Mecanica Solidos Beer-03

8/20/2019 Mecanica Solidos Beer-03

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MECHANICS OFMATERIALS

Third Edition

Ferdinand P. Beer

E. Russell Johnston, Jr.

CHAPTER

Torsion .

Lecture Notes:

J. Walt Oler

Texas Tech University

© 2002 The McGraw-Hill Companies, Inc. All rights reserved.


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Beer • Johnston • DeWolf

Contents

Introduction

Torsional Loads on Circular Shafts

Net Torque Due to Internal StressesAxial Shear Components

Shaft Deformations

Statically Indeterminate Shafts

Sample Problem 3.4

Design of Transmission ShaftsStress Concentrations

Plastic Deformations

© 2002 The McGraw-Hill Companies, Inc. All rights reserved. 3 - 2

ear ng ra n

Stresses in Elastic Range

Normal Stresses

Torsional Failure Modes

Sample Problem 3.1

Angle of Twist in Elastic Range

as op as c a er a s

Residual Stresses

Example 3.08/3.09

Torsion of Noncircular Members

Thin-Walled Hollow Shafts

Example 3.10


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Torsional Loads on Circular Shafts

• Interested in stresses and strains of

circular shafts subjected to twisting

couples or torques

• Shaft transmits the torque to the

• Turbine exerts torque T on the shaft


• Generator creates an equal and

opposite torque T ’


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Net Torque Due to Internal Stresses

( )∫ ∫== dAdF T τ ρ

• Net of the internal shearing stresses is an

internal torque, equal and opposite to the

applied torque,

• Although the net torque due to the shearing


is not

• Unlike the normal stress due to axial loads, the

distribution of shearing stresses due to torsional

loads can not be assumed uniform.

• Distribution of shearing stresses is statically

indeterminate – must consider shaft

deformations


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Axial Shear Components

• Torque applied to shaft produces shearing

stresses on the faces perpendicular to the

axis.

• Conditions of equilibrium require the

existence of equal stresses on the faces of the

two planes containing the axis of the shaft


• The existence of the axial shear components is

demonstrated by considering a shaft made up

of axial slats.

The slats slide with respect to each other when

equal and opposite torques are applied to the

ends of the shaft.


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• From observation, the angle of twist of the

shaft is proportional to the applied torque and

to the shaft length.

L

T

∝

∝

φ

φ

Shaft Deformations

• -


,

of a circular shaft remains plane andundistorted.

• Cross-sections of noncircular (non-

axisymmetric) shafts are distorted when

subjected to torsion.

• Cross-sections for hollow and solid circular

shafts remain plain and undistorted because a

circular shaft is axisymmetric.


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Shearing Strain

• Consider an interior section of the shaft. As a

torsional load is applied, an element on the

interior cylinder deforms into a rhombus.

• Since the ends of the element remain planar,

the shear strain is equal to angle of twist.


• Shear strain is proportional to twist and radius

maxmax and γ ρ

γ φ

γ c L

c==

L L

φ γ ρφ γ == or

• It follows that


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Stresses in Elastic Range

41 c J π =

• Multiplying the previous equation by theshear modulus,

maxγ γ Gc

G =

maxτ τ c

=

From Hooke’s Law, τ G=

, so

The shearing stress varies linearly with the


J cdAcdAT max2max τ

ρ

τ

ρτ ∫ =∫ ==

• Recall that the sum of the moments from

the internal stress distribution is equal to

the torque on the shaft at the section,

41

422

1 cc J −= π

and max J

T

J

Tcτ τ ==

• The results are known as the elastic torsion

formulas,

.


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Normal Stresses

• Elements with faces parallel and perpendicular

to the shaft axis are subjected to shear stresses

only. Normal stresses, shearing stresses or a

combination of both may be found for other

orientations.

( ) 0max0max 245cos2 τ τ == A AF

• Consider an element at 45o to the shaft axis,


max0

0max45 2

2o τ τ σ === A

A AF

• Element a is in pure shear.

