# A hollow shaft of 16 mm outside diameter and 12 mm inside diameter is subjected to a torque of 40 N-m. The shear stress at the outside and inside of the material of the shaft are respectively

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ESE Civil 2014 Paper 1: Official Paper
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1. 62.75 N/mm2 and 50.00 N/mm2
2. 72.75 N/mm2 and 54.54 N/mm2
3. 79.75 N/mm2 and 59.54 N/mm2
4. 80.00 N/mm2 and 40.00 N/mm2

Option 2 : 72.75 N/mm2 and 54.54 N/mm2
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CT 3: Building Materials
2962
10 Questions 20 Marks 12 Mins

## Detailed Solution

Concept:

Shear stress-induced in the shaft at any location (r) from the centre is given by,

$$\frac{\tau }{r} = \frac{T}{{{I_P}}}$$

Where, r = Radial distance from centre to the location at which shear stress is calculated (m), IP = Polar moment of inertia (m4) and τ = Shear stress N/m2

Polar moment of inertia for hollow circular shaft,

$${I_p} = \frac{\pi }{{32}} \times \left( {D_o^4 - D_i^4} \right)$$

Where, Do = External diameter and Di = Internal diameter of shaft

Calculation:

External diameter (do) = 16 mm

Internal Diameter (di) = 12 mm

Torque Applied (T) = 40 N-m

Polar moment of inertia of hollow circular shaft –

$${I_{P}} = \frac{\pi }{{32}}\left( {{{16}^4} - {{12}^4}} \right) = 4396\;m{m^4}$$

Shear stress at inside of material,

$${\tau _{inside}} = \frac{T}{{{I_P}}} \times {r_i}$$

$$= \frac{{40}}{{4396}} \times \frac{{0.012}}{2}$$

$$= 54.595\; \times {10^{ - 6}}\;N/{m^2} \approx 54.595\;MPa$$

Shear stress at outside of material,

$${\tau _{inside}} = \frac{T}{{{I_P}}} \times {r_o}$$

$$= \frac{{40}}{{4396}} \times \frac{{0.016}}{2}$$

$$= 72.793 \times {10^{ - 6}} \approx 72.793\;MPa$$