Finding normal stress in composite bar

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SUMMARY

The discussion focuses on calculating the tensile normal stress in a composite bar consisting of aluminum and steel. The aluminum bar has a modulus of elasticity of 70 GPa and a length of 500 mm, while the steel bar has a modulus of elasticity of 210 GPa and a length of 1500 mm. The tensile normal strain measured in the aluminum bar is εAl = 1000 × 10-6, resulting in a calculated tensile normal stress of σAl = 0.07 GPa. The stress in the steel bar is equal to the stress in the aluminum bar due to the equal cross-sectional area, confirming that the force in each piece is a function of the applied load P.

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Homework Statement


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
composite bar.

ba1mD.png



Homework Equations


ε_x=σ_x/E

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.
 
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The cross section area of each section is the same. If you know the stress in the alum piece, what must be the stress in the steel piece? Hint: what is the force in each piece as a function of P?
 
PhanthomJay said:
The cross section area of each section is the same. If you know the stress in the alum piece, what must be the stress in the steel piece? Hint: what is the force in each piece as a function of P?

So the stress would be the same in the steel piece?
 
PhanthomJay said:
Yes! and welcome to PF!

A most excellent welcome indeed, thanks again!
 

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