Moment area: slopes at A, B, deflection at D

Assumptions: simply supported beam $AB$ of span $L = 18$ m, $P = 10$ kN at $C$ (mid-span, 9 m from $A$); $D$ is taken 6 m from $A$ because its exact position is not clear in the figure. $EI$ is constant.

Reactions and BM diagram

$R_A = R_B = 5$ kN. Maximum BM at $C$: $M_C = 5\times 9 = 45$ kN m. At $D$: $M_D = 5\times 6 = 30$ kN m.

  A o-----6-----D---3---C--------9--------o B
                        10 kN
  BMD: triangle, peak 45 kN m at C

Slope at A and B

The loading is symmetrical, so the tangent at $C$ is horizontal. By the first moment-area theorem, the slope at $A$ equals the area of the $M/EI$ diagram between $A$ and $C$:

$$\theta_A = \frac{1}{EI}\times\frac12\times 9\times 45 = \frac{202.5}{EI}$$ $$\theta_B = \theta_A = \frac{202.5}{EI}\ \text{(opposite sense)}$$

(Check: $PL^2/16EI = 10\times 18^2/16 = 202.5$.)

Deflection at D

By the second theorem, the deviation of $A$ from the tangent at $C$ is the moment of the $M/EI$ area $AC$ about $A$. Since the tangent at $C$ is horizontal and $A$ has no deflection, this is the deflection at $C$:

$$y_C = \frac{1}{EI}\left[\frac12\times 9\times 45\right]\times\frac{2}{3}\times 9 = \frac{1215}{EI}$$

Deviation of $D$ from the tangent at $C$ (moment about $D$ of the $M/EI$ area between $D$ and $C$; rectangle $30\times 3$ plus triangle $\frac12\times 3\times 15$):

$$t_D = \frac{1}{EI}\left[30\times 3\times 1.5 + \tfrac12\times 3\times 15\times 2\right] = \frac{135+45}{EI} = \frac{180}{EI}$$ $$y_D = y_C - t_D = \frac{1215-180}{EI} = \frac{1035}{EI}$$

(Check: $y = \dfrac{Px(3L^2-4x^2)}{48EI} = 1035/EI$.)

Answer: $\theta_A = \theta_B = 202.5/EI$ kN m² (rad); $y_D = 1035/EI$ kN m³ downward (D at 6 m from A).