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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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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
© 2002 The McGraw-Hill Companies, Inc. All rights reserved. 3 - 3
• 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
© 2002 The McGraw-Hill Companies, Inc. All rights reserved. 3 - 4
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
© 2002 The McGraw-Hill Companies, Inc. All rights reserved. 3 - 5
• 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
• -
© 2002 The McGraw-Hill Companies, Inc. All rights reserved. 3 - 6
,
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.
© 2002 The McGraw-Hill Companies, Inc. All rights reserved. 3 - 7
• 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
© 2002 The McGraw-Hill Companies, Inc. All rights reserved. 3 - 8
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,
© 2002 The McGraw-Hill Companies, Inc. All rights reserved. 3 - 9
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
© 2002 The McGraw-Hill Companies, Inc. All rights reserved. 3 - 10
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
© 2002 The McGraw-Hill Companies, Inc. All rights reserved. 3 - 11
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
© 2002 The McGraw-Hill Companies, Inc. All rights reserved. 3 - 12
( )
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
© 2002 The McGraw-Hill Companies, Inc. All rights reserved. 3 - 13
( ) ( ) ( )[ ]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
© 2002 The McGraw-Hill Companies, Inc. All rights reserved. 3 - 14
• 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
© 2002 The McGraw-Hill Companies, Inc. All rights reserved. 3 - 15
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
© 2002 The McGraw-Hill Companies, Inc. All rights reserved. 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 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
© 2002 The McGraw-Hill Companies, Inc. All rights reserved. 3 - 17
( )
( )
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
© 2002 The McGraw-Hill Companies, Inc. All rights reserved. 3 - 18
( )
( )
( )
( )
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
© 2002 The McGraw-Hill Companies, Inc. All rights reserved. 3 - 19
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
© 2002 The McGraw-Hill Companies, Inc. All rights reserved. 3 - 20
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.
© 2002 The McGraw-Hill Companies, Inc. All rights reserved. 3 - 21
( ) ∫=∫=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 =
© 2002 The McGraw-Hill Companies, Inc. All rights reserved. 3 - 22
• 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
© 2002 The McGraw-Hill Companies, Inc. All rights reserved. 3 - 23
• Residual stresses found from principle of superposition
( ) 0=∫ dAτ
J
Tc
m =′
τ
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T
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
© 2002 The McGraw-Hill Companies, Inc. All rights reserved. 3 - 25
( )( )
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
© 2002 The McGraw-Hill Companies, Inc. All rights reserved. 3 - 26
( )( )( )( )
( )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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T h
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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
© 2002 The McGraw-Hill Companies, Inc. All rights reserved. 3 - 28
( ) ( )
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.
© 2002 The McGraw-Hill Companies, Inc. All rights reserved. 3 - 29
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
© 2002 The McGraw-Hill Companies, Inc. All rights reserved. 3 - 30
( )( )
( ) 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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