# Homework Help: Stress/Strain Calculations

1. Feb 17, 2016

### FritoTaco

1. The problem statement, all variables and given/known data
A ¼ in. diameter rod must be machined on a lathe to a smaller diameter for use as a specimen in a tension test. The rod material is expected to break at a normal stress of 63,750 psi. If the tensile testing machine can apply no more than 925 lb of force to the specimen, calculate the maximum rod diameter that should be used for the specimen.

2. Relevant equations
σ = F / A (simga = Force / Area) Sigma also means stress
A = π*r2
d = 2*r

3. The attempt at a solution
If I'm trying to find the max diameter the rod can be so the machine is just strong enough to break the rod. Am I first suppose to find the area as I did and got 0.049 and then calculate for sigma (stress)? Then what I haven't shown is trying to calculate for max diameter. So, first thing that's going through my mind, I look at d = 2*r. And I say I need to find a new r to get my new d. I look at A = π*r2, and think that if I plug in 0.125(1/4th) in for r and get A = 0.049. I don't know if that will help me or not. I next look at σ = F / A, I have σ (63,750 psi) and F (925 lb). I guess all that's left is A (area). So rearrange the equation to become A = F / σ to get 925 / 63,750 = 0.0145. Do I now take A = π*r2 and get r by itself to become r = A / π2 and that is 0.0145 / π2 which is 0.00147? Then that is the radius and multiply by 2 to get the diameter which becomes 0.00294? This seems unlikely and incorrect. Can someone point me in the right direction please and thank you. The work shown in the picture is a little different then what I also had in mind.

2. Feb 17, 2016

### SteamKing

Staff Emeritus
First of all, get in the habit of carrying units along with all your calculations. It will be less confusing for you and certainly less confusing for anyone reviewing your work.

Second, learn what the formulas you are using mean. If σ = F / A and you are given σ and F, then A must be the cross sectional area which gives σ when F is applied. All of this existential doubt running through your post is unnecessary and distracting. All you are doing is confusing yourself.

You calculated that the new A for the rod must be 0.0145. 0.0145 what? This is where units are vital.

Since the rod will have a circular cross section, then A = πr2 . When you tried to solve this equation for r, you got r = A / π2.

Third, check your algebra and arithmetic at all steps of your calculations. You had a simple equation for the area of a circle, and you solved it incorrectly to find r.

3. Feb 18, 2016

### FritoTaco

Sorry, I forgot about the units. 0.0145 in2. Also, yeah I won't start doubting myself anymore I'll just state what I have, thanks. Also, what do you mean I solved r incorrectly? Are you saying this part: r = A / π2 which equals 0.0145in2/ π = 0.00294in2. I just rearranged the equation to get r by itself and plug in A which was previously solved for.

4. Feb 18, 2016

### SteamKing

Staff Emeritus
If A = π ⋅ r2, then r ≠ A / π2. For a circular section, A = π ⋅ r2, not A = (π r)2

r2 = A / π. What do you have to do next to get r ?

5. Feb 19, 2016

### FritoTaco

Oh, yeah that seems correct I don't know why I had it flipped. So I got r2 = π/0.0145in2. r2 = √216.6. r = 14.72. Then I plugged r into d = 2*r. Which came out to be 29.42in2. Shouldn't the diameter be a lot smaller than that?

6. Feb 19, 2016

### SteamKing

Staff Emeritus
A = π ⋅ r2. So how did you get r2 = π / A ? When you get a non-sensical result, you must get in the habit of checking your work.

7. Feb 19, 2016

### FritoTaco

Oh sorry, I realized that was a mistake, I shouldn't be rushing. r2 = A / π. So now, r2 = 0.049in2 / π = 0.0156in2. √r = √0.0156in2 = 0.0124in2. That is r so now I should plug it into d = 2r. So d = 2(0.124in2) = 0.25in.

Last edited: Feb 19, 2016
8. Feb 19, 2016

### SteamKing

Staff Emeritus
Congratulations! You have calculated the original diameter of the un-machined rod.

Your calculations have gone somewhat awry after having dealt with the area of a circle calculation. The original A you calculated was 0.0145 in2.
From this, you calculated r2 = 0.049 in2, which is not correct. Always double-check your arithmetic for careless mistakes in copying results from your calculator to your paper.

9. Feb 20, 2016

### FritoTaco

So once I found 0.0145 in2. Do I plug that back in r2 = A / π which equals √0.0145in2 and r = 0.068in2. Multiply by 2 for diameter to get 0.136in2.

10. Feb 20, 2016

### SteamKing

Staff Emeritus
A / π = 0.0145 / π = r2. The diameter of 0.136 in < the original diameter of 0.25 in, which seems reasonable.

11. Feb 20, 2016

### FritoTaco

Alright, thank you for the help. I am now able to understand all the variables and formulas being used. I'm completely new to this so this helped, thanks!