Building and Construction - Residential Seismic Design

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Residential Seismic Design

Wayne Whitney

2006-04-13, 6:21 pm

Hello,

I'm having a little trouble understanding how the base shear in
residential seismic design is distributed across the structure. Wind
design seems much simpler, as if I understand correctly, you just get
a constant pressure applied over a face of the building. So can
someone point me to a good discussion of how to distribute the base
shear across the various stories? Alternatively, I have a few
questions below.

Thanks,
Wayne

1) First, all the discussions I've read say that seismic forces act at
various heights along the building. It seems to me that what is
really happening is that the earth moves one way, acting on the
foundation and through the foundation to the wood frame. Then the
earth moves the other way while the house is still moving the first
way. So the force the wood frame really sees is inertial. Is that
right? It doesn't make any difference in the calculations, it just
helps with my understanding.

2) Secondly, the discussions I've read assume that the seismic forces
act on the different diaphragms of the house (roof and floors)
separately, and not on any other part of the house directly. What
is the basis of this assumption? Is it that the weight of the
house is concentrated in the diaphragms? Then should each wall
segment be split in half vertically and assigned to the adjoining
diaphragm? Also, if the seismic forces are assumed to act on the
diaphragms directly, does that mean that the walls never experience
any out-of-plane forces in seismic design?

3) Thirdly, can someone point me to a really good discussion of
collectors? I haven't quite understood what they are.

Phil Scott

2006-04-13, 9:21 pm

--
Phil Scott
Ideas are bullet proof.
"Wayne Whitney" <whitney@post.harvard.edu> wrote in message
news:slrne3tesr.7gj.whitney@pizza.private...
> Hello,
>
> I'm having a little trouble understanding how the base shear
> in
> residential seismic design is distributed across the
> structure. Wind
> design seems much simpler, as if I understand correctly, you
> just get
> a constant pressure applied over a face of the building. So
> can
> someone point me to a good discussion of how to distribute
> the base
> shear across the various stories? Alternatively, I have a
> few
> questions below.
>
> Thanks,
> Wayne
>
> 1) First, all the discussions I've read say that seismic
> forces act at
> various heights along the building. It seems to me that
> what is
> really happening is that the earth moves one way, acting
> on the
> foundation and through the foundation to the wood frame.
> Then the
> earth moves the other way while the house is still moving
> the first
> way. So the force the wood frame really sees is inertial.
> Is that
> right? It doesn't make any difference in the
> calculations, it just
> helps with my understanding.

The forces and effects will be as difficult to calculate as
Katrina was on New Orleans. Senior issues would be some
flexibility in the building. Your calcs will not be able to
cover wide range and combinations of frequency, amplitide and
direction,

Historical knowledge of what lasted in the area and what didnt
might be the most valuable.

>
> 2) Secondly, the discussions I've read assume that the
> seismic forces
> act on the different diaphragms of the house (roof and
> floors)
> separately, and not on any other part of the house
> directly. What
> is the basis of this assumption? Is it that the weight of
> the
> house is concentrated in the diaphragms? Then should each
> wall
> segment be split in half vertically and assigned to the
> adjoining
> diaphragm? Also, if the seismic forces are assumed to act
> on the
> diaphragms directly, does that mean that the walls never
> experience
> any out-of-plane forces in seismic design?
>
> 3) Thirdly, can someone point me to a really good discussion
> of
> collectors? I haven't quite understood what they are.

Bobk207

2006-04-14, 2:21 am

Wayne Whitney wrote:
> Hello,
>
> I'm having a little trouble understanding how the base shear in
> residential seismic design is distributed across the structure. Wind
> design seems much simpler, as if I understand correctly, you just get
> a constant pressure applied over a face of the building. So can
> someone point me to a good discussion of how to distribute the base
> shear across the various stories? Alternatively, I have a few
> questions below.
>
> Thanks,
> Wayne
>
> 1) First, all the discussions I've read say that seismic forces act at
> various heights along the building. It seems to me that what is
> really happening is that the earth moves one way, acting on the
> foundation and through the foundation to the wood frame. Then the
> earth moves the other way while the house is still moving the first
> way. So the force the wood frame really sees is inertial. Is that
> right? It doesn't make any difference in the calculations, it just
> helps with my understanding.
>
> 2) Secondly, the discussions I've read assume that the seismic forces
> act on the different diaphragms of the house (roof and floors)
> separately, and not on any other part of the house directly. What
> is the basis of this assumption? Is it that the weight of the
> house is concentrated in the diaphragms? Then should each wall
> segment be split in half vertically and assigned to the adjoining
> diaphragm? Also, if the seismic forces are assumed to act on the
> diaphragms directly, does that mean that the walls never experience
> any out-of-plane forces in seismic design?
>
> 3) Thirdly, can someone point me to a really good discussion of
> collectors? I haven't quite understood what they are.

Wayne-

1) >>>.So the force the wood frame really sees is inertial. Is that
> right? <<<<<

yes, you are correct about the "real" forces

get a copy of the 1985 Fine Home Building article on seismic retrofit

See if you can find a used copies:
Peace of Mind in Earthquake Country
Terra Non Firma - Portable Stanford (I wont tell anyone)
http://www.amazon.com/gp/product/07...glance&n=283155

2) In order to make CE design "simplier" & more "uniform", codes were
developed. One result of this, is that code based design has gotten
further & further away from basic engineering principles but has become
simplier. Simplier to execute but necessarily to understand.
Understanding the code is sometimes more difficult that understanding
the underlying principles.

somewhat but depends on the specifics of the home geometry, you can do
the calcs based on the actual dimensions & construction materials
[color=darkred]
>Then should each wall segment be split in half vertically and assigned to the adjoining diaphragm?

yes, something like that

>Also, if the seismic forces are assumed to act on the diaphragms directly, does that mean that the walls never experience any out-of-plane forces in seismic design?

No, they do experience out-of-plane forces in a seismic event.
However, in wood framed construction, walls are restrained in
out-of-plane movement by their connection at the mud sill and the roof
or floor diaphragm. Walls never (at least I've never seen them) "blow
out" in the out-of-plane. The bottom line is that, in the
out-of-plane direction walls are "load" not "resistance".

