JAWAHARLAL NEHRU TECHNOLOGICAL UNIVERSITY HYDERABAD B.Tech II STRENGTH OF MATERIALS - II Previous Question Paper (July/August - 2021 - R16)

To derive the torsion formula, consider a circular shaft of radius R and length L subjected to a torque T. We assume the shaft is homogeneous, isotropic, and undergoes pure torsion where plane sections remain plane and radii remain straight.

Let a point at a radial distance r from the center undergo a shear strain γ. From the geometry of deformation, the arc length subtended by the angle of twist θ at radius r is rθ. The shear strain is defined as the ratio of this deformation to the length L:

Using Hooke's Law for shear, the shear stress q at radius r is related to the shear modulus N by q = Nγ. Substituting the expression for γ:

Rearranging this gives the relationship between stress and twist:

Next, consider an elemental ring of area dA at radius r. The shear force on this element is dF = q dA. The torque dT produced by this force about the central axis is:

Integrating over the entire cross-section to find the total torque T:

Since the polar moment of inertia J is defined as ∫ r² dA, we have T = (Nθ/L)J. Rearranging this yields:

Combining this with the earlier expression q/r = Nθ/L, we obtain the complete torsion formula:

Given data:

• Stiffness (k) = 1 N/mm
• Maximum load (W) = 45 N
• Maximum shear stress (τ) = 126 N/mm²
• Solid length (L_s) = 45 mm
• Modulus of rigidity (N) = 4.2 × 10⁴ N/mm²

Let d be the diameter of the wire, D be the mean coil diameter, and n be the number of turns.

1. Using the stiffness formula for a close-coiled helical spring:

2. Using the shear stress formula:

From the stress equation, we find the spring index (C = D/d):

3. The solid length is given by L_s = n × d = 45 mm, so n = 45/d.

4. Substitute D = C × d and n = 45/d into the stiffness equation:

Substitute C = 1.1d²:

The diameter of the wire is approximately 2.47 mm.