Conjugate beam: rotation at A, deflection at C

Data: $AB = 6$ m ($2I$), $BC = 2$ m ($I$) overhang, $w = 60$ kN/m on the whole beam. Units: kN, m, $EI$ taken for the section $I$.

Real beam: reactions and BM

$$\begin{aligned} R_B\times 6 &= 60\times 8\times 4 \Rightarrow R_B = 320\ \text{kN}, \quad R_A = 480-320 = 160\ \text{kN}\\ M_x &= 160x - 30x^2 \ (0\le x\le 6), \quad M_B = -60\times\tfrac{2^2}{2} = -120\ \text{kN m} \end{aligned}$$

The loading on the conjugate beam is $M/EI$, so in $AB$ it is $M/(2EI)$ and in $BC$ it is $M/EI$.

Conjugate beam

Real hinge $A$ gives a simple support. Real roller $B$ (with overhang) gives an internal hinge. Real free end $C$ gives a fixed end.

 Real:   A o=======o B========= C (free)
 Conj.:  A o=======o hinge====== C (fixed)
         load = M/EI (sagging down, hogging up)

Span AB (simply supported at A, hinge at B)

Load $q' = \dfrac{160x-30x^2}{2EI} = \dfrac{80x-15x^2}{EI}$; total load $=\int_0^6 q'\,dx = \dfrac{360}{EI}$.

Taking moments about the hinge $B$ (moment of conjugate beam is zero there):

$$R'_A = \frac{1}{6}\int_0^6 (80x-15x^2)(6-x)\,dx = \frac{1260}{6EI} = \frac{210}{EI}$$ $$R'_B = \frac{360-210}{EI} = \frac{150}{EI}$$

Span BC (cantilever, fixed at C)

$M/EI$ in $BC$ is negative (hogging), $M = -30(2-x')^2$ with $x'$ from $B$. Its load is upward on the conjugate beam: total $= \int_0^2 30(2-x')^2dx' = 80/EI$, with moment about $C$ equal to $\int_0^2 30(2-x')^3dx' = 120/EI$.

The hinge $B$ passes the force $150/EI$ downward onto the cantilever. Moment at the fixed end $C$:

$$M'_C = 150\times 2 - 120 = \frac{180}{EI}\ (\text{hogging})$$

Results

• Rotation at $A$ = shear in conjugate beam at $A$: $\theta_A = \dfrac{210}{EI}$ (clockwise).
• Deflection at $C$ = moment in conjugate beam at $C$: $y_C = \dfrac{180}{EI}$, and since the conjugate moment is hogging it is upward.

Answer: $\theta_A = 210/EI$ kN m² (clockwise), $y_C = 180/EI$ kN m³ (upward).