===

0=ip

0=tiσ

From Equation (n) i20.9

( ) ( )[ ] psirrg

s io

o

t

222

134

µµρω

−++=

3284.0 inlb=ρ

30.0=µ

inro 4=

inri 1=

sec386 ingo =

ttith s+= σσ

( )( )

( )( ) ( )( )[ ]222

13.0143.033864

284.00000,30 −+++=

ω

sec1746 rad=ω

( )rpmRPM 673,16

2

174660

2

60===

ππ

ω

(b) Solving for the maximum stress in the disk and the pressure within the interface.

+−

+=

s

ists

h

ihthi

E

p

E

pDi

µσµσ

( )

×

+−−

×

+=

1030

3.0

1030

3.02002.0

6

iiith pppσ

ith p+= σ000,30

ith p−= 000,30σ

ttith s+= σσ

( )ii

io

ioiti pp

rr

rrp

15

17

14

1422

22

22

22

=

−

+=

−

+=σ

( ) ( )[ ] psirrg

s io

o

t

222

134

µµρω

−++=


Page 25 of 25

( )sec1047

60

000,102rad==

πω

( )( )( )

( )( ) ( )( )[ ] psist 788,1013.0143.033864

1047284.0 222

=−++=

ttith s+= σσ

788,1015

17000,30 +=−= iith ppσ

psipi 9000= (interface pressure)

psipith 000,219000000,30000,30 =−=−=σ (maximum stress)

- end -

Solucionario faires

Engineering

Transcript of Solucionario faires