Conjugate beam: slope and deflection at free end C

Given data

Beam ABC: A hinge, B roller, $AB = 6$ m ($2EI$) with UDL 15 kN/m, overhang $BC = 2$ m ($EI$) with 3 kN at C.

   15 kN/m (2EI)               3 kN
 A =============== B ----------v C
 ^        6 m      ^   2 m (EI)

Step 1: Reactions

$R_{A} = 44.00$ kN, $R_{B} = 49.00$ kN (positive = upward; check: $\sum F_y = 93.00$ kN equals the total load 93.00 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 $x = 6$ m (conjugate moment is zero there):

$$R'_A\times6 = (126.00)(3.071) = 387.00$$ $$R'_A = \frac{64.500}{EI}\ \text{(slope at A, clockwise)}$$

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

$$V' = R'_A - \sum A = 64.500 - [(126.00) + (-6.00)] = \frac{-55.500}{EI}\ \text{(slope, anticlockwise)}$$ $$M' = R'_A\,x - \sum A(x-\bar x) = 516.000 - [(126.00)(5.071) + (-6.00)(1.333)] = \frac{-115.000}{EI}\ \text{(deflection, upward)}$$

Answer: slope at C $= \dfrac{55.5}{EI}$ (anticlockwise) and deflection at C $= \dfrac{115}{EI}$ upward (the hogging moment from the UDL span lifts the overhang more than the 3 kN load pushes it down).