Maximum deflection with unsymmetrical loading

$EI = 200\times10^9\times1.5\times10^{-4} = 30\times10^6\ \text{N m}^2 = 30\,000\ \text{kN m}^2$.

Bending moment diagram

$$R_A = \frac{40\times3}{5} = 24\ \text{kN}, \qquad R_B = 16\ \text{kN}, \qquad M_C = 24\times2 = 48\ \text{kN m}$$

The $M/EI$ diagram is a triangle of base 5 m and height $48/EI$, with its apex at C. Total area $= \frac12\times5\times48 = 120$ (in kN m$^2$, divided by $EI$).

 A      C(2 m)            B
 /\    48
/  \_____
 triangle area = 120/EI

(a) Slopes at the supports

Deviation of B from the tangent at A (moment of the area about B). The centroid of a triangle lies at the mean position of its three corners: from A, $\bar x = (0+2+5)/3 = 2.333$ m, so its distance from B is $5 - 2.333 = 2.667$ m.

$$t_{B/A} = 120\times2.667 = 320/EI$$ $$\theta_A = \frac{t_{B/A}}{L} = \frac{320}{5\,EI} = \frac{64}{EI} = \frac{64}{30000} = 2.13\times10^{-3}\ \text{rad}$$

Similarly, for the deviation of A from the tangent at B, the centroid is $2.333$ m from A:

$$t_{A/B} = 120\times2.333 = 280/EI \;\Rightarrow\; \theta_B = \frac{280}{5EI} = \frac{56}{EI} = 1.87\times10^{-3}\ \text{rad}$$

(b) Deflection under the load C

At C, the tangent at A lies $\dfrac{2}{5}t_{B/A} = \dfrac{128}{EI}$ below the line joining the supports. The deviation of C from the tangent at A is the moment of the area from A to C about C: area $\frac12\times2\times48 = 48$, centroid $2/3$ m from C:

$$t_{C/A} = 48\times\frac23 = \frac{32}{EI}$$ $$y_C = \frac{128 - 32}{EI} = \frac{96}{30000} = 3.2\times10^{-3}\ \text{m} = 3.2\ \text{mm}$$

(c) Maximum deflection

The point of zero slope D lies in the longer segment CB (3 m), at a distance $x$ from B. The slope at B equals the area of the $M/EI$ diagram from B to D, where $M = R_Bx = 16x$:

$$\theta_B = \frac12\,x\,\frac{16x}{EI} \;\Rightarrow\; 8x^2 = 56 \;\Rightarrow\; x = \sqrt7 = 2.646\ \text{m}\ (<3\text{ m, so valid})$$

So D is 2.646 m from B, i.e. 2.354 m from A. The maximum deflection equals the deviation of B from the horizontal tangent at D:

$$y_{max} = 8x^2\times\frac{2x}{3}\Big/EI = \frac{16x^3}{3EI} = \frac{16\times18.52}{3\times30000} = 3.29\times10^{-3}\ \text{m}$$

Answer: $\theta_A = 2.13\times10^{-3}$ rad, $\theta_B = 1.87\times10^{-3}$ rad; $y_C = 3.20$ mm; $y_{max} = 3.29$ mm at 2.35 m from A.