Lecture Mechanics of materials (Third edition) - Chapter 3: Torsion

Third Edition

CHAPTER

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

Torsion

Lecture Notes:
J. Walt Oler
Texas Tech University

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MECHANICS OF MATERIALS

Beer • Johnston • DeWolf

Contents
Introduction

Statically Indeterminate Shafts

Torsional Loads on Circular Shafts

Sample Problem 3.4

Net Torque Due to Internal Stresses

Design of Transmission Shafts

Axial Shear Components

Stress Concentrations

Shaft Deformations

Plastic Deformations

Shearing Strain

Elastoplastic Materials

Stresses in Elastic Range

Residual Stresses

Normal Stresses

Example 3.08/3.09

Torsional Failure Modes

Torsion of Noncircular Members

Sample Problem 3.1

Thin-Walled Hollow Shafts

Angle of Twist in Elastic Range

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
• Turbine exerts torque T on the shaft
• Shaft transmits the torque to the
generator
• Generator creates an equal and

opposite torque T’

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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,
T = ∫ ρ dF = ∫ ρ (τ 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
• Unlike the normal stress due to axial loads, the
distribution of shearing stresses due to torsional
loads can not be assumed uniform.
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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
• 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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Shaft Deformations
• From observation, the angle of twist of the
shaft is proportional to the applied torque and
to the shaft length.
φ ∝T
φ∝L

• 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 (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
Lγ = ρφ or γ =

ρφ
L

• Shear strain is proportional to twist and radius
γ max =

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cφ
ρ
and γ = γ max
L
c

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Stresses in Elastic Range
• Multiplying the previous equation by the
shear modulus,
Gγ =

ρ
c

Gγ max

From Hooke’s Law, τ = Gγ , so
τ=

ρ
c

τ max

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

J = 12 π c 4

• Recall that the sum of the moments from
the internal stress distribution is equal to
the torque on the shaft at the section,
τ
τ
T = ∫ ρτ dA = max ∫ ρ 2 dA = max J
c
c

(

J = 12 π c24 − c14

)

• The results are known as the elastic torsion
formulas,
τ max =

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Tc
Tρ
and τ =
J
J
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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.
• Consider an element at 45o to the shaft axis,
F = 2(τ max A0 )cos 45 = τ max A0 2

σ

45o

=

F τ max A0 2
=
= τ 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
• 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 analysis 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
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.
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• 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
SOLUTION:Problem 3.1
• Cut sections through shafts AB and BC
and perform static equilibrium analysis
to find torque loadings

∑ 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

J=

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

π
(
c24 − c14 ) = [(0.060 )4 − (0.045)4 ]
2
2

π

= 13.92 × 10− 6 m 4

τ max = τ 2 =

TBC c2 (20 kN ⋅ m )(0.060 m )
=
J
13.92 × 10 − 6 m 4

τ max =

Tc
Tc
=
J π c4
2

65MPa =

6 kN ⋅ m
π c3
2

c = 38.9 × 10 −3 m
d = 2c = 77.8 mm

= 86.2 MPa

τ min c1
=
τ max c2

τ min
86.2 MPa

τ min = 64.7 MPa

=

45 mm
60 mm

τ max = 86.2 MPa
τ 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,
γ max =

cφ
L

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

τ max
G

=

Tc
JG

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

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

Ti Li
i J i Gi

φ =∑

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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,
φ = φ1 + φ2 =

TA L1 TB L2
−
=0
J1G J 2G

LJ
TB = 1 2 TA
L2 J1

• Substitute into the original equilibrium equation,
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

∑ M B = 0 = F (0.875 in.) − T0

• Apply a kinematic analysis to relate
the angular rotations of the gears

rBφ B = rCφC
rC
2.45 in.
φC =
φC
rB
0.875 in.

∑ M C = 0 = F (2.45 in.) − TCD

φB =

TCD = 2.8 T0

φ B = 2 .8 φ 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

φA / B =

τ max =

TAB c
T (0.375 in.)
8000 psi = 0
π (0.375 in.)4
J AB
2
TCD c
2.8 T0 (0.5 in.)
8000 psi =
π (0.5 in.)4

J CD
2

T0 = 561lb ⋅ in.

