You did something wrong here; you might want to show your work. Since the load is perfectly symmetrical about the center of the beam, you might want to venture an educated guess as to how the applied loads distribute to each end reaction. But first note that the sum of the end reactions must equal the sum of the applied vertical loads. You've got two concentrated loads of 15 kN each, and a distributed load of 5 kN/m over 6 m, that's another 30 kN, so all together, you have 60 kN acting down. So the end reactions have to sum to 60 kN acting up, from Newton 1. Also, when you sum moments about A to determine the external end reaction at D, note that the distributed load of 5 kN/m over the 6 m length can be represented as a 30 kN load acting at the c.g. of the loading (which in this case is the center of the beam).
The problem asks for the (internal) shear and bending moment at the center. So just expose the section from the left end to the center. At the center, there wiil be an unknown vertical load (the shear) and an unknown moment (the bending moment). You'll need to solve for these unknowns using sum of F_y = 0 , and sum of moments = 0.
Also said:
yes, correct.
You already did, but it's good to check your work.
No, you must sum ALL forces to solve for the unknown shear, V, in your FBD of the left part of the beam. That includes the concentrated load of 15 kN down, the end reaction of 30 kN up, the downward contribution of the distributed load, and V, all summed to 0 to solve for V.
to solve for the bending moment, sum moments about the right end of your FBD. It will be easier to assist you if you show your work.
Looks Good!
You're good at them now, no? You're bending moment is correct, except it should be acting counterclockwise on the left part of the beam.