Recent content by ziziu

Oh nice one. Thank you guys. Very helpful site. Seems pretty simple once you know how.. Thanks again.

I'm still a bit lost. Max allowable load = Working stress x Area ? Pi x 12.5^2 = 490.87mm^2 Is the area of the 25mm diameter. 221.181 MPa x 490.87mm^2 = 108.56 kN Max allowable load = 108.56 kN? Pretty much a guess.

Thanks for your reply although i still don't fully understand. I understand what you are saying about the ult stress divided by the safety factor to give me working stress. But is the question not asking me to do this for the new diameter? 774.1326 mm^2 is for the diameter 12.5 in part A...

Sorry yes that wasn't meant to have 2 there.

"Determine working stress and greatest allowable load" Homework Statement The maximum load in a tensile test on a mild steel specimen of diameter 12.5 mm is 95 kN, calculate the ultimate tensile stress. Also, determine the working stress and greatest allowable load on a rod of the same...