mechanics of materials seventh edition ferdinand p. beer e. russell johnston, jr. john t. dewolf...

MECHANICS OF MATERIALS

Seventh Edition

Ferdinand P. Beer

E. Russell Johnston, Jr.

John T. DeWolf

David F. Mazurek

Lecture Notes:

Brock E. Barry

U.S. Military Academy

CHAPTER

Copyright © 2015 McGraw-Hill Education. Permission required for reproduction or display.

3Torsion



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Beer • Johnston • DeWolf • Mazurek

Contents

3 - 2

Torsional Loads on Circular Shafts

Net Torque Due to Internal Stresses

Axial Shear Components

Shaft Deformations

Shearing Strain

Stresses in Elastic Range

Normal Stresses

Torsional Failure Modes

Sample Problem 3.1

Angle of Twist in Elastic Range

Statically Indeterminate Shafts

Sample Problem 3.4

Design of Transmission Shafts

Stress Concentrations

Plastic Deformations

Elastoplastic Materials

Residual Stresses

Concept Application 3.8/3.9

Torsion of Noncircular Members

Thin-Walled Hollow Shafts

Concept Application 3.10



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Torsional Loads on Circular Shafts

3 - 3

• Stresses and strains in members of circular cross-section are subjected to twisting couples or torques

• Generator creates an equal and opposite torque T’

• Shaft transmits the torque to the generator

• Turbine exerts torque T on the shaft

Fig. 3.2 (a) A generator provides power at a constant revolution per minute to a turbine through shaft AB. (b) Free body diagram of shaft AB along with the driving and reaction torques on the generator and turbine, respectively.



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Net Torque Due to Internal Stresses

3 - 4

dAdFT

• 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 stresses is known, the distribution of the stresses is not.

• Unlike the normal stress due to axial loads, the distribution of shearing stresses due to torsional loads cannot be assumed uniform.

• Distribution of shearing stresses is statically indeterminate – must consider shaft deformations.

Fig. 3.24 (a)Free body diagram of section BC with torque at C represented by the representable contributions of small elements of area carrying forces dF a radius from the section center. (b) Free-body diagram of section BC having all the small area elements summed resulting in torque T.

Fig. 3.3 Shaft subject to torques and a section plane at C.



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Axial Shear Components

3 - 5

• 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 slats slide with respect to each other when equal and opposite torques are applied to the ends of the shaft.

• The existence of the axial shear components is demonstrated by considering a shaft made up of slats pinned at both ends to disks.

Fig. 3.6 Model of shearing in shaft (a) undeformed; (b) loaded and deformed.

Fig. 3.5 Small element in shaft showing how shear stress components act.



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Shaft Deformations

3 - 6

• From observation, the angle of twist of the shaft is proportional to the applied torque and to the shaft length.

L

T

• When subjected to torsion, every cross-section of a circular shaft remains plane and undistorted.

• Cross-sections for hollow and solid circular shafts remain plain and undistorted because a circular shaft is axisymmetric.

• Cross-sections of noncircular (non-axisymmetric) shafts are distorted when subjected to torsion.

Fig. 3.8 Comparison of deformations in circular (a) and square (b) shafts.

Fig. 3.7 Shaft with fixed support and line AB drawn showing deformation under torsion loading: (a) unloaded; (b) loaded.



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Shearing Strain

3 - 7

• Consider an interior section of the shaft. As a torsional load is applied, an element on the interior cylinder deforms into a rhombus.

• Shear strain is proportional to twist and radius

maxmax and cL

c

LL

or

• It follows that

• Since the ends of the element remain planar, the shear strain is equal to angle of twist.

Fig. 3.13 Shearing Strain Kinematic definitions for torsion deformation. (a) The angle of twist (b) Undeformed portion of shaft of radius with (c) Deformed portion of the shaft having same angle of twist,and strain, angles of twist per unit length, .



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Stresses in Elastic Range

3 - 8

Jc

dAc

dAT max2max

• Recall that the sum of the moments of the elementary forces exerted on any cross section of the shaft must be equal to the magnitude T of the torque:

and max J

T

J

Tc

• The results are known as the elastic torsion formulas,

• Multiplying the previous equation by the shear modulus,

max Gc

G

maxc

From Hooke’s Law, G , so

The shearing stress varies linearly with the distance from the axis of the shaft.

