Conjugate beam: vertical deflection and rotation at C

Given data

Overhanging beam ABC: $AB = 5$ m with UDL 5 kN/m, overhang $BC = 2$ m with 10 kN at C. $EI$ constant.

Step 1: Reactions

$R_{A} = 8.50$ kN, $R_{B} = 26.50$ kN (positive = upward; check: $\sum F_y = 35.00$ kN equals the total load 35.00 kN).

Step 2: Bending moment diagram (sagging +), as $M/EI$ pieces

In AB, $M = 8.5x - 2.5x^2$. This is the free (simply supported) parabola of the UDL plus a hogging triangle caused by the overhang moment $M_B = -10\times2 = -20$ kNm. In BC, $M = 10x - 70$.

(Areas are in units of $1/EI$; the net area on AB is $+2.083$.)

Step 3: Conjugate beam

The real beam has a hinge at A, a roller at B and a free end at C, so the conjugate beam has a simple support at A, an internal hinge at B and a fixed end at C. 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 span AB about the hinge B (conjugate moment is zero there):

$$R'_A\times5 = (52.083)(2.500) + (-50.000)(1.667) = 130.208 - 83.333 = 46.875$$ $$R'_A = \frac{9.375}{EI}\ \text{(slope at A, clockwise)}$$

At C (free end) ($x = 7$ m):

$$V' = R'_A - \sum A = 9.375 - [(52.083) + (-50.000) + (-20.000)] = \frac{27.292}{EI}\ \text{(slope, clockwise)}$$ $$M' = R'_A\,x - \sum A(x-\bar x) = 65.625 - [(52.083)(4.500) + (-50.000)(3.667) + (-20.000)(1.333)] = 65.625 - 24.375 = \frac{41.250}{EI}\ \text{(deflection, downward)}$$

Answer: deflection at the free end C $= \dfrac{41.25}{EI}$ downward; rotation at C $= \dfrac{27.29}{EI}$ rad (clockwise) (kNm$^3$/EI and kNm$^2$/EI).