Interior wood stud bearing wall - typical tract home

[AELLC]

This is a very typical situation - we have 3-point bearing roof trusses, the spans being 33.2' and 5.5'. The interior bearing wall is only 2x4 studs. The truss calculations say there is a huge reaction at the interior bearing wall, because of the large difference of the 2 spans. It is at least 2.18 greater than if the trusses were broken at the interior bearing wall, and each truss was simple supported instead of 3-pt brg (see attached).

Does the truss design software assume that the supporting G.T. 4 is a hinge condition with no capability of up or down deflection? In reality, I would imagine GT4 is lifted (deflected) upwards and as a result, the reaction on the interior bearing wall is much less.

 

[BAretired]

Some software may consider lifting or settlement of the girder truss, but from experience, I know that some does not. In this case, three point support is not a good option.

Use a simple truss 33.2' long and stick frame the 5' span.

BA

 

[AELLC]

That would be easy, but this project is a Code Update for houses that have been built for at least a decade. Another Engineer had stamped the previous editions of this.

It is a lot easier to deal with custom houses, but when dealing with cookie-cutter tract homes, and this is a very typical plan, we can't re-invent the wheel for fear of being fired by the Builder. The trusses material and labor cost are much cheaper when they are 3-pt brg, and there is less worry of ceiling board cracking, etc.

But wait - it gets even worse - I rechecked my numbers, and it is more like 2.3 times instead of 2.18.

 

[AELLC]

Continuing along these lines, it is common in both custom and tract homes to have a 3-pt brg situation with equal spans.

Now when I do my calcs, I assume the interior reaction is 1.25 x trib width.

However, when the truss calcs come in a la truss waiver (deferred until bldg. dept submittal), they show a much higher reaction. (see attached).

Now I am assuming this is due to the truss overall moment of inertia is much higher at the central portion than near the ends, due to pitch of truss.

 

[dhengr]

AELLC:
BA’s right on the money, his is the way I’d do it. But, I can also understand the less expensive idea that you crane the entire truss into place in one shot. The truss design doc’s. and calcs. should show that you actually have two primary supports and a 5.5' canti. and uplift out at the tip of the canti. Sometimes they will cut the top chord at the middle support and fix it enough to provide lifting ability/strength, but it ends up acting as two spans because of this cut. If what they are doing isn’t causing ceiling cracking problems at the canti. tip, I’d accept it and move on. Look at the forces/stresses and details in the top chord right over the middle support, and the calcs. should show some upward deflection out at the canti. tip.

 

[boo1]

Why have the load interior bearing wall? Looks like the truss configuration is the same the great room truss. The cost savings in the truss are less the the interior grade beam and interior header.

 

[AELLC]

I don't think they are cutting the top chord. There are no problems with how these are being built. This is in the Phoenix area, no snow, and that 20 psf roof live load is mostly "imaginary" - construction live.

When the original model was built years ago, supposedly there was a sagging ceiling deflection problem because the wall was not designated as bearing by the original EOR, so the Framer made the wall bearing and the truss mfr probably field-modded the trusses.

The only problem I have now is a Plans Checker who is looking at all the calcs now and demanding to know why the interior brg wall isn't at least 2x4 at 12" o.c.

It is not easy to convince a tract home Builder to build the stud wall at 12 oc when it has ben built at 16 oc for at least a decade.

 

[AELLC]

boo1-

See my previous post. There actually was a deflection problem when the model was first built, with that wall being non-bearing. I imagine the deflection of GT4 added to the deflection at the midspan of the common trusses 24" oc to cause a problem.

The 2-2x8 header doesn't cost much, and there is zero cost increase to the foundation because it is all California-style flat post-tensioned slab.

 

[mijowe]

Are they addressing the uplift in the girder truss?

 

[AELLC]

mijowe,

The truss Mfr always adds a note to the effect the EOR (being me) needs to design truss-to-wall or post connx with appropriate hardware to resist uplift.

I posted in an another unrelated thread that this load is frequently so huge it is impossible to find a Simpson strap that even works. I suspect there must be large difference between reality and what the truss software churns out.

 

[mijowe]

I was wondering if the girder truss was designed with uplift, or did the assume some other loading?

I ran the above spans as a two span continuous beam, and did in fact get a reaction that was about 2.44 x what it would be as a trib width. The uplift at the cantilever end was about 45% of the downward reaction.

 

[mijowe]

If you don't connect the end of the 5' span, treat it as a cantilever, the reaction becomes reasonable.

 

[AELLC]

mijowe,

To me, it appears they run a wind uplift case during construction (no concrete roof tiles), and a regular calculation with full gravity loads.


The only way a 2x4 stud wall at 16 oc even works is if I assume the load = 1.86 x trib, and I don't even have a problem with that.

 

[AELLC]

mijowe -

The common truss ends HAVE to be connected to the GT4.

 

[CANPRO]

AELLC, I just want to clarify what you mean when you say the reaction is at least 2.18 times the trib width. Do you mean the load is approximately the load on the single truss (lbs/ft) by a trib width of 2.18 x 1/2(33.2'+5.5') = 42.18' ?

I checked a multi-span beam in a simple spreadsheet. With two spans of 33.2' and 5.5' and a unit load of 1 lb/ft, I get an interior reaction of 44.5 lbs which would agree with your approximation of 2.18. I think they're analysis is correct.

 

[CANPRO]

mijowe, I just noticed after posting that you had already run the simple beam check...should have read through a bit closer.

 

[AELLC]

mijowe and CANEIT,

The multiplier is exactly 2.3, not 2.44.

2.44 is for 33.2' and 5', while we have 33.2' and 5.5'.


Therefore, the reaction is 2.3 x 1/2(33.2'+5.5') = 44.5, which is exactly CANEIT's calc.