Machine Design MD By S K Mondal T&Q.pdf

7/25/2019 Machine Design MD By S K Mondal T&Q.pdf

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Design of Joint

S K Mondals Chapter 1

1. Design of Joint

Theory at a glance (IES, IAS, GATE & PSU)

Cotters In machinery, the general term shaftrefers to a member, usually of circular cross-

section, which supports gears, sprockets, wheels, rotors, etc., and which is subjected to

torsion and to transverse or axial loads acting singly or in combination.

An axle is a non-rotating member that supports wheels, pulleys, and carries notorque. A spindle is a short shaft. Terms such as line-shaft, head-shaft, stub shaft,

transmission shaft, countershaft, and flexible shaft are names associated with special

usage.

A cotter is a flat wedge-shaped piece of steel as shown in figure below.This is used to connect

rigidly two rods which transmit motion in the axial direction, without rotation. These joints

may be subjected to tensile or compressive forces along the axes of the rods.

Examples of cotter joint connections are: connection of piston rod to the crosshead of a steam

engine, valve rod and its stem etc.

Figure- A typical cotter with a taper on one side only

A typical cotter joint is as shown in figure below. One of the rods has a socket end into which

the other rod is inserted and the cotter is driven into a slot, made in both the socket and the

rod. The cotter tapers in width (usually 1:24)on one side only and when this is driven in, the

rod is forced into the socket. However, if the taper is provided on both the edges it must be less

than the sum of the friction angles for both the edges to make it self locking i.e. + < +1 2 1 2

where1 2, are the angles of taper on the rod edge and socket edge of the cotter respectively

and 1 2, are the corresponding angles of friction. This also means that if taper is given on

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Design of Joint

S K Mondals Chapter 1one side only then < +1 2 for self locking. Clearances between the cotter and slots in the

rod end and socket allows the driven cotter to draw together the two parts of the joint until the

socket end comes in contact with the cotter on the rod end.

t

3d 1

d 2dd

Figure- Cross-sectional views of a typical cotter joint

Figure- An isometric view of a typical cotter joint

Design of a cotter joint

If the allowable stresses in tension, compression and shear for the socket, rod and cotter be

,t c and respectively, assuming that they are all made of the same material, we may write

the following failure criteria:

1. Tension failure of rod at diameter d

=2

4td P

Fig. Tension failure of the rod

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Design of Joint

S K Mondals Chapter 12. Tension failure of rod across

slot

=

2

1 14 td d t P

Fig. Tension failure of rod across slot

3. Tensile failure of socket

across slot

( ) ( )

=

2 2

2 1 2 14

td d d d t P

Fig. Tensile failure of socket across slot

4. Shear failure of cotter

=2bt P

Fig. Shear failure of cotter

5. Shear failure of rod end

=1 1

2l d P

Fig. Shear failure of rod end

6. Shear failure of socket end

( ) =3 12l d d P

Fig. Shear failure of socket end

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Design of Joint

S K Mondals Chapter 17. Crushing failure of rod or

cotter

=1 c

d t P

Fig. Crushing failure of rod or cotter

8. Crushing failure of socket or

rod

( ) =3 1 cd d t P

Fig.Crushing failure of socket or rod

9. Crushing failure of collar

( )

=2 24 1

4cd d P

Fig. Crushing failure of collar

10. Shear failure of collar

=1 1d t P

Fig. Shear failure of collar

Cotters may bend when driven into position. When this occurs, the bending moment cannot be

correctly estimated since the pressure distribution is not known. However, if we assume a

triangular pressure distribution over the rod, as shown in figure below, we may approximate theloading as shown in figure below.

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Design of Joint


Figure-Bending of the cotter

This gives maximum bending moment

= +

3 1 1

2 6 4

d d dPand

The bending stress, ( )

+ +

= =

3 1 3 11 1

3 2

32 6 4 2 6 4

12

b

d d d dd dP bP

tb tb

Tightening of cotter introduces initial stresses which are again difficult to estimate.

Sometimes therefore it is necessary to use empirical proportions to design the joint. Some

typical proportions are given below:

A design based on empirical relation may be checked using the formulae based on failure

Mechanisms.

Question:Design a typical cotter joint to transmit a load of 50 kN in tension or compression.Consider that the rod, socket and cotter are all made of a material with the following

allowable stresses:

Allowable tensile stress ( y)= 150 MPa

Allowable crushing stress (c)= 110 MPa

Allowable shear stress (y)= 110 MPa.

Answer: Refer to figures

d1= 1.21.d d4 =

1.5.d

l = l1= 0.75d

d2= 1.75.d t = 0.31d t1= 0.45d

d3 = 2.4 d b = 1.6d c = clearance

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Design of Joint


t

3d 1

d 2dd

Axial load( )

= 2

4

yP d . On substitution this gives d=20 mm. In general

Standard shaft size in mm is

6 mm to 22 mm diameter 2 mm in increment






We therefore choose a suitable rod size to be 25 mm.

Refer to figure

For tension failure across slot = 2

14

yd d t P . This gives

d1t = 1.58x10-4m2. From empirical relations we may take t=0.4d i.e. 10 mm and this

gives d1= 15.8 mm. Maintaining the proportion let d1= 1.2 d = 30 mm.

Refer to figure

The tensile failure of socket across slot ( )

=

2 2

2 1 2 14

yd d d d t P

This gives d2= 37 mm. Let d2= 40 mm

Refer to figure above

For shear failure of cotter 2bt = P. On substitution this gives b = 22.72 mm.

Let b = 25 mm.

Refer to figure

For shear failure of rod end 2l1d1 = Pand this gives l1= 7.57 mm. Let l1= 10 mm.

Refer to figure

For shear failure of socket end2l(d2-d1) = Pand this gives l = 22.72 mm. Let l = 25

mm.

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Design of Joint

S K Mondals Chapter 1Refer to figure

For crushing failure of socket or rod (d3-d1)tc = P. This gives d3= 75.5 mm. Let d3= 77mm.

Refer to figure

For crushing failure of collar ( )

=2 24 1

4c

d d P. On substitution this gives d4= 38.4

mm. Let d4= 40 mm.

Refer to figure

For shear failure of collar d1t1= P which gives t1= 4.8 mm. Let t1= 5 mm.

Therefore the final chosen values of dimensions are

d = 25 mm; d1= 30 mm; d2= 40 mm; d3= 77 mm; d4= 40 mm; t = 10 mm; t1= 5 mm; l =

25 mm; l1= 10 mm; b = 27 mm.

Knuckle Joint

A knuckle joint (as shown in figure below) is used to connect two rods under tensile load. This

joint permits angular misalignment of the rods and may take compressive load if it is guided.

d

2t

3d

2d

1d

t

1t

1t

2t

d

Figure- A typical knuckle joint

These joints are used for different types of connections e.g. tie rods, tension links in bridge

structure. In this, one of the rods has an eye at the rod end and the other one is forked with

eyes at both the legs. A pin (knuckle pin) is inserted through the rod-end eye and fork-end eyes

and is secured by a collar and a split pin.

Normally, empirical relations are available to find different dimensions of the joint and they

are safe from design point of view. The proportions are given in the figure above.d = diameter of rod

d1= d t = 1.25d

d2= 2d t1= 0.75d

d3= 1.5.d t2= 0.5d

Mean diameter of the split pin = 0.25 d

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Design of Joint

S K Mondals Chapter 1However, failures analysis may be carried out for checking. The analyses are shown below

assuming the same materials for the rods and pins and the yield stresses in tension,

compression and shear are given by t, cand .

1. Failure of rod in tension:

=2

4t

d P

2. Failure of knuckle pin in double sheer:

=2

12

4d P

3. Failure of knuckle pin in bending (if the pin is loose in the fork)

Assuming a triangular pressure distribution on the pin, the loading on the pin is shown in

figure below.

Equating the maximum bending stress to tensile or compressive yield stress we have

+ =

1

3

1

163 4

t

t tP

d

Figure- Bending of a knuckle pin

4. Failure of rod eye in shear:(d2 - d1) t = P

5. Failure of rod eye in crushing:

d1t c= P

6. Failure of rod eye in tension:

(d2 - d1)t t= P

7. Failure of forked end in shear:

2(d2 - d1)t1 = P

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Design of Joint

S K Mondals Chapter 18. Failure of forked end in tension:

2(d2 - d1)t1 t= P

9. Failure of forked end in crushing:

2d1t1c= P

The design may be carried out using the empirical proportions and then the analytical

relations may be used as checks.

