Bridge Guy Needs Building Help (simple lateral theory)

[bassplayer45]

This will probably seem trivial to alot, but having extremely limited building experience makes studying for part 2 of the test seem like college all over again. I will try and attach the example problem after the fact, but explain the question the best i can so you can explain the theory to me.

NCEES Structural questions and solutions (blue book)

Tilt up concrete wall warehouse with roof measuring 200 x 200 x 100 x 100.

Roof 26 ft A.F.F

Panels 26 A.F.F

Roof DL 15 psf

Panel DL 70 psf

Cs = 0.15

What is the chord force midway within the 200 foot wall.

Now, knowing my limited lateral theory, the chord force in the 200 foot section is equal to the distributed load moment divided by the 100 foot wall (couple)

My first question is. Why arent all 4 walls included in the total weight to be multiplied by Cs to get base shear?

My second question is, what is the theory behind (26/2)? (see below) as if it were a simple span wall? Wouldnt the dead weight of each wall simply be their area multiplied by 70 psf?

W = 100 * 200 * 15 (roof) + (2 * 200 * (26/2) * 70 psf) (2 walls?) = 664 kips

Appreciate any help

 

[mkaiser82]

1. Because the walls parallel to the load do not contribute to the load in the diaphragm. The shear due their mass stays in the wall.

2. Because 1/2 the wall weight is tributary to the roof level and will be transmitted to the diaphragm and the lower 1/2 is resisted by the connection(s) at the base of the wall. You are correct in thinking of it as a simple span between the base and roof level.

 

[nateluke]

This will probably seem trivial to alot, but having extremely limited building experience makes studying for part 2 of the test seem like college all over again. I will try and attach the example problem after the fact, but explain the question the best i can so you can explain the theory to me.

NCEES Structural questions and solutions (blue book)

Tilt up concrete wall warehouse with roof measuring 200 x 200 x 100 x 100.

Roof 26 ft A.F.F

Panels 26 A.F.F

Roof DL 15 psf

Panel DL 70 psf

Cs = 0.15

What is the chord force midway within the 200 foot wall.

Now, knowing my limited lateral theory, the chord force in the 200 foot section is equal to the distributed load moment divided by the 100 foot wall (couple)

My first question is. Why arent all 4 walls included in the total weight to be multiplied by Cs to get base shear?

My second question is, what is the theory behind (26/2)? (see below) as if it were a simple span wall? Wouldnt the dead weight of each wall simply be their area multiplied by 70 psf?

W = 100 * 200 * 15 (roof) + (2 * 200 * (26/2) * 70 psf) (2 walls?) = 664 kips

Appreciate any help

I am also a bridge guy and this question threw me off at first too. mkaiser nailed the answer to this one. These are two pretty important bits of information that seem to be glossed over frequently in books (no mention of this in SERM...about 1 paragraph explaining this in Lindeburg "Seismic Design of Building Structures" )

 

[bassplayer45]

Appreciate the quick response. I was racking my brain because i knew the theory of proportioning out shear forces to walls that are parallel to the seismic force, but couldn't grasp why only half the walls would be included for the diaphragm force. Thanks for the help.

 

[mkaiser82]

Any time. Good luck in your studying.

 

[bassplayer45]

Just signed up for the test (cue ominous music)