Moment area: deflection at D

Given data

Beam ABCD: A hinge at 0, B at 2.5 m (20 kN), C roller at 5 m, D free end at 7 m with 10 kN. $EI$ constant.

      20 kN                     10 kN
 A ----v------------- C --------v D
 ^ 2.5 m B   2.5 m    ^   2 m

Step 1: Reactions

$R_{A} = 6.00$ kN, $R_{C} = 24.00$ kN (positive = upward; check: $\sum F_y = 30.00$ kN equals the total load 30.00 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 $x = 5$ m from the tangent at A (Theorem II, moment of the $M/EI$ area about that support):

$$t = (18.75)(3.333) + (-6.25)(-1.667) = \frac{72.92}{EI}$$

That support does not deflect, so $\theta_A = -t/5$:

$$\theta_A = \frac{-14.583}{EI}\ \text{(clockwise)}$$

At B ($x = 2.5$ m):

$$\theta = \theta_A + \sum A = -14.583 + [(18.75)] = \frac{4.167}{EI}\ \text{(anticlockwise)}$$ $$y = \theta_A\,x + t_{P/A} = -36.458 + [(18.75)(0.833)] = \frac{-20.833}{EI}\ \text{(downward)}$$

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

$$\theta = \theta_A + \sum A = -14.583 + [(18.75) + (-6.25) + (-20.00)] = \frac{-22.083}{EI}\ \text{(clockwise)}$$ $$y = \theta_A\,x + t_{P/A} = -102.083 + [(18.75)(5.333) + (-6.25)(0.333) + (-20.00)(1.333)] = \frac{-30.833}{EI}\ \text{(downward)}$$

Answer: vertical deflection at D $= \dfrac{30.83}{EI}$ (kNm$^3$/EI), downward.