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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.
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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?
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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.
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Quote:
Thanks and here are a couple pix of what I have so far. ![]() ![]()
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^ 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.
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Quote:
Thanks that is what I thought. What about section inertia. How do I figure that out?
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