Moment area: B, C, maximum deflection

Given data

Simply supported beam ABCD, span $AD = 10$ m: A hinge, D roller, $AB = 3$ m (200 kN at B), $BC = 5$ m (300 kN at C), $CD = 2$ m. $EI$ constant. $x$ is measured from A.

       200 kN         300 kN
 A ------v--------------v----- D
 ^  3 m  B     5 m      C 2 m  ^

Step 1: Reactions

$R_{A} = 200.00$ kN, $R_{D} = 300.00$ kN (positive = upward; check: $\sum F_y = 500.00$ kN equals the total load 500.00 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 $x = 10$ m from the tangent at A (Theorem II, moment of the $M/EI$ area about that support):

$$t = (900.00)(8.000) + (3000.00)(4.500) + (600.00)(1.333) = \frac{21500.00}{EI}$$

That support does not deflect, so $\theta_A = -t/10$:

$$\theta_A = \frac{-2150.000}{EI}\ \text{(clockwise)}$$

At B ($x = 3$ m):

$$\theta = \theta_A + \sum A = -2150.000 + [(900.00)] = \frac{-1250.000}{EI}\ \text{(clockwise)}$$ $$y = \theta_A\,x + t_{P/A} = -6450.000 + [(900.00)(1.000)] = \frac{-5550.000}{EI}\ \text{(downward)}$$

At C ($x = 8$ m):

$$\theta = \theta_A + \sum A = -2150.000 + [(900.00) + (3000.00)] = \frac{1750.000}{EI}\ \text{(anticlockwise)}$$ $$y = \theta_A\,x + t_{P/A} = -17200.000 + [(900.00)(6.000) + (3000.00)(2.500)] = \frac{-4300.000}{EI}\ \text{(downward)}$$

At the point of maximum deflection ($x = 5.08333$ m):

$$\theta = \theta_A + \sum A = -2150.000 + [(900.00) + (1250.00)] = \frac{0.000}{EI}\ \text{(zero)}$$ $$y = \theta_A\,x + t_{P/A} = -10929.167 + [(900.00)(3.083) + (1250.00)(1.042)] = \frac{-6852.083}{EI}\ \text{(downward)}$$

Maximum deflection

The deflection is maximum where the slope is zero. The slope at B is $-1250/EI$ (clockwise) and at C it is $+1750/EI$, so the zero lies between B and C. In BC, $M/EI = 600/EI$ is constant, so the slope changes by $600/EI$ per metre:

$$-1250 + 600\,(x - 3) = 0 \;\Rightarrow\; x = 3 + 2.083 = 5.083\ \text{m from A}$$

The deflection there (Step 3 tangent offset, from the same formula) is

$$y_{max} = \theta_A x + t_{P/A} = \frac{-6852.08}{EI}$$

Answer: deflection at B $= 5550/EI$ downward; at C $= 4300/EI$ downward (kNm$^3$/EI). Maximum deflection $= 6852.1/EI$ downward at $5.083$ m from A (and 4.917 m from D).