L7 Arches and Cables contd

Añadido: 2026-05-31

[00:00:00]okay so in that last video you all saw a demonstration of the behavior of arches and chains and the relationship between those two systems now I want to show you all my own demonstration just to prove to you that nothing in that video was made up and also I want to ask you further questions about the distinction between arches and chains this is an advanced structures course so hopefully we'll be able to answer more advanced questions about these systems so here you can see I have a bunch of elements that I've linked together with pinned joints that is when I apply twisting motion like this the pin offers no restraint it is able to rotate freely about the joint and so if I hold it up like this we have a chain and if I flip it around like this and place its ends into this cardboard here so I'm applying some boundary conditions and these would be pinned conditions because I'm giving a vertical reaction Against Gravity I'm also providing a horizontal reaction against the thrust of this Arc but there is again no restraint against rotation at this joint and so it is it pinned m so now if I align this Arch properly you can see that it's able to hold up its own self weight however if I take an external load and apply it at one of these joints you can see you get a very very quickly collapses conversely if I go back to the chain configuration take it out and hold it up in my hands and now I have to do this in steps I'm running out of hands if I place my weight here and move my hands roughly as far apart as the boundary conditions in the cardboard you can see that the chain was able to reconfigure itself and hold up that weight without collapsing and if I put that weight down you can see that the chain has now reconfigured itself back to what we saw earlier see if I can hook the weight there we go so it's able to reconfigure itself and now if I mirror this configuration back to our Arch condition foreign my end conditions make sure that it's stable at the ends and now if I mirror that configuration we saw earlier so this end was raised up and now if I reapply the load really make sure that our ends are stable now you can see that the arch stands so now my question to all of you why is the arch not able to reconfigure itself from its natural position when we apply an external load but the chain is able to reconfigure itself I'm going to give you guys the answer in just a few moments but I want you to try to think about it yourself you have all the information that you need to answer this question and I'll give you a hint it has something to do with internal forces and what we learned in week one about trigonometry and what we learned in week two about static equilibrium so if it helps maybe draw yourself a diagram of each condition with an external load applied at a joint and then draw the internal forces and the force components in each of these different members and see if you can come up with an answer to why the tension system is able to realign itself to an external load but the compression system the arch does not so let's try to answer this question of why the tension system the cable is able to realign itself to an externally applied load but the arch is not and let's start with some free body diagrams here I've drawn the arch first so let's start there so our simplified Arch here starts off in a balanced condition note that this is not a true Arch we've actually made some simplifications for this demonstration a true Arch would have full curvature along its length from one end to the other with no bending in any of the elements here we have this straight horizontal piece and so there will be some bending in this piece due to its own self-weight but we're not concerned about this middle piece for this exercise we're really only going to concentrate on the two end pieces we're also going to neglect self-weight for this discussion as the self-weight of the structure is insignificant and small compared to our externally applied load so let's look at the internal forces in these two pieces these end pieces when we apply an external load so if we apply a load at the center of this Arch that load would be divided equally between the two end pieces and it would generate equal internal forces in the end pieces equal to half of the applied load so if this were load P each of these vertical reactions would be p over 2.

[00:07:32]looking at our sum of Y equilibrium equation we would see that we have equilibrium here but we also know that these pieces at either end are pinned at both ends there's a pin here and pins down here and so they must start and end with zero bending moment and if they must start and end with zero moment and there are no external perpendicular forces applied to them we know that the moment throughout the member will be zero in other words there will be no bending forces in this member so the forces through this member must be axial in other words the internal forces must be aligned with the member this vertical component we've just drawn is not aligned with the axis of the member so the total force in this member must look like this which means that we will have some horizontal component to the internal force in this member we call this horizontal component thrust and that's why we need pins at either end down at the bottom here to resist that thrust that's pushing out so these end conditions must provide restraint in the opposite directions similar to how they provide restraint against the vertical component and since we applied the load at the center of the arch these internal forces both the Y component and the X component are equal and opposite on both sides so the thrust on the left side is equal to the thrust on the right side and once again if we looked at R sum of moment in the X Direction we called this one FX this would also be f x and so we would have f x minus f x equals zero but now let's try something different now let's try taking this load and moving it closer to one of our sides rather than placing it in the middle now if we consider our internal forces things change significantly as we've seen in our free body diagram of beams more of this externally applied load is going to end up in the closer support so this reaction here we'll call it fy1 will be larger than this reaction on this side FY 2.

[00:10:59]and so we have a larger vertical component in this member on the right a larger vertical component of course also means a larger horizontal component or thrust as well compare that to the other side which must have a smaller horizontal thrust so this will be f x one this will be f x 2 and FX1 is greater than FX2 so now looking at our equilibrium equations we can see that for the sum of forces in the Y we are actually still okay fy1 and fy2 will end up being equal to p however if we look at the horizontal Direction now we have this FX1 to the left plus FX2 to the right and these two are not equal to each other and so adding them together cannot give us zero so our structure therefore is not in equilibrium in other words it is moving so now let's ask ourselves how would this structure need to move in order to become stable to reach static equilibrium while there are two options either this joint on the right here moves up and conversely the one on the left moves down or vice versa where this joint on the Right Moves down and the other moves up if this joint here on the right closer to the applied load were to move up than this member on the right would become more vertical and therefore it's horizontal component would decrease on the left side this joint would move down and this member would become more horizontal and so its horizontal component would grow and eventually we would come to a point where FX2 is equal to FX1 and we would reach equilibrium so that actually sounds like exactly what we need we needed those horizontal components to balance out and if these joints realign this way eventually that's exactly what would happen they would balance out but there's one small problem with that in order for this joint to move up on the right here this heavy weight the supplied load would also need to accelerate Up Against Gravity and how often do you see weights move upward on their own well hopefully never because that's not how gravity works so the only other option is for this weight to come down until it reaches the ground and the system becomes stable again and you can see that's exactly what happened in our demonstration now compare that to the tension system everything in this system is the same so if I apply this load here we're going to have some large vertical reaction here and a smaller vertical reaction here a large horizontal reaction and a small horizontal reaction but there is one key difference now for this system to become stable this joint needs to move down if this joint moves down this right piece becomes more vertical our horizontal thrust decreases on the other side this joint moves up and our horizontal component increases and eventually we will get minus f x 1 plus FX2 equals zero we will achieve static equilibrium and we achieved that with the joint moving down and so our weight also moved down it moved with gravity so in order for the tension system to become stable the weight moves in the same direction as gravity moving in the same direction as gravity is a much easier task for a weight than moving Against Gravity and so our tension system realigns itself and becomes stable our Arch however cannot realign itself and so it collapses