ah, yes you are right. Unfortunately I don't have the correct answer in front of me anymore so I can't check if that's all that is wrong.
Hopefully the other stuff looks right though?
Here is what I tried:
Fx=500lbf
Fy=-800lbf
T=Fy14in
c=r (radius)
J=pi*r4/4
which gives torsion shear=-16900psi
V=Fy
Q=y'A'
y'=r
A'=pi*r2/2
t=r
with beam shear=-2844psi
total shear (xy)=torsion shear-beam shear
It's been a while since I was actually in Mechanics of...
Homework Statement
I have encountered this problem several times in class, but have never really understood it. There is a solid round bar that is fixed on one end, and that bends 90 degrees a ways down. At the free end of the bar there are forces applied in each Cartesian direction.
Like...
The problem is very similar to #9.74, here: http://books.google.com/books?id=o4-5K8xqlY0C&pg=PA504&lpg=PA504&dq=product+of+inertia+L5+angle&source=web&ots=KMDPgSTiT9&sig=rYBjTLwd-bTiA-IrMRCXB5XNycs&hl=en&sa=X&oi=book_result&resnum=1&ct=result#PPA504,M1"
I understand how the parallel axis...
I apparently read your first post before you edited it or something, so I was talking about something else. The edited one clarified it for me though.
Thanks
I have reduced the problem involving different forces and a couple in a 2D plane to an equivalent force-couple system. The force F is -27.5lbi + 11.65lbj, and the couple is -66.8inlbk.
I need the single offset resultant that accounts for the couple.
The way I thought to solve it was to take...