Moment area: overhang beam free-end deflection

Given data

Overhanging beam ABC: A hinge, B roller, $AB = 4$ m ($2EI$, UDL 10 kN/m), overhang $BC = 2$ m ($EI$) with 100 kN at C.

   10 kN/m (2EI)                100 kN
 A =============== B ------------v C
 ^       4 m       ^    2 m (EI)

Step 1: Reactions

$R_{A} = -30.00$ kN, $R_{B} = 170.00$ kN (positive = upward; check: $\sum F_y = 140.00$ kN equals the total load 140.00 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: Tangent at A

Deviation of the support at $x = 4$ m from the tangent at A (Theorem II, moment of the $M/EI$ area about that support):

$$t = (-173.33)(1.231) = \frac{-213.33}{EI}$$

That support does not deflect, so $\theta_A = -t/4$:

$$\theta_A = \frac{53.333}{EI}\ \text{(anticlockwise)}$$

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

$$\theta = \theta_A + \sum A = 53.333 + [(-173.33) + (-200.00)] = \frac{-320.000}{EI}\ \text{(clockwise)}$$ $$y = \theta_A\,x + t_{P/A} = 320.000 + [(-173.33)(3.231) + (-200.00)(1.333)] = \frac{-506.667}{EI}\ \text{(downward)}$$

Answer: vertical deflection at the free end C $= \dfrac{506.67}{EI}$ (kNm$^3$/EI), downward.