Int wall 90 deg to roof trusses non-bearing?
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Int wall 90 deg to roof trusses non-bearing?
I want to remove a good portion of the wall separating the living room and dining room in our home 1970's vintage wood frame on concrete slab house. The wall runs down the center of the house, and is perpendicular to the roof trusses. It has a 4 ft wide opening between the rooms, but I'm not sure what kind of header, if any, is above the opening. Of course, I'll run this past the local building dept., but I had heard that with truss roof construction, all roof loads bear on the exterior walls. Is this likely to be the case?
There is 14" of wall above the door opening. If taking the wall out on one side of the 4 ft opening is doable... then I would like to know if it is best to remove all of the wall material and patch the void in the drywall left where the wall was removed ... or leave a short stub of wall, say 6 ", extending down from the ceiling (faux beam?). My concern is that there might not be anything to nail the dry wall patch to excepth the 24 " on center trusses. Any input will be appreciated!
Grommet
Tempe, AZ
There is 14" of wall above the door opening. If taking the wall out on one side of the 4 ft opening is doable... then I would like to know if it is best to remove all of the wall material and patch the void in the drywall left where the wall was removed ... or leave a short stub of wall, say 6 ", extending down from the ceiling (faux beam?). My concern is that there might not be anything to nail the dry wall patch to excepth the 24 " on center trusses. Any input will be appreciated!
Grommet
Tempe, AZ
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The wall you are describing is most certainly a load bearing wall. You didn't mention the total width of the house, and premade trusses can only handle just so much span without support. Your house was built before the common use of Silent Floor, and other type spanning beams, so your only other alternative is to run LVL or other beam types to replace the wall you will be removing. As you stated, however, it will be best to run this past your local inspector to get their advice, since they will be looking at what you are looking at.
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Hi Chandler,
Thanks for the input. That's the problem with some of those rules of thumb huh....The house is 30 ft wide, with a 4 in 12 pitch roof. The trusses are 2x4 W-types on 24" centers. The span I would need to support is about 14 ft. The house is single story, roofed with asphalt shingles. I need some LVL for a garage header, so I guess I'll look for some for this project at the same time.
Grommet
Thanks for the input. That's the problem with some of those rules of thumb huh....The house is 30 ft wide, with a 4 in 12 pitch roof. The trusses are 2x4 W-types on 24" centers. The span I would need to support is about 14 ft. The house is single story, roofed with asphalt shingles. I need some LVL for a garage header, so I guess I'll look for some for this project at the same time.
Grommet
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OK,
Let me know if I'm on the right track here. Looking at typical composite roof loads per the codes (20 lbs psf live load, 15 psf dead load, 0 snow=35 psf Total Load), and 30 ft truss span with 1.5 ft overhangs (31.5 ft span x 2 ft width of Tributary Area=63 sq ft total Tributary Area), means each truss is supporting 2,205 lbs. If the center wall is load bearing, it can only be carrying 1/3 or 735 lbs for each of the trusses which cross on 2 ft centers. Each foot of wall is carrying half that or 367 lbs per lineal ft. Pretty high loads for dimensional lumber, but 2 pieces of 1 3/4 x 9 1/2 LVL 1.9E or 2.0E, or 2 1 3/4 x 11 3/4 LVL 1.7E should exceed that in capacity. What do you think the cost and availability of the headers would be?
Grommet
Let me know if I'm on the right track here. Looking at typical composite roof loads per the codes (20 lbs psf live load, 15 psf dead load, 0 snow=35 psf Total Load), and 30 ft truss span with 1.5 ft overhangs (31.5 ft span x 2 ft width of Tributary Area=63 sq ft total Tributary Area), means each truss is supporting 2,205 lbs. If the center wall is load bearing, it can only be carrying 1/3 or 735 lbs for each of the trusses which cross on 2 ft centers. Each foot of wall is carrying half that or 367 lbs per lineal ft. Pretty high loads for dimensional lumber, but 2 pieces of 1 3/4 x 9 1/2 LVL 1.9E or 2.0E, or 2 1 3/4 x 11 3/4 LVL 1.7E should exceed that in capacity. What do you think the cost and availability of the headers would be?
