meeshu-
Thanks for the reply. In looking at the formula that I was using, w is the distributed load, not the weight of the beam. The units are lb/in and when I checked the units, they seem to work out to me. I'll have to check when I get home later today.
I will have to take a look at the...
Hello meeshu-
Thanks for your reply...here are my comments...
I got the same for the variables that I was using...with the exception of the density of the steel (I was using 0.283 lb/in^3) and the modulus of elasticity for steel (I used 30 x 10^6 lb/in^2)
The figures that I got were...
Hey-
I tried to reply to this earlier with my cell, but I guess that it didn't work. The I and A seem inconsistent with each other...interesting...
The actual dimensions of the beam are as follows. I'm planning on using a 5" tall x 2" wide steel channel welded together such that the...
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Okay...now that AlephZero showed me my first mistake...I'll try again...
Using the corrected value for w:
w=8.34 lb/ft (or 100.08 lb/in) for a uniformly distributed load
I = 23.05 in^4 (same as above)
Using the equation for deflection from above, and solving for y yields...
AlephZero-
You're right...it should be 29.5 * 0.283...my bad...
The area that I'm looking at is a hollow beam. I took the area of the larger rectangle and subtracted the area of the smaller rectangles, which should leave the area of the thin walled section. Is that incorrect?
AlephZero-
Thanks for the reply. The 104 that I came up with was from figuring the cross sectional area, multiplying by the density:
V = lwh
V = lA
V = (12)(2.46)
V = 29.5 in^3
If I take the volume and divide by the density, I get the weight per unit length:
29.5 in^3 / (0.283...
Hello-
This isn't a homework problem, but a real life problem. I'm trying to size a beam for a set of stairs that I'm trying to design. I have some material picked out, and it looks strong enough, but when I try to run some numbers to confirm, they don't appear to match what I've experienced in...