All things being equal, the 7" square tubing will have a beam stress about 1.5 times lower than the 8" square, assuming the beams are not loaded over their yield points.

8" is will bend less.

For deflection you divide by I, the moment of inertia & it is about 73 in^4 for the 7 in & about 77.6 for the 8. Dividing by the bigger number will give you a smaller deflection.

I'm assuming A36 material, you want somebody to work this out & figure the deflections & stress?

max deflection for a beam fixed at one end with a point load on the other can be found with the following equation:

(P*L^3)/(3EI)

Where P is the load, L is the length, E is the elasticity modulus, and I is the section inertia perpendicular to the neutral axis.

Pretty simple eq. Note that all the units have to be correct, for example, P in pounds, L in inches, E in PSI, and I in inches^4. You end up with (lbs*in^3)/(3*lbs/in^2*in^4) which all cancels out to be just inches. Which is the maximum deflection at that point where the load is applied, AKA how much the beam droops at the end.

What does ^ mean? How do I figure out I for dummies?

Thanks and here are a couple pix of what I have so far.


http://forums.pelicanparts.com/uploa...1233507275.jpg
http://forums.pelicanparts.com/uploa...1233507333.jpg

^ mean 'to the power'. I have no other way of writing it on the forums.

So 4^2 is '4 squared' = 16

L^3 is 'L cubed' etc.

Thanks that is what I thought. What about section inertia. How do I figure that out?