TAB_HOO said:
the way you have it drawn the moments don't sum to zero
I suppose fundamentally you are correct, we are assuming an analysis for shear only. What's the magnitude of the torsional load onto such a member, and is the local pull-out a bigger concern than global torsional stability of the built-up member? Generally speaking, the torsional load is limited.
If we assume a worst case scenario, where joists are hangered into the side of one of the plies, we induce an eccentric load, but the joists that are hangered also provide torsional restraint. The built-up beam should also be braced for LTB checks by the designer.
TAB_HOO said:
Are the P's that are directed upward in your drawing internal shear forces?
The 'P' at the top is the force we are calculating for, the 'P' at the bottom is the reaction that is assumed to be resolved through each member. For the 2-ply member of matching thickness, I assume each ply has the same EI, and therefore, half the load is resolved as a reaction in the left ply, and half the load is resolved in a reaction on the right ply.
For the second example where one ply is larger, I ratio the reaction based on the EI, since the right member is twice as thick, it carries twice as much load.
The 'P' that is directed upward is intended to show the load that stays within the ply, and is not transferred through the fasteners to the adjacent ply.
TAB_HOO said:
In order for the multi-ply beam to behave like a single beam, each ply has to have the same deflection.
Correct. The load is transferred as shear through the fasteners. The fasteners are only responsible to transfer the load that is to be carried by the unloaded plies. The fasteners do not need to transfer all of the load experienced by the outer ply, since the outer ply will support some of the loading with or without the fasteners.