# Homework Help: Stress Problem

1. Nov 11, 2006

### jubaitca

have uploaded the question on the web. Please have a look at:

I have tried doing this problem, but i am not getting the right answer. This is what i did:

y(unknown)=40cos60 = 7 mm
x (unknown) = 40 cos 30 - 20 == 14.641mm

Then

Pcompression = 2(75x7x(34.641/2)) + 2(((20x75)/2)(14.641)
= 40149 N

But the answer is somehow... 73.92 KN

Is there anything wrong with my calculations?

Thanks

Last edited by a moderator: May 2, 2017
2. Nov 11, 2006

### PhanthomJay

You might note first off that this beam does not obey Hookes Law, that is, it is beyond its elastic range in the outer fibers and in a semi-plastic state, ( or else the stress distribution would vary linearly rather than uniformly over the top section). I think the problem may have been written that way to make the calculations less tedious. Anyway, you have proceeded correctly, but have made 2 math errors. 40(cos60) is 20 (not 7!). And your conversions are off: 75MPa is 75*10^6 N/m^2, and you should convert the mm distances to meters (I'm glad i work in psi and inches!). Once you make those corrections, you will get the correct compressive (and tension) force of 73.92kN. To get the moment , you've got to find the c.g. of the tension and compression load, then determine the couple.

Last edited by a moderator: May 2, 2017
3. Nov 12, 2006

### jubaitca

thanks a lot...im so stupid at times....by the way, i don't need to convert it to meters because MPa = N/mm^2.

but thanks a lot, i have wasted so many hours thinking on this and the error just was that i could use my calculator properly...lol

4. Nov 12, 2006

### PhanthomJay

Yes, thanks, I'm so used to USA units that metric leaves me cold. The use of SI in beam anaysis and design in the States is virtually non existent, and seing as how attempts over the past 40 years to convert have gone nowhere, I suspect that it always will be.