T0 = 561lb ⋅ in

(

)

= 0.387 rad = 2.22o

φC / D =

T0 = 663 lb ⋅ in.

τ max =

(561lb ⋅ in.)(24in.)
TAB L
=
J ABG π (0.375 in.)4 11.2 × 106 psi
2
2.8 (561lb ⋅ in.)(24in.)
TCD L
=
J CDG π (0.5 in.)4 11.2 × 106 psi
2

(

= 0.514 rad = 2.95o

(

)

)

φ B = 2.8φC = 2.8 2.95o = 8.26o
φ A = φ B + φ A / B = 8.26o + 2.22o

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φ A = 10.48o
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Design of Transmission Shafts
• Principal transmission shaft
performance specifications are:
- power

- speed
• Designer must select shaft
material and cross-section to
meet performance specifications
without exceeding allowable
shearing stress.

• Determine torque applied to shaft at
specified power and speed,
P = Tω = 2πfT
T=

P

ω

=

P
2πf

• Find shaft cross-section which will not
exceed the maximum allowable
shearing stress,
τ max =

Tc
J

T

J π 3
= c =
τ max
c 2

(

(solid shafts )

)

T
π 4 4
J
c2 − c1 =
=
τ max
c2 2c2

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(hollow shafts )

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Stress Concentrations
• The derivation of the torsion formula,
τ 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
τ max = K

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Tc
J

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

Tc
J

• 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
section,
c

c

0

0

T = ∫ ρτ (2πρ dρ ) = 2π ∫ ρ 2τ dρ

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Elastoplastic Materials
• At the maximum elastic torque,
TY =

J
τ Y = 12 πc3τ Y
c

φY =

Lγ Y
c

• As the torque is increased, a plastic region
ρ
τ

=
τY )
τ
=
τ
(
Y ) develops around an elastic core (
ρY =

ρY

Lγ Y

φ

⎛

ρY3 ⎞⎟

T=

2 πc 3τ ⎜1 − 1
Y⎜
3
4

T=

3⎞
⎛

4 T ⎜1 − 1 φY ⎟
3 Y⎜
4 3⎟

⎝

⎝

c ⎟⎠
3

=

⎛

4 T ⎜1 − 1
3 Y⎜
4

⎝

ρY3 ⎞⎟
c3 ⎟⎠

φ ⎠

• As ρY → 0, the torque approaches a limiting value,
TP = 43 TY = plastic torque

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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
• On a T-φ curve, the shaft unloads along a straight line
to an angle greater than zero
• Residual stresses found from principle of superposition

Tc
′ =
τm
J
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∫ ρ (τ dA) = 0
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Example 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 T = 4.6 kN ⋅ m at each end.
Assuming that the shaft is made of an
elastoplastic material with τ 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.

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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
• Find the residual stress distribution by
a superposition of the stress due to
twisting and untwisting the shaft

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Example
SOLUTION: 3.08/3.09
• Solve Eq. (3.32) for ρY/c and
evaluate the elastic core radius
⎛ 1 ρY3 ⎞
4
T = 3 TY ⎜1 − 4 3 ⎟ ⇒
⎜
c ⎟⎠
⎝
J=

1 πc 4
2

=

1π
2

ρY

⎛
T
= ⎜⎜ 4 − 3
c
TY
⎝

(25 ×10 m)

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

1

3

−3

= 614 × 10−9 m 4

τY =

TY c
J

τ J
⇒ TY = Y
c

(
150 × 106 Pa )(614 × 10−9 m 4 )
TY =
25 × 10

−3

m

φ
ρ
= Y
c
φY

⇒ φ=

φY
ρY c

(

)

TY L
3.68 × 103 N (1.2 m )
=
φY =
JG
614 × 10-9 m 4 (77 × 10 Pa )

(

)

φY = 93.4 ×10−3 rad
93.4 × 10−3 rad
= 148.3 × 10−3 rad = 8.50o
φ=
0.630

φ = 8.50o

= 3.68 kN ⋅ m

ρY

4.6 ⎞
⎛
= ⎜4 −3
⎟
c

3.68 ⎠
⎝

1

3

= 0.630

ρY = 15.8 mm
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