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Inmanystructuresthemembersarerequiredtoresistmorethanonekindofloading

(combinedloading).Thesecanoftenbeanalyzedbysuperimposingthestressesand

strains

cause

by

each

load

acting

separately.

Superpositionofstressesandstrainsispermissibleonlyunderthefollowing

conditions:

a.Thestresses

and

the

strains

must

be

alinear

function

of

the

applied

loads

(Hookes

lawmustbeobeyedandthedisplacementsmustbesmall).

b.Theremustbenointeractionbetweenthevariousloads.

COMBINED LOADS

Examples: wideflangebeamsupportedbyacable(combinedbendingandaxial

load),

cylindricalpressure

vessel

supported

as

abeam,

and

shaft

in

combined

torsion

andbending.

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1.Selectthepointonthestructurewherethestressesandthestrainsaretobedetermined.

2.Foreachloadonthestructure,determinethestressresultantatthecrosssection

containingtheselectedpoint..

3.Calculatethenormalandshearstressesattheselectedpointduetoeachofthestress

resultant.

4.Combinethe

individual

stresses

to

obtain

the

resultant

stresses at

the

selected

point.

5.Determinetheprincipalstressesandmaximumshearstressesattheselectedpoint.

6.DeterminethestrainsatthepointwiththeaidofHookeslawforplanestress.

7.Selectadditionalpointsandrepeattheprocess.

t

pr

Ib

VQ

I

My

A

P

==

=

==

Method of Analysis:

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Thebarshownissubjectedtotwotypesofloads:a

torqueTand

avertical

load

P.

Letusselectarbitrarilytwopoints.PointA(topof

thebar)andpointB(sideofthebarinthesame

crosssection).

Theresulting

stresses

acting

across

the

section

are

thefollowing:

AtwistingmomentequaltothetorqueT.

A

bending

moment

Mequal

to

the

load

Ptimes

thedistanceb.

AshearforceVequalstotheloadP.

Illustration of the Method

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Thetwistingmoment

producesatorsional shearstresses

Thestress

1

actshorizontallytotheleftatpointAand

verticallydownwardsatpointB.

Thebending

moment

Mproduces

atensile

stress

at

point

A

However,thebendingmomentproducesnostressatpointB,becauseBis

locatedontheneutralaxis.

3

2

r

T

I

Tr

Polar

torsion

==

3

4

r

M

I

Mrbending

==

TheshearforceVproducesnoshearstressatthetopofthe

bar(point

A),

but

at

point

Bthe

shear

stress

is

as

follows:

A

and

1

areactinginpointA,whilethe

1

and

2

areacting

inpointB.

A

V

Ib

VQshear

3

4==

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Notethattheelementisinplanestresswith

x

=

A

,

y

= 0 , and xy

= -

1

.

Astress

element

in

point

Bis

also

in

plane

stress

and

the

onlystressesactingonthiselementaretheshear

stresses

1

and

2

. Therefore

x

=

y

= 0 and

xy

= -

1

+

2

).

At point A:

x

=

A

,

y

= 0 , and

xy

= -

1

At point B

x

=

y

= 0 and

xy

= -

1

+

2

).

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Ofinterestarethepointswherethestressescalculatedfromtheflexureandshear

formulashavemaximumorminimumvalues,calledcriticalpoints.

Iftheobjectiveoftheanalysisistodeterminethelargeststressesanywhereinthe

structure,then

the

critical

points

should

be

selected

at

cross

sections

where

the

stressresultantshavetheirlargestvalues.

Furthermore,withinthosecrosssections,thepointsshouldbeselectedwhere

eitherthenormalstressesortheshearstresseshavetheirlargestvalues.

Selection

of

Critical

Areas

and

Points

Forinstance,

the

normal

stresses

due

to

bending

arelargestatthecrosssectionofmaximum

bendingmoment,whichisatthesupport.

Therefore,pointsCandDatthetopandbottom

ofthe

beam

at

the

fixed

ends

are

critical

points

wherethestressesshouldbecalculated.

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Stress at which point?

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The rotor shaft of an helicopter drives the rotor blades that provide the lifting

force to support the helicopter in the air. As a consequence, the shaft is

subjected to a combination of torsion and axial loading.

