Undersized wood header

[jeffhed]

I have a project with an 8'-0" wide opening for a sliding glass door that leads out to a back patio. Originally the plans showed a beam supporting the patio trusses. The contractor changed the roof trusses to cantilever, thus increasing the load beyond the header capacity that I originally designed. However, the header is framed with a single 2x4 plate on top and bottom of the member. If I look at the whole section with both plates included, the member is sufficient. I need some ideas on connecting the 2x4 plates to the existing header. From my numbers, I'm thinking it can't be done, but I kind of want a crazy check before I make them replace the beam. The header is (2) 2x10 DF #2 and both the top and bottom plate are also DF #2. Deflection is not a problem, only bending stresses. My design moment is 4958 lb-ft = 59496 lb-in. I then divide the design moment by the original header depth to give me the resulting force couple 59496/9.25 = 6432 lbs. With a force that high, it seems it may not be possible to bond these plates to the beam as I only have 4'-0" of beam to bond to on either side of the center of the header (connection force of 6432/4 = 1608 plf).Any other suggestions in strengthening this beam?

 

[msquared48]

If the header does not fill the space, why not add an LVL to the side of the same depth. Or shore and replace the header with multiple LVL's.

Otherwise, you will have to perform a shear flow analysis to determine what connections are needed to the top and bottom plates to get the composite action you are seeking.

Mike McCann
MMC Engineering

 

[jeffhed]

Mike, I realize the method I posted is not completely accurate, but we use it as a quick way to see if the forces are in the ball park of being able to remedy the problem or if a new beam is required. I also wanted to give people an idea on the forces I am dealing with. After running a shear flow analysis on the beam I have ended up with a much better force, however still a high force of 3984 lbs. The window is in, and exterior water proofing is on. So I am looking for ideas on how to strengthen the beam since connecting the plate seems to be unlikely. If there are no creative ideas, then we are back to where I was before, removing and replacing the beam.

 

[msquared48]

No way of nailing Simpson A34's or LPT4's to the top and bottom plates to develop the section?

Mike McCann
MMC Engineering

 

[jeffhed]

LTP4 plates look like they will probably be the way to go. Although distributing 3984 lbs over 4'-0" of header comes out to approximately 6 LTP4s on each side of the window center line. I only need one of the 2x plates to make my section work, so it probably won't seem too excessive. Using 2 plates only ended up dropping the connection force about 1000 lbs. Seems like kind of a lot, but I'm sure its going to cost less than replacing the beam. Thanks Mike

 

[msquared48]

No worries.

Mike McCann
MMC Engineering

 

[JAE]

jeffhed - how did you arrive at 3984 lbs?

I back calculated your uniform load = 619.75 plf
This gives a shear of 2479 lbs at the end

The Ix = 536 in^4
Q for the top 2x4 = 28.2 in^3

q = VQ/I = 130.4 plf

This horizontal shear diminishes from a high at the ends = 130.4 plf to zero at midspan.

The total shear across the 1/2 span on the top would be 130.4 x 4 / 2 = 261 lbs

Perhaps I screwed up my calcs (it's late and it's been a long day).

 

[FSS]

Did you verify actual design loads? Truss guys are usually higher than code. Design may still work.

 

[jeffhed]

JAE,
Your reaction is essentially the same as mine, I had 2486 lbs since the load is slightly tapered. But unless I am misunderstanding the calculations (it's been a while), your units for Q should be lb/in? Even though I am using 3.5 wide plate, for simplicity I used 1.5"x3" plate for a total cross sectional area of 3"x10.75".
I = 3*(10.75)^3/12 = 311 in^4.
A = (1.5*3)=4.5 in^2.
Ybar = (10.75/2)-(1.5/2) = 4.625 in.
Q = A*Ybar = (4.5)*(4.625) = 20.8125 in^3
VQ/I = 2486*20.8125/311 = 166.36 lb/in ==> (your units said plf)
Q tapers to zero at the mid span so I created a spreadsheet that is attached to my beam sizing spreadsheet. My spread sheet calculated the shear at each increment (.01 feet) and summed them up which gave me the 3984 lbs. I thought it seemed really high for a beam that was overstressed by 20% with the new loads.

FSS
I have got the loads down as far as I can. I determined the loads myself, I did not get them from the truss calculations. But I compared my loads and the truss manufacturers loads because of your comment and the are very similar.

 

[JAE]

I knew I was tired...yes - inches and not feet.

Your Ybar should be 5.375" right? (to the centroid of the area)

Total depth of (2)2x10 plus 2x4 top and bottom is 9 1/4" + 2(1.5) = 12.25"

Ybar = (12.25/2) - 1.5/2 = 5.375"

Q then is 1.5 x 3 x 5.375 = 28.2 in^3

But your overall numbers are generally in the range. I just was half-asleep so pardon my mistake.

 

[jeffhed]

JAE,
Ybar is still 4.625, once I ran the numbers and saw the magnitude of the forces, I realized that using the top 2x4 plate didn't help as much as I thought it might (3100 lb connection force on top and bottom instead of 3900 lbs on bottom). The section still works with only the bottom plate, so I progressed with only the bottom plate being considered, thus the section is only 10.75" deep. Thanks for your input though, it was nice to get some feedback since the force seemed so high.

 

[MiketheEngineer]

Could you add a 1/2''x 9.25'' plywood section to the 2x10's??

That might give you enough??

And maybe it is already there??

 

[jeffhed]

Mike,
With tying the plate to the bottom of the (2) 2x10s I have the sufficient strength. However, this was also an option I looked at, but I wasn't so sure how to approach the nailing. I assume just nail it like a multi-ply beam? Anyway, I assumed I could get the nailing to work, but I didn't think the minimal additional width would give me what I needed.

 

[MiketheEngineer]

Assuming the plywood is as good as the DF - and you can use a 15% short term duration - I get you need an Fb of about 1035 psi - a bit shy for DF #2 North.

Re-justify your loads. You might be able to consider some small reduction there that would make it work.

1/8'' or 1/4'' Steel plate will most likely work??