• Note that all stresses for elements a and c have

the same magnitude

• Element c is subjected to a tensile stress on

two faces and compressive stress on the other

two.


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Torsional Failure Modes

• Ductile materials generally fail in

shear. Brittle materials are weaker in

tension than shear.

• When subjected to torsion, a ductile


maximum shear, i.e., a planeperpendicular to the shaft axis.

• When subjected to torsion, a brittle

specimen breaks along planesperpendicular to the direction in

which tension is a maximum, i.e.,

along surfaces at 45o to the shaft

axis.


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Sample Problem 3.1

SOLUTION:

• Cut sections through shafts AB

and BC and perform static

equilibrium analysis to findtorque loadings

• Apply elastic torsion formulas to


Shaft BC is hollow with inner and outer

diameters of 90 mm and 120 mm,

respectively. Shafts AB and CD are solid

of diameter d . For the loading shown,

determine (a) the minimum and maximum

shearing stress in shaft BC , (b) the

required diameter d of shafts AB and CD

if the allowable shearing stress in these

shafts is 65 MPa.

• Given allowable shearing stress

and applied torque, invert the

elastic torsion formula to find the

required diameter

stress on shaft BC


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SOLUTION:

• Cut sections through shafts AB and BC

and perform static equilibrium analysis

to find torque loadings

Sample Problem 3.1


( )

CD AB

AB x

T T

T M

=⋅=

−⋅==∑mkN6

mkN60 ( ) ( )

mkN20

mkN14mkN60

⋅=

−⋅+⋅==∑ BC

BC x

T

T M


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• Apply elastic torsion formulas tofind minimum and maximum

stress on shaft BC

• Given allowable shearing stress andapplied torque, invert the elastic torsion

formula to find the required diameter

Sample Problem 3.1


( ) ( ) ( )[ ]46

444142

m1092.13

045.0060.022

−×=

−=−=

π cc J

( )( )

MPa2.86m1092.13

m060.0mkN2046

22max

=

×

⋅===

−

J

cT BC τ τ

MPa7.64

mm60

mm45

MPa2.86

min

min

2

1

max

min

=

==

τ

τ

τ

τ

c

c

MPa7.64

MPa2.86

min

max

=

=

τ

τ

m109.38

mkN665

3

3

2

4

2

max

−×=

⋅===

c

c MPa

cTc

J Tc

π π τ

mm8.772 == cd


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Angle of Twist in Elastic Range

• Recall that the angle of twist and maximum

shearing strain are related,

L

cφ γ =max

• In the elastic range, the shearing strain and shearare related by Hooke’s Law,

JG

Tc

G== maxmax

τ γ


• Equating the expressions for shearing strain andsolving for the angle of twist,

JG

TL=φ

• If the torsional loading or shaft cross-section

changes along the length, the angle of rotation isfound as the sum of segment rotations

∑=i ii

ii

G J

LT φ


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• Given the shaft dimensions and the applied

torque, we would like to find the torque reactions

at A and B.

Statically Indeterminate Shafts

• From a free-body analysis of the shaft,

which is not sufficient to find the end torques.

ftlb90 ⋅=+ B A T T


e pro em s s a ca y n e erm na e.

ftlb9012

21 ⋅=+ A A T J L

J LT

• Substitute into the original equilibrium equation,

A B B A T

J L

J LT

G J

LT

G J

LT

12

21

2

2

1

121 0 ==−=+= φ φ φ

• Divide the shaft into two components which

must have compatible deformations,


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Sample Problem 3.4

SOLUTION:

• Apply a static equilibrium analysis on

the two shafts to find a relationship

between T CD and T 0

• Apply a kinematic analysis to relate

the angular rotations of the gears


Two solid steel shafts are connected

by gears. Knowing that for each shaft

G = 11.2 x 106 psi and that the

allowable shearing stress is 8 ksi,

determine (a) the largest torque T 0that may be applied to the end of shaft

AB, (b) the corresponding angle

through which end A of shaft AB

rotates.

