Angle of twist: Definition, Formula, Units, Examples [with Pdf]

What is angle of twist?

angle of twist

The angle of twist is defined as the angular deformation in an object due to a couple of twisting torques.

If the twisting moment or torque is applied along the longitudinal axis of the shaft, then the angular deformation is observed in the shaft, this angular deformation is considered as the angle of twist.
The term angle of twist is denoted by the symbol θ and it is expressed by the unit of degree or radian.

Generally, the shafts and axles are designed in such a way that it has a minimum angle of twist.

Angle of twist equation:

1) For the shaft with a length of L, Modulus of rigidity G, and Polar moment of inertia of J, the angle of twist is given by,

`\theta =\frac{T L}{GJ}`

2) For the shaft with different cross-section steps and with n number of twisting torque, the angle of twist is given by,

shaft with different cross-section
Shaft with different cross-section

θ =`\frac{T_{1}L_{1}}{G_{1}J_{1}}+\frac{T_{2}L_{2}}{G_{2}J_{2}}`+`\cdots \frac{T_{n}L_{n}}{G_{n}J_{n}}`

θ =`\sum_{i=1}^{n}\frac{T_{i}L_{i}}{G_{i}J_{i}}`

3) For the shaft with a varying cross-section the angle of twist is given by,

beam with a varying cross-section
Shaft with a varying cross-section

θ =`\int_{0}^{L}\frac{Tdx}{GJ}`

Angle of twist examples:

The following numerical will help you to understand the angle of twist.

1) The aluminum shaft of length 1 m, G= 24 GPa, and having a diameter of 40 mm is twisted by the torque of 100 N.m. Find the angle of twist in the shaft.


G = 24 GPa= 24 × 10³ N/mm²
d = 40 mm
T = 100 N.m = 100 × 10³
L = 1 m = 1000 mm

Step 1] Polar moment of inertia:

`J=(\frac{\pi }{32})d^{4}=(\frac{\pi }{32})40^{4}`

J = 251327 `mm^{4}`

Step 2] Angle of twist:

The angle of twist in the shaft is given by,

θ =`\frac{TL}{JG}`

θ =`\frac{(100\times 10^{3})\times 1000}{251327\times (24\times 10^{3})}`

Angle of twist = θ = 0.016 radian

2) For the aluminum shaft, the limiting value of angle of twist is 0.1 radian. the shaft has a length of 1m with a diameter of 50 and calculate a minimum permissible value for torque. (Assume, G = 25 GPa).


L = 1 m = 1000 mm
θ = 0.1 radian
d = 50 mm
G = 25 GPa = 25 × 10³ N/mm²

Step 1] Polar moment of inertia:

J =`\frac{\pi }{32}d^{4}`

J =`\frac{\pi }{32}\times 50^{4}`

J = 613592 `mm^{4}`

Step 2] Maximum torque (T):

θ = `\frac{TL}{JG}`

`0.1=\frac{T\times 1000}{613592\times (25\times 10^{5})}`

T = 1533980
T = 1533.980 N.m

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Pratik is a Graduated Mechanical engineer. He enjoys sharing the engineering knowledge learned by him with people.

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