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Re: Diaphragm Calculations

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Dennis:

You ask:
> What am I missing???????

Here is the problem with your calc.  You say:
> Since the basic formula for I is (base * height^3)/12  we can assume
> the base to be the span between shear elements or gridlines (L) and
> the height to be the depth of the diaphragm (b). Therefore the
> formula changes to I = (L b^3)/12. 

You are calculating the moment of inertia of the PLAN view of the 
diaphragm.  That is incorrect.  You must calculate the moment of 
inertia of the section that is bending which looks like a very 
deep TJI laying on its side.  It is b deep by t thick (say 1/2") with 
chords that are double top plates (or similar; chord area is A). 
Going back to basic mechanics, the moment of inertia of this "I" 
section is (t*b^3)/12 + A*(b/2)^2 + A*(b/2)^2.  The first term is the 
correct form of the "basic formula" you referred to and the second 
and third terms are the contributions of each chord to the moment of 
inertia.  However, because the "web" is so thin and deep and is 
jointed in both directions (composed of 4' x 8' pieces) it doesn't 
make sense to assume that plane sections remain planar and its 
FLEXURAL stiffness is generally considered negligible. So the first 
term in the moment of inertia calc goes away.  What you are left with 
is the contributions of the chords which produce

I = 2 * A*(b/2)^2 = (A*b^2)/2

Similarly, the second term of the diaphragm deflection calculation 
(v*h)/(G*t) accounts for the shear deflection of the diaphragm.  It 
is based on the "web" only and ignores the contribution of the 
chords.

These two definitions of the appropriate stiffnesses of the diaphragm 
(for flexure and shear) are consistent with the common design 
approach that assumes that the "web" carries all of the shear and the 
chords carry all of the flexure.

Hopefully this clarifies things.

-Mike

********************
> Mike, I was reviewing your calculation for the conversion of the diaphragm 
> deflection (simple span part) and came up with a different solution than you. 
> Can you please review my notes below and let me know where I may have gone 
> wrong?
> 
> In a message dated 7/8/99 6:54:19 PM Pacific Daylight Time, mtv(--nospam--at)skilling.com 
> writes:
> 
> << Dennis:
>  
>  The deflection term you are asking about represents the flexural 
>  deformation of the diaphragm.  Here is how the equation is derived.
>  
>  Assume:
>  Simple span beam
>  Uniformly distributed load
>  Base diaphragm moment of inertia on chords only
>  
>  Maximum deflection = ( 5 w L^4 ) / ( 384 E I )
>  Maximum shear force, V = w L / 2
>  Maximum unit shear, v = V /  b
>  
>  I = sum (A d^2), where d = b / 2
>  
>  So, I = (A b^2 ) / 2
>  
>  Substituting (and simplifying),
>  
>  Maximum deflection = (5 v L^3 ) / ( 96 E A b ) 
>  Given the units noted in the code, the deflection is in feet.
> *******************************************************
> <<Dennis>> Mike, I may be confused by your Moment of Inertia term I which you 
> define as: 
> 
> I = (A b^2 ) / 2
> 
> Since the basic formula for I is (base * height^3)/12  we can assume the base 
> to be the span between shear elements or gridlines (L) and the height to be 
> the depth of the diaphragm (b). Therefore the formula changes to I = (L 
> b^3)/12. 
> If the area is L*b the formula reduces to:  I = (A*b^2)/12
> 
> combining all of the terms:
> 
> Deflection = 5*2 v b L^4 12  / (384 E A b^2) which then reduces to:
>  
> Deflection = 5 v L^3 / (16 E A b)
> 
> This still does not balance to match the deflection formula 5 v L^3 / (8 E A 
> b)
> 
> What am I missing???????
> 
> Dennis
> 
>  *************************************************************
>  Max defl (inch) = (5 v L^3 ) / ( 8 E A b )
>  
>  or,  = (5/8) (v L^3 ) / ( E A b )
>  
>  As a point of interest, this also highlights the limitations 
>  (assumptions) of the formula.  If we say that for all load conditions 
>  and all boundary conditions,
>  
>  Max defl (inch) = X (v L^3 ) / ( E A b )
>  
>  X is 5/8 for single, simple span with uniformly dist load, but can 
>  range from 1/8 (fixed-fixed, unif load) to 1 (pin-roller, centered 
>  point load).
>  
>  The second term of the deflection is based on a similar derivation 
>  for the deflection due to shearing of the "web".
>  
>  The equation for shear wall deflection (flexural and shear) is 
>  derived based on a cantilever column with a concentrated load at the 
>  top.
>  
>  -Mike
>   >>

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Michael Valley                                   E-mail: mtv(--nospam--at)skilling.com
Skilling Ward Magnusson Barkshire Inc.                  Tel:(206)292-1200
1301 Fifth Ave, #3200,  Seattle  WA 98101-2699          Fax:        -1201