Shear flow question

[BadgerPE]

Purely theoretical question here. Imagine a situation where an existing (2) ply 2x10 beam is overstressed under new loading conditions. The decision is made that the "simple" fix is to sister an additional 2x10 ply alongside the existing beam. For sake of discussion lets say that the additional ply does not carry load through bearing of the joists above so all load transfer to the new ply needs to be through mechanical fasteners.

This is where I get hung up calculating how to transfer the load to the new ply. Because the new 2x10 ply is the same height as the existing plies, the distance from the neutral axis of the new ply to the centroid of the overall section is 0. Does this mean there is no shear flow between the new and existing members? If the new ply needs to take 100 plf of load, do the fasteners only need to be designed to take that load?

I attached a sketch of the situation I was envisioning. Not too creative though.

Again, this is not a real world situation, but I am a little confused as to if there is actually shear flow present. I am leaning toward no, because the stress profile is the same across the cross section.

 

[rb1957]

if the fasteners weren't there, how would the sister panel deflect ?

so the FBD has 300 lbs/ft applied at the top of the existing beam, 200 lbs/ft being reacted by the existing beam, and 100 lbs/ft being sheared into the sister.

no?

another day in paradise, or is paradise one day closer ?

 

[JAE]

This is not really a "shear flow" issue using VQ/I. This is more of a direct load transfer (vertically) as rb1957 suggests.

So just design your bolts to transfer 100 lb/ft in direct vertical shear along the length.



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[KootK]

Yeah, I'd say that you pretty much nailed it in your second paragraph Badger. One thing to keep in mind is that some load would be locked into the original beam by virtue of it's deflected state, unless you stress relieved it. The new ply would only share in the incremental load applied after that new ply is installed. Of course, that's more a concern for a real world problem.

If you needed torsional or lateral torsional buckling resistance, that would require some nominal horizontal shear transfer capacity in the fasteners to make the plies behave compositely. That's just theoretical fun though. Nobody who derives their living from pay checks would got to that much trouble.

The greatest trick that bond stress ever pulled was convincing the world it didn't exist.

 

[BadgerPE]

Thanks for the sanity check all!

 

[rb1957]

careful ... you don't want to go to an asylum for a sanity check ...

another day in paradise, or is paradise one day closer ?

 

[BAretired]

I agree that this is not a shear flow issue, but it would be wrong to say there is no shear flow present. Shear flow exists in the two ply beam before adding a third ply.

There is a slight torsional problem because load is applied eccentrically to the shifted centroid. Torsion could be eliminated by supporting the original beam, attaching the new ply and blocking above it, thus ensuring that the load is applied at the centroid of the three ply beam.

Shear flow will still be present in the new three ply beam, but it flows vertically, not across the connectors.

BA

 

[KootK]

The effective eccentricity of load is an interesting point. Since there is presumably no torsion resisting mechanism at the ends of the beam, the resultant of the bearing reactions would have to be coplanar with the centroid of the applied load. That could limit the vertical load that can be effectively transferred to the new ply.

Here's another quirk that I'm not at all sure about. The primary job of the fasteners is to force the curvature of the new ply to match the curvature of the existing plies. There is more than one option for making that happen:

1) Transmit vertical shear to the new ply, through the fasteners, so that the vertical deflection (curvature) of the new ply is compliant.

2) Transmit horizontal shear to the new ply, through the fasteners, so that the vertical deflection (curvature) is compliant.

3) Some combination of #1 and #2.

For reasons of stiffness and fastener slip, I'm sure that mechanism #1 predominates. It would, however, be theoretically be possible to get the job done without transmitting any vertical force to the new ply. I could use another dozen classes in mechanics of materials I'm afraid.

The greatest trick that bond stress ever pulled was convincing the world it didn't exist.

 

[dhengr]

BadgerPE:
The real world trick here is that you have to (unload) jack-up the existing beam before you apply the third 2x, or the three 2x’s will not share the load equally. You imply that the existing beam is pretty fully stressed at 200lbs./ft. and you want to add another 100lbs./ft. of loading. If you do this without shoring the existing, the two existing 2x’s (already fully stressed) will deflect some more (strain some more) and be over stressed when the load is added. Over jack the existing a bit (camber it, your or my engineering judgement here), to account for the fact that there will be some movement btwn. the new fasteners and the new and old 2x’s, to really bring the fasteners into play and transfer load, before things really settle in. The thought is that you want the three members to all deflect the same (strain the same), then we can assume they are sharing the load equally. Take a look at a good wood design text book and/or the NDS for how dowel type fasteners work and their failure modes: bolts in holes allow a fair amount of movement before the bolt crushes some wood and is bearing; common nails bend and crush some wood as they cantilever to transfer the load; the newer structural screw may be best here because they also draw the three members into intimate contact. Otherwise, I agree with the above, you must transfer 100lbs./ft., as a shear loading to the new member. This same stress/strain thinking would apply if you were trying to reinforce an existing steel or concrete beam.

 

[jfmann]

I have wrestled with this classic......and very practical...... joist-reinforcement problem (after first trying VQ / I and becoming befuddled)........the answer to which ends up being essentially the same answer to the following question (which should arise at some point during many years dealing with wood design issues);

When does a ledger board become part of the beam?

Various answers proposed in this thread are not quite adequate and some are misleading.

Key requirement (which is identified in some posts) is that......as for any beam-element to resist bending moment....... the reinforcement member (RM) must deform in bending if it is to resist any load. Connection of the RM to main beam (MB) must therefore be strong enough (and configured properly) to make such deformation occur.

First, consider RM)......which is only partial-length....... to be connected to MB only at each end of the RM with a simple dowel-type connector. When MB deflects in response to load, the RM will not bend........and therefore will not resist any load. At that point, the RM is merely a ledger board, hanging off the MB (main beam).

If we now add one additional connector.....at mid-length of RM........a simple elevation diagram should make clear that the RM must now deform (as a beam-element) along with MB......even though RM will not deform in exactly the same curvature of MB. Free-body diagram can now be drawn for each member........with vertical forces (from each of the 3 connectors) acting on each member (of course in opposite directions).

Try some calculations and you will see that, in general, as long as the RM is not too short, 2 longitudinal rows of connectors (nails for wood) usually is adequate to ensure that the RM acts compositely with the MB.........such that you can apportion load in according to stiffness as usual for built-up assemblies. However, unusual conditions require careful analysis.

See post (by another) from way back.......thread507-168270........that note the key requirement of deformation compatibility for this problem.




John F Mann, PE

http://www.structural101.com

 

[Brad805]

Presumably you have sufficient bearing. Most of the cases I find with shear problems have insufficient bearing.