3_torsion_MKM.pdf

MECHANICS OFMECHANICS OF MATERIALS

CHAPTER

MATERIALS

Torsion

Gunawan

MECHANICS OF MATERIALSContents

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

Example 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

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MECHANICS OF MATERIALSTorsional 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’

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MECHANICS OF MATERIALSNet Torque Due to Internal Stresses

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

( )∫ ∫== dAdFT τρρ

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

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loads can not be assumed uniform.

MECHANICS OF MATERIALSAxial 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 is

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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MECHANICS OF MATERIALSShaft Deformations

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

L

T∝

φ

φ

L∝φ

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

di t t dundistorted.

• Cross-sections for hollow and solid circular shafts remain plain and undistorted because a

• Cross-sections of noncircular (non-axisymmetric) shafts are distorted when

circular shaft is axisymmetric.

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subjected to torsion.

MECHANICS OF MATERIALSShearing 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 binterior cylinder deforms into a rhombus.

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

ρφ• It follows that

the shear strain is equal to angle of twist.

• Shear strain is proportional to twist and radius

LL ρφγρφγ == or

maxmax and γργφγcL

c==

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MECHANICS OF MATERIALSStresses in Elastic Range

• Multiplying the previous equation by the shear modulus,

maxγργ Gc

G =c

maxτρτc

=

From Hooke’s Law, γτ G= , so

• Recall that the sum of the moments from

421 cJ π= The shearing stress varies linearly with the

radial position in the section.

JdAdAT max2max ττ∫∫

• 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 ρρτ ∫ =∫ ==

( )41

42

1 ccJ −= π

• The results are known as the elastic torsion formulas,

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( )122 ccJ −= π and max J

TJ

Tc ρττ ==

MECHANICS OF MATERIALSNormal 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.

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

max0max

45

0max0max

22

245cos2

o ττσ

ττ

===

==

AA

AF

AAF

045 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.

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the same magnitude

MECHANICS OF MATERIALSTorsional 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

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gaxis.

MECHANICS OF MATERIALSSample Problem 3.1

SOLUTION:

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

equilibrium analysis 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

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

MECHANICS OF MATERIALSSOLUTION:Sample Problem 3.1• Cut sections through shafts AB and BC

and perform static equilibrium analysis to find torque loadings

( )

CDAB

ABx

TT

TM

=⋅=

−⋅==∑mkN6

mkN60

( ) ( )kN14kN60∑ TM

Gunawan 3 - 12

( ) ( )mkN20

mkN14mkN60

⋅=

−⋅+⋅==∑

BC

BCx

T

TM

MECHANICS OF MATERIALS

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−=−=

ππ ccJ mkN665 34max⋅

=== MPaTcJ

Tcππ

τ

46 m1092.13

22−×=

( )( )109213

m060.0mkN2046

22max

⋅=== −J

cTBCττm109.38 3

32

42

−×=c

ccJ ππ

mm8772 == cd

MPa2.86m1092.13 46

=

× −J

mm45min1min ==ττ c

mm8.772 == cd

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MPa7.64

mm60MPa2.86

min

2max

=τ

τ c

MPa7.64

MPa2.86

min

max

=

=

τ

τ

MECHANICS OF MATERIALSAngle of Twist in Elastic Range

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

Lcφγ =max L

γmax

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

TcτJGTc

G== max

maxτγ

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

JGTL

=φ

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

∑= iiGJLTφ

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i iiGJφ

MECHANICS OF MATERIALSStatically 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,ftlb90 ⋅=+ BA TT

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

• Divide the shaft into two components which

ABBA T

JLJLT

GJLT

GJLT

12

21

2

2

1

121 0 ==−=+= φφφ

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

ftlb9021 ⋅=+ AA TJLJLT

• Substitute into the original equilibrium equation,

JLGJGJ 1221

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12JL

MECHANICS OF MATERIALSSample 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 = 11.2 x 106 psi and that the allowable shearing stress is 8 ksi,

• 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

gdetermine (a) the largest torque T0that may be applied to the end of shaft AB, (b) the corresponding angle

rotation of end A

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through which end A of shaft ABrotates.

MECHANICS OF MATERIALSSample 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 gearsthe two shafts to find a relationship

between TCD and T0

the angular rotations of the gears

rr φφ =( )( )

0

in.45.20

in.875.00

TFM

TFM

CDC

B

−==

−==

∑

∑CC

B

CB

CCBB

rr

rr

φφφ

φφ

in.875.0in.45.2

==

=

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08.2 TTCD = CB φφ 8.2=


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

( )( )( ) ( )64

2/

psi102.11in.375.0.24in.lb561

×

⋅==

AB

ABBA

inGJLTφ

π

( )( )in.375.0

in.375.08000 42

0max ==

TpsiJ

cT

AB

ABπ

τ ( )( )( ) ( )64/

o

psi10211in50.24in.lb5618.2

2.22rad387.0

×

⋅==

==

CD

CDDC

inGJLTφ

π

( )( )i50

in.5.08.28000

in.lb663

40

max

0

2

==

⋅=

TpsiJ

cT

T

CDπ

τ

( ) ( )

( ) oo

o

2

2689528282

95.2rad514.0

psi102.11in.5.0

===

==

×

CB

CDGJ

φφ

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( )

in.lb561

in.5.0

0

42

⋅=T

JCD π

inlb5610 ⋅=T

( )oo

/ 2.2226.8

26.895.28.28.2

+=+=

===

BABA

CB

φφφ

φφ

o48.10=Aφ

MECHANICS OF MATERIALSDesign of Transmission Shafts

• Principal transmission shaft performance specifications are:

po er

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

fTTP 2- power- speed

fPPT

fTTP

πω

πω

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

π

τ

TJJ

Tc=

without exceeding allowable shearing stress.

