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I saw this analysis online but I have noticed contradicting values from this method of analysis. The drawing of the simply supported beam comes from the crane structure above. All we desire is the reaction forces as well as the external forces being experienced at each of the three points which represent the joints of part of the truss. The person has placed $150N$ acting in the middle of the beam creating a moment of $22500 Nmm$ and creating reactions of $75N$ at each support.

But I was told on one of my previous questions that this is innorect due to it being part of a truss and not just a beam. Is the correct way to split up the $150N$ into $50N$ acting upon points $P_1$ $P_2$ and $P_3$ which would result in a total moment of $30000 Nmm$ and reactions of $100N$ on the right end and $50N$ on the left end? Which method is correct in this situation?

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  • $\begingroup$ is the concentrated load of 150 N the result of a distributed load over 300 mm? $\endgroup$
    – NMech
    Dec 16, 2021 at 11:26
  • $\begingroup$ Yes it is the result of a distributed load over 300 mm $\endgroup$ Dec 16, 2021 at 11:28

3 Answers 3

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(I don't know which question you are referring to but) IMHO -if you take the usual assumption (i.e. non deformable structure)- the correct way would be the following :

enter image description here


Note that in terms of moment this will have the same effect, however, the loads on the truss's rods will be different than if you used a different scheme.

So, for the loads on the support either the concentrated load or the distributed load or the scheme I am proposing should have no difference.


The whole concept behind this, is the resultant of parallel forces (well actually the opposite). I.e.: if you have two or more parallel forces then which is the resultant force that can substitute those forces.

Essentially for each segment you have:

Original enter image description here
Splitting UDL to each side enter image description here
Concentrated Load enter image description here

So if you put all segment next to each other

enter image description here

which results to the first image above

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  • $\begingroup$ Thanks for the explanation. Is there a way to justify/ prove that for the given scenario that this is the correct way to convert the UDL to points loads. If this were just a beam with no truss attached to it we would use the $150N$ acting in the middle for our calculations wouldn't we? So why is it different when the beam is part of a truss? $\endgroup$ Dec 16, 2021 at 11:42
  • $\begingroup$ I've updated the answer. However, I am only answering on how to break the load to 25, 50,50 and 25. I have not yet understood what you mean by "If this were just a beam with no truss attached to it we would use the 150N acting in the middle for our calculations wouldn't we?". Beams and the rods of a truss may seem similar but they behave (most of the time) in different ways. The essence is that rods transfer only axial load, while beams transfer axial, shear and bending forces. However, unless there is a concrete example I can't really help you. $\endgroup$
    – NMech
    Dec 16, 2021 at 14:52
  • $\begingroup$ Well I saw this video youtube.com/watch?v=VIva-QRumpU involving a truss with a UDL. But in this video the person has placed $1000N$ in the middle of the UDL span. According to your diagrams should it have been $500N$ at each end of the member? If that were the case then at point B, the reaction would be $500N$ acting up but there would still be $500N$ acting down but then for equilibrium $F_2$ would need to have no vertical component? $\endgroup$ Dec 17, 2021 at 3:13
  • $\begingroup$ Also I'm not really understanding how you derived that the forces are 25, 50, 50 and 25. By looking at each segment there is a total of $50N$ acting over the segment or $25N$ per $50$ mm as you have split the UDL into two sections I am assuming from the diagram. But how does this result in $25N$ acting on each end of the segment? $\endgroup$ Dec 17, 2021 at 6:31
  • $\begingroup$ On a truss rod you can only place forces on the edges of the rod (aka nodes). I will try to explain it intuitively (without using equations of equilibrium). Since you have a total load of 50 N, and you need to split it at two, then 25 goes to one end, and 25 on the other. $\endgroup$
    – NMech
    Dec 17, 2021 at 7:27
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I agree with the answers given with the exception that the moment of the concentrated load of 150N should not be applied at a distance of 150mm. it should be broken into two loads of 75N and each applied respectively at 100mm and 200mm.

As far as the stress in members we also should work the horizontal component of the cable tension as an axial force to all three horizontal members.

So the members are under combined moment and normal stresses.

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For reactions' concern, the UDL, the point loads, and the concentrated load (Pe) are all equivalent.

Let's check the forces at the joint "K":

  • Fk = 50 N/100 mm * 300 mm = 150 N (using UDL)

  • Fk = 225 N + 425 N = 150 N (using Joint loads)

  • Fk = 150 N (Concentrated load)

  • Mk = 50 N/100 mm * 300 mm * 150 mm = 22500 N-mm (using UDL)

  • Mk = 25*(0) + 50100 + 50200 +25*300 = 22500 N-mm (using Joint loads)

  • Mk = 150*150 = 22500 N-m (using Concentrated load)

Conclusion - the equivalent loads will produce identical forces/reactions at any point of the truss. However, as mentioned previously, the internal forces on the truss members will be different. Now for the beam, note the differences in bending and shear forces on the diagrams below.

enter image description here

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