Conjugate beam: free end of overhang, UDL and point load

Given data

Simply supported span $AB = 4$ m with an overhang $BC = 2$ m. UDL 2 kN/m over AB; 3 kN at the free end C. $EI$ constant.

   2 kN/m                  3 kN
 A ============ B --------v C
 ^    4 m       ^   2 m

Step 1: Reactions

$R_{A} = 2.50$ kN, $R_{B} = 8.50$ kN (positive = upward; check: $\sum F_y = 11.00$ kN equals the total load 11.00 kN).

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

In AB, $M = 2.5x - x^2 = (4x - x^2) - 1.5x$: the free (simply supported) parabola of the UDL plus a hogging triangle caused by the overhang moment $M_B = -3\times2 = -6$ kNm. In BC, $M = 3x - 18$.

(Areas are in units of $1/EI$; the net area on AB is $-1.333$.)

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):

$$R'_A\times4 = (10.667)(2.000) + (-12.000)(1.333) = 21.333 - 16.000 = 5.333$$ $$R'_A = \frac{1.333}{EI}\ \text{(slope at A, clockwise)}$$

At support B ($x = 4$ m):

$$V' = R'_A - \sum A = 1.333 - [(10.667) + (-12.000)] = \frac{2.667}{EI}\ \text{(slope, clockwise)}$$

(The conjugate moment at B is zero, so the deflection at B is zero, as it must be.)

At the free end C ($x = 6$ m):

$$V' = R'_A - \sum A = 1.333 - [(10.667) + (-12.000) + (-6.000)] = \frac{8.667}{EI}\ \text{(slope, clockwise)}$$ $$M' = R'_A\,x - \sum A(x-\bar x) = 8.000 - [(10.667)(4.000) + (-12.000)(3.333) + (-6.000)(1.333)] = 8.000 + 5.333 = \frac{13.333}{EI}\ \text{(deflection, downward)}$$

(Check: $\theta_B\times2 + \dfrac{Pa^3}{3EI} = 2.667\times2 + \dfrac{3\times8}{3} = 13.33$.)

Answer: deflection at the free end C $= \dfrac{13.33}{EI}$ downward (kNm$^3$/EI). Slope at support A $= \dfrac{1.333}{EI}$ (clockwise); slope at support B $= \dfrac{2.667}{EI}$ (clockwise) (kNm$^2$/EI). The slope at the free end C is $\dfrac{8.667}{EI}$ (clockwise).