An Engineering Question! (1 Viewer)

[Silverstone]

The maximum force allowed in member GF is 100 kN.
If a factor of safety of 2.5 is applied to calculations, what is the minimum cross-sectional
area of member GF?
(A) 191 mm 2
(B) 477 mm 2
(C) 1190 mm 2
(D) 2976 mm 2

Anyone have any ideas? The answer is C, but how do I solve it?

I know F.o.S = UTS/Stress

Greatly Appreciate it.

 

[Xayma]

That is the factor of saafety for brittle memebers anyway.

You need more information such as the maximum stress it can hold.

After all if a steel member and an aluminium member both had a maximum force allowed of 100kN, then the aluminium member would need a larger area.

σ=F/A
A=F/σ

Where F is 100kN*2.5

EDIT: Hmm can we get HTML turned on in here if it isn't too much trouble.

 

[A2RAYA]

The maximum force allowed in member GF is 100 kN.
If a factor of safety of 2.5 is applied to calculations, what is the minimum cross-sectional
area of member GF?
(A) 191 mm 2
(B) 477 mm 2
(C) 1190 mm 2
(D) 2976 mm 2

Anyone have any ideas? The answer is C, but how do I solve it?

I know F.o.S = UTS/Stress

Greatly Appreciate it.

ok firstly u forget about f.o.s till the end

u got stress= F/A

the stress is given right?...and unless im wrong i've done this question and the stress given is something like 210 MPa right?

anyway, assuming it is we have 210*10^6= 100*10^3/A

therefore we get A= 100*10^3/210*10^6
= 476*10^-6 (rounded off)

now times this by 2.5 to take into account ur f.o.s and bingo u get 1190*10^-6 m^2....or 1190mm^2

hope this makes sense...its pretty hard getting the subscripts n stuff here but yeh...its like 3 lines of working