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## Q. 16.1

Determine the reactions and draw the shear and bending moment diagrams for the two-span continuous beam shown in $Fig. 16.5(a)$ by the slope-deflection method.

## Verified Solution

Degrees of Freedom From Fig. 16.5(a), we can see that only joint $B$ of the beam is free to rotate. Thus, the structure has only one degree of freedom, which is the unknown joint rotation, $\theta _B$.

Fixed-End Moments By using the fixed-end moment expressions given inside the back cover of the book, we evaluate the fixed-end moments due to the external loads for each member:

$FEM_{AB}=\frac{Pab^2}{L^2}=\frac{18(10)(15)^2}{(25)^2}=64.8 k-ft \circlearrowleft$        or       $+64.8 k-ft$

$FEM_{BA}=\frac{Pa^2b}{L^2}=\frac{18(10)^2(15)}{(25)^2}=43.2 k-ft \circlearrowright$        or         $-43.2 k-ft$

$FEM_{BC}=\frac{wL^2}{12}=\frac{2(30)^2}{12}=150 k-ft \circlearrowleft$            or           $+150 k-ft$

$FEM_{CB} = 150 k-ft \circlearrowright$        or        $-150 k-ft$

Note that in accordance with the slope-deflection sign convention, the counterclockwise fixed-end moments are considered as positive, whereas the clockwise fixed-end moments are considered to be negative.

Chord Rotations Since no support settlements occur, the chord rotations of both members are zero; that is, $\psi _{AB} = \psi _{BC} = 0$.

Slope-Deflection Equations To relate the member end moments to the unknown joint rotation, $\theta _B$, we write the slope-deflection equations for the two members of the structure by applying Eq. ($M_{nf}=\frac{2EI}{L}(2\theta_n + \theta_f – 3\psi) + FEM_{nf}$). Note that since the supports $A$ and $C$ are fixed, the rotations $\theta _A = \theta _C = 0$. Thus the slope-deflection equations for member $AB$ can be expressed as

$M_{AB}=\frac{2EI}{25}(\theta _B) + 64.8 = 0.08EI\theta _B + 64.8$              (1)

$M_{BA}=\frac{2EI}{25}(2\theta _B) – 43.2 = 0.16EI\theta _B – 43.2$              (2)

Similarly, by applying Eq. ($M_{nf}=\frac{2EI}{L}(2\theta_n + \theta_f – 3\psi) + FEM_{nf}$) for member $BC$, we obtain the slope-deflection equations

$M_{BC}=\frac{2EI}{30}(2\theta _B) + 150 = 0.133EI\theta _B + 150$             (3)

$M_{CB}=\frac{2EI}{30}(\theta _B) – 150 = 0.0667EI\theta _B – 150$                 (4)

Equilibrium Equation The free-body diagram of joint $B$ is shown in Fig. 16.5(b). Note that the member end moments, which are assumed to be in a counterclockwise direction on the ends of the members, must be applied in the (opposite) clockwise direction on the free body of the joint, in accordance with Newton’s third law. By applying the moment equilibrium equation $\sum{M_B}=0$ to the free body of joint $B$, we obtain the equilibrium equation

$M_{BA} + M_{BC} = 0$                                 (5)

Joint Rotation To determine the unknown joint rotation, $\theta _B$, we substitute the slope-deflection equations (Eqs. (2) and (3)) into the equilibrium equation (Eq. (5)) to obtain

$(0.16EI\theta _B – 43.2) + (0.133EI\theta _B + 150) = 0$

or

$0.293EI\theta _B = -106.8$

from which

$EI\theta _B = -364.5 k-ft^2$

Member End Moments The member end moments can now be computed by substituting the numerical value of $EI\theta _B$ back into the slope-deflection equations (Eqs. (1) through (4)). Thus,

$M_{AB} = 0.08(-364.5) + 64.8 = 35.6 k-ft \circlearrowleft$

$M_{BA} = 0.16(-364.5) – 43.2 = -101.5 k-ft$          or     $101.5 k-ft \circlearrowright$

$M_{BC} = 0.133(-364.5) + 150 = 101.5 k-ft \circlearrowleft$

$M_{CB} = 0.0667(-364.5) – 150 = -174.3 k-ft$         or        $174.3 k-ft \circlearrowright$

Note that a positive answer for an end moment indicates that its sense is counterclockwise, whereas a negative answer for an end moment implies a clockwise sense. Since the end moments $M_{BA}$ and $M_{BC}$ are equal in magnitude but opposite in sense, the equilibrium equation, $M_{BA} + M_{BC} = 0$, is indeed satisfied.

Member End Shears The member end shears, obtained by considering the equilibrium of each member, are shown in Fig. 16.5(c).

Support Reactions The reactions at the fixed supports $A$ and $C$ are equal to the forces and moments at the ends of the members connected to these joints. To determine the reaction at the roller support $B$, we consider the equilibrium of the free body of joint $B$ in the vertical direction (see Fig. 16.5(c)), to obtain

$B_y = S_{BA} + S_{BC} = 9.84 + 27.57 = 37.41 k \uparrow$

The support reactions are shown in Fig. 16.5(d).                                   Ans.

Equilibrium Check To check our calculations of member end shears and support reactions, we apply the equations of equilibrium to the free body of the entire structure. Thus (see Fig. 16.5(d)),

$\uparrow \sum{F_y}=0$

$8.16 – 18 + 37.41 – 2(30) + 32.43 = 0$                           Checks

$+\circlearrowleft \sum{M_C}=0$

$35.6 – 8.16(55) + 18(45) – 37.41(30) + 2(30)(15) – 174.3 = 0.2 \approx 0$              Checks

Shear and Bending Moment Diagrams The shear and bending moment diagrams can now be constructed by using the beam sign convention described in Section 5.1. These diagrams are shown in Fig. 16.5(e) and (f ).                    Ans.