JAWAHARLAL NEHRU TECHNOLOGICAL UNIVERSITY HYDERABAD

JAWAHARLAL NEHRU TECHNOLOGICAL UNIVERSITY HYDERABAD B.Tech II STRENGTH OF MATERIALS II Previous Question Paper (May - 2019 - R13)

B.Tech | II | II | R13 | May - 2019

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

Note: This question paper contains two parts A and B. Part A is compulsory which carries 25 marks. Answer all questions in Part A. Part B consists of 5 Units. Answer any one full question from each unit. Each question carries 10 marks and may have a, b, c as sub questions.

Part A

125 marks

State the assumptions made in the theory of pure torsion.

2 marks

Answer

The assumptions made in the theory of pure torsion are:

The material of the shaft is homogeneous and isotropic.
The shaft is subjected to pure torsion, meaning the twisting moment is applied at the ends and no shear force is present.
The shaft is circular in cross-section (solid or hollow).
Plane sections perpendicular to the axis of the shaft remain plane after twisting.
The cross-sections rotate as a rigid whole about the axis of the shaft.
The stress distribution is uniform across the cross-section.
The shear stress is proportional to the radius.

Determine the energy stored in a close coiled helical spring of mean diameter 120 mm, the diameter of the spring wire is 8 mm and the number of coils in the spring is 12 subjected to an axial load of 10 kN. Take the modulus of rigidity is 0.8 × 10^5 N/mm^2.

3 marks

Answer

Given:

Mean diameter of the spring, D = 120 mm

Diameter of the spring wire, d = 8 mm

Number of coils, n = 12

Axial load, W = 10 kN = 10 × 10³ N

Modulus of rigidity, G = 0.8 × 10⁵ N/mm²

The energy stored in a close coiled helical spring is given by the formula:

U = (1/2) * W * δ

where δ is the deflection of the spring. The deflection of a close coiled helical spring is given by:

δ = (8 * W * D³ * n) / (G * d⁴)

Substituting the given values:

δ = (8 * 10 × 10³ N * (120 mm)³ * 12) / (0.8 × 10⁵ N/mm² * (8 mm)⁴)

δ = (8 * 10000 * 1728000 * 12) / (80000 * 4096)

δ = 16588800000000 / 3276800000

δ = 50625 mm

Now, calculate the energy stored:

U = (1/2) * W * δ

U = (1/2) * (10 × 10³ N) * 50625 mm

U = 5000 * 50625

U = 253125000 N-mm

Therefore, the energy stored in the spring is 253125000 N-mm or 253.125 J.

What is the significance of Rankine’s empirical formula?

2 marks

Answer

What are the limitations of Euler’s column theory?

3 marks

Answer

Define core of a section and draw the core of a rectangular section ‘B×D’.

2 marks

Answer

What are the conditions to be checked for the stability of a dam?

3 marks

Answer

Indicate the location of shear center of a symmetrical channel section.

2 marks

Answer

Define the principal axes of a cross-section.

3 marks

Answer

What is the necessity of shrinkage allowance in the process of fabrication of compound cylinders?

2 marks

Answer

A thin cylindrical shell of internal diameter 1250 mm and 10 mm thick is subjected to an internal fluid pressure of 5 N/mm^2. Determine the maximum shear stress.

3 marks

Answer

Part B

Design a hollow steel shaft of length 3 m to transmit 120 kW at 300 rpm. The shear stress is to be limited to 75 N/mm^2 and the angle of twist in not to exceed 0.5° per meter length of shaft. Assume the internal diameter of the shaft is half the external diameter. Use the modulus of rigidity is 0.85 × 10^5 N/mm^2.

10 marks

Answer

A solid circular shaft is subjected to a bending moment of 300 kNm in addition to a twisting moment of 125 kNm. If the yield stress of the shaft material is 325 N/mm^2, design the shaft according to the (a) maximum shearing stress and (b) maximum principal stress theories of failures.

10 marks

Answer

A cast iron column of hollow circular section has outside diameter 300 mm, thickness 22 mm and length 6.6 m is fixed at both ends. Determine the safe load carrying capacity of the column with a factor of safety of 3. The Rankine’s constants are f_c = 550 MPa and α = 1/1600. Also find the Euler’s buckling load. Adopt E = 1 × 10^5 N/mm^2.

10 marks

Answer

A curved beam semi-circular in plan has 5 m radius and supported on three equally spaced supports. The beam carries a uniformly distributed load of 50 kN/m of the circular length. Analyse the beam and draw the twisting moment diagram.

10 marks

Answer

A steel hinged-hinged I- strut, 50 mm × 100 mm × 6 mm, of length 3 m is subjected to an axial load of 60 kN in addition to a transverse uniformly distributed load 20 kN/m length over its entire span. Determine the bending moment.

10 marks

Answer

A 50 m high RC chimney is subjected to the horizontal wind pressure of 3 kN/m^2. The external diameter of the chimney tapers from 4.5 m at the base to 2.5 m at the top. The internal diameter at the base is 3 m. draw the stress distribution across the base-section.

10 marks

Answer

A simply supported steel beam of span 3.6 m has I-section 125 mm × 200 mm × 10 mm is subjected to equal and opposite moments of 10 kNm acting at the ends of the beam in a plane making an angle 30° with respect to vertical and passing through the centroid of the section. Determine the maximum stresses induced at any cross-section of the beam.

10 marks

Answer

A simply supported steel beam of span 3 m has an unequal angle section 90 mm × 60 mm × 12 mm with the longer leg vertically downward. The beam is subjected to uniformly distributed load of 5 kN/m over its entire span in a vertical plane passing through the centroid of the cross-section. Determine the maximum deflection.

10 marks

Answer

1010 marks

Derive the expressions for the longitudinal and circumferential stresses for a thin cylinder of length ‘L’, internal diameter ‘d’ and thickness ‘t’ subjected to an internal fluid pressure of ‘p’.

5 marks

Answer

A thin steel spherical shell has 750 mm internal diameter and 12 mm thickness. Find the change in the diameter and volume of the spherical shell if it is subjected to an internal fluid pressure of 5 N/mm^2.

5 marks

Answer

A steel compound cylinder is composed of a tube of 300 mm external diameter and 250 mm internal diameter shrunk on to a tube of 250 mm external diameter and 200 mm internal diameter. The radial pressure at the junction of the two tubes is 20 N/mm^2. Draw the variation of hoop stress across the wall thickness of the compound cylinder if it is subjected to an internal fluid pressure of 100 N/mm^2.

10 marks

JAWAHARLAL NEHRU TECHNOLOGICAL UNIVERSITY HYDERABAD B.Tech II STRENGTH OF MATERIALS II Previous Question Paper (May - 2019 - R13)

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Part B

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