HORIZONTAL DISTRIBUTION OF DESIGN BASE SHEAR - ASCE 7-05 12.8.4

[kngpenn]

I HAVE A 2 STORY BUILDING
14' TO THE SECOND STORY LEVEL
14' TO THE ROOF LEVEL
AND 3' OF PARAPET

THE DESING BASE SHEAR IS 398 K

DOES ANY ONE HAVE A SIMPLE METHOND OF DETERMINING THE HORIZONTAL DISTRIBUTION?

 

[Lion06]

The horizontal distribution is dependent on the relative stiffnesses of your lateral resisting elements (if you have a rigid diaphragm), and just by trib width for a flexible diaphragm.
The Vx is the Fx from 12.8.3

 

[kngpenn]

I HAVE METAL DECK WHICH MAKES A RIGID DIAPHRAGM. CAN YOU POINT ME TO AN EXAM PERHAPS?

 

[kngpenn]

AN EXAMPLE I MEAN...

THIS IS WHAT HAPPENS WHEN YOU TRY TO WORK AND TYPE AT THE SAME TIME.

 

[Lion06]

I don't know of a textbook example, but you need to calc the relative rigidity of each of your lateral resisting elements and the location of the center of rigidity. Then assign a direct shear based on the relative stiffnesses. You will also have a torsional moment which will be resisted by all of the shearwalls in both directions. This will add some shear to the direct shear in some cases, in other cases it will work to reduce the direct shear.

 

[KBVT]

I hope your "metal deck" refers to a composite system or you may have some trouble achieving your rigid diaphragm requirements.

 

[COEngineeer]

kngpenn, the horizontal distribution depends on where the location of your shear wall / lateral frame are. If you have a rectangular building with 4 walls around it, then it is easy. Just divide the shear with the length of the wall parallel to the load. But more than likely your building wont be that simple so the center of mass is not equal to center of rigidity so you will create torsion (your building is twisting). So you will have shear due to torsion on the top of your direct shear. Good luck! And like KBVT said, you need concrete on top of your metal deck to be considered rigid.

Never, but never question engineer's judgement

 

[whyun]

I find doing it by hand by far the quickest way to distribute a single level horizontal distribution to LFRS below. This gets you a "rough" idea how the story shear will distribute. Understanding of direct shear and torsional shear is essential.

Creating a computer model may get you more accurate results but it is time consuming to create the model. Only when one can predict the behavior shall one create a computer model to verify the anticipated results.

 

[JLNJ]

You might need to include accidental torsion even if your building is perfectly symmetrical.

Had a question like that on the SEII...