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Seismic Load Combinations

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Mark1921

Structural
Feb 9, 2016
11
In reference to this problem:

For the given values of : ρ = 1.3, SDS = 1.10, the seismic load combinations are:

1.2D + 1.3QE + (0.2)(1.1)D + 0.5L = 1.42D + 1.3QE + 0.5L (Comb. 5)
when the signs of QE and D are the same, and
1.2D + 1.3QE − (0.2)(1.1)D + 0.5L = 0.98D + 1.3QE + 0.5L (Comb. 6)
when the signs of QE and D are opposite.
0.9D + 1.3QE + (0.2)(1.1)D = 1.12D + 1.3QE (Comb. 7)
when the signs of QE and D are the same, and
0.9D + 1.3QE − (0.2)(1.1)D = 0.68D + 1.3QE (Comb. 8)
when the signs of QE and D are opposite.

Among these combinations, Comb. 5 & 8 governs (according to my reference).

My question is: Between combinations 6 and 8, why does combination 8 governs, when both have 1.3 QE, and Comb. 6 having 0.98D compared to 0.68D of Comb. 8?

I just got stuck here! Thank you so much!
 
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Basic seismic is a lateral force. The lower dead load in combination 8 will do less to restrain the overturning effect on the building caused by the lateral seismic than the higher dead load in combination 6.
 
Think of a simple 2D Frame.

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Just thinking lateral for a second. If you push on top left corner you'll be pushing downward on right support and pulling upwards on left support. When you then add gravity loads (dead/live) in, the gravity loads will pushing down on both supports. So your gravity load is in the same direction as your lateral for the right support, but opposite for your left. That's why the two answers that are correct are the one with the most gravity load (1.42D+0.5L) and the least gravity load (0.68D). They'll give you the worst case downwards and uplift reactions, respectively.

Either combination 5 or 8 (as you've numbered them, numbering is different in the example you provided) will always control for lateral earthquake cases. 6 and 7 (again as you've numbered them) never will.
 
You should specify the element you are designing. If your lateral resisting system consists of shear walls or columns, then most probably the combination with the less axial force (Comb 8) will govern. Have a look on the interaction diagram (M-N) of the element. You will notice that big axial forces are beneficial since they reduce the tension stresses in the section, so you will need less tension reinforcement. At the limit if the axial force is zero, you will be in pure tension, and your section will require the maximal reinforcement.

Kamal Safa, PhD
Ecole Polytechnique
 
I second ajh1. That case would generally govern for overturning and possibly sliding as well depending on your foundation type.

To edit my response, I was referring to your load combination 8. I generally wouldn't do the load combinations 6 and 7 that you have shown. I normally only add the 0.2SdsDL to the 1.2DL case and subtract it in the 0.9DL case.
 
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