ILD: reaction at A and BM at C

Data: simply supported beam $AD$ ($A$ hinge, $D$ roller). $AB = 2$ m with 80 kN at $B$; $BC = 1$ m; $CD = 5$ m with 20 kN/m UDL. So $L = 8$ m and $C$ is 3 m from $A$.

Reaction at A

ILD for $R_A$: ordinate $\dfrac{8-x}{8}$, i.e. 1 at $A$ and 0 at $D$.

• Ordinate at $B$ ($x=2$): $\dfrac{6}{8} = 0.75$.
• Ordinates under the UDL: $0.625$ at $C$ and $0$ at $D$, so the area $=\tfrac12\times 5\times 0.625 = 1.5625$ m.

$$R_A = 80(0.75) + 20(1.5625) = 60 + 31.25 = 91.25\ \text{kN}$$

Bending moment at C

ILD for $M_C$: a triangle with peak $\dfrac{ab}{L} = \dfrac{3\times 5}{8} = 1.875$ m under $C$; at $B$ the ordinate is $\dfrac{2\times 5}{8} = 1.25$ m.

  A o--B--C-----------o D
  ILD M_C:   /\ 1.875
           1.25 at B

Area of the ILD under the UDL (triangle from $C$ to $D$) $=\tfrac12\times 5\times 1.875 = 4.6875$ m².

$$M_C = 80(1.25) + 20(4.6875) = 100 + 93.75 = 193.75\ \text{kN m}$$

Check: $M_C = R_A\times 3 - 80\times 1 = 273.75 - 80 = 193.75$ kN m.

Answer: $R_A = 91.25$ kN, $M_C = 193.75$ kN m (sagging).