Fig. 3.14 Distribution of shearing stresses in a torqued shaft; (a) Solid shaft, (b) hollow shaft.



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Normal Stresses

3- 9

• 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.

• 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.

max0

0max45

0max0max

2

2

245cos2

o

A

A

A

F

AAF

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

• Element a is in pure shear.

Fig. 3.17 Circular shaft with stress elements at different orientations.

Fig. 3.18 Forces on faces at 45° to shaft axis.

Fig. 3.19 Shaft elements with only shear stresses or normal stresses.



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Torsional Failure Modes

3 - 10

• Ductile materials generally fail in shear. Brittle materials are weaker in tension than shear.

• When subjected to torsion, a ductile specimen breaks along a plane of maximum shear, i.e., a plane perpendicular to the shaft axis.

• When subjected to torsion, a brittle specimen breaks along planes perpendicular to the direction in which tension is a maximum, i.e., along surfaces at 45o to the shaft axis.

Photo 3.2 Shear failure of shaft subject to torque.



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Sample Problem 3.1

3 - 11

Shaft BC is hollow with inner and outer diameters of 90 mm and 120 mm, respectively. Shafts AB and CD are solid and 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.

SOLUTION:

• Cut sections through shafts AB and BC and perform static equilibrium analyses to find torque loadings.

• Given allowable shearing stress and applied torque, invert the elastic torsion formula to find the required diameter.

• Apply elastic torsion formulas to find minimum and maximum stress on shaft BC.



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3 - 12

Sample Problem 3.1SOLUTION:• Cut sections through shafts AB and BC

and perform static equilibrium analysis to find torque loadings.

CDAB

ABx

TT

TM

mkN6

mkN60 mkN20

mkN14mkN60

BC

BCx

T

TM

Fig. 1 Free-body diagram for section between A and B.

Fig. 2 Free-body diagram for section between B and C.



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3 - 13

Sample Problem 3.1• Apply elastic torsion formulas to

find minimum and maximum stress on shaft BC.

46

4441

42

m1092.13

045.0060.022

ccJ

MPa2.86

m1092.13

m060.0mkN2046

22max

J

cTBC

MPa7.64

mm60

mm45

MPa2.86

min

min

2

1

max

min

c

c

MPa7.64

MPa2.86

min

max

• Given allowable shearing stress and applied torque, invert the elastic torsion formula to find the required diameter.

m109.38

mkN665

3

32

42

max

c

cMPa

c

Tc

J

Tc

mm8.772 cd

Fig. 3 Shearing stress distribution on cross section.

Fig. 4 Free-body diagram of shaft portion AB.



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Angle of Twist in Elastic Range

3 - 14

• Recall that the angle of twist and maximum shearing strain are related,

L

c max

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

JG

Tc

G max

max

• Equating the expressions for shearing strain and solving for the angle of twist,

JG

TL

• If the torsional loading or shaft cross-section changes along the length, the angle of rotation is found as the sum of segment rotations

i ii

ii

GJ

LT

Fig. 3.20 Torque applied to fixed end shaft resulting angle of twist .

Fig. 3.21 Shaft with multiple cross-section dimensions and multiple loads.



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Statically Indeterminate Shafts

3 - 15

• Given the shaft dimensions and the applied torque, we would like to find the torque reactions at A and B.

• From a free-body analysis of the shaft,

which is not sufficient to find the end torques. The problem is statically indeterminate.

ftlb90 BA TT

ftlb9012

21 AA TJL

JLT

• Substitute into the original equilibrium equation,

ABBA T

JL

JLT

GJ

LT

GJ

LT

12

21

2

2

1

121 0

• Divide the shaft into two components which must have compatible deformations,

Fig. 3.25 (a) Shaft with central applied torque and fixed ends. (b) free-body diagram of shaft AB. (c) Free-body diagrams for solid and hollow segments.



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Sample Problem 3.4

3 - 16

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 T0 that may be applied to the end of shaft AB, (b) the corresponding angle through which end A of shaft AB rotates.

SOLUTION:

• Apply a static equilibrium analysis on the two shafts to find a relationship between TCD and T0 .

