Conjugate beam: slope and deflection at C

Given data and assumptions

Simply supported beam AB, span 12 m. The portion from $0$ to $7$ m (A to D) has rigidity $3EI$ and the portion from $7$ to $12$ m (D to B) has $EI$. A 120 kN load acts 3 m from A; a UDL of 10 kN/m acts over the last 3 m (from 9 m to 12 m). The point C is taken at the start of the UDL, $x = 9$ m (the figure is partly illegible; this is the reading used).

      120 kN                      10 kN/m
 A -----v--------- 3EI -------D--- EI ---C=========B
 ^<-3 m->|<------- 4 m ------>|<- 2 m ->|<- 3 m ->^

Step 1: Reactions

$R_{A} = 93.75$ kN, $R_{B} = 56.25$ kN (positive = upward; check: $\sum F_y = 150.00$ kN equals the total load 150.00 kN).

Step 2: Bending moment diagram (sagging +), as $M/EI$ pieces

(Stiffness is in terms of $EI$; "Area" is $\int M/EI\,dx$ in units of $1/EI$; $\bar x$ is the centroid distance from A.)

Step 3: Conjugate beam

The conjugate of a simply supported beam is a simply supported beam of the same span. It carries the $M/EI$ diagram of Step 2 as load (sagging $M/EI$ acts downward, hogging acts upward). Conjugate shear $V'$ = slope (positive = clockwise) and conjugate moment $M'$ = deflection (positive = downward).

Step 4: Conjugate reaction at A (slope at A)

Taking moments of the conjugate beam about the support at $x = 12$ m (conjugate moment is zero there):

$$R'_A\times12 = (140.62)(10.000) + (305.00)(7.153) + (300.00)(4.058) + (208.12)(1.946) = 5210.42$$ $$R'_A = \frac{434.201}{EI}\ \text{(slope at A, clockwise)}$$

At the 120 kN load ($x = 3$ m):

$$V' = R'_A - \sum A = 434.201 - [(140.62)] = \frac{293.576}{EI}\ \text{(slope, clockwise)}$$ $$M' = R'_A\,x - \sum A(x-\bar x) = 1302.604 - [(140.62)(1.000)] = \frac{1161.979}{EI}\ \text{(deflection, downward)}$$

At C ($x = 9$ m):

$$V' = R'_A - \sum A = 434.201 - [(140.62) + (305.00) + (300.00)] = \frac{-311.424}{EI}\ \text{(slope, anticlockwise)}$$ $$M' = R'_A\,x - \sum A(x-\bar x) = 3907.812 - [(140.62)(7.000) + (305.00)(4.153) + (300.00)(1.058)] = \frac{1339.271}{EI}\ \text{(deflection, downward)}$$

Answer: at C, slope $= \dfrac{311.42}{EI}$ (anticlockwise) and deflection $= \dfrac{1339.27}{EI}$ downward (kNm$^2$/EI and kNm$^3$/EI). Slope at A $= 434.20/EI$ (clockwise).