The strength of material play and important role in any structure. The effect of Torsion is always considerable.

1. Government Engineering College,
Bhavnagar.
Civil Engineering Department

2. Strength Of Materials
Topic:-
Torsion
By,
Bhavik Shah – 130210106049

3. Contents
 Introduction
 Torsion Formula
 Assumptions
 Power Transmitted by shaft
 Torsional Rigidity

4. Introduction
 Torsion is a moment that twists/deforms a member about its
longitudinal axis.
 By observation, if angle of rotation is small, length of shaft and its
radius remain unchanged.

5. 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.

6. The Torsion Formula
• If the shaft has a solid circular cross section,
• If a shaft has a tubular cross section,

7. Example
• 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.

8. 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

9. Assumptions
 The material is homogeneous, i.e. of uniform elastic
properties throughout.
 The material is elastic, following Hooke's law with shear
stress proportional to shear
strain.
 The stress does not exceed the elastic limit or limit of
proportionality.
 Circular sections remain circular.

10. Assumptions
 Cross-sections remain plane. (This is certainly not the case
with the torsion of non-circular sections.)
 Cross-sections rotate as if rigid, i.e. every diameter rotates
through the same angle.
 Practical tests carried out on circular shafts have shown
that the theory developed below on the basis of these
assumptions shows excellent correlation with experimental
results.

11. Power Transmission
 Power is defined as work performed per unit of time
 Instantaneous power is
 Since shaft’s angular velocity  = d/dt, we can also
express power as
 Frequency f of a shaft’s rotation is often reported. It
measures the number of cycles per second and since 1
cycle = 2 radians, and  = 2f T, then power
P = T (d/dt)
P = T
P = 2f T

12. Power Transmission
Shaft Design
 If power transmitted by shaft and its frequency of rotation is known,
torque is determined from Eqn 5-11
 Knowing T and allowable shear stress for material, allow and applying
torsion formula,
 For solid shaft, substitute J = (/2)c4 to determine c
 For tubular shaft, substitute J = (/2)(co
2  ci
2) to determine co and ci

13. Example
Solid steel shaft shown used to transmit 3750 W from
attached motor M. Shaft rotates at  = 175 rpm and
the steel allow = 100 MPa.
Determine required diameter of shaft to nearest mm.

14. Solution
Torque on shaft determined from P = T,
Thus, P = 3750 N·m/s
Thus, P = T,T = 204.6 N·m
Since 2c = 21.84 mm, select shaft with diameter of d = 22 mm

15. Torsional Rigidity
The angle of twist per unit length of shafts is given by the torsion theory as:

L
T
G J

The quantity G J is called the torsional rigidity of the shaft and is thus given as:
G J
T
L

 / (12)
i.e. the torsional rigidity is the torque divided by the angle of twist (in radians) per unit
length.

16. Thank You For Bearing