Moment of Inertia quest

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EM_PS

shining like a lighter...
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Ok, so i have a problem w/ one ofthe NCEES prac probs involving selecting a beam.

using the beam deflection formulas for the conditions stated, i get a calculated Ix value of 103 in4 - They get a value of 120 in4 and I'm just not seeing how they got it.

Any insights as to what i'm doing wrong? I can post my calcs & their solution, but i wanna see what others may arrive at unbiasedly.

thanx, EM

 
Jeeez, nevermind I found the problem - Pa^2/6EI * (3L-a) - i was taking the 3L part for the first deflection calc to be 3L/2 - Recalc i got 120.13 in4

Sorry, glad i caught it first i guess :blush:

 
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^^Great! I just spent about 20 minutes calculating this myself and I came up w/120. So I was trying to help!

I did confirm that 120 is correct so maybe that helps.

 
oops sorry - Yeah for whatever reason, i used L/2 (for L) for calculating the deflection from the 1st loading scenario (1 kip, half-length) - the 'a' value is L/2, but where L is called for in that calc (max deflection value) , L = L, no matter what. Part of the problem is i could not follow along with their hodgepodge solution equation, but having duplicated the answer, feel good about it. I try bouncing this stuff off our cat, but sadly he mainly rebuffs me.

 
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This is what i did to setup problem:

[1kip*(6' x 12in/1ft) ^2/(6*29000ksi*X) *(3L-a)]+[1kip*(12' x 12in/1ft) ^2/(6*29000ksi*X) *(3L-a)]∶= 0.375in

Where:

(3L-a) left side = (3*144” – 72”) {originally I was using (3*72”-72”)}

And

(3L-a) right side = (3*144”-144”)

The utility of the Casios here is great. Just punch in equation, use SOLV for X variable, which is the moment of inertia, Ix. 4 minute solution easy enough (hopefully!)

Their solution, they solved for max def values for the 2 loadings w/out moment of inertia values, then took the sum of them divided by the 0.375 in to get their Ix value. It works the same, just didn't really follow that particular method.

and i'm sure nobody gives a flaming rip. . . i'll shaddup now

 
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