Mechanical Engineering - Strength of Materials - Discussion
Discussion Forum : Strength of Materials - Section 8 (Q.No. 4)
4.
Two closely coiled helical springs 'A' and 'B' are equal in all respects but the diameter of wire of spring 'A' is double that of spring 'B' The stiffness of spring 'B' will be __________ that of spring 'A'
Discussion:
4 comments Page 1 of 1.
Kumara said:
9 years ago
k = GD^4 / 8D^3 N.
Here d is small dia mean wire dia, as Stiffness is directly proortinal to d^4. so the answer is 1/16 of another spring.
Here d is small dia mean wire dia, as Stiffness is directly proortinal to d^4. so the answer is 1/16 of another spring.
Anand said:
10 years ago
Stiffness, K = Load/Deflection.
So, K = W/[(8WD^3.n)/(Gd^4)].
K = G.D^4.
= 8.D^3.n.
So, K = W/[(8WD^3.n)/(Gd^4)].
K = G.D^4.
= 8.D^3.n.
Shri said:
1 decade ago
k = p(D^3)N/(Gd^4).
Thus stiffness proportional to d^4.
Thus stiffness proportional to d^4.
DILIP KUMAR said:
1 decade ago
Can any one explain this?
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