# Elasticity limit of copper Problem

• loudgrrl4_ever
In summary, the problem is to determine the minimum diameter of a copper wire under a given load without exceeding its elastic limit of 1.50E8 N/m2. The elastic limit is the maximum stress that can be applied while staying within the parameters of Hooke's law. The equation used to solve for the diameter is (pi/4)d^2 = F/stress, with the given values of force and elastic limit. The unit strain or change in length of the wire is not needed for this problem.
loudgrrl4_ever
Okay, here is the problem:
If the elastic limit of copper is 1.50E8 N/m2, determine the minimum diameter a copper wire can have under a load of 10.5 kg if its elastic limit is not to be exceeded.
Obviously I am looking for d in this problem. I am given the downward force (102.9 N = ma); A = (pi * d2)/4.
I am having trouble identifying the correct symbol for the elastic limit. My professor has a habit of hading out worksheets full of formulas with no explanation on them. For some reason I know that when given modulus, it will be called modulus, so this is the TSmax? I'm quite confused.
Trying it with 1.50E8 = stress, I get stress = F/A; A = F/stress; (pi/4)d*d = F/stress; d = (sqrt)[(4*102.9)/(pi * 1.5E8)]; d=.000934m; d=0.93400mm
If I knew the unit strain, or the length and change in length this problem would be much easier. I haven't done any with so little information before.
I haven't tried that answer yet (I just checked and I answered 0.093400 earlier when I thought I used this) but I would just like to know that I have done it correctly. My textbook spends all of a page and a half talking about this section, and it makes little sense to me unless the problems are simple.
Thanks :)

loudgrrl4_ever said:
Trying it with 1.50E8 = stress, I get stress = F/A; A = F/stress; (pi/4)d*d = F/stress; d = (sqrt)[(4*102.9)/(pi * 1.5E8)]; d=.000934m; d=0.93400mm
The elastic limit is the maximum stress (just check the units) that can be applied while staying within the parameters of Hooke's law. So your thinking here is just fine. (I did not check your arithmetic.)

If I knew the unit strain, or the length and change in length this problem would be much easier.
All that is irrelevant.
I haven't done any with so little information before.
This problem is probably easier than you are used to.

Following up on this, do not confuse the elastic stress limit of a material, often called yield stress, and quite often denoted as $$\sigma_y$$, which has units of $$force/length^2$$, with the modulus of elasticity of a material, often called $$E$$, which also has units of $$force/length^2$$. Both are properties of the material that are independent of each other. The yield stress denotes the stress level above which point the material will no longer obey Hookes law. The modulus of elasticity is the slope of the stress - strain curve within the elastic range, and is a measure of how much the material will strain (e) or elongate/compress under a given stress, per $$\sigma = eE$$. As Doc Al noted, you have the correct equation for determining the required diameter; the strain in the material is irrelevant in this problem.

## 1. What is the elasticity limit of copper?

The elasticity limit of copper is the maximum stress that the material can withstand before it undergoes permanent deformation. This value varies depending on factors such as temperature and the type of copper alloy.

## 2. How is the elasticity limit of copper determined?

The elasticity limit of copper is usually determined through tensile testing, where a sample of the material is pulled until it reaches its maximum stress point. Other methods such as hardness testing and fatigue testing can also be used.

## 3. What factors affect the elasticity limit of copper?

The elasticity limit of copper can be affected by factors such as temperature, strain rate, and the presence of impurities or defects in the material. Alloying elements can also influence the elasticity limit of copper.

## 4. What is the importance of knowing the elasticity limit of copper?

Knowing the elasticity limit of copper is crucial in engineering and construction applications, as it helps determine the maximum load that the material can withstand before it permanently deforms. This information is also essential in designing safe and efficient structures using copper.

## 5. How does the elasticity limit of copper compare to other metals?

The elasticity limit of copper is relatively low compared to other metals such as steel and titanium. However, copper has excellent ductility and can withstand high levels of strain before failure, making it suitable for a wide range of applications.

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