# Torsional Shear Stress – Definition, Formula [Gate Notes]

By BYJU'S Exam Prep

Updated on: September 25th, 2023

Shear stress can be of two types: direct shear stress and torsional shear stress. Direct shear stress is produced due to the effect of direct shearing force. On the other hand, **torsional shear stress** is indirect shear stress produced inside the body subjected to torsion/twisting.

In an engineering context, the elements or members that are generally subjected to twisting have a circular cross-section, for example, shafts to transfer power from engines or motors, bolted connections, etc. For that reason in this article, we will discuss torsional shear stress acting on a circular cross-section.

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Table of content

## Definition of Torsional Shear Stress

Before understanding torsional shear stress, we need to understand what causes it. As explained earlier, torsional shear stress is generated in a body that is subjected to a twisting moment. The figure below shows a circular bar subjected to M_{X}, M_{Y} and M_{Z} moments.

The moments M_{X} and M_{Y} cause rotations about lateral axes; hence, they are bending moments. Whereas the moment M_{Z} will cause rotation about the longitudinal axis (z-axis). Hence, M_{Z} is a twisting moment. The twisting moment is also referred to as torque. Torque will tend to cause angular displacement in the body. Therefore, shear stress is generated in the body, referred to as torsional shear stress. Basically, torsional shear stress is the internal resistance of the body to resist twisting.

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## Assumptions for the Theory of Pure Torsion

When a circular shaft is subjected to torque, it will only generate torsional shear stress in the shaft (pure torsion). Our analysis makes certain assumptions to determine the torsional shear stress that will be produced. These assumptions are listed below –

- The material should be linearly elastic, isotropic and homogeneous.
- Hooke’s law is valid.
- The cross-section remains plane and circular before and after twisting.
- The shaft should be of uniform cross-section
- The shaft should be perfectly straight

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## Torsional Shear Stress Equation

Consider a circular shaft fixed at one end and subjected to torque (T) on the other end. Due to this torque, shear stresses are produced in two mutually perpendicular planes: in the plane of the cross-section in the circumferential direction and normal to the cross-sectional plane in the longitudinal direction.

Due to the circular cross-section, the torsional shear stress generated will be zero at the central axis and maximum at the outer surface, as shown in the figure above. The torsional shear stress a distance r from the centre of the cross-section is given by –

τ/r=τ_{max}/R

where τ_{max} = maximum torsional shear stress in the shaft

R = radius of the shaft

The torsional shear stress equation is given by-

T/I_{P}=τ_{max}/R=GΘ/L

where I_{P} = polar moment of inertia

G = modulus of rigidity

θ = angle of twist in radian

L = length of the shaft

## Torsional Shear Stress in Different Circular Sections

The torsional shear stress distribution will differ for a solid circular shaft, hollow circular shaft, and thin circular tube. The distribution and polar section modulus for each of these cases has been explained in this section.

### Solid Circular Shaft

Torsional stress distribution for a solid circular shaft subjected to a torque (T) is shown in the figure below –

I_{P}=(π/32)D^{4}

Polar section modulus,

Z_{P}=I_{P}/R_{max}=π×D^{4}/(32D/2)=πD^{3}/16

Therefore, the maximum torsional shear stress will be

τ_{max}=T/Z_{P}=16T/πD^{3}

### Hollow Circular Shaft

Torsional stress distribution for a hollow circular shaft subjected to a torque (T) is shown in the figure below –

I_{P}=π/32(D_{o}^{4}-D_{i}^{4})

Polar section modulus,

Z_{P}=I_{P}/R_{max}=π(D_{o}^{4}-D_{i}^{4})/32×D_{o}/2=π(D_{o}^{4}-D_{i}^{4})/16D_{o}

Therefore, the maximum torsional shear stress can be calculated from –

τ_{max}=T/Z_{P}

### Thin Circular Tube With Mean Radius R

Torsional stress distribution for a thin circular tube of mean radius R and thickness t when subjected to a torque (τ) is shown in the figure below –

I_{P}=2πR^{3}t

Polar section modulus,

Z_{P}=I_{P}/R_{max}=2πR^{3}t/R=2πR^{2}t

For a thin circular tube, the shear stress is assumed to be uniform across the whole thickness as shown in the figure above. Therefore, the torsional shear stress throughout the thickness will be

τ=T/Z_{P}