Mechanical Engineering - Strength of Materials - Discussion
Discussion Forum : Strength of Materials - Section 4 (Q.No. 23)
23.
Two closely-coiled helical springs 'A' and 'B' of the same matenal, same number of turns and made from same wire are subjected to an axial load W. The mean diameter of spring 'A' is double the mean diameter of spring 'B'. The ratio of deflections in spring 'B' to spring 'A' will be
Discussion:
7 comments Page 1 of 1.
Roshan said:
10 years ago
Deflection = (8*max load*D^3)/G*diameter of spring wire.
Here D is mean diameter of spring.
Using this answer comes out to be option A.
Here D is mean diameter of spring.
Using this answer comes out to be option A.
Ganesh said:
9 years ago
As per deflection formula the ratio is 2.
SURESH DARSI said:
9 years ago
Deflection = (8FD^3)/(Gd^4).
Sagar said:
8 years ago
Here, the formula is (8WD^3)/(Cd^4).
Ppradeep gk said:
8 years ago
Here, Delt=6(4wR^(3)n)/cd^4.
Aakash said:
8 years ago
X = (8WD^3)/(Gd^4).
From this formula, the ratio will be 1/8.
From this formula, the ratio will be 1/8.
Virat said:
5 years ago
@Akash.
In question, the mean diameter is twice.
In question, the mean diameter is twice.
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