Test: Helical Springs - 2

8 Questions MCQ Test Topicwise Question Bank for Mechanical Engineering | Test: Helical Springs - 2

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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?


From the equation of torsion


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


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

Where δ = Deflection of spring θ = angle of twist

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



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




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


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.


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


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.


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


Stiffness of the spring will be same in each case


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


Angular twist in spring is

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


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


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