Conjugate beam: deflection at B and D

Given data

Beam ABCD as labelled: A hinge at 0, D at 5 m (20 kN down), C roller at 8 m ($DC = 3$ m), overhang $CB = 3$ m with UDL 8 kN/m and free end B at 11 m. $AC$ has $2EI$ and the overhang $CB$ has $EI$. Required: deflection at B (free end) and D.

        20 kN                 8 kN/m (EI)
 A -------v---------- C =============== B
 ^   5 m  D   3 m (2EI) ^      3 m

Step 1: Reactions

$R_{A} = 3.00$ kN, $R_{C} = 41.00$ kN (positive = upward; check: $\sum F_y = 44.00$ kN equals the total load 44.00 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 real beam has a hinge at A, a roller at the second support and a free end at the overhang tip, so the conjugate beam has a hinge at A, an internal hinge at the support and a fixed end at the tip. 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 $x = 8$ m (conjugate moment is zero there):

$$R'_A\times8 = (18.75)(4.667) + (-15.75)(0.286) = 83.00$$ $$R'_A = \frac{10.375}{EI}\ \text{(slope at A, clockwise)}$$

At D (20 kN) ($x = 5$ m):

$$V' = R'_A - \sum A = 10.375 - [(18.75)] = \frac{-8.375}{EI}\ \text{(slope, anticlockwise)}$$ $$M' = R'_A\,x - \sum A(x-\bar x) = 51.875 - [(18.75)(1.667)] = \frac{20.625}{EI}\ \text{(deflection, downward)}$$

At B (free end) ($x = 11$ m):

$$V' = R'_A - \sum A = 10.375 - [(18.75) + (-15.75) + (-36.00)] = \frac{43.375}{EI}\ \text{(slope, clockwise)}$$ $$M' = R'_A\,x - \sum A(x-\bar x) = 114.125 - [(18.75)(7.667) + (-15.75)(3.286) + (-36.00)(2.250)] = \frac{103.125}{EI}\ \text{(deflection, downward)}$$

Answer: deflection at D $= \dfrac{20.625}{EI}$ downward; deflection at the free end B $= \dfrac{103.125}{EI}$ downward (kNm$^3$/EI).