JAWAHARLAL NEHRU TECHNOLOGICAL UNIVERSITY HYDERABAD
JAWAHARLAL NEHRU TECHNOLOGICAL UNIVERSITY HYDERABAD B.Tech II STRENGTH OF MATERIALS I Previous Question Paper (August/September - 2022 - R15)
Answer any five questions
All questions carry equal marks
Main
Derive the relationship between elastic modulus and shear modulus.
Answer
Consider a square element of side length a subjected to pure shear stress τ on its faces. This state of pure shear is equivalent to a state of tensile stress τ and compressive stress τ acting on the planes at 45° to the shear planes. The shear strain γ is given by τ/G, where G is the shear modulus.
The diagonal of the square, which has length a√2, undergoes a change in length δL due to the normal stresses. The strain along the diagonal is:
Also, the shear strain γ is related to the change in the right angle of the element. For small strains, the change in length of the diagonal is δL = (γ/2) * a√2. The strain along the diagonal is also δL / (a√2) = γ/2 = τ/(2G).
Equating the two expressions for strain:
Simplifying this gives the relationship:
A copper rod of 16 mm diameter and 0.9 m long is heated through 50°C. Determine the stress and its nature if (i) the extension is prevented by gripping it at both ends, and (ii) if one of the grips yield by 0.5 mm. Assume E for Copper = 120 GPa and α = 18 × 10-6 /°C.
Answer
Given: Diameter d = 16 mm, Length L = 0.9 m = 900 mm, Temperature change ΔT = 50°C, E = 120 GPa = 120,000 N/mm², α = 18 × 10-6 /°C.
Free expansion δL = α * L * ΔT = 18 × 10-6 * 900 * 50 = 0.81 mm.
(i) If extension is prevented, the thermal strain is fully converted to compressive stress:
(ii) If the grip yields by δyield = 0.5 mm, the net expansion prevented is δnet = δL - δyield = 0.81 - 0.5 = 0.31 mm.
The stress induced is due to this restricted expansion:
A stepped bar, consisting of Brass rod (L = 2 m and A = 500 mm2) is rigidly attached between a steel rod (L = 2.5 m and A = 750 mm2) and a Copper rod (L = 1.5 m and A = 250 mm2), is subjected to axial loads as shown in Figure 1. If the stresses in Steel, Brass and Copper are limited to 150 MPa, 125 MPa and 100 MPa respectively, determine the safe value of P. Also find the corresponding elongation of the bar. Adopt, E for Steel = 200 GPa, E for Brass = 100 GPa and E for Copper = 120 GPa.
Answer
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Draw the shear force and bending moment diagrams for a beam supported and loaded as shown in Figure 2.
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Draw the shear force and bending moment diagrams for a beam supported and loaded as shown in Figure 3.
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A Simply supported steel beam of span 5 m is subjected to two concentrated loads of 50 kN and 75 kN at a distance 3 m and 2 m, respectively from the right support in addition to uniformly distributed load of 30 kN/m over its entire span. The cross-section of the beam is shown in Figure 4. Draw the flexural stress distribution across the section subjected to maximum bending moment.
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A simply supported steel beam of span 6 m carries a uniformly distributed load of 60 kN/m over the entire span, in addition to a concentrated load of 150 kN at its mid-span. The cross-section of the beam is a T-section with flange width 150 mm, flange thickness 20 mm, web thickness 16 mm and the overall depth of the section is 200 mm. Draw the distribution of shear stress across the depth of a section subjected maximum shear force. Also calculate the ratio of maximum shear stress to the mean shear stress.
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The state of stress at a point in a strained member is shown in Figure 5, using the Mohr’s circle of stresses, determine the:
magnitude and the direction of principal stresses
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magnitude of the maximum shear stress
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Stresses acting on a plane θ = 30°.
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A simply supported beam of span 6 m subjected to loading as shown in Figure 6, find the maximum deflection and the slope at the supports.
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