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

Concept: A key is a mechanical component inserted between a rotating machine shaft and the hub of a pulley or gear to facilitate torque transmission. During operation, the transmitted torque \(T\) creates a tangential shearing force \(F\) at the outer surface boundary of the shaft. The relationship connecting the torque \(T\), the tangential force \(F\), and the shaft radius (\(r = \frac{d}{2}\)) is written as: \[ T = F \cdot \left(\frac{d}{2}\right) \quad \Rightarrow \quad F = \frac{2T}{d} \] This tangential force \(F\) acts directly on the critical shear plane of the embedded key. Let the key have width \(w\), thickness \(h\), and length \(l\). For a square key, the width matches the thickness: \[ w = h = \frac{d}{4} \]

Step 1: Determining the shearing area of the key.
Shear failure occurs when the key splits along its mid-plane parallel to the tangent of the shaft surface. The area resisting this shearing force is the product of the key's width (\(w\)) and its inserted length (\(l\)): \[ A_{\text{shear}} = w \cdot l \] Substituting the given geometric dimension \(w = \frac{d}{4}\): \[ A_{\text{shear}} = \left(\frac{d}{4}\right) \cdot l = \frac{l \cdot d}{4} \]

Step 2: Calculating the average shear stress (\(\tau\)).
The average shear stress developed within the key is defined as the total tangential force divided by the resisting shear area: \[ \tau = \frac{F}{A_{\text{shear}}} \] Substitute the expressions for \(F\) and \(A_{\text{shear}}\) into this formula: \[ \tau = \frac{\left(\frac{2T}{d}\right)}{\left(\frac{l \cdot d}{4}\right)} \]

Step 3: Simplifying the algebraic expression.
To simplify this fraction, we multiply the numerator by the reciprocal of the denominator: \[ \tau = \frac{2T}{d} \times \frac{4}{l \cdot d} = \frac{2 \times 4 \times T}{l \cdot d \cdot d} = \frac{8T}{l \cdot d^2} \] The derived average shear stress matches Option (3).