For example using the 2nd equation we have

=2

1

2P

d. We may now put value of d1 from

empirical relation and then find FACTOR OF SAFTEY,(F.S.)=

ywhich should be more

than one.

Q. Two mild steel rods are connected by a knuckle joint to transmit an axialforce of 100 kN. Design the joint completely assuming the working stresses for

both the pin and rod materials to be 100 MPa in tension, 65 MPa in shear and

150 MPa in crushing.

d

2t

3d

2d

1d

t

1t

1t

2t

d

Refer to figure above

For failure of rod in tension,

= 24

yP d . On substituting P = 100 kN, y= 100 MPa we

have d= 35.6 mm. Let us choose the rod diameter d = 40 mm which is the next standard size.

We may now use the empirical relations to find the necessary dimensions and then

check the failure criteria.

d1= 40 mm t= 50 mm

d2= 80 mm t1= 30 mm;

d3= 60 mm t2= 20 mm;

Split pin diameter = 0.25 d1= 10 mm

To check the failure modes:

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Design of Joint


1. Failure of knuckle pin in shear:

=

2

124

yP d ,which gives y= 39.8

MPa. This is less than the yield shear stress.

2. For failure of knuckle pin in bending:

+

=

1

3

1

163 4

y

t tP

d. On substitution

this gives y= 179 MPa which is more than the allowable tensile yield stress of 100

MPa. We therefore increase the knuckle pin diameter to 55 mm which gives y= 69

MPa that is well within the tensile yield stress.

3. For failure of rod eye in shear: (d2 - d1)t= P. On substitution d1= 55mm = 80

MPa which exceeds the yield shear stress of 65 MPa. So d 2should be at least 85.8

mm. Let d2be 90 mm.

4. for failure of rod eye in crushing: d1 t c= Pwhich gives c= 36.36 MPa that is

well within the crushing strength of 150 MPa.

5. Failure of rod eye in tension: (d2 - d1)tt= P.Tensile stress developed at the rod

eye is then t= 57.14 MPa which is safe.

6. Failure of forked end in shear: 2(d2-d1)t1= P.Thus shear stress developed in

the forked end is = 47.61 MPa which is safe.

7. Failure of forked end in tension: 2(d2 -d1)t1y= P. Tensile strength developed in

the forked end is then y= 47.61 MPa which is safe.

8. Failure of forked end in crushing: 2d1t1c= Pwhich gives the crushing stress

developed in the forked end as c = 42 MPa. This is well within the crushing

strength of 150 MPa.

Therefore the final chosen values of dimensions are:

d1 = 55 mm t = 50 mm

d2= 90 mm t1= 30 mm; and d = 40 mm

d3= 60 mm t2= 20 mm;

Keys

In the assembly of pulley, key and shaft, Key is made theWeakest so that it is cheap and easy

to replace in case of failure.

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S K

F

r

g

r

Recta

1. Re

pr

W

Th

ond

igure- (a)

ounded; als

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ectangular

etric sizes.

ngular

ctangular

portions of

Width

Thick

ere d = Di

e key has t

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quare or

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) Woodru

key; = hu

(f)Tapere

and sq

sunk key:

this key ar

of key, (w

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with both

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= 2w/ 3 =

iameter of

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f Joi

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with fla

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d/ 6

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ctangular

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er same as

in figure

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angular ke

om. (e) T

or 1 for 1

in (e).

below. The

er 1

e end

with

pered

00 for

usual

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2.

k

3.

is

a

T

If

o

Q

S

K M

Square s

y is that it

Gib-head

usually pr

d its use i

he usual pr

a square k

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rl

=

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50 mmSelect

lution:C

figure b

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s width an

key: It is

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shown in

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idth, (

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thickness

rectangul

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the gib he

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the

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t rotatesriate key

ctangular

ey may fail

Desi

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are equal, i.

w = t = d/

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removal

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Figure- G

d key are:

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l

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at 1000 ror the ge

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(a) in shea

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with a he

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b-head key

= d

t ) =2

t case, for s

=2

x

dT

ss in shear

ield stren

m and trr.

h (w), thic

r or (b) in c

Joint

key.

angular su

d at one e

head key

.

/ 4

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hear failur

and l is th

gth of 350

ansmits 1

ness (t) a

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is shown i

6

we have

key length

MPa has

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hapte

a square s

sgib hea

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diamete

ough a g

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1

nk

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low

of

ar.

in

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S K

Sh

W

Nan

yie4m

Cr

Flat k

A flat ke

friction fotight fit. T

Saddl

A saddle

is concave

used for h

stress con

Tang

ond

ear failur

ere torqu

eing in rpyis the y

ld stress an2.

shing fai

y

, as shown

r the grip.heses keys

key

key, show

to fit the

eavy loads.

centration

nt Key

ls

: The fail

e transmit

, w, L andield stress

d substitut

ure =

in figure

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Flat key

in figure

haft surfa

A simple

ccurs in th

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Figure

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ted is, (

d are then shear of

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) =Pow

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s are paralickness of s

ry similar t

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f Joi

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2/

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equations

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2

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eter of theng yto be

and we ha

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rip and th

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Chapt

shaft respehalf of the

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epend entir

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above.

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9 x 10-

ely on

for a

side

not be

little

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A

t

A

s

k

u

d

T

K M

eather

feather ke

the hub or

oodru

woodruff

own in fig

yway is cu

ed in mac

pth.

he main a

ndal

key

is used w

the shaft a

ff key

key is a f

re below.

t in the sh

hine tools

vantages

s

en one com

nd the key

orm of sun

t is usuall

aft using a

nd autom

of a wood

Desi

Figure-T

ponent sli

ay usually

Figure-f

k key whe

used for

milling cu

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Figure-

ruff key a

n of

angent Key

es over ano

has a slidi

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e the key

hafts less

ter, as sho

to the extr

oodruff ke

e as follo

Joint

ther. The k

g fit.

hape is th

han about

wn in the

advantag

s:

ey may be

at of a tru

60 mm dia

igure belo

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hapte

astened eit

cated disc,

meter and

. It is wi

rom the e

1

her

as

the

ely

tra

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Design of Joint

S K Mondals Chapter 11. It accommodates itself to any taper in the hub or boss of the mating piece.

2. It is useful on tapering shaft ends. Its extra depth in the shaft prevents any tendency

to turn over in its keyway.

The main dis-advantages of a woodruff key are as follows:

1. The depth of the keyway weakens the shaft.

2. It can not be used as a feather.

Circular (Pin) Keys

Significantly lower stress concentration factors result from this type of key as compared to

parallel or tapered keys. A ball end mill can be used to make the circular key seat.

Splines

Splinesare essentially stub by gear teeth formed on the outside of the shaft and on the inside

of the hub of the load-transmitting component. Splines are generally much more expensive tomanufacture than keys, and are usually not necessary for simple torque transmission. They

are typically used to transfer high torques. One feature of a spline is that it can be made

with a reasonably loose slip fit to allow for large axial motion between the shaft and

component while still transmitting torque. This is useful for connecting two shafts where

relative motion between them is common, such as in connecting a power takeoff (PTO) shaft of

a tractor to an implement.

Stress concentration factors are greatest where the spline ends and blends into the shaft, but

are generally quite moderate.

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K M

ndal

s

Desi

n of

Joint

Splines c

as a s

keyways

keys ma

shaft.

There are

of spli

industry:

splines,

splines.

Splines

uniform

transfershaft tha

n be thou

eries of

with m

chined on

two major

es used

1) straight-

nd 2) in

rovide a

circumfer

f torque ta key.

hapte

ht of

axial

ating

to a

types

in

sided

olute

more

ntial

o the

1

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Design of Joint


Spline Manufacturing Methods

Splines are either cut (machined) or rolled. Rolled splines are stronger than cut splines due

to the cold working of the metal. Nitriding is common to achieve very hard surfaces which

reduce wear.

Welded jointsA welded joint is a permanent joint which is obtained by the fusion of the edges of the twoparts to be joined together, with or without the application of pressure and a filler material.

