ILD: RA, RB, SF and BM at C

Data: beam with $A$ hinge and $B$ roller, $AB = 6$ m, $C$ at 3 m from $A$. A 15 kN/m UDL acts on $CB$ (3 m). An overhang of 3 m beyond $B$ carries a 1.5 kN point load at its free end $D$ ($x = 9$ m from $A$). Let the unit load be at $x$ from $A$ ($0\le x\le 9$).

ILDs

$$R_A = \frac{6-x}{6}, \qquad R_B = \frac{x}{6}$$

• $R_A$: 1 at $A$, 0 at $B$, $-0.5$ at $D$.
• $R_B$: 0 at $A$, 1 at $B$, $1.5$ at $D$.
• $V_C$: $-\dfrac{x}{6}$ for $x<3$ (ordinate $-0.5$ at $C^-$); $\dfrac{6-x}{6}$ for $x>3$ (ordinate $+0.5$ at $C^+$, 0 at $B$, $-0.5$ at $D$).
• $M_C$: $R_B\times 3 = 0.5x$ for $x\le 3$; $R_A\times 3 = 0.5\,(6-x)$ for $x\ge 3$; peak $1.5$ m at $C$ and $-1.5$ at $D$.

Reactions

Area of $R_A$ ILD under the UDL ($C$ to $B$): ordinates $0.5$ and $0$, so area $=\tfrac12\times 3\times 0.5 = 0.75$ m.

$$R_A = 15(0.75) + 1.5(-0.5) = 11.25 - 0.75 = 10.5\ \text{kN}$$

Area of $R_B$ ILD under the UDL: ordinates $0.5$ and $1$, area $=\tfrac12(0.5+1)\times 3 = 2.25$ m.

$$R_B = 15(2.25) + 1.5(1.5) = 33.75 + 2.25 = 36\ \text{kN}$$

(Check: $10.5 + 36 = 46.5 = 15\times 3 + 1.5$.)

Shear force at C

The UDL starts at $C$ and there is no point load at $C$, so the ordinates of $V_C$ under the UDL are $0.5$ at $C^+$ and $0$ at $B$ (area $=0.75$ m).

$$V_C = 15(0.75) + 1.5(-0.5) = 10.5\ \text{kN}$$

Bending moment at C

Area of the $M_C$ ILD under the UDL (triangle from $C$ to $B$): $\tfrac12\times 3\times 1.5 = 2.25$ m². Ordinate at $D$: $-1.5$.

$$M_C = 15(2.25) + 1.5(-1.5) = 33.75 - 2.25 = 31.5\ \text{kN m}$$

Answer: $R_A = 10.5$ kN, $R_B = 36$ kN, $V_C = 10.5$ kN, $M_C = 31.5$ kN m (sagging).