Conjugate beam: stepped symmetric beam

Given data

Symmetrical beam ABCD, simply supported at A and D, span 6 m, in three portions: AB (1.5 m, $I$), BC (3 m, $2I$), CD (1.5 m, $I$) as stated in the question (so B and C are 1.5 m from the ends). Loads 20 kN at B and at C. $E = 200$ kN/mm$^2 = 2\times10^8$ kN/m$^2$, $I = 2\times10^{10}$ mm$^4 = 0.02$ m$^4$, so $EI = 4,000,000$ kNm$^2$.

        20 kN        20 kN
 A -----v------------v----- D
 ^ 1.5 m B   3 m (2I) C 1.5 m ^
    (I)                  (I)

Step 1: Reactions

$R_{A} = 20.00$ kN, $R_{D} = 20.00$ kN (positive = upward; check: $\sum F_y = 40.00$ kN equals the total load 40.00 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: Conjugate beam

The conjugate of a simply supported beam is a simply supported beam of the same span. It carries the $M/EI$ diagram of Step 2 as load (sagging $M/EI$ acts downward, hogging acts upward). Conjugate shear $V'$ = slope (positive = clockwise) and conjugate moment $M'$ = deflection (positive = downward).

Step 4: Conjugate reaction at A (slope at A)

Taking moments of the conjugate beam about the support at $x = 6$ m (conjugate moment is zero there):

$$R'_A\times6 = (22.50)(5.000) + (45.00)(3.000) + (22.50)(1.000) = 270.00$$ $$R'_A = \frac{45.000}{EI}\ \text{(slope at A, clockwise)}$$

At A ($x = 0$ m):

$$V' = R'_A - \sum A = 45.000 - [0] = \frac{45.000}{EI}\ \text{(slope, clockwise)}$$ $$M' = R'_A\,x - \sum A(x-\bar x) = 0.000 - [0] = \frac{0.000}{EI}\ \text{(deflection, zero)}$$

At B ($x = 1.5$ m):

$$V' = R'_A - \sum A = 45.000 - [(22.50)] = \frac{22.500}{EI}\ \text{(slope, clockwise)}$$ $$M' = R'_A\,x - \sum A(x-\bar x) = 67.500 - [(22.50)(0.500)] = \frac{56.250}{EI}\ \text{(deflection, downward)}$$

At C ($x = 4.5$ m):

$$V' = R'_A - \sum A = 45.000 - [(22.50) + (45.00)] = \frac{-22.500}{EI}\ \text{(slope, anticlockwise)}$$ $$M' = R'_A\,x - \sum A(x-\bar x) = 202.500 - [(22.50)(3.500) + (45.00)(1.500)] = \frac{56.250}{EI}\ \text{(deflection, downward)}$$

At D ($x = 6$ m):

$$V' = R'_A - \sum A = 45.000 - [(22.50) + (45.00) + (22.50)] = \frac{-45.000}{EI}\ \text{(slope, anticlockwise)}$$ $$M' = R'_A\,x - \sum A(x-\bar x) = 270.000 - [(22.50)(5.000) + (45.00)(3.000) + (22.50)(1.000)] = \frac{0.000}{EI}\ \text{(deflection, zero)}$$

Results (symmetry check)

By symmetry $\theta_D = -\theta_A$ and $y_C = y_B$. With $EI = 4,000,000$ kNm$^2$:

Slopes: A is clockwise, B is clockwise, C is anticlockwise, D is anticlockwise.

Answer: $\theta_A = \theta_D = \dfrac{45}{EI} = 1.125\times10^{-5}$ rad, $\theta_B = \theta_C = \dfrac{22.5}{EI} = 5.625\times10^{-6}$ rad (opposite senses), and $y_B = y_C = \dfrac{56.25}{EI} = 0.0141$ mm downward; $y_A = y_D = 0$.

If the figure's dimensions are used instead (AB = BC = CD = 2 m, with $I$, $2I$, $I$): $\theta_A = \dfrac{60}{EI} = 1.500\times10^{-5}$ rad, $\theta_B = \dfrac{20}{EI} = 5.000\times10^{-6}$ rad, $y_B = y_C = \dfrac{93.33}{EI} = 0.0233$ mm.