The heat required for the fusion of the material may be obtained by burning of gas (in case of

gas welding) or by an electric arc (in case of electric arc welding). The latter method is

extensively used because of greater speed of welding.

Welding is extensively used in fabrication as an alternative method for casting or forging and

as a replacement for bolted and riveted joints. It is also used as a repair medium e.g. to reunite

metal at a crack, to build up a small part that has broken off such as gear tooth or to repair a

worn surface such as a bearing surface.

Types of welded joints:

Welded joints are primarily of two kinds:

(a) Lap or fillet joint

Obtained by overlapping the plates and welding their edges. The fillet joints may be single

transverse fillet, double transverse fillet or parallel fillet joints (see figure below).

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(

T

K M

) Butt

Formed

(for thick

butt joint Squa Singl Singl Singl Singl

hese are sc

ndal

Joints

y placing t

plates) on

s may be ofe butt joint

V-butt joi

U-butt joi

J-butt join

bevel-butt

ematically

s

Figur

he plates e

the edges

different t

t, double

t, double

t, double J-

joint, doub

shown in f

Desi

e- Differen

dge to edg

efore wel

pes, e.g.

-butt joint

-butt joint

butt joint

le bevel bu

gure below

n of

types of la

and weldi

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Joint

p joints

g them. G

ing to the

rooves are

shape of t

hapte

sometimes

e grooves,

1

cut

the

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S K

There ar

Co Ed T-j

Weldi

A welding

1.

2.

3.

4.5.

6.

7.

8.

These wel

ond

other ty

ner joint (s

ge or seal j

oint (see fig

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symbol ha

eference li

rrow.

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imensionsupplemen

inish sym

ail.

pecificatio

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ls

es of wel

ee figure b

int ( see fig

ure below)

bol

following

ne.

ymbols (li

ary symbol

ols

processes

ls are place

De

Figure- Di

ed joints,

low)

ure below)

asic eleme

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s.

d in standa

ign o

ferent type

for examp

nts:

t joints etc.

rd location

f Joi

of butt joi

le,

(see figure

t

ts

below)

Chapt

er 1

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E

w

T

b

F

(s

K M

xample: If

ith convex

esign

he main fai

tt joint is

here t=

t =t

l =le

or a square

ee figure b

ndal

the desire

ontour, th

of a bu

lure mecha

allowable t

ickness of

ngth of the

butt joint

low).

s

weld is a f

weld symb

tt joint

nism of we

ensile stren

the weld

weld.

is equal t

Desi

illet weld o

ol will be a

Figure-

ded butt jo

=Pgth of the

the thickn

n of

f size 10 m

s following

Tee joint

int is tensil

tt

eld materi

ess of the p

Joint

to be don

e failure. T

al.

lates. In g

e on each s

herefore th

neral, this

hapte

de of Tee j

e strength

need not b

1

int

f a

so

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S K

Desig

Consider

the weld

loadP act

a fillet we

throat is

transvers

Wheres

Athroat= th

For a dou

Desig

Consider

to see froarea (try

throat are

ond

n of tra

single tra

etal is ver

s as shear

ld. If the fil

asily seen

fillet weld

= allowable

roat area.

le transver

n of pa

parallel fi

the streno prove it).

a =2

t

lhA .

ls

nsver

nsverse joi

complicat

orce on the

let weld ha

o be2

hl .

is

shear stre

se fillet joi

Fig

rallel fi

let weld as

th of mateThe allow

he total al

De

Figu

e fillet

t as shown

ed. In desi

throat are

s equal bas

ith the ab

P =

s

t the allow

re- Enlarg

llet joi

shown in f

ial approaable load c

lowable loa

ign o

re-butt join

joint

in figure b

n, a simple

a, which is

e and heig

ove conside

s .Athroa

able load is

e view of fil

t

gure below

h that therried by e

d isP = 2s

f Joi

elow. The

procedure

the smalles

t, (h, say),

ration the

t

twice that

let welding

. Each weld

maximumch of the j

At.

t

eneral stre

is used ass

t area of th

then the c

ermissible

of the singl

carries a l

hear occuroint is

s

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ss distribu

ming that

e cross sec

oss section

load carrie

e fillet join

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along thetA ,whe

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ion in

entire

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of the

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.

s easy

throatre the

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I

a

fi

C

b

a

di

t

K M

designing

ove design

gure below)

here

esign

onsider a c

low. If the

in parall

ameter of

rque the m

here

ndal

a weld joi

criteria.

the allowa

At

tA = thro

of circ

ircular sha

shaft is su

l fillet join

he shaft, t

aximum sh

Td= outer

throatt = t

pJ = pola

=

32

s

t the desig

hen a com

le load is

= throat a

t area alon

lar fil l

ft connecte

jected to a

t. Assumin

e maximu

ear stress i

max

torque apiameter of

hroat thick

moment o

( + throat2td

Desi

n variables

ination of

2 sP =ea along t

g the trans

Figuret weld

d to a plat

torque, sh

g that the

shear st

the weld i

2

p

dT

J

+

lied.the shaft

ness

f area of th

) 4 4d

n of

are hand

transverse

t s tA A + e longitudi

verse direc

subje

e by mean

ear stress

weld thick

ess occurs

s

throat

throat sec

Joint

l. They can

and paralle

nal directio

ion.

ted to

of a fillet

evelops in

ness is ver

in the thro

tion.

be selecte

l filled join

n.

torsio

joint as s

the weld in

y small co

at area. Th

hapte

based on

t required

own in fig

a similar

mpared to

us, for a gi

1

the

see

ure

ay

the

en

Page 22 of 263


23/263

S K

When tt

The throa

Q. A pl

mea

the l

Solution.

a loa

= th

25

56

allo

(Not

Q. Two

tran

the

Solution:

assu

speci

MPa

weld

ond

roat

,d

t dimension

ate 50 m

ns of para

ength of t

In a parall

d of 502

P=

oat length

3

3

10 25

2.5 10

=

ance for s

that the a

plates 2

sverse we

ize of the

According

ming maxi

fied the sh

. Assuming

is calculat

ls

=max

4and hence

wide an

llel fillet

e weld. A

el fillet we

N 25kN= .

12.5

2mm .

.5mm . Ho

arting or s

llowance h

0 mm wi

ds at the

weld assu

to the desi

um shear

ear strengt

there are t

d from

35

Or l

De

3

2

throat

dT

t d

weld dime

2

throt

d 12.5 m

elds. The

ssume all

ding two li

Maximum

inces

p

lt

ever som

topping of

s no conne

de and 1

ends. If th

ming the

gn principl

stress occ

h may be c

wo welds, e

3 110 l

=

= 141.42 m

ign o

2

throat

T

t

sion can b

2

td

=

Fig.

thick is

plates ar

wable sh

nes of weld

shear stres

656 10 . H

extra len

the bead.

tion with t

mm thi

e plates a

llowable

of fillet (t

rs along th

alculated a

ach weld c

30 103

2

m.

f Joi

2

selected fr

ax

to be we

subjecte

ar streng

ing are to b

in the par

ence the m

gth of the

usual all

he plate thi

k are to

e subject

ensile str

ransverse)

e throat ar

half of te

rries a loa

65 10

t

om the equ

lded to a

to a loa

h to be 56

e provided.

allel fillet

inimum len

weld is t

owance of

ckness)

be welde

d to a loa

ss 70 MP

joint the w

ea. Since t

sile streng

of 35 kN

Chapt

ation

other pla

of 50 kN.

MPa.

Each line

eld is plt

,

gth of the

be provi

12.5 mm is

d by mea

d of 70 k

.

eld is de

ensile stre

th, i.e. ,s

nd the size

er 1

te by

Find

hares

here t

eld is

ed as

kept.

ns of

, find

igned

gth is

= 35

of the

Page 23 of 263


24/263

Design of Joint


Adding an allowance of 12.5 mm for stopping and starting of the bead, the length of the

weld should be 154 mm.

Q. A 50 mm diameter solid shaft is to be welded to a flat plate and is required to

carry a torque of 1500 Nm. If fillet joint is used foe welding what will be the

minimum size of the weld when working shear stress is 56 MPa.

