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Re: Allowable Deflection ??

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Be sure to account for shear deflection as well, if applicable.
5wl^4/384EI + wl^2/GA * k
k = shear deflection constant; 6/5 for rectangular sections, 10/9 for circular sections
G = shear modulus or modulus of rigidity
A = cross-sectional area
This is discussed in Wood Engineering by G. Gurfinkel and (I presume) most other wood books.

>>> kspeoples(--nospam--at) 12/10/02 07:00AM >>>
If this is your first time doing it, just make sure you work through the units to see what w acutally is - pounds per inch or pounds per foot.  I usually work in kips and feet, so with my E being in psi, I have a 1728 factor at the top of my equation.  I have seen quite a few deflection calcs from the younger guys where they are off by a factor of 12.
Kenneth S. Peoples, P. E.
Lehigh Valley Technical Associates
1584 Weaversville Road
Northampton, PA 18067-9039
Phone: (610) 262-6345
Fax: (610) 262-8188
e-mail: kpeoples(--nospam--at)
----- Original Message -----
Sent: Monday, December 09, 2002 7:16 PM
Subject: Re: Allowable Deflection ??


.25"=5w(60)^4/(384*E*I)  for unifomly loaded span

susbt. E and I (E in psi and I in in^4 )

solve for w to get allowable load

 ken kirkland <kenmail234(--nospam--at)> wrote:

If I have the allowable load 40psf for span = 60 inches,  how can fingure out the allowable load under l/240?


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