Conjugate beam: slope at A, deflection at C

Given data

Beam ABC: A hinge, 30 kN at 3 m from A, B roller 3 m beyond the load (at 6 m), overhang $BC = 2$ m with 20 kN at C. $EI$ constant.

Step 1: Reactions

$R_{A} = 8.33$ kN, $R_{B} = 41.67$ kN (positive = upward; check: $\sum F_y = 50.00$ kN equals the total load 50.00 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. In the portion 3 to 6 m the moment changes sign, so its net area is small and negative and its "centroid" falls outside the portion; it is used only to give the correct first moment of that portion.)

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 $x = 6$ m (conjugate moment is zero there):

$$R'_A\times6 = (37.50)(4.000) + (-22.50)(-0.667) = 165.00$$ $$R'_A = \frac{27.500}{EI}\ \text{(slope at A, clockwise)}$$

At the 30 kN load ($x = 3$ m):

$$V' = R'_A - \sum A = 27.500 - [(37.50)] = \frac{-10.000}{EI}\ \text{(slope, anticlockwise)}$$ $$M' = R'_A\,x - \sum A(x-\bar x) = 82.500 - [(37.50)(1.000)] = \frac{45.000}{EI}\ \text{(deflection, downward)}$$

At C (free end) ($x = 8$ m):

$$V' = R'_A - \sum A = 27.500 - [(37.50) + (-22.50) + (-40.00)] = \frac{52.500}{EI}\ \text{(slope, clockwise)}$$ $$M' = R'_A\,x - \sum A(x-\bar x) = 220.000 - [(37.50)(6.000) + (-22.50)(1.333) + (-40.00)(1.333)] = \frac{78.333}{EI}\ \text{(deflection, downward)}$$

Answer: slope at A $= \dfrac{27.5}{EI}$ (clockwise); deflection at C $= \dfrac{78.33}{EI}$ downward (kNm$^2$/EI and kNm$^3$/EI).