Moment area: slope and deflection at E and at 4 m

Given data

Beam A to E: A hinge at 0, C at 2 m, D at 6 m, B roller at 8 m, E free end at 9 m. UDL 10 kN/m on CD (from 2 m to 6 m), 15 kN at E. $EI$ constant.

          10 kN/m
 A --- C =========== D --- B -- v E
 ^  2 m      4 m       2 m  ^ 1 m 15 kN

Step 1: Reactions

$R_{A} = 18.12$ kN, $R_{B} = 36.88$ kN (positive = upward; check: $\sum F_y = 55.00$ kN equals the total load 55.00 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 $x = 8$ m from the tangent at A (Theorem II, moment of the $M/EI$ area about that support):

$$t = (36.25)(6.667) + (183.33)(4.055) + (13.75)(2.061) = \frac{1013.33}{EI}$$

That support does not deflect, so $\theta_A = -t/8$:

$$\theta_A = \frac{-126.667}{EI}\ \text{(clockwise)}$$

At the point 4 m from A ($x = 4$ m):

$$\theta = \theta_A + \sum A = -126.667 + [(36.25) + (95.42)] = \frac{5.000}{EI}\ \text{(anticlockwise)}$$ $$y = \theta_A\,x + t_{P/A} = -506.667 + [(36.25)(2.667) + (95.42)(0.943)] = \frac{-320.000}{EI}\ \text{(downward)}$$

At E (free end) ($x = 9$ m):

$$\theta = \theta_A + \sum A = -126.667 + [(36.25) + (183.33) + (13.75) + (-7.50)] = \frac{99.167}{EI}\ \text{(anticlockwise)}$$ $$y = \theta_A\,x + t_{P/A} = -1140.000 + [(36.25)(7.667) + (183.33)(5.055) + (13.75)(3.061) + (-7.50)(0.667)] = \frac{101.667}{EI}\ \text{(upward)}$$

Answer: at E: slope $= \dfrac{99.17}{EI}$ (anticlockwise) and deflection $= \dfrac{101.67}{EI}$ upward; at 4 m from A: deflection $= \dfrac{320}{EI}$ downward (kNm$^2$/EI and kNm$^3$/EI).