Calculating the maximum load that can be carried & extension

Ive been given a question which i'm stuck on and cannot answer. Ive only been able to calculate the maximum allowable working stress and other then that im 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 section of 10.5mm x 12.5mm and a length of 0.9m (900mm). On testing this specimen it was found that the ultimate tensile stress was 420 N/mm^2. Calculate:

i) the maximum load that can be carried using a Factor of Safety of 2.5
ii) the extension under this load

So what i currently have worked out is the maximum allowable working stress which is 420/2.5=168 KN/m^2 (is this right?) and the cross section area which is presume is 10.5mm x12.5mm = 131.25mm^2 (is this right too?)

Thank you in advance

 

Thank you but what do i do after is what im stuck on

 

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? im not sure whether this is right

 

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

 

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 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 im assuming i did this correctly i acquired the force, but is the force "the maximum load that can be carried" as the question is asking? or do i do something else from here?

radius= 22.05/pie x 168= 0.4177817256
0.4177817256 square rooted = 0.2044m

Diameter = 2x0.2044=0.4088m=408.8mm
408.8m/2=204.4
pie x (204.4^2) = 131253.7304

extension = E = FxL / A x change in Length
so Change in length = 22.05 x 900/ 131253,7304 x 192.31 (this is the youngs modulus i got from my previous question)
= 0.000786mm extension

Now i feel as ive gone wrong somewhere here but im not sure

 

Thank you for that, so is 22.05 kN (the force) the maximum load that can be applied? Im 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 ive managed to complete)

And how did you determine that the specimen was of rectangular shape from the cross section? My lecturer told me that to find the diameter of a mild steel bar is to find the radius and then times the radius by 2 to get the diameter?

 

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

 

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?

 

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 ive provided?

 

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