Finding normal stress in composite bar

In summary, the problem involves a composite bar consisting of an aluminum bar and a steel bar. 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. After a force is applied, a tensile normal strain of 1000 × 10^-6 is measured in the aluminum bar. Using the equation ε_x=σ_x/E, the tensile normal stress in the aluminum bar is calculated to be 0.07 GPa and the change in length is 0.5 mm. As the cross-sectional area of each section is the same, the stress in the steel bar
  • #1
newhere
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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.

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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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  • #2
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?
 
  • #3
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?
 
  • #4
Yes! and welcome to PF!
 
  • #5
PhanthomJay said:
Yes! and welcome to PF!

A most excellent welcome indeed, thanks again!
 

FAQ: Finding normal stress in composite bar

1. What is a composite bar?

A composite bar is a structure made up of two or more different materials that are bonded together to form a single unit. These materials have different mechanical properties and are arranged in a specific way to achieve desired strength and stiffness.

2. What is normal stress?

Normal stress is the force per unit area that acts on a material in a direction perpendicular to the cross-sectional area. It is a measure of the internal forces that a material experiences when subjected to an external load.

3. How do you find normal stress in a composite bar?

To find normal stress in a composite bar, you need to know the applied load, the cross-sectional area of each material, and the modulus of elasticity of each material. You can then use the equation σ = F/A, where σ is the normal stress, F is the applied load, and A is the cross-sectional area of the material.

4. What factors can affect the normal stress in a composite bar?

The normal stress in a composite bar can be affected by various factors, such as the type and arrangement of materials, the magnitude and direction of the applied load, and the temperature. Other factors that can affect normal stress include the presence of defects or imperfections in the materials and the duration of the applied load.

5. How is normal stress distributed in a composite bar?

In a composite bar, the normal stress is distributed among the different materials based on their individual properties and the arrangement of the materials. The materials with higher stiffness and strength will experience a greater share of the normal stress, while the materials with lower stiffness and strength will experience a lower share of the stress.

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