Cantilever: slope and deflection under load P1

Given: $P_1 = 20$ kN at $a = 3$ m from the fixed end; $P_2 = 30$ kN at the free end ($L = 5$ m). $E = 2\times10^5$ N/mm², $I = 2\times10^8$ mm⁴.

$$EI = 2\times10^5\times2\times10^8 = 4\times10^{13}\ \text{N mm}^2 = 4\times10^4\ \text{kNm}^2$$

Use superposition. Take $x$ from the fixed end; the moment due to $P_2$ at the free end is $M = P_2(L - x)$.

Effect of $P_1$ (at its own position)

$$\theta_1 = \frac{P_1 a^2}{2EI} = \frac{20\times10^3\times3000^2}{2\times4\times10^{13}} = 0.00225\ \text{rad}$$ $$y_1 = \frac{P_1 a^3}{3EI} = \frac{20\times10^3\times3000^3}{3\times4\times10^{13}} = 4.5\ \text{mm}$$

Effect of $P_2$ (at free end) at the position of $P_1$

Integrate $EI\,y'' = P_2(L-x)$ with $y = y' = 0$ at the fixed end, then put $x = 3$ m:

$$EI\,\theta_2 = P_2\left[Lx - \frac{x^2}{2}\right]_{x=3} = 30\,(15 - 4.5) = 315\ \text{kNm}^2 \Rightarrow \theta_2 = 0.007875\ \text{rad}$$ $$EI\,y_2 = P_2\left[\frac{Lx^2}{2} - \frac{x^3}{6}\right]_{x=3} = 30\,(22.5 - 4.5) = 540\ \text{kNm}^3 \Rightarrow y_2 = 13.5\ \text{mm}$$

Total under $P_1$

$$\theta = 0.00225 + 0.007875 = 0.0101\ \text{rad}$$ $$y = 4.5 + 13.5 = 18.0\ \text{mm}$$

Answer: slope under $P_1$ = 0.0101 rad (about 0.58°); deflection under $P_1$ = 18.0 mm downward.