prepered Eng Abdiqaadir Abdi Abdi

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