Seventh Edition
CHAPTER
3
MECHANICS OF
MATERIALS
Ferdinand P. Beer
E. Russell Johnston, Jr.
John T. DeWolf
David F. Mazurek
Torsion
Lecture Notes:
Brock E. Barry
U.S. Military Academy
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Contents
Torsional Loads on Circular Shafts
Statically Indeterminate Shafts
Net Torque Due to Internal Stresses
Sample Problem 3.4
Axial Shear Components
Design of Transmission Shafts
Shaft Deformations
Stress Concentrations
Shearing Strain
Plastic Deformations
Stresses in Elastic Range
Elastoplastic Materials
Normal Stresses
Residual Stresses
Torsional Failure Modes
Concept Application 3.8/3.9
Sample Problem 3.1
Torsion of Noncircular Members
Angle of Twist in Elastic Range
Thin-Walled Hollow Shafts
Concept Application 3.10
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Torsional Loads on Circular Shafts
• Stresses and strains in members of
circular cross-section are subjected
to twisting couples or torques
• Turbine exerts torque T on the shaft
• Shaft transmits the torque to the
generator
• Generator creates an equal and
opposite torque T’
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
Fig. 3.3 Shaft subject to torques and a
section plane at C.
• Net of the internal shearing stresses is an
internal torque, equal and opposite to the
applied torque,
T    dF     dA
• 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.
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.
• Unlike the normal stress due to axial loads, the
distribution of shearing stresses due to torsional
loads cannot be assumed uniform.
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Axial Shear Components
• Torque applied to shaft produces shearing
stresses on the faces perpendicular to the
axis.
Fig. 3.5 Small element in shaft showing how
shear stress components act.
• Conditions of equilibrium require the
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 slats pinned at both ends to disks.
• The slats slide with respect to each other
when equal and opposite torques are applied
to the ends of the shaft.
Fig. 3.6 Model of shearing in shaft
(a) undeformed; (b) loaded and deformed.
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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 length.
 T
L
Fig. 3.7 Shaft with fixed support and line
AB drawn showing deformation under
torsion loading: (a) unloaded; (b)
loaded.
• 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.
Fig. 3.8 Comparison of deformations in
circular (a) and square (b) shafts.
• Cross-sections of noncircular (nonaxisymmetric) 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
interior cylinder deforms into a rhombus.
• Since the ends of the element remain planar,
the shear strain is equal to angle of twist.
• It follows that
L   or  

L
• Shear strain is proportional to twist and radius
 max 
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, .
c

and    max
L
c
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Stresses in Elastic Range
• Multiplying the previous equation by the
shear modulus,
G 

c
G max
From Hooke’s Law,   G , so

Fig. 3.14 Distribution of shearing stresses
in a torqued shaft; (a) Solid shaft, (b)
hollow shaft.

c
 max
The shearing stress varies linearly with the
distance  from the axis of the shaft.
• 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:


T    dA  max   2 dA  max J
c
c
• The results are known as the elastic torsion
formulas,
 max 
Tc
T
and  
J
J
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Normal Stresses
Fig. 3.17 Circular shaft with stress elements at
different orientations.
• 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,
F  2 max A0  cos 45   max A0 2
 45o 
Fig. 3.18 Forces on faces at 45° to shaft axis.
Fig. 3.19 Shaft elements with only shear
stresses or normal stresses.
F  max A0 2

  max
A
A0 2
• Element a is in pure shear.
• Element c is subjected to a tensile stress on
two faces and compressive stress on the other
two.
• Note that all stresses for elements a and c have
the same magnitude.
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Torsional Failure Modes
Photo 3.2 Shear failure of shaft subject to torque.
• 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.
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Sample Problem 3.1
SOLUTION:
• Cut sections through shafts AB
and BC and perform static
equilibrium analyses to find
torque loadings.
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.
• 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.
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Sample Problem 3.1
SOLUTION:
• Cut sections through shafts AB and BC
and perform static equilibrium analysis
to find torque loadings.
Fig. 1 Free-body diagram for section between A and B.
Fig. 2 Free-body diagram for section between B and C.
 M x  0  6 kN  m   TAB
 M x  0  6 kN  m   14 kN  m   TBC
TAB  6 kN  m  TCD
TBC  20 kN  m
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Sample Problem 3.1
• Apply elastic torsion formulas to
find minimum and maximum
stress on shaft BC.
Fig. 3 Shearing stress distribution on cross section.
J 


c24  c14   0.060 4  0.045 4 
2
2
• Given allowable shearing stress and
applied torque, invert the elastic torsion
formula to find the required diameter.
Fig. 4 Free-body diagram of shaft portion AB.

