Timber Framers Guild Forums Tools Timber Engineering Calculating Members Notched at Bottom

I am brushing up on different joinry over the holidays and am interested in notched members. I am working with V=(2Hbd(d/h))/3. Where V=vertical Shear
b=width of beam
d=depth of tenon
h=depth fo beam
H=Working Stress in horizontal shear, shear parallel to grain.

My question is about H and there is no real value for angling my notch cut in this equation. Is H total load of member divided into half? And do I compare my answer of V to the Shear in psi. ie White oak of 1250.

I'm not entirely clear on your question, but I beleive Steve Chappell looks at the question of beveled notches... do you have "A Timberframer's Workshop"?
If so, look at page 63....

Thanks for your help Mark. I have the old book and found a better example on page 29. Now a new question. Total load to compair with V/8. Is this half the load of the bearing member? ie. 4x12 joist spaning 14' on 2' centers. Don't I compair the total load of bearing member divided in half since each end bears the load equally.

The equation you listed is an older one. The current standard equation (from the 2001 NDS) for a bending member notched on the tension face (typically the bottom side) is

V'r = (2/3 * F'v * b * dn) * (dn/d)^2

Where
F'v = adjusted allowable horizontal shear stress value
b = member width
d = full member depth
dn = depth of member at notch (what's left)

This results in an allowable shear capacity in the timber. For simple beams with symmetric loading, this is one half of the load to the beam.

Fv is the allowable shear stress, not the average ultimate value listed in the Wood Handbook.

The currently published value of Fv for white oak (not grade dependent) is 205psi, not 1250psi.

While the code does say (and research and historical precedent has shown) a gradual notch does decrease the stress concentration, BUT, not in a manner that is easily computed. Standard practice is to assume a square notch.

Note with either the old or current equation, the allowable capacity (and witnessed in physical testing) of a notched beam is LESS than a smaller unotched beam of depth dn.

Joe

I would recommend consulting an engineer. If you are fuzzy on what means what, notching is not a place you want to error. The previously mentioned references will also be of help. I would recommend really studying the topic before proceeding if you choose not to consult an engineer. It is clear that some fundamental engineering knowledge is absent in this case. For instance your comment, "Is this half the load of the bearing member? ie. 4x12 joist spanning 14' on 2' centers. Don't I compare the total load of bearing member divided in half since each end bears the load equally." shows that you don't understand simple loading of a beam. Each bearing does not necessarily carry half the load. Only in situations of symmetrical loading is this the case. This may come across a bit harsh, but if you don't understand the equations you are working with, then take the time to learn them, or consult someone who has.

Thanks Mark for the new formula. I have been following the general guidelines for shouldered dovetails. I was wondering by following the joist or purlin ¾ rule I would have to use the notched member calculation and this isn’t mentioned in the formula. Then looking closer at the 5/8 rule in the book I began to wonder about the meat under the shoulder. Would this be verified with the single shear formula (AxS) and would I use the Fv Shear parallel to the grain.