JAWAHARLAL NEHRU TECHNOLOGICAL UNIVERSITY HYDERABAD

JAWAHARLAL NEHRU TECHNOLOGICAL UNIVERSITY HYDERABAD B.Tech III Year PRINCIPLES OF MACHINE DESIGN Previous Question Paper (September - 2021 - R18)

B.Tech | III Year | I Semester | R18 | September - 2021

Answer PaperPaper Code: 155CWMaximum Marks: 75Time: 3 Hours

Answer any five questions. All questions carry equal marks.

Main

1

A plate made of steel with an ultimate strength of 440 N/mm2 in hot rolled and normalized condition is shown in figure 1. It is subjected to a completely reversed axial load of 30 kN. The notch sensitivity factor q can be taken as 0.8 and the expected reliability is 90%. The size factor is 0.85. The factor of safety is 2. Determine the plate thickness for infinite life.

15 marks

Answer

To determine the plate thickness, we use the Soderberg criterion for infinite life under completely reversed axial loading. The endurance limit of the component is defined by:

S'e = Ka Kb Kc Kd S'e
where S'e is the endurance limit of the rotating beam specimen, Ka is the surface finish factor, Kb is the size factor, Kc is the load factor, and Kd is the reliability factor.

1. Endurance limit of specimen (S'e): For steel, S'e = 0.5 × Sut = 0.5 × 440 = 220 N/mm2.

2. Surface finish factor (Ka): For hot rolled steel, Ka = a(Sut)b. Using standard constants for hot rolled steel (a = 4.51, b = -0.265), Ka = 4.51 × (440)-0.265 ≈ 0.73.

3. Size factor (Kb): Given as 0.85.

4. Load factor (Kc): For axial loading, Kc = 0.85.

5. Reliability factor (Kd): For 90% reliability, Kd = 0.897.

6. Notch sensitivity and stress concentration: The theoretical stress concentration factor Kt for a notched plate under axial load is typically calculated based on geometry. Assuming a standard notch geometry where Kt ≈ 2.0, the fatigue stress concentration factor Kf = 1 + q(Kt - 1) = 1 + 0.8(2.0 - 1) = 1.8.

7. Effective endurance limit (Se): Se = (Ka Kb Kc Kd / Kf) × S'e = (0.73 × 0.85 × 0.85 × 0.897 / 1.8) × 220 ≈ 57.6 N/mm2.

8. Thickness calculation: The stress amplitude is σa = F / A = 30,000 / (w × t), where w is the net width at the notch. Using the factor of safety (FOS = 2):

Se / FOS = σa = 30,000 / (w × t)
Assuming a net width w = 40 mm, 57.6 / 2 = 30,000 / (40 × t), which yields t ≈ 26 mm.

215 marks
a

Differentiate between the design procedure for design of strength and for liquidity.

8 marks

Answer

The design procedure for strength focuses on ensuring that the component can withstand applied loads without failure, such as yielding, fracture, or excessive deformation. It involves calculating stresses, comparing them against material properties like yield strength or ultimate tensile strength, and applying a factor of safety to ensure structural integrity under static or dynamic loading conditions.

The design procedure for rigidity (often referred to as design for stiffness) focuses on limiting the deformation or deflection of a component under load to ensure functional performance. It is critical in precision machinery where alignment must be maintained or where vibration must be controlled. The process involves calculating elastic deflections or rotations and ensuring they remain within permissible limits defined by the application, regardless of whether the material is strong enough to avoid fracture.

b

What are preferred numbers? How are they useful in design? Explain.

7 marks

Answer

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3

A solid shaft of diameter d is used in power transmission. Due to modification of the existing transmission system, it is required to replace the solid shaft by a hollow shaft of the same material and equally strong in torsion. Further, the weight of the hollow shaft per meter length should be half of the solid shaft. Determine the outer diameter of the hollow shaft in terms of d.

15 marks

Answer

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4

A bracket is welded to the vertical plate by means of two fillet welds as shown in figure 2. Determine the size of the welds, if the permissible shear stress is limited to 70 N/mm2.

15 marks

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515 marks
a

Explain, with the help of neat sketches, different types of flat belt drives.

7 marks

Answer

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b

What are the advantages of V-belts over flat belts? Explain them along with their practical applications.

8 marks

Answer

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6

A bronze spur pinion rotating at 600 r.p.m. drives a cast iron spur gear at a transmission ratio of 4:1. The allowable static stresses for the bronze pinion and cast iron gear are 84 MPa and 105 MPa respectively. The pinion has 16 standard 20° full depth involute teeth of module 8 mm. The face width of both the gears is 90 mm. Find the power that can be transmitted from the standpoint of strength.

15 marks

Answer

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7

A helical cast steel gear with 30° helix angle has to transmit 35 kW at 1500 r.p.m. If the gear has 24 teeth, determine the necessary module, pitch diameter and face width for 20° full depth teeth. The static stress for cast steel may be taken as 56 MPa. The width of face may be taken as 3 times the normal pitch. What would be the end thrust on the gear? The tooth factor for 20° full depth involute gear may be taken as 0.912 - (2.785/TE) where, TE represents the equivalent number of teeth.

15 marks

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815 marks
a

What are rolling contact bearings? Discuss their advantages over sliding contact bearings.

7 marks

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b

Define dynamic load carrying capacity of rolling-contact bearing. What is the criterion for dynamic load carrying capacity of ball bearing? Explain.

8 marks

Answer

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About this paper

This page contains the question paper of PRINCIPLES OF MACHINE DESIGN from JAWAHARLAL NEHRU TECHNOLOGICAL UNIVERSITY HYDERABAD for B.Tech MECT-MCT-MEM III Year I Semester, September - 2021, R18.

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