Question Paper: Strength of Materials : Question Paper May 2016 - Mechanical Engineering (Semester 3) | Mumbai University (MU)
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## Strength of Materials - May 2016

### Mechanical Engineering (Semester 3)

TOTAL MARKS: 80
TOTAL TIME: 3 HOURS
(1) Question 1 is compulsory.
(2) Attempt any three from the remaining questions.
(3) Assume data if required.
(4) Figures to the right indicate full marks.

### Answer any four from the following

1(a) A uniformly tapering rod of length l and diameters d1 and d2 is subjected to an axial pull P. Prove that the total extension of the rod is $$\delta l=\frac{4Pl}{\pi E d_1 d_2}$$(5 marks) 1(b) Write the assumptions in simple bending and hence derive the bending formula $$\dfrac{M}{I}=\dfrac{\sigma }{y}=\dfrac{E}{R}$$(5 marks) 1(c) Draw the shear force and bending moment diagram for a simply supported beam of length l, subtecded to a clockwise couple M at the centre of the beam.(5 marks) 1(d) Derives an expression for the maximum and the minimum stresses at the base of a column of rectangular section, when it is subjected to a load which is eccentric to both axes.(5 marks) 1(e) In a hollow circular shaft, the outer and inner diameters are 200 mm and 100 mm respectively. If the shear is not ot exceed 40 N/mm2, find the maximum torque, which the safely can safely transmit.(5 marks) 1(f) What is equivalent length of a column? Give the ratio of equivalent length and actual length of columns, with various end conditions. Also write the expression for crippling liad, P for various end conditions.(5 marks) 2(a) Draw axial force, shear force and bending moment diagrams for the beam loaded as shown in the figure. Also, determine the points of contra-flexure and maximum moments. (10 marks) 2(b) A point load of 10 kN applied to a simply supported beam at mid-span, produces a deflection of 6 mm and a maximum bending stress of 20 N/mm2. Calculate the maximum value of the momentary stress produced, when a weight of 5kN is allowed to fall through a height of 18 mm on the beam at the middle of the span.(10 marks) 3(a) A short column od rectangular cross section 80mm × 60mm carries a load of 40kN at a point 20mm from longer side and 35mm from the shorter side. Determine the maximum compression and tensile stresses in the section.(10 marks) 3(b) Find the Euler Crushing load for a hollow cylindrical cast iron column 200mm external diameter and 25mm thick, if it is 6m long and hinged at both ends.
Take, E = 1.2 × 106 N/mm2
Compare the load with the crushing load as given by Rankine formula, taking fc=550 N/mm2 and α;=1/1600.
(10 marks)
4(a) Determine the position and the amount of maximum deflection for the beam shown in the figure. Take, EI = 1.8 × 104 kNm2. (10 marks)
4(b) A hollow shaft of external diameter 120 mm transmits 300 kW power at 200 rpm. Determine the maximum internal diameter, if maximum stress in the shaft is not to exceed 60 N/mm2(10 marks) 5(a) A closed cylindrical verses, made of steel plates 4 mm thick with plane ends carries fluid under a pressure of 3 N/mm2. The diameter of the cylinder os 250mm and length is 750mm. Calculate the longitudinal and hoop stresses in the cylinder wall and determine changes in diameter, length and volume of the cylinder.
Take, if E = 2.1 × 105 N/mm2, 1/m = 0.286
(10 marks)
5(b) A simply supported beam, with a span of 1.3 m and a rectangular cross section of 150 mm wide and 250 mm deep, carries a concentrated load of W at the centre. If the allowable stresses are 7 N/mm2 for bending and 1 N/mm2 for shear, what is the value of the safe load W? Also draw bending stress and shear stress distribution diagrams(10 marks) 6(a) Find the principle moments of inertia and directions of principle axes for the angle section shown below. (10 marks)
6(b) A flitched beam consists of a wooden joint 100mm wide and 200mm deep strengthened by two plates 10mm thick and 200mm deep as shown in figure. If the maximum stress in the wooden joint is 7 N/mm2, find the corresponding maximum attained in steel. Find also moment of resistance of composite section.
Take, Es = 2 × 105 N/mm2 and Ew = 1 × 104 N/mm2. (10 marks)

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 written 2.3 years ago by Team Ques10 ♦♦ 410
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