Solution.According to the procedure for calculating strength in the weld joint,

=s

throat

T

t d22

Where the symbols have usual significance. For given data, the throat thickness is 6.8

mm. Assuming equal base and height of the fillet the minimum size is 9.6 mm. Therefore

a fillet weld of size 10 mm will have to be used.

Q. A strap of mild steel is welded to a plate as shown in the following figure.

Check whether the weld size is safe or not when the joint is subjected to

completely reversed load of 5 kN.

Fig.

Solution.As shown in the figure the joint is a parallel fillet joint with leg size as 9 mm and

the welding is done on both sides of the strap. Hence the total weld length is 2(50) = 100

mm.

In order to calculate the design stress the following data are used

k1 = 2.7 (parallel fillet joint, refer table 3) (there is required a table to solve theproblem)

w = 0.9 cm

K = -1 for completely reversed loading

The value of the allowable fatigue stress (assuming the weld to be a line) is then

1

358 0.9214.8 kgf/cm = 214800N/m

1.5

= = (approx).The design stress is Therefore

Page 24 of 263


25/263

S K

=1,214

2.d

fluctuatin

Threa

B

th

se

o

S

in

in

Bolts

Machi

ond

=800

795567

g load allo

ed fa

lt - Thre

rough hole

cured by

posite the

rew - Th

serted thro

to a thread

ne Scr

ls

N/m . Sinc

able for th

teners

ded faste

s in matin

ightening

ead of the

eaded fas

ugh a hole

ed hole in a

ws

De

the tota

joint is 79

er designe

members

a nut fro

bolt.

ener desig

in one m

mating me

ign o

l length o

55.6 N. The

d to pass

and to be

the end

ned to be

mber and

mber.

f Joi

f the wel

joint is the

t

is 0.1 m

refore safe.

Fig.

Fig.

Chapt

, the ma

er 1

imum

Page 25 of 263


26/263

T

T

n

K M

heet

hread

he pitch lin

hread

hread Ser

mber of th

ndal

etal an

F

Profile

e or diamet

Series

es- group

reads per i

s

d Lag

igure-Shee

er is locate

of diamet

ch applied

Desi

Screw

metal scre

at the h

r-pitch co

to a specifi

n of

ws are ofte

eight of th

binations

c diameter.

Joint

n self-tappi

theoretica

istinguish

ng.

l sharp v-th

d from eac

hapte

read profil

h other by

1

.

the

Page 26 of 263


27/263

Design of Joint


M-SeriesMetric system of diameters, pitches, and tolerance/allowances.

Tensile Stress AreaThe average axial stress in a fastener is computed using a tensile stress area.

=ave

t

F

A

+ =

2

4 2

r p

t

D DA

F Axial Force

DrRoot Diameter

DpPitch Diameter

At Tensile Stress Area

aveAverage axial stress

Length of Engagement (Equal Strength Materials)

If the internal thread and external thread material have the same strength, then

Tensile Strength (External

Thread)

maxt

t

F

A =

Shear Strength (Internal

Thread)

max

,

0.5 ts i e

F

A L =

Where

max ,0.5

t t t s i eF A A L = =

=,

2 te

s i

AL

A

Bolt/Nut Design PhilosophyANSI standard bolts and nuts of equal grades are designed to have the bolt fail before the

threads in the nut are stripped.

The engineer designing a machine element is responsible for determining how something

should fail taking into account the safety of the operators and public. Length of engagement is

an important consideration in designing machine elements with machine screws.

Objective Questions (For GATE, IES & IAS)

Page 27 of 263


28/263

Design of Joint


Previous 20-Years GATE Questions

KeysGATE-1. Square key of side "d/4" each and length l is used to transmit torque "T"

from the shaft of diameter "d" to the hub of a pulley. Assuming the length

of the key to be equal to the thickness of the pulley, the average shear

stress developed in the key is given by [GATE-2003]

2 2 3

4T 16T 8T 16T(a) (b) (c) (d)

ld ld ld d

GATE-1. Ans. (c)If a square key of sides d/4 is used then. In that case, for shear failure we

have xd d

l T4 2

=

x 2

8Tor

ld = [Where x is the yield stress in shear and l is the key length.]

GATE-2. A key connecting a flange coupling to a shaft is likely to fail in[GATE-1995]

(a) Shear (b) tension (c) torsion (d) bending

GATE-2. Ans. (a)Shear is the dominant stress on the key

Welded joints

GATE-3. A 60 mm long and 6 mm thick fillet weld carries a steady load of 15 kN

along the weld. The shear strength of the weld material is equal to 200

MPa. The factor of safety is [GATE-2006]

(a) 2.4 (b) 3.4 (c) 4.8 (d) 6.8

GATE-3. Ans. (b)

Strengthof materialFactorofsafety

Actualloadorstrengthonmaterial=

3

6

o

200(in MPa) 200(in MPa)3.4

58.91(in MPa)15 10

660 10 (in MPa)

cos45

= =

Threaded fasteners

Page 28 of 263


29/263

Design of Joint

S K Mondals Chapter 1GATE-4. A threaded nut of M16, ISO metric type, having 2 mm pitch with a pitch

diameter of 14.701 mm is to be checked for its pitch diameter using two or

three numbers of balls or rollers of the following sizes [GATE-2003]

(a) Rollers of 2 mm (b) Rollers of 1.155 mm

(c) Balls of 2 mm (d) Balls of 1.155 mm GATE-4. Ans. (b)

Previous 20-Years IES Questions

CottersIES-1. Assertion (A): A cotter joint is used to rigidly connect two coaxial rods carrying

tensile load.

Reason (R): Taper in the cotter is provided to facilitate its removal when it fails

due to shear. [IES-2008]

(a) Both A and R are true and R is the correct explanation of A(b) Both A and R are true but R is NOT the correct explanation of A

(c) A is true but R is false

(d) A is false but R is true

IES-1. Ans. (b) A cotter is a flat wedge shaped piece of rectangular cross-section and its

width is tapered (either on one side or both sides) from one end to another for an

easy adjustment. The taper varies from 1 in 48 to 1 in 24 and it may be increased up

to 1 in 8, if a locking device is provided. The locking device may be a taper pin or a

set screw used on the lower end of the cotter. The cotter is usually made of mild

steel or wrought iron. A cotter joint is a temporary fastening and is used to connect

rigidly two co-axial rods or bars which are subjected to axial tensile or compressive

forces.

IES-2. Match List I with List II and select the correct answer using the code given

below the Lists: [IES 2007]

List I List II

(Application) (Joint)

A. Boiler shell 1. Cotter joint

B. Marine shaft coupling 2. Knuckle joint

C. Crosshead and piston road 3. Riveted joint

D. Automobile gear box 4. Splines

(gears to shaft) 5. Bolted Joint

Code: A B C D A B C D

(a) 1 4 2 5 (b) 3 5 1 4

(c) 1 5 2 4 (d) 3 4 1 5

IES-2. Ans. (b)

IES-3. Match List-I (Parts to be joined) with List-II (Type of Joint) and select the

correct answer using the code given below: [IES-2006]

List-I List -II

A. Two rods having relative axial motion 1. Pin Joint

B. Strap end of the connecting rod 2. Knuckle Joint

C. Piston rod and cross head 3. Gib and Cotter Joint

D. Links of four-bar chain 4. Cotter Joint

A B C D A B C D

(a) 1 3 4 2 (b) 2 4 3 1

Page 29 of 263


30/263

I

I

I

I

I

I

I

I

I

K M(c)

S-3. Ans.

S-4. M

Li

A.

B.

C.

D.

Code

(a

(c)

S-4. Ans.

S-5. In

(a

(c)

S-5. Ans.

S-6. Mco

Li

A.

B.

C

D.

(a

(c)

S-6. Ans.

