Government Engineering College, Bhavnagar.

Civil Engineering Department

Strength Of Materials

Topic:- Torsion

By,

Bhavik Shah – 130210106049

Contents

Introduction Torsion Formula Assumptions Power Transmitted by shaft Torsional Rigidity

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.

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.

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

• If a shaft has a tubular cross section,

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.

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

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.

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.

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 = 2f T, then power

P = T (d/dt)

P = T

P = 2f T

Power TransmissionShaft 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)(co2 ci

2) to determine co and ci

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.

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

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 JT

L /

(12)

i.e. the torsional rigidity is the torque divided by the angle of twist (in radians) per unit length.

Thank You For Bearing