Jawaharlal Nehru Technological University Hyderabad B.Tech II STRENGTH OF MATERIALS - II Previous Question Paper (November/December - 2020 - R16)

The torsion equation relates the applied torque, the polar moment of inertia, the shear stress, and the angle of twist for a circular shaft. Consider a circular shaft of length L, radius R, subjected to a torque T. Under pure torsion, plane sections remain plane, and radii remain straight.

Let θ be the angle of twist in radians and N be the modulus of rigidity. For a fiber at radius r, the shear strain γ is given by:

From the definition of shear modulus, the shear stress q at radius r is:

Rearranging this gives the relationship:

Consider an elemental ring of radius r and thickness dr. The area of this element is dA = 2πr dr. The force on this element is dF = q dA. The torque dT provided by this element is:

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

The integral term represents the polar moment of inertia J, where J = πR⁴ / 2. Thus:

Rearranging the terms, we obtain the complete torsion equation:

Given: 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 known 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 = (64 × 50000 × 1000 × d³ × n) / (85000 × d⁴), which leads to 100 = (3200000000d³n) / (85000d⁴), resulting in 100 = 37647.06n / d, or d = 376.47n.

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

Using d = 184.26 mm, calculate the mean radius R = 10d = 1842.6 mm. Now, find the number of turns n using the previously derived relation d = 376.47n: n = 184.26 / 376.47 = 0.489 turns. Since n must be an integer for a practical design, we round up to n = 1 turn. Recalculating with n = 1: d = 376.47 mm, R = 3764.7 mm.