Question

A solid cantilever shaft of diameter 0.1 m and length 2 m is subjected to a torque of 10 kN-m at the free end (shear modulus is 82 GPa). The maximum induced shear stress is .................... N/mm\(^2\) (round off to the nearest integer).

Hint:

When solving mechanics problems, unit consistency is critical. Converting all quantities to a base set of units (like N and mm for stress calculations in MPa) at the beginning is a good practice to avoid errors. The formula \( \tau_{max} = 16T / (\pi D^3) \) is a very useful shortcut for solid circular shafts.

Answer 1

Step 1: Understanding the Concept:
This problem requires the calculation of the maximum shear stress in a solid circular shaft subjected to a pure torsional load. The maximum shear stress occurs at the outer surface of the shaft.
Step 2: Key Formula or Approach:
The torsion formula relates the shear stress (\(\tau\)) at a radial distance \(r\) from the center to the applied torque (\(T\)) and the polar moment of inertia (\(J\)) of the shaft's cross-section: \[ \frac{T}{J} = \frac{\tau}{r} \] The maximum shear stress (\(\tau_{max}\)) occurs at the outer radius (\(r=R\)): \[ \tau_{max} = \frac{T . R}{J} \] For a solid circular shaft of diameter D, the radius is \(R = D/2\) and the polar moment of inertia is \( J = \frac{\pi D^4}{32} \). Substituting these into the formula gives: \[ \tau_{max} = \frac{T . (D/2)}{\frac{\pi D^4}{32}} = \frac{16T}{\pi D^3} \] The length of the shaft and the shear modulus are not needed to calculate the maximum shear stress; they would be needed to calculate the angle of twist.
Step 3: Detailed Calculation:
Given values:
- Torque, \(T = 10\) kN-m
- Diameter, \(D = 0.1\) m
It's important to use consistent units. The question asks for the answer in N/mm\(^2\) (which is the same as MPa). Let's convert the given values to N and mm.
- Torque, \(T = 10 \text{ kN-m} = 10 \times 10^3 \text{ N} \times 10^3 \text{ mm} = 10 \times 10^6 \text{ N-mm}\)
- Diameter, \(D = 0.1 \text{ m} = 100 \text{ mm}\)
Now, apply the formula for maximum shear stress: \[ \tau_{max} = \frac{16T}{\pi D^3} = \frac{16 \times (10 \times 10^6)}{\pi \times (100)^3} \] \[ \tau_{max} = \frac{160 \times 10^6}{\pi \times 10^6} = \frac{160}{\pi} \] \[ \tau_{max} \approx 50.9295 \text{ N/mm}^2 \] Rounding off to the nearest integer: \[ \tau_{max} = 51 \text{ N/mm}^2 \] Step 4: Final Answer:
The maximum induced shear stress is 51 N/mm\(^2\).