For a 50mm diameter shaft transmitting a torque

= 2.4kN.m anda tensile force P = 125kN, determine the maximum tensile stress,

maximum compressive stress, and maximum shear stress in the

shaft.

Solution

The stresses in the rotor shaft are produced by the combinedaction of the axial force P and the torque . Therefore the

stresses at any point on the surface of the shaft consist of a

tensile stresso and a shear stress o.

( ) MPa

m

kNAP 66.63

05.04

1252===

The tensile

stress

The shear stress to is obtainedfrom the torsion formula

( )

( ) MPa

mkN

ITrP

Torsion 78.97

32

05.02

05.0.4.2

4 =

==

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( ) MPaxyyx

MAX 1032

2

2

=+

=

33.2103

2480

==MPa

MPa

SFMSST

Knowing the stresses o and o, we can now obtain the principalstresses and maximum shear stresses . The principal stresses are

obtained from

( )2

2

2,122

xy

yxyx

+

+

=

The maximum in-plane shearstresses are obtained using the

formula

Because the principal stresses 1 and 2 have opposite signs, the maximum in-planeshear stresses are larger than the maximum out-of-plane shear stresses. Therefore,

the maximum shear stress in the shaft is 103MPa.

( )

MPa

MPa

71

135

78.97

2

66.630

2

66.630

2

1

2

2

2,1

=

=

+

+=

Will it fail if yield=480MPa?

( ) ( )( ) ( )

65.22.181

480

2.181717113513522

==

=+=

Pa

MPaSFDET

MPaVM

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A thin wall cylindrical pressure vessel with a circular cross section is

subjected to internal gas pressure p and simultaneously compressed by an

axial load P = 12k. The cylinder has inner radius r = 2.1in. And wall thickness t

= 0.15in. Determine the maximum allowable internal pressure pallow basedupon an allowable shear stress of 6500psi in the wall of the vessel.

Solution

The stresses on the wall of the pressure vessel are caused by a combined action of

the internal pressure and the axial force. We can isolate a stress element in pointA.The x-axis is parallel to the longitudinal axis of the pressure vessel and the y-axis is

circumferential. Note that there are no shear stresses acting on the element.

The longitudinal stress x is equal to the tensile stress

produced by the internal pressure minus thecompressive stress produced by the axial force.

rt

P

t

pr

A

P

t

pr

x

222==

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The circumferential stressy is equal to the tensile

stress produced by the internal pressure.

Note that

y

>

x

.t

pry =

Since no shear stresses act on the element the

above stresses are also the principal stresses

rt

P

t

pr

tpr

x

y

222

1

==

==

( )

( )( ) ( )( )[ ] psip

inink

in.in.p

rtP

tpr

pin.

in.p

t

pr

60630.715.01.22

12150212

22

0.14150

12

2

1

===

===

substituting numerical values

In-Plane Shear Stresses

The maximum in-plane shear stress is( ) ( ) ( )( )

psippsipp

Max 30325.32

60630.70.14

2

21 +=

=

=

Since max is limited to 6500psi thenpsippsippsi allowed 99030324.365000 =+=

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Out-of-Plane Shear Stresses

The maximum out-of-plane shear stress is either

Comparing the three calculated values for the allowable pressure, we see

that (pallow)3 = 928psi governs.

At this pressure, the principal stresses are

1 = 13000psi and2 = 430psi.

These stresses have the same signs, thus confirming that one of the out-

of-plane shear stresses must be the largest shear stress.

2

2

2

1

=

=

Max

Max

From the first equation we get

From the second equation we

get:

psip

psippsi

allowed 2720

30325.365000

=

=

psipppsi allowed 9280.765000 ==

Allowable internal pressure

A i f di i 2 0 1 2 i t d b

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A sign of dimensions 2.0mx1.2m is supported by a

hollow circular pole having outer diameter 220mm and

inner diameter 180mm (see figure). The sign offset

0.5m from the centerline of the pole and its lower edgeis 6.0m above the ground.

Determine the principal stresses and maximum shear

stresses at pointsA and B at the base of the pole due

to wind pressure of 2.0kPa against the sign.