3)
ATC is a not for profit technology transfer org
SEAOC - Structural Engineers Association of California
http://www.atcouncil.org/pdfs/bp3b.pdf
http://www.seaoc.org/Pages/committe...BCh4_101903.pdf
http://www.shearwalls.com/frm_main.html

cheers
Bob

Bobk207

2006-04-14, 3:21 am

Bob Morrison

2006-04-14, 12:21 pm

In a previous post Wayne Whitney wrote...
> I'm having a little trouble understanding how the base shear in
> residential seismic design is distributed across the structure. Wind
> design seems much simpler, as if I understand correctly, you just get
> a constant pressure applied over a face of the building. So can
> someone point me to a good discussion of how to distribute the base
> shear across the various stories?

The building code and ASCE 7-02 describe how to do this

Basically, you take the weight at each diaphragm level and multiply it
times the height to the level. add those numbers up to get a total. The
base shear is apportioned according to the weight times height ratio

Fx = V x [WxHx/(W1H1 + W2H2 + ...)]

Unless the building is more than 3 stories then you add 0.07 V to the top
story or roof.

The story shear (Fx) can be applied to the diaphragm as a uniform load
acting in a horizontal direction if your structure is rectangular with no
large holes in it. However, it is not uncommon to have a 2-story high
atrium space, so it becomes more complicated by the fact that there is no
floor at that location.

Look for APA Publication X-305, Introduction to Lateral Design. You can
find it here, but you have to register:

http://www.apawood.org/level_b.cfm?content=pub_main

--
Bob Morrison, PE, SE
R L Morrison Engineering Co
Structural & Civil Engineering
Poulsbo WA
bob at rlmorrisonengr dot com

Wayne Whitney

2006-04-14, 1:21 pm

On 2006-04-14, Bob Morrison <SpamFighter@junk.com> wrote:

> The story shear (Fx) can be applied to the diaphragm as a uniform load
> acting in a horizontal direction if your structure is rectangular with no
> large holes in it.

Just to clarify--wind loads are typically applied as a uniform
pressure across the face of a building, so for wind design, the walls
must be checked for the ability to transfer this out-of-plane load to
the diaphragms. Whereas in the simplified seismic design present in
the codes, the load is applied directly to the diaphragms, so there is
no need to check any wall segments for out-of-plane loads. Is that
right?

> Look for APA Publication X-305, Introduction to Lateral Design.

Thanks!

Cheers, Wayne

Bob Morrison

2006-04-14, 2:21 pm

In a previous post Wayne Whitney wrote...
> Just to clarify--wind loads are typically applied as a uniform
> pressure across the face of a building, so for wind design, the walls
> must be checked for the ability to transfer this out-of-plane load to
> the diaphragms. Whereas in the simplified seismic design present in
> the codes, the load is applied directly to the diaphragms, so there is
> no need to check any wall segments for out-of-plane loads. Is that
> right?
>

In general that is true. The one exception would be if you have a heavy
masonry veneer applied over the wood framing and the walls are 2 stories
or more high with no intermediate floor(s). But, you would also be
checking this wall for wind load resistance, so depending on the weight of
the veneer and the seismic force, it might be that wind governs even in
this case.

--
Bob Morrison, PE, SE
R L Morrison Engineering Co
Structural & Civil Engineering
Poulsbo WA
bob at rlmorrisonengr dot com

Wayne Whitney

2006-04-14, 2:21 pm

On 2006-04-14, Bob Morrison <SpamFighter@junk.com> wrote:

> In general that is true. The one exception would be if you have a
> heavy masonry veneer applied over the wood framing and the walls are
> 2 stories or more high with no intermediate floor(s). But, you
> would also be checking this wall for wind load resistance, so
> depending on the weight of the veneer and the seismic force, it
> might be that wind governs even in this case.

I'm in a location where the 2% 50 year 0.2 second spectral
acceleration is estimated by USGS to be 2.5g. So I've always assumed
that in my situation the seismic design controls the lateral force
resisting system design. I guess for thoroughness I should calculate
the wind loads and make the comparison, although right now that just
seems like one more thing to learn.

Thanks for all your help.

Cheers, Wayne

Bob Morrison

2006-04-14, 2:21 pm

In a previous post Wayne Whitney wrote...
> I'm in a location where the 2% 50 year 0.2 second spectral
> acceleration is estimated by USGS to be 2.5g. So I've always assumed
> that in my situation the seismic design controls the lateral force
> resisting system design. I guess for thoroughness I should calculate
> the wind loads and make the comparison, although right now that just
> seems like one more thing to learn.
>

For the structure as a whole you are probably correct in assuming that
seismic controls. However, when you get into components and cladding
portion of wind design, there are situations where wind can govern.

For checking exterior stud design pay particular attention to the
tributary area and the location. The "C&C" load factor goes up for small
areas or for corners. Also, if you have 2 story high walls with lots of
glass, be sure to check the jamb framing that runs from sill to top plate.
Think of the jamb framing as a vertical beam (or beam/column if on a
bearing wall) and design accordingly.

This catches a lot of people off guard.

As you are learning, houses are not necessarily the "simple" structures
some people think they are. I've been doing this kind of structural
design for 35 years and still get surprised at how many different load
conditions can be generated by a "simple" structure.

--
Bob Morrison, PE, SE
R L Morrison Engineering Co
Structural & Civil Engineering
Poulsbo WA
bob at rlmorrisonengr dot com

Wayne Whitney

2006-04-18, 2:21 pm

On 2006-04-14, Bob Morrison <SpamFighter@junk.com> wrote:

> Basically, you take the weight at each diaphragm level and multiply it
> times the height to the level. add those numbers up to get a total. The
> base shear is apportioned according to the weight times height ratio
>
> Fx = V x [WxHx/(W1H1 + W2H2 + ...)]

OK, how about a simple example:

One-story house with 9' ceilings over 3.5' cripple walls, no
structural irregularities. Roof diaphragm and half of the first story
walls weigh 25 klbs; the floor diaphragm, cripple walls and half of
the first story walls also weighs 25 klbs. So we have:

W1 = 25 klbs ; H1 = 4' ; W1H1 = 100 klbs-ft
W2 = 25 klbs ; H2 = 14' ; W2H2 = 350 klbs-ft
------------------
Total = 450 klbs-ft

Thus the shear load experienced by the roof diaphragm is 7/9 of the
base shear; the floor diaphragm experiences the full base shear, 7/9
of which is coming down through the first story shear walls, and 2/9
of it developed with the structure from the seismic forces.

Sounds right?

