# Fourth Edition MECHANICS OF MATERIALS OF Fourth Edition MECHANICS OF MATERIALS CHAPTER Ferdinand P. Beer E. Russell Johnston, Jr. John T. DeWolf Torsion Lecture Notes: J. Walt Oler

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• MECHANICS OF Fourth Edition

MECHANICS OF MATERIALS

CHAPTER

MATERIALS Ferdinand P. Beer E. Russell Johnston, Jr. John T. DeWolf

Torsion

Lecture Notes: J. Walt Oler Texas Tech Universityy

• MECHANICS OF MATERIALS

Fourth Edition

Beer • Johnston • DeWolf

Contents

Introduction

Statically Indeterminate Shafts

Sample Problem 3.4

Net Torque Due to Internal Stresses

Axial Shear Components

Design of Transmission Shafts

Stress Concentrations

Shaft Deformations

Shearing Strain

Stresses in Elastic Range

Plastic Deformations

Elastoplastic Materials

Residual StressesStresses in Elastic Range

Normal Stresses

Torsional Failure Modes

Residual Stresses

Examples 3.08, 3.09

Torsion of Noncircular Members

Sample Problem 3.1

Angle of Twist in Elastic Range

Thin-Walled Hollow Shafts

Example 3.10

• MECHANICS OF MATERIALS

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• Interested in stresses and strains of circular shafts subjected to twisting

lcouples or torques.

• Turbine exerts torque T on the shaft.

• Generator creates an equal and

• Shaft transmits the torque to the generator.

• Generator creates an equal and opposite torque T’.

• MECHANICS OF MATERIALS

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Net Torque Due to Internal Stresses

• 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.

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

• Unlike the normal stress due to axial loads, the distribution of shearing stresses due to torsional l d t b d if

• MECHANICS OF MATERIALS

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Axial Shear Components

• Torque applied to shaft produces shearing stresses on the faces perpendicular to the

iaxis.

• Conditions of equilibrium require the existence of equal stresses on the faces of the

• The existence of the axial shear components

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.

• MECHANICS OF MATERIALS

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Shaft Deformations

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

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Shearing Strain

• Consider an interior section of the shaft. As a torsional load is applied, an element on the i t i li d d f i t h b

interior cylinder deforms into a rhombus.

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

• For small value of , It follows that

the shear strain is equal to angle of twist.

AA L 

AA   or AA L

L       (and are in radians)

AA  

AA 

• Shear strain is proportional to twist and radius

maxmax and  

L c



L

(Shown next page)

cL tan AA

L  

  

• MECHANICS OF MATERIALS

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Shearing Strain

c

At shaft surface, Max shearing strain

max c L  

Shearing strain at a distance  from the shaft axisShearing strain at a distance  from the shaft axis

Shear strain inside the shaft, L L   

   

max maxShear strain at the shaft's surface,

Lc L c

LL

  

 

  

max max

LL c c

    

    

maxc  

• MECHANICS OF MATERIALS

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Stresses in Elastic Range • Multiplying previous eq. by the shear modulus,

max  G c

G  c

From Hooke’s Law,  G , so

The shearing stress varies linearly4 2 1 cJ  max

 c

 The shearing stress varies linearly with the radial position in the section.

F h ll li d

max max 2

From or c c      

• For hollow cylinder

 414221 ccJ  

2

1 1 min max

2

Substitute cc c

    

J = Polar moment of inertia min 1

max 2

c c

 

• MECHANICS OF MATERIALS

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Stresses in Elastic Range

• Recall that the sum of the moments from the internal stress distribution is equal to the torque on the shaft at the section

J c

dA c

dAT max2max    

the torque on the shaft at the section,

Th lt k th l ti t i

dAJ  2max c

• The results are known as the elastic torsion formulas,

andmax TTc  

and max JJ 

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Normal Stresses • Elements with faces parallel and perpendicular

to the shaft axis are subjected to shear stresses only (Element a). Normal stresses, shearing stresses or a combination of both may be found for other orientations (Element b).

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

max 0max

45

0max0max

2 2

245cos2

o  







A A

A F

AAF

0 2AA

• Element a is in pure shear. • Element c is subjected to a tensile stress on

• Note that all stresses for elements a and c have

j two faces and compressive stress on the other two.

the same magnitude.

• MECHANICS OF MATERIALS

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Torsional Failure Modes

• Ductile materials generally fail in shear. Brittle materials are weaker inshear. 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 planesspecimen breaks along planes perpendicular to the direction in which tension is a maximum, i.e., along surfaces at 45o to the shaft

g axis.

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• MECHANICS OF MATERIALS

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• MECHANICS OF MATERIALS

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Sample Problem 3.1

SOLUTION:

• Cut sections through shafts AB d BC d f iand BC and perform static

Shaft BC is hollow with inner and outer

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

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,

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

determine (a) the minimum and maximum shearing stress in shaft BC, (b) the required diameter d of shafts AB and CD

required diameter.

if the allowable shearing stress in these shafts is 65 MPa.