Solution

Stress Resultant: The wind pressure against the sign

produces a resultant force W that acts at the midpointof the sign and it is equal to the pressure p times the

areaA over which it acts:

( )( ) kNmmkPapAW 8.42.10.20.2 ===

The line of action of this force is at height h = 6.6m above the ground and at

distance b = 1.5m from the centerline of the pole.

The wind force acting on the sign is statically equivalent to a lateral force Wand a torque

acting on the pole.

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The torque is equal to the force W

times the distance b:

( )( )mkNT

mkNWbT

=

==

2.7

5.18.4

The stress resultant at the base of the pole consists of a bending moment M, a

torque and a shear force V. Their magnitudes are:M = Wh = (4.8kN)(6.6m) = 31.68kN.m

= 7.2kN.m

V = W = 4.8kN

Examination of these stress resultants shows that maximum bending stressesoccur at pointA and maximum shear stresses at point B.

Therefore,A and B are critical points where the stresses should be determined.

Stresses at points A and B

The bending moment M produces a tensile stressa at pointA, but no stress at

point B (which is located on the neutral axis)

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( )( )( )

( ) MPamkN

dd

dM

a 91.54

64

18.022.0

11.068.31

64

2444

1

4

2

2

=

=

=

The torque

produces shear stresses1 at pointsA and B.

Finally, we need to calculate the direct shear stresses at pointsA and B

due to the shear force V.

( )( )( )

( ) MPammkN

dd

dT

Torsion 24.6

32

18.022.0

11.0.2.7

32

2444

142

2

=

=

=

VQ

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The shear stress at pointA is zero, and the shear stress at

point B (2) is obtained from the shear formula for a circular

tube

The stresses acting on the cross section at pointsA and B

have now been calculated.

( )

( )

)(2

3

2

64

12

3

1

3

2

4

1

4

2

2

rrb

rrQ

ddI

Ib

VQ

=

=

=

=

( ) MPamAVMax 7637.001257.0 480022 2,2 ===

Stress Elements

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Stress Elements

For both elements the y-axis is parallel to the longitudinal

axis of the pole and the x-axis is horizontal.

Point A :x = 0

y = a = 54.91MPa

xy = 1 = 6.24MPa

( )22

2,122

xy

yxyx

+

+=Principal stresses at PointA

Substituting 1,2 = 27.5MPa +/- 28.2MPa

1

= 55.7MPa and

2

= - 0.7MPa

The maximum in-plane shear

stresses can be obtained from

the equation( ) MPaxyyxMAX 2.282

2

2

=+

=

Because the principal stresses have opposite signs, the maximum in-plane

shear stresses are larger than the maximum out-of-plane shear stresses.Then,

max

= 28.2MPa.

Point B :

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Point B :

x = y = 0

xy = 1 + 2

xy = 6.24MPa + 0.76MPa = 7.0MPaPrincipal stresses at point B are

1 = 7.0MPa 2 = - 7.0 MPa

And the maximum in-plane shear stress is

max = 7.0MPa

The maximum out-of-plane shear stresses are half of

this value.Note

If the largest stresses anywhere in the pole are needed, then we must alsodetermine the stresses at the critical point diametrically opposite pointA,

because at that point the compressive stress due to bending has its largest

value.

The principal stresses at that point are

1 = 0.7MPa and 2 = - 55.7MPa

The maximum shear stress is 28.2MPa.

(In this analysis only the effects of wind pressure are considered. Other loads, such asweight of the structure, also produce stresses at the base of the pole).

A tubular post of square cross section supports a

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A tubular post of square cross section supports a

horizontal platform (see figure).

The tube has outer dimension b = 6in.

And wall thickness t = 0.5in.The platform has dimensions 6.75in x 24.0in and

supports an uniformly distributed load of 20psi acting

over its upper surface.

The resultant of this distributed load is a verticalforce

P1 = (20psi)(6.75in x 24.0in) = 3240lb

This force acts at the midpoint of the platform, whichis at distance

d = 9in. from the longitudinal axis of the post.

A second load P2 = 800lb acts horizontally on the

post at height h = 52in above the base.