( )

( ) ( )shafts hollow2

shaftssolid2

max

41

42

22

max

3

τπ

τ

Tcccc

J

cc

=−=

==

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2 max22 τcc

MECHANICS OF MATERIALSStress Concentrations

• The derivation of the torsion formula,

JTc

=maxτ

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

JTcK=maxτ

concentration factors are applied as

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MECHANICS OF MATERIALSPlastic Deformations

• With the assumption of a linearly elastic material,

JTc

=maxτ

• 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 ρτρπρπρρτ

section,

Gunawan 3 - 21

MECHANICS OF MATERIALSElastoplastic Materials

• At the maximum elastic torque,

YYY ccJT τπτ 3

21==

cL Y

Yγφ =

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

Yτ

ρρτ =

γ YL

⎟⎟⎠

⎞⎜⎜⎝

⎛−=⎟

⎟⎠

⎞⎜⎜⎝

⎛−= 3

3

41

34

3

3

413

32 11

cT

ccT Y

YY

Yρρτπ

φγρ YL

Y =

⎠⎝⎠⎝ cc

⎟⎟⎠

⎞⎜⎜⎝

⎛−= 3

3

41

34 1

φφY

YTT

• As , the torque approaches a limiting value,0→Yρ

torque plastic TT YP == 34

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MECHANICS OF MATERIALSResidual 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-φ curve, the shaft unloads along a straight line to an angle greater than zero

• Residual stresses found from principle of superpositionp p p p

Gunawan 3 - 23

( ) 0=∫ dAτρ

JTc

m =′τ

MECHANICS OF MATERIALSExample 3.08/3.09

SOLUTION:

• Solve Eq. (3.32) for ρY/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=YτGPa77=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

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of residual stresses.

MECHANICS OF MATERIALSSOLUTION:Example 3.08/3.09• Solve Eq. (3.32) for ρY/c and

evaluate the elastic core radius3

1314 ⎟

⎞⎜⎛⎟

⎞⎜⎛ YY Tρρ

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

=⇒=φφρφ YY

341 341

34

⎟⎟⎠

⎞⎜⎜⎝

⎛−=⇒⎟

⎟⎠

⎞⎜⎜⎝

⎛−=

Y

YYY T

Tcc

TT ρρ

( )m1025 3214

21 ×== −cJ ππ

( )( )( )( )49-

3

Pa1077m10614m2.1N1068.3

××

×==φ

ρφ

φ

YY

YY

JGLT

cc

m10614 49

=⇒=

×= −

YY

YY c

JTJ

cT ττ

( )( )

o33

3

8 50rad103148rad104.93

rad104.93

=×=×

=

×=

−−

−

φ

φY

( )( )m1025

m10614Pa101503

496

×

××= −

−

YT

cJ 8.50rad103.148630.0

=×==φ

o50.8=φ

mkN68.3 ⋅=

630.068.36.434

31

=⎟⎠⎞

⎜⎝⎛ −=

cYρ

Gunawan 3 - 25

mm8.15=Yρ

MECHANICS OF MATERIALSExample 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

×

×⋅×==′

−

JTcτ

( )( )3 m21mN1064 ⋅×

=′φJGTL MPa3.187=

( )( )( )( )

3

949

rad108.116

Pa1077m1014.6m2.1mN106.4

×=

××

×=

−

( )33 rad108.116108.116 ×−×=

′−=

−−

φφpφ

Gunawan 3 - 26

o1.81=o81.1=pφ

MECHANICS OF MATERIALSTorsion 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== φτ

• For uniform rectangular cross-sections,

linearly

• At large values of a/b, the maximum

Gabcabc 32

21

max == φτ

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

Gunawan 3 - 27

MECHANICS OF MATERIALSThin-Walled Hollow Shafts

• Summing forces in the x-direction on AB,( ) ( )

flowshear

0

====

Δ−Δ==∑qttt

xtxtF

BBAA

BBAAx

τττ

ττ

shear stress varies inversely with thickness

qBBAA

( ) ( ) dAqpdsqdstpdFpdM 20 ==== τ

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

tAT

qAdAqdMT

2

220

=

=== ∫∫

τtA2

∫=t

dsGA

TL24

φ

• Angle of twist (from Chapt 11)

Gunawan 3 - 28

tGA4

MECHANICS OF MATERIALSExample 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) 0.120 in. on AB and CD and 0.200 in. on CD and BDBD.

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

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MECHANICS OF MATERIALSExample 3.10SOLUTION:

• Determine the shear flow through the tubing walls

• Find the corresponding shearing stress with each wall thickness

tubing wallswith a uniform wall thickness,

in1600in.kip335.1

==tqτ

in.160.0tksi34.8=τ

ith i bl ll thi k

( )( ) in.986.8in.34.2in.84.3 2==A

with a variable wall thickness

in.120.0in.kip335.1

== ACAB ττ

( ) in.kip335.1

in.986.82in.-kip24

2 2 ===A

Tq

in.200.0in.kip335.1

== CDBD ττ

ksi13.11== BCAB ττ

Gunawan 3 - 30

ksi68.6== CDBC ττ


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3_torsion_MKM.pdf

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