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Re: Rigid plywood diaphragms

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In a message dated 11/22/98 5:13:56 AM EST, itsekson(--nospam--at)jps.net writes:

See inserted comments:

<< I will in fact use tributary area method as the base for my horizontal
shear
 distribution (it does not take that much time and it seems to be still
 working).  It should not be that difficult to put together the spreadsheet
 using UBC Standards formulas for shear wall deflection either.  I am
 wondering though if it would give me the realistic shear distribution ( I
 don't think so)  and would it be any closer to the actual performance of the
 system than the customary method above.>>>

Putting together the spread sheet is not a problem for shear walls in the same
line, except how you want to handle the the holdown assembly uplift.  I found
that to solve this you have to use an iterative solution until all shear walls
along a given line have approximately the same deflection (we use +/- .005
inches).  As you keep adjusting shear loads between shear walls, you keep
changing holdown sizes (we added a data base of Simpson holdown sizes and
deflections under maximum loads for the program to use in the iterative
solution.)

You may want to include in your program the ability to say what size holdowns
you want to use on each wall instead of letting your program pick the size,
the nail spacing to use and also plywood on two sides of the wall.  This will
help if you are trying to evaluate existing wall conditions.  You may also
want to consider tie-rods.

I believe the UBC formula for shear wall deflection is fairly accurate up to
first yield of the shear wall panel.  I am not sure about it though from the
first yield point to the wall's ultimate strength.  John Rose would be able to
answer this question more accurately.   

The distribution of shear to individual walls in the same line will vary
depending upon wall length initially (Your program might give the following
results: short walls 250 plf while longer walls are 500 plf).  When the slack
in the holdown assembly is taken up by overturning (you increased the loads on
these wall while keeping the same size holdowns and nailing), then the shear
in all the walls will approach the same (say 500 plf).
 
 <<<Let's not forget that we are talking about the RELATIVE RIGIDITY of shear
 walls along different lines of resistance.  For the sake of the argument
 let's look at the direction when framing is parallel to shear walls.  So the
 overturning moment and boundary member forces are the function of the unit
 shear times the height of the wall.  I can drop out several components of
 the total deflection from my relative rigidity calcs that would cancel out.
 These elements may include deflection due to oversized holes in holdowns
 (which in fact are 2 to 3 times more than the holdown deformations itself),
 plywood shear deformations, slip in sill bolts.
  >>

As you mentioned above, what is the relative rigidity of shear walls along
different lines of resistance. I would still calculate the deflection of all
shear wall lines using the UBC formula, I don't think you will be able to drop
out certain components, in fact I think you would have to use the rigidity of
the entire shear wall line since all walls along that line have the same
deflection (tied together by the double top plate).  There would be less work
by using wall lines instead of individual walls in terms or calculating
rigidity (similar to rigidity of walls with openings) and probably more
accurate. 

Depending upon the deflection of each shear wall line, I would redistribute
the lateral loads to adjacent shear walls (consider semi-rigid diaphragm).  If
the deflection in one wall line exceeds the deflection of an adjacent wall
line by more than 20% (an arbutary value) I would redistribute the loads.
Proposed additional force to add to wall line which deflects less:
        
                            (D1 - D2) 
additional force  =  -------------  x  Vwall(i) x 0.5
                               D1
D1  =  Larger shear wall line deflection 
D2   = Smaller shear wall line deflection
Vwall(i) = Total shear force from shear wall line with larger deflection

(0.5 since taking some of the lateral force away from one wall line, it will
now deflect less, hopefully less than 20% difference in deflection between
this wall and the  adjacent shear wall which is receiving the additional load
).

I would then recalculate the deflection for the wall line which deflected less
with the additional force added. ( I would not recalculate the wall line which
I took the force away from, since it will now have more reserve capacity and
should already meet code deflection requirements).  The 20% may seem high, but
if one shear wall line of resistance deflects 0.4 inches, the next adjacent
shear wall line would have to deflect more than 0.48 inches before you would
consider a redistribution.  0.8 inches of additional deflection for a shear
wall line which say is offset by 15 feet or more from the wall which deflects
0.4 inches is relatively small.  I would think that the 20% value, or some
other value, is something which would have to be agreed upon by others.

Just some additional thoughts,

Michael Cochran