Moment area: mid-span deflection under UDL

Given data

Simply supported beam, span $L$, rigidity $EI$, UDL $w$ over the whole span. The deflection at mid-span C is required.

  w kN/m
 A ===================== B
 ^          L            ^

Bending moment

$$R_A = R_B = \frac{wL}{2},\qquad M = \frac{wLx}{2} - \frac{wx^2}{2}$$

Maximum moment at mid-span: $M_C = \dfrac{wL^2}{8}$. The $M/EI$ diagram is a parabola with peak $\dfrac{wL^2}{8EI}$.

Using the symmetry

By symmetry the tangent at mid-span C is horizontal. The deflection at C below the support level is therefore equal to the deviation of A from the tangent at C (Theorem II).

Area and its moment about A (portion A to C)

$$\text{Area} = \int_0^{L/2}\frac{M}{EI}dx = \frac{w}{2EI}\left[\frac{Lx^2}{2} - \frac{x^3}{3}\right]_0^{L/2} = \frac{wL^3}{24EI}$$

Using the parabola property, the area under a parabola from its vertex over base $L/2$ with height $h$ is $\tfrac23 \cdot \tfrac{L}{2}\cdot h = \tfrac23\cdot\tfrac{L}{2}\cdot\dfrac{wL^2}{8EI} = \dfrac{wL^3}{24EI}$ (check). Its centroid is $\dfrac{5L}{16}$ from A.

Deflection at mid-span

$$t_{A/C} = \text{Area}\times\bar{x}_A = \frac{wL^3}{24EI}\times\frac{5L}{16}$$ $$\boxed{y_C = \frac{5wL^4}{384\,EI}}\ \text{(downward)}$$

Check with the slope: $\theta_A = \dfrac{wL^3}{24EI}$ (clockwise), the area between A and C.