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Test: Helical Springs - 2 - Mechanical Engineering MCQ


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8 Questions MCQ Test - Test: Helical Springs - 2

Test: Helical Springs - 2 for Mechanical Engineering 2024 is part of Mechanical Engineering preparation. The Test: Helical Springs - 2 questions and answers have been prepared according to the Mechanical Engineering exam syllabus.The Test: Helical Springs - 2 MCQs are made for Mechanical Engineering 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Helical Springs - 2 below.
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Test: Helical Springs - 2 - Question 1

Which one of the following is the correct expression for maximum shear stress induced in the wire of a closed-coiled helical spring of wire diameter d and mean coil radius R which carries an axial load W?

Detailed Solution for Test: Helical Springs - 2 - Question 1

From the equation of torsion

Test: Helical Springs - 2 - Question 2

Given that,
d = diameter of spring, R = mean radius of coils, n = number of coiis and G = modulus of rigidity, the stiffness of the close-coiied helical spring subject to an axial load W is equal to

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

The deflection of a close coil helical spring subjected to axial  force ‘W’ is 

Where δ = Deflection of spring θ = angle of twist
= TL/GJ
∴  

Spring constant or stiffness factor is defined as the axial force required to produce unit deflection

∴ 

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Test: Helical Springs - 2 - Question 3

A closely coiled helical spring of radius R, contains n turns and is subjected to an axial load W. if the radius of the coil wire is r and modulus of rigidity of the coil material is G, the deflection of the coil is

Detailed Solution for Test: Helical Springs - 2 - Question 3

∴  
⇒ 

Test: Helical Springs - 2 - Question 4

Consider the following types of stresses:
1. Torsional shear
2. Transverse direct shear
3. Bending stress

The stresses that are produced in the wire of a closed coiled helical spring subjected to an axial load, would include

Detailed Solution for Test: Helical Springs - 2 - Question 4

When axial load is applied to the spring stress are 3etup due to:
(i) Torsion
(ii) Direct shear and
(iii) Bending
Where ever the stresses due to direct shear and bending are very small and may be neglected.

Test: Helical Springs - 2 - Question 5

A closed-coil helical spring is subjected to a torque about its axis. The spring wire would experience a

Detailed Solution for Test: Helical Springs - 2 - Question 5

When a closed coil helical spring, fixed at one end is subjected to a twisting couple about the central axis of the spring, then bending moment will be produced at every cross-section of the spring. The twisting couple will try to wind up or unwind the spring.

Test: Helical Springs - 2 - Question 6

A weighing machine consists of a 2 kg pan resting on a spring, in this condition, with the pan resting on the spring, the length of the spring is 200 mm. When a mass of 20 kg is placed on the pan, the length of the spring becomes 100 mm? For the spring, the un-deformed length L0 and the spring constant k (stiffness) are

Detailed Solution for Test: Helical Springs - 2 - Question 6


Stiffness of the spring will be same in each case

Test: Helical Springs - 2 - Question 7

A close-coiled helical spring shown below in figure-I is to be cut into two equal pieces and combined as a parallel spring as shown in figure-II. The ratio of the maximum angular twist of the situation shown in figure-II to that of figure-I, due to the same load W will be

Detailed Solution for Test: Helical Springs - 2 - Question 7

Angular twist in spring is

Where,
d is diameter of wire of spring
G is modulus of rigidity
W is axial pull on spring
R is mean radius of coil
L is length of spring
∴ 

Test: Helical Springs - 2 - Question 8

A close-coiled helical spring has wire diameter 10 mm and spring index 5. If the spring contains 10 turns, then the length of the spring wire would be

Detailed Solution for Test: Helical Springs - 2 - Question 8

Wire diameter, d = 10 mm
Spring index = 5 = D/d
D = 5d; D = 50 mm
Number of turns n = 10
Length of wire = πD x n = π x 50 x 10 = 3.14 x 500 = 1570 mm

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