Grommet
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Wrong track. It depends on the truss. I'll give a couple examples to show some discrepancies in your reasoning.
The easy case is to consider is pretending that your truss is two simply supported beams. Imagine two beams, one spanning from each wall to the center so each beam is half the length of your truss. The center wall would carry 1/2 the weight of each beam, or 1/2 the total load - not 1/3.
If the beam is continuous (one long beam), the load on the center support can be higher. So you can intuitively understand, imagine a perfectly rigid beam that does not flex. Theoretically, it could balance on the center with the end supports removed and the center would carry all the load. Nothing is perfectly rigid; a beam wants to deflect and part of the load is carried by the end supports. Without explaining the calculations for the load on the center support, a beam supported in the exact center with a uniformly distibuted load will theoretically have the center supporting 5/8 of the total load.
Now, if you want to really complicate calculations, you might consider the fact that the LVL support will also deflect (more than the outside walls will compress). Now, imagine the center of the truss is being supported by a very stiff spring which will cause more load to be transfered to the outside walls.
On top of all of this, trusses aren't beams. They act differently. You need to find a table that covers headers in interior bearing walls or consult an engineer. The tables are designed to work for almost all cases. Or you can try asking a lumberyard that sells LVLs; they may have someone that can size the LVL for you (they sell more that way).
The easy case is to consider is pretending that your truss is two simply supported beams. Imagine two beams, one spanning from each wall to the center so each beam is half the length of your truss. The center wall would carry 1/2 the weight of each beam, or 1/2 the total load - not 1/3.
If the beam is continuous (one long beam), the load on the center support can be higher. So you can intuitively understand, imagine a perfectly rigid beam that does not flex. Theoretically, it could balance on the center with the end supports removed and the center would carry all the load. Nothing is perfectly rigid; a beam wants to deflect and part of the load is carried by the end supports. Without explaining the calculations for the load on the center support, a beam supported in the exact center with a uniformly distibuted load will theoretically have the center supporting 5/8 of the total load.
Now, if you want to really complicate calculations, you might consider the fact that the LVL support will also deflect (more than the outside walls will compress). Now, imagine the center of the truss is being supported by a very stiff spring which will cause more load to be transfered to the outside walls.
On top of all of this, trusses aren't beams. They act differently. You need to find a table that covers headers in interior bearing walls or consult an engineer. The tables are designed to work for almost all cases. Or you can try asking a lumberyard that sells LVLs; they may have someone that can size the LVL for you (they sell more that way).
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Hi Phil,
Thanks for pointing out my deviations from the true path. In the simplified case for a continuous beam supported in the center bearing 5/8 of the total load as you outlined above, I would need a beam to handle 689 lbs per lineal ft. Allowable uniform roof loads for two pieces of 1 3/4 x 11 3/4 2.0E LVL exceeds 900 lbs per lineal ft, so that might be ccol....but. Since, as you mention, there is more to consider, I will look for a table specific to this application, or will consult an engineer. I located one table specific to headers in interior walls, but it assumes various combined floor and roof loads, hence..... still surfing.
Grommet
Thanks for pointing out my deviations from the true path. In the simplified case for a continuous beam supported in the center bearing 5/8 of the total load as you outlined above, I would need a beam to handle 689 lbs per lineal ft. Allowable uniform roof loads for two pieces of 1 3/4 x 11 3/4 2.0E LVL exceeds 900 lbs per lineal ft, so that might be ccol....but. Since, as you mention, there is more to consider, I will look for a table specific to this application, or will consult an engineer. I located one table specific to headers in interior walls, but it assumes various combined floor and roof loads, hence..... still surfing.
Grommet
Last edited by Grommet; 03-25-07 at 04:37 PM.