Question

Consider a metal bar of initial length L and cross-sectional area A. The Young's modulus of the material of the bar is Y. Find the "spring constant" k of such a bar for low values of tensile strain. Express your answer in terms of Y, L, and A.

Answer 1

Answer:

k = Y*A/L

Step-by-step explanation:

We can apply the Law of Hooke in order to explain the problem.

If we define k = F / ΔL  and the Y = S / δ

Where S is the uniaxial stress: S =  F / A   (i)

and δ is the strain: δ = ΔL / L   (ii)

ΔL is the change in length

we can combine the equations i and ii as follows

Y = (F / A) / (ΔL / L) = (F * L) / (A * ΔL)   (iii)

if k = F / ΔL the equation iii results

Y = k * (L / A)  ⇒   k = Y*(A / L)