Conjugate beam: slope and deflection at free end

Given data

Beam ABC: A hinge at the left end, D at 4 m from A (24 kN down), B roller support at 6 m from A, overhang BC = 4 m (stiffness $2EI$) with UDL 10 kN/m. Portion AB has $EI$. Required: slope and deflection at the free end C.

        24 kN                    10 kN/m
 A ------v------- B ======================= C
 ^   4 m   D 2 m  ^        4 m (2EI)
   (EI)

Step 1: Reactions and bending moments

Moments about B: $R_A\times6 = 24\times2 - 40\times2$, so $R_A = -5.333$ kN (downward); $R_B = 24 + 40 + 5.333 = 69.33$ kN.

$$M_D = -5.333\times4 = -21.33\ \text{kNm},\qquad M_B = -5.333\times6 - 24\times2 = -80\ \text{kNm (hogging)}$$

(Check from the overhang: $M_B = -10\times4^2/2 = -80$ kNm.)

Step 2: $M/EI$ diagram (hogging, so conjugate loads act upward)

Step 3: Conjugate beam

Real hinge A stays a hinge; the interior support B becomes an internal hinge; the free end C becomes a fixed end.

Moment about the hinge B of the part AB (moment of the conjugate beam at B = 0):

$$R'_A\times6 = 42.67(6-2.667) + 42.67(6-5) + 58.67(6-5.333) = 142.2 + 42.67 + 39.11 = 224$$ $$R'_A = \theta_A = \frac{37.33}{EI}\ \text{(anticlockwise)}$$

Step 4: Slope at C (shear at fixed end C)

Total upward load on AB $= 42.67 + 42.67 + 58.67 = 144$; shear at B (continuous through the internal hinge):

$$\theta_B = 37.33 - 144 = -\frac{106.67}{EI}$$

Shear at C: $\theta_C = -106.67 - 53.33 = -\dfrac{160}{EI}$

Step 5: Deflection at C (moment at fixed end C)

$$y_C = \theta_B\times4 + (-53.33)(3) = -426.67 - 160 = -\frac{586.67}{EI}$$

The negative sign means downward.

Answer: slope at C $= 160/EI$ (clockwise); deflection at C $= 586.67/EI$ downward (kNm$^2$ and kNm$^3$ per $EI$ respectively; with $EI$ in kNm$^2$ the answers are in rad and m).