Question

A shaft diameter 'd' is connected to the hub through a square key each of side 'd/4' and length '\(l\)' and transmits a torque 'T'. Assume the length of key is equal to thickness of the pulley, the average shear stress developed in the key is

• A. \(\frac{\text{T}}{l\text{d}^2}\)
• B. \(\frac{\text{2T}}{l\text{d}^2}\)
• C. \(\frac{\text{8T}}{l\text{d}^2}\)
• D. \(\frac{\text{4T}}{l\text{d}^2}\)

Hint:

For any standard rectangular or square key: - Tangential Force: \(F = \frac{2T}{d}\) - Shear Area: \(A_s = w \cdot l\) - Shear Stress: \(\tau = \frac{2T}{w \cdot l \cdot d}\) Substituting the square key width \(w = \frac{d}{4}\) directly yields: \(\tau = \frac{2T}{(d/4) \cdot l \cdot d} = \frac{8T}{l d^2}\).

Answer 1

The Correct Option is C

Step 1: Relate the shear force on the key to the transmitted torque.
The torque is carried across the key at the shaft surface, a radius \(d/2\) from the centre, so the tangential shearing force is \[ F = \frac{T}{d/2} = \frac{2T}{d} \]

Step 2: Write the torque directly in terms of the shear stress.
The key shears on a plane of width \(w = d/4\) and length \(l\), so the shearing force can also be written as \(F = \tau \times (w \times l) = \tau \times \dfrac{d}{4} \times l\). Since \(T = F \times d/2\), substituting gives: \[ T = \tau \times \frac{d}{4} \times l \times \frac{d}{2} = \frac{\tau \, l \, d^2}{8} \]

Step 3: Solve for the shear stress.
Rearranging the above equation directly for \(\tau\): \[ \tau = \frac{8T}{l d^2} \]
This matches option (3).