Flexural rigidity EI = 50000 kN·m² is fixed. The UDL w acts over the full length (L_AB + c).
Top = beam model (supports A, B with UDL and tip load) / Middle = shear force diagram / Bottom = bending moment diagram (negative at the overhang root)
Supports A (x=0) and B (x=L_AB) are pinned and an overhang of length c extends from B to the free end D. The following formulas give reactions and moments under a UDL w acting over the whole length and a tip load P at D.
Right reaction from moment equilibrium about A:
$$R_B = \frac{w\,L_{AB}^{2}/2 + w\,c\,(L_{AB}+c/2) + P\,(L_{AB}+c)}{L_{AB}}$$Left reaction from vertical equilibrium:
$$R_A = w\,(L_{AB}+c) + P - R_B$$Negative bending moment at the overhang root B:
$$M_B = -\left(\frac{w\,c^{2}}{2} + P\,c\right)$$Tip deflection at D (cantilever approximation rooted at B):
$$\delta_D \approx \frac{w\,c^{4}}{8\,EI} + \frac{P\,c^{3}}{3\,EI}$$The maximum positive moment in the main span occurs where shear vanishes, at x = R_A / w. Increasing the overhang c or the tip load P enlarges the magnitude of M_B (negative), so designers must pay particular attention to the tension on the top fiber at the root.