Conjugate beam: overhanging beam free end

Given data

Beam ABCD: A hinge at 0, B at 6 m (300 kN), C roller at 12 m, D free end at 14 m with 150 kN. $AC$ ($= 12$ m) has $2EI$; overhang $CD = 2$ m has $EI$.

      300 kN                          150 kN
 A ------v------------------ C --------v D
 ^   6 m B      6 m   (2EI)  ^  2 m (EI)

Step 1: Reactions

$R_{A} = 125.00$ kN, $R_{C} = 325.00$ kN (positive = upward; check: $\sum F_y = 450.00$ kN equals the total load 450.00 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 $x = 12$ m (conjugate moment is zero there):

$$R'_A\times12 = (1125.00)(8.000) + (675.00)(5.333) = 12600.00$$ $$R'_A = \frac{1050.000}{EI}\ \text{(slope at A, clockwise)}$$

At B (300 kN) ($x = 6$ m):

$$V' = R'_A - \sum A = 1050.000 - [(1125.00)] = \frac{-75.000}{EI}\ \text{(slope, anticlockwise)}$$ $$M' = R'_A\,x - \sum A(x-\bar x) = 6300.000 - [(1125.00)(2.000)] = \frac{4050.000}{EI}\ \text{(deflection, downward)}$$

At D (free end) ($x = 14$ m):

$$V' = R'_A - \sum A = 1050.000 - [(1125.00) + (675.00) + (-300.00)] = \frac{-450.000}{EI}\ \text{(slope, anticlockwise)}$$ $$M' = R'_A\,x - \sum A(x-\bar x) = 14700.000 - [(1125.00)(10.000) + (675.00)(7.333) + (-300.00)(1.333)] = \frac{-1100.000}{EI}\ \text{(deflection, upward)}$$

Answer: at the free end D the deflection is $\dfrac{1100}{EI}$ upward (the 150 kN load is small compared with the span load, so the overhang rises), and the rotation is $\dfrac{450}{EI}$ anticlockwise (kNm$^3$/EI and kNm$^2$/EI).