Torsional shear stress: Definition, Formula, Calculation

What is Torsional shear stress?

Torsional shear stress is the shear stress offered by the body against torsional load or twisting load. it is denoted by the symbol โ€˜๐œโ€™.

The value of torsional shear stress varies within the cross-section of the object. The value for shear stress is minimum at the neutral axis of the cross-section while it is maximum at the outermost surface of the cross-section of the object.

The units of torsional shear stress are N/mยฒ in the SI system while lb/ftยฒ in the FPS system.

Torsional shear stress equation:

Torsional shear stress can be found by using the torsional equation.

Therefore,

`\tau= \frac{T}{J}\times r`

Where,
T = Applied torque
J = Polar moment of inertia
r = Distance between the neutral axis and the point where shear stress is to be calculated

Torsional shear stress formula for circular shaft:

A] For solid shaft:-

Torsional shear stress for circular solid shaft

The above diagram shows the torsional shear stress distribution in a hollow circular shaft. In that figure, the value for ๐œ is minimum at the neutral axis while it is maximum at r = d/2

For the solid circular shaft, the shear stress at any point in the shaft is given by,

`\tau= \frac{T}{J}\times r`

But for solid shaft,

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

Therefore the torsional shear stress for the circular shaft is given by,

`\tau= \frac{32T}{\pi\times d^{4} }\times r`

B] For hollow shaft:-

Torsional shear stress for circular hollow shaft

The above diagram shows the torsional shear stress distribution in a hollow circular shaft. In that figure, the value for ๐œ is minimum at r = di/2 while it is maximum at r = do/2

For hollow circular shaft with outer diameter (do) and inner diameter (di), the polar moment of inertia is given by,

`J= (\frac{\pi }{32})\times [do^{4}-di^{4}]`

Therefore the torsional shear stress for hollow shaft is given by,

`\tau= \frac{32T}{\pi\times (do^{4}-di^{4}) }\times r`

Maximum torsional shear stress for circular shaft:

A] For solid shaft:-

The maximum shear stress in a solid circular shaft is observed at the outermost surface where r = d/2

`\tau_{max}= \frac{32T}{\pi\times d^{4} }\times \frac{d}{2}`

`\tau_{max}= \frac{16T}{\pi\times d^{3} }`

B] For hollow shaft:-

The maximum shear stress in a hollow circular shaft is observed at the outer diameter (do).
โˆด r = do/2

`\tau_{max}= \frac{32T}{\pi\times (do^{4}-di^{4}) }\times\frac{do}{2}`

`\tau_{max}= \frac{16T\times do}{\pi\times (do^{4}-di^{4}) }`

Torsional shear stress solved examples:

1] The shaft of the motor is rotating with a maximum torque of 6 N.m. If the shaft has a diameter of 25 mm, find the maximum shear stress acting onto the shaft.

Given:
T = 3 N.m
d = 25 mm = 0.025 m

Solution:-

The maximum shear stress acting on the solid circular shaft is given by,

`\tau_{\text{max}} = \frac{16T}{\pid^{3}}`

`\tau_{\text{max}} = \frac{16\times3}{\pi \times 0.025^{3}}`

`\therefore \tau_{\text{max}}= 977847.97\ \text{N/m}^ยฒ`


2] The hollow circular shaft shown in the below figure is under the torque of 5 N.m. Find the shear stress at the radius of r = 20 mm.

hollow shaft of external diameter 60 mm

Given:
T = 5 N.m
`d_{i}` = 30 mm = 0.03 m
`d_{o}` = 60 mm = 0.06 m
r = 20 mm = 0.02 m

Solution:-

For the above hollow shaft, the polar moment of inertia is given by,

`J = \frac{\pi}{32}(d_{o}^{4} โ€“ d_{i}^{4})`

`J = \frac{\pi}{32}(0.06^{4} โ€“ 0.03^{4})`

`J = 1.192\times10^{-6} m^{4}`

The shear stress at r = 0.02 m is given by,

`\tau_{r=0.02m} = \frac{T}{J}\times r`

`\tau_{r=0.02m} = \frac{5}{1.192\times10^{-6}}\times 0.02`

`\therefore \tau_{r=0.02m}= 83892.61\ \text{N/m}^ยฒ`

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