Torsion: Shear Stress &
Twist (3.1-3.5)
MAE 314 – Solid Mechanics
Yun Jing
Torsion: Shear Stress & Twist
1
Torsion of Circular Shafts

In this chapter, we will examine uniaxial bars subject to torque.

Where does this occur?
Transmission Shaft
Force Couples
Torsion: Shear Stress & Twist
2
Torsion of Circular Shafts

We assume



Bar is in pure torsion
Small rotations (the length and radius will not change)
How does the bar deform?

Cross-section of the bar remains the same shape, bar is simply rotating.

Cross-section remains perpendicular to axis of cylinder (cylinder does not
warp).
Not true for most
non-circular bars
Torsion: Shear Stress & Twist
3
Angle of Twist

Deformation of a circular shaft subjected to pure torsion




What are the boundary conditions on ?



Fix left end of shaft
A moves to A’
 = angle of twist (in radians)
x
(x) = 0 at x = 0
(x) =  at x = L
For pure torsion,  is linear.
 ( x) 
x
L
Torsion: Shear Stress & Twist
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Shearing Strain


Calculate the surface shear strain in the
cylinder.
For pure torsion (x) = x / L, so


L
Torsion: Shear Stress & Twist
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Shearing Strain

The maximum shear strain on the surface of
the cylinder occurs when ρ=c.
 max

Maximum shear strain on surface
c

L
We can express the shearing strain at any
distance from the axis of the shaft as


c
 max
Torsion: Shear Stress & Twist
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Shearing Strain

We can also apply the equation for maximum
surface shear strain to a hollow circular tube.
 min

c1

L
 max
c2

L
c
c
This applies for all types of materials: elastic, linear, non-linear,
plastic, etc.
Torsion: Shear Stress & Twist
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Elastic Shearing Stress


Calculate shear stress in a bar made of linearly elastic material.
Recall Hooke’s Law for shearing stress: τ=Gγ
 max  G max 

Gc
    max
c
L
Torsion: Shear Stress & Twist
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Torque


We still need to relate τ to the applied torque T, which is generally the
known, applied load.
First, find the resultant moment acting on a cross-section and set this
equal to T.


c
 max
2
dA 
T 
A

c
2
c
 maxdA
 maxdA 
 max
c
2

 dA
c
A
Torsion: Shear Stress & Twist
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Torque

Continuing from previous slide:
T


 max
c
2

 dA 
 max
A
c
J   max 
Tc
J
Where J is the polar moment of inertia of the cross section of the bar (see
Appendix A.3 in your textbook).
Plug this into the equation for τmax.
 max
Gc


L
Gc Tc


L
J
Torsion: Shear Stress & Twist

TL
GJ
10
Torque

For a non-uniform bar
n
n
Ti Li
  i  
i 1
i 1 Gi J i

For a continuously varying bar
L
T ( x)

dx
GJ ( x)
0
Torsion: Shear Stress & Twist
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Inclined Plane

Cut a rectangular element along the plane at an angle θ.
Torsion: Shear Stress & Twist
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Inclined Plane

Sum forces in x-direction.
y
x
 A0 sec A0 sin  A0 tan cos  0
   sin  cos   sin  cos
   2 sin  cos   sin 2

Sum forces in y-direction.
 A0 sec A0 cos  A0 tan sin   0
   cos2   sin 2 
   cos2
Torsion: Shear Stress & Twist
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Inclined Plane

