I am computing the deflection of a beam.

If the length of the beam is 10 m and I have load at the position 4 m from left with force 5 N.

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Now I add another load at the exact same position with force 4 N.

Can I just sum up the forces and write 9 N instead of 5 N?

It's a school exercise in which I'm told to plot the deflected beam and show all the loads. The only clever way to show the loads is to plot arrows at the positions and annotate the arrows with the load force. But if I have many loads at the same position, it will be quite messed up.


2 Answers 2


Yes, you can simply add the loads together. Imagine taking an actual beam, and stacking some weights on that point. Then add more weights on top of that. There's no way to tell the difference between 9 N, 5+4 N, 3+6 N, etc. Make sure you're reading the problem correctly though, because that sounds like an almost trivial case. Beam deflection problems will often have distributed loads, which are different. For some purposes (end reactions only) you can treat those as equivalent point loads, but for others (internal force/moment reactions) you have to consider the distribution itself.

  • $\begingroup$ Thank you! I hope I'm reading the problem correctly. I have computed the deflection at selected positions with multiple loads (with the superposition principle). I don't know how to double check if I have computed it correctly since I don't know nothing about what reasonable deflection values could be. Now I am asked to plot the beam with position on the x-axis and deflection on the y-axis and clearly show the loads on the plot. Do you think it is a good idea to plot the loads as described in my question? $\endgroup$
    – Jamgreen
    May 6, 2015 at 17:19
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    $\begingroup$ I think you're right in saying that showing the loads individually, instead of combined (where appropriate), would be messy and confusing. If you have a 5N and 4N load at the same point, I would plot that as 9N, unless otherwise directed. And in fact, it's a good idea to ask a question like this to your professor/TA, as they're the ones writing and grading the assignment. $\endgroup$ May 6, 2015 at 17:44
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    $\begingroup$ @TrevorArchibald Small quibble, but assuming a distributed load is a concentrated point load at its center will change the answer when calculating deflection, right? I think you meant that when calculating reactions you can consider the distributed load to act at a point. $\endgroup$
    – Ethan48
    May 6, 2015 at 19:59
  • $\begingroup$ I thought the deflection would be the same for either case, but it's been a while since I've done beam theory. I might have been confusing the fact that end support reactions won't change, but internal reactions will. $\endgroup$ May 6, 2015 at 20:22
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    $\begingroup$ If you imagine a scaffold plank with piles of cinder blocks all spread evenly over it, and them move them all to the center of the plank, it will certainly sag more when they are all at the center. This is because they create a larger bending moment, and deflection is proportional to the bending moment. $\endgroup$
    – Ethan48
    May 6, 2015 at 20:50

If the two loads are applied at the same point and in the same direction, they can be added together and represented as a single load in the same direction. I can't think if any exception to this case for constant loads.

Regarding distributed loads, you can represent them with a point load for the purposes of determining reaction forces, but not for calculating internal moment or shear.


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