Conjugate beam: slope and deflection at free end C

Given data

Beam ABC: A hinge at 0, B roller at 8 m. UDL 2 kN/m over AB; 10 kN at D (4 m from A); overhang $BC = 4$ m with 5 kN at 2 m from B (at 10 m from A). C is the free end at 12 m. $EI$ constant.

        10 kN   2 kN/m          5 kN
 A ------v============ B ------v------- C
 ^  4 m  D     4 m     ^  2 m     2 m

Step 1: Reactions

$R_{A} = 11.75$ kN, $R_{B} = 19.25$ kN (positive = upward; check: $\sum F_y = 31.00$ kN equals the total load 31.00 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):

$$R'_A\times8 = (72.67)(5.431) + (52.67)(3.038) = 554.67$$ $$R'_A = \frac{69.333}{EI}\ \text{(slope at A, clockwise)}$$

At D (10 kN) ($x = 4$ m):

$$V' = R'_A - \sum A = 69.333 - [(72.67)] = \frac{-3.333}{EI}\ \text{(slope, anticlockwise)}$$ $$M' = R'_A\,x - \sum A(x-\bar x) = 277.333 - [(72.67)(1.431)] = \frac{173.333}{EI}\ \text{(deflection, downward)}$$

At C (free end) ($x = 12$ m):

$$V' = R'_A - \sum A = 69.333 - [(72.67) + (52.67) + (-10.00)] = \frac{-46.000}{EI}\ \text{(slope, anticlockwise)}$$ $$M' = R'_A\,x - \sum A(x-\bar x) = 832.000 - [(72.67)(9.431) + (52.67)(7.038) + (-10.00)(3.333)] = \frac{-190.667}{EI}\ \text{(deflection, upward)}$$

Answer: slope at C $= \dfrac{46}{EI}$ (anticlockwise); deflection at C $= \dfrac{190.67}{EI}$ upward (kNm$^2$/EI and kNm$^3$/EI). (The deflection under the 10 kN load at D is $173.33/EI$ downward.)