Question 12 A torque of 5 KNm is applied to a solid round shaft of 100 mm diameter. Calculate the maximum shear stress in the shaft. (A) 32.12 MPa (B) 50.93 MPa (C) 42.56 MPa (D) 25.46 MPa

To calculate the maximum shear stress in the shaft, you can use the formula for shear stress due to torque: \[ \tau_{\text{max}} = \frac{T \cdot r}{J} \] where \( T \) is the torque (5 KNm = 5000 Nm), \( r \) is the radius (0.1 m / 2 = 0.05 m), and \( J \) is the polar moment of inertia for a circular shaft, given by: \[ J = \frac{\pi \cdot d^4}{32} \] \[ J = \frac{\pi \cdot (0.1)^4}{32} = \frac{\pi \cdot 0.0001}{32} \approx 9.82 \times 10^{-6} \, \text{m}^4 \] Now substituting into the shear stress formula: \[ \tau_{\text{max}} = \frac{5000 \cdot 0.05}{9.82 \times 10^{-6}} \approx 25.46 \, \text{MPa} \] So the correct answer is (D) 25.46 MPa. Incredibly, this problem taps into the principles of torsion and material science—absolutely essential in designing everything from bridges to car axles! Understanding shear stress helps engineers ensure structures can handle twisting forces without failing. If you’re keen to dive deeper into the fascinating world of mechanics, you might want to explore topics like material strength, fatigue, and failure criteria. Great resources include engineering textbooks or even online platforms such as Coursera or Khan Academy, which provide comprehensive courses on mechanics of materials!