# NCEES Structural Blue Book, Lateral question 110



## bassplayer45 (Mar 1, 2013)

Hi again, appreciate the input. Just want a verification of the theory on problem 110 of the lateral portion of the blue book to just make sure i understand it correctly. As a bridge guy with the test approaching, my building theory is still lacking.

It is looking for the drag force in member A. Based on the way the building is set up, all walls running in the north and south direction do NOT contribute to the total seismic force in the diaphragm because their forces stay within their own walls. Therfore, the walls that DO contribute to the total diaphragm force are the 100 foot long walls running east and west.

In the solution, they calculate the weight at diaphragm level using simple beam equations, then use the tributary weight of the roof for the seismic force in member A.

For the shear force due to seismic, they multiply the weight of the wall by (40/2) * 2 for the left portion and (60/2) * 2 for the right. Is this stating that half the shear in the "bay" goes to the wall side and the other half to the drag strut?

Second, when they proportion out the loads, they state the full 7818 shear goes to the strut and (40/60) * 13528 goes to it as well. Is the (40/60) proportioning becuase the diaphragm does not extend the full 100 feet and only takes that portion of the shear?

Appreciate the help. Slowly getting there, these message boards have been a life saver.


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## Ble_PE (Mar 1, 2013)

To me it helps visualize what's happening when you call the drag strut by its other name, collector. The purpose of this member is "collect" the forces from the flexible diaphragms and transfer them into the shear wall. That's one thing to note is that the end walls are shear walls, as well as the N-S wall that connects the diaphragms.

You are correct in your assumption that the diaphragm forces are split in "half" at the drag strut. Flexible diaphragms work in the same way simply supported beams work. Under a uniform load, half of the shear is transferred to each support. The reasoning behind how they sum up the shear loads for the total load in the drag strut takes into account the wall at the south end of the building. That wall supports a proportional amount of shear from the 60' wide diaphragm equal to its length divided by the overall length of the diaphragm, i.e. 20'/60'.

I hope that this helps, it's a bit hard to type out an explanation to a problem like this.


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## ajk244 (Mar 1, 2013)

In case you still have some confusion on the last part of your question, hopefully this helps. Here's a quick sketch of an elevation view of that collector and that inside corner wall.

To visualize what's happening, you can draw an axial diagram for the collector (The hatched area on the sketch). For the distributed load applied at the roof level, the axial load is increasing unimpeded from left to right until the wall provides the resistance and the axial diagram goes back down to zero. The max. axial in the collector occurs right where it meets the wall. At that point, the full 40'of the left diaphragm is contributing load and 40' of the total 60' of the right diaphragm have contributed load before the wall provides the resistance.


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## bassplayer45 (Mar 3, 2013)

Thanks alot, these helped out quite a bit!


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