beam stress and deflection for a wood beam

I will make the assumption you are talking a simply supported or some other horizontal type beam, loaded between the supports. The shear parallel to grain is what you are looking for. The upper fibers are in compression, the lower in tension. There is a shearing action across the beam, between fibers from top-to-bottom. It is this shear that keeps the beam together--the grain usually runs lengthwise in a standard 2x4. Is that perfectly muddy?

It gets clearer if you are in an older Black Angus restaurant, where they have such a beam prominently displayed... split across the length, where the shear is a maximum.

David A. Smith

The modulus of rupture is the ultimate allowable for bending.

The tension strength parallel may not be practical to test for.

Shear strength parallel to grain should be lower than across grain. Shearing across the fibers is harder than with the fibers.

Check both bending and shear.

- CJF.

wood is anisotropic ( I think that is the proper "tropic"). That basically means that what you find in one axis is not able to be used to determine what you will find in another axis - and concepts such as shear strength and tensile strength and bending strength and all the parameters we use reliably in steel are not always related (there are no cellular filaments or knots in metals, for example)

A couple of things to keep in mind -

Thus, most wood design uses deflection criteria and rules of thumb rather than the typical metal-beam analysis models. Those rules of thumb are often rigorous enough, given the variation in strengths encountered.in wood, even in the same species.

A Minnesota old growth hickory or a desert pine is a far different animal than a new growth Alabama farm-aspen or a balsa. Damp wood is different than dry.

I have heard 700 psi as the tensile parallel to the grain, but if that is the test strength, 300 psi would be the 10 year reliable. IMHE.

etc.

I see that the link had all the info already. Sorry.

Chris, You've gotten some pretty good advice from the earlier posts, but here's the quick and easy solution. The ultimate failure point of the scenario you just described can be calculated by the following formula:

ultimate allowable tensile stress

I have a text that can answer most, if not all, of your wood design questions.:

"Timber Design and Construction Handbook", by Timber Engineering Company, An affiliate of the National Lumber Manufacturers Association, McGraw-Hill Book Company, ISBN 07-064606-6

My copy is dated 1956. (It lists a 2x4 as 1-5/8 x 3-5/8, so a newer edition would be better.) I know the text was out of print for a time. It may be hard to find that is why I included the above details. Check a city library for it. The book is considered by many to be a classic.

Jim Y

a book called "Wood

States Department of

book in pdf format

leave me confused as

a load in the

allowable stress of

numbers for

compression and both

parallel to the

of thing with wood

Materials like glass and carbon fiber are not only anisotropic, but have a brittle (even explosive) failure mode. White woods are sometimes estimated at 1200 psi ultimate and Youngs of 1.3E6 psi in beam applications, BUT the design factor is increased to keep away from brittle failure. Wooden aircraft wing spars often use a load factor of X6 and a design factor of 1.5X

These loads are considered short term, because a wood plane spends most of its life in the hangar.

Brian Whatcott Altus OK