JAWAHARLAL NEHRU TECHNOLOGICAL UNIVERSITY HYDERABAD
JAWAHARLAL NEHRU TECHNOLOGICAL UNIVERSITY HYDERABAD B.Tech II MECHANICS OF SOLIDS Previous Question Paper (March - 2021 - R13)
Answer any five questions. All questions carry equal marks.
Main
Explain about different types of stresses and strains.
Answer
Stresses are internal resisting forces per unit area developed in a body when subjected to external loads. The primary types are: 1) Normal Stress (Tensile or Compressive), acting perpendicular to the cross-section. 2) Shear Stress, acting tangentially to the cross-section. 3) Volumetric Stress, occurring when a body is subjected to uniform pressure from all sides.
Strains represent the deformation per unit length. The primary types are: 1) Linear Strain, defined as the change in length divided by the original length. 2) Lateral Strain, the strain in a direction perpendicular to the applied load. 3) Shear Strain, the change in the angle between two lines originally perpendicular to each other. 4) Volumetric Strain, the ratio of change in volume to the original volume.
The temperature of a steel ring is raised through 1500°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: Diameter of wheel (D) = 1.2 m = 1200 mm, Temperature rise (ΔT) = 150 °C (assuming 150°C as 1500°C is physically unrealistic for steel), E = 2 × 105 N/mm2, α = 12 × 10-6/°C.
1. Original Diameter (d): The ring must expand to 1200 mm at the elevated temperature. Using the thermal expansion formula:
Substituting values: 1200 = d(1 + 12 × 10-6 × 150). Solving for d:
2. Thermal Stress (σ): When the ring cools, it is constrained by the wheel, preventing contraction. The thermal strain (ε) is αΔT.
The stress developed is:
The original diameter of the ring is 1197.84 mm, and the induced tensile stress upon cooling is 360 N/mm2.
A bar of length ‘l’ is supported at A and B which are at distances ‘a’ from the ends. Find the distance ‘a’ such that maximum moment is least. Using the above results find the most economical length of railway sleeper if the rails are 1.6 m apart.
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Briefly explain Point of contra flexure.
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Derive relation between S.F., B.M and rate of loading at a section of a beam.
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Derive the expression for finding shear stress in a beam in the form
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State of stress at a point in a material is as shown in the figure. Determine: a) principal stresses b) maximum shear stress c) plane of maximum shear stress and d) the resultant stress on the plane of maximum shear stress.
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A circular log of timber has diameter D. Find the dimensions of the strongest rectangular section one can cut from this.
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A hollow shaft, having an internal diameter 40% of its external diameter, transmits a power of 500 KW at 100 R.P.M. Determine the external diameter of the shaft if the shear stress is not to exceed 60 N/mm2 and the twist in a length of 3 m should not exceed 1.5 degrees. Assume Maximum Torque is equal to 1.25 times the Mean torque and Modulus of Rigidity = 8×104 N/mm2.
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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. If, in addition to the internal pressure, the cylinder is subjected to a torque of 200 N m, find the magnitude and nature of the principal stresses set up in the cylinder. E = 200 GN/m2, v = 0.3.
Answer
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