Cantilever deflection by integration and superposition

(a) Integration

Take the origin at the fixed end A and $x$ towards the free end B. The bending moment at a section is hogging, from the load on the part to its right:

$$M = -\frac{w(L-x)^2}{2}$$ $$EI\frac{d^2y}{dx^2} = -\frac{w(L-x)^2}{2}$$ $$EI\frac{dy}{dx} = \frac{w(L-x)^3}{6} + C_1, \qquad EI\,y = -\frac{w(L-x)^4}{24} + C_1x + C_2$$

At the fixed end $x = 0$: slope $= 0$ and $y = 0$.

$$C_1 = -\frac{wL^3}{6}, \qquad C_2 = \frac{wL^4}{24}$$

At the free end $x = L$:

$$EI\,\theta_B = -\frac{wL^3}{6} \;\Rightarrow\; |\theta_B| = \frac{wL^3}{6EI}, \qquad EI\,y_B = -\frac{wL^4}{6}+\frac{wL^4}{24} = -\frac{wL^4}{8} \;\Rightarrow\; |y_B| = \frac{wL^4}{8EI}$$

The negative signs mean the free end slopes and deflects downward.

(b) Superposition

Total deflection is the sum of the deflections due to each load acting alone (valid for linear elastic behaviour and small deflections).

$EI = 200\times10^9\times8\times10^{-5} = 16\times10^6\ \text{N m}^2 = 16\,000\ \text{kN m}^2$.

Answer: slope $= 7.03\times10^{-3}$ rad $(0.40^\circ)$, deflection $= 14.8$ mm downward at the free end.