horizontal shear in wood beams

[alan131]

Client wants to open up an interior bearing wall in a home. Was checking on the new concentrated loads on the existing house girder using a software program. Conditions are pretty standard: 24' deep house, (3) 2 x 10 center girder supporting First Floor, Second Floor & storage Attic with support walls; lally columns @ 8'-6" oc.


Program failed the beam for shear, not only for the proposed concentrated loads which was expected, but also for the existing conditions. Horizontal shear in wood beams is usually considered critical for high concentrated loads on short spans, whereas for longer spans, bending is usually primary.

Why is length of beam not a consideration in the calculation for shear in the sizing of wood beams? At what point is a beam considered short with excessive loads?
Also, for the existing uniform loading conditions, the program called for a 16" deep beam rather than the existing 10" which on the face of it seems ridiculous.

 

[JAE]

I don't know how anyone can help you directly on this as we don't have access to your program and don't know what you entered into the program.

That said, horizontal shear does control in wood sometimes.

It also depends on if the program is using an older code. The value of C_H is a factor used in older wood codes. In the 1997 NDS the design values were reduced to account for splits and shakes. The designer had to include the C_H factor to increase allowable stresses if there weren't any splits or shakes. In the 2001 and after codes that factor was eliminated. Your software might be using a lower value and thus showing a shear problem.

What sort of loading are you using?
Live loads for residential floors are typically 40 psf in the states and in some recent codes the upper floors are 30 psf.

 

[hokie66]

Just wondering why you need a program to analyze a wood beam, and why you would trust it to do so. Are you an engineer?

 

[alan131]

Thanks JAE for the reply,

Loading is Live loads @ 40PSF for 1st floor, 30 PSF for 2nd and 10 PSF for limited storage attic; figured 10 PSF per floor for dead loads. Program is StrucCalc which has been useful for residential calculations.

Any thoughts on why length isn't factored in for shear calculations . . .and yet shear becomes a critical factor for short beams?

 

[JAE]

What shear are you getting at the ends of your beams?

 

[alan131]

Hokie66,

I'm an architect but when registering for the forum it did not have that category but I do spec so I chose that. Find the program saves time when viewing options between steel, wood and flitch plates.

But I did go to Tech. . . Go Hokies!

 

[alan131]

5826# at the ends

 

[msquared48]

The length does figure into shear for the beam for uniform loads and affects the distribution for point loads. I do not know why you are saying it doesn't.

Mike McCann
MMC Engineering
Motto: KISS
Motivation: Don't ask

 

[JAE]

At that shear I get a unity of 1.55 for Spruce Pine Fir No. 2 with a (3)2x10 beam. That is overstressed.

The moment unity is worse - 2.41 based on simple span beams and based on continuous beams it is 1.60.

So it appears that you have a problem with the original design.

(unity = 1.0 means just safe per code)

 

[hokie66]

To try to explain why shear is often critical for short beams, take a long beam and a short beam, both simple spans, and apply the same total load to each. Use W as the total load, and say 8' and 16' for the lengths. The bending moments are then (W x8/8) = W for the short beam and (W x 16/8) = 2W for the long beam, but the reactions (end shears) are the same. So if you select a beam for bending, the short beam would be smaller, but would still have to carry the same shear. Clear as mud?

 

[alan131]

At the risk of sounding like. . .

Understand intuitively why a short header with a heavy concentrated load would fail in horizontal shear as compared to a longer span, and that shear is more critical than bending on short beams.

However, the equation for max. horizontal shear stress: 1.5 V/bd doesn't take into account length of span. If a 5000# conc. load is located midspan over a 3' window header as compared to a 12' wall opening header, assuming both headers are the same, the max. horizontal shear remains the same. While the shear values along the beam will vary, the max. remains the same. In my example, max. hor. shear would be the same for both the 3' and 12' spans. For a doug fir header, the 95 PSI hor. shear strength along the grain remains the same. I'm missing the calculation link that makes this more critical for the 3' span.

Thanks for humoring me. . .

 

[BAretired]

alan,

It is not more critical for the 3' span. Both beams in your example are equally critical in shear.

Short beams are more likely to be governed by shear stress while long beams are more likely to be governed by deflection or bending stress.

BA

 

[BAretired]

Actually, come to think of it, it is more critical in the 12' span because the critical section for shear is depth 'd' away from the support, so the critical shear stress is higher in the 12' span.

BA

 

[BAretired]

Wrong!!! It is the same for both beams as the critical shear is 2500# in each case.

BA

 

[a2mfk]

Do yourself a favor and hire a structural engineer who will answer all those questions plus the ones you don't know to ask... Sorry I had to be that guy, but you seem to be operating outside of your area of expertise and wood beam design of this nature is not considered "incidental to the nature of your work"..

 

[hokie66]

I think he is getting it now. Actually, it is refreshing to find an architect who will listen.

 

[alan131]

Sorry for the brain lock. I think I have confused the categories. Breyer had put it this way: "As a general guide, shear is critical on relatively short, heavily loaded spans." I had taken critical to refer to short vs. long spans, rather than referring to shear vs. bending or deflection.

Thanks for all your patience and responses. . . alan

a2mfk. . . your suggestion is duly noted

 

[dhengr]

Alan:
Generally speaking, for a given total load W=P, either uniformly distributed (W) or concentrated (P) at the beam center, a longer beam will force you to use a deeper header or a double 2x header to satisfy bending and deflection criteria; and the shear near the reaction is basically W/2. That’s what hokie was saying, 19FEB11 0:01, in slightly different words. And, the shorter beam allows you to use a smaller header for bending and deflection, again same W or P and shear near the reaction, but now the horiz. shear stress 1.5V/bd is greater, because the smaller member gives a smaller “bd” in the denominator. With the centered point load the shear is essentially constant along the beam length, except for variation due to the uniform weight of the beam. If you move the point load toward one of the reactions, the max. bending moment will start to reduce, implying a smaller member; and in the extreme with P only “d” from one reaction the moment will be very small, but the shear will now be almost P, not P/2. Look at some of the beam formulas in one of your ref. books.

That being said, I fully agree with a2mfk. And, I suspect you get madder than hell when your local Structural Engineers practice any Architecture on a job which is predominantly a structural job. And, even happier when they come to you for architectural advice on that job. These damn computer programs make many people pretend to be proficient at things that they should be very careful doing, for lack of a fuller understanding of the problem.

 

[hokie66]

Come on guys, he is an architect, and in many places allowed to perform engineering work on a limited range of projects. Besides, he is a Hokie, so a good guy.

I like to try to teach architects who will listen. In this vein, alan131, do you know why shear stress, particularly in wood beams, is used as 1.5V/bd rather than just V/bd as usually used for steel and concrete?

 

[alan131]

From what I understand 1.5 is to provide for maximum shear not just average shear.

BTW. . . UVa beat the Hokies in mens BB today. . . nuts.