Maximum Rod Diameter Calculation for Tension Test Specimen

[FritoTaco]

Homework Statement

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.

Homework Equations

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

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.

 

[SteamKing]

Homework Statement

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.

Homework Equations

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

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.
View attachment 96032

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.

 

[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.

 

[SteamKing]

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.

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 ?

 

[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?

 

[SteamKing]

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?

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.

 

[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.

 

[SteamKing]

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.

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.

 

[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.

 

[SteamKing]

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.

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

 

[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!