Determine the principal stresses and maximum

shear stresses at points A and B at the base of the

post due to the loads P1 and P2.

Solution

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Stress Resultants

The force P1 acting on the platform is statically

equivalent to a force P1 and a momentM1 = P1d acting on the centroid of the cross section

of the post.

The load P2

is also shown.

Solution

The stress resultant at the base of the post due to the

loads P1 and P2 and the moment M1 are as follows:

(A) An axial compressive force P1 = 3240lb

(B) A bending moment M1 produced by the force P1:

M1 = P1d = (3240lb)(9in) = 29160lb-in

(C) A shear force P2 = 800lb

(D) A bending moment M2 produced by the force P2:M2 = P2h = (800lb)(52in) = 41600lb.in

Examinations of these stress resultants shows that both M1 and M2 produce

maximum compressive stresses at point A and the shear force produces

maximum shear stresses at point B. Therefore, A and B are the crit ical points

where the stresses should be determined.

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Stresses at points A and B(A) The axial force P1 produces uniform compressive stresses throughout the

post. These stresses areP1 = P1 / A whereA is the cross section area of the

post

A = b2 (b 2t)2 = 4t(b-t) = 4 (0.5in)(6in 0.5in) = 11.0in2

P1 = P1 / A = 3240lb / 11.00in2 = 295psi

(B) The bending moment M1 produces compressive stresses M1 at pointsA

and B. These stresses are obtained from the flexure formula

M1 = M1 (b / 2) /

where is the moment of inertia of the cross section. The moment of inertia is

= [b4

- (b -2t)4

] / 12 = [(6in)4

(5in)4

] / 12 = 55.92in4

Thus,M1 = M1 b / 2 = (29160lb.in)(6in) / (2)(55.92in

4) = 1564psi

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(C) The shear force P2 produces a shear stress at point B but not at pointA.We know that an approximate value of the shear stress can be obtained by

dividing the shear force by the web area.

P2 = P2 / Aweb =P2 /(2t(b 2t)) =800lb / (2)(0.5in)(6in1in)= 160psi

The stress p2 acts at point B in the direction shown in the above figure.We can calculate the shear stress

P2 from the more accurate formula. The

result of this calculation isP2 = 163psi, which shows that the shear stress

obtained from the approximate formula is satisfactory.

D) The bending moment M2 produces a compressive stress at pointA but no

stress at point B. The stress atA is

M2 = M2 b / 2 = (41600lb.in)(6in) / (2)(55.92in4) = 2232psi.

This stress is also shown in the above figure.

Stress Elements

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Each element is oriented so that the y-axis is

vertical (i.e. parallel to the longitudinal axis of the

post) and the x-axis is horizontal axisPoint A : The only stress in pointA is a

compressive stressa in the y direction

A = P1 + M1 + M2A = 295psi + 1564psi + 2232psi = 4090psi(compression)

Thus, this element is in uniaxial stress.

Principal Stresses and Maximum Shear Stressx = 0

y = -a = - 4090psi

xy = 0

Since the element is in uniaxial stress,

1 = x and 2 = y = - 4090psi

And the maximum in-plane shear stress is

max = (1 -2) / 2 = (4090psi) = 2050psiThe maximum out-of-plane shear stress has the same magnitude.

Point B:

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Here the compressive stress in the y direction is

B = P1 + M1

B = 295psi + 1564psi = 1860psi (compression)And the shear stress is

B = P2 = 160psi

The shear stress acts leftward on the top face and

downward on the x face of the element.Principal Stresses and Maximum Shear Stress

x =0

y = - B = - 1860psi

xy = - P2 = - 160psi

Substituting

1,2

= - 930psi +/- 944psi

1

= 14psi and

2

= - 1870psi

( )22

2,1

22 xy

yxyx

+

+=

The maximum in-plane shear stresses can be obtained from the equation

Because the principal stresses have opposite signs, the maximum in-plane

shear stresses are larger than the maximum out-of-plane shear stresses.

Then, max

= 944psi. ( ) psixyxMAX 9442

22

=+

=

Three forces are applied to a short steel post as

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Three forces are applied to a short steel post as

shown. Determine the principle stresses,

principal planes and maximum shearing stress

at point H.

