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I have the above situation with a Warren truss and a UDL acting along the bottom of the truss. The UDL is $60 kN$ per $3.75$ metres. The distance between the joints along the bottom is $3.75$ m. I am looking to determine the forces in each member of the truss. In my engineering class I have been told that an easier way of dealing with these UDLs is to find the total force acting on the truss and then make that force a point load acting directly in the middle of the truss. But does this method have repercussions or make assumptions? I would imagine that having 4 forces at each joint along the bottom would give a different result? Or do both methods give the same results for the forces all the members?

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Unless specifically required, truss members are usually assumed to be capable of carrying axial load only. You should replace the UDL with the equivalent concentrated load applied at each panel point (joint).

However, for heavy-duty truss, sometimes the chord member is made continuous that is to be designed as a beam for the resulting bending moment and shear. It is rare though.

If considered as a truss (discrete chord members), the two systems below will yield different member forces, though the support reactions are the same.

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Yes, you are right.

We can not replace the four forces with only one equal to the sum of four, at the center.

The more joints a truss has with the same overall span and same UDL the less max moment or max stresses will be.

Let's compare it to a simply supported beam with a length L and UDL of $P/L$.

If we handle this beam by applying UDL along the length, L, we get:

$P_{total}= \frac{P}{L}L=P$

$M= \frac{\frac{P}{L}L^2}{8} =\frac{PL}{8}$

But if we apply the sum of the UDL at the middle of the beam we get:

$M=\frac{PL}{4}$

So by applying a concentrated equal load we get double the correct M.

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  • $\begingroup$ Thanks for the response. I am confused. On multiple examples that I have seen on UDLs on simply supported beams the UDL is treated as a point load in the middle of the UDL span in order to find moments and reaction forces. But you have said that this method results in getting the wrong value for M? $\endgroup$ Dec 15, 2021 at 11:38

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