Maybe I was too hasty. The units of I are in^4/ft. The expression for [λ] is (12k/4EI)1/4. I think the factor 12 is converting feet to inches, so the units work out okay.
Using the final expression for P, with f'c = 4000 psi, te = 10 in. and k = 100 #/in^3, I find P = 1375 lb/lin ft.
This compares to Table 3-2 values of 1340 to 1605 for flexural strengths of 550 to 700 psi which seems reasonable, but I am not sure how flexural strength relates to f'c.
Table 3-2 on page 3-5 gives a value of 1340#/ft when k = 100, t = 10 and flexural strength = 550.
The value of f'c corresponding to the above expression is:
f'c = (550/9)2 = 3734 psi which, when used in the final expression on page B-1, attached gives a value of 1340.
Agree with you. I had a brain fart and was ending up with (lb/in) instead of (lb-in) in the denominator. I guess you overlook really simple, stupid things and need someone else to point it out to you. Thanks.
I did something similar the first time, too. It seemed right to consider I in units of in^4 whereas it should be in^4 per foot. I also was not sure of the relationship between flexural strength and f'c, but it seems that 9[√]f'c is what is normally used.
I still feel more comfortable reinforcing the slab to spread the wall load over a width determined on the basis of allowable soil bearing than to rely on unreinforced concrete to do the job.
Perhaps this would be a useful technique to use when adding a heavy wall load to an existing slab of known thickness and very little reinforcement. Is that your situation?
Yes, my situation is the same. The client wants to add an electrical room with a roof slab inside an existing building and I was trying to check the slab on grade.
The drawings indicated a 5" slab with 6x6 WWR, but the cores came out to anywhere between 6.5" to 7".
I had realized the 12 came from converting in^4/ft to in^4/in, but I made a math error in the denominator and did not get it work dimensionally.
