Download - Fourth Edition MECHANICS OF MATERIALS · PDF fileMECHANICS OF Fourth Edition MECHANICS OF MATERIALS CHAPTER Ferdinand P. Beer E. Russell Johnston, Jr. John T. DeWolf Torsion Lecture

MECHANICS OFFourth Edition

MECHANICS OF MATERIALS

CHAPTER

MATERIALSFerdinand P. BeerE. Russell Johnston, Jr.John T. DeWolf

Torsion

Lecture Notes:J. Walt OlerTexas Tech Universityy

© 2006 The McGraw-Hill Companies, Inc. All rights reserved.


FourthEdition

Beer • Johnston • DeWolf

Contents

Introduction

Torsional Loads on Circular Shafts

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

© 2006 The McGraw-Hill Companies, Inc. All rights reserved. 3 - 2


FourthEdition


Torsional Loads on Circular Shafts

• 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’.



FourthEdition


Net Torque Due to Internal Stresses

• Net of the internal shearing stresses is an internal torque, equal and opposite to the

dAdFT

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


loads cannot be assumed uniform.


FourthEdition


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.



FourthEdition


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.



FourthEdition


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

tan in radianxy xy

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)


cLtan AA

L


FourthEdition


Shearing Strain

c

At shaft surface, Max shearing strain

maxcL

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

Shear strain inside the shaft, LL

maxmaxShear strain at the shaft's surface, Lc

L cLL

maxmax LL

c c


maxc


FourthEdition


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

max Gc

G c

From Hooke’s Law, G , so

The shearing stress varies linearly421 cJ max

c

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

F h ll li d

max max2

From or c c

• For hollow cylinder

41

422

1 ccJ

2

11 min max

2

Substitute ccc


J = Polar moment of inertiamin 1

max 2

cc


FourthEdition


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

Jc

dAc

dAT max2max

the torque on the shaft at the section,

Th lt k th l ti t i

dAJ 2maxc

• The results are known as the elastic torsion formulas,

andmaxTTc

dAdFT

and max JJ



FourthEdition


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,

max0max

45

0max0max

22

245cos2

o

AA

AF

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

jtwo faces and compressive stress on the other two.


the same magnitude.


FourthEdition


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


gaxis.


FourthEdition




FourthEdition


Sample Problem 3.1

SOLUTION:

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

equilibrium analyses to find torque loadings.

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.


FourthEdition


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

and perform static equilibrium analysis to find torque loadingsto find torque loadings.

CDAB

ABx

TT

TM

mkN6

mkN60 mkN20

mkN14mkN60

BC

BCx

T

TM



FourthEdition


A l l ti t i f l t Gi ll bl h i t dSample 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.

4441

42 045.0060.0

22

ccJ mkN665 34max

MPaTcJ

Tc

46 m1092.13

22

109213

m060.0mkN2046

22max

J

cTBCm109.38 3

32

42

c

ccJ

mm8772 cd

MPa2.86m1092.13 46

J

mm45min1min c

mm8.772 cd

dAB = dBC


MPa7.64

mm60MPa2.86

min

2max

c

MPa7.64

MPa2.86

min

max


FourthEdition


290 N-m

Fixed end



FourthEdition


2.8 kN-m

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



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FourthEdition



Why ? Show on next page !!!CCD AB

B

rT Tr


FourthEdition


Gear CGear C Gear BGear B

rc=240 mm FFTCD

rB=80 mm

BCT=TAB= 900 N-m

FCD

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

CD ABT TF C B

CCD AB

r rrT Tr


Br


FourthEdition




FourthEdition


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

shearing strain are related,

Lc max and max J

TJ

Tc L

max

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

T

max JJ

maxmax

Tc cG JG L

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

JGTL

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.


FourthEdition


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

iiGJLT



FourthEdition




FourthEdition


(A to B)

(A to C)


See from A to C


FourthEdition


A/BB/C

A to B B to C

B/C

A/B

A to C


A to C


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FourthEdition


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.


FourthEdition


Sample Problem 3.4SOLUTION:

• 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

CCD AB

B

rT Tr

T T


222.73

CD O

CD O

T T

T T


FourthEdition


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


FourthEdition


D = 0 (Fix)



FourthEdition




FourthEdition


Design of Transmission Shafts

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

fTTP 2

fPPT

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

TJJ

Tc

without exceeding allowable shearing stress.

Angular velocity 2 f

1 Hz(rpm) Hz60 rpm

f

shafts hollow2

shaftssolid2

max

41

42

22

max

3

Tcccc

J

cc

Angular velocity, 2 f


p 2 max22 cc


FourthEdition


Watts = N-m/s



FourthEdition


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,



FourthEdition


Stress Concentrations

• The derivation of the torsion formula,

JTc

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.

JTcKmax

concentration factors are applied as



FourthEdition




FourthEdition


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

JTc

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

ddT0

2

022

section,



FourthEdition


Elastoplastic Materials• At the maximum elastic torque,

YYY ccJT 3

21

cL Y

Y

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

Y

YL

3

3

41

34

3

3

413

32 11

cT

ccT Y

YY

Y

YL

Y

cc

3

3

41

34 1

Y

YTT

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

torque plastic TT YP 34



FourthEdition


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

JTc

m


FourthEdition


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

MPa150YGPa77G

• 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.


FourthEdition


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

evaluate the elastic core radius13

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

YY

31

341 3

3

41

34

Y

YYY T

Tcc

TT

m1025 3214

21 cJ

49-

3

Pa1077m10614m2.1mN1068.3

YY

YY

JGLT

cc

m10614

m1025

49

22

YY

YY

JTcT

cJ

o33

3

8.50rad103.148rad104.93

rad104.93

Pa1077m10614

Y

m1025

m10614Pa101503

496

Y

YY

T

cT

J 630.0

o50.8

mkN68.3

630.06836.434

31

Y


68.3 c

mm8.15Y


FourthEdition


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

m1025mN106.449-

33max

JTc

3 21N1064

JGTL

MPa3.187

3

949

3

6.69rad108.116

Pa1077m1014.6m2.1mN106.4

o1 81

69.650.8

pφ


1.81o81.1p


FourthEdition


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

21

max

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



FourthEdition


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

tAT

qAdAqdMT

2

220

tA2

t

dsGA

TL24

• Angle of twist (from Chapter 11)


tGA4


FourthEdition


Example 3.10Extruded 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 gwith each wall thickness .



FourthEdition


Example 3.10SOLUTION:

• 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


Download - Fourth Edition MECHANICS OF MATERIALS · PDF fileMECHANICS OF Fourth Edition MECHANICS OF MATERIALS CHAPTER Ferdinand P. Beer E. Russell Johnston, Jr. John T. DeWolf Torsion Lecture

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