Wood Shearwall Deflection/Stifness & Load Distribution Assumption 4

[Jerehmy]

I have no structural coworkers besides my boss, who is 70. So whenever I talk about lateral design of wood he raises an eyebrow/rolls his eyes says something along the lines of "been doing this for 50 years and never had any problems!". So this is my only place to talk wood lateral theory so sorry for an abundance of posts on the subject. Anyways-

Breyer 6th edition section 9.6:

"... It is generally assumed that the unit shear (lb/ft) in each of the full-height segments is the same; this is generally reasonable because all of the wall piers are constrained by the collector to deflect the same amount under lateral loading"

This is how we design, but this doesn't make sense to me (partially).

A shearwalls bending stiffness is mostly in the chords, i.e flanges. So if we are calculating the stiffness of a member's flanges, it is:

~ Ad² which would be Ab² for a shearwall

So the length of shear wall's bending stiffness is proportional to the length of the wall SQUARED. BUT, you when looking at the deflection equations in the SDPWS, the bending portion of the deflection equation is:

8vh³/EAb

If we have a single full length shearwall, v = V/b, so:

8Vh³/EAb²

So the first term is proportional to the square of the length of the shearwall. Makes sense.

Assuming h = 8th, for short shearwalls (4ft) the shear stiffness and bending stiffness are of the same order of magnitude.
For longer shearwalls (12ft) the bending stiffness is about 5x the shear stiffness.

So longer shearwall stiffness is more proportional to b² than just b. So that initial assumption is invalid. This bugs me. If we assume the shear is distributed based on length instead of stiffness, doesn't this give a much lower collector force than is actually there?

Let's say we have shearwall 1 with b₁ = 30ft and shearwall 2 with b₂ = 10ft and an opening of 10ft between them (total wall length of 50ft).

Shearwall 1 will have a stiffness approximately 8-9 times that of shearwall 2, but the load is distributed assuming shearwall 1 has 3 times the stiffness.

Let's say the shear is 10kip. So diaphragm shear is 200plf and shearwall shear is 250plf

The calculated max collector force would be (40ft)(200plf) - (30ft)(250plf) = 500# &
(20ft)*(200pl)-(10ft)*(250plf) = 1500#

If we use stiffness, shearwall 1 gets ~ all the load (~90%). So the max collector force is more like ~(20ft)(200plf) = 4000#

So my question is why do we used this method if it can't lead to such erroneous results? Or am I missing something. Halp.

 

[msquared48]

As I understand the problem, the max collector force to each shear wall would be 200 X 10/2 = 1000#, 1/2 of the distance between the walls times the diaphragm edge shear. The rest if the diaphragm force goes directly to the shear wall it engages.

However, the maximum chord force is likely to be much larger.

All shear walls with a common top plate will deflect the same laterally. I envision the top plate a
s a force distribution mechanism to take the force from weaker walls to thecstiffervones.

Mike McCann, PE, SE (WA)

 

[Jerehmy]

That's not the max collector force I got. I drew a free body diagram as recommended in example 9.12 of Breyer 6th. I made the cut right before the end of the opening. You have 10ft of shear wall and 10ft of opening.

Diaphragm shear is 10kip/50ft = 200plf
Shear wall shear = 10kip/40ft = 250plf

Max collector force based on the above distribution would be:
200plf*20ft = 4kip applied shear from diaphragm
250plf*10ft = 2.5kip resistance from shearwall

This gives a minimum of 1.5kip collector shear. That's assuming that the shear distribution is uniform to each shear wall. I have major gripes with this because a shearwalls stiffness is much more proportional to the length squared than linearly with the length for shearwalls 10-12ft or longer.

Doing a rough calculation, the 30ft shearwall has 8-9 times the stiffness as the short one. This leads to a much much higher collector force because the 10ft shearwall essentially resists nothing (I got about of shear ~10%).

So instead of 1500# collector force, you have:
(10ft)*(200plf) = 2kip for the 10ft shearwall and
(10ft)(200plf) = 2kip for the opening
~4 kips max collector force

Why do we assume even distribution when it isn't the case. I feel like I'm missing something

 

[KootK]

I agree with your conclusion completely Jerehmy. To some extent, I think that the conventional shear distribution algorithm hearkens back to a time when we were a little less aggressive in our use of two foot long shear walls etc.

