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# Test: Design Of Helical Springs

## 15 Questions MCQ Test Machine Design | Test: Design Of Helical Springs

Description
This mock test of Test: Design Of Helical Springs for Mechanical Engineering helps you for every Mechanical Engineering entrance exam. This contains 15 Multiple Choice Questions for Mechanical Engineering Test: Design Of Helical Springs (mcq) to study with solutions a complete question bank. The solved questions answers in this Test: Design Of Helical Springs quiz give you a good mix of easy questions and tough questions. Mechanical Engineering students definitely take this Test: Design Of Helical Springs exercise for a better result in the exam. You can find other Test: Design Of Helical Springs extra questions, long questions & short questions for Mechanical Engineering on EduRev as well by searching above.
QUESTION: 1

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

Solution:

Explanation: There are no inactive coils in plain ends.

QUESTION: 2

### Spring having square ends has 1 inactive coil.

Solution:

Explanation: There are 2 inactive coils.

QUESTION: 3

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

Solution:

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

QUESTION: 4

Martin’s factor compensates for curvature effect in springs.

Solution:

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

QUESTION: 5

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

Solution:

Explanation: Deflection=θxD/2.

QUESTION: 6

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

Solution:

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

QUESTION: 7

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

Solution:

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

QUESTION: 8

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

Solution:

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

QUESTION: 9

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

Solution:

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

QUESTION: 10

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

Solution:

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

QUESTION: 11

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

Solution:

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

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.

Solution:

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

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.

Solution:

Explanation: D=Cd.

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.

Solution:

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

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.

Solution:

Explanation: mv²/2=Pxdeflection/2.