Question

A solid circular torsional member OPQ is subjected to torsional moments as shown in the figure (not to scale). The yield shear strength of the constituent material is 160 MPa.
 

The absolute maximum shear stress in the member (in MPa, round off to one decimal place) is \(\underline{\hspace{1cm}}\).

Hint:

For torsional members, the shear stress is calculated using the formula \( \tau = \frac{T \cdot r}{J} \), where \( J \) is the polar moment of inertia.

Answer 1

Correct Answer: 14 - 16

Step 1: Torque diagram and internal torque

From equilibrium of the shaft:

Internal torque in segment OP:

T_OP = 2 + 1 = 3 kN·m

Internal torque in segment PQ:

T_PQ = 1 kN·m

Step 2: Maximum shear stress formula

For a solid circular shaft:

τ_max = (16T) / (πD³)

Step 3: Shear stress in segment OP

T_OP = 3 kN·m = 3 × 10⁶ N·mm

D = 0.1 m = 100 mm

τ_OP = (16 × 3 × 10⁶) / (π × 100³)

τ_OP = 15.278 MPa

Step 4: Shear stress in segment PQ

T_PQ = 1 kN·m = 1 × 10⁶ N·mm

D = 0.08 m = 80 mm

τ_PQ = (16 × 1 × 10⁶) / (π × 80³)

τ_PQ = 9.94 MPa

Step 5: Absolute maximum shear stress

Absolute maximum shear stress occurs in the segment having the higher value.

τ_abs = max(τ_OP, τ_PQ)

τ_abs = 15.278 MPa

Final Answer:

Absolute maximum shear stress in the member = 15.3 MPa (rounded to one decimal place)