Girder Design Question 1

[LoftyAce]

It’s been a while since I last worked on wood design. I’m working on a girder/beam design. The girder will be three (3) 2X10 placed right next to each other. When calculating Fb’, since I have three 2X10s do I calculate Fb’ for one 2X10 and then multiply by 3 to account for all three 2X10?

I attached a photo that shows an example of what the three 2X10s look like.

I appreciate any help.

 

[271828]

I wouldn't think about it that way. The applied stress, fb, is 3x lower than it would be if you had one 2x10. The allowable stress is the same as if you had one 2x10.

In other words:

1. Compute fb = M / S, where S is 3x the S for one 2x10.

2. Compute Fb' for one 2x10.

3. The triple 2x10 is ok for bending if fb <= Fb'.

 

[Eng16080]

I mostly agree with the comments above. I would stress (no pun intended) that the value Fb' is for the entire beam (all 3 plies). The adjustment fact, Cr, in particular will give a slight increase for a repetitive member, which a 3-ply beam is considered to be. Had you only calculated Fb' for one 2x10, you would be missing this increase. Also, the beam stability factor, CL could potentially differ between a single 2x10 and a 3-ply 2x10, although in your case it's probably 1.0 regardless assuming the compression side is braced (which I'm guessing it is).

 

[LoftyAce]

I really appreciate both of your responses. I understand now. Essentially your fb is reduced because S would be larger when accounting for 3 2X10s in the following equation:

fb = M/S

Thus, fb is the stress that the beam or all three 2x10s are experiencing at the same time. I’m rambling but explaining it helps me internalize what I’ve learned.

Thank you!

 

[phamENG]

Make sure the plies of your beam are fastened together appropriately. Just because they're next to each other doesn't mean they'll share the load. Inadequate fastening could result in higher stress in the loaded ply (side loaded, eccentric - real or incidental - load from above, or minor stiffness variations between the plies due to grain variations, knots, etc. that cause the true E to be different than the published minimums).

 

[lexpatrie]

Lofty - Inline photos are highly preferred for the future - use the camera icon and you can attach a jpeg directly.

For your use, which appears to be indoor, the three plies can share load equally, provided there's a valid way for the load to get into all three equally (some of the LVL manufacturers list connections for multi-ply LVLs that can be used as a guide.) It's possible that your element above adequately distributes the load to the (3) 2x10s (like if it's a truss with a mid-span bearing it was designed for), or a wall that's roughly a thickness match for the 4.5" wide (3) 2x10s.

The rest of youse, you can use subscripts you know? It's a pain on a phone, I'll admit, but it helps readability down the line.

To your actual question, Lofty, it's six of one, half a dozen of the other, you can look at the total load on the (3) 2x10s, or 1/3 the load on one 2x10. [Again, providing the load distributes evenly to all three in a rational fashion], you can try ForteWeb to design dimensional lumber, if you need something to check your work, they don't present results quite the way you're likely to run your calculations, but if it works on your calculation AND theirs, then that's a positive sign, if not, you can work backwards from their allowable moment to determine how they arrived at the allowable stress (F[sub]b[/sub]') and F[sub]v[/sub]'. ForteWeb presents element level capacities (M[sub]allowable[/sub], V[sub]allowable[/sub]), not stress limits. But you can eventually figure out what C[sub]L[/sub] they used.

Although technically, I think they figure out a bracing spacing requirement that will make it work rather than straight up calculating a C[sub]L[/sub], they seem to iterate on this to figure out what the bracing space limit is for the loads given. C[sub]L[/sub] would somewhat typically be 1.0 for an interior girder or beam or joist with subfloor above it.

If you really want, there are some wood design calculations in college level architecture courses, those can be helpful to refresh your knowledge.