• 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.

• Apply a kinematic analysis to relate the angular rotations of the gears.



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3 - 17

Sample Problem 3.4

SOLUTION:

• Apply a static equilibrium analysis on the two shafts to find a relationship between TCD and T0 .

0

0

8.2

in.45.20

in.875.00

TT

TFM

TFM

CD

CDC

B

• Apply a kinematic analysis to relate the angular rotations of the gears.

CB

CCB

CB

CCBB

r

r

rr

8.2

in.875.0

in.45.2

Fig. 1 Free-body diagrams of gears B and C.

Fig. 2 Angles of twist for gears B and C.



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3 - 18

• Find the T0 for the maximum allowable torque on each shaft – choose the smallest.

in.lb561

in.5.0

in.5.08.28000

in.lb663

in.375.0

in.375.08000

0

42

0max

0

42

0max

T

Tpsi

J

cT

T

Tpsi

J

cT

CD

CD

AB

AB

inlb5610 T

Sample Problem 3.4• Find the corresponding angle of twist for each

shaft and the net angular rotation of end A.

oo

/

oo

o

642

/

o

642

/

2.2226.8

26.895.28.28.2

95.2rad514.0

psi102.11in.5.0

.in24in.lb5618.2

2.22rad387.0

psi102.11in.375.0

.in24in.lb561

BABA

CB

CD

CDDC

AB

ABBA

GJ

LT

GJ

LT

o48.10A

Fig. 3 Free-body diagram of shaft AB.

Fig. 4 Free-body diagram of shaft CD.

Fig. 5



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Design of Transmission Shafts

3 - 19

• Principal transmission shaft performance specifications are:

power Speed of rotation

• Determine torque applied to shaft at specified power and speed,

f

PPT

fTTP

2

2

• Find shaft cross-section which will not exceed the maximum allowable shearing stress,

shafts hollow2

shafts solid2

max

41

42

22

max

3

max

Tcc

cc

J

Tc

c

J

J

Tc

• Designer must select shaft material and dimensions of the cross-section to meet performance specifications without exceeding allowable shearing stress.



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Stress Concentrations

3 - 20

• The derivation of the torsion formula,

assumed a circular shaft with uniform cross-section loaded through rigid end plates.

J

Tcmax

J

TcKmax

• Experimental or numerically determined concentration factors are applied as

• The use of flange couplings, gears and pulleys attached to shafts by keys in keyways, and cross-section discontinuities can cause stress concentrations

Fig. 3.28 Plot of stress concentration factors for fillets in circular shafts.

Fig. 3.26 Coupling of shafts using (a) bolted flange, (b) slot for keyway.



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Plastic Deformations

3 - 21

• With the assumption of a linearly elastic material,

J

Tcmax

cc

ddT0

2

022

• The integral of the moments from the internal stress distribution is equal to the torque on the shaft at the section,

• Shearing strain varies linearly regardless of material properties. Application of shearing-stress-strain curve allows determination of stress distribution.

• If the yield strength is exceeded or the material involved is a brittle materials with a nonlinear shearing-stress-strain curve, these relationships cease to be valid.

Fig. 3.29 Distribution of shearing strain for torsion of a circular shaft.

Fig. 3.30 Nonlinear, shear stress-strain diagram.

Fig. 3.31 Shearing strain distribution for shaft with nonlinear stress-strain response.



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Elastoplastic Materials

3 - 22

• At the maximum elastic torque,

YYY cc

JT 3

21

c

L YY

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

)Y Y

Y

3

3

41

34

3

3

413

32 11

cT

ccT Y

YY

Y

3

3

41

34 1

Y

YTT

YL

Y

• As , the torque approaches a limiting value,

0Y

torque plastic TT YP 34

• Valid only for a solid circular shaft made of an elastoplastic material.

Fig. 3.34 Stress-strain distribution for elastic-perfectly plastic shaft at different stages of loading: (a) elastic, (b) impending yield, (c) partially yielded, and (d) fully yielded.



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Residual Stresses

3 - 23

• When the torque is removed, the reduction of stress and strain at each point takes place along a straight line to a generally non-zero residual stress.

