Mechanical Engineering - Strength of Materials - Discussion

Discussion Forum : Strength of Materials - Section 9 (Q.No. 6)
6.
A closely-coiled helical spring is cut into two halves. The stiffness of the resulting spring will be
same
double
half
one-fourth
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
11 comments Page 1 of 2.

Hari said:   1 decade ago
STIFFNESS OF SPRING = [(G*d)/(8* C^3 * N)].

Where N - No.of coils.
As you can see stiffness is inversely proportional to No of coils. When you cut the spring into half the stiffness will get doubled.

Amit said:   1 decade ago
Concept is correct but formula used is wrong.

Correct formula is [(G.d^4)/(64.R^3.n)].

Saurabh shukla said:   10 years ago
Both formula are same dear C = D/d.

Dhavalo said:   9 years ago
Yes, Both are same. Agree @Shukla.

Crdmistry said:   8 years ago
Cd^4 / 8 D^3 n this formula also A correct first one is wrong.

SHEKHAR said:   7 years ago
But if, the stiffness of a material is measured in terms of Young's modulus which itself is a materialistic property (and remains same for a material) then how the stiffness varies by varying the length?

Please explain.

Santosh Naganur said:   7 years ago
@Shekhar.

Obviously, the young's Modulus is same for a material but there is a two parts of same material therefore resultant stiffness will be k+k=2K.

Murali said:   6 years ago
(G.d^4)/64R^3n is the formula for stiffness. if n becomes half stiffness becomes 2(Gd^4)/(64R^3n). It means stiffness doubled.

Ani said:   5 years ago
@Santosh.

If the springs are used in parallel connection then what will be the stiffness?

Jimit said:   5 years ago
N = n/2 so that cd^4/8D^3(n/2) = 2(cd^4/8D^3n) = 2K.

So, the stiffness is double.


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