τmax occurs at θ = 0º, ±90º

σmax occurs at θ = ±45º

τmax = σmax

When σθ is max, τθ = 0, and when τθ is max, σθ =0.
Torsion: Shear Stress & Twist
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Example Problem
Part 1. For the 60 mm diameter solid cylinder and loading shown,
determine the maximum shearing stress.
Part 2. Determine the inner diameter of the hollow cylinder , of 80 mm
outer diameter, for which the maximum stress is the same as in part 1.
Torsion: Shear Stress & Twist
15
Example Problem
Part 1. For the aluminum shaft shown (G = 27 GPa), determine the torque
T that causes an angle of twist of 4o.
Part 2. Determine the angle of twist caused by the same torque T in a solid
cylindrical shaft of the same length and cross-sectional area.
Torsion: Shear Stress & Twist
16
Torsion: Statically
Indeterminate Problems and
Transmission Shafts (3.6-3.8)
MAE 314 – Solid Mechanics
Yun Jing
Torsion: Statically Indeterminate Problems and
Transmission Shafts
17
Statically Determinate Problems
T3
T2
T1
Find the maximum shearing stress in each bar.
Torsion: Statically Indeterminate Problems and
Transmission Shafts
18
Statically Indeterminate Problems

Method for torsion is the same as the method for statically
indeterminate axial load deflection problems.

Apply what you’ve already learned:

M=R–N

M = number of compatibility equations needed

R = number of unknown reactions (or internal stresses)

N = number of equilibrium equations

Compatibility equations for a torsion problem are based on angle of
twist.
Torsion: Statically Indeterminate Problems and
Transmission Shafts
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Statically Indeterminate Problems
LCD
dCD
dAB
LAB
rC
rB
Find the largest torque T0 that can be applied to the end of shaft
AB and the angle of rotation of the end A of shaft AB. Allowable
shearing stress is  allow
Torsion: Statically Indeterminate Problems and
Transmission Shafts
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A circular shaft AB consists of a 10-in.-long, 7/8-in.diameter steel cylinder, in which a 5-in.long,5/8-in.diameter cavity has been drilled from end B. The shaft is
attached to fixed supports at both ends, and a 90 lb.ft
torque is applied at its midsection. Determine the torque
Torsion: Shear
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exerted on the shaft by each
of Stress
the& Twist
supports.
Transmission Shafts

In a transmission, a circular shaft transmits mechanical power from
one device to another.
Generator
Turbine




ω = angular speed of rotation of the shaft
The shaft applies a torque T to another device
To satisfy equilibrium the other device applies torque T to the shaft.
The power transmitted by the shaft is
P  T
Torsion: Statically Indeterminate Problems and
Transmission Shafts
22
Transmission Shafts

Units for P=Tω






ω = rad/s
T = N·m (SI)
T = ft·lb (English)
P = Watts (1 W = 1 N·m/s) (SI)
P = ft·lb/s (1 horsepower = hp = 550 ft·lb/s) (English)
We can also express power in terms of frequency.
  2f
f  Hz  s 1
P  2fT
Torsion: Statically Indeterminate Problems and
Transmission Shafts
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Example Problem
A 1.5 meter long solid steel shaft of 22 mm diameter is to
transmit 12 kW. Determine the minimum frequency at which the
shaft can rotate, knowing that G = 77.2 GPa, that the allowable
shearing stress is 30 MPa, and that the angle of twist must not
exceed 3.5o.
Torsion: Statically Indeterminate Problems and
Transmission Shafts
24
Stress Concentrations in Circular Shafts

Up to now, we assumed that transmission shafts are loaded at the ends
through solidly attached, rigid end plates.

In practice, torques are applied through flange couplings and fitted
keyways, which produce high stress concentrations.
Fitted keyway
Flange coupling

One way to reduce stress concentrations is through the use of a fillet.
Torsion: Statically Indeterminate Problems and
Transmission Shafts
25
Stress Concentrations in Circular Shafts

Maximum shear stress at the fillet
 max  K


Tc
J
Fillet
Tc/J is calculated for the smaller-diameter shaft
K = stress concentration factor
Torsion: Statically Indeterminate Problems and
Transmission Shafts
26
Example Problem
The stepped shaft shown rotates at 450 rpm. Knowing that r = 0.25 in,
determine the maximum power that can be transmitted without
exceeding an allowable shearing stress of 7500 psi.
Torsion: Statically Indeterminate Problems and
Transmission Shafts
27