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# Re: Plywood rigid diaphragms

• To: seaint(--nospam--at)seaint.org
• Subject: Re: Plywood rigid diaphragms
• From: Mlcse(--nospam--at)aol.com
• Date: Sat, 12 Sep 1998 01:10:54 EDT

```In a message dated 9/11/98 7:32:32 PM EST, chuckuc(--nospam--at)dnai.com writes:

<< >
> The .005H is from the Los Angeles City Division 93 Ordinance.  This was the
> maximum limit in the 1994 UBC for structures with a period less than 0.7
> seconds, which could be even less if  0.04/Rw governed.
>
> I am don't know how this now correlates to the 1997 UBC, so hopefully
someone
> can explain the comparison, if there is still one.  If you use the 1997 UBC
> the maximum deflection limit would be based upon section 1630.9.2.  The
> maximum inelastic drift is (delta m) = 0.7R(delta s).  Per section
1630.10.2
> the (delta m) drift can not exceed 0.025H for structures with a period less
> than 0.7 seconds which most bearing wall wood structures will be (R= 5.5).
>
> The (delta m) drift is then = (0.7)(5.5)(0.025)(H) = 0.0963H.  The
questions I
> have are do you somehow still account for the 1.4 conversion (97UBC Rw =
5.5
> vs 94 UBC Rw = 8).  Is (delta m) suppose to be similar to the (94 UBC
> deflection )(3Rw/8).
>
> Michael Cochran
.
Michael-
I think your math is wrong.  You should divide by .7R not multiply.  For
a shearwall that reaches ultimate load at 2.5" deflection (.025H) the
allowable capacity will now be the test load at at a deflection of .65"
--as compared with the old citeria of .5" (.005H).  As far as I know
there's nothing magical about .005H or .025H, its just a way to set a
reasonable deflection limit which will allow us to achieve deflection
compatabily between the shearwalls at each floor level.
If all the walls are the same height and construction, Ed Diekmann's
suggestion to apportion load based on wall length x capacity makes for a
pretty reasonable approach. (In effect, the wall's stiffness is assummed
proportional to the number of nails. Probably as good as we can do for
now, given the paucity of test data on tie down deflection.)
Chuck Utzman,P.E.
>>

Thanks Chuck, I see where I made my mistake in back calculating (delta s) from
(delta m).  The 1997 UBC elastic deflection (delta s) limitation is
essentially  the same as in the 1994 UBC as I would expect, except it seems
that the allowable deflection for bearing walls without plywood sheathing is
now greater.

Structure with period less than 0.7 seconds:

1994 UBC:  deflection = 0.005H max.
0.04/8 = 0.005H using Rw = 8
0.04/6 = 0.0067H, must limit deflection to 0.005H when us
Rw = 6

1997 UBC:   deflection = 0.025H max
(delta m) = 0.7 R (delta s)
Rw = 5.5
(delta s) = 0.025H / (0.7)(5.5)(1.4 conversion factor) =
0.0046 H

Rw = 4.5
(delta s) = 0.025H / (0.7)(4.5)(1.4 conversion factor) =
0.0057 H

I agree that there is nothing magical about the drift limit of 0.025H or
0.005H, and both are just a reasonable limit for determining shear wall
deflections.  Mr. Ed Diekmann's proposed method (assuming construction and
wall heights are all the same) of wall length x capacity (number of nails per
wall) appears similar to how I understand one method of shear distribution was
used years ago for walls on a given floor .  Add up the total length of all
walls on a given floor level in a given direction, divide by the story shear
to determine the shear on a pounds/foot basis.  If the shear was less than
180plf or 175plf, then stucco or gypboard were considered to be adequate to
handle all the shear at that time.  No check for overturning and holdowns
since forces were considered to be so low.  In this case the nailing is the
same for all walls, which I assume could vary using the (wall length x
capacity) proposed method.

What I am not sure about is how you would distribute the stiffness based upon
the number of nails.  The more nails you add, the more load the wall takes,
but does the average nail stress remain the same as for a shorter wall which
takes less load.  When we use the tributary area method, all walls along a
given wall line are often designed having the same average force per foot of
wall length, therefore the individual nail slip would be the same between
walls.  I would think by the proposed capacity method that you are
distributing the story shear based upon wall bending and shear deflection
which incorporate wall length, igoring nail slip and holdown deflection.   Is
the proposed method (using stiffness from number of nails) actually to
distribute forces between parallel wall lines instead of to individual walls
along a given wall line.

Ignoring the deflection of holdowns, I think some redistribution of forces
between parallel wall lines needs to be considered, especially if there are
longer walls on one wall line as compared to the adjacent wall line.  But this
redistribution of forces has to be by a simple method in order for engineers
to incorporate this in their design practice.

Thanks Again,

Michael Cochran

```