Recent content by Mido999

Have I completed this question or do I still need to work out the maximum load?

Thank you very much for your and everybody that posed on this thread's help! Would you say that I've finished this question with the answers I've provided?

Disregard my previous question, what about this for the extension: 22.05 x 900/131,25 x 192.31 (young's modulus from before) = 0.786mm extension?

What is the formula to find the diameter of a rectangular specimen?

Thank you for that, so is 22.05 kN (the force) the maximum load that can be applied? I am not too familiar with this subject The young's modulus calculation was stress/strain so 0.2/1.04 x10 to the power of -3 = 192.307692 = 192.31 kN/mm^2 (this was the previous question which I've managed to...

I gave it another try and got the following: Cross section = 10.5mm x 12.5mm = 131.25mm^2 0.9m length =900mm Ultimate tensile stress = 420/2.5=168N/mm^2 168N/mm^2=168,000,000 Pa 168,000,000 x 0.00013125m^2 (131.25mm^2) = 22050N = 22.05 Kn Force = 22.05 Kn At this moment I am assuming i did...

I think it's either N/mm^2 or Kn/mm^2? - I'm sorry for being so slow by the way guys, I'm pretty new to this field so I hope you guys can bear with me

I don't know whether I'm allowed to bump this thread but my deadline is on Friday and I really need to get this done am I on track or far off? Continuing from my previous post

The young's modulus from my previous question was 192.31 Kn/mm^2 To find out the load i suppose i would need to maximum tensile force right? is this done by dividing the maximum allowable stress by the cross sectional area so 168/131.25=1.28 Kn? I am not sure whether this is right

Thank you but what do i do after is what I am stuck on

Ive been given a question which I'm stuck on and cannot answer. I've only been able to calculate the maximum allowable working stress and other then that I am stuck on how to answer the following question: A specimen of the same material that was used in the above test (mild steel), had a cross...