Now suppose that for a given axis, we have 25' of shear walls on the
first floor and 40' of shear walls on the cripple walls (fewer
windows). We use the simple method of distributing the shear forces
equally to the available shear wall length. So each foot of first
story wall must handle (7/)*(1/25) = 7/225 of the base shear, and each
foot of the cripple walls must handle 1/40 of the base shear. Now
7/225 > 1/40, so the first story shear walls must have a higher
capacity in plf than the cripple wall shear walls.

Is this correct and in fact typical? The floor diaphragm will spread
the higher load per foot of shear from the first story shear walls
above out among the greater length of cripple wall shear walls?

Thanks, Wayne

Bob Morrison

2006-04-18, 4:21 pm

In a previous post Wayne Whitney wrote...
> OK, how about a simple example:
>
<snip>
>
> Thus the shear load experienced by the roof diaphragm is 7/9 of the
> base shear; the floor diaphragm experiences the full base shear, 7/9
> of which is coming down through the first story shear walls, and 2/9
> of it developed with the structure from the seismic forces.
>
> Sounds right?

Depending on the shear wall arrangement, the floor will not carry any of
the roof diaphragm loads. If the only shear walls are on the exterior
then all the forces pass through the diaphragm at the edges and do not
load the floor diaphragm. If you have interior shear walls that do not
stack with wall sin the crawl space then the diaphragm will transfer the
loads above to the walls below in a similar fashion to a beam with a point
load applied at the shear wall above.
>
> Now suppose that for a given axis, we have 25' of shear walls on the
> first floor and 40' of shear walls on the cripple walls (fewer
> windows). We use the simple method of distributing the shear forces
> equally to the available shear wall length. So each foot of first
> story wall must handle (7/)*(1/25) = 7/225 of the base shear, and each
> foot of the cripple walls must handle 1/40 of the base shear. Now
> 7/225 > 1/40, so the first story shear walls must have a higher
> capacity in plf than the cripple wall shear walls.

Assuming only exterior shear walls, the diaphragm edge shear at the roof
level (top of first floor walls) is equal to half the story shear. So,
for the 25 feet of wall along each edge the shear force in the wall is
[(7/9)*0.5] or 7/18 of the base shear.
>
> Is this correct and in fact typical? The floor diaphragm will spread
> the higher load per foot of shear from the first story shear walls
> above out among the greater length of cripple wall shear walls?

For only exterior shear walls at the base level (cripple wall, the load to
the 40' of cripple wall is half the base shear. Since the ratio of the
lengths of shear wall is less than the ratio of shears applied, the load
per foot of shear wall will be higher for the first floor walls than it is
for the cripple walls. This is pretty common.

Did that answer your question?

--
Bob Morrison, PE, SE
R L Morrison Engineering Co
Structural & Civil Engineering
Poulsbo WA
bob at rlmorrisonengr dot com

Wayne Whitney

2006-04-18, 4:21 pm

On 2006-04-18, Bob Morrison <SpamFighter@junk.com> wrote:

> Assuming only exterior shear walls, the diaphragm edge shear at the roof
> level (top of first floor walls) is equal to half the story shear.

That's half the story shear, per edge, right? I was speaking about
both edges at once, sorry if that wasn't clear. If the two exterior
edges have different lengths of shear panel, should each edge still be
assigned half the load, so that the shear panels on the edge with less
shear panel have a higher design shear load in plf?

Now what if there is a single interior shear wall line, how is that
handled?

> Since the ratio of the lengths of shear wall is less than the ratio
> of shears applied, the load per foot of shear wall will be higher
> for the first floor walls than it is for the cripple walls. This is
> pretty common.

OK, thank you, I just wanted to double check that. Now a followup
question on the tension/compression loads in the squash posts:

My understanding is that ignoring eccentricity, a shear wall segment
acted on by a given shear load in plf from above will experience the
same load in plf in the side members. E.g. a 4'x4' segment with a
shear of 500 plf will have a tension/compression load of 500plf * 4' =
2000lbs, and an 4'x9' segment experiencing a shear of 500 plf will
have a tension/compression load of 500 plf * 9' = 4500 lbs.

Now suppose that those two segments are stacked on top of each other,
a 4' tall cripple wall segment and a 9' tall first story segment.
[Suppose the design shear load in plf is the same because the greater
total load in the lower level is exactly balanced by a greater length
of shear wall.] I'm confused by the idea that the squash post in the
first floor segment could experience 4500 lbs, while the squash post
load in the cripple wall comes out to only 2000 lbs. If this is
right, do the floor-to-floor ties need to be rated for 2000 lbs or
4500 lbs?

Thanks again for all the wonderful information and guidance.

Yours, Wayne

Verizon

2006-04-18, 7:21 pm

As the first two questions have been answered rather thoroughly, I'll take a poke at the third.
The APA has an excellent (if not short) "Technical Topic" (or paper) on IBC & IRC collectors.
Very easy read.

http://www.apawood.org/level_c.cfm?...pub_tch_libmain

Look for the topic entitled "Collector Design for Bracing in Conventional Construction" about 1/3 of the way down.

Lot of excellent material on this site,
Dennis

"> 3) Thirdly, can someone point me to a really good discussion of
> collectors? I haven't quite understood what they are.

Bob Morrison

2006-04-19, 12:21 pm

In a previous post Wayne Whitney wrote...
> That's half the story shear, per edge, right? I was speaking about
> both edges at once, sorry if that wasn't clear. If the two exterior
> edges have different lengths of shear panel, should each edge still be
> assigned half the load, so that the shear panels on the edge with less
> shear panel have a higher design shear load in plf?
>

That is correct. The diaphragm is treat as a simply supported beam
(flexible diaphragm).

> Now what if there is a single interior shear wall line, how is that
> handled?

Not allowed when designing with flexible wood diaphragms. You must have
at least 2 shear lines in each direction. There are some exceptions for
small 3-sided structures (like garages) where diaphragm deflections can be
tolerated.

--
Bob Morrison, PE, SE
R L Morrison Engineering Co
Structural & Civil Engineering
Poulsbo WA
bob at rlmorrisonengr dot com

Bob Morrison

2006-04-19, 12:21 pm

In a previous post Wayne Whitney wrote...
> My understanding is that ignoring eccentricity, a shear wall segment
> acted on by a given shear load in plf from above will experience the
> same load in plf in the side members. E.g. a 4'x4' segment with a
> shear of 500 plf will have a tension/compression load of 500plf * 4' =
> 2000lbs, and an 4'x9' segment experiencing a shear of 500 plf will
> have a tension/compression load of 500 plf * 9' = 4500 lbs.

