JAWAHARLAL NEHRU TECHNOLOGICAL UNIVERSITY HYDERABAD

JAWAHARLAL NEHRU TECHNOLOGICAL UNIVERSITY HYDERABAD B.Tech II STRENGTH OF MATERIALS II Previous Question Paper (July/August - 2021 - R13)

B.Tech | II | II | R13 | July/August - 2021

Answer PaperPaper Code: 114DPMaximum Marks: 75Time: 3 hours

Answer any five questions. All questions carry equal marks.

Main

1

Design a solid shaft required to transmit 250 kW power at 150 rpm. The maximum torque is not to exceed the mean by 30% and the shear stress is limited to 75 N/mm2. What percentage saving in weight would be obtained if the solid shaft were replaced by a hollow shaft with internal diameter is 0.8 times the external diameter, the length, material and the maximum shear stress being the same?

15 marks

Answer

First, calculate the mean torque (Tmean) transmitted by the shaft using the power formula:

P = (2πNTmean) / 60000

Substituting P = 250 kW and N = 150 rpm:

250 = (2 × π × 150 × Tmean) / 60000
Tmean = 15915.5 N·m = 15.9155 × 106 N·mm

The maximum torque (Tmax) is 30% above the mean:

Tmax = 1.3 × Tmean = 1.3 × 15.9155 × 106 = 20.69 × 106 N·mm

For a solid shaft, the torque equation is:

Tmax = (π / 16) × τ × d3

Substituting τ = 75 N/mm2:

20.69 × 106 = (π / 16) × 75 × d3
d3 = 1404886.5
d = 112.01 mm

For a hollow shaft with di = 0.8do, the torque equation is:

Tmax = (π / 16) × τ × (do4 - di4) / do
20.69 × 106 = (π / 16) × 75 × (do4 - (0.8do)4) / do
20.69 × 106 = 14.726 × do3 × (1 - 0.4096)
do3 = 2381273.6
do = 133.57 mm, di = 106.86 mm

The weight of the shaft is proportional to the cross-sectional area (A). Percentage saving in weight is calculated as:

% Saving = ((Asolid - Ahollow) / Asolid) × 100
Asolid = π/4 × (112.01)2 = 9853.5 mm2
Ahollow = π/4 × (133.572 - 106.862) = 5045.2 mm2
% Saving = ((9853.5 - 5045.2) / 9853.5) × 100 = 48.8%
2

Derive the expression for the deformation of a close-coiled helical spring subjected to an axial pull.

15 marks

Answer

A close-coiled helical spring is defined by a small helix angle (typically less than 10 degrees), which allows us to assume that the coils lie in planes perpendicular to the spring axis. When subjected to an axial load W, the wire is primarily subjected to torsion.

Let D be the mean coil diameter, d be the wire diameter, n be the number of turns, C be the modulus of rigidity, and R be the mean radius (D/2).

The length of the wire L is given by:

L = π D n

The torque T acting on the wire is:

T = W × (D / 2)

From the torsion equation, the shear stress τ and the angle of twist θ are related to the torque. The strain energy U stored in the spring due to torsion is:

U = T² L / (2 G J)

Where J is the polar moment of inertia for a solid circular section:

J = (π d⁴) / 32

Substituting T, L, and J into the strain energy equation:

U = (W² D² / 4) × (π D n) / (2 G × (π d⁴ / 32))
U = (4 W² D³ n) / (G d⁴)

According to Castigliano's theorem, the deflection δ is the partial derivative of the strain energy with respect to the axial load W:

δ = ∂U / ∂W = ∂ / ∂W [ (4 W² D³ n) / (G d⁴) ]

Differentiating with respect to W yields the final expression for deformation:

δ = (8 W D³ n) / (G d⁴)
3

Derive the Euler’s critical load for a column (L, A, I and E) with one end fixed and the other end is free.

15 marks

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4

A semi-circular beam of radius 5 m and uniform cross-section is supported on three symmetrically placed columns. The beam is subjected to a uniformly distributed load of intensity 25 kN/m. Analyse the beam and draw the bending moment diagram.

15 marks

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5

A 3 m long steel hinged-hinged tubular strut with external and internal diameters of 150 mm and 125 mm respectively, is subjected to an axial compressive load of 75 kN and a transverse load 25 kN at its mid-span. Determine the maximum bending moment and stresses.

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6

A masonry dam of trapezoidal section, 6.6 m high, has a vertical water face and retains water to a depth of 6 m. The width of the dam at the top is 1 m and 4 m at the base. Determine the maximum and minimum stresses intensities at the base. The unit weight of masonry is 20 kN/m3.

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7

A beam of rectangular cross-section, 75 mm × 150 mm, is subjected to a bending moment of 20 kNm in a plane making an angle 450 (Anti-Clockwise) with respect to a vertical axis passing through the centroid of the section. Determine the neutral axis of the section and also calculate the maximum bending stress induced in the section.

15 marks

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8

A cylinder has 1.8 m diameter, 12.5 mm wall thickness and 3 m long containing a fluid at a pressure of 3.5 N/mm2. Determine the circumferential and longitudinal stresses due to the fluid pressure.

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This page contains the question paper of STRENGTH OF MATERIALS II from JAWAHARLAL NEHRU TECHNOLOGICAL UNIVERSITY HYDERABAD for B.Tech CE II II, July/August - 2021, R13.

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