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## Q. 16.2

Determine the reactions and draw the shear and bending moment diagrams for the three-span continuous beam shown in $Fig. 16.6(a)$ by the slope-deflection method.

## Verified Solution

Degrees of Freedom $\theta_B$ and $\theta_C$.

Fixed-End Moments

$FEM_{AB} = \frac{3(18)^2}{30}= 32.4 k-ft \circlearrowleft$          or       $+32.4 k-ft$

$FEM_{BA} = \frac{3(18)^2}{20}= 48.6 k-ft \circlearrowright$          or        $-48.6 k-ft$

$FEM_{BC} = \frac{3(18)^2}{12}= 81 k-ft \circlearrowleft$            or          $-81 k-ft$

$FEM_{CB} = 81 k-ft \circlearrowright$           or        $+81 k-ft$

$FEM_{CD} = \frac{3(18)^2}{20}= 48.6 k-ft \circlearrowleft$          or        $+48.6 k-ft$

$FEM_{DC} = \frac{3(18)^2}{30}= 32.4 k-ft \circlearrowright$          or       $-32.4 k-ft$

Slope-Deflection Equations Using Eq. ($M_{nf}=\frac{2EI}{L}(2\theta_n + \theta_f – 3\psi) + FEM_{nf}$) for members $AB, BC$, and $CD$, we write

$M_{AB}=\frac{2EI}{18}(\theta_B) + 32.4 = 0.111EI \theta_B + 32.4$               (1)

$M_{BA}=\frac{2EI}{18}(2\theta_B) – 48.6 = 0.222EI \theta_B – 48.6$               (2)

$M_{BC}=\frac{2EI}{18}(2\theta_B+\theta_C) + 81 = 0.222EI \theta_B +0.111EI \theta_C + 81$               (3)

$M_{CB}=\frac{2EI}{18}(\theta_B+2\theta_C) – 81 = 0.111EI \theta_B +0.222EI \theta_C – 81$               (4)

$M_{CD}=\frac{2EI}{18}(2\theta_C) + 48.6 = 0.222EI \theta_C + 48.6$               (5)

$M_{AB}=\frac{2EI}{18}(\theta_C) – 32.4 = 0.111EI \theta_C – 32.4$               (6)

Equilibrium Equations See Fig. 16.6(b).

$M_{BA} + M_{BC} = 0$                (7)

$M_{CB} + M_{CD} = 0$                (8)

Joint Rotations By substituting the slope-deflection equations (Eqs. (1) through (6)) into the equilibrium equations (Eqs. (7) and (8)), we obtain

$0.444EI\theta_B + 0.111EI\theta_C = -32.4$                 (9)

$0.111EI\theta_B + 0.444EI\theta_C = 32.4$                 (10)

By solving Eqs. (9) and (10) simultaneously, we determine the values of $EI\theta_B$ and $EI\theta_C$ to be

$EI\theta_B = -97.3 k-ft^2$

$EI\theta_C = 97.3 k-ft^2$

Member End Moments To compute the member end moments, we substitute the numerical values of $EI\theta_B$ and $EI\theta_C$ back into the slope-deflection equations (Eqs. (1) through (6)) to obtain

$M_{AB} = 0.111(-97.3) + 32.4 = 21.6 k-ft \circlearrowleft$          Ans.

$M_{BA} = 0.222(-97.3) – 48.6 = -70.2 k-ft$      or     $70.2 k-ft \circlearrowright$                Ans.

$M_{BC} = 0.222(-97.3) + 0.111(97.3) + 81 = 70.2 k-ft \circlearrowleft$         Ans.

$M_{CB} = 0.111(-97.3) + 0.222(97.3) – 81$

$= -70.2 k-ft$          or         $70.2 k-ft \circlearrowright$                    Ans.

$M_{CD} = 0.222(97.3) + 48.6 = 70.2 k-ft \circlearrowleft$                Ans.

$M_{DC} = 0.111(97.3) – 32.4 = -21.6 k-ft$        or      $21.6 k-ft \circlearrowright$          Ans.

Note that the numerical values of $M_{BA} , M_{BC} , M_{CB}$, and $M_{CD}$ do satisfy the equilibrium equations (Eqs. (7) and (8)).

Member End Shears and Support Reactions See Fig. 16.6(c) and (d). Ans.

Equilibrium Check The equilibrium equations check.

Shear and Bending Moment Diagrams See Fig. 16.6(e) and (f ).        Ans.