Test: Solid Mechanics- 2 - Civil Engineering (CE) MCQ

The simple torsion equation is written as

This states that the shearing stress varies directly as the distance ‘r' from the axis of the shaft and the following is the stress distribution

Hence the maximum shear stress occurs on the outer surface of the shaft where r = R and at the centre the shear stress is zero.

A-3, B-1, C-4

Beam A is cantilever so maximum bending moment occurs at the fixed support

Beam B is simply supported so maximum bending moment occurs at midspan

Beam C is cantilever so maximum bending moment occurs at fixed support
M_C= 

S_sy = Torsional yield strength

Both shaft will rotate through same degree.

Hence 

R₁ + R₂ = 0

∑M_A = 0

⇒ R₂ × 3 = -M


So maximum shear force = M/3

Polar modulus, 

Let the diameter of solid shaft is ‘d’

M = 5x² + 10x

Shear Force = dM/dx = 10x + 10

Shear force at X = 10 mm

= 10 x 10 + 10 = 110 N

Using the equation of deflection:

W = 40 kN/m

∑F_V = 0

R_A + R_B = 3 + 2 × 2

R_A + R_B = 7 kN

∑F_H = 0

H_A = 0

∑Mz = 0

R_A × 4 – 3 × 3 - 2 × 2 × 1= 0

R_A = 3.25 kN

R_B = 3.75 kN

BM_D = R_A × 2 - 3 × 1

BM_D = 3.25 × 2 - 3 = 3.5 kN