seismic dead load 2

You need to include all the building's weight in the determination of the seismic forces on a building. Excluding walls just because they are parallel to some theoretical earthquake direction seems like courting disaster to me.

GraemeL is absolutely right. There are at least two ways to handle this.

One way is to sum all the weight of walls (using proper tributary heights) divide by the floor area to estimate weight per area that should be added to the floor dead weight.

Alternate is to only account for the perpendicular walls and when parallel walls are being designed, apply the appropriate seismic load due to self-weight. This may be applied at the center of the wall OR conservatively be applied at the top.

Which method to use depends on they type of building under consideration. First method is common for light framed building with plywood/OSB sheathing. Second method is used for concrete or masonry wall structures.

WHYUN:

I am a bit confused by your response.

If, for example, I was analyzing a tall, load bearing CMU gymnasium (1 story), the base shear value used for analysis would be highly dependant upon whether the walls parallel to the direction considered were included in 'W'. Am I missing something here?

Would it be safe to say:

Regardless of what direction the seismic load is being applied, the ENTIRE structure weight is to be used to determine the BASE SHEAR, V. The only variable then, would be the Response Modification Factor,(R), of the lateral load resisting framing being designed. ?

I appreciate all opinions and responses!

Thank you

Assuming that the same R factor applies in each direction, the total base shear is indeed the same for each direction. The issue is how this base shear is distributed to the diaphragms and frames/walls.

Say, for example, that you have a one-story rectangular building with a flexible diaphragm and masonry shear walls on four sides. The roof diaphragm acts as a deep beam to deliver the loads from the roof itself and the upper portion of the transverse walls to the longitudinal walls. The longitudinal wall forces need not be included in the diaphragm analysis. The longitudinal wall forces are applied directly at the wall where they originate. The diaphragm loads may be different in each direction, but when you sum the forces at the base of the structure, the total will be equal for each direction.

Here is how the examples in TI 809-04 (the Army Corps of Engineers' "seismic design for buildings&quot handle this question.

Say that your tall CMU load bearing wall gymnasium has a flexible roof diaphragm. In this case, the base shear in each direction would be determined (using the appropriate R value for the lateral system in that direction) based on the roof diaphragm's self weight and the weight of the perpendicular walls. Then, when designing the shear wall for in-plane forces, that wall's own inertial forces would be added to the distributed base shear forces. You add the inertial forces AFTER you've distributed the base shear because there's no way that the inertial forces in one end wall could suddenly jump out of the wall and be transmitted through the flexible diaphragm and into the other end wall.

Now, for a rigid diaphragm, there is the possibility that the inertial forces in a wall could be distributed to another parallel wall. In the case of a rigid diaphragm, the base shear is calculated based on the weight of the entire building including both parallel walls and perpendicular walls. (In this case, WHYUN, you are correct, the only difference between the base shears in either direction would be the R values.) This total base shear is then distributed based on the walls' relative rigidities.

So, in both cases, the inertial forces in the walls are being accounted for SOMEWHERE. In the inital post by wailuku, it appears this person was asking if the forces due to parallel walls can be omitted. Obviously, they must be accounted for somewhere. It is really just a matter of answering the question, "Can the inertial weight of a wall be distributed to other parallel walls, or must the wall be designed to resist it's own entire inertial weight?"

I hope this helps!

Shemp:

I have been trying print the USACOE 809-04 and can't seem to get Chap 7 part 4 and part 6...anyway, where is this example you speak of?

I am still confused by the term "Seismic Base Shear" then.
How would one differentiate between a single base shear value based upon the entire building weight (assuming R's are the same) and a "Directional Base Shear" that does not account for the wall weights normal to plane being analyzed?

In accordance with the newly adopted NYS Building Code, we must LIST the "Design Base Shear", and up until this point, I have been listing two (2) seperate values...North-South, East-West and basing these values under the assumption that the walls normal to the plane are ignored. This is substantial considering, for example, long load bearing walls where all of the roof snow load and dead load is applied, verses the non-load bearing walls where only a small fraction of roof dead load and the walls self-weight are applied!

Again, appreciate the help!

383CUDA,
The examples that I was looking at are in Appendix H of TI 809-04. You can download it at this website:


In your question, you mentioned a directional base shear that does not account for the wall weights NORMAL to the plane being analyzed. Walls that are normal to the plane being analyzed should always be included in the base shear calculation. This is because the shaking will cause the normal wall to vibrate in its out of plane direction, which will produce horizontal reactions that go straight into the diaphragms.

Since you now have to show the base shear values on the plans, here is what I would do. If you have the case that I described in my last post where you have a flexible diaphragm and you choose to calculate 2 different "directional" base shears that omit parallel wall weights, then I would put a note saying something to the effect of, "weights of parallel walls are not included in base shear values, however, are considered in the design of shear walls."

Clear as mud?

Shemp:

OK, how's this!

Would it be CLEARER and SIMPLER to state the DESIGN BASE SHEAR based upon the TOTAL building weight, regardless of direction? This is how it is shown in several older references I have, but this was the V=ZIKCsW days!

And thanks again for the help, I appreciate it!

Yeah, that would definitely be the simplest way and I'm sure no one would ever have a problem with that.
But, I'm thinking that for the design of the shear walls, it may not be the most conservative way to go. I'm kind of thinking out loud here... Say that you have a flexible diaphragm building that's 100ft long with three shear walls all in the same direction. One shear wall is at the end of the building, the next happens to be only 10' away, and the third is 90' from the second wall at the other end of the building. In this case, one end wall would receive (by tributary widths) 5% of the lateral forces, the second would receive 50% and the third would receive 45%. If you were to take the entire base shear (meaning the self weights of the parallel walls are included) and then distribute it according to these percentages, then the wall that only receives 5% would probably be designed for less force than the actual self weight inertial force.
Now, I don't think the wall would fall down just because it wasn't designed for self weight inertial forces, but technically it would be underdesigned.

Shemp:

Thanks again for the responses!

If I could throw 1 more thing at you, I'll let my portion of this thread pass on quietly...

Is it always correct to apply the lateral load as point loads to the floor or roof level? For example, I have seen examples which apply the lateral force as an inverted triangle, therfore inducing out of plane bending to the wall normal to the direction being investigated. Again, for a simple rectangular masonry wall constructed building, 1 story, would I apply the base shear along the roof line or would I assume the inverted triangular load mentioned above and design the masonry for out of plane bending?

The load distribution on a building can be represented by an inverted triangle because this shape is somewhat close to the fundamental mode shape of the building. These forces are then assumed to be lumped together and act at the diaphragms. The base shear is NOT distributed uniformly to the walls. Only the self weight of the wall will induce out of plane inertial forces on the wall. Now, think how the wall would vibrate. If it is supported at the top (by the roof diaphragm) and at the bottom, then the fundamental mode shape would be a parabola - like the deflected shape of a simply supported beam. This shape leads to a force distribution that is most easily represented by a uniform loading. Therefore, for out of plane seismic loads on a simply supported wall, you wouldn't use an inverted triangular load distribution.

In IBC 2000 (nased on 97 Nehrp)

w (effective seismic weight) is calculated the following ways:

W= total dead load + ....

Warehouses 25% live
BUildings with partitions 10 psf
Snow Load> 30 psf 20% snow
Permanent Equipment 100% dead