Overhang beam deflection and slope at C

Given data

Beam ABC: A hinge, B roller at $AB = 10$ m, overhang $BC = 5$ m, with $P = 10$ kN downward at C. $EI = 100\,000$ kNm$^2$.

    A               B       C
    ^---------------^-------+ 10 kN
         10 m           5 m

Real loading

$$R_A\times10 = -10\times5 \;\Rightarrow\; R_A = -5\ \text{kN (down)},\quad R_B = 15\ \text{kN}$$

Hogging moments: $M = 5x$ in AB ($x$ from A), $M = 10s$ in BC ($s$ from C). $M_B = 50$ kNm.

Deflection at C: unit vertical load at C

Virtual moments (hogging): $m = 0.5x$ in AB, $m = s$ in BC.

$$\Delta_C = \frac{1}{EI}\left[\int_0^{10} 5x\,(0.5x)\,dx + \int_0^5 10s\cdot s\,ds\right] = \frac{1}{EI}\left[\frac{2.5\times1000}{3} + \frac{10\times125}{3}\right]$$ $$\Delta_C = \frac{833.33 + 416.67}{EI} = \frac{1250}{EI} = \frac{1250}{100\,000} = 0.0125\ \text{m}$$

Slope at C: unit couple at C

Virtual moments: $m = x/10$ in AB, $m = 1$ in BC (same sense as $M$).

$$\theta_C = \frac{1}{EI}\left[\int_0^{10} 5x\,\frac{x}{10}dx + \int_0^5 10s\cdot1\,ds\right] = \frac{1}{EI}\left[166.67 + 125\right] = \frac{291.67}{EI}$$ $$\theta_C = \frac{291.67}{100\,000} = 2.917\times10^{-3}\ \text{rad (clockwise)}$$

Answer: deflection at C $= 12.5$ mm downward; slope at C $= 2.917\times10^{-3}$ rad (clockwise). (Check: $\Delta_C = \dfrac{Pa^2(L+a)}{3EI} = 12.5$ mm, $\theta_C = \dfrac{Pa(2L+3a)}{6EI}$.)