Jawaharlal Nehru Technological University Hyderabad
Jawaharlal Nehru Technological University Hyderabad B.Tech II STRENGTH OF MATERIALS - II Previous Question Paper (July/August - 2021 - R16)
Answer any five questions. All questions carry equal marks.
Main
Derive the torsion equation for a circular shaft:
Answer
Consider a circular shaft of length L, radius R, subjected to a torque T at one end while the other end is fixed. Under the action of torque, a line on the surface of the shaft rotates by an angle γ (shear strain), and the cross-section rotates by an angle θ (angle of twist).
Assumptions: The material is homogeneous, isotropic, and elastic; plane cross-sections remain plane after twisting; and radii remain straight.
1. From the geometry of the deformed shaft, the shear strain γ at a radius r is related to the angle of twist θ by:
2. According to Hooke's Law for shear, the shear stress q at radius r is:
where N is the modulus of rigidity. Rearranging gives:
3. 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. The resisting torque dT provided by this element is:
4. Integrating over the entire cross-section from 0 to R gives the total resisting torque T:
where J = ∫ r2 dA = πR4 / 2 is the polar moment of inertia. Thus, T / J = Nθ / L.
Combining this with the stress relation, we obtain the torsion equation:
A close coiled helical spring is to have a stiffness of 1N/mm of compression under a maximum load of 45N and a maximum shearing stress of 126N/mm2. The solid length of the spring (when the coils are touching) is to be 45mm. Find the diameter of the wire. Take N = 4.2 × 104 N/mm2 (modulus of rigidity).
Answer
Given data:
Stiffness (k) = 1 N/mm
Maximum load (W) = 45 N
Maximum shear stress (τ) = 126 N/mm2
Solid length (Ls) = 45 mm
Modulus of rigidity (N) = 4.2 × 104 N/mm2
Let d be the wire diameter, D be the mean coil diameter, and n be the number of turns.
1. From the formula for solid length: Ls = n × d = 45. Thus, n = 45/d.
2. From the stiffness formula: k = (G × d4) / (8 × D3 × n) = 1. Substituting G = 4.2 × 104 and n = 45/d:
3. From the shear stress formula: τ = (8 × W × D) / (π × d3) = 126.
4. Substitute D into the stiffness equation:
The required diameter of the wire is approximately 3.06 mm.
Find an expression for crippling load for a long column when one end of the column is fixed and other end is free.
Answer
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Find Euler’s critical load for a hollow cylindrical cast iron column 200 mm external diameter and 25 mm thick, if it is 6m long and hinged at both ends. Take E = 8 × 104 N/mm2.
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A masonry dam of trapezoidal section is 12m high with a top width of 2m. The water face has a batter of 1 in 12. Find the minimum bottom width necessary so that tensile stresses are not induced on the base section. Masonry weighs 22500 N/m3 and water weighs 9810 N/m3.
Answer
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A curved beam, semi circular in plan and supported on three equally spaced supports. The beam carries a U.D.L of ‘w’ per unit circular length. Determine the bending moment and twisting moments.
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A thick spherical shell of 125 mm internal radius and 50 mm thick is subjected to an internal pressure of 5 Mpa. Determine the variation of hoop and radial stresses in the shell. Take E = 200 GPa, poissions ratio (υ) = 0.30.
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Explain the concept of un-symmetrical bending. What are the conditions that should be satisfied for a beam to bend without twisting?
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