JAWAHARLAL NEHRU TECHNOLOGICAL UNIVERSITY HYDERABAD

JAWAHARLAL NEHRU TECHNOLOGICAL UNIVERSITY HYDERABAD B.Tech III DESIGN OF MACHINE MEMBERS - I Previous Question Paper (May - 2018 - R13)

B.Tech | III | III-I | R13 | May - 2018

Answer PaperPaper Code: 115DVMaximum 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. Illustrate your answers with NEAT sketches wherever necessary.

Part A

125 marks
a

Define ‘Endurance limit’.

2 marks

Answer

The endurance limit (or fatigue limit) is the maximum stress amplitude that a material can withstand for an infinite number of stress cycles without failure.

b

What are the advantages of hole – basis system over shaft – basis system?

3 marks

Answer

The hole-basis system is preferred because it is more economical and practical. Its primary advantages include:

  • It allows the use of standard drills, reamers, and broaches to produce the hole, which are fixed-size tools.
  • It reduces the number of expensive internal cutting tools required.
  • Different types of fits can be achieved by simply varying the size of the shaft, which is easier to machine than the hole.
c

What is ‘caulking’? What is its objective?

2 marks

Answer

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d

What is the cause of residual stresses in welded joints? How are they relieved?

3 marks

Answer

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e

Where do you use a Knuckle joint? Give practical examples.

2 marks

Answer

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f

Sketch the following types of keys: Taper sunk key, Feather key, Woodruff key.

3 marks

Answer

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g

Define equivalent torsional moment and equivalent bending moment for a shaft.

2 marks

Answer

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h

Give at least three practical applications of Couplings.

3 marks

Answer

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i

What is helical torsion spring? How does it differ from helical compression spring?

2 marks

Answer

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j

What is pulsating shear stress? Why are springs subjected to pulsating shear stress?

3 marks

Answer

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Part B - Unit I

210 marks
a

What is ‘Preferred numbers or Preferred series’? What are its advantages?

5 marks

Answer

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b

The stresses at a point in a body are sx = 90 N/mm2, sy = 20 N/mm2, and sxy = 80 N/mm2. The material tests syp = 280 N/mm2. Find the factor of safety according to the: (i) Maximum principal stress theory of failure, (ii) Maximum shear stress theory of failure (iii) Maximum strain energy theory of failure.

5 marks

Answer

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OR
310 marks
a

Explain the effects of Stress concentration in Fatigue loading.

5 marks

Answer

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b

A stepped shaft transmits a torque varying from 800 N-m to 1200 N-m. The ratio of diameters is 1.5 and the stress concentration factor is 1.2. Determine the diameter of the shaft for infinite life for a design factor of safety 1.8. The value of sut = 600 N/mm2, and syt = 450 N/mm2.

5 marks

Answer

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Part B - Unit II

410 marks
a

Discuss the methods of failure of riveted joints.

5 marks

Answer

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b

Show that the plane of maximum shear stress occurs at 45° for a parallel load on a fillet weld of equal legs. Neglect bending. Determine the allowable force P per cm of weld length, if the allowable shear stress is 95 N/mm2.

5 marks

Answer

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OR
510 marks
a

The maximum pull in the tie rods of a turnbuckle used in the roof truss is 4.5 kN. The tie rods are made of steel 40C8 (syt = 380 N/mm2) and the factor of safety is 5. Determine the nominal diameter of the threads on the tie rod on the basis of Maximum principal stress theory. Assume dc = 0.8 d.

5 marks

Answer

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b

What is efficiency of riveted joint? How do you find it?

5 marks

Answer

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Part B - Unit III

610 marks
a

What are the basic functions of keys? What are the factors on which the selection of the type of key for a given application depends?

5 marks

Answer

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b

Two rods are connected by means of a knuckle joint. The axial force acting on the rods is 25 kN. The rods and pin are made of plain carbon steel 45C8 (syt = 380 N/mm2), and the factor of safety is 2.5. The yield strength in shear is 57.7% of the yield strength in tension. Calculate the diameter of the rods and the diameter of the pin.

5 marks

Answer

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OR
710 marks
a

What are the applications of a cotter joint? Explain.

5 marks

Answer

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b

A 45 mm diameter shaft is made of steel with a yield strength of 400 MPa. A parallel key of size 14 mm width and 9 mm thickness made of steel with a yield strength of 340 MPa is to be used. Find the required length of key, if the shaft is loaded to transmit the maximum permissible torque. Use maximum shear stress theory and assume a factor of safety of 2.

5 marks

Answer

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Part B - Unit IV

810 marks
a

Distinguish between Shaft, Axle and Spindle.

5 marks

Answer

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b

A sleeve-coupling is used on a shaft of 40 mm diameter, delivering a torque of 50 N–m. Calculate the diameter of the pin required to hold the coupling, if the design stress for the pin material in shear is 100 MPa.

5 marks

Answer

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OR
910 marks
a

Explain the design procedure for Muff coupling.

5 marks

Answer

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b

A hollow circular shaft of outer and inner diameters do and di respectively is subjected to a torsional moment of Mt over a length l. The permissible angle of twist is θ degrees. Prove that the shaft diameter is given by: do = [ (584 Mt l) / (G θ (1 - C4)) ]1/3, where C = di/do.

5 marks

Answer

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Part B - Unit V

1010 marks
a

How do you determine the natural frequency of helical springs?

5 marks

Answer

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b

A helical compression spring is subjected to an initial pre-load of 50 N, and the maximum force during the load cycle is 300 N. The wire diameter is 5 mm, and spring index is 5. The spring is made of oil-hardened and tempered steel wire of Grade-SW (syt = 1440 N/mm2). Determine the factor of safety against fluctuating stresses.

5 marks

Answer

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OR
11

A concentric spring consists of two helical compression springs, one inside the other. The free length of the outer spring is 25 mm greater than that of the inner spring. The wire diameter and mean coil diameter of the inner spring are 8 mm and 64 mm respectively. Also, the wire diameter and mean coil diameter of the outer spring are 10 mm and 80 mm respectively. The number of active coils in the inner and outer springs are 10 and 15 respectively. Assume the same material for both the springs. The modulus of rigidity of spring material is 81370 N/mm2. Calculate the stiffness of spring when the deflection is more than 25 mm.

10 marks

Answer

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About this paper

This page contains the solved question paper with answers of DESIGN OF MACHINE MEMBERS - I from JAWAHARLAL NEHRU TECHNOLOGICAL UNIVERSITY HYDERABAD for B.Tech AME III III-I, May - 2018, R13.

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