As shown above, two identical rectangular beams are placed and NO pinned together. I know each beam has its own neutral surface but why they carry half of the bending moments? I thought the bending moments they carry are same because each of the beam is subjected same force P at the end.
Thanks
Let's consider the two beams as two springs which will deflect under the load $P$ the amount:
$$ D = \frac {P*L^3}{3EI}$$
So K or stiffness of these two springs is:
$$ k = \frac{P}{d} = \frac{3EI}{L^3} $$
These two springs are working in parallel, each sharing half of the force, therefore half of the momentum (if they were to act in series they should have been attached lengthwise).
If they were pinned together the combined composite $I$ of the new beam would have been $8I$ because the height of the beam would have doubled and we know $I$ is related to $H^3$.
Or consider initially the top beam is taking all the load and deflects fully under the load, then we add the second beam right next to it and the new additional support decreases the deflection by half. Now if we slide the second beam under the first, it is the exact same situation, each beam shares half of the force. This is similar to old car leaf springs.
That they share half the force or half the momentum isn't immediately obvious; there is no physical law saying so.
But it is through geometric constraints (under the assumption of zero vertical compression of the beams!) that we know they must have the same deflection and thus stress-state.
But imagine having a big stack of short stubby rubbery beams (say pencil erasers). If you make the stack big enough the lowest beam won't feel anything. And if your two beams are of any realistic material (not Eulerium) the first beam will carry ever slightly so much more load than the second.
