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

Answer:

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

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))