Determine internal forces in Section EFG.

Solution

( )( ) ( )( )

( )( ) mkN3m100.0kN300

mkN5.8

m200.0kN75m130.0kN50

kN75kN50kN30

===

=

====

zy

x

x

zx

MM

M

M

VPV

Note: Section properties, ( )( ) 23m106.5m140.0m040.0 ==A

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Note: Section properties,

( )( )

( )( ) 463121

463121

m10747.0m040.0m140.0

m1015.9m140.0m040.0

==

==

z

x

I

I

( )( )

( )( )

( ) MPa66.0MPa2.233.8093.8

m1015.9

m025.0mkN5.8

m10747.0

m020.0mkN3

m105.6

kN50

46

4623-

=+=

+

=

++=

x

x

z

z

y I

bM

I

aM

A

P

Evaluate the stresses at H.

Shear stress at H.

( )( )[ ]( )

( )( )

( )( )MPa52.17m040.0m1015.9

m105.85kN75

m105.85

m0475.0m045.0m040.0

46

36

36

11

=

==

=

==

tI

QV

yAQ

x

zyz

Normal stress at H.

Calculate principal stresses and maximum shearing stress.

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p p g

=

===

===

=+=+=

=+==

98.13

96.2720.33

52.172tan

MPa4.74.370.33

MPa4.704.370.33

MPa4.3752.170.33

pp

min

max

22max

p

CDCY

ROC

ROC

R

=

=

=

=

98.13

MPa4.7

MPa4.70

MPa4.37

min

max

max

p

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The cantilever tube shown is to be

made of 2014 aluminum alloytreated to obtain a specified

minimum yield strength of

276MPa. We wish to select a stock

size tube (according to the table

below). Using a design factor of

n=4.

The bending load isF=1.75kN, the axial tension isP=9.0kNand thetorsion is T=72N.m. What is the realized factor of safety?

Consider the critical area ( top surface).

GPaGPaSy

06900276.0

==

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IMc

APx +=

Maximum bending moment = 120F

I

dkNxmm

A

kNx

+=2

75.11209

J

d

J

d

J

Trzx

36272

=

==

( ) 21

22 3 zxxVM +=

GPaGPan

VM 0690.04

==

For the dimensions of that tube

57.406043.0

276.0===

VM

ySn

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A certain forceF is applied atD near the end of the 15-in lever,

which is similar to a socket wrench. The bar OABCis made ofAISI

1035 steel, forged and heat treated so that it has a minimum (ASTM)yield strength of 81kpsi. Find the force (F) required to initiate

yielding. Assume that the leverDCwill not yield and that there is no

stress concentration atA.

Solution:

1) Find the critical sectionThe critical sections will be

either pointA or Point O. As the

moment of inertia varies withr4then pointA in the 1in diameter

is the weakest section.

d 2) Determine the stresses at the

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Fd

inF

d

dM

I

Myx 6.142

1432

64

234

=

=

==

2) Determine the stresses at the

critical section

Fin

inF

d

dT

J

Trzx 4.76

)1(

1516

32

234

=

=

==

3) Chose the failurecriteria.

The AISI 1035 is a

ductile material. Hence,we need to employ the

distortion-energy

theory.

lbf

S

F

F

VM

y

zxxxyyxyxVM

4165.194

81000

5.19433 22222

===

=+=++=

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Apply the MSS theory. For a point undergoing plane stress with

only one non-zero normal stress and one shear stress, the two non-zero principal stresses (A and B) will have opposite signs (Case 2).

( ) ( )( )lbfFFF

S

S

zxxzxx

yBA

zxxyBA

3884.7646.14281000

42

2

222

21

22

222

2

2

2

max

=

+=

+=+

=

+

==

=

A round cantilever bar is subjected to torsion plus a transverse load at

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A round cantilever bar is subjected to torsion plus a transverse load at

the free end. The bar is made of a ductile material having a yield

strength of50000psi. The transverse force (P) is500lb and the torqueis 1000lb-in applied to the free end. The bar is5in long (L) and a

safety factor of 2 is assumed. Transverse shear can be neglected.

