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.