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Test: Design Of Helical Springs - Mechanical Engineering MCQ


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15 Questions MCQ Test Design of Machine Elements - Test: Design Of Helical Springs

Test: Design Of Helical Springs for Mechanical Engineering 2024 is part of Design of Machine Elements preparation. The Test: Design Of Helical Springs questions and answers have been prepared according to the Mechanical Engineering exam syllabus.The Test: Design Of Helical Springs MCQs are made for Mechanical Engineering 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Design Of Helical Springs below.
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Test: Design Of Helical Springs - Question 1

 If a spring has plain ends then number of inactive coils is?

Detailed Solution for Test: Design Of Helical Springs - Question 1

Explanation: There are no inactive coils in plain ends.

Test: Design Of Helical Springs - Question 2

Spring having square ends has 1 inactive coil.

Detailed Solution for Test: Design Of Helical Springs - Question 2

Explanation: There are 2 inactive coils.

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Test: Design Of Helical Springs - Question 3

 The angle of twist for the equivalent bar to a spring is given by? (Symbols have their usual meaning)

Detailed Solution for Test: Design Of Helical Springs - Question 3

Explanation: θ=Ml/GJ where M=PD/2, l=πDN and J=πd⁴/32.

Test: Design Of Helical Springs - Question 4

Martin’s factor compensates for curvature effect in springs.

Detailed Solution for Test: Design Of Helical Springs - Question 4

Explanation: Wahl’s factor accommodates curvature effect while designing spring.

Test: Design Of Helical Springs - Question 5

 The axial deflection of spring for the small angle of θ is given by?

Detailed Solution for Test: Design Of Helical Springs - Question 5

Explanation: Deflection=θxD/2.

Test: Design Of Helical Springs - Question 6

A spring of stiffness constant k is cut in two equal parts. The stiffness constant of new spring will be k/2.

Detailed Solution for Test: Design Of Helical Springs - Question 6

Explanation: k=Gd⁴/8DᵌN, hence k is inversely proportional to number of coils. Thus result will be 2k.

Test: Design Of Helical Springs - Question 7

 For two spring connected in series, the force acting on each spring is same and equal to half of the external force.

Detailed Solution for Test: Design Of Helical Springs - Question 7

Explanation: Te force on each spring is equal to the external force.

Test: Design Of Helical Springs - Question 8

For two springs connected in series, the net deflection is equal to the sum of deflection in two springs.

Detailed Solution for Test: Design Of Helical Springs - Question 8

Explanation: The net deflection is sum of the deflection of sprigs connected in series.

Test: Design Of Helical Springs - Question 9

 For two springs connected in parallel, net force is equal to the sum of force in each spring.

Detailed Solution for Test: Design Of Helical Springs - Question 9

Explanation: Net force applied is distributed in the two springs.

Test: Design Of Helical Springs - Question 10

Patenting is defined as the cooling below the freezing point of water.

Detailed Solution for Test: Design Of Helical Springs - Question 10

Explanation: Patenting is heating steel above critical range followed by rapid cooling.

Test: Design Of Helical Springs - Question 11

Find the Wahl’s factor if spring index is 6.

Detailed Solution for Test: Design Of Helical Springs - Question 11

Explanation: K=[4C-1/4C-4]+0.615/C.

Test: Design Of Helical Springs - Question 12

Find the shear stress in the spring wire used to design a helical compression sprig if a load of 1200N is applied on the spring. Spring index is 6, and wire diameter 7mm.

Detailed Solution for Test: Design Of Helical Springs - Question 12

Explanation: τ=K x 8PC/πd² where K=[4C-1/4C-4]+0.615/C.

Test: Design Of Helical Springs - Question 13

 Find the mean coil diameter of a helical compression sprig if a load of 1200N is applied on the spring. Spring index is 6, and wire diameter 7mm.

Detailed Solution for Test: Design Of Helical Springs - Question 13

Explanation: D=Cd.

Test: Design Of Helical Springs - Question 14

 Find total number coils in a spring having square and ground ends. Deflection in the spring is 6mm when load of 1100N is applied. Modulus of rigidity is 81370N/mm². Wire diameter and pitch circle diameter are 10mm and 50mm respectively.

Detailed Solution for Test: Design Of Helical Springs - Question 14

Explanation: Deflection=8PDᵌN/Gd⁴ or N=4.4 or 5. Total coils=5+2(square grounded ends).

Test: Design Of Helical Springs - Question 15

A railway wagon moving with a speed of 1.5m/s is brought to rest by bumper consisting of two springs. Mass of wagon is 100kg. The springs are compressed by 125mm. Calculate the maximum force acting on each spring.

Detailed Solution for Test: Design Of Helical Springs - Question 15

Explanation: mv²/2=Pxdeflection/2.

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