S-7. In

th

sh

(a

S-7. Ans.

ndal 1 4

(d)

atch List I

st I (Type

Riveted j

Welded jo

Bolted joi

Knuckle j

s: A B

4 3

2 4

(c)

a gib and

Single she

Single she

(d)

atch Listrrect ans

st I

Bolts in

cylinder

Cotters i

Rivets in l

Bolts hol

a flange

A B

4 1

3 1

(a)

a cotter

ickness is

earing str

120 N/mm

(d) It i

Shear str

s3

with List

of joints)

int

int

nt

oint

C

2

3

cotter joi

ar only

r and crus

(Items ier using

olted join

cover plat

cotter joi

ap joints

ing two fl

oupling

C

3

4

joint, the

12 mm.

ess develo2

s a case of

ssLoad

2 Are=

Desi

2 (d)

II and sel

List

1. P

2. S

3. L

4. F

D

1 (b)

1 (d)

t, the gib

hing

joints)he codes

ts of engi

e

nt

anges in

D

2 (b)

2 (d)

width of

he load

ped in the

(b) 100 N/

ouble shea360 10

2 50 12

=

n of

2 3

ct the cor

II (An ele

in

rap

ock washe

illet

A

2 3

2 4

and cotte

(b) doub

(d) doub

ith List IIiven belo

L

e 1

2

3

4

A

4 2

3 2

the cotter

cting on

cotter?

m2 (

r.

250N/mm=

Joint

4

rect answ

ment of th

r

C

4

1

are subje

e shear onl

e shear an

(Type ofthe List

ist II

.Doubletr

. Torsiona

Single tr

. Tension

C

3

4

at the c

he cotter

) 75 N/mm

1

er.

e joint)

D

1

3

cted to

y

crushing

failure) a:

nsverse s

l shear

nsverse s

D

1

1

ntre is 5

is 60 kN.

(d)

hapte

[IES-19

[IES-20

d select[IES-20

ear

ears

mm and

What is

[IES-20

50 N/mm2

1

4]

06]

he04]

its

he

04]

Page 30 of 263


31/263

Design of Joint

S K Mondals Chapter 1IES-8. The spigot of a cotter joint has a diameter D and carries a slot for cotter.

The permissible crushing stress is x times the permissible tensile stress for

the material of spigot where x > 1. The joint carries an axial load P. Which

one of the following equations will give the diameter of the spigot?

[IES-2001]

(a)t

P x 1D 2

x

=

(b)

t

P x 1D 2

x

+=

(c)

t

2 P x 1D

x

+=

(d)

t

2PD x 1= +

IES-8. Ans. (b)

IES-9. Match List-l (Machine element) with List-II (Cause of failure) and select

the correct answer using the codes given below the lists: [IES-1998]

List-I List-II

A. Axle 1. Shear stress

B. Cotter 2. Tensile/compressive stress

C. Connecting rod 3. Wear

D. Journal bearing 4. Bending stress

Code: A B C D A B C D(a) 1 4 2 3 (b) 4 1 2 3

(c) 4 1 3 2 (d) 1 4 3 2

IES-9. Ans. (b)

In machinery, the general term shaft refers to a member, usually of circularcross-section, which supports gears, sprockets, wheels, rotors, etc., and which is

subjected to torsion and to transverse or axial loads acting singly or in combination.

An axle is a non-rotating member that supports wheels, pulleys, and carries notorque.

A spindle is a short shaft. Terms such as line-shaft, head-shaft, stub shaft,transmission shaft, countershaft, andflexible shaft are names associated with

special usage.

IES-10. The piston rod and the crosshead in a steam engine are usually connected

by means of [IES-2003]

(a) Cotter joint (b) Knuckle joint (c) Ball joint (d) Universal joint

IES-10. Ans. (a)

IES-11. A cotter joint is used when no relative motion is permitted between the

rods joined by the cotter. It is capable of transmitting [IES-2002]

(a) Twisting moment (b) an axial tensile as well as compressive load

(c) The bending moment (d) only compressive axial load

IES-11. Ans. (b)

IES-12. Match List I with List II and select the correct answer using the codesgiven below the lists: [IES-1995]

List I List II

(Different types of detachable joints) (Specific use of these detachable joints)

A. Cotter joint 1. Tie rod of a wall crane

B. Knuckle joint 2. Suspension bridges

C. Suspension link joint 3. Diagonal stays in boiler

D. Turn buckle (adjustable joint) 4. Cross-head of a steam engine

Codes: A B C D A B C D

(a) 4 2 3 1 (b) 4 3 2 1

(c) 3 2 1 4 (d) 2 1 4 3

Page 31 of 263


32/263

I

I

I

I

I

I

K MS-12. Ans

S-13. M

gi

A.

B.

C.

D.

1.

2.

3.

4.

5.

C

S-13. Ans

t

m

C

et

S-14. A

th

u

R

lo(a

(b

(c)

(d

S-14. Ans

eysS-15. In

(a

ndal. (a)

atch List

ven below

List I (Ty

Cotter joi

Knuckle j

Turn buc

Riveted j

List II (

Connects

Rigidly co

Connects

Permane

Connects

des: A

(a) 5

(c) 5. (b)A cott

o rods whic

y be subje

nnection o

are exam

sertion (

e connectin

on the dire

ason (R):

ding and rBoth A an

Both A an

A is true b

A is false

. (b)

the asse

pulley is

s

I with Li

the lists:

pe of join

nt

oint

le

int

ode of joi

two rods

nnects tw

two rods

t fluid-tig

two shafts

B

1

3er is a flat

h transmit

ted to tens

piston rod

les of cotte

):When t

g rods eith

ction of rot

A turn bu

quiring sud R are ind

d R are ind

ut R is fals

ut R is tru

bly of pul

ade the we

Desi

t II and

)

ting me

r bars pe

o member

aving thr

ht joint b

and tran

C D

3 2

2 4wedge-sha

motion in t

le or comp

to the cros

joint.

e coupler

er move cl

tion of the

kle is used

sequent avidually tr

vidually tr

e

Fig. Tu

ley, key a

akest

n of

elect the

bers)

mitting s

s

eaded end

tween tw

mits torq

(b) 2

(d) 2ed piece of

he axial dir

essive forc

s-head of a

f a turn b

oser or mo

coupler.

to connect

justment fe and R is

e but R is

rnbuckle

d shaft

(b) key i

Joint

correct a

all amou

s

flat piec

e

B

1

3steel. This

ection, wit

s along the

steam engi

ckle is tur

e away fr

two round

r tightenithe correct

otthe cor

made the

swer usi

t of flexi

s

C D

3 4

1 4is used to

out rotatio

axes of the

ne, valve ro

ned in one

m each ot

rods subje

g or loosenexplanatio

ect explan

[IE

weakest

hapte

g the co

[IES-19

ility

connect rig

n. These joi

rods.

d and its s

direction b

her depend

[IES-19

cted to ten

ing.of A

tion of A

S-1993; 19

1

es

93]

dly

nts

em

oth

ing

96]

sile

8]

Page 32 of 263


33/263

S K

IES-15.

IES-16.

IES-16.A

IES-17.

IES-17. A

IES-18.

IES-18. A

IES-19.

IES-19. A

ond(c) Key is

ns. (b)Ke

failure.

Match Li

correct a

List-I

A. Woodr

B. Kenne

C. Feathe

D. Flat ke

Code:

(a)

(c)

ns. (b)A fe

be fastene

Match Li

given bel

List-I (Ke

A. Gib he

B. Woodr

C. Parall

D. Spline

Code:

(a)

(c)

ns. (c)

A spur g

rectangul

(a) Shear s

(c) Both sh

ns. (c)Key

Assertion

capacity a

Reason (

(a) Both A

(b) Both A

(c) A is tru

(d) A is fal

ns. (d)

lsade the st

is made

st-I (Type

swer usi

ff key

y key

r key

y

B

2 3

2 3

ather key i

either to t

st-I with

w the list

/splines)

d key

ff key

l key

B

1 2

2 1

ar trans

ar section

tress alone

ear and be

develops b

(A): The

d to increa

):Highly l

and R are i

and R are i

e but R is f

e but R is

De

ongest

he weakes

of keys)

g the cod

L

1

2

3

4

C

1 4

4 1

s used whe

he hub or t

Fig.

