JAWAHARLAL NEHRU TECHNOLOGICAL UNIVERSITY HYDERABAD
JAWAHARLAL NEHRU TECHNOLOGICAL UNIVERSITY HYDERABAD B.Tech II MECHANICS OF SOLIDS Previous Question Paper (November/December - 2020 - R15)
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Main
Show that volumetric strain in a bar is equal to the sum of strains in three mutually perpendicular directions.
Answer
Consider a rectangular element of a bar with dimensions dx, dy, and dz along the x, y, and z axes respectively. When subjected to normal stresses, the dimensions change to dx', dy', and dz', where dx' = dx(1 + εx), dy' = dy(1 + εy), and dz' = dz(1 + εz), with εx, εy, and εz representing the linear strains in the respective directions.
The initial volume of the element is V = dx · dy · dz. The final volume V' after deformation is:
Expanding the product and neglecting the higher-order terms of strains (εxεy, εyεz, εzεx, and εxεyεz) as they are infinitesimally small, the expression simplifies to:
The volumetric strain ev is defined as the change in volume per unit original volume:
Thus, the volumetric strain of a bar is equal to the algebraic sum of the linear strains in three mutually perpendicular directions.
The temperature of a steel ring is raised through 150°C in order to fit it on a wooden wheel of 1.2 m diameter. Find the original diameter of the steel ring and also the stresses developed in the ring, when it cools back to normal temperature. Assume Es = 2 × 105 N/mm2 and αs = 12× 10-6/°C.
Answer
Given data:
Temperature rise (ΔT) = 150°C
Diameter of wooden wheel (D) = 1.2 m = 1200 mm
Modulus of elasticity (Es) = 2 × 105 N/mm2
Coefficient of thermal expansion (αs) = 12 × 10-6 /°C
Let d be the original diameter of the steel ring. When heated, the ring expands to the diameter of the wooden wheel (D). The relationship is given by:
Substituting the values:
The original diameter of the steel ring is 1197.84 mm.
When the ring cools back to normal temperature, it is prevented from shrinking to its original diameter by the wooden wheel. The strain (ε) developed in the ring is equal to the thermal strain that would have occurred due to cooling:
The stress (σ) developed in the ring is calculated using Hooke's Law:
The stress developed in the ring upon cooling is 360 N/mm2 (tensile stress).
Draw the shear force and bending moment diagram for the given beam. Find the values of maximum shear force and maximum bending moment. (Figure 1)
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Draw the shear force and bending moment diagram for the cantilever shown in Figure 2.
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What are the assumptions made in simple theory of bending? Derive relation between bending moment and bending stress.
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The below figure 3 shows the cross-section of a cantilever beam of 2.5 m span. Material used is steel for which maximum permissible stress is 150 N/mm2. What is the maximum uniformly distributed load this beam can carry? (Figure 3)
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Derive the expressions for principal planes, principal stresses and maximum shear stress.
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A thin cylinder 75 mm internal diameter, 250 mm long with walls 2.5 mm thick is subjected to an internal pressure of 7 MN/m2. Determine the change in internal diameter and the change in length. E = 200 GN/m2, ν = 0.3.
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