• Find the corresponding angle of twist

for each shaft and the net angular

rotation of end A

• Find the maximum allowable torque

on each shaft – choose the smallest


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SOLUTION:

• Apply a static equilibrium analysis on

the two shafts to find a relationship

between T CD

and T 0

• Apply a kinematic analysis to relate

the angular rotations of the gears

Sample Problem 3.4


( )

( )

0

0

8.2

in.45.20

in.875.00

T T

T F M

T F M

CD

CDC

B

=

−==

−==

∑

∑

C B

C C B

C B

C C B B

r

r r r

φ φ

φ φ φ

φ

8.2

in.875.0

in.45.2

=

==

=


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• Find the T 0 for the maximumallowable torque on each shaft –

choose the smallest

• Find the corresponding angle of twist for eachshaft and the net angular rotation of end A

( )( ).24in.lb561 ⋅== AB

in LT

Sample Problem 3.4


( )

( )

( )

( )

in.lb561

in.5.0

in.5.08.28000

in.lb663

in.375.0

in.375.08000

0

4

2

0max

0

4

2

0max

⋅=

==

⋅=

==

T

T psi

J

cT

T

T psi

J

cT

CD

CD

AB

AB

π

π

τ

τ

inlb5610 ⋅=T

( )

( )( )

( )

( )

( )

oo /

oo

o

64

2

/

o

2

2.2226.8

26.895.28.28.2

95.2rad514.0

psi102.11in.5.0

.24in.lb5618.2

2.22rad387.0

psi102.11in.375.0

+=+=

===

==

×

⋅==

==

×

B A B A

C B

CD

CD DC

AB

in

G J

LT

G J

φ φ φ

φ φ

φ π

π

o48.10= Aφ


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Design of Transmission Shafts

• Principal transmission shaft

performance specifications are:

- power

- speed

• Determine torque applied to shaft at

specified power and speed,

f

PPT

fT T P

π ω

π

2

2

==

==

• Find shaft cross-section which will not• Designer must select shaft

material and cross-section to


excee e max mum a owa e

shearing stress,

( )

( ) ( )shaftshollow2

shaftssolid2

max

41

42

22

max

3

max

τ

π

τ

π

τ

T cc

cc

J

T

cc

J

J

Tc

=−=

==

=

meet performance specificationswithout exceeding allowable

shearing stress.


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Stress Concentrations

• The derivation of the torsion formula,

assumed a circular shaft with uniform

cross-section loaded through rigid end

plates.

J

Tc=maxτ

• The use of flange couplings, gears and


J

TcK =maxτ

• Experimental or numerically determined

concentration factors are applied as

pulleys attached to shafts by keys inkeyways, and cross-section discontinuities

can cause stress concentrations

T E


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Plastic Deformations

• With the assumption of a linearly elastic material,

J

Tc=maxτ

• Shearing strain varies linearly regardless of material

• If the yield strength is exceeded or the material has

a nonlinear shearing-stress-strain curve, this

expression does not hold.


( ) ∫=∫=cc

d d T 0

2

0

22 ρ τ ρ π ρ πρ ρτ

• The integral of the moments from the internal stress

distribution is equal to the torque on the shaft at the

section,

properties. Application of shearing-stress-straincurve allows determination of stress distribution.

