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JNTU B.Tech III DESIGN OF MACHINE MEMBERS - I Previous Question Paper (February - 2022 - R18)

B.Tech | III | I | R18 | February - 2022

Answer PaperPaper Code: 155AXMaximum Marks: 75Time: 3 hours

Answer any five questions. All questions carry equal marks.

Main

115 marks
a

Explain the design considerations in machine design and explain the design process.

5 marks

Answer

Design considerations include: Type of load (static, dynamic, or impact), material properties (strength, ductility, fatigue resistance), kinematics of the mechanism, manufacturing limitations, cost, safety factors, and environmental conditions. The design process follows a systematic sequence: 1. Identification of the need or problem. 2. Selection of possible mechanisms and materials. 3. Analysis of forces and stresses. 4. Determination of component dimensions based on failure theories. 5. Modification of design for manufacturability and assembly. 6. Final detailing and preparation of production drawings.

b

A 25 mm diameter shaft is made of forged steel 30C8 (Sut = 600 N/mm2). There is a step in the shaft and the theoretical stress concentration factor at the step is 2.1. The notch sensitivity factor is 0.84. Determine the endurance limit of the shaft if it is subjected to a reversed bending moment.

10 marks

Answer

Given: Diameter d = 25 mm, Sut = 600 N/mm2, Kt = 2.1, q = 0.84. First, calculate the fatigue stress concentration factor (Kf):

Kf = 1 + q(Kt - 1) = 1 + 0.84(2.1 - 1) = 1.924

The endurance limit of the material (Se') for steel is typically 0.5 × Sut:

Se' = 0.5 × 600 = 300 N/mm2

The actual endurance limit (Se) of the shaft considering the stress concentration is given by:

Se = Se' / Kf = 300 / 1.924 = 155.93 N/mm2

Thus, the endurance limit of the shaft under reversed bending is 155.93 N/mm2.

215 marks
a

How can the strength of low carbon steel be increased?

5 marks

Answer

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b

The force acting on a bolt consists of two components—an axial pull of 12 kN and a transverse shear force of 6 kN. The bolt is made of steel FeE 310 (Syt = 310 N/mm2) and the factor of safety is 2.5. Determine the diameter of the bolt using the maximum shear stress theory of failure.

10 marks

Answer

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315 marks
a

Explain the difference between endurance limit and fatigue strength of a material.

5 marks

Answer

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b

Determine the thickness of a 120 mm wide uniform plate for safe continuous operation if the plate is to be subjected to a tensile load that has a maximum value of 250 kN and a minimum value of 100 kN. The endurance limit stress is 225 MPa, yield point stress is 300 MPa and factor of safety is 1.5.

10 marks

Answer

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415 marks
a

How will you reduce stress concentration in threaded parts?

5 marks

Answer

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b

A machine component is subjected to fluctuating stress that varies from 40 to 100 N/mm2. The corrected endurance limit stress for the machine component is 270 N/mm2. The ultimate tensile strength and yield strength of the material are 600 and 450 N/mm2 respectively. Find the factor of safety using (i) Gerber theory and (ii) Soderberg line.

10 marks

Answer

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515 marks
a

Describe about the welding inspection. What are the methods of welding inspection?

5 marks

Answer

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b

How much length of a 10 mm fillet weld is required to weld the long side of an ISA angle 150 × 75 × 10 to a steel plate with side welds only? A static load of 125 kN acts through the centre of gravity of the angle section which is 53.2 mm from the short side. The allowable load per mm of the weld length is 665 N.

10 marks

Answer

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615 marks
a

Explain the methods of increasing the shock absorbing capacity of the bolt.

5 marks

Answer

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b

An eccentrically loaded bracket is welded to the support. The permissible shear stress for the weld material is 55 N/mm2 and the load is static. Determine the throat and leg dimensions for the welds.

10 marks

Answer

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715 marks
a

Explain the construction of Knuckle joint with the help of neat sketch.

7 marks

Answer

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b

Why is cotter provided with a taper? Why is a taper provided only on one side?

8 marks

Answer

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815 marks
a

Explain why hollow shaft has greater strength and stiffness than solid shaft of equal weight?

8 marks

Answer

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b

How will you design shaft on the basis of torsional rigidity?

7 marks

Answer

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