List-II an

s:

C

3 4

3 4

itting po

. The type

ring stress

th shear a

effect of k

se its torsio

calized str

ndividually

ndividually

lse

rue

ign o

(d) a

stren

so that it

with List

s given b

ist-II

. Loose fit

. Heavy d

. Self-alig

. Normal i

(b)

(d)

one comp

e shaft an

feather key

d select t

List-

1. Sel

2. Fa

3. Mo

4. Ax

(b)

(d)

er is co

(s) of stre

(b) be

es (d) sh

d bearing

yways on

nal rigidity

esses occur

true and R

true but R

f Joi

ll the thr

th

is cheap a

-II (Char

low the L

ing, light

ty

ing

dustrial

A B

3 2

3 2

nent slides

the keyw

e correc

I (Applica

f aligning

ilitates r

stly used

ial movem

A B

1 2

2 1

nected t

ses devel

aring stres

earing, bea

stresses.

a shaft is

.

at or near

is the corr

is notthe

t

e are de

d easy to

cteristic)

sts:

duty

se

C

1

4

over anoth

y usually h

answer

tion)

moval

ent possib

C

4

4

the shaf

ped in th

alone

ring and be

to reduce

he corners

ct explana

orrect expl

Chaptigned for

eplace in

and sele

[IES

D

4

1

er. The key

as a slidin

using the

[IES

le

D

3

3

with a

e key is fa

[IES-

nding stres

its load ca

[IES-

of keyways.

tion of A

anation of

er 1equal

ase of

t the

1997]

may

fit.

code

2008]

ey of

e.

1995]

ses.

rrying

1994]

Page 33 of 263


34/263

Design of Joint

S K Mondals Chapter 1IES-20. Which key is preferred for the condition where a large amount of impact

torque is to be transmitted in both direction of rotation? [IES-1992]

(a) Woodruff key (b) Feather key (c) Gib-head key (d) Tangent key

IES-20. Ans. (d)

IES-21. What is sunk key made in the form of a segment of a circular disc of

uniform thickness, known as? [IES-2006]

(a) Feather key (b) Kennedy key (c) Woodruff key (d) Saddle key

IES-21. Ans. (c)

IES-22. What are the key functions of a master schedule? [IES-2005]

1. To generate material and capacity requirements

2. To maintain valid priorities

3. An effective capacity utilization

4. Planning the quantity and timing of output over the intermediate time

horizons

Select the correct answer using the code given below:

(a) 1, 2 and 3 (b) 2, 3 and 4 (c) 1, 3 and 4 (d) 1, 2 and 4IES-22. Ans. (b)

IES-23. A square key of side d/4 is to be fitted on a shaft of diameter d and in the

hub of a pulley. If the material of the key and shaft is same and the two are

to be equally strong in shear, what is the length of the key? [IES-2005]

(a)d

2

(b)

2 d

3

(c)

3 d

4

(d)

4 d

5

IES-23.Ans. (a)

IES-24. Which one of the following statements is correct? [IES-2004]

While designing a parallel sunk key it is assumed that the distribution of

force along the length of the key(a) Varies linearly (b) is uniform throughout

(c) varies exponentially, being more at the torque input end

(d) varies exponentially, being less at torque output end

IES-24. Ans. (c)Parallel sunk key. The parallel sunk keys may be of rectangular or square

section uniform in width and thickness throughout. It may be noted that a parallel

key is a taperless and is used where the pulley, gear or other mating piece is

required to slide along the shaft. In designing a key, forces due to fit of the key are

neglected and it is assumed that the distribution of forces along the length of key is

uniform.

IES-25. Match List-I (Device) with List-II (Component/Accessory) and select the

correct answer using the codes given below the Lists: [IES-2003]

List-I List-II

(Device) (Component/Accessory)

A. Lifting machine 1. Idler of Jockey pulley

B. Fibre rope drive 2. Sun wheel

C. Differential gear 3. Sheave

D. Belt drive 4. Power screw

Codes: A B C D A B C D

(a) 4 3 1 2 (b) 3 4 1 2

(c) 4 3 2 1 (d) 3 4 2 1

IES-25. Ans. (c)

Page 34 of 263


35/263

S K

IES-26.

IES-26. A

IES-27.

IES-27.A

ond

A pulley

means of

key is ta

the key is

the lengt

(a)4

ns. (c)

Shearings

Assertion

Reason (

of the mati(a) Both A

(b) Both A

(c) A is tru

(d) A is fal

ns. (b)

The main

1. It accom

2. It is use

tendency t

The main

1. The dep

2. It can n

ls

is connec

a rectang

en as d/4.

equal to

of the ke

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trengthof

(A):A Wo

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ng piece.and R are i

and R are i

e but R is f

e but R is

dvantages

modates it

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turn over

is-advanta

h of the ke

t be used a

De

ted to a

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For full

he torsio

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2

dey:F .

4

Torque(

Torsiona

or T d

For sam

d. .l .

4

lor

d 2

=

=

=

druff key i

druff key

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rue

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strength

d16

=

=

an easily

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true but R

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2003]

Page 35 of 263


36/263

Design of Joint

S K Mondals Chapter 1IES-28. The key shown in the above

figure is a

(a) Barth key

(b) Kennedy key

(c) Lewis key

(d) Woodruff key[IES-2000]

IES-28. Ans. (a)

IES-29. Match List I (Keys) with List II (Characteristics) and select the correct

answer using the codes given below the Lists: [IES-2000]

List I List II

A. Saddle key 1. Strong in shear and crushing

B. Woodruff key 2. Withstands tension in one direction

C. Tangent key 3. Transmission of power through frictionalresistance

D. Kennedy key 4. Semicircular in shape


(a) 3 4 1 2 (b) 4 3 2 1

(c) 4 3 1 2 (d) 3 4 2 1

IES-29. Ans. (d)

IES-30. Match List-I with List-II and select the correct answer using the code

given below the Lists: [IES-2009]

List-I List-II

(Description) (shape)

A. Spline 1. InvoluteB. Roll pin 2. Semicircular

C. Gib-headed key 3. Tapered on on side

D. Woodruff key 4. Circular


(a) 1 3 4 2 (b) 2 3 4 1

(c) 1 4 3 2 (d) 2 4 3 1

IES-30. Ans. (c)

IES-31. The shearing area of a key of length 'L', breadth 'b' and depth 'h' is equal to

(a) b x h (b) Lx h (c) Lx b (d) Lx (h/2) [IES-1998]

IES-31. Ans. (c)

SplinesIES-32. Consider the following statements: [IES-1998]

A splined shaft is used for

1. Transmitting power

2. Holding a flywheel rigidly in position

3. Moving axially the gear wheels mounted on it

4. Mounting V-belt pulleys on it.

Of these statements

(a) 2 and 3 are correct (b) 1 and 4 are correct

Page 36 of 263


37/263

Design of Joint

S K Mondals Chapter 1(c) 2 and 4 are correct (d) 1 and 3 are correct

IES-32. Ans. (d)

Welded jointsIES-33. In a fillet welded joint, the weakest area of the weld is [IES-2002]

(a) Toe (b) root (c) throat (d) face

IES-33. Ans. (c)

IES-34. A single parallel fillet weld of total length L and weld size h subjected to a

tensile load P, will have what design stress? [IES 2007]

(a) Tensile and equal toP

0.707Lh (b) Tensile and equal to

P

Lh

(c) Shear and equal toP

0.707Lh (d) Shear and equal to

P

Lh

IES-34. Ans. (c)

Throat, t = h cos450=2

1h

v= 0.707h

T =P

Lt=

P

0.707Lh

IES-35. Two metal plates

of thickness t

and width 'w' are

joined by a fillet

weld of 45 as

shown in given

figure. [IES-1998]

When subjected to a pulling force 'F', the stress induced in the weld will be

(a)o

F

wtsin45 (b)

F

wt (c)

oFsin45

wt (d)

2F

wt

IES-35. Ans. (a)

IES-36.A butt welded joint, subjected to

tensile force P is shown in the

given figure, l = length of theweld (in mm) h = throat of the

butt weld (in mm) and H is the

total height of weld including

reinforcement. The average

tensile stress t, in the weld is

given by [IES-1997]

( ) ( ) ( ) ( )t t t tP P P 2P

a b c d Hl hl 2hl Hl

= = = =

IES-36. Ans. (b)

Page 37 of 263


38/263

Design of Joint

S K Mondals Chapter 1IES-37. In the welded joint shown in the given

figure, if the weld at B has thicker fillets

than that at A, then the load carrying

capacity P, of the joint will

(a) increase

(b) decrease

(c) remain unaffected

(d) exactly get doubled

[IES-1997]

IES-37. Ans. (c)

IES-38. A double fillet welded joint with parallel fillet weld of length L and leg B is

subjected to a tensile force P. Assuming uniform stress distribution, the

shear stress in the weld is given by [IES-1996]

(a)2P

B.L

(b)P

2.B.L

(c)P

2.B.L (d)

2P

B.L

IES-38. Ans. (c)

IES-39. The following two figures show welded joints (x x x x x indicates welds),

for the same load and same dimensions of plate and weld. [IES-1994]

The joint shown in

(a) fig. I is better because the weld is in shear and the principal stress in the weld is

not in line with P

(b) fig. I is better because the load transfer from the tie bar to the plate is not direct

(c) fig. II is better because the weld is in tension and safe stress of weld in tension is

greater than that in shear

(d) fig. II is better because it has less stress concentration.

