ILD for SF and BM at C of beam

Assumptions (figure partly unclear): beam $A$–$B$–$C$ is supported by a hinge at $A$ and a roller at $C$ ($AB = BC = 2$ m, span $AC = 4$ m), with an overhang $C$–$D$–$E$–$F$ of 3 m (1 m each). Loads: 40 kN at $B$; 5 kN/m UDL from $C$ to $E$ (2 m); 60 kN at $F$ (free end). The section is taken at the support $C$.

Reactions (for checking)

$\Sigma M_A = 0$: $R_C\times 4 = 40\times 2 + 10\times 5 + 60\times 7 = 550 \Rightarrow R_C = 137.5$ kN; $R_A = 110-137.5 = -27.5$ kN (uplift).

ILD for bending moment at C

Load at distance $s$ from $A$. For $s\le 4$, $M_C = 0$ (the load is carried directly by the supports). For $s>4$, $M_C = -(s-4)$ (hogging).

 A o---B-----o C---D---E---F
 ILD(M_C):  0  0   0  -1 -2 -3

$$M_C = 5\times 2\times(-1) + 60\times(-3) = -10 - 180 = -190\ \text{kN m}$$

ILD for shear at C

Just left of C: ordinate $=-s/4$ for $0\le s<4$ (so $-0.5$ at $B$ and $-1$ as the load approaches $C$); $-(s-4)/4$ for the overhang ($-0.25$ at $D$, $-0.5$ at $E$, $-0.75$ at $F$).

Using ordinates: UDL area from $C$ to $E$ $=\tfrac12(0+(-0.5))\times 2 = -0.5$, so

$$V_{C^-} = 40(-0.5) + 5(-0.5) + 60(-0.75) = -20 - 2.5 - 45 = -67.5\ \text{kN}$$

Just right of C: ordinate 0 for loads on the span, and 1 for loads on the overhang:

$$V_{C^+} = 5\times 2\times 1 + 60\times 1 = 70\ \text{kN}$$

Check by statics: $V_{C^-} = R_A - 40 = -27.5-40 = -67.5$ kN and $V_{C^+} = -67.5 + 137.5 = +70$ kN.

Answer: $M_C = -190$ kN m (hogging); $V_C = -67.5$ kN just left and $+70$ kN just right of $C$.