Conjugate beam: slopes at supports, deflection at E

Given data

Beam ABC: $AB = 20$ m (hinge at A, roller at B) with UDL 5 kN/m; overhang $BC = 3$ m with 3 kN at C. E is 5 m from A. $EI$ constant.

Step 1: Reactions

$R_{A} = 49.55$ kN, $R_{B} = 53.45$ kN (positive = upward; check: $\sum F_y = 103.00$ kN equals the total load 103.00 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 real beam has a hinge at A, a roller at the second support and a free end at the overhang tip, so the conjugate beam has a hinge at A, an internal hinge at the support and a fixed end at the tip. 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 $x = 20$ m (conjugate moment is zero there):

$$R'_A\times20 = (3243.33)(10.092) = 32733.33$$ $$R'_A = \frac{1636.667}{EI}\ \text{(slope at A, clockwise)}$$

At support A ($x = 0$ m):

$$V' = R'_A - \sum A = 1636.667 - [0] = \frac{1636.667}{EI}\ \text{(slope, clockwise)}$$ $$M' = R'_A\,x - \sum A(x-\bar x) = 0.000 - [0] = \frac{0.000}{EI}\ \text{(deflection, zero)}$$

At E (5 m from A) ($x = 5$ m):

$$V' = R'_A - \sum A = 1636.667 - [(515.21)] = \frac{1121.458}{EI}\ \text{(slope, clockwise)}$$ $$M' = R'_A\,x - \sum A(x-\bar x) = 8183.333 - [(515.21)(1.751)] = \frac{7281.250}{EI}\ \text{(deflection, downward)}$$

At support B ($x = 20$ m):

$$V' = R'_A - \sum A = 1636.667 - [(3243.33)] = \frac{-1606.667}{EI}\ \text{(slope, anticlockwise)}$$ $$M' = R'_A\,x - \sum A(x-\bar x) = 32733.333 - [(3243.33)(10.092)] = \frac{0.000}{EI}\ \text{(deflection, zero)}$$

At C (free end) ($x = 23$ m):

$$V' = R'_A - \sum A = 1636.667 - [(3243.33) + (-13.50)] = \frac{-1593.167}{EI}\ \text{(slope, anticlockwise)}$$ $$M' = R'_A\,x - \sum A(x-\bar x) = 37643.333 - [(3243.33)(13.092) + (-13.50)(2.000)] = \frac{-4793.000}{EI}\ \text{(deflection, upward)}$$

Answer: slope at A $= \dfrac{1636.67}{EI}$ (clockwise); slope at B $= \dfrac{1606.67}{EI}$ (anticlockwise); deflection at E $= \dfrac{7281.25}{EI}$ downward (kNm$^2$/EI and kNm$^3$/EI). (At C the deflection is $4793/EI$ upward, which is not asked.)