Question

4. The bar has cross-sectional area A and modulus of elasticity E. If an axial force F directed toward the right is applied at C: (a) What is the normal stress in the part of the bar to the left of C? [15 pts] (b) What is the resulting displacement of the point C? [15 pts]

Answer 1

a) ΔL/L = F / (E A), b)
`L_(f)` = L (1 + L F /(EA) )

Step-by-step explanation:

Let's write the formula for Young's module

E = P / (ΔL / L)

Let's rewrite the formula, to have the pressure alone

P = E ΔL / L

The pressure is defined as

P = F / A

Let's replace

F / A = E ΔL / L

F = E A ΔL / L

ΔL / L = F / (E A)

b) To calculate the elongation we must have the variation of the length, so the length of the bar must be a fact. Let's clear

ΔL = L [F / EA]

`L_(f)` -L = L (F / EA)

`L_(f)` = L + L (F / EA)

`L_(f)` = L (1 + L (F / EA))