# Solve Stress-Strain Question: 3in Copper Rod Reduces to 2in Diameter

• Eastonc2
In summary, the problem involves reducing a 3-in diameter copper rod to 2-in diameter by pushing it through an opening. To account for elastic strain, the diameter of the opening should be 1.995 in. However, there were some errors in the calculations that led to a slightly different answer.
Eastonc2
[1] Problem Statement:
A 3-in. diameter copper rod is to be reduced to 2-in dameter by being pushed through an opening. to account for elastic strain, what should the diameter of the opening be? Modulus of Elasticity (E) =17e6 psi, yield strength (YS) = 40,000psi.

[2] My work:
I've tried to relate the change in area to the change in length assuming constant volume, (Ao/A = L/Lo) and got final length = 2.25*Lo. using 1 for Lo, Lf = 2.25 in. I then just used the modulus of elasticity to calculate how much strain would be present before stress was removed, using YS=E(x-1.25), giving me x = ~1.2524 in. Adding that to Lo, gives me a final length under stress of 2.2524 in. I then related this back to (Ao/A = L/Lo), and got A= ~3.13825. diving by pi, and taking the sqrt, I get r= ~.99947 in., so Diameter = ~1.9989 in.

The book lists an answer of 1.995 in.

I am very confident that I'm doing something wrong here but these are all the methods discussed in the book, and in class thus far, so if anyone can point me in the right direction i'd appreciate it.

Thank you for posting your work and question. It seems like you have a good understanding of the concepts and equations involved in this problem. However, there are a few things that could be corrected to get a more accurate answer.

Firstly, when relating the change in area to the change in length, the equation should be Ao/A = (Lo+ΔL)/(Lo). This takes into account the initial length (Lo) and the change in length (ΔL).

Secondly, when calculating the final length under stress, the equation should be Lf = Lo + ΔL, where ΔL is the change in length due to the stress. In your calculation, you used 1.25 as the value for ΔL, but this should be the change in length at the yield strength, which in this case is 40,000psi.

Finally, when relating the change in area to the change in length, the equation should be Ao/A = (Lo+ΔL)/(Lo). This takes into account the initial length (Lo) and the change in length (ΔL).

By correcting these errors, the final diameter should be 1.995 in, as listed in the book. I hope this helps and good luck with your studies!

## 1. What is stress-strain and how is it related to the reduction of a copper rod?

Stress-strain is a measure of the relationship between the force applied to a material and the resulting deformation or change in shape. In the case of the copper rod, the stress-strain is referring to the amount of force applied to the rod and how it causes the rod to reduce in diameter from 3 inches to 2 inches.

## 2. How is the stress-strain question solved?

To solve the stress-strain question, we need to use the formula for strain, which is the change in length divided by the original length. In this case, the change in length is 1 inch (from 3 inches to 2 inches) and the original length is 3 inches. Plugging these values into the formula gives us a strain of 1/3 or 0.33.

## 3. What factors can affect the stress-strain relationship in a copper rod?

There are several factors that can affect the stress-strain relationship in a copper rod, including the purity of the copper, the temperature at which the force is applied, and any previous deformation or damage to the rod. These factors can alter the rod's ability to withstand stress and may affect the resulting strain.

## 4. How is stress calculated in a stress-strain question?

Stress is calculated by dividing the force applied to the material by the cross-sectional area of the material. In the case of the copper rod, if a force of 100 pounds is applied and the cross-sectional area is 1 square inch, the stress would be 100 pounds per square inch (psi).

## 5. What are some real-world applications of stress-strain relationships?

Stress-strain relationships are important in many industries, such as engineering, construction, and manufacturing. They are used to design and test materials for various applications, such as building structures, bridges, and vehicles. They are also used in the medical field to understand the strength and durability of materials used in medical devices and implants.

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