An elastic bar (young’s modulus e) of initial length l is fixed at one end and is axially loaded at the other end with a force p (fig. 1a). (a) derive the expression to obtain the displacement ∆ of the loaded end of the bar. (b) calculate the displacement if e = 70 gpa, l = 100 mm, and p = 1 kn.

 ΔL = 1.43 10 -9 m

Explanation:

a) Let's start from Newton's second law, the force in a spring is elastic

    F = - k Δx

Let's divide the two sides by the area

    F / A = -k Δx / A

In general area is long by wide, the formulated pressure is

    P = F / A

    P = - k Δx / l x

    P = (-k / l) Δx/x

Call us at Young's constant module

    P = E Δx / x

Let's change x for L

   E = P / (ΔL/L)

b) we cleared

   ΔL = P L / E

Let's reduce the magnitudes to the SI system

   E = 70 GPa = 70 109 Pa

   L = 100mm (1m / 1000mm) = 0.100m

   P = 1 kN = 1 103 N

   ΔL = 1 103  0.100/ 70 109

   ΔL = 1.43 10 -9 m

connect the batteries in series and hope