STRUCTURAL MECHANICS: CE203
Chapter 5
Torsion
Notes are based on Mechanics of Materials: by R. C. Hibbeler, 7th Edition, Pearson
Dr B. Achour & Dr Eng. K. El-kashif
Civil Engineering Department, University of Hail, KSA
(Spring 2011)
Chapter 5: Torsion
Torsional Deformation of a Circular Shaft
Torque is a moment that twists a member about
its longitudinal axis.
 If the angle of rotation is small, the length of the
shaft and its radius will remain unchanged.

Chapter 5: Torsion
The Torsion Formula
When material is linear-elastic, Hooke’s law applies.
 A linear variation in shear strain leads to a
corresponding linear variation in shear stress
along any radial line on the cross section.

Chapter 5: Torsion
The Torsion Formula

If the shaft has a solid circular cross section,

If a shaft has a tubular cross section,
Chapter 5: Torsion
Example 5.2

The solid shaft of radius c is subjected to a torque T. Find the fraction of T that is
resisted by the material contained within the outer region of the shaft, which has
an inner radius of c/2 and outer radius c.
Solution:
Stress in the shaft varies linearly, thus
The torque on the ring (area) located within
the lighter-shaded region is
For the entire lighter-shaded area the torque is
Chapter 5: Torsion
Solution:

Using the torsion formula to determine the maximum stress in the shaft, we have
Substituting this into Eq. 1 yields
Chapter 5: Torsion
Example 5.3

The shaft is supported by two bearings and is subjected to three torques.
Determine the shear stress developed at points A and B, located at section a–a of
the shaft.
Solution:
From the free-body diagram of the left segment,
The polar moment of inertia for the shaft is
Since point A is at ρ = c = 75 mm,
Likewise for point B, at ρ =15 mm, we have
Chapter 5: Torsion
Power Transmission

Power is defined as the work performed per unit of time.
For a rotating shaft with a torque, the power is

Since

For shaft design, the design or geometric parameter is

, the power equation is
Chapter 5: Torsion
Example 5.5

A solid steel shaft AB is to be used to transmit 3750 W from the motor M to which it is
attached. If the shaft rotates at w =175 rpm and the steel has an allowable shear stress of
allow τallow =100 MPa, determine the required diameter of the shaft to the nearest mm.
Solution:
The torque on the shaft is
Since
As 2c = 21.84 mm, select a shaft having a diameter of 22 mm.
Chapter 5: Torsion
Angle of Twist

Integrating over the entire length L of the shaft, we have
Φ = angle of twist
T(x) = internal torque
J(x) = shaft’s polar moment of inertia
G = shear modulus of elasticity for the material

Assume material is homogeneous, G is constant, thus

Sign convention is
determined by right hand rule,
Chapter 5: Torsion
Example 5.8

The two solid steel shafts are coupled together using the meshed gears. Determine the
angle of twist of end A of shaft AB when the torque 45 Nm is applied. Take G to be 80 GPa.
Shaft AB is free to rotate within bearings E and F, whereas shaft DC is fixed at D. Each shaft
has a diameter of 20 mm.
Solution:
From free body diagram,
Angle of twist at C is
Since the gears at the end of the shaft are in mesh,
Chapter 5: Torsion
Solution:

Since the angle of twist of end A with respect to end B of shaft AB caused by the
torque 45 Nm,
The rotation of end A is therefore
Chapter 5: Torsion
Example 5.10

The tapered shaft is made of a material having a shear modulus G. Determine the
angle of twist of its end B when subjected to the torque.
Solution:
From free body diagram, the internal torque is T.
Thus, at x,
For angle of twist,
Chapter 5: Torsion
Example 5.11

The solid steel shaft has a diameter of 20 mm. If it is subjected to the two torques,
determine the reactions at the fixed supports A and B.
Solution:
By inspection of the free-body diagram,
Since the ends of the shaft are fixed,
Using the sign convention,
Solving Eqs. 1 and 2 yields TA = -345 Nm and TB = 645 Nm.
Chapter 5: Torsion
Solid Noncircular Shafts

The maximum shear stress and the angle of twist
for solid noncircular shafts are tabulated as below:
Chapter 5: Torsion
Example 5.13
The 6061-T6 aluminum shaft has a cross-sectional area in the shape of an equilateral
triangle. Determine the largest torque T that can be applied to the end of the shaft if
the allowable shear stress is τallow = 56 MPa and the angle of twist at its end is restricted
to Φallow = 0.02 rad. How much torque can be applied to a shaft of circular cross section
made from the same amount of material? Gal = 26 GPa.
Solution:
By inspection, the resultant internal torque at any cross section
along the shaft’s axis is also T.
By comparison, the torque is limited due to the angle of twist.
Chapter 5: Torsion
Solution:

For circular cross section, we have
The limitations of stress and angle of twist then require
Again, the angle of twist limits the applied torque.
Chapter 5: Torsion
Thin-Walled Tubes Having Closed Cross Sections

Shear flow q is the product of the tube’s thickness and
the average shear stress.

Average shear stress for thin-walled tubes is
= average shear stress
T = resultant internal torque at the cross section
t = thickness of the tube
Am = mean area enclosed boundary

For angle of twist,
Chapter 5: Torsion
Example 5.14

Calculate the average shear stress in a thin-walled tube having a circular cross
section of mean radius rm and thickness t, which is subjected to a torque T. Also,
what is the relative angle of twist if the tube has a length L?
Solution:
The mean area for the tube is
For angle of twist,
Chapter 5: Torsion
Example 5.16

A square aluminum tube has the dimensions. Determine the average shear stress
in the tube at point A if it is subjected to a torque of 85 Nm. Also compute the
angle of twist due to this loading. Take Gal = 26 GPa.
Solution:
By inspection, the internal resultant torque is T = 85 Nm.
The shaded area is
For average shear stress,
Chapter 5: Torsion
Solution:

For angle of twist,
Integral represents the length around the centreline boundary of the tube, thus
Chapter 5: Torsion
Stress Concentration


Torsional stress concentration factor, K, is used to
simplify complex stress analysis.
The maximum shear stress is then determined from the
equation
Chapter 5: Torsion
Example 5.18
The stepped shaft is supported by bearings at A and B. Determine the maximum stress in the
shaft due to the applied torques. The fillet at the junction of each shaft has a radius of r = 6
mm.
Solution:
By inspection, moment equilibrium about the axis
of the shaft is satisfied
The stress-concentration factor can be determined
by the graph using the geometry,
Thus, K = 1.3 and maximum shear stress is
Chapter 5: Torsion
Inelastic Torsion

Considering the shear stress acting on an element of
area dA located a distance p from the center of the shaft,

Shear–strain distribution over a radial line on a shaft is
always linear.
Perfectly plastic assumes the shaft will continue to twist
with no increase in torque.
It is called plastic torque.


Chapter 5: Torsion
Example 5.20

A solid circular shaft has a radius of 20 mm and length of 1.5 m. The material has an
elastic–plastic diagram as shown. Determine the torque needed to twist the shaft Φ = 0.6
rad.
Solution:
The maximum shear strain occurs at the surface of
the shaft,
The radius of the elastic core can be obtained by
Based on the shear–strain distribution, we have
Chapter 5: Torsion