# load-bearing or non load-bearing wall?



## dgsquare (Apr 22, 2010)

Hi,

I have a little reno project going on right now. I am just about to take a wall down and still wondering if it is load-bearing or not. I had the opinion of two contractors already but they disagreed on the matter... The fact is that the framing shows characteristics of both a load-bearing and non load-bearing wall.

Here is what you need to know:

- 8' partition wall between kitchen and living room, located on first floor (unfinished basement below)
- Runs perpendicular to the joists and trusses
- No point of contact with trusses in the attic (meaning that the web doesn't sit directly on the top plate, offset by a foot or so)
- Also offset with footing in the basement (main beam in the basement is aligned with the footing) 
- Double top plate, 2X4
- Had a pocket door (now removed)
- Header above pocket door is a 2X4 on the flat with criple studs (no beam or plywood)
- Trimer studs are present and made of finger-jointed studs (vertical use only)

Can you help?

Cheers,

dgsquare


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## tpolk (Nov 7, 2009)

truss above probably nonbearing right there


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## kwikfishron (Mar 11, 2010)

I would say it’s not load bearing. Full span trusses are designed to be bearing on the outside walls. If it was load bearing that should be a solid and not a box header.


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## Yoyizit (Jul 11, 2008)

Stretch a level line at right angles to the partition as close to the ceiling as possible.

A non load bearing partition will have the max. ceiling sag in the middle of the ceiling span. A load bearing partition will be raising the ceiling slightly.

In a 10' span you might be looking for a sag of 3/8".

That single horiz. 2x4x5' laying on its flat side above the opening can't bear much load, if that is its purpose. See how much sag it has in the middle. 0.1" sag means its carrying about 0.6# per inch of length, unless it is already warped. 
It's like having a built-in strain gage.

Using Excel prevents math mistakes \/

b=	3.5	>enter value
h=	1.5	>enter value
I=	0.984375	=calc'd value

E=	1000000	>enter value
d=	0.1	>enter value
L=	60	>enter value

w = d384EI/5(L^4) 

d384EI=	37800000	=calc'd value
5(L^4)=	64800000	=calc'd value
w=	0.583333333	=calc'd value


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## Big Bob (Jul 27, 2007)

agree with tpolk... from info you provide..

a pic from top side would confirm your statements...cheaper than an Engr...:thumbsup:


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## dgsquare (Apr 22, 2010)

Thanks for the quick replies.

Here is more info and a couple of pics from the attic. In the first picture, you see the 2X4 that makes up the top plate of the wall, and you can clearly see that there is no contact point with the web. In the second picture you see a little bit more of the trusses, again with the 2X4 that makes up the top plate of the wall (to the right completely).

@kwikfishron, I agree with you that the header should be solid or at least strengthened with plywood to make it load bearing.

@Yoyizit, the 2X4X5' that makes the bottom part of the header is sagging a little (no more than a 1/4"), with the lowest point in the middle. I think this is due to the fact that trimers are found below that 2X4 on each side but not above it. It is definitely not designed to support a load. I still have to do the level line exercise to see if the whole ceiling is sagging.


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## kwikfishron (Mar 11, 2010)

If that walls in your way, get rid of it.


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## Big Bob (Jul 27, 2007)

kwikfishron said:


> if that walls in your way, get rid of it.


have at it ! Totally agree...


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## dgsquare (Apr 22, 2010)

Goin' down!

Thanks,

dgsquare


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