• Residual stresses found from principle of superposition 0 dA

J

Tcm

• Plastic region develops in a shaft when subjected to a large enough torque.

• On a T- curve, the shaft unloads along a straight line to an angle greater than zero.Fig. 3.37 Shear stress-strain

response for loading past yield reversing until compressive yield occurs.

Fig. 3.38 Torque-angle of twist response for loading past yield, followed by unloading. Fig. 3.39 Superposition of elastic-plastic state (a) plus

linear elastic unloading (b) equals residual (c) sharing stress distributions.



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3 - 24

A solid circular shaft is subjected to a torque at each end. Assuming that the shaft is made of an elastoplastic material with and determine (a) the radius of the elastic core, (b) the angle of twist of the shaft. When the torque is removed, determine (c) the permanent twist, (d) the distribution of residual stresses.

MPa150YGPa77G

mkN6.4 T

SOLUTION:

• Solve Eq. (3.29) for Y/c and evaluate the elastic core radius

• Find the residual stress distribution by a superposition of the stress due to twisting and untwisting the shaft

• 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

• Solve Eq. (3.15) for the angle of twistFig. 3.36 Loaded circular shaft.



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3 - 25

SOLUTION:• Solve Eq. (3.29) for Y/c and

evaluate the elastic core radius3

1

3413

3

41

34

Y

YYY T

T

ccTT

mkN68.3

m1025

m10614Pa10150

m10614

m1025

3

496

49

3214

21

Y

YY

YY

T

c

JT

J

cT

cJ

630.068.3

6.434

31

cY

mm8.15Y

• Solve Eq. (3.15) for the angle of twist

o33

3

49-

3

8.50rad103.148630.0

rad104.93

rad104.93

Pa1077m10614

m2.1mN1068.3

Y

YY

Y

YY

Y

JG

LT

cc

o50.8




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3 - 26

• Evaluate Eq. (3.15) 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

o

3

949

3

1.81

69.650.8

6.69rad108.116

Pa1077m1014.6

m2.1mN106.4

pφ

JG

TL

o81.1p


• 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

Tc

Fig. 3.40 Superposition of stress distributions to obtain residual stresses.



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Torsion of Noncircular Members

3 - 27

• Planar cross-sections of noncircular shafts do not remain planar and stress and strain distribution do not vary linearly

• Previous torsion formulas are valid for axisymmetric or circular shafts

Gabc

TL

abc

T3

22

1max

• For uniform rectangular cross-sections,

• At large values of a/b, the maximum shear stress and angle of twist for other open sections are the same as a rectangular bar.

Fig. 3.41 Twisting of shaft with square cross section.

Fig. 3.44 Shaft with rectangular cross section, showing the location of maximum shearing stress.



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Thin-Walled Hollow Shafts

3 - 28

• Summing forces in the x-direction on AB,

shear stress varies inversely with thickness

flowshear

0

qttt

xtxtF

BBAA

BBAAx

t

ds

GA

TL24

• Angle of twist (from Chapter 11)

tA

T

qAdAqdMT

dAqpdsqdstpdFpdM

2

22

2

0

0

• Compute the shaft torque from the integral of the moments due to shear stress

Fig. 3.47 Thin-walled hollow shaft subject to torsional loading.

Fig. 3.51 Shear flow in the member wall.

Fig. 3.53 Area for shear flow.

Fig. 3.48 Segment of thin-walled hollow shaft.



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3 - 29

Structural 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) 0.120 in. 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 .

Fig. 3.54 Square thin-walled aluminum tubing having: (a) uniform thickness, (b) non-uniform thickness, (c) median area line (next slide)



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3 - 30

• Find the corresponding shearing stress with each wall thickness.

With a uniform wall thickness,

ksiinkip

tA

T35.8

)in.in.)(8.986160.0(2

.24

2 2

ksi35.8

With a variable wall thickness

in.120.0

in.kip335.1 ACAB

in.200.0

in.kip335.1 CDBD

ksi13.11 BCAB

ksi68.6 CDBC


SOLUTION:

• Determine the shear flow through the tubing walls.

2in.986.8in.34.2in.84.3 A

mechanics of materials seventh edition ferdinand p. beer e. russell johnston, jr. john t. dewolf...

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