JAWAHARLAL NEHRU TECHNOLOGICAL UNIVERSITY HYDERABAD

JAWAHARLAL NEHRU TECHNOLOGICAL UNIVERSITY HYDERABAD B.Tech II Year DESIGN OF MACHINE MEMBERS I Previous Question Paper (July/August - 2021 - R16)

B.Tech | II Year | II-II | R16 | July/August - 2021

Answer PaperPaper Code: 134AQMaximum Marks: 75Time: 3 Hours

Answer any Five Questions. All Questions Carry Equal Marks.

Main

1

A bolt is subjected to an axial pull of 12 kN and a transverse shear force of 6 kN. The yield strength of bolt material is 300 MPa, considering a factor of safety of 3, find the diameter of bolt using:

a) Maximum principal stress theory

b) Maximum shear stress theory

15 marks
a
Maximum principal stress theory calculation
7 marks

Answer

Given: Axial load P = 12,000 N, Shear load V = 6,000 N, Yield strength σy = 300 MPa, Factor of safety FOS = 3. The allowable stress σ = σy / FOS = 300 / 3 = 100 MPa. The bolt is subjected to normal stress σ = 4P / (πd2) and shear stress τ = 4V / (πd2). According to the maximum principal stress theory, the principal stress σ1 must be less than or equal to the allowable stress σ.

σ1 = (σ / 2) + √((σ / 2)2 + τ2) = σ

Substituting σ = 4P / (πd2) and τ = 4V / (πd2):

σ1 = (2P / (πd2)) + √((2P / (πd2))2 + (4V / (πd2))2) = 100
σ1 = (4 / (πd2)) * (P/2 + √((P/2)2 + V2)) = 100

Substituting P = 12,000 and V = 6,000:

σ1 = (4 / (πd2)) * (6,000 + √(6,0002 + 6,0002)) = 100
σ1 = (4 / (πd2)) * (6,000 + 8,485.28) = 57,941.12 / d2 = 100
d2 = 579.41 => d ≈ 24.07 mm
b
Maximum shear stress theory calculation
8 marks

Answer

According to the maximum shear stress theory (Guest's theory), the maximum shear stress τmax must be less than or equal to the allowable shear stress, where τallow = σy / (2 * FOS) = 300 / (2 * 3) = 50 MPa.

τmax = √((σ / 2)2 + τ2) = τallow

Substituting σ = 4P / (πd2) and τ = 4V / (πd2):

τmax = (4 / (πd2)) * √((P / 2)2 + V2) = 50

Substituting P = 12,000 and V = 6,000:

τmax = (4 / (πd2)) * √(6,0002 + 6,0002) = 50
τmax = (4 / (πd2)) * 8,485.28 = 33,941.12 / d2 = 50
d2 = 678.82 => d ≈ 26.05 mm
215 marks
a

What are the theories of failure and explain about maximum principal stress theory.

7 marks

Answer

Sign in to view the complete answer paper.

b

An axle of 1 m length is simply supported at its ends. It carries a flywheel of 25 kN weight at its centre. Determine the diameter of axle if the permissible normal (bending) stress is not to exceed 60 MPa.

8 marks

Answer

Sign in to view the complete answer paper.

3

Determine the thickness of 200 mm wide plate for safe continuous operation, if it is subjected to a tensile load varying from 120 kN to 260 kN. The yield strength of the plate material is 350 MPa and the endurance limit is 220 MPa. Take the factor of safety as 2.

15 marks

Answer

Sign in to view the complete answer paper.

4

A rod is subjected to an alternating tensile force varying from 25 kN to 75 kN. Determine the diameter of the rod according to the Goodman method, when the ultimate tensile strength of the rod is 950 MPa and yield strength 600 MPa, factor of safety is 2. Neglect the stress concentration effect and correction factors.

15 marks

Answer

Sign in to view the complete answer paper.

515 marks
a

What is uniform bolt strength? Explain.

7 marks

Answer

Sign in to view the complete answer paper.

b

The cylinder head of a compressor is secured by means of 12 studs of yield stress 320 MPa and the initial tension in the bolts is equally loaded, such that a cylinder pressure of 3 MPa is required for the joint to be on the point of opening. Determine the size of bolts if the factor of safety is 2 and stress concentration factor is 3.

8 marks

Answer

Sign in to view the complete answer paper.

6

Two plates of 20 mm thickness are joined by a double riveted lap joint. The pitch of each row of rivets is 100 mm. The permissible stresses are 160 MPa in tension, 100 MPa in shear, and 200 MPa in crushing. Determine how the joint fails and find the efficiency of the joint.

15 marks

Answer

Sign in to view the complete answer paper.

715 marks
a

How is the key designed?

7 marks

Answer

Sign in to view the complete answer paper.

b

A spindle transmits 12 kW at 1000 rpm. The angular deflection is not to exceed 0.25° per metre length of spindle. If the modulus of rigidity is 84 kN/mm², find the diameter of the spindle and shear stress in the spindle.

8 marks

Answer

Sign in to view the complete answer paper.

8

Design and draw a C.I muff coupling when a shaft transmits 100 kW at 200 rpm. If the allowable shear and crushing stresses for the shaft and key material are 40 N/mm² and 60 N/mm², the permissible shear stress in the muff is 20 N/mm². The maximum torque transmitted is 30% higher than the mean torque.

15 marks

Answer

Sign in to view the complete answer paper.

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 MSNT II Year II-II, July/August - 2021, R16.

For the question paper without answers, switch to the Question Paper tab.

Other papers in DESIGN OF MACHINE MEMBERS I

Loading Question Papers

Fetching data from database...