That is correct. BTW, a wood shear wall with 500 plf is pretty heavily loaded.
This will require the use of 15/32" Struct I plywood, 3x framing and
0.131" x 2-1/2" (8d Common)@ 3" o/c. on the edges.

> Now suppose that those two segments are stacked on top of each other,
> a 4' tall cripple wall segment and a 9' tall first story segment.
> [Suppose the design shear load in plf is the same because the greater
> total load in the lower level is exactly balanced by a greater length
> of shear wall.] I'm confused by the idea that the squash post in the
> first floor segment could experience 4500 lbs, while the squash post
> load in the cripple wall comes out to only 2000 lbs. If this is
> right, do the floor-to-floor ties need to be rated for 2000 lbs or
> 4500 lbs?
>

Generally, the loads are stackable. Draw the wall panel as a simple free-
body diagram with the forces from the upper wall and the horizontal story
shear applied in the appropriate locations. You will have the horizontal
component at the top and bottom of the lower wall panel resisted by the
sheathing nailing (and/or anchor bolts). You will have an overturning
moment(load couple) from the upper wall applied to the top of the lower
wall and you will have the overturning moment from the horizontal shear
applied at the top of the lower wall.

One way to eliminate the overturing couple from the upper wall to the
lower is to have tension/compression resisting devices aligned with the
ends of the upper wall. This is often not practical, especially if there
is a window or door opening below the upper wall segment.

The key to understanding this is the free body diagram of the lower wall
panel.

--
Bob Morrison, PE, SE
R L Morrison Engineering Co
Structural & Civil Engineering
Poulsbo WA
bob at rlmorrisonengr dot com

Wayne Whitney

2006-04-19, 1:21 pm

On 2006-04-19, Bob Morrison <SpamFighter@junk.com> wrote:

> In a previous post Wayne Whitney wrote...
>
>
> That is correct. The diaphragm is treat as a simply supported beam
> (flexible diaphragm).

Ahh, the diaphragm is treated as a simply supported beam, so the loads
on the shear panel lines are the equivalent of the beam reaction
forces, and if the diaphragm is uniformly loaded, then the reaction
forces will be equal.

>
> Not allowed when designing with flexible wood diaphragms. You must have
> at least 2 shear lines in each direction.

Sorry I wasn't clear, I meant one interior shear wall line in addition
to the two exterior wall lines. I had proposed to divide the shear
load from the diaphragm into thirds, and apply one third to each wall
line. Is this OK, or should one instead look at the diaphragm as a
uniformly loaded beam, and assign the shear loads based on the
location of the interior shear wall line?

Thanks, Wayne

Bob Morrison

2006-04-19, 2:21 pm

In a previous post Wayne Whitney wrote...
> Ahh, the diaphragm is treated as a simply supported beam, so the loads
> on the shear panel lines are the equivalent of the beam reaction
> forces, and if the diaphragm is uniformly loaded, then the reaction
> forces will be equal.

That is correct. However, when considering seismic forces it is not
unusual to have a large hole in your floor diaphragm for stairs or similar
features. Since the weight of the floor is now not uniformly distributed
the diaphragm "beam" is no longer uniformly loaded. You can also run into
this problem when considering wind loads because of projections, differing
roof heights, etc.

> Sorry I wasn't clear, I meant one interior shear wall line in addition
> to the two exterior wall lines. I had proposed to divide the shear
> load from the diaphragm into thirds, and apply one third to each wall
> line. Is this OK, or should one instead look at the diaphragm as a
> uniformly loaded beam, and assign the shear loads based on the
> location of the interior shear wall line?
>

In this case you would have two diaphragms -- each segment is treated as a
single simple span beam. So the middle shear wall will have half the load
and the two exterior walls will each have their tributary reaction. So if
the "middle" wall is really at the 1/3 point of the building your
diaphragm edge reactions (or shear wall loads) would be 1/6, 1/2, 1/3 of
the story shear, respectively.

I typically ignore interior walls that are on the top floor unless the
building has a flat roof or the roof framing is in some way connected to
the wall. This is because the shear wall must connect to the roof
diaphragm in order to carry any load. It is difficult, but not
impossible, to extend an interior wall up to the bottom of the roof
sheathing. But if it's not needed, why bother.

--
Bob Morrison, PE, SE
R L Morrison Engineering Co
Structural & Civil Engineering
Poulsbo WA
bob at rlmorrisonengr dot com

Wayne Whitney

2006-04-19, 3:21 pm

On 2006-04-19, Bob Morrison <SpamFighter@junk.com> wrote:

> I typically ignore interior walls that are on the top floor unless the
> building has a flat roof or the roof framing is in some way connected to
> the wall.

In the case of a one-story building with a second story on the rear
half only, is it proper to treat these as two sepearate buildings (the
one-story section and the two-story section) and simply combine the
loads from each side when designing the common wall? In this case an
interior shear wall segment is supported an exterior shear wall
segment above.

Thanks, Wayne

Wayne Whitney

2006-04-19, 3:21 pm

On 2006-04-19, Wayne Whitney <whitney@post.harvard.edu> wrote:

> In this case an interior shear wall segment is supported an exterior
> shear wall segment above.

And I meant to add, is in turn supported by its own continuous
foundation wall.

Cheers, Wayne

Wayne Whitney

2006-04-19, 5:21 pm

On 2006-04-19, Bob Morrison <SpamFighter@junk.com> wrote:

> That is correct. BTW, a wood shear wall with 500 plf is pretty heavily loaded.
> This will require the use of 15/32" Struct I plywood, 3x framing and
> 0.131" x 2-1/2" (8d Common)@ 3" o/c. on the edges.

If the sheathing and nailing is applied to both sides of the wall, the
rating can be doubled, yes?

The ratings of the individual segments obviously depend foremost upon
the design base shear V, which in turn depends on the coefficient used
to relate V to the dead weight of the building, W. If I have read the
2001 CBC (1997 UBC) correctly, the most it requires for a 1 or 2 story
residence is V=0.22W.

On the other hand, the 2000 "Residential Structural Design Guide" from
www.pathnet.org, which you recommended to me a while ago, suggests
that one should use V = (0.8 a / R) W, where a is the 2% 50-year 2
second spectral acceleration from the USGS, in g. [0.8 = 1.2 * (2/3),
where SDS in their formula = (2/3) a.] For a = 2.5g (my location) and
R = 5.5, this gives V = 0.364 W, which is about 5/3 as great as the
2001 CBC figure.

