# A solid shaft of 2 m diameter is used to transmit torque. The maximum shear stress induced to the shaft is 100 N/m2 . Determine the maximum torque transmitted by the shaft.

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DSSSB JE ME 2019 Official Paper Shift - 2 (Held on 06 Nov 2019)
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1. 146 N-m
2. 188 N-m
3. 165 N-m
4. 157 N-m

Option 4 : 157 N-m

## Detailed Solution

Concept:

Using torsion equation:

$$\frac{T}{{{I_P}}} = \frac{{{{\rm{τ }}_{max}}}}{R} = \frac{{G\theta }}{L}$$

Where,

T = applied torque, IP = Polar section modulus, τmax = Maximum shear stress in the shaft material, R = Radius of the shaft, G = Modulus of Rigidity, θ = Angle of twist, L = Length of the shaft

$${I_P} = \frac{π }{{32}}{D^4}$$

R = $$\frac{D}{2}$$

From shear stress criteria:

$$\frac{{{{\rm{τ }}_{{\rm{max}}\left( {allowable} \right)}} \times 2}}{D} = \frac{T}{{{I_P}}}$$

Substituting the values of $${I_P}$$ and R we get

Torque (T) = $$\frac{π }{{16}}τ {d^3}$$

Calculation:

Given:

d = 2 m, τ = 100 N/m2

T = $$\frac{π }{{16}}τ {d^3}$$

Taking π = 3.14

T = $$\frac{3.14\;\times\;100\;\times\;2^3}{16}$$ = 157 N-m