Area of a rectangle is length * width
So at section 1, the area A would be 30mm*5mm = 150mm = 0.15m ?
Section 2, the area A would be 10mm*5mm = 50mm = 0.05m ?
Then I would just plug those areas into the normal stress = P/A equation for both sections?
Thanks for the quick reply
"The axial load for a given test sample carries 1590 N. Calculate tensile stress at sections (1) and (2) assuming the sample thickness is 5mm. (Rectangular cross section).
Ah excellent! Thank you very much. The equations for Ra and Rb were a little more complicated but I got the answers the back of the book has.
I'm surprised I didn't see that I was cancelling out both A's when they clearly aren't equal to each other.
Thank you very much. This forum has...
I didn't assume the cross-sectional areas were the same. The fat piece is 6 in^2 and the skinnier piece is 3 in^2.
In my thinking:
Stress of AC should = 44.44 kips/6 in^2 = 7.41
Stress of BC should = 55.56 kips/3 in^2 = 18.52
However those are not the answers. In the back of the...
I've been having difficulty with this problem for the past 5 hours. I am understanding some part, but I can't get the final answer.
This is what I got so far:
EFx = Ra + Rb - P = 0
Ra + Rb = P
Sab = Sac + Scb =...
I've been having trouble with the homework problem attached as an image.
I need to find the tension in the cable and the reactions at A (Ay and Ax). In previous homeworks, I've been able to find the reactions at similar points A by finding the moment about A around a few other points. Now...