Calculating Load Magnitude for Normal Stress in a Uniform Cross-Sectional Area

[positive infinity]

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)

http://img137.imageshack.us/img137/9845/problmezf4.png

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 I am going down the right track any help is appreciated thanks!

 

[Pyrrhus]

Where is A, B, D?

 

[positive infinity]

sorry about that i updated the picture

 

[positive infinity]

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 that's the force or do I have to calculate the cross sectional area of ABC if so how do I go about doing that?

 

[Bradracer18]

You will soon learn in this class, that MPA * mm^2 = N One is x10^6 and the other is x10^-6

 

[positive infinity]

alright I see that so MPA is N/mm^2 so I don't 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?

 

[PhanthomJay]

alright I see that so MPA is N/mm^2 so I don't 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?

I don't believe you are correctly understanding the definition of 'cross sectional area' and 'normal stress'. The given 800mm^2 cross sectional area of BD is its thickness times its width, and its plane is perpendicular to the normal (axial) force in member BD. You seem to be trying to calculate the area of member ABC (which, by the way, is not needed in this problem) by assuming that the cross section area is width times length. This is incorrect. Once you understand this, your next step is to draw a free body diagram of link ABC, as was suggested in the problem statement. In so doing, you must identify the magnitude and direction of the member force BD acting on joint B, and use statics/equilibrium equations to solve for P.