JAWAHARLAL NEHRU TECHNOLOGICAL UNIVERSITY HYDERABAD
JAWAHARLAL NEHRU TECHNOLOGICAL UNIVERSITY HYDERABAD B.Tech II Year STRENGTH OF MATERIALS - II Previous Question Paper (November/December - 2020 - R13)
Answer any Five Questions. All Questions Carry Equal Marks.
Main
A hollow circular shaft is subjected to a bending moment of 3000 kNm and a torque of 4500 kNm. Design the shaft according to maximum shear stress theory. Adopt the stress at elastic limit as 240 N/mm2, Poisson’s ratio as 0.3 and the factor of safety as 2.
Answer
To design the hollow circular shaft using the Maximum Shear Stress Theory (Tresca's Theory), we first determine the allowable shear stress. Given the stress at the elastic limit (yield stress) σ_y = 240 N/mm2 and a factor of safety (FOS) = 2, the allowable tensile/compressive stress is σ_all = σ_y / FOS = 120 N/mm2. According to the Maximum Shear Stress Theory, the maximum allowable shear stress τ_max is σ_y / (2 × FOS) = 60 N/mm2.
The equivalent twisting moment T_e is calculated using the formula:
Substituting the given values M = 3000 kNm = 3 × 106 Nm and T = 4500 kNm = 4.5 × 106 Nm:
The relationship between the equivalent torque and the shaft dimensions for a hollow shaft is:
Assuming a ratio of inner diameter to outer diameter k = d/D = 0.5, we substitute d = 0.5D into the equation:
Solving for D:
Thus, the inner diameter d = 0.5 × 788 = 394 mm. The shaft should be designed with an outer diameter of 788 mm and an inner diameter of 394 mm.
An open coiled helical spring is made of 16 mm diameter steel rod coiled to a mean diameter of 150 mm. If the number of coils is 20 and the angle of helix is 12°, calculate the deflection under an axial load of 900 N and also determine the maximum bending and shear stresses induced at any section of the wire.
Answer
Given data: wire diameter d = 16 mm, mean coil diameter D = 150 mm, number of coils n = 20, helix angle α = 12°, axial load W = 900 N. Assuming modulus of rigidity G = 84 GPa and Young's modulus E = 210 GPa.
The mean radius R = D / 2 = 75 mm. The length of the wire L = (πDn) / cos(α) = (π × 150 × 20) / cos(12°) = 9646.6 mm.
The deflection δ is given by:
Substituting the values:
The bending moment M = WR cos α = 900 × 75 × cos 12° = 65925 N-mm. The twisting moment T = WR sin α = 900 × 75 × sin 12° = 14013 N-mm.
Maximum bending stress σ = (32M) / (πd³) = (32 × 65925) / (π × 16³) = 82.25 N/mm².
Maximum shear stress τ = (16T) / (πd³) = (16 × 14013) / (π × 16³) = 8.74 N/mm².
The maximum principal stress σ_max = (σ / 2) + sqrt((σ / 2)² + τ²) = 41.125 + sqrt(41.125² + 8.74²) = 83.08 N/mm².
A hinged-hinged column of length 6 m has a square hollow section with outside dimensions 150 mm × 150 mm. The column is subjected to an axial compressive load of 450 kN. If the working load is not to exceed 50% of the crippling load or crushing load whichever is less, find the thickness of the column section. Adopt E = 2×105 N/mm2 and the maximum stress in crushing failure is 250 N/mm2 and α = 1/7500.
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A beam, semi-circular in plan of radius equal to 6 m, is simply supported on three equally spaced supports. The beam is subjected to uniformly distributed load of 25 kN/m of curved length. Draw the twisting moment diagram.
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A steel fixed beam-column of I-section, 150 mm × 200 mm × 10 mm, is 4.2 m long. It is subjected to an axial load of 250 kN and a vertical lateral uniformly distributed load of 20 kN/m over the entire span. Determine the maximum bending moment.
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A square column of size 500 mm × 500 mm is subjected to a compressive load of 1500 kN at an eccentricity of 100 mm and 150 mm with respect to the centroidal axes as shown in Figure. Determine the stresses at the four corners of the column section.
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A cantilever beam, span 2 m, consists of an I-section 100 mm × 150 mm × 12 mm is subjected to a concentrated load of 60 kN at its free end. The line of action of the load is passing through the centroid and making an angle 30° with respect to vertical. Determine the maximum stress.
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A steel cylindrical shell has wall thickness 12 mm, internal diameter 1.75 m and length 4.2 m is carrying a fluid at a pressure of 4 N/mm2. Find the change in diameter, length and volume of the cylinder. Assume E = 2×105 N/mm2 and Poisson’s ratio = 0.25.
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