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Chapter 16

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.

Step-by-Step

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.