IES-39. Ans. (c)Figure II is better because the weld is in tension and safe stress of weld in

tension is greater than shear.

IES-40. Assertion (A): In design of double fillet welding of unsymmetrical sections with

plates subjected to axial loads lengths of parallel welds are made unequal.

Reason (R): The lengths of parallel welds in fillet welding of an unsymmetrical

section with a plate are so proportioned that the sum of the resisting moments ofwelds about the centre of gravity axis is zero. [IES-2008]

(a) Both A and R are true and R is the correct explanation of A

(b) Both A and R are true but R is NOT the correct explanation of A



IES-40. Ans. (a)Axially loaded unsymmetrical welded joints

Page 38 of 263


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Design of Joint


1

1 1

1 1

2 2

1 1 2 2

1 1 2 2

1 1 2 2

P

A

P A

P t I

P t I

P y P y

tI y tI y

I y I y

=

=

=

=

=

=

=

IES-41. Two plates are joined together by means of

single transverse and double parallel fillet

welds as shown in figure given above. If the size

of fillet is 5 mm and allowable shear load per

mm is 300 N, what is the approximate length of

each parallel fillet?(a) 150 mm

(b) 200 mm

(c) 250 mm

(d) 300 mm[IES-2005]

IES-41. Ans. (b) ( )300 100 2l 15000 or l 200 + = =

IES-42. A circular rod of diameter d is welded to a flat plate along its

circumference by fillet weld of thickness t. Assuming was the allowable

shear stress for the weld material, what is the value of the safe torque that

can be transmitted? [IES-2004]

(a) 2 wd .t. (b)2

w

d.t.

2

(c)

2

w

d.t.

2 2

(d)

2

w

d.t.

2

IES-42. Ans. (b)

( )

W

W

2

W W

Shear stress

Shear fore dt

d dTorque T dt .t

2 2

=

=

= =

IES-43. A circular solid rod of diameter d welded to a rigid flat plate by a circular

fillet weld of throat thickness t is subjected to a twisting moment T. The

maximum shear stress induced in the weld is [IES-2003]

(a)2

T

td (b)

2

2T

td (c)

2

4T

td (d)

3

2T

td

IES-43. Ans. (b)3 2

dT.

T.r 2T2

J td td

4

= = =

Page 39 of 263


40/263

Design of Joint

S K Mondals Chapter 1IES-44. The permissible stress in a filled weld is 100 N/mm2. The fillet weld has

equal leg lengths of 15 mm each. The allowable shearing load on weldment

per cm length of the weld is [IES-1995]

(a) 22.5 kN (b) 15.0 kN (c) 10.6 kN (d) 7.5 kN.

IES-44. Ans. (c)Load allowed = 100 x 0.707 x 10 x15 = 10.6 kN

Threaded fastenersIES-45. A force F is to be transmitted through a square-threaded power screw

into a nut. If t is the height of the nut and d is the minor diameter, then

which one of the following is the average shear stress over the screw

thread? [IES 2007]

(a)2f

dt (b)

F

dt (c)

F

2 dt (d)

4F

dt

IES-45. Ans. (b)

IES-46. Consider the case of a square-threaded screw loaded by a nut as

shown in the given figure. The

value of the average shearing

stress of the screw is given by

(symbols have the usual meaning)

( )

r r

2F F(a) (b)

d h d h

2F F(c) d

dh dh

[IES-1997]

IES-46. Ans. (b)

IES-47. Assertion (A): Uniform-strength bolts are used for resisting impact loads.

Reason (R): The area of cross-section of the threaded and unthreaded parts is

made equal. [IES-1994]

(a) Both A and R are individually true and R is the correct explanation of A

(b) Both A and R are individually true but R is notthe correct explanation of A



IES-47. Ans. (c)A is true and R is false.

IES-48. How can shock absorbing capacity of a bolt be increased? [IES 2007]

(a) By tightening it property

(b) By increasing the shank diameter

(c) By grinding the shank

(d) By making the shank diameter equal to the core diameter of thread

IES-48. Ans. (d)

IES-49. The number of slots is a 25 mm castle nut is [IES-1992]

(a) 2 (b) 4 (c) 6 (d) 8

IES-49. Ans. (c)

Page 40 of 263


41/263

Design of Joint


Answers with Explanation (Objective)

Page 41 of 263


42/263

Design of Friction Drives


2. Design of Friction Drives

Theory at a glance (GATE, IES, IAS & PSU)

Couplings

Introduction

Couplings are used to connect two shafts for torque transmission in varied applications. It may

be to connect two units such as a motor and a generator or it may be to form a long line shaftby connecting shafts of standard lengths say 6-8m by couplings. Coupling may be rigid or they

may provide flexibility and compensate for misalignment. They may also reduce shock loading

and vibration. A wide variety of commercial shaft couplings are available ranging from a

simple keyed coupling to one which requires a complex design procedure using gears or fluid

drives etc.

However there are two main types of couplings:

Rigid couplings.

Flexible couplings.

Rigid couplingsare used for shafts having no misalignmentwhile the flexiblecouplings

can absorb some amount of misalignment in the shafts to be connected. In the next

section we shall discuss different types of couplings and their uses under these two broad

headings.

Types and uses of shaft couplings

Rigid couplings

Since these couplings cannot absorb any misalignment the shafts to be connected by a rigidcoupling must have good lateral and angular alignment. The types of misalignments are

shown schematically in figurebelow.

Page 42 of 263


43/263



Figure- Types of misalignments in shafts

Sleeve coupling

One of the simple types of rigid coupling is a sleeve coupling which consists of a cylindrical

sleeve keyed to the shafts to be connected. A typical sleeve coupling is shown in figure below

0d

L

id

Figure- A typical sleeve coupling

Normally sunk keys are used and in order to transmit the torque safely it is important to

design the sleeve and the key properly. The key design is usually based on shear and bearingstresses. If the torque transmitted is T, the shaft radius is r and a rectangular sunk key of

dimension b and length L is used then the induced shear stress (figure below) in the key is

given by

2

TL

b r

=

And for safety

( )2 / yT bLr <

Page 43 of 263


44/263



Where yis the yield stress in shear of the key material. A suitable factor of safety must be

used. The induced crushing stress in the key is given as

2 2

br

Tb L r =

And for a safe design

4T /(bLr) < c

Where cis the crushing strength of the key material.

Figure- Shear and crushing planes in the key.

The sleeve transmits the torque from one shaft to the other. Therefore if d i is the inside

diameter of the sleeve which is also close to the shaft diameter d (say) and d0 is outside

diameter of the sleeve, the shear stress developed in the sleeve is

( )

=

0

4 4

0

16sleeve

i

Td

d d and the shear stress in the shaft is given by

=3

16shaft

i

T

d. Substituting yield shear stresses of the sleeve and shaft materials for

sleeveand shaftboth diand d0may be evaluated.

However from the empirical proportions we have:

d0= 2di+ 12.5 mm and L=3.5d.

These may be used as checks.

Sleeve coupling with taper pinsTorque transmission from one shaft to another may also be done using pins as shown in figure

below.

Page 44 of 263


45/263



Figure- A representative sleeve coupling with taper pins

The usual proportions in terms of shaft diameter d for these couplings are:

d0= 1.5d, L = 3d and a = 0.75d.

The mean pin diameter dmean= 0.2 to 0.25 d. For small couplings dmeanis taken as 0.25d and for

large couplings dmeanis taken as 0.2d. Once the dimensions are fixed we may check the pin forshear failure using the relation

=

22

4 2mean

dd T.

