Design for Bending

While trying to prove a point I was reviewing the Timber framer's guild workbook in sizing timbers. The section on bending S=M/Fb. M is calculated in feet, but no reference is given to converting this to inch pounds for the S=M/Fb comparision of timber to load review. This Does have to be converted doesn't it? The inspector also was trying to calculate M with WLsquared/8. Is this something new?

FS:

I'm not sure what your inspector was trying to calculate, but the maximum moment (Mmax) for a uniformly loaded beam is WL/8. Convert L to inches, if necessary. Use inches for everything. The units for Mmax are in-lbs. The units for Fb are psi (lbs/in^2). Thus the units for S are in^3.

CB.

Free State

Daiku was on the right track. However, M= wL^2/8, for a uniformly loaded beam , where w = load per lineal inch (or foot) and L = length of beam (in.). To calculate w, multiply the total load (in pounds per square inch) by the spacing of the beams (in inches) or if the beam is used singly, use the supported width of floor area. [Note: the bending moment formula depends upon the type of loading condition. Here it is given for a uniformly loaded beam only.]

Once M (the bending moment) is known, use the material property Fb (psi) to calculate the section property S (in.^3). S is given as the (width x depth^2)/6.

This should give you the dimensions of the piece required. I hope this helps.

Mark Gillis

Mark and I are in agreement. I was using (big) W as the total load in pounds. He is using (little) w as the load in pounds per inch. w = W/L, thus the two equivalent formulas are:

Mmax = wL^2/8
Mmax = WL/8

As Mark points out, this formula is valid only for the uniformly loaded case.

CB.

Thank you very much. I was using the Big W and my inspector was using the little w.