JNTU B.Tech II STRENGTH OF MATERIALS - II Previous Question Paper (November/December - 2020 - R16)

Consider a circular shaft of length L, radius R, and polar moment of inertia J, subjected to a twisting moment T. Under pure torsion, a plane section remains plane, radii remain straight, and the angle of twist θ is proportional to the distance from the axis. Let a point at radius r from the center undergo a shear strain γ.

From the geometry of deformation, the arc length displacement is rθ, which equals Lγ. Thus, the shear strain is given by:

Using Hooke's Law for shear, where N is the modulus of rigidity:

Rearranging this gives the relationship between shear stress and the angle of twist:

To relate this to the applied torque T, consider an elemental ring of radius r and thickness dr. The area of this element is dA = 2πr dr. The shear force on this element is dF = q dA = (Nθr/L)(2πr dr). The torque contribution dT is:

Integrating from 0 to R to find the total torque T:

Since the polar moment of inertia J = ∫ r² dA = ∫(0 to R) r² (2πr dr) = πR⁴ / 2, the integral simplifies to:

Rearranging the terms yields the final torsion equation:

Given parameters: Mean radius R = 10d, where d is the wire diameter. Energy U = 2.5 kNm = 2500000 Nmm. Extension δ = 100 mm. Maximum shear stress τ = 75 N/mm². Modulus of rigidity G = 0.85 × 10⁵ N/mm².

The energy absorbed by a spring is given by U = (1/2) × W × δ, where W is the axial load. Substituting the values: 2500000 = (1/2) × W × 100, which gives W = 50000 N.

The extension of a close-coiled helical spring is given by δ = (64WR³n) / (Gd⁴), where n is the number of turns. Substituting R = 10d: 100 = (64 × 50000 × (10d)³ × n) / (0.85 × 10⁵ × d⁴). Simplifying this: 100 = (3200000 × 1000d³ × n) / (85000 × d⁴), which leads to 100 = 37647.06 × n / d, or d = 376.47n.

The maximum shear stress in the spring is given by τ = (16WR) / (πd³). Substituting the known values: 75 = (16 × 50000 × 10d) / (πd³). Simplifying: 75 = 8000000 / (πd²), which gives d² = 8000000 / (75π) = 33953.05. Thus, d = 184.26 mm.

Using the relation d = 376.47n, we find the number of turns: n = 184.26 / 376.47 = 0.489 turns. Since n must be a practical value, we typically round up or adjust the design; however, based on the provided constraints, the wire diameter is 184.26 mm and the mean radius R = 1842.6 mm with n ≈ 0.49 turns.