Structural Systems ncarb practice exam question #6

[ebiejohn]

I am nearing the end of my studying and I think my brain is mush. i need help in understanding why ncarb's answer is 57.3. and why I'm getting two separate answers, for that matter.

Let me show you my process.

We know that I = bd(cubed)/12 and S=I/C and (in this case) C=d/2

If we substitute bd(cubed)/12 in and the C/2 the section modulus formula, you get bd(cubed)/12/d/2

or bd(squared)/6

When i calculate using the revised formula (bd(squared/6), i get 46.7.

for illustration purposes:
(6)(10)(10)/6 - (2)(2.5)(8)(8)/6 = 100 - 53.3 = 47.7


If i go the long way, i.e. calculating for I of larger rectangle and subtracting 2 times the I of the void rectangles and then substituting that for I in the S=I/C formula, I get what ncarb gets.

I = (6)(10)(10)(10)/12 - (2)(2.5)(8)(8)(8)/12 = 500 - 213.4 = 286.7
S = I/c/2 = 286.7/5 = 57.3

Mathematically this should not happen and for the life of me, I can't figure out why it is happening that way. Maybe it is is something really simple but my brain is fried and I'm at a loss to see it.
e
any insight would be appreciated.

Thanks!

[Jeaston1972]

Don't over think it.

I think you are looking at question #5

The basic shape is a rectangle, therefore, 1/12 (b(d^3). DO the overall shape, then subtract out the two voids (also rectangles). Your formula should be 1/12 (b(d^3) - (2) 1/12 (b(d^3). The overall shape is the first part, the smaller void areas are the second. Your answer should be 57.33 in^3.

Question #6, just calc the tributary area, multiply by the stated live load...

[bench350]

I made the same mistake and noticed several comments regarding "Subtracting S" from the main body. This is not correct. You do not subtract the Section Moduli from each other on this type of problem but rather the Moment of Inertias. The formulas for S and I on a simple rectangular beam are so similar that it's easy to confuse S and I. The solution you show that provided the correct answer is the correct way to solve this problem but you can also use an elaborate formula (BH^2/6) - (bh^3/6H) (See Wikipedia).

[ErnestoCED]

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