Thanks
Based on your suggestions, I looked around and found
http://www.mech.utah.edu/~bamberg/research/BeamBendingSimulation/BeamBendingApplet.html
It appears that doubling the width ( depth ) of the beam causes the  
deflection to be reduced about 8-fold.
Michael


> On Sep 30, 2008, at 9:18 PM, Sean Breheny wrote:
>
> Hi,
>
> This would be a good start:
>
> http://structsource.com/analysis/types/beam.htm
>
> On this page they write these formulas in terms of the moment of
> inertia of the beams. I'm not sure exactly which moment of inertia
> they mean but I'd guess that it is around the axis which is defined by
> the cross product of the length and the applied force, and computed at
> the center of mass of the beam.
>
> Your answer may depend upon how the ends are fixed (whether they are
> allowed to pivot or not).
>
> It is not entirely clear to me how you are defining width and
> thickness and the direction of the applied force. I think that the
> dimension which is not length and is also not in the direction of your
> force will make a linear change in the stiffness. I think that the
> dimension in the direction of the applied force makes a cubed change,
> and the length makes a change which is not a pure power but a
> polynomial.
>
> Sean
>
>
> On Tue, Sep 30, 2008 at 10:14 PM, agscal -AGSCalabrese <agscal@gmail.com 
> > wrote:
>>
>>
>> You have a beam that is ( theoretically )  x long, y wide and z
>> thick.  Weight is applied perpendicular to the xz plane. Beam is
>> supported at both ends of the length.
>> What happens if you compare it to a beam of the same material x long,
>> 2y wide and z thick ?
>> ( Presume the weight of the beam itself is not an issue and the beams
>> are stabilized from sway)
>>
>>
>> For example:   If you have a beam made of 7/16 inch OSB that is 21
>> feet long and 8 inches wide.
>> How much stronger will a similar beam be if it is 16 inches wide ?
>>
>> TIA
>>
>> Michael Algernon
>>
>>
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