JNTU
JNTU B.Tech II ELEMENTS OF MECHANICAL ENGINEERING Previous Question Paper (October - 2020 - R13)
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Main
A steel bar of 30 mm square in section is subjected to an axial compressive load of 80 kN. Find the percentage change in volume of the bar of 450 mm long. Find modulus of rigidity and bulk modulus if young’s modulus is 200 GPa and poisson’s ratio is 0.3.
Answer
Given data: Side of square section (a) = 30 mm, Axial compressive load (P) = 80 kN = 80,000 N, Length (L) = 450 mm, Young's modulus (E) = 200 GPa = 200,000 N/mm², Poisson's ratio (μ) = 0.3.
1. Calculation of stress (σ):
Area (A) = a² = 30 × 30 = 900 mm².
Stress (σ) = P / A = 80,000 / 900 = 88.89 N/mm² (compressive).
2. Percentage change in volume (dV/V):
Volumetric strain (εv) = (σ / E) × (1 - 2μ).
εv = (88.89 / 200,000) × (1 - 2 × 0.3) = 0.000444 × 0.4 = 0.0001778.
Since the load is compressive, the volume decreases. Percentage change in volume = 0.0001778 × 100 = 0.01778% (decrease).
3. Modulus of rigidity (G):
Formula: G = E / (2 × (1 + μ))
G = 200,000 / (2 × (1 + 0.3)) = 200,000 / 2.6 = 76,923.08 N/mm² or 76.92 GPa.
4. Bulk modulus (K):
Formula: K = E / (3 × (1 - 2μ))
K = 200,000 / (3 × (1 - 2 × 0.3)) = 200,000 / (3 × 0.4) = 200,000 / 1.2 = 166,666.67 N/mm² or 166.67 GPa.
A load of 21 kN rests on a rigid beam and is being taken equally by three vertical streets each of 20 mm diameter and equi spaced in vertical plane. The outer two rods are of steel and central one is copper. An additional load of 30 kN is added midway between rods. Determine stresses in each bar, if the rigid horizontal bar remains horizontal. Neglect the weight of bar. Take Es=205 GPa and Ec=110 GPa.
Answer
Let the three vertical rods be denoted as 1 (steel), 2 (copper), and 3 (steel). Since the beam remains horizontal, the elongation of all three rods must be equal, denoted by δ. The relationship between force P, area A, modulus E, and length L is given by δ = (P × L) / (A × E). Since all rods have the same length L and diameter d = 20 mm, the area A = π/4 × (20)2 = 314.16 mm2.
Equating the deformations:
This implies Ps1 = Ps3 = Ps and Ps / Es = Pc / Ec. Thus, Ps = Pc × (205 / 110) = 1.8636 Pc.
The total load is 21 kN + 30 kN = 51 kN. Summing the vertical forces: 2Ps + Pc = 51,000 N. Substituting the expression for Ps:
The stresses are calculated as σ = P / A:
The stress in the copper rod is 34.34 MPa and the stress in each steel rod is 64.00 MPa.
A horizontal cantilever of 4 m long carries UDL of 2.5 kN/m which extends over the middle 2 m of beam and acts downwards. A concentrated load of 5 kN acts downwards at a point 1m from fixed end. Another concentrated load of 3 kN acts upwards at the free end. Draw BM and SF diagrams.
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A simply supported beam of 10 m span carries UDL of 6 kN/m over the left half of the beam and a point load of 15 kN is placed 2m away from the right support. Draw the BM and SF diagrams.
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A spherical shell of 1 m diameter is subjected to an internal pressure of 2 N/mm2. Taking maximum allowable stress as 100 N/mm2. Find the thickness of plate. Take joint efficiency as 80% and E=200 GPa.
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A cast iron pipe has 20 cm internal diameter and a thickness of 40 mm and carries water under a pressure of 5 N/mm2. Calculate the maximum and minimum intensities of circumferential stress. Sketch the circumferential and radial stresses across the section.
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Sketch and explain the construction and working of four stroke diesel engine.
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Distinguish between S.I and C.I engines.
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Derive an equation to find the length of cross belt.
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What is the importance of initial tension and centrifugal tension in the belt drives.
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