James G. MacGregor 3rd Edition, Design of Beam when b and d not Known 1

[CWEngineer]

I have been trying to figuere out Example 4-6 on page 126 and I am having a hard time understanding a particular part.

In this example, on part 3: b and d are computed using the following equation. (Msub_u/phi*ksub_n)=((bd^2)/12000).

Then on part 8: it shows Msub_n= (Asub_s)*(fsub_y)(d-a/d). When the numbers are plugged into this equation it is divided by 12000. I know this has to do with the equation in part 3, but I am having a hard time trying to figuere out why.

I would appreciate if someone can help me figure this out, hope my explanation is not to confusing.

THANKS

 

[rholder98]

The 12000 is a unit conversion between in-lb and ft-kip. From part 8: 5.58 (in^2) x 60000 (psi) * [33.0 in - (7.03/2) in] = 9871578 in-lb --> *(1 ft/12 in)*(1k/1000#) = 823 ft-k.

Hope that helps.

 

[CWEngineer]

Thanks, I just finished figuering that out. SORRY. I spend some time trying to find out in the text and I was trying to somehow relate that 12000 in part 8 with the 1200 in equation part 3. Make sense now.

THANKS