## strength of materials problem

Knowing the portion of the link BD has a uniform cross-sectional area of 800mm^2, determine the magnitude of the load P for which the normal stress in that portion of BD is 50 Mpa. (Hint draw a free body diagram of the link ABC)

sorry for the paint sketch I don't have a scanner, anyway I approached the problem as σ = F/A and we already have σ and A so I rearranged the problem as σ*A=F which came out to 40000N or 40Kn I'm not sure as to where to go from here or if Im going down the right track any help is appreciated thanks!
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 sorry about that i updated the picture

## strength of materials problem

ok i redid the math in order for mpa to be converted to force I would have to convert all the measurements to meters so now I get 40N. but I am still stuck I'm not sure if thats the force or do I have to calculate the cross sectional area of ABC if so how do I go about doing that?
 You will soon learn in this class, that MPA * mm^2 = N One is x10^6 and the other is x10^-6
 alright I see that so MPA is N/mm^2 so I dont have to convert anything so it would be 40KN, 50MPa or 50N/mm^2 * 800mm^2 = 40,000N or 40KN, butI'm still stuck on the other link would it have the same thickness as link BD? link BD 800mm^2/510mm = 1.57mm??? is it the same for the rest of the problem?

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