Jawaharlal Nehru Technological University Hyderabad

Jawaharlal Nehru Technological University Hyderabad B.Tech II MECHANICS OF SOLIDS Previous Question Paper (October - 2020 - R13)

B.Tech | II | I | R13 | October - 2020

Answer PaperPaper Code: 113ACMaximum Marks: 75Time: 2 hours

Answer any five questions. All questions carry equal marks.

Main

1

A vertical bar 3.5 meter long and 1500 mm2 cross sectional area is fixed at the upper ends and has a collar at the lower end. Determine the maximum stress induced when a weight of:

a) 2500 N falls through a height of 15 cm on the collar

b) 25 kN falls through a height of 3 cm on the collar. Take E = 2.0 × 105 N/mm2.

15 marks

Answer

The maximum stress induced in a bar due to an impact load is given by the formula:

σ = (P/A) * [1 + √(1 + (2AEh / PL))]

Given parameters: Length (L) = 3500 mm, Area (A) = 1500 mm2, Modulus of Elasticity (E) = 2.0 × 105 N/mm2.

a) Weight (P) = 2500 N, Height (h) = 150 mm:

  1. Calculate the static stress (σs) = P/A = 2500 / 1500 = 1.667 N/mm2.
  2. Calculate the term (2AEh / PL) = (2 * 1500 * 200000 * 150) / (2500 * 3500) = 90,000,000 / 8,750,000 = 10.286.
  3. Maximum stress (σ) = 1.667 * [1 + √(1 + 10.286)] = 1.667 * [1 + 3.359] = 7.266 N/mm2.

b) Weight (P) = 25,000 N, Height (h) = 30 mm:

  1. Calculate the static stress (σs) = P/A = 25000 / 1500 = 16.667 N/mm2.
  2. Calculate the term (2AEh / PL) = (2 * 1500 * 200000 * 30) / (25000 * 3500) = 18,000,000 / 87,500,000 = 0.2057.
  3. Maximum stress (σ) = 16.667 * [1 + √(1 + 0.2057)] = 16.667 * [1 + 1.098] = 34.967 N/mm2.
2

A rectangular plate made of steel is 4 m long and 20 mm thick and is subjected to an axial tensile load of 40 kN. The width of the plate varies from 30 mm at one end to 80 mm at the other end. Find the elongation, if E = 2 × 105 N/mm2.

15 marks

Answer

To find the total elongation of a rectangular plate with varying width, we consider a small elemental strip of length dx at a distance x from the narrower end. Given: length L = 4000 mm, thickness t = 20 mm, load P = 40000 N, E = 200000 N/mm2, width at x=0 (b1) = 30 mm, and width at x=L (b2) = 80 mm.

The width bx at any distance x is given by the linear relation:

bx = b1 + (b2 - b1) * (x / L) = 30 + (80 - 30) * (x / 4000) = 30 + 0.0125x

The elongation dδ of the elemental strip is:

dδ = (P * dx) / (E * Ax) = (P * dx) / (E * bx * t)

Integrating from x = 0 to L:

δ = ∫ (P / (E * t)) * (dx / (30 + 0.0125x))
δ = (P / (E * t * 0.0125)) * [ln(30 + 0.0125x)]04000
δ = (40000 / (200000 * 20 * 0.0125)) * ln(80 / 30)
δ = (40000 / 50000) * ln(2.667) = 0.8 * 0.9808 = 0.7846 mm

The total elongation of the plate is 0.7846 mm.

3

A cantilever of length 7m carries a gradually varying load, zero at the free end to 5 kN/m at the fixed end. Draw the S.F and B.M diagrams for the cantilever.

15 marks

Answer

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4

Draw Shear force and bending moment diagram. Locate the position of maximum bending moment in the beam.

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Answer

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5

Derive the equation:

M/I = σ/y = E/R

Where M = B.M, I = Plane M.I, σ = bending stress, y = distance of a layer from N.A, E = elastic modulus and R = radius of curvature.

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6

For a T–section with dimensions flange width 100 mm, depth = 200 mm and uniform thickness of 40 mm, obtain shear stress distribution and calculate maximum and average shear stresses if it is subjected to a S.F. = 100 kN.

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7

An element is subjected to tensile stresses of 60 N/mm2 and 20 N/mm2 acting on two perpendicular planes and is also accompanied by shear stress of 20 N/mm2 on these planes. Draw the Mohr’s circle of stresses and determine the magnitudes and directions of principal stresses and also the greatest shear stress.

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8

A cylindrical shell is 3m long, having 1cm internal diameter and 1.5mm thickness. Calculate the maximum intensity of the shear stress and also the change in the dimensions of the shell if it is subjected to an internal fluid pressure of 1.5 N/mm2.

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This page contains the question paper of MECHANICS OF SOLIDS from Jawaharlal Nehru Technological University Hyderabad for B.Tech AE II I, October - 2020, R13.

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