normal stress of section 1 = 1590N / 0.15m^2 = 10600 N/m^2
normal stress of section 2 = 1590N / 0.05m^2 = 31800 N/m^2
Could it be that easy? Feel like I'm missing something still
Area of a rectangle is length * width
So at section 1, the area A would be 30mm*5mm = 150mm = 0.15m ?
and
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
Homework Statement
"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).
http://img208.imageshack.us/img208/121/tensilestressby1.th.jpg [Broken]
Homework Equations...
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
and
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.
http://img223.imageshack.us/img223/1103/1015zz1.th.jpg [Broken]
This is what I got so far:
EFx = Ra + Rb - P = 0
Ra + Rb = P
Sab = Sac + Scb =...
Says it attached and I can access it by clicking on it, but I uploaded it elsewhere as well.
http://img265.imageshack.us/img265/5564/363no0.th.jpg [Broken]
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...