Beam with two symmetrical loads by moment-area

Set-up

    P            P
    |            |
    v____________v
  A   a   |  L-2a   a  B
 R_A = P   R_B = P

By symmetry $R_A = R_B = P$. Bending moment: rises linearly from 0 at A to $Pa$ at the load, then constant $Pa$ between the loads. The $M/EI$ diagram is a trapezoid, symmetrical about mid-span C.

Slope at support A

By symmetry the slope at mid-span C is zero (the tangent at C is horizontal). By Theorem I, the slope at A equals the area of the $M/EI$ diagram between A and C:

$$\theta_A = \frac1{EI}\left[\frac12\,a\,(Pa) + \left(\frac L2-a\right)(Pa)\right] = \frac{Pa(L-a)}{2EI}$$

Deflection at mid-span

The deflection at C equals the deviation of A from the tangent at C, i.e. the moment of the area between A and C about A (Theorem II):

$$y_C = \frac{1}{EI}\left[\frac{Pa^2}{2}\cdot\frac{2a}{3} + Pa\left(\frac L2-a\right)\left(\frac{L}{4}+\frac a2\right)\right]$$

The first term is the triangle (centroid $2a/3$ from A); the second is the rectangle (centroid at $a + \frac12(\frac L2 - a) = \frac L4+\frac a2$).

$$y_C = \frac{1}{EI}\left[\frac{Pa^3}{3} + \frac{PaL^2}{8} - \frac{Pa^3}{2}\right] = \boxed{\frac{Pa\,(3L^2-4a^2)}{24EI}}$$

Numerical values

$$\theta_A = \frac{30\times2\times(6-2)}{2\times25000} = 4.8\times10^{-3}\ \text{rad}$$ $$y_C = \frac{30\times2\times(3\times36-4\times4)}{24\times25000} = \frac{60\times92}{600000} = 9.2\times10^{-3}\ \text{m}$$

Answer: $\theta_A = \theta_B = 4.8\times10^{-3}$ rad $(0.275^\circ)$; maximum deflection at mid-span $= 9.2$ mm downward.