Anyway, my calculations so far on my house show that some key shear
wall segments may need to be rated at 800 plf to 1000 plf, using the
base shear coefficient of 0.364. This is based on a dead weight of
27000 lbs for the 728 ft^2 single story portion of my structure, and a
dead weight of 37400 lbs for the 576 ft^2 2 story portion of my
structure.

Still in the thick of it,
Wayne

Wayne Whitney

2006-04-19, 5:21 pm

On 2006-04-19, Bob Morrison <SpamFighter@junk.com> wrote:

>
> Generally, the loads are stackable.

Does this mean that the upper squash post may experience a load of
4500 lbs in tension, so that it needs a floor-to-floor tie rated at
4500 lbs to transfer that load to the lower squash post; and that the
lower squash post may experience an additional 2000 lbs in tension
load, so that it needs a 6500 lbs hold-down to the foundation?

> One way to eliminate the overturing couple from the upper wall to the
> lower is to have tension/compression resisting devices aligned with the
> ends of the upper wall. This is often not practical, especially if there
> is a window or door opening below the upper wall segment.

I don't understand what you mean by "tension/compression resisting
devices". Something other than floor-to-floor ties and squash posts?
And do you mean aligned with the ends of the upper wall segment, or
the very corners of the full upper wall?

Thanks, Wayne

Bob Morrison

2006-04-19, 6:21 pm

In a previous post Wayne Whitney wrote...
> In the case of a one-story building with a second story on the rear
> half only, is it proper to treat these as two sepearate buildings (the
> one-story section and the two-story section) and simply combine the
> loads from each side when designing the common wall? In this case an
> interior shear wall segment is supported an exterior shear wall
> segment above.
>

Yes. That is exactly how it is to be done.

> And I meant to add, is in turn supported by its own continuous
> foundation wall.

That is not necessary, but it is helpful. The bottom of the interior wall
could load onto the first floor diaphragm with the load then being
distributed to the shear walls supporting the first floor diaphragm. If
there are overturning forces on this wall then they can be resisted by
installing a beam under the wall attaching the HD's to the beam. The beam
ends must then be "held-down" where it is supported.

Again, as long as you have a continuous (but not necessarily contiguous)
load path from roof to foundation, then all is well.

BTW, I would like to comment to any others who have been following this
series of posts: this is a crash course in the design of light frame
structures for lateral forces. The ideas Bob K, Wayne and I have been
discussing apply to both wind and seismic loads. This whole series of
posts probably should have done in sci.engr.civil, but perhaps some of you
builders will learn a little about the mechanics of how one goes about
designing these systems and why HD hardware and controlled nailing is
required.

--
Bob Morrison, PE, SE
R L Morrison Engineering Co
Structural & Civil Engineering
Poulsbo WA
bob at rlmorrisonengr dot com

Bob Morrison

2006-04-19, 6:21 pm

In a previous post Wayne Whitney wrote...
> If the sheathing and nailing is applied to both sides of the wall, the
> rating can be doubled, yes?
>
> The ratings of the individual segments obviously depend foremost upon
> the design base shear V, which in turn depends on the coefficient used
> to relate V to the dead weight of the building, W. If I have read the
> 2001 CBC (1997 UBC) correctly, the most it requires for a 1 or 2 story
> residence is V=0.22W.
>
> On the other hand, the 2000 "Residential Structural Design Guide" from
> www.pathnet.org, which you recommended to me a while ago, suggests
> that one should use V = (0.8 a / R) W, where a is the 2% 50-year 2
> second spectral acceleration from the USGS, in g. [0.8 = 1.2 * (2/3),
> where SDS in their formula = (2/3) a.] For a = 2.5g (my location) and
> R = 5.5, this gives V = 0.364 W, which is about 5/3 as great as the
> 2001 CBC figure.
>
> Anyway, my calculations so far on my house show that some key shear
> wall segments may need to be rated at 800 plf to 1000 plf, using the
> base shear coefficient of 0.364. This is based on a dead weight of
> 27000 lbs for the 728 ft^2 single story portion of my structure, and a
> dead weight of 37400 lbs for the 576 ft^2 2 story portion of my
> structure.
>

Yes you can double the shear value for sheathing two sides. Remember that
any wall with a load of more than 350 plf requires 3x sill plates and 3x
framing at panel edges.

The CBC is based on Seismic "Zone" values and does not reflect the latest
thinking from most structural engineers, which is the "Probabilistic"
method described in ASCE 7-02, IBC2003, and the 2000 "Guide" described
above. BTW, in light frame structures such as yours you can use an "R" of
6.5 under IBC2003/ASCE 7-02. This will reduce your load to 0.308W under
the "Simplified" Analysis Procedure described in ASCE 7-02.

For loads as high as 800-1000 plf you might consider using manufactured
shear wall panels such as those developed by Simpson Strong-tie or
Hardyframe.

--
Bob Morrison, PE, SE
R L Morrison Engineering Co
Structural & Civil Engineering
Poulsbo WA
bob at rlmorrisonengr dot com

Bob Morrison

2006-04-19, 6:21 pm

In a previous post Wayne Whitney wrote...
>
> Does this mean that the upper squash post may experience a load of
> 4500 lbs in tension, so that it needs a floor-to-floor tie rated at
> 4500 lbs to transfer that load to the lower squash post; and that the
> lower squash post may experience an additional 2000 lbs in tension
> load, so that it needs a 6500 lbs hold-down to the foundation?

Yes, unless the ends of the walls do not align with each other. In this
case you need to a little calculating to distribute the overturning force
to the ends of the lower wall.

>
> I don't understand what you mean by "tension/compression resisting
> devices". Something other than floor-to-floor ties and squash posts?
> And do you mean aligned with the ends of the upper wall segment, or
> the very corners of the full upper wall?

These would align with the upper wall segments. For example, you could
extend a rod for a hold-down from the upper floor wall through the lower
floor wall into the foundation. This eliminates the overturning force
from upper wall being applied to the lower wall. Posts/studs (squash
posts) must also align with the wall segment(s) above to make the load
transfer mechanism work. This is similar to the detail I e-mailed you off-
list.

Just to throw a wrinkle in your considerations, there is a newer design
methodology called a "perforated shear wall". This works pretty well if
there are not too many openings and the forces are not too high. The
design methodology can reduce the amount of hardware required, but often
requires increased nailing (nails closer together).

