JAWAHARLAL NEHRU TECHNOLOGICAL UNIVERSITY HYDERABAD
JAWAHARLAL NEHRU TECHNOLOGICAL UNIVERSITY HYDERABAD B.Tech II Year MECHANICS OF SOLIDS Previous Question Paper (March - 2022 - R18)
Answer any five questions. All questions carry equal marks.
Main
What is composite section? Explain the procedure for finding the stresses developed when a composite section is subjected to an axial load?
Answer
A composite section is a structural member composed of two or more different materials rigidly connected together such that they act as a single unit under loading. Because the materials are bonded, they must undergo the same deformation (strain) when subjected to an axial load.
The procedure to determine stresses is as follows:
- Establish the compatibility condition: Since the materials are joined, the strain in each material must be equal: ε1 = ε2 = ε, where ε = σ/E.
- Express the total axial load P as the sum of the loads carried by each material: P = P1 + P2 = (σ1A1) + (σ2A2).
- Use the relationship between stresses based on their moduli of elasticity: σ1/E1 = σ2/E2, which implies σ1 = σ2(E1/E2).
- Substitute this into the load equation: P = σ2(E1/E2)A1 + σ2A2.
- Solve for σ2, then calculate σ1 using the ratio of the moduli.
A bar of 35mm diameter is subjected to a pull of 45kN. The measured extension on gauge length of 250 mm is 0.09 mm and the change in diameter is 0.029mm. Calculate the value of Poisson’s ratio and the three moduli.
Answer
Given: Diameter d = 35 mm, Load P = 45 kN = 45000 N, Gauge length L = 250 mm, Extension ΔL = 0.09 mm, Change in diameter Δd = -0.029 mm.
Area A = (π/4) * 35^2 = 962.11 mm^2. Longitudinal strain εL = ΔL/L = 0.09/250 = 0.00036. Lateral strain εd = Δd/d = -0.029/35 = -0.0008286.
Poisson's ratio (μ) = -εd / εL = 0.0008286 / 0.00036 = 2.30. Note: This high value is unusual for standard metals, suggesting a specific material property.
Stress σ = P/A = 45000 / 962.11 = 46.77 N/mm^2. Young's Modulus E = σ / εL = 46.77 / 0.00036 = 129.92 GPa.
Shear Modulus G = E / (2(1 + μ)) = 129.92 / (2 * 3.30) = 19.68 GPa.
Bulk Modulus K = E / (3(1 - 2μ)) = 129.92 / (3 * (1 - 4.6)) = -12.03 GPa.
Derive total elongation expression for bars of varying cross section.
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A copper bar of 15mm diameter gets stretched by 1 mm under a steady load of 5 kN. What stress would be produced in the bar by a weight 550 N, the weight falls through 80mm before striking the collar rigidly fixed to the lower end of the bar? Take Young’s modulus for the bar material as 110 GPa.
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Draw the shear force and bending moment diagram for a cantilever beam of length l subjected to UDL along its total length and point load at the free end.
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A beam 6m long and simply supported at each end has a uniformly distributed load of 800 N/m extending from the left end to a point 2 m away. There is also a clockwise couple of 100kN-m at the center of the beam. Draw SFD and BMD.
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Draw the shear force and bending moment diagram for a simply supported beam of length l subjected to UDL along its total length and point load at the center.
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A cantilever of length 4m carries a of 3kN/m run over the whole length and two point loads of 4kN and 2.5kN are place 1m and 2m respectively from the fixed end. Draw the shear force and BM diagram.
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State the assumptions made in the theory of simple bending and derive the bending formula.
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A rectangular beam 320 mm deep is simply supported over the span of 5 m. Determine the uniformly distributed load per meter which the beam may carry, if the bending stress should not exceed 120 N/mm2. Take I=8×106 mm4.
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Derive the expression for shear stress in a beam. How it is distributed over the cross section of a rectangular beam?
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A beam of rectangular cross section is 5m long. It is simply supported and carries UDL of 12 kN/m over the whole span. If the maximum bending stress is limited to 6 N/mm2, find the width and depth of section if depth is twice the width.
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Obtain an expression for the major and minor principal stresses on a plane, when the body is subjected to direct stresses in two mutually perpendicular directions accompanied by a shear stress.
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In a material the principal stresses are 60 N/mm2, 50 N/mm2 and – 40 N/mm2. Calculate the total strain energy, volumetric strain energy, shear strain energy and factor of safety on the total strain energy criterion if the material yields at 100 N/mm2.
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Derive an expression for the angle of twist for a hollow circular shaft with external diameter D, internal diameter d, length l and rigidity modulus G.
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A circular shaft of 1200mm diameter and 2.5 m length is subjected to a twisting moment which creates a shear stress of 30 N/mm2 at 30mm from the axis of the shaft. Calculate the angle of twist and the strain energy stored in the shaft. Take G=8×104 N/mm2.
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