Determine the minimum diameter to avoid yielding using both MSSand DET criteria.

Solution

1) Determine the critical section

The critical section occurs at thewall.

dPL

dT

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34

32

64

2

d

PL

d

PL

I

Mcx

=

==

34

16

32

2

d

T

d

T

J

Tcxy

=

==

( ) ( )

( )

( )

+=

+=

+

=

+

=+

+=

22

32,1

22

3

2

3

2

332,1

2

2

2

2

2,1

10005500550016

16161616

2222

d

TPLPLdd

T

d

PL

d

PL

xy

xx

xy

yxyx

26450 8.980 The stresses are in the wrong

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indn

S

dd

y

MAX

MAX

031.1

000,25

2

500002

4.13715

2

)8.980(26450

2

31

3331

===

=

=

=

31d

=32

d=

g

order.. Rearranged to

31

26450

d=

33

8.980

d=

MSS

indn

S

d

dddd

y

VM

VM

025.12

5000026950

8.980264508.98026450

3

33

2

3

2

3312

32

1

==

+=+=

DET

The factor of safety for a machine

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The factor of safety for a machine

element depends on the particular

point selected for the analysis. Basedupon the DET theory, determine the

safety factor for points A and B.

This bar is made of AISI 1006 cold-drawn steel (Sy=280MPa) and it is

loaded by the forces F=0.55kN, P=8.0kN and T=30N.m

Solution:

Point A 2324432

464

2

d

P

d

Fl

d

P

d

dFl

Area

P

I

Mcx +=+

=+=

( )( )( )( ) ( )( )( ) MPax 49.95

02.01084

02.01.01055.032

2

3

3

3

=+=

( )( )

MPaTTr

xy 10.19301616

33 ====

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( )dJxy 020.0 33

( ) ( )[ ]77.2

1.101

280

1.1011.19349.953 21

2222

===

=+=+=

VM

y

xyxVM

Sn

MPa

Point B ( )( )

( )( )

( )( )( )

( )[ ]

22.602.45

280

02.4543.21347.25

43.21

02.04

3

1055.04

02.0

3016

3

416

47.2502.0

10844

21

22

2

3

33

2

3

2

==

=+=

=

+=+=

===

n

MPa

MPaA

V

d

T

MPad

P

VM

xy

x

The shaft shown in the figure below is supported by two bearings and carries two V-

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g pp y g

belt sheaves. The tensions in the belts exert horizontal forces on the shaft, tending to

bend it in the x-z plane. Sheaves B exerts a clockwise torque on the shaft when

viewed towards the origin of the coordinate system along the x-axis. Sheaves C

exerts an equal but opposite torque on the shaft. For the loading conditions shown,

determine the principal stresses and the safety factor on the element K, located on the

surface of the shaft (on the positive z-side), just to the right of sheave B. Consider

that the shaft is made of a steel of a yield strength of 81ksi

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Shearing force = 165lb

Bending Moment = -1540lb-in( )

( ) ksi

r

rM

I

Mc

x

031.8625.0

15404

4

34

=

===

( )( ) ksi

rT

JTrxz 868.2

625.0110022 33 ====

( ) ( )xy

xxxy

yxyx

2222

2

2

2

2

2,1 +

=+

+=

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( )ksiksi 92.095.8

868.22

03.8

2

03.8

21

2

2

2,1

==

+

=

( )

2.8935.4 2

81

2..

935.42

92.095.8

2..... 31

====

=

=

=

Max

Y

Max

S

nFactorSafety

ksiMSS

( ) ( ) ( )( )

58.844.9

81..

44.992.095.892.095.8......22

31

2

3

2

1

====

=+=+=

VM

y

VM

SnFactorSafety

ksiDET

A horizontal bracketABC consists of two perpendicular armsAB and BC, of

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1.2m and 0.4m in length respectively. The ArmAB has a solid circular cross

section with diameter equal to 60mm. At point C a load P1=2.02kN acts

vertically and a load P2=3.07kN acts horizontally and parallel to armAB.

For the points p and q, located at supportA, calculate:

(1)The principal stresses.

(2) the maximum in-plane shear stress.

1.2m

q