--
Bob Morrison, PE, SE
R L Morrison Engineering Co
Structural & Civil Engineering
Poulsbo WA
bob at rlmorrisonengr dot com

Wayne Whitney

2006-04-19, 7:21 pm

On 2006-04-19, Bob Morrison <SpamFighter@junk.com> wrote:

> Yes you can double the shear value for sheathing two sides. Remember that
> any wall with a load of more than 350 plf requires 3x sill plates and 3x
> framing at panel edges.

OK, is a double 2x acceptable for the sill plate and panel edges,
instead of a 3x? And on the panel edge, can one of the double 2x be a
jack stud for an opening, as we discussed previously?

Wayne

Wayne Whitney

2006-04-19, 7:21 pm

On 2006-04-19, Bob Morrison <SpamFighter@junk.com> wrote:

> Just to throw a wrinkle in your considerations, there is a newer design
> methodology called a "perforated shear wall". This works pretty well if
> there are not too many openings and the forces are not too high.

Right, I'm aware of it, it is the 2000 "Guide" from www.pathnet.org.
My situation does involve high loads and quite a few openings, though.
So I'll probably not try out the calculations.

One thing I've wondered about in the "segmented" shear wall approach:
for two succesive segments in a single shear wall line with an opening
in between, the shear wall edge members on either side of the opening
(in the two different segments) will always have opposite loads. Is
there no way to somehow tie these edge members together in the space
above the opening to partially cancel the opposing loads? This would
reduce the hold down hardware required on the interior shear panel
edges, much as the "perforated" shear wall approach does.

Cheers, Wayne

Bob Morrison

2006-04-19, 9:21 pm

In a previous post Wayne Whitney wrote...
> OK, is a double 2x acceptable for the sill plate and panel edges,
> instead of a 3x? And on the panel edge, can one of the double 2x be a
> jack stud for an opening, as we discussed previously?
>

No. See footnote 3 in UBC97 Table 23-II-I-1.

--
Bob Morrison, PE, SE
R L Morrison Engineering Co
Structural & Civil Engineering
Poulsbo WA
bob at rlmorrisonengr dot com

Bob Morrison

2006-04-19, 9:21 pm

In a previous post Wayne Whitney wrote...
> One thing I've wondered about in the "segmented" shear wall approach:
> for two succesive segments in a single shear wall line with an opening
> in between, the shear wall edge members on either side of the opening
> (in the two different segments) will always have opposite loads. Is
> there no way to somehow tie these edge members together in the space
> above the opening to partially cancel the opposing loads? This would
> reduce the hold down hardware required on the interior shear panel
> edges, much as the "perforated" shear wall approach does.
>

No. The segmented approach assumes the openings extend from top of floor
to top of plate. This is consistent with the "flexible" diaphragm
approach. In order to transfer the load to an adjacent panel you would
need to have a moment connection at the top of the opening. The portal
frame method allowed by APA (and as an amendment to IRC2003)for garages
accomplishes this somewhat, but hold-downs are still required at the edges
of the wall segments.

BTW, I forgot to have you check your H/L ratios for the shear panels. For
seismic loads you are limited to H/L <= 2. For an 8' plate height this
means your panel must be 4 feet long. This is not quite consistent with
the alternate braced wall panel provisions for the prescriptive method,
but who said the code has to make sense. APA has a method to reduce the
wall capacity on a sliding scale for 2<H/L<=3.5.

If you cannot meet this requirement then it is definitely time to start
looking at manufactured SW panels. They can get around the H/L<=2 ratio
problem because of test data.

--
Bob Morrison, PE, SE
R L Morrison Engineering Co
Structural & Civil Engineering
Poulsbo WA
bob at rlmorrisonengr dot com

Wayne Whitney

2006-04-19, 10:21 pm

On 2006-04-20, Bob Morrison <SpamFighter@junk.com> wrote:

> BTW, I forgot to have you check your H/L ratios for the shear panels. For
> seismic loads you are limited to H/L <= 2. For an 8' plate height this
> means your panel must be 4 feet long. This is not quite consistent with
> the alternate braced wall panel provisions for the prescriptive method,
> but who said the code has to make sense. APA has a method to reduce the
> wall capacity on a sliding scale for 2<H/L<=3.5.

Yes, the 2000 Design guide from www.pathnet.org has an aspect ratio
factor of 1/sqrt(0.5 * a) for aspect ratios a between 2 and 4. Are
you saying that this is not recognized by the 2001 CBC (1997 UBC)?
Speaking of which, the 2004 CBC (2001 UBC, I think) is due to come
into effect soon, does the 2001 UBC recognize this?

Thanks, Wayne

Wayne Whitney

2006-04-19, 10:21 pm

On 2006-04-20, Bob Morrison <SpamFighter@junk.com> wrote:

>
> No. See footnote 3 in UBC97 Table 23-II-I-1.

"In Seismic Zones 3 and 4, where allowable shear values exceed 350
pounds per foot (5.11 N/mm), foundation sill plates and all framing
members receiving edge nailing from abutting panels shall not be less
than a single 3-inch (76 mm) nominal member and foundation sill plates
shall not be less than a single 3-inch (76 mm) nominal member. In
shear walls where total wall design shear does not exceed 600 pounds
per foot (8.76 N/mm), a single 2-inch (51mm) nominal sill plate may be
used, provided anchor bolts are designed for a load capacity of 50
percent or less of the allowable capacity and bolts have a minimum of
2-inch-by-2-inch-by-3/16-inch (51 mm by 51 mm by5 mm) thick plate
washers. Plywood joint and sill plate nailing shall be staggered in
all cases."

I got this from the errata at www.iccsafe.org. Is it me, or is the
first sentence redundant in mentioning foundation sill plates twice?
Or perhaps I have misunderstood the marked up errata.

If I understand this correctly, this requires, among other things, 3x
studs where two panel abut when the shear panel rating exceeds 350
plf. I thought it was common to use a double 2x in this situation,
but you say a double 2x isn't "not less than a single 3-inch nominal
member"?

Cheers, Wayne

Bob Morrison

2006-04-19, 10:21 pm

Organization: R L Morrison Engineering Co
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In a previous post Wayne Whitney wrote...
> Yes, the 2000 Design guide from www.pathnet.org has an aspect ratio
> factor of 1/sqrt(0.5 * a) for aspect ratios a between 2 and 4. Are
> you saying that this is not recognized by the 2001 CBC (1997 UBC)?
> Speaking of which, the 2004 CBC (2001 UBC, I think) is due to come
> into effect soon, does the 2001 UBC recognize this?
>

UBC97 does not use the formula, but does allow "alternate design methods".
You just have to convince a building official that your "Alternate Design
Method" has merit.

