GATE Mechanical Engineering (ME) 2020 Shift 2 Solved Paper

Question : 40 of 65

Marks: +1, -0

A helical spring has spring constant

{k}

. If the wire diameter, spring diameter and the number of coils are all doubled then the spring constant of the new spring becomes

Solution:

Given

{K}_{3}= {K}

d	D	n
2d	2D	2n

{d} →

coil diameter

{D} →

spring diameter

{K}=\;{ {Gd}^{ {G}}}/{ {BD}^{3} {n}}

{K} ∝ \;{ {d}^{4}}/{ {D}^{3} {n}} ⇒ \;{ {K}_{2}}/{ {K}_{1}}=(\;{ {d}_{2}}/{ {d}_{1}})^{4} ×(\;{ {D}_{1}}/{ {D}_{2}})^{3} ×(\;{ {n}_{1}}/{ {n}_{ {x}}})

\;{ {K}_{2}}/{ {K}_{1}}=(2)^{4} × \;{1}/{(2)^{3}} × \;{1}/{2}

{K}_{2}= {K}_{1}

Go to Question:

Prev Question

Next Question