# Need help understanding this Stress/Strain equation

• Girn261
Does this make sense? By the way, the minus sign indicates that the copper is in compression, while the steel is in tension.
Girn261

## Homework Statement

Please see the uploaded picture, need help understanding a few things. First off, If you look at the equation Strain(c)/E(c)+Strain(s)/E(s) = Ac(delta T) - As(Delta T)

and below that they wrote

2x Stress(s)/100x10^9 + Stress(s)/210x10^9

My question is, how did they go from the top part of the equation, to the bottom? Strain is gone and stress magically appeared. also if you look at the picture I uploaded, the 100x10^9 magically disappears and also they somehow get 5xStress(s)

I wish I had someone to ask for help, but I am stuck and I'm not in school and my co-workers are not very helpful.

## Homework Equations

Stress=P/area , Strain = coeff. of linear expansion x delta T , modulus of elasticity = stress / strain

## The Attempt at a Solution

#### Attachments

• 20170802_131432_HDR.jpg
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I can't read much of what is on that image . Please post a better copy .

Sorry, hope this helps

#### Attachments

• 20170802_142144_HDR.jpg
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I think the equation directly after the Strain_c+Strain_s should read Stress_c and Stress_s in place of the Strains. The units don't even make any sense with what the equation gives. Strain/E does not give a unitless number, but Stress/E would, which matches the other side's unitless answer. I think it was just a simple misprint.

As for the 5xStress_s, they got that from multiplying both sides of the equation by 210x10^9. However, that should give 2.1x2xStress_s+Stress_s, which is 5.2xStress_s. Maybe they forgot the 0.1?

Girn261
It is clear to me that this problem has not been solved correctly by whomever provided the answer. So, let's go back and solve it correctly. For a rod that experiences a combination of thermal expansion and mechanical strain, the correct equation for the stress is:
$$\sigma=E(\epsilon-\alpha \Delta T)$$where ##\epsilon## is the actual strain experienced by the rod, and ##\alpha \Delta T## is the strain that the rod would experience if it were free to expand unconstrained (i.e., under conditions where the axial stress is zero).

So, for the steel we have: $$\sigma_S=E_S(\epsilon-\alpha_S \Delta T)\tag{1}$$and for the copper we have:
$$\sigma_C=E_C(\epsilon-\alpha_C \Delta T)\tag{2}$$
Note that, in these equations, the actual strain ##\epsilon## is the same for both materials (as indicated by the problem statement). If A is the cross sectional area of the copper and 2A is the cross sectional area of the steel, then the total tensile force F in the bar is given by:
$$F=(2A)\sigma_S+A\sigma_C=0\tag{3}$$Note that in this equation, the tensile force F is equal to zero since the bar is not under tensile load. Eqns. 1-3 can be combined to solve for the actual strain ##\epsilon## of both bars. What do you get (algebraically) for ##\epsilon##?

CivilSigma and Girn261
Chestermiller said:
It is clear to me that this problem has not been solved correctly by whomever provided the answer. So, let's go back and solve it correctly. For a rod that experiences a combination of thermal expansion and mechanical strain, the correct equation for the stress is:
$$\sigma=E(\epsilon-\alpha \Delta T)$$where ##\epsilon## is the actual strain experienced by the rod, and ##\alpha \Delta T## is the strain that the rod would experience if it were free to expand unconstrained (i.e., under conditions where the axial stress is zero).

So, for the steel we have: $$\sigma_S=E_S(\epsilon-\alpha_S \Delta T)\tag{1}$$and for the copper we have:
$$\sigma_C=E_C(\epsilon-\alpha_C \Delta T)\tag{2}$$
Note that, in these equations, the actual strain ##\epsilon## is the same for both materials (as indicated by the problem statement). If A is the cross sectional area of the copper and 2A is the cross sectional area of the steel, then the total tensile force F in the bar is given by:
$$F=(2A)\sigma_S+A\sigma_C=0\tag{3}$$Note that in this equation, the tensile force F is equal to zero since the bar is not under tensile load. Eqns. 1-3 can be combined to solve for the actual strain ##\epsilon## of both bars. What do you get (algebraically) for ##\epsilon##?

Thank you I will try that out when I get a chance

I get +26.25 MPa for the stress in the steel and -52.5 MPa for the stress in the copper (assuming the initial temperature of the bar was 20 C). So the copper exhibits a compressive stress, while the steel is in tension. This is because the copper wants to thermally expand more than the steel, so the copper is causing the steel to develop tensile stress, while the steel is holding back, and causing the copper to develop compressive stress.

CivilSigma

## 1. What is the stress/strain equation used for?

The stress/strain equation is used to measure the relationship between the force applied to a material and the resulting deformation or change in shape of the material.

## 2. How is stress and strain defined in the equation?

Stress is defined as the force applied to a material per unit area, while strain is defined as the change in length of a material per unit length. In the equation, stress is represented by the Greek letter sigma (σ) and strain is represented by the Greek letter epsilon (ε).

## 3. What are the units of measurement for stress and strain in the equation?

The units of measurement for stress are typically in pascals (Pa) or newtons per square meter (N/m2), while the units for strain are typically in meters per meter (m/m) or millimeters per millimeter (mm/mm).

## 4. How is the stress/strain equation used in materials testing?

In materials testing, the stress/strain equation is used to determine the strength and stiffness of a material. By applying a known force to a sample of the material, the resulting strain can be measured and used to calculate the stress. This information can then be used to evaluate the material's performance and durability under different conditions.

## 5. Are there any limitations to the stress/strain equation?

While the stress/strain equation is a useful tool for understanding the behavior of materials, it does have some limitations. It assumes that the material is homogeneous and isotropic, meaning that its properties are the same in all directions. It also does not take into account factors such as temperature, time, and fatigue, which can also affect a material's behavior under stress.

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