 13 .92  10  6 m 4
 max   2 
TBC c2 20 kN  m 0.060 m 

J
13 .92 10  6 m 4
 max 
Tc
Tc

J  c4
2
 min
86.2 MPa
 min  64.7 MPa

45 mm
60 mm
6 kN  m
 c3
2
c  38.9 10 3 m
 86 .2 MPa
 min c1

 max c2
65 MPa 
d  2c  77.8 mm
 max  86.2 MPa
 min  64.7 MPa
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Angle of Twist in Elastic Range
• Recall that the angle of twist and maximum
shearing strain are related,
 max 
c
L
• In the elastic range, the shearing strain and shear
are related by Hooke’s Law,
Fig. 3.20 Torque applied to fixed end shaft
resulting angle of twist .
 max 
 max
G

Tc
JG
• Equating the expressions for shearing strain and
solving for the angle of twist,

TL
JG
• If the torsional loading or shaft cross-section
changes along the length, the angle of rotation is
found as the sum of segment rotations
Fig. 3.21 Shaft with multiple cross-section
dimensions and multiple loads.
Ti Li
i J i Gi
 
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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.
• From a free-body analysis of the shaft,
TA  TB  90 lb  ft
which is not sufficient to find the end torques.
The problem is statically indeterminate.
• Divide the shaft into two components which
must have compatible deformations,
  1  2 
TAL1 TB L2

0
J1G J 2G
LJ
TB  1 2 TA
L2 J1
• Substitute into the original equilibrium equation,
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.
LJ
TA  1 2 TA  90 lb  ft
L2 J1
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Sample Problem 3.4
SOLUTION:
• Apply a static equilibrium analysis on
the two shafts to find a relationship
between TCD and T0 .
• Apply a kinematic analysis to relate
the angular rotations of the gears.
• Find the maximum allowable torque
on each shaft – choose the smallest.
Two solid steel shafts are connected
by gears. Knowing that for each shaft
• Find the corresponding angle of twist
G = 11.2 x 106 psi and that the
for each shaft and the net angular
allowable shearing stress is 8 ksi,
rotation of end A.
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.
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Sample Problem 3.4
SOLUTION:
• Apply a static equilibrium analysis on
the two shafts to find a relationship
between TCD and T0 .
• Apply a kinematic analysis to relate
the angular rotations of the gears.
Fig. 2 Angles of twist for gears B and C.
Fig. 1 Free-body diagrams of gears B and C.
 M B  0  F 0.875 in.   T0
rB B  rCC
rC
2.45 in.
C 
C
rB
0.875 in.
 M C  0  F 2.45 in.   TCD
B 
TCD  2.8 T0
 B  2. 8  C
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Sample Problem 3.4
• Find the T0 for the maximum
• Find the corresponding angle of twist for each
allowable torque on each shaft –
shaft and the net angular rotation of end A.
choose the smallest.
Fig. 5
Fig. 3 Free-body
diagram of shaft
AB.
 max 
Fig. 4 Free-body
diagram of shaft
CD.
T 0.375 in. 
TAB c
8000 psi  0
 0.375 in. 4
J AB
2
T0  663 lb  in.
 max 
TCD c
2.8 T0 0.5 in. 
8000 psi 
 0.5 in. 4
J CD
2
T0  561 lb  in.
T0  561 lb  in
A/ B 
C / D
561lb  in. 24 in .
TAB L

J AB G  0.375 in. 4 11.2  106 psi
2

 0.387 rad  2.22o
T L
2.8 561lb  in. 24 in .
 CD 
J CD G  0.5 in. 4 11.2  106 psi

2
 0.514 rad  2.95o




 B  2.8C  2.8 2.95o  8.26o
 A   B   A / B  8.26o  2.22o
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 A  10.48o
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Design of Transmission Shafts
• Principal transmission shaft
performance specifications are:
- power
- Speed of rotation
• Designer must select shaft
material and dimensions of the
cross-section to meet
performance specifications
without exceeding allowable
shearing stress.
• Determine torque applied to shaft at
specified power and speed,
P  T  2fT
T
P


P
2f
• Find shaft cross-section which will not
exceed the maximum allowable
shearing stress,
 max 
Tc
J
J  3
T
 c 
c 2
 max

solid shafts 

J
 4 4
T

c2  c1 
c2 2c2
 max
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hollow
shafts 
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Stress Concentrations
• The derivation of the torsion formula,
Fig. 3.26 Coupling of
shafts using (a) bolted
flange, (b) slot for
keyway.
 max 
Tc
J
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
can cause stress concentrations
• Experimental or numerically determined
concentration factors are applied as
 max  K
Tc
J
Fig. 3.28 Plot of stress concentration factors
for fillets in circular shafts.
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Plastic Deformations
• With the assumption of a linearly elastic material,
 max 
Fig. 3.29 Distribution of shearing
strain for torsion of a circular
shaft.
Fig. 3.30 Nonlinear, shear stressstrain diagram.
Tc
J
• 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.
• 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
section,
c
c
0
0
T    2 d   2   2 d
Fig. 3.31 Shearing strain distribution for
shaft with nonlinear stress-strain response.
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Elastoplastic Materials
• At the maximum elastic torque,
TY 
J
 Y  12 c3 Y
c
Y 
L Y
c
• As the torque is increased, a plastic region

