Stress/Strain with two rods joined together

  • Thread starter aal0315
  • Start date
In summary, a steel and brass rod is pulled resulting in a total change in length of 1.20 mm. Knowing that the total length is the sum of the lengths of each section and that the force and cross-sectional area are the same, we can use the equations for strain, stress, and change in length to determine that the brass section deforms twice as much as the steel section. This means that the length of the steel section increases by 0.4 mm and the length of the brass section increases by 0.8 mm.
  • #1
aal0315
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Homework Statement


A rod is made of two sections joined end to end. The sections are identical, except the one is steel and the other is brass. While one end is held fixed, the other is pulled to result in change in length of 1.20 mm. By how much does the length of each section increase?


Homework Equations


Strain = stress / E
Stress = F/A
delta L = 1/E x L


The Attempt at a Solution


I know that E(steel) = 200x10^9 and E(brass) = 100x10^9
I know that the total length is L= L1 +L2, where L1 is the length of the steel and L2 is the length of the brass
I know that the force and cross-sectional area is the same, so those don't matter.
I feel that the delta L = 1/(E(steel)+E(brass))xL would be right, but i don't know where to find the value of L.
or is it that delta L = 1/E(steel)xL1, where delta L = 1.20mm? but i feel that this is wrong because it says in the question that the whole rod lengthens that much.
any suggestions?
 
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  • #2
the total change in length of the rod is equal to the sum of the change in lengths of its component parts. If E_B = 1/2E_S, which piece deforms more, and how much more?
 
  • #3
the brass deforms and it is double the amount of the steel, so is this right:
1.2 = x+2x , where x is the length of steel
x = 0.4mm and the length of brass increases by 0.8mm?
 
  • #4
aal0315 said:
the brass deforms and it is double the amount of the steel, so is this right:
1.2 = x+2x , where x is the length of steel
x = 0.4mm and the length of brass increases by 0.8mm?
Yes, very good.
 
  • #5
thank you :eek:)
 

FAQ: Stress/Strain with two rods joined together

What is stress/strain and how is it measured?

Stress is the amount of force applied to a material per unit area, while strain is the resulting deformation or change in length of the material. Stress can be measured in units of force divided by area, such as Pascals (Pa) or pounds per square inch (psi). Strain is typically measured as a percentage change in length or as a dimensionless ratio.

What happens to the stress/strain of two rods when they are joined together?

When two rods are joined together, the stress and strain are distributed across both rods. This means that the total stress and strain on each rod will be less than if they were separate, as the load is now shared between them. However, the exact distribution of stress and strain will depend on the material properties and geometry of the rods.

How does the material properties of the rods affect stress/strain?

The material properties of the rods, such as their Young's modulus and yield strength, will determine how much stress and strain the rods can withstand before they permanently deform or break. Materials with higher Young's modulus values are stiffer and will experience less strain for a given stress, while materials with higher yield strengths are stronger and can handle higher stresses before failure.

What is the difference between tensile and compressive stress/strain?

Tensile stress/strain occurs when a material is being pulled or stretched, while compressive stress/strain occurs when a material is being pushed or compressed. In both cases, the material experiences a change in length, but the direction of the deformation is opposite. The magnitude of stress and strain may also differ between tensile and compressive loading.

How can we use stress/strain data to design structures or products?

By understanding the stress and strain behavior of different materials, we can select the most appropriate materials for a given application and design structures or products that can withstand expected loads without failing. This information can also help engineers optimize the shape and dimensions of a structure to minimize stress and strain and improve overall performance and longevity.

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