Here T is the torque and the shear stress must not exceed the shear yield stress of the pin

material. A suitable factor of safety may be used for the shear yield stress.

Clamp coupling

A typical clamp coupling is shown in figure below. It essentially consists of two half cylinderswhich are placed over the ends of the shafts to be coupled and are held together by through

bolt.

Figure- A representative clamp coupling

The length of these couplings L usually vary between 3.5 to 5 times the and the outside

diameter d0 of the coupling sleeve between 2 to 4 times the shaft diameter d. It is assumed

that even with a key the torque is transmitted due to the friction grip. If now the number of

bolt on each half is n, its core diameter is dcand the coefficient of friction between the shaft

and sleeve material is we may find the torque transmitted T as follows.

The clamping pressurebetween the shaft and the sleeve is given by

Page 45 of 263


46/263



( )2 / 22 4

c t

np d dL

=

Where nis the total number of bolts, the number of effective bolts for each shaft is n/2and tis the allowable tensile stress in the bolt. The tangential force per unit area in the shaft

periphery is F = p. The torque transmitted can therefore be given by

=

2 2

dL dT p

Ring compression type couplings

The coupling (figure below) consists of two cones which are placed on the shafts to be coupled

and a sleeve that fits over the cones. Three bolts are used to draw the cones towards each

other and thus wedge them firmly between the shafts and the outer sleeve. The usualproportions for these couplings in terms of shaft diameter d are approximately as follows:

d1= 2d + 15.24 mm L1= 3d

d2= 2.45d + 27.94 mm L2= 3.5d + 12.7 mm

d3= 0.23d + 3.17 mm L3= 1.5d

And the taper of the cone is approximately 1 in 4 on diameters.

Figure- A representative ring compression type coupling.

.

Oldham couplingThese couplings can accommodate both lateral and angular misalignmentto some extent.

An Oldham coupling consists of two flanges with slots on the faces and the flanges are keyed

or screwed to the shafts. A cylindrical piece, called the disc, has a narrow rectangular raised

portion running across each face but at right angle to each other. The disc is placed between

the flanges such that the raised portions fit into the slots in the flanges. The disc may be made

of flexible materials and this absorbs some misalignment. A schematic representation is

shown in figure below.

Page 46 of 263


47/263

S K

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Page 47 of 263


48/263

T

(

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fl

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p

b

o

T

t

c

p

(

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Page 48 of 263


49/263


S K Mondals Chapter 2(3) Hub length, L = 1.5d

But the hub length also depends on the length of the key. Therefore this length L must be

checked while finding the key dimension based on shear and crushing failure modes.

(4) Key dimensions:

If a square key of sides bis used then b is commonly taken as4

d. In that case, for shear failure

we have

=

k4 2

y

d dL T

Where yis the yield stress in shear and Lkis the key length.

This gives

=k 28TL d

If Lkdetermined here is less than hub length L we may assume the key length to be the same

as hub length.

For crushing failure we have

=

k8 2

c

d dL TWhere cis crushing stress induced in the key. This gives

=2

k

16c

T

L d

And if c < cy , the bearing strength of the key material ,the key dimensions chosen are in

order.

(5) Bolt dimensions:

The bolts are subjected to shear and bearing stresses while transmitting torque.

Considering the shear failuremode we have

cb yb

dn d2

4 2=

Where n is the number of bolts,dbnominal bolt diameter, Tis the torque transmitted, b is

the shear yield strength of the bolt material and dc is the bolt circle diameter. The bolt

Page 49 of 263


50/263


S K Mondals Chapter 2diameter may now be obtained if n is known. The number of bolts nis often given by the

following empirical relation:

43150n d= +

Whered is the shaft diameter in mm. The bolt circle diameter must be such that it should

provide clearance for socket wrench to be used for the bolts. The empirical relation takes care

of this.

Considering crushing failurewe have

= cb cybd

n d t2

.2

Where2

t is the flange width over which the bolts make contact and cybis the yield crushing

strength of the bolt material. This gives t2. Clearly the bolt length must be more than 2 2t and

a suitable standard length for the bolt diameter may be chosen from hand book.

(6) A protecting flange is provided as a guard for bolt heads and nuts. The thickness t3is less

than 22

tthe corners of the flanges should be rounded.

(7)The spigot depth is usually taken between 2-3mm.

(8)Another check for the shear failure of the hub is to be carried out. For this failure mode we

may write

11 2

2yf

dd t =

Where d1is the hub diameter and yf is the shear yield strength of the flange material.

Knowing yf we may check if the chosen value of t2is satisfactory or not.

Finally, knowing hub diameter d1, bolt diameter and protective thickness t2

We may decide the overall diameter d3.

Flexible rubber bushed couplings

Flexible coupling

Page 50 of 263


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S KAs discus

variety of

will be dis

This is si

below.

In a rigidthe bolts

However i

above) pr

Because o

(1) Be

Rubber b

bushes ar

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Thickness

Where db

thickness

onded earlier t

flexible co

cussed her

plest type

Fi

coupling thnd in this

n the bush

vide flexibi

f the rubbe

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shings are

mostly av

ness for lae

may be ta

is the diam

of rubber b

Desi

lshese coupli

plings are

.

of flexible

gure- A typ

e torque isrrangemen

d coupling

lity and th

bushing t

tress

available

ailable in t

ger bores.

en to be 1.

eter of the

shing. We

gn of

ngs can acc

available

oupling an

ical flexible

transmittet shafts ne

the rubber

se couplin

e design fo

for differen

ickness be

Brass slee

mm. The o

dr =

olt or pin,

may now

Frict

ommodate

commercial

d a typical

coupling w

from oned be aligne

bushings o

can accom

r pins shou

t inside an

tween 6 m

es are ma

utside dia

b+2 tbr+2

tbris the t

rite

ion D

some misal

ly and prin

coupling of

ith rubber

alf of the cd very well

ver the pin

odate so

ld be consid

d out side

to 7.5 mm

de to suit

eter of rub

tr

ickness of

ives

ignment an

cipal featu

this type i

ushings.

oupling to.

s (bolts) (a

e misalign

ered carefu

diameters.

for bores

the require

er bushin

he brass sl

Chaptd impact.

res of only

shown in

he other t

shown in

ent.

lly.

However

pto 25 mm

ments. Ho

dris given

eeve and tr

er 2large

a few

igure

rough

igure

ubber

and 9

ever,

by

is the

Page 51 of 263


52/263



2.

2c

r b

dn d t p T =

Where dcis the bolt circle diameter and t2the flange thickness over the bush contact area. A

suitable bearing pressure for rubber is 0.035 N/mm2 and the number of pin is given by

325

dn= + where d is in mm.

The dc here is different from what we had for rigid flange bearings. This must be judged

considering the hub diameters, out side diameter of the bush and a suitable clearance. A rough

drawing is often useful in this regard.

From the above torque equation we may obtain bearing pressure developed and compare this

with the bearing pressure of rubber for safely.

(2) Shear stress

The pins in the coupling are subjected to shear and it is a good practice to ensure that the

shear plane avoids the threaded portion of the bolt. Unlike the rigid coupling the shear stress

due to torque transmission is given in terms of the tangential force F at the outside diameter

of the rubber bush.

Shear stress at the neck area is given by

2

2

4

b rb

neck

p t d

d =

Where dneckis bolt diameter at the neck i.e. at the shear plane.

Bending Stress

The pin loading is shown in Figure below.

Page 52 of 263


53/263

S K

Clearly t

bending o

Knowing

using app

We may a

Hub diam

Hub lengt

Pin diam

ond

e bearing

the pin. C

he shear a

ropriate th

lso assume

eter = 2d

h = 1.5d

ter at the

Desi

ls

Figure- L

pressure t

nsidering

d bending

ories of fail

the followi

eck =0.

gn of

ading on a

at acts as

n equivale

b =

stresses w

ure.

g empirica

d

Frict

pin suppor

distribute

nt concentr

( 23

2

br

F t

d

may chec

l relations:

ion D

ing the bus

d load on

ted load F

)/ 2

the pin di

ives

hings.

rubber bus

pt2d the b

ameter fo

Machine Design MD By S K Mondal T&Q.pdf

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