Cantilever: slope and deflection at free end

Given: $AB = 2$ m with 50 kN at B; $BC = 3$ m with 40 kN at free end C; total length 5 m. $E = 200\ \text{kN/mm}^2 = 2\times10^5\ \text{N/mm}^2$.

$$I = \frac{50\times100^3}{12} = 4.167\times10^6\ \text{mm}^4,\qquad EI = 8.333\times10^{11}\ \text{N mm}^2 = 833.3\ \text{kNm}^2$$

 fixed       50kN        40kN
 |A----2m----B-----3m-----C

Use superposition with standard cantilever results (point load $P$ at distance $a$ from the fixed end: $\theta = Pa^2/2EI$, $y = Pa^3/3EI$ at the load point, and the portion beyond the load stays straight).

Slope at C

$$\theta_C = \frac{40\times10^3\times(5000)^2}{2EI} + \frac{50\times10^3\times(2000)^2}{2EI} = 0.600 + 0.120 = 0.72\ \text{rad}$$

Deflection at C

• Due to 40 kN at C: $\dfrac{40\times10^3\times5000^3}{3EI} = 2000$ mm
• Due to 50 kN at B: $\dfrac{50\times10^3\times2000^3}{3EI} = 160$ mm at B, plus slope $0.12$ rad $\times$ 3000 mm $= 360$ mm further to C

$$y_C = 2000 + 160 + 360 = 2520\ \text{mm}$$

Answer: slope at free end = 0.72 rad; deflection = 2520 mm (2.52 m) downward. These values are far beyond the small-deflection limit because the section (50 × 100 mm) is very flexible for these loads; the formulas still give the theoretical result.