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Old 03-06-2010, 08:11 AM   #1
 
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Strengthing floor joists


I have an old farm house that has 2 x 6 joists covering a 14' span. There is a section that has two of the joists cut out so heating duct could be run so naturally the floor is sagging. I was going to sister in a replacement in this section but my research shows this span needs 2 x 10 joists. I was told that I can sandwich in plywood and another 2 x 6 on each beam and this would be equal to putting in the 2 x 10's. Is this correct?
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Old 03-06-2010, 08:52 AM   #2
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In order to determine the required size of the beam, you need to list the C-C distance of the joists as well as the span, plus you need to know the code required floor loading. You can get that from your local code enforcement official. If there is no code, you are going to need to estimate the loading based on the use condition of the floor.

As for equivalence of a 2x10 to 2x6 sandwich beams:

A nominal 2x10 with true dimensions of 1.75 x 9.25 inches has a moment of inertia (I) of approximately 115 in^4. A 2x6 has a moment of inertia of 21 in^4, therefore you would need more than 5 2x6's to have the equivalent strength of a single 2x10. So in order to be equal to the 2x10 in strength (we are only talking strength here, not deflection), you would need a beam of size 9.5 inches wide by 5.25 inches deep.

It does not matter whether the beam is made of plywood or lumber grade wood for the computation of I. It is true that plywood typically has a higher allowable extreme fiber bending stress than lumber grade wood, however in a laminated type beam the lower allowable bending stress of the lumber in the beam would control.

So the short answer is that in theory you could build up an equivalent beam, in practice it is not going to work too well. If you have headroom issues, you may want to look into steel beams. Much higher strength than wood for an equal sized beam. due to the much greater allowable extreme fiber bending stress.
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Old 03-06-2010, 02:01 PM   #3
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Quote:
Originally Posted by Daniel Holzman View Post
In order to determine the required size of the beam, you need to list the C-C distance of the joists as well as the span, plus you need to know the code required floor loading. You can get that from your local code enforcement official. If there is no code, you are going to need to estimate the loading based on the use condition of the floor.

As for equivalence of a 2x10 to 2x6 sandwich beams:

A nominal 2x10 with true dimensions of 1.75 x 9.25 inches has a moment of inertia (I) of approximately 115 in^4. A 2x6 has a moment of inertia of 21 in^4, therefore you would need more than 5 2x6's to have the equivalent strength of a single 2x10. So in order to be equal to the 2x10 in strength (we are only talking strength here, not deflection), you would need a beam of size 9.5 inches wide by 5.25 inches deep.

It does not matter whether the beam is made of plywood or lumber grade wood for the computation of I. It is true that plywood typically has a higher allowable extreme fiber bending stress than lumber grade wood, however in a laminated type beam the lower allowable bending stress of the lumber in the beam would control.

So the short answer is that in theory you could build up an equivalent beam, in practice it is not going to work too well. If you have headroom issues, you may want to look into steel beams. Much higher strength than wood for an equal sized beam. due to the much greater allowable extreme fiber bending stress.
What would be the case if he's referring to old-growth true 2x6?
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Old 03-06-2010, 02:58 PM   #4
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Tb, I think he did---- "A nominal 2x10 with true dimensions of 1.75 x 9.25" Already multiplied by 1.47 to get the total.

Daniel, wouldn't 2- 2x6's. D/F #2, single use- 1250 fb, 1,700,000 E, gang nailed next to each cut one give him 750# / 14' span= 54# total minus 10# dead load, leave 44# live load per ft.? Which would work for him with a uniform distributed floor load and 1/2-" deflection?

Be safe, Gary
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Old 03-06-2010, 03:23 PM   #5
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My computations for a pair of nominal 2x6 joists nailed together adequately, assuming finished size 1.75 x 5.25 inches, E = 1.7 million, Fb = 1250 psi, and a total distributed load of 54 psf (44 live, 10 dead) yields as follows:

Factor safety bending = 0.95
Maximum deflection = 0.9 inches
d/L = 190

This is a bit on the light side, especially the d/L is less than 240, which I generally treat as a minimum for floors. The factor of safety less than one is also not a good idea, if I were designing and stamping the plans it could not go out this way, however it is likely that the code live load is only 40 psf, which would yield a factor of safety of 1.03. This all assumes 16 inches on center. Regardless, I would not design for such a low Fs, and the d/L is a non starter.

However, that was not the OPS question, so this should be considered a theoretical discussion at this point.
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