Question 16.2: Determine the reactions and draw the shear and bending momen...
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.
Degrees of Freedom θ_B and θ_C
Fixed-End Moments
FEM_{AB} = \frac{27(6)^2}{30} = 32.4 kN-m\circlearrowleft or + 32.4 kN-m
FEM_{BA} = \frac{27(6)^2}{20} = 48.6 kN-m\circlearrowright or – 48.6 kN-m
FEM_{BC} = \frac{27(6)^2}{12} = 81 kN-m\circlearrowleft or + 81 kN-m
FEM_{CB} = 81 kN-m \circlearrowright or -81 kN-m
FEM_{CD} = \frac{27(6)^2}{20} = 48.6 kN-m\circlearrowleft or + 48.6 kN-m
FEM_{DC} = \frac{27(6)^2}{30} = 32.4 kN-m\circlearrowright or – 32.4 kN-m
Slope-Deflection Equations Using Eq. (16.9) for members AB, BC, and CD, we write
M_{nf} = \frac{2EI}{L}(2θ_n + θ_f – 3ψ) + FEM_{nf} (16.9)
M_{AB} = \frac{2EI}{6}(θ_B) + 32.4 = 0.33EIθ_B + 32.4 (1)
M_{BA} =\frac{2EI}{6}(2θ_B) – 48.6 = 0.67EIθ_B – 48.6 (2)
M_{BA} =\frac{2EI}{6}(2θ_B + θ_C) + 81 = 0.67EIθ_B + 0.33EIθ_C + 81 (3)
M_{CB} =\frac{2EI}{6}(θ_B + 2θ_C) – 81 = 0.33EIθ_B + 0.67EIθ_C – 81 (4)
M_{CD} =\frac{2EI}{6}(2θ_C) + 48.6 = 0.67EIθ_C + 48.6 (5)
M_{DC} =\frac{2EI}{6}(θ_C) – 32.4 = 0.33EIθ_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
1.34EIθ_B + 0.33EIθ_C = -32.4 (9)
0.33EIθ_B + 1.34EIθ_C = 32.4 (10)
By solving Eqs. (9) and (10) simultaneously, we determine the values of EIθ_B and EIθ_C to be
EIθ_B = -32.08 kN-m^2EIθ_C = 32.08 kN-m^2
Member End Moments To compute the member end moments, we substitute