Determine the reactions on supports A and C and write the functions of the internal forces for the beam sections AB

This is a statically determinate structure (four reactions minus one release equals three static equilibrium equations).

So we should be able to do this using simple statics:

$$begin{align} sum F_x &= A_x + C_x = 0 therefore A_x &= -C_x sum F_y &= A_y + C_y - 300cdot6 = 0 therefore C_y &= 300cdot6 - A_y sum M_A &= -300cdot6cdotdfrac{6}{2} + 600 + 2.5C_x = 0 therefore C_x &= dfrac{300cdot6cdotdfrac{6}{2} - 600}{2.5} = 1920text{ N} therefore A_x &= -1920text{ N} end{align}$$

But how can we determine $A_y$ and $C_y$? Well, thankfully we have the hinge at B, which we can use to ensure that the bending moment to the left of B is zero:

$$begin{align} sum M_B^- &= -6A_y + 300cdot6cdotdfrac{6}{2} + 600 = 0 therefore A_y &= dfrac{300cdot6cdotdfrac{6}{2} + 600}{6} = 1000text{ N} therefore C_y &= 300cdot6 - 1000 = 800text{ N} end{align}$$

As for the internal force equations, it's worth noting that AB has no concentrated forces (just a bending moment). We therefore know that the shear diagram will be continuous and therefore we can have one shear equation for all of $xin[0, 6]$.

Since shear is the integral of the distributed load, we can easily get that

$$begin{align} q(x) &= -300 V(x) &= int q(x) = -300x + C V(0) &= C = 1000 therefore V(x) &= -300x + 1000 end{align}$$

And we also know that the bending moment is the integral of the shear diagram, so we get:

$$begin{align} V(x) &= -300x + 1000 M(x) &= int V(x) = -150x^2 + 1000x + C end{align}$$

But here we must remember that the bending moment diagram will have a discontinuity due to the concentrated moment at D. So:

$$begin{align} M(x) &= -150x^2 + 1000x + C M(0) &= C_{[0,3)} = 0 M(6) &= -5400 + 6000 + C_{(3,6]} = 0 therefore C_{(3,6]} &= -600 therefore M(x) &= begin{cases} -150x^2 + 1000x &xin[0, 3) -150x^2 + 1000x - 600 &xin(3, 6] end{cases} end{align}$$