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.

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

So, the stiffness is double.

@Santosh.

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

(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.

@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.

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.

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

Yes, Both are same. Agree @Shukla.

Both formula are same dear C = D/d.

Concept is correct but formula used is wrong.

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