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JNTU B.Tech III DESIGN OF MACHINE MEMBERS I Previous Question Paper (October - 2020 - R16)

B.Tech | III | III-I | R16 | October - 2020

Answer PaperPaper Code: 135AGMaximum Marks: 75Time: 2 hours

Answer any five questions. All questions carry equal marks. Illustrate your answers with NEAT sketches wherever necessary.

Main

115 marks
a

Explain any two theories of failure.

7 marks

Answer

The two common theories of failure for ductile materials are:

  1. Maximum Principal Stress Theory (Rankine's Theory): This theory states that failure occurs when the maximum principal stress in a complex stress system reaches the yield point stress of the material in a simple tension test. It is primarily used for brittle materials. Mathematically, σ1 = σyt.
  2. Maximum Shear Stress Theory (Guest's Theory): This theory states that failure occurs when the maximum shear stress in a complex stress system reaches the shear stress at the yield point in a simple tension test. It is widely used for ductile materials. Mathematically, τmax = (σ1 - σ2) / 2 = σyt / 2.
b

A beam of rectangular cross section is fixed at one end and carries an electric motor weighing 400 N at a distance of 300 mm from the fixed end. The maximum bending stress in the beam 40 MPa. Find the width and depth of the beam, if the depth is twice that of width.

8 marks

Answer

Given: Load (W) = 400 N, Length (L) = 300 mm, Bending Stress (σb) = 40 MPa (N/mm²), Depth (d) = 2b.

The maximum bending moment (M) at the fixed end is:

M = W × L = 400 × 300 = 120,000 N-mm

The section modulus (Z) for a rectangular beam is:

Z = (b × d²) / 6 = (b × (2b)²) / 6 = (4b³) / 6 = (2/3)b³

Using the bending stress formula σb = M / Z:

40 = 120,000 / ((2/3)b³)
40 = 180,000 / b³
b³ = 180,000 / 40 = 4,500
b = ∛4,500 ≈ 16.51 mm

Calculating depth (d):

d = 2 × 16.51 = 33.02 mm

Thus, the width of the beam is 16.51 mm and the depth is 33.02 mm.

215 marks
a

Explain the concept of stiffness in tension, bending, and torsion.

7 marks

Answer

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b

A mild steel shaft of 60 mm diameter is subjected to a bending moment of 3000 N-m and a torque T. If the yield point of steel in tension is 200 MPa, find the maximum value of this torque without causing yielding of the shaft according to the maximum distortion energy theory of failure.

8 marks

Answer

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315 marks
a

Illustrate the methods of reducing stress concentration.

7 marks

Answer

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b

A 50 mm diameter shaft is made from carbon steel having ultimate tensile strength of 600 MPa. It is subjected to a torque which fluctuates between 2000 N-m to (-900) N-m. Using Solderberg method, calculate the factor of safety.

8 marks

Answer

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415 marks
a

Explain the effect of the following factors on the type of fatigue failure: (i) Stress distribution (ii) Surface treatment.

7 marks

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b

A rectangular plate, 15 mm thick, made of a brittle material is shown in figure. Calculate the stresses at each of the three holes of 3 mm, 5 mm, and 10 mm diameter holes. (All dimensions in the figure are in mm).

8 marks

Answer

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5

A double riveted lap joint with chain riveting is made for joining two plates having thickness of 12 mm. If the allowable tensile stress, allowable shear stress, and allowable compressive stress are 65 MPa, 55 MPa and 80 MPa respectively, find the rivet diameter, pitch of rivets, distance between the rows of rivets, and efficiency of the joint.

15 marks

Answer

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615 marks
a

What is leg of fillet weld and throat of fillet weld? And what is the relationship between the leg and throat of fillet weld?

7 marks

Answer

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b

Prove that the plane, where maximum shear stress is induced, is inclined at 45 degrees to the leg dimension in the case of parallel fillet weld of equal legs.

8 marks

Answer

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7

Two rods made of plain carbon steel 40C8 (Syt = 380 N/mm^2) are connected by means of a cotter joint. The diameter of each rod is 50 mm, and the cotter is made from a steel plate of 15 mm thickness. Calculate the dimensions of the socket end, making the following assumptions: a) The yield strength in compression is twice the tensile yield strength, and b) The yield strength in shear is 50 % of the tensile yield strength. Take the factor of safety as 6.

15 marks

Answer

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8

Design a cast iron flange coupling to transmit 15 KW at 900 rpm. The service factor may be assumed as 1.35. The following permissible stresses may be used: Shear stress for shaft, bolt and key material = 40 MPa; Crushing stress for bolt and key = 80 MPa; Shear stress for cast iron = 8 MPa.

15 marks

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