• MECHANICS OF MATERIALS

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

 

CDAB

ABx

TT

TM



 mkN6

mkN60     mkN20

mkN14mkN60





BC

BCx

T

TM

• MECHANICS OF MATERIALS

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A l l ti t i f l t Gi ll bl h i t d Sample Problem 3.1 • Apply elastic torsion formulas to

find minimum and maximum stress on shaft BC.

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

      444142 045.0060.022   ccJ mkN665 34max

  MPaTc

J Tc

 

46 m1092.13

22 

   109213

m060.0mkN20 46

2 2max

  J

cTBC m109.38 3

3 2

4 2

c

ccJ 

mm8772  cd

MPa2.86 m1092.13 46

 J

mm45min1min   c

mm8.772  cd

dAB = dBC

MPa7.64

mm60MPa2.86

min

2max



 c

MPa7.64

MPa2.86

min

max

• MECHANICS OF MATERIALS

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290 N-m

Fixed end

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2.8 kN-m

Free end (T = 0)(T = 0)

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Why ? Show on next page !!!CCD AB B

rT T r

 

• MECHANICS OF MATERIALS

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Gear CGear C Gear BGear B

rc=240 mm F

FTCD rB=80 mm

BC T=TAB= 900 N-m

FCD

TCD = (F)(rc) TAB = 900 = (F)(rB)

CD ABT TF   C B

C CD AB

r r rT T r

Br

• MECHANICS OF MATERIALS

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• MECHANICS OF MATERIALS

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Angle of Twist in Elastic Range • Recall that the angle of twist and maximum

shearing strain are related,

L c max and max J

T J

Tc   L

max

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

T 

max JJ

max max

Tc c G JG L     

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

JG TL



The relation shows that, within the elastic range, the angle of twist  is proportional to the torque T applied to the shaft. This is accordance with the experimental evidence. The

previous eqs also provides us with a convenient method for determining the modulus of rigidity of a given materials.

• MECHANICS OF MATERIALS

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Angle of Twist in Elastic Range

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

 i ii

ii GJ LT

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(A to B)

(A to C)

See from A to C

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A/B B/C

A to B B to C

B/C

A/B

A to C

A to C

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Sample Problem 3.4

SOLUTION:

• Apply a static equilibrium analysis on th t h ft t fi d l ti hithe two shafts to find a relationship between TCD and T0 .

• Apply a kinematic analysis to relate

Two solid steel shafts are connected • Find the maximum allowable torque

on each shaft – choose the smallest.

the angular rotations of the gears.

Two solid steel shafts are connected by gears. Knowing that for each shaft G = 77 GPa and that the allowable shearing stress is 55 MPa, determine

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

o eac s a t c oose t e s a est.

g (a) the largest torque T0 that may be applied to the end of shaft AB, (b) the corresponding angle through which

rotation of end A.

end A of shaft AB rotates.

• MECHANICS OF MATERIALS

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Sample Problem 3.4 SOLUTION:

• Apply a static equilibrium analysis on the two shafts to find a relationship

• Apply a kinematic analysis to relate the angular rotations of the gears.the two shafts to find a relationship

between TCD and T0 . the angular rotations of the gears.

60

C CD AB

B

rT T r

T T

22 2.73

CD O

CD O

T T

T T

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Fi d th T f th i Fi d th di l f t i t f h

Sample Problem 3.4 • Find the T0 for the maximum

allowable torque on each shaft – choose the smallest.

• Find the corresponding angle of twist for each shaft and the net angular rotation of end A.

D = 0 (Fix)

8.05°

10.2°

61.8

inlb5610 T

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D = 0 (Fix)

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Design of Transmission Shafts

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

fTTP 2

f PPT

fTTP





2

2





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

exceed the maximum allowable shearing stress,

• Designer must select shaft material and cross-section to meet performance specifications without exceeding allowable g

 h ftlid3

max

TJ J

Tc 

without exceeding allowable shearing stress.

Angular velocity 2 f 

1 Hz(rpm) Hz 60 rpm

f  

 

   shafts hollow 2

shaftssolid 2

max

4 1

4 2

22

max

3

 

Tcc cc

J

c c



Angular velocity, 2 f 

p 2 max22 cc

• MECHANICS OF MATERIALS

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Watts = N-m/s

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Statically Indeterminate Shafts

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

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

• Substitute into the original equilibrium equation,

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Stress Concentrations

• The derivation of the torsion formula,

J Tc

max

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

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

• Experimental or numerically determined t ti f t li d

keyways, and cross section discontinuities can cause stress concentrations.

