Mechanical Engineering - Strength of Materials - Discussion

Answer is A.

When you cut a closely coiled helical spring into two halves, each half will essentially behave like an independent spring. However, the stiffness of the resulting springs will not change. The stiffness, or spring constant, is determined by the material and the physical dimensions of the spring, such as the wire diameter, coil diameter, and number of coils, which remain unchanged when the spring is cut. Therefore, the stiffness of each resulting spring will be the same as the original spring.

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.

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.

@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.

(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.

@Santosh.

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

Concept is correct but formula used is wrong.

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

N = n/2 so that cd^4/8D^3(n/2) = 2(cd^4/8D^3n) = 2K.

So, the stiffness is double.

Cd^4 / 8 D^3 n this formula also A correct first one is wrong.

Both formula are same dear C = D/d.