T E


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Elastoplastic Materials

• As the torque is increased, a plastic region( ) develops around an elastic core ( )Y τ τ = Y Y

τ ρ

τ =

φ ρ Y

LY =

• At the maximum elastic torque,

Y Y Y cc

J T τ π τ 3

21==

c

L Y Y φ =


• As , the torque approaches a limiting value,0→Y ρ

torque plasticT T Y P == 34

−=

−= 341343413

32 11c

T c

cT Y Y Y Y ρ

τ π

−=

3

3

41

34 1

φ

φ Y Y T T

T E


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Residual Stresses

• Plastic region develops in a shaft when subjected to a

large enough torque

• On a T-φ curve, the shaft unloads along a straight line

to an an le reater than zero

• When the torque is removed, the reduction of stress

and strain at each point takes place along a straight lineto a generally non-zero residual stress


• Residual stresses found from principle of superposition

( ) 0=∫ dAτ

J

Tc

m =′

τ


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E


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SOLUTION:

• Solve Eq. (3.32) for ρ Y /c and

evaluate the elastic core radius

31

3413

3

41

34

−=⇒

−=

Y

Y Y Y

T

T

cc

T T ρ ρ

m10614

m1025

49

3214

21

×=

×==

−

−c J π π

• Solve Eq. (3.36) for the angle of twist

( )( )( )( )

3

49-

3

rad104.93

Pa1077m10614

m2.1N1068.3

×=

××

×==

=⇒=

−φ

φ

ρ

φ φ

φ

Y

Y Y

Y

Y Y

Y

JG

LT

cc

Example 3.08/3.09


( )( )

mkN68.3

m1025

m10614Pa101503

496

⋅=

×

××=

=⇒=

−

−

Y

Y Y Y Y

T

c J T

J cT τ τ

630.068.3

6.434

31

=

−=

c

Y ρ

mm8.15=Y

o33 8.50rad103.148630.0

rad104.93=×=

×=

−−φ

o50.8=φ

T h

E d


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• Evaluate Eq. (3.16) for the angle

which the shaft untwists when

the torque is removed. The

permanent twist is the difference

between the angles of twist and

untwist

=′φ TL

• Find the residual stress distribution by

a superposition of the stress due to

twisting and untwisting the shaft

MPa3.187

m10614

m1025mN106.449-

33

max

=

×

×⋅×==′

−

J

Tcτ

Example 3.08/3.09


( )( )( )( )

( )o

33

3

949

3

1.81

rad108.116108.116

rad108.116

Pa1077m1014.6

m2.1mN106.4

=

×−×=

′−=

×=

××

⋅×=

−−

−

φ φ pφ

o

81.1=

pφ


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Thin-Walled Hollow Shafts

• Summing forces in the x-direction on AB,

shear stress varies inversely with thickness

( ) ( )

flowshear

0

====

∆−∆==∑

qt t t

xt xt F

B B A A

B B A A x

τ τ τ

τ τ

• Compute the shaft torque from the integral


( ) ( )

tA

T

qAdAqdM T

dAq pdsqdst pdF pdM

2

22

2

0

0

=

===

====

∫∫

τ

τ

∫=t

ds

G A

TL24

φ

• Angle of twist (from Chapt 11)

T h

E d


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Example 3.10

Extruded aluminum tubing with a rectangular

cross-section has a torque loading of 24 kip-

in. Determine the shearing stress in each of

the four walls with (a) uniform wall thickness

of 0.160 in. and wall thicknesses of (b) 0.120in. on AB and CD and 0.200 in. on CD and

BD.


SOLUTION:

• Determine the shear flow through the

tubing walls

• Find the corresponding shearing stress

with each wall thickness

T h

E d

B J h t D W lf


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SOLUTION:

• Determine the shear flow through the

tubing walls

• Find the corresponding shearingstress with each wall thickness

with a uniform wall thickness,

in.160.0

in.kip335.1==

t

qτ

ksi34.8=τ

Example 3.10


( )( )

( ) in.kip335.1

in.986.82in.-kip24

2

in.986.8in.34.2in.84.3

2

2

===

==

AT q

A

with a variable wall thickness

in.120.0

in.kip335.1== AC AB τ τ

in.200.0

in.kip335.1== CD BD τ τ

ksi13.11== BC AB τ τ

ksi68.6== CD BC τ τ

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Mecanica Solidos Beer-03