J TcKmax

concentration factors are applied as

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• MECHANICS OF MATERIALS

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Beer • Johnston • DeWolf

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Plastic Deformations • With the assumption of a linearly elastic material,

J Tc

max

• If the yield strength is exceeded or the material has a nonlinear shearing-stress-strain curve, this expression does not hold.

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

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

ti

curve allows determination of stress distribution.

   cc

ddT 0

2

0 22 

section,

• MECHANICS OF MATERIALS

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Elastoplastic Materials • At the maximum elastic torque,

YYY cc JT  32

1 c

L Y Y

 

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

Y 

  

 YL

  

   

 

 

   

  3

3

4 1

3 4

3

3

4 13

3 2 11

c T

c cT YYYY



  YLY 

 cc

  

   

  3

3

4 1

3 4 1

 Y

YTT

• As , the torque approaches a limiting value,0Y torque plastic TT YP  3

4

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Residual Stresses

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

Wh th t i d th d ti f t• 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.

• On a T- curve, the shaft unloads along a straight line to an angle greater than zero.

• Residual stress found from principle of superposition.p p p p

  0 dA J

Tc m 

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Example 3.08/3.09

SOLUTION:

• Solve Eq. (3.32) for Y/c and evaluate the elastic core radius.

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

A solid circular shaft is subjected to a torque at each end. Assuming that the shaft is made of an l l i i l i h

mkN6.4 T • Evaluate Eq. (3.16) for the angle

which the shaft untwists when the torque is removed. The permanent t i t i th diff b t thelastoplastic material with

and determine (a) the radius of the elastic core, (b) the angle of twist of the shaft When the

MPa150Y GPa77G

• Find the residual stress distribution by

twist is the difference between the angles of twist and untwist.

angle of twist of the shaft. When the torque is removed, determine (c) the permanent twist, (d) the distribution of residual stresses

a superposition of the stress due to twisting and untwisting the shaft.

of residual stresses.

• MECHANICS OF MATERIALS

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SOLUTION Example 3.08/3.09 SOLUTION: • Solve Eq. (3.32) for Y/c and

evaluate the elastic core radius 13 

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

  YY31

341 3 3

4 1

3 4

 

  

 

 

   

 

Y

YY Y T

T cc

TT 

 m1025 321421  cJ    

  49- 3

Pa1077m10614 m2.1mN1068.3

 





 

Y Y

YY

JG LT

cc

  m10614

m1025

49

22



 

Y Y

Y Y

JTcT

cJ



   

o3 3

3

Pa1077m10614

 



 

Y

   m1025

m10614Pa10150 3

496

 



Y

YY

T

c T

J  630.0

o50.8

mkN68.3 

630.0 683 6.434

3 1

  

   Y

68.3 c

mm8.15Y

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Example 3.08/3.09

• Evaluate Eq. (3.16) for the angle which the shaft untwists when the torque is removed The

• Find the residual stress distribution by a superposition of the stress due to twisting and untwisting the shaftthe torque is removed. The

permanent twist is the difference between the angles of twist and untwist.

twisting and untwisting the shaft

   m10614

m1025mN106.4 49-

33 max

 

J Tc

  3 21N1064 

JG TL

MPa3.187

     

3

949

3

Pa1077m1014.6 m2.1mN106.4



 

o1 81

69.650.8



 pφ

1.81 o81.1p

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Torsion of Noncircular Members • Previous torsion formulas are valid for

axisymmetric or circular shafts.

Pl ti f i l• Planar cross-sections of noncircular shafts do not remain planar and stress and strain distribution do not vary linearly

TLT  

• For uniform rectangular cross-sections,

linearly.

• At large values of a/b, the maximum

Gabcabc 32 2

1 max  

shear stress and angle of twist for other open sections are the same as a rectangular bar.

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Thin-Walled Hollow Shafts • Summing forces in the x-direction on AB,

    flowshear

0



 qttt

xtxtF

BBAA

BBAAx





shear stress varies inversely with thickness

qBBAA

    dAqpdsqdstpdFpdM 20  

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

tA T

2

220

 

 tA2

 t ds

GA TL

24 

• Angle of twist (from Chapter 11)

tGA4

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Example 3.10 Extruded aluminum tubing with a rectangular cross-section has a torque loading of 2.7 kN . m. Determine the shearing stress in each of the four walls with (a) uniform wall thickness of 4 mm and wall thicknesses of (b) 3 mm on AB and CD and 5 mm on CD and BD.

SOLUTION:

D t i th h fl th h th• Determine the shear flow through the tubing walls.

• Find the corresponding shearing stress p g g with each wall thickness .

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Example 3.10 SOLUTION:

• Determine the shear flow through the tubing walls

• Find the corresponding shearing stress with each wall thickness.

tubing walls.

With a uniform wall thickness,

With a variable wall thicknessWith a variable wall thickness