Simply supported beam question 1

[AlpineEngineer]

I have a very simple question:

I have a series of 28' long floor joists that have bearing at each end and an LVL beam for bearing at the midpoint (14'). Typically I would specify a 'flush' LVL beam; hence the joists would be cut at 14' and hung with simpson hangers off the LVL. In determining my loads on the LVL beam I assume each 14' floor joist is a simply supported beam and each end of the floor joist carries 1/2 the load (assuming uniform loading).

Now, lets say I decide to drop the LVL beam, not cut the floor joists, and just rest the floor joists on top of the LVL. Now, I have an indeterminate beam situation with my floor joists. My Statics book says for this uniformly loaded indeterminate beam that my middle reaction is 1.25*w*L where w is my load in lbs/ft and L is my length. Thus, the middle reaction is greater and upsizes my LVL. I am trying to make sure I have this correct and intuitively I am not sure that I understand why the indeterminate situation has more loading at the middle reaction.

Thanks so much,

 

[swearingen]

To help your intuition, cut the beam in half with one end fixed and one end pinned. As long as you have a uniform load, you can assume your total beam is two of these with the fixed ends connected to each other.

Considering just the fixed/pinned half beam, which end will see more load in a uniform load situation? Intuition (and calculation) tell you it's the fixed end. This in turn means that the center of your full beam would have more load.

Doing this will increase the size of your LVL beam, but the moments in the joist go down, so you can downsize the joist. The question is which is more economical considering the expense of the different connections as well.

If you "heard" it on the internet, it's guilty until proven innocent.

 

[Lion06]

If you think of fixing the end of the joist at the lvl and pinning it at the other support, you will find that the reaction at the lvl will be (5/8)*w*L. If you do this on both sides (which is essentially what you are doing by making the beam continuous), you will (10/8)*w*L, which is 1.25*w*L. That being said, your LVL will never be stiff enough to actually see that load, but designing them for it is fine.
As far as the reactions on the LVL go (assuming the joists are continuous over it), I look at this a little differently than most. That reaction given above is assuming a rigid support that will NOT deflect. We know that your LVL will deflect. This means that you are being conservative by using this reaction for the LVL-and I would do that. However, when looking at the end reactions that are NOT continuous over a support, you should account for the fact the reaction at the LVL will be lower than given above and, as a result, the non-continuous end reactions will be higher. This will really only matter for required bearing distance, but I would check it.

 

[UcfSE]

It comes down to load following the stiffest path, which for the continuous joists is the middle support that is the LVL. You can always go back to mechanics and derive the equations for the reactions and let the math tell you the "why". That's probably the best way to go if you really want to know.

I agree with using the simply-supported reactions for the joist ends, as long as it doesn't kill your framing plans, which I doubt it will do.

 

[CJJS]

Or if you have a lot of time on your hands, you could figure out the stiffness of the LVL and model it as a spring. Then, your reactions will be more accurate. But, who has time for that?

 

[UcfSE]

Or model it in a 3D program...

 

[JAE]

...or use finite elements!

 

[csd72]

Just remember that if only one of your spans is loaded then it will behave closer to a simple span. You should therefore always design for the simply supported reaction as a minimum (wl/2).

 

[dik]

If you fit it in snug, you'll be safe with your 1.25 increase in loading... you can jack the system to act like a 28' clear span with a point load in the middle that offsets the total weight... and you don't really want that.

Dik

 

[miecz]

The problem, as I see it, is determiing the shear at the end of the joists. If the LVL were completely rigid, the end shear would be no higher than .281wL, where L is 28ft.
If the LVL were flimsy, the end shear would be .5wL, again L=28ft. The end shear lies somewhere in between those two values. Unless one performs a stiffness analysis, one must use the conservative solution, or else guess.

 

[DonPhillips]

The continuous condition transfers some of the moments from the mid span of the joists to the joists at the LVL and then the shear at the ends of the joists are reduced. You have to look at the shear diagram to see it and is not intuitive unless you have created some shear diagrams.

Don Phillips

http://worthingtonengineering.com

 

[MWPC]

Alpine,

Leave the joists continuous if you can, it will help the stiffness of the floor.

Swearingen,

Assuming fully loaded, the continuous joist moment over the support will be equal to the simple span midpoint moment.

Ken