After UBC97 the code becomes IBC2000, then IBC2003 (Washington use this
one) and coming this summer will be IBC2006. I think most jurisdictions
in Washington will switch to IBC2006 when it becomes available. IBC2006
is based on ASCE 7-05 (a much better publication than ASCE 7-02) and the
most current versions of AISC and ACI318.

--
Bob Morrison, PE, SE
R L Morrison Engineering Co
Structural & Civil Engineering
Poulsbo WA
bob at rlmorrisonengr dot com

Bob Morrison

2006-04-19, 11:21 pm

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In a previous post Wayne Whitney wrote...
> "In Seismic Zones 3 and 4, where allowable shear values exceed 350
> pounds per foot (5.11 N/mm), foundation sill plates and all framing
> members receiving edge nailing from abutting panels shall not be less
> than a single 3-inch (76 mm) nominal member and foundation sill plates
> shall not be less than a single 3-inch (76 mm) nominal member. In
> shear walls where total wall design shear does not exceed 600 pounds
> per foot (8.76 N/mm), a single 2-inch (51mm) nominal sill plate may be
> used, provided anchor bolts are designed for a load capacity of 50
> percent or less of the allowable capacity and bolts have a minimum of
> 2-inch-by-2-inch-by-3/16-inch (51 mm by 51 mm by5 mm) thick plate
> washers. Plywood joint and sill plate nailing shall be staggered in
> all cases."
>
> I got this from the errata at www.iccsafe.org. Is it me, or is the
> first sentence redundant in mentioning foundation sill plates twice?
> Or perhaps I have misunderstood the marked up errata.

Yes the quote is redundant. BTW, IBC2003 requires the use of 3x3x1/4"
square washers on the anchor bolts instead of the smaller 2x2x3/16.

Note that you may use a 2x foundation sill plate if you use twice as many
anchor bolts for walls up to 600 plf.

> If I understand this correctly, this requires, among other things, 3x
> studs where two panel abut when the shear panel rating exceeds 350
> plf. I thought it was common to use a double 2x in this situation,
> but you say a double 2x isn't "not less than a single 3-inch nominal
> member"?

In my opinion you should be able to use 2-2x if you fasten them together
with enough nails to provide complete shear transfer. So, if your shear
wall is a 600 plf shear wall then you must be able to transfer 600 plf
between the double studs. In other words you must be able to transfer the
panel edge shear between the two pieces. It seems to me that you could
field laminate with nails and glue to give a full 3-inch nailing surface.
Some Building Officials accept this and others don't.

--
Bob Morrison, PE, SE
R L Morrison Engineering Co
Structural & Civil Engineering
Poulsbo WA
bob at rlmorrisonengr dot com

Wayne Whitney

2006-04-20, 1:21 am

On 2006-04-20, Bob Morrison <SpamFighter@junk.com> wrote:

> Note that you may use a 2x foundation sill plate if you use twice as many
> anchor bolts for walls up to 600 plf.

Just to clarify, if you have a 500 plf shear wall segment on the
foundation, you can still use a 2x sill plate if you restrain the sill
plate at 1000 plf, e.g. one 1500 shear anchor every 18". Yes? What
is the thinking behind this?

Thanks, Wayne

Bob Morrison

2006-04-20, 1:21 am

In a previous post Wayne Whitney wrote...
> Just to clarify, if you have a 500 plf shear wall segment on the
> foundation, you can still use a 2x sill plate if you restrain the sill
> plate at 1000 plf, e.g. one 1500 shear anchor every 18". Yes? What
> is the thinking behind this?
>

If you have a 500 plf shear wall, you must design the anchor bolts for
1000 plf. Or the other way around is to size the anchor bolts for 500 plf
and then cut the spacing in half.

The thinking on this was never very clear to me, but I believe it has
something to do splitting of the plate around the anchor bolts at high
shear loads. The 2x2x3/16 or 3x3x1/4" washers are supposed to help with
this, but I've not seen the research.

Here's one for consideration:

If your openings are windows then the sill plate is continuous underneath
the window and you take the shear per foot of sill plate into account when
sizing anchor bolts.

For example: total wall length = 20 feet & shear panel length = 10 feet
load in shear panel = 500 plf. load in sill = 250 plf

I don't do this, but some engineers I know do.

--
Bob Morrison, PE, SE
R L Morrison Engineering Co
Structural & Civil Engineering
Poulsbo WA
bob at rlmorrisonengr dot com

Wayne Whitney

2006-04-20, 1:21 am

On 2006-04-20, Bob Morrison <SpamFighter@junk.com> wrote:

> If your openings are windows then the sill plate is continuous
> underneath the window and you take the shear per foot of sill plate
> into account when sizing anchor bolts. . . . I don't do this, but
> some engineers I know do.

Do you not do this because you doubt its validity? It seems
reasonable, it's like a reverse collector.

Speaking of collectors, when checking the capacity of a top plate
collector with a story above, is it appropriate to only consider the
additional shear load coming for the immediately attached diaphragm,
as the shear load from the story above will have already been
"collected"? This is assuming that shear wall segments are stacked,
so that the shear load from one segment above goes directly into the
segment below without interacting with the collector.

Cheers, Wayne

Bob Morrison

2006-04-20, 1:21 am

In a previous post Wayne Whitney wrote...
> Do you not do this because you doubt its validity? It seems
> reasonable, it's like a reverse collector.
>
>

I don't do it because sometimes windows have a tendency to turn into doors
when the wall is being framed or at some future time when a deck gets
added.

--
Bob Morrison, PE, SE
R L Morrison Engineering Co
Structural & Civil Engineering
Poulsbo WA
bob at rlmorrisonengr dot com

Bob Morrison

2006-04-20, 1:21 am

In a previous post Wayne Whitney wrote...
> Speaking of collectors, when checking the capacity of a top plate
> collector with a story above, is it appropriate to only consider the
> additional shear load coming for the immediately attached diaphragm,
> as the shear load from the story above will have already been
> "collected"? This is assuming that shear wall segments are stacked,
> so that the shear load from one segment above goes directly into the
> segment below without interacting with the collector.
>

"Collectors" that are a continuous set of top plates are usually not a
problem, but you should check the nailing where the plate is spliced. And
yes the "collected" load is only for the attached diaphragm as long as the
shear walls above are stacked on those below.

