Question

A hollow circular shaft has an outer diameter of \(100\,\mathrm{mm}\) and a wall thickness of \(25\,\mathrm{mm}\). Allowable shear stress in the shaft is \(125\,\mathrm{MPa}\). The maximum torque the shaft can transmit is

• A. \(46\,\mathrm{kN\!-\!m}\)
• B. \(24.5\,\mathrm{kN\!-\!m}\)
• C. \(23\,\mathrm{kN\!-\!m}\)
• D. \(11.5\,\mathrm{kN\!-\!m}\)

Hint:

For a hollow circular shaft, \[ \boxed{ T= \frac{\pi}{16}\tau \frac{D^4-d^4}{D}. } \]

Answer 1

The Correct Option is C

Step 1: Work out the polar section modulus of the hollow shaft.
With outer diameter $D = 100\,\mathrm{mm}$ and wall thickness $t = 25\,\mathrm{mm}$, the inner diameter is $d = D - 2t = 50\,\mathrm{mm}$. The polar modulus is \[ Z_p = \frac{\pi (D^4 - d^4)}{16D} = \frac{\pi (100^4 - 50^4)}{16 \times 100} \approx 184{,}078\,\mathrm{mm^3}. \]
Step 2: Use the torsion relation directly with the polar modulus.
\[ T = \tau \times Z_p. \]
Step 3: Substitute and convert to kN.m.
\[ T = 125 \times 184{,}078 \approx 23 \times 10^6\,\mathrm{N\cdot mm} = 23\,\mathrm{kN\cdot m}. \]
\[ \boxed{23\,\mathrm{kN\cdot m}} \]