Question

. 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.

Answer 1

ΔL = 1.43 10 -9 m

Step-by-step 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