In my opinion, there ought to be some limitation to the tune of: any shear wall segment having an aspect ratio more than twice that of the lowest aspect shear wall in a given line shall be assumed to carry no shear.

The latest generation wood codes contain numerous aspect ratio limitations and discounts that, frankly, I've been having a hard time keeping on top of. Some of those may address this issue to some degree. Although, to my knowledge, the codes do no prescribe the method of shear wall distribution directly.

I like to debate structural engineering theory -- a lot. If I challenge you on something, know that I'm doing so because I respect your opinion enough to either change it or adopt it.

 

[msquared48]

Personally, I think the approach on Amrhein for CMU walls is more appropriate for wood shear walls, only modifying the top and bottom for partial fixity to no fixity. Time marches on...


Mike McCann, PE, SE (WA)

 

[Jerehmy]

Thanks for taking the time to read through my narrative.

Kootk - glad I'm not alone

Mike - that books at my office. Any section in particular? I've used it the book, but for out of plane flexure. We don't do a ton of masonry work.

 

[manstrom]

I have spent too much time playing with this topic. It's a pet project.

If you model the deflection in a wood shear wall using the 3 part equation in the SDPWS, it breaks down to essentially this

SW deflection = (bending deflection) + (shear deflection) + (hold down deflection)

If you run some sample numbers on a traditional wall, you'll find that the bending deflection is very small as long as the wall has a h/w ratio of about 3 or better. In my example calc of a "typical" mulitfamily building, a 30' gyp wall has bending deflection of 7E-4", and shear deflection of 0.12". I have no uplift, so hold down deflection is 0.

Based on this, I can drop out the bending and hold down deflection (eliminating hold down deflection simplifies my example, you can add it in but things get more complicated). The point is that the bending factor is very small. So your defleciton becomes

SW deflection = shear deflection (including nail slip) = vh/1000Ga

So it is linear. delta is proportional to unit shear, v.

What is also interesting is that allowable unit shear v / Ga is almost a constant for both gyp and osb. It is around 18 to 22 for most common shear wall configurations. So, if you take it a step further (still a simplification), for an appropriately sized shear wall with v / Ga = 20 (avg)

SW deflection = (20) * h / 1000 = 0.02h.





When I am working on a problem, I never think about beauty but when I have finished, if the solution is not beautiful, I know it is wrong.

-R. Buckminster Fuller

 

[Jerehmy]

Masntrom - I see what you are saying. The bending stiffness is a lot higher, but the wall is going to deflect a good portion anyways because of shear deflection.

Man I made a bad assumption.

The shearwall stiffness is governed by whatever deflects the most. I'm talking specifically about shear and bending. Shear is far and away a lot less stiff than bending. So for a wood shearwall, it's overall stiffness is impacted very little by bending. So for shearwalls in succession that are separated, the load really is linearly distributed based on shearwall length.

 

[manstrom]

Yes

It's a common assumption in wood, but it is never really explained. It's a good question.

With concrete, there is a huge difference since most of the deflection is in bending, not shear deflection. With wood, it is the opposite.



When I am working on a problem, I never think about beauty but when I have finished, if the solution is not beautiful, I know it is wrong.

-R. Buckminster Fuller

 

[KootK]

Man I made a bad assumption.

Not necessarily. It's all based on wall aspect ratio. If you follow Manstrom's 3:1 aspect ratio guidance, any wall in a four story building stands a good chance of being significantly influenced by bending stiffness if its length is 10' or less. Party walls wouldn't fall into this category but your exterior frontage walls often would.

With concrete, there is a huge difference since most of the deflection is in bending, not shear deflection.

Yes and no. For low rise buildings, shear deflection is important for cmu and concrete as well, depending on the wall aspect ratio. It's common practice in low rise buildings designed by hand to distribute lateral force based on wall length, just like with wood. Certainly, this would be reasonable for any shear wall design using the squat shear wall provisions.

I like to debate structural engineering theory -- a lot. If I challenge you on something, know that I'm doing so because I respect your opinion enough to either change it or adopt it.