Test: Torsion of Shafts - 3 - Mechanical Engineering MCQ

From the equation of torsion

The strain energy in the shaft is 1.73 N mm.

Given-

• Torque = 10 Nm
• Modulus of rigidity = 80 GPa

From the figure we can say that

T₁ = T₂ = T = 10 Nm

Here

J₁ = π/32 d⁴

By substituting the value we get

J₁ = (π/32) × 25⁴ × 10⁻¹²

G₁ = 80 × 10⁹

l = 0.1 m

θ₁ = 0.33 × 10⁻³

So,

T₁ = J₁G₁θ₁/l₁ ⇒ θ₁ = T₁l₁/J₁G₁

Hence,

θ₁/θ₂ = (J₂/J₁) = l₁/l₂ = G₂/G₁

Now according to the question

θ₁ = 4 θ₂ So,

θ = θ₁ + θ₂ = 5/4 × 0.33 × 10⁻³

Now strain energy can be given by

Strain energy = 1/2 Tθ

By substituting the value we get

Strain energy = 1/2 × 10 × 5/4 × 0.33 × 10⁻³ = 1.73

For solid shaft
 ............... (i)
for hollow shaft
.................. (ii)
From (i) and (ii), we get

Hence reduced torque

When shaft diameter is doubled then


 = 30 MPa

Maximum bending stress, 
Maximum shear stress, 

Principal stresses,

Maximum principal stress is obtained by choosing '+' sign and minimum principal stress is be obtained by '-' sign.

Let the diameter of solid shaft be d_s. Let the external and internal diameter of the hollow shaft be d₀ and d_i respectively.
It is given that both the shafts are made of same material, have same weight and length and are subjected to equal torsional force.
∴ Weight of solid shaft = Weight of hollow shaft

Now, we know that, torsional stiffness is given as

For same value of G and L


Now, the quantity is always greater than 1
∴ K_h > K_s

Maximum shear stress 

Normal stress = 0

The symmetry of the shaft shows that there is no torsion on section AB.
∴ Rotation,