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Problem -

enter image description here

I have interpreted "by hinging them together at B" as following -

enter image description here

Now if this is the case, then applying a load on the right beam will have no effect, in my understanding, on left beam. Right beam will press against the left support B and in return the support will apply a reaction force on the right beam.

The left beam is simply resting on the pin so there shouldn't be any "stressing" in the left beam and hence the BM at A should be zero.

enter image description here

As in the diagram above, the beam BC applies reactions P/4 and P/4 (a total of P/2) on the pin. However these reactions will be balanced by surface reactions, so the beam AC as such should experience no force.

But turns out the answer is $PL/2$

What's going wrong?

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4 Answers 4

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The hinged region at B is not a support, actually. Thats just a connection (or to be more precise, a pinned connection). A pinned support is something different; a pinned support is something what you see on the right end C. The support itself is connected to the ground (for example), and cannot translate in any of the directions (which means that the right end of beam at C cannot translate in any direction at all). However, the hinged connection at B can translate anywhere it wants (because it is not connected to the ground, so the ground is not providing any reaction force in the vertical or horizontal direction over there at all). The hinge at B means that both the beams have to translate together in vertical and horizontal axis, but can rotate freely there (with respect to each other), so it means that they will transfer the forces but no bending moment.

When you apply a load P at the direction shown in the first figure, the right beam is trying to move down, and hence hinge at B is supposed to move down (since there is no attachment to ground at that location, which will resist it from moving down). When B moves down, it is asking the left beam also to move down (because that hinge at B includes the right end of left beam pinnedly connected to the left end of right beam). This means that both supports, i.e. A and C wants to move down but cannot because of the supports applied at them (since those are connected to the grounds, which basically provides a reaction force by resisting their down motion). This is the reason why you will see stresses in both the beams, and not just the right beam.

So, now lets see the force reactions at each. Isolate the right beam, do statics and you will find P/2 as the force on the point C and on the left end of right beam (which is hinged with the left beam). This force is transferred to the other left beam, and at the reaction A, this shear force should be resisted.

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  • $\begingroup$ That makes it very clear, thank you. $\endgroup$
    – Harshit Rajput
    Commented Jan 20, 2022 at 16:21
  • $\begingroup$ Hi Rameez, I thought of this question again today, and I have something to ask, if you could help. What if the load P was applied on the beam AB(the cantilever)? I think the cantilever beam would've bent, while the right beam (BC) would have simply rotated, developing no BM in it (i.e. no BM or any stresses in BC, in the case if load P was applied to AB) $\endgroup$
    – Harshit Rajput
    Commented Mar 12, 2022 at 11:31
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    $\begingroup$ @HarshitRajput, imagine there is no hinge connection at right end of left beam. The right end is free to move, so it won't develop any force reactions there since there is no support or connection at all. But as soon as it tries to move down (because it will since you are applying a down force on the left beam), it will make then try to make the right beam move down with it as well. But the right beam cannot move down, since it is pinned at its right end. So it means that both beams should experience forces at their supports. $\endgroup$
    – Rameez Ul Haq
    Commented Mar 12, 2022 at 19:14
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    $\begingroup$ And yes you are right that the right beam will only rotate without any reaction moment, but that is also gonna be in the case of the original problem. $\endgroup$
    – Rameez Ul Haq
    Commented Mar 12, 2022 at 19:19
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Like you said at point B there will be a reaction force from the left.

However, if you look that system in isolation, that reaction needs to be supported at the end. And that force at B creates a reaction moment on B.

Since the reaction at B will be equal to P/2 (just take the FBD of the right part and calculate the reactions), then the bending moment at the fixed support would be $\frac{P}{2} \cdot L$, like the answer is telling you.

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  • $\begingroup$ I have made an edit to the question. Please have a look Nmech. $\endgroup$
    – Harshit Rajput
    Commented Jan 20, 2022 at 14:11
  • $\begingroup$ Ohh wait, is it like there is no surface to offer those P/4 and P/4 reaction, so ultimately the cantilever beam has to bear it. $\endgroup$
    – Harshit Rajput
    Commented Jan 20, 2022 at 14:27
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    $\begingroup$ @HarshitRajput yes, those P/4 and P/4 on the pin will need to be provided by the beam (the beam supports the pin - if that makes better sense). So that is producing the P/2 force. $\endgroup$
    – NMech
    Commented Jan 20, 2022 at 15:18
  • $\begingroup$ @HarshitRajput did you delete the question on the Biot number? I was planning to answer it over the weekend if I managed found some time. $\endgroup$
    – NMech
    Commented Jan 22, 2022 at 14:52
  • $\begingroup$ Oh yes I did. I actually found a different method to reach up to the concept of Biot Number, which made sense as to why Bi can be less than 1. Should I undelete it though? it would be nice to get your perspective $\endgroup$
    – Harshit Rajput
    Commented Jan 23, 2022 at 10:51
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One way to imagine the force on the beam AB is to substitute the BC with just its tributary load P/2, (the support it gets from the hinge).

Now we have only a cantilever beam AB with a concentrated load of $\ \frac{P}{2} $ applied at the left hand. So the reaction moment is as the book says:

$\ M=\frac{P}{2}*L=\frac{PL}{2}$

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An internal hinge is a special device used to link two beam segments. Similar to the typical pin support, it can generate reactions in the direction opposite to the load (applied force), with the condition that structural equilibrium must be maintained within the support, $\sum F_x = 0$ and $\sum F_y = 0$.

enter image description here

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Both the systems above are considered as "structurally determinate", however, it becomes "structurally indeterminate to the first degree if an inclined force is applied. I leave it to you to figure out why.

enter image description here

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  • $\begingroup$ A good question is gonna be what would be the reaction force in vertical direction at the fixed support when the pinned connection is replace by a bolted connection. Because if we isolate the right beam and find the reaction force at B, it would still be P/2 i.e. this is the force which is being transferred to the left beam. $\endgroup$
    – Rameez Ul Haq
    Commented Jan 21, 2022 at 8:24
  • $\begingroup$ @RameezUlHaq You can put the good question on your answer. I don't understand your question, so no comment. Sorry. $\endgroup$
    – r13
    Commented Jan 22, 2022 at 15:48

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