You should also check your double top plate as a diaphragm chord. For
most residential projects this is typically not a problem, but for a
highly loaded diaphragm it might be.

BTW, typical construction uses double top plates, but if you stack your
joists/rafters over the studs and provide a strap at the plate splice you
might be able to go to a single top plate (check chord forces!).

--
Bob Morrison, PE, SE
R L Morrison Engineering Co
Structural & Civil Engineering
Poulsbo WA
bob at rlmorrisonengr dot com

Wayne Whitney

2006-04-20, 12:21 pm

On 2006-04-20, Bob Morrison <SpamFighter@junk.com> wrote:

> You should also check your double top plate as a diaphragm chord. For
> most residential projects this is typically not a problem, but for a
> highly loaded diaphragm it might be.

I don't understand how checking the top plate as a diaphragm chord
differs from checking it as a collector. Do you mean just considering
the entire length of top plate, and looking at the net/tension
compression on it at each point, given the loads from the diaphragm
and from above, and the reactions from the shear walls segments below?

> BTW, typical construction uses double top plates, but if you stack your
> joists/rafters over the studs and provide a strap at the plate splice you
> might be able to go to a single top plate (check chord forces!).

Well, the areas in my house that are of primary concern engineering
wise are already framed with a single continuous top plate, no
splices. The 1908 house is 26' x 52', and all the members in the
short dimension are fully 26' long. The long dimension does have a
double top plate.

Cheers, Wayne

RicodJour

2006-04-20, 1:21 pm

Hey, Bob. How much are you charging for this course in seismic design?
;)

R

Bob Morrison

2006-04-20, 2:21 pm

In a previous post RicodJour wrote...
> Hey, Bob. How much are you charging for this course in seismic design?
> ;)
>

For now let's just say it's for the benefit of the community

--
Bob Morrison, PE, SE
R L Morrison Engineering Co
Structural & Civil Engineering
Poulsbo WA
bob at rlmorrisonengr dot com

Bob Morrison

2006-04-20, 2:21 pm

In a previous post Wayne Whitney wrote...
> I don't understand how checking the top plate as a diaphragm chord
> differs from checking it as a collector. Do you mean just considering
> the entire length of top plate, and looking at the net/tension
> compression on it at each point, given the loads from the diaphragm
> and from above, and the reactions from the shear walls segments below?
>

They are different. The "collector" forces are when the load is parallel
to the plate and the "chord" forces occur when the load is perpendicular
to the plate in question -- sort of like the flanges of a beam.

Refer to Section 6.5.3.2 Diaphragm Design in the Residential Structural
Design Guide.

The chord force for a uniformly loaded diaphragm is M/d, where
M=wl^2/8 , w = seismic (or wind) force applied as a uniform load over
the diaphragm length "l", and d = diaphragm depth

--
Bob Morrison, PE, SE
R L Morrison Engineering Co
Structural & Civil Engineering
Poulsbo WA
bob at rlmorrisonengr dot com

Wayne Whitney

2006-04-20, 4:21 pm

On 2006-04-20, Bob Morrison <SpamFighter@junk.com> wrote:

> For now let's just say it's for the benefit of the community

Well, it is very generous of you. If you ever need any help in
abstract mathematics (my training), or even applied math, or anything
done in Berkeley, CA, I'd be happy to help.

Cheers, Wayne

Wayne Whitney

2006-04-20, 7:21 pm

On 2006-04-20, Bob Morrison <SpamFighter@junk.com> wrote:

> They are different. The "collector" forces are when the load is
> parallel to the plate and the "chord" forces occur when the load is
> perpendicular to the plate in question -- sort of like the flanges
> of a beam.

OK, so what confuses me is that the top plate isn't a chord of the
diaphragm per se, the rim joist or band joist is. But the point is
that the rim joist or band joist is connected to the top plate, and so
the top plate and this connection needs to be rated to withstand the
"chord" force?

> Refer to Section 6.5.3.2 Diaphragm Design in the Residential
> Structural Design Guide.

Right, that and Example 6.4 later in the chapter make this clear. For
posterity, this is the Residential Structural Design Guide available
from www.pathnet.org.

Thanks, Wayne

Bob Morrison

2006-04-20, 9:21 pm

In a previous post Wayne Whitney wrote...
> OK, so what confuses me is that the top plate isn't a chord of the
> diaphragm per se, the rim joist or band joist is. But the point is
> that the rim joist or band joist is connected to the top plate, and so
> the top plate and this connection needs to be rated to withstand the
> "chord" force?
>

Typically one does not use the rim or "band" joist unless it is proven to
be continuous for the full length of the diaphragm or can made to be
continuous. The main reason for double top plates is that this then
becomes a continuous member.

You "net" section however, is a single plate member unless you
specifically design a top plate splice detail that provides continuity of
both members at the splice.

Again, for residential structure this typically not much of a problem:

2x4 DF Stud Grade has an allowable tension value of 450 psi. This can be
multiplied by a load duration factor of 1.6 for seismic loads giving an
allowable of 720 psi. Multiply this by the area of 1.5"x3.5" and you get
an allow chord force of 3780 pounds at the splice point.. Since allowable
compression parallel to grain stress of 850 psi is higher than 450 psi,
tension controls.

Here's an example:

Story shear is 20,000 pounds, diaphragm is 40'x20'
Thus:
diaphragm load is 20,000/40' = 500 plf (pretty high but possible)
chord force = (500)(40^2)/(8)(20)= 5000 pounds

So a single 2x4 DF-L Stud grade isn't good enough. 3780# < 5000#
A 2x6 plate will give 5940 pounds > 5000# OK!
or switch to construction grade (Ft=650psi) = 5460 pounds allowable
or use a triple plate with staggered splices so that there are always 2
full members in tension.

If we look at the diaphragm the other direction:
diaphragm load = 20,000/20 = 1000 plf
chord force = (1000)(20^2)/(8)(40) = 1250# Not much of a problem!

As you can see it takes a pretty big diaphragm with a pretty heavy load to
get beyond a single stud grade 2x4.

--
Bob Morrison, PE, SE
R L Morrison Engineering Co
Structural & Civil Engineering
Poulsbo WA
bob at rlmorrisonengr dot com

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