(   Y ) develops around an elastic core (    Y )
Y 
Y
L Y


Y3 
T
2 c 3 1  1
Y
3
4
T
3

4 T 1  1 Y 
3 Y
4 3


c 
3


4 T 1  1
3 Y
4

Y3 
c3 
 
• As Y  0, the torque approaches a limiting value,
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.
TP  43 TY  plastic torque
• Valid only for a solid circular shaft made of an
elastoplastic material.
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Residual Stresses
• Plastic region develops in a shaft when subjected to a
large enough torque.
• 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.
Fig. 3.37 Shear stress-strain response
for loading past yield reversing until
compressive yield occurs.
• On a T- curve, the shaft unloads along a straight line
to an angle greater than zero.
• Residual stresses found from principle of superposition
 
m
Fig. 3.38 Torque-angle of twist response
for loading past yield, followed by
unloading.
Tc
J
   dA  0
Fig. 3.39 Superposition of elastic-plastic state (a) plus linear elastic
unloading (b) equals residual (c) sharing stress distributions.
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Concept Application 3.8/3.9
SOLUTION:
• Solve Eq. (3.29) for Y/c and evaluate
the elastic core radius
Fig. 3.36 Loaded circular shaft.
A solid circular shaft is subjected to a
torque T  4.6 kN  m at each end.
Assuming that the shaft is made of an
elastoplastic material with  Y  150 MPa
and G  77 GPa 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.
• Solve Eq. (3.15) for the angle of twist
• 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
• Find the residual stress distribution by
a superposition of the stress due to
twisting and untwisting the shaft
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Concept Application 3.8/3.9
SOLUTION:
• Solve Eq. (3.29) for Y/c and
evaluate the elastic core radius
T
3

4 T 1  1 Y
3 Y
4 3

 
c 


 614 10
Y 
TY c
J
m
1
 3
Y

T
  4  3 
c 
TY 
J  12 c 4  12  25 10 3 m
9
• Solve Eq. (3.15) for the angle of twist

4
 J
 TY  Y
c
 Y

Y
c
 

Y
Y c

TY L
3.68  103 N  m 1.2 m 
Y 

JG
614  10-9 m 4 77  10 Pa 


Y  93.4  103 rad
93.4  103 rad

 148.3  103 rad  8.50o
0.630

150 10 6 Pa 614 10 9 m 4 
TY 
  8.50o
25 10 3 m
 3.68 kN  m
Y
4.6 

 4 3

c 
3.68 
1
3
 0.630
Y  15.8 mm
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Concept Application 3.8/3.9
• 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
 
• Find the residual stress distribution by
a superposition of the stress due to
twisting and untwisting the shaft


Tc
4.6  10 3 N  m 25  10 3 m

 max


J
614  10 -9 m 4

 187 .3 MPa
TL
JG

4.6  103 N  m 1.2 m 

6.14  109 m4 77  109 Pa 
 116.8  103 rad  6.69
φp     
 8.50  6.69
 1.81o
 p  1.81o
Fig. 3.40 Superposition of stress distributions to obtain residual stresses.
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Seventh
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MECHANICS OF MATERIALS
Beer • Johnston • DeWolf • Mazurek
Torsion of Noncircular Members
• Previous torsion formulas are valid for
axisymmetric or circular shafts
Fig. 3.41 Twisting of shaft with square cross section.
• Planar cross-sections of noncircular
shafts do not remain planar and stress
and strain distribution do not vary
linearly
• For uniform rectangular cross-sections,
Fig. 3.44 Shaft with rectangular cross section, showing
the location of maximum shearing stress.
 max 
T
c1ab2

TL
c2 ab3G
• 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.
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MECHANICS OF MATERIALS
Beer • Johnston • DeWolf • Mazurek
Thin-Walled Hollow Shafts
Fig. 3.47 Thinwalled hollow shaft
subject to torsional
loading.
• Summing forces in the x-direction on AB,
 Fx  0   A t Ax    B t B x 
 At A  Bt B   t  q  shear flow
shear stress varies inversely with thickness
Fig. 3.48
Segment of thinwalled hollow
shaft.
• Compute the shaft torque from the integral
of the moments due to shear stress
dM0  p dF  p t ds  q pds  2q dA
T   dM0   2q dA  2qA
Fig. 3.51 Shear flow in
the member wall.

T
2tA
• Angle of twist (from Chapter 11)
Fig. 3.53 Area for
shear flow.

TL
ds

4 A2G t
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MECHANICS OF MATERIALS
Beer • Johnston • DeWolf • Mazurek
Concept Application 3.10
Structural aluminum tubing with a rectangular
cross-section has a torque loading of 24 kipin. 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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MECHANICS OF MATERIALS
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Concept Application 3.10
SOLUTION:
• Determine the shear flow through the
tubing walls.
• Find the corresponding shearing
stress with each wall thickness.
With a uniform wall thickness,

T
24kip  in.

 8.35ksi
2
2tA 2(0.160in.)(8.986in. )
  8.35ksi
With a variable wall thickness
A  3.84in.2.34in.  8.986in.2
 AB   AC 
1.335 kip in.
0.120 in.
 AB   BC  11 .13 ksi
 BD   CD 
1.335 kip in.
0.200 in.
 BC   CD  6.68 ksi
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