Conjugate beam: deflection at B, slope at D

Given data

Beam ABCD: A hinge at 0, B at 4 m (50 kN down), C roller at 8 m, D free end at 11 m. $AB$ and $BC$ have $EI$; $BC$ carries 20 kN/m; overhang $CD = 3$ m has $2EI$ with 30 kN at D. Required: deflection at B and slope at D.

      50 kN    20 kN/m                30 kN
 A ----v---------------------- C ---------v D
 ^ 4 m B        4 m (EI)       ^   3 m (2EI)

Step 1: Reactions

$R_{A} = 33.75$ kN, $R_{C} = 126.25$ kN (positive = upward; check: $\sum F_y = 160.00$ kN equals the total load 160.00 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 C 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):

$$R'_A\times8 = (270.00)(5.333) + (196.67)(3.525) = 2133.33$$ $$R'_A = \frac{266.667}{EI}\ \text{(slope at A, clockwise)}$$

At B ($x = 4$ m), deflection needed:

$$V' = R'_A - \sum A = 266.667 - [(270.00)] = \frac{-3.333}{EI}\ \text{(slope, anticlockwise)}$$ $$M' = R'_A\,x - \sum A(x-\bar x) = 1066.667 - [(270.00)(1.333)] = \frac{706.667}{EI}\ \text{(deflection, downward)}$$

At D ($x = 11$ m), slope needed:

$$V' = R'_A - \sum A = 266.667 - [(270.00) + (196.67) + (-67.50)] = \frac{-132.500}{EI}\ \text{(slope, anticlockwise)}$$ $$M' = R'_A\,x - \sum A(x-\bar x) = 2933.333 - [(270.00)(8.333) + (196.67)(6.525) + (-67.50)(2.000)] = \frac{-465.000}{EI}\ \text{(deflection, upward)}$$

Answer: deflection at B $= \dfrac{706.67}{EI}$ downward; slope at D $= \dfrac{132.5}{EI}$ (anticlockwise, i.e. the free end tilts upward). (Values in kNm$^3$/EI and kNm$^2$/EI; with $EI$ in kNm$^2$ they are metres and radians.)