PEin2010
Well-known member
So I tried to solve Hiner Problem #12 by using the equations which have Qe instead of E in the load combinations and I miraculously got the same answer as what the solution is describing. My question is that I didn't bother applying a plus or a minus sign, so I'm afraid I may have just got the right answer by fluke. For the maximum axial force in column B, I used, (1.2+0.2Sds)D + rho(Qe) and for the minimum I used (0.9-0.2Sds)D - rho(Qe). The only reason I used them was because I got the maximum and minimum answers from those. Please provide comments as to whether there is any merit to my method!!!
Thank you!
I attach the problem here for reference. The solution calculated E = Eh+/- Ev first and then uses the load combinations based on that.
Thank you!
I attach the problem here for reference. The solution calculated E = Eh+/- Ev first and then uses the load combinations based on that.
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