Moment area: slope and deflection under 50 kN

Given data

Simply supported beam AB, $L = 12$ m. $AC = CD = DB = 4$ m. 50 kN at D (8 m from A). $EI$ constant.

                       50 kN
 A ------- C ------- D v ----- B
 ^   4 m     4 m         4 m   ^

Step 1: Reactions

$R_{A} = 16.67$ kN, $R_{B} = 33.33$ kN (positive = upward; check: $\sum F_y = 50.00$ kN equals the total load 50.00 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 $x = 12$ m from the tangent at A (Theorem II, moment of the $M/EI$ area about that support):

$$t = (533.33)(6.667) + (266.67)(2.667) = \frac{4266.67}{EI}$$

That support does not deflect, so $\theta_A = -t/12$:

$$\theta_A = \frac{-355.556}{EI}\ \text{(clockwise)}$$

At C ($x = 4$ m):

$$\theta = \theta_A + \sum A = -355.556 + [(133.33)] = \frac{-222.222}{EI}\ \text{(clockwise)}$$ $$y = \theta_A\,x + t_{P/A} = -1422.222 + [(133.33)(1.333)] = \frac{-1244.444}{EI}\ \text{(downward)}$$

At D ($x = 8$ m):

$$\theta = \theta_A + \sum A = -355.556 + [(533.33)] = \frac{177.778}{EI}\ \text{(anticlockwise)}$$ $$y = \theta_A\,x + t_{P/A} = -2844.444 + [(533.33)(2.667)] = \frac{-1422.222}{EI}\ \text{(downward)}$$

At the point of maximum deflection ($x = 6.532$ m):

$$\theta = \theta_A + \sum A = -355.556 + [(355.56)] = 0\ \text{(zero slope)}$$ $$y = \theta_A\,x + t_{P/A} = -2322.489 + [(355.56)(2.177)] = \frac{-1548.319}{EI}\ \text{(downward)}$$

Slope at the supports

$\theta_A = 355.56/EI$ (clockwise), as found above. At B ($x = 12$ m): $\theta_B = \theta_A + \text{total area} = -355.56 + 800 = 444.44$, so $\theta_B = \dfrac{444.44}{EI}$ (anticlockwise). (Total area of the $M/EI$ triangle $= \tfrac12\times12\times133.33 = 800$.)

Maximum deflection

The slope is zero in the portion AD, where $M = 16.667x$. Slope: $\theta(x) = -355.56 + 8.333x^2$. Setting it to zero:

$$x = \sqrt{\frac{355.56}{8.333}} = 6.532\ \text{m from A}$$

This lies in AD, so the assumption is valid. The deflection there, from the same formula, is $y_{max} = \dfrac{1548.3}{EI}$ downward.

Answer: $\theta_A = 355.56/EI$ (clockwise), $\theta_B = 444.44/EI$ (anticlockwise); deflection at C $= 1244.4/EI$, at D $= 1422.2/EI$ (both downward); maximum deflection $= 1548.3/EI$ downward at $6.532$ m from A.