Jawaharlal Nehru Technological University Hyderabad

Jawaharlal Nehru Technological University Hyderabad B.Tech III Year PRINCIPLES OF MACHINE DESIGN Previous Question Paper (February - 2022 - R13)

B.Tech | III Year | I Semester | R13 | February - 2022

Answer PaperPaper Code: 115BDMaximum Marks: 75Time: 3 hours

Answer any FIVE Questions. All Questions carry Equal Marks.

Main

1

The principal stresses induced at a point in a machine component made of steel 50C4 (Syt = 460 N/mm2) are as follows: Maximum principal stress = 200 N/mm2 and Minimum principal stress = 150 N/mm2. Calculate the factor of safety by:

15 marks
a

Maximum shear stress theory

7 marks

Answer

According to the Maximum Shear Stress Theory (Guest's Theory), the factor of safety (FOS) is calculated by equating the maximum shear stress in the component to the shear yield strength of the material. For a two-dimensional stress state where the principal stresses are σ1 = 200 N/mm2 and σ2 = 150 N/mm2, the maximum shear stress is given by τmax = (σ1 - σ2) / 2 = (200 - 150) / 2 = 25 N/mm2. The shear yield strength is τy = Syt / (2 * FOS). Alternatively, the condition is σ1 - σ2 = Syt / FOS. Substituting the values: 200 - 150 = 460 / FOS. Therefore, FOS = 460 / 50 = 9.2.

b

Distortion energy theory

8 marks

Answer

According to the Distortion Energy Theory (von Mises-Hencky Theory), the factor of safety is determined by the von Mises stress (σ') which must be less than or equal to the yield strength divided by the factor of safety. The formula for von Mises stress in a 2D state is σ' = sqrt(σ12 - σ1σ2 + σ22). Substituting the given values: σ' = sqrt(2002 - (200 * 150) + 1502) = sqrt(40000 - 30000 + 22500) = sqrt(32500) ≈ 180.28 N/mm2. The factor of safety is FOS = Syt / σ' = 460 / 180.28 ≈ 2.55.

215 marks
a

Briefly explain about Stress concentration, Theoretical stress Concentration factor, Fatigue stress concentration factor and notch sensitivity.

7 marks

Answer

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b

Explain Goodman and Soderberg’s methods for combination of stresses.

8 marks

Answer

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315 marks
a

Explain the design of welded joints subjected to axial load.

5 marks

Answer

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b

Two plates 16 mm thick are joined by a double U butt weld and a single U butt weld. Determine the strength of the welded joint in tension in each case. Effective length of weld is 150 mm. Allowable stress in butt weld in tension is 142 N/mm2.

10 marks

Answer

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415 marks
a

Explain the design procedure for sleeve and cotter joint.

10 marks

Answer

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b

What are the types of Keys? Explain them.

5 marks

Answer

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515 marks
a

What are the advantages and disadvantages of V-belt drive over flat belt drives?

5 marks

Answer

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b

A pulley of 0.9m diameter revolving at 200 rpm is to transmit 7.5 kW find the width of a leather belt if the maximum tension is not to exceed 145 N in 10mm width. The tension in the tight side is twice that in the slack side. Determine the diameter of the shaft and the dimensions of the various parts of the pulley, assuming it to have six arms. Maximum shear stress is not to exceed 63 MPa.

10 marks

Answer

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6

Select a wire rope for a vertical mine hoist to lift a load of 55kN from a depth 300 metres. A rope speed of 500 metres/min is to be attained in 10 seconds.

15 marks

Answer

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7

Design a spur gear for transmitting 8kW power pinion speed = 764 rpm; speed ratio = 2; the gears are made up of C45 steel; life = 10000 hrs.

15 marks

Answer

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815 marks
a

Explain radial ball bearing in brief.

7 marks

Answer

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b

A radial ball bearing has a basic load rating of 50 kN. If the desired rating life of the bearing is 6000 hrs. What equivalent radial load can the bearing carry at 500 rev/min.

8 marks

Answer

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About this paper

This page contains the solved question paper with answers of PRINCIPLES OF MACHINE DESIGN from Jawaharlal Nehru Technological University Hyderabad for B.Tech MECT-MCT-MEM III Year I Semester, February - 2022, R13.

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