Castigliano: slope at A, deflection at D

Beam ABCD: A hinged, C roller (AC = 10 m), loads 30 kN at B (4 m from A) and 10 kN at the end D of the 2 m overhang CD. EI is constant. Apply a dummy couple $M'$ (clockwise) at A, and treat the load at D as a variable $P$ (equal to 10 kN at the end).

        30 kN                  10 kN
          v                      v
 A o------B---------------------C o--D
 hinge                         roller
 |<- 4 ->|<------- 6 ------->|<- 2 ->|

Reactions (with $M'$ and $P$)

Moments about A:

$$R_C = \frac{30(4) + 12P + M'}{10},\qquad R_A = 30 + P - R_C$$

At $M' = 0$, $P = 10$: $R_C = 24$ kN, $R_A = 16$ kN.

Bending moments ($x$ from A for AC; $s$ from D for the overhang)

$$\begin{aligned} AB\ (0 \le x \le 4):\ & M = M' + R_Ax, \\ BC\ (4 \le x \le 10):\ & M = M' + R_Ax - 30(x-4), \\ DC\ (0 \le s \le 2):\ & M = -Ps \end{aligned}$$

so $R_A = 30 + P - R_C = 18 - 0.2P - 0.1M'$, with $\dfrac{\partial R_A}{\partial M'} = -0.1$ and $\dfrac{\partial R_A}{\partial P} = -0.2$.

At $M' = 0$ and $P = 10$: $R_A = 16$ kN, $M = 16x$ in AB, $M = 120 - 14x$ in BC, $M = -10s$ in the overhang.

Slope at A

$$EI\,\theta_A = \int_0^4 16x\,(1-0.1x)\,dx + \int_4^{10}(120-14x)(1-0.1x)\,dx = 93.87 + 64.80$$ $$\theta_A = \frac{158.67}{EI}\ \text{rad (clockwise, the positive sense of } M'\text{)}$$

Deflection at D

$$EI\,\delta_D = \int_0^4 16x(-0.2x)\,dx + \int_4^{10}(120-14x)(-0.2x)\,dx + \int_0^2(-10s)(-s)\,ds$$ $$EI\,\delta_D = -68.27 - 134.40 + 26.67 = -176.00$$

The negative sign shows that D moves opposite to the 10 kN load.

$$\delta_D = \frac{176}{EI}\ \text{upward}$$

(The 30 kN load at B tilts the beam at C so that the overhang rises; this outweighs the downward bending of the overhang under its own 10 kN load, which is only $160/EI$.)

Answer: slope at A $= \dfrac{158.67}{EI}$ rad (clockwise); deflection at D $= \dfrac{176}{EI}$ (upward).