The 2000 mm long composite bar shown in Fig. 1 consists of an aluminum bar having
a modulus of elasticity EAl = 70 GPa and length LAl = 500 mm, which is securely fastened
to a steel bar having modulus of elasticity ESt = 210 GPa and length LSt = 1500 mm. After
the force P is applied, a tensile normal strain of εAl = 1000 × 10-6 is measured in the
aluminum bar. Find the tensile normal stress in each bar and the total elongation of the
The Attempt at a Solution
So I first took the equation and rearranged it, such that ε_x*E=σ_x and got σ_AL=0.07 GPa. Then for the changed in distance of the aluminum bar, dl=ε*l_0 = (1000*10^-6)(500mm) = 0.5mm change for Al bar.
Assuming that's correct, I'm stuck on how to find the stress for the steel bar and the length it changed.