The rigid case. We start with the middle support. The structure is symmetrical and carries the same load on both spans, so the response of the beam will also be symmetrical. This means that the beam's rotation over the middle support is zero, since one side of the support balances the other.
Now, a fixed support achieves the same result, which means each span behaves exactly like step 3's 6 m long propped cantilever: fixed at the middle support and propped at the outer ends.
Therefore, each end reaction is step 3's prop reaction,
Rend​=83wL​=83×20×6​=45 kN The hogging moment over the middle support is step 3's moment at the fixed support,
Mmid​=8wL The middle support resists whatever applied load the two ends don't. The total load is 2wL=240 kN and the ends take 2×45=90 kN of it, so,
Rmid​=2wL−2Rend​= which is the solver's 150.0 kN, and the classic 5wL/4. Another way to look at it is as follows: in step 3 the fixed support resisted 75 kN of shear, and here the middle support is the fixed support for both spans at once, so it takes 2×75=150 kN. That's the 150 kN jump we read in the shear diagram.
Finally, the sagging moment peak sits where the shear is zero. Measured from an end support, the shear is Rend​−wx, which is zero at x=R m, and the moment there is,
Msag​=Rend​ The solver gives us 50.6 kN m at 2.25 m, or 9wL2/128 from above. The same expression is zero at x=2Rend​/w m, where the moment changes from sagging to hogging.
The springs. With a spring, the beam still doesn't rotate over the middle support, but it does deflect down, so a single span can no longer be modelled as a propped cantilever. The simplest alternative is to treat the beam as one 12 m simply supported span, with the middle reaction R pushing up at its centre. On its own, with no midspan support, the uniform load would cause the centre to deflect by,
δw​=384EI5 where EI=210×106×14,100×10 kN m². This is the 182.37 mm we get when the middle support is deleted. The reaction lifts the centre by , and the spring lets it sink by , so for the beam and the spring to meet, compatibility says,
384EI5w(2L) For k=50,000 kN/m, (2L)3/48EI=1,728/( m/kN and m/kN, so kN. The middle sinks mm, as the table says. The moment over the middle support is the 12 m span's kN m of sagging, less the that the reaction takes off it,
Mmid​=360−3×147.6=−82.7 kN m The 10,000 and 5,000 kN/m rows come out the same way, at 138.6 and 128.8 kN. And the formula holds both limits: a rigid support is 1/k=0, which gives R=150 kN and the 90 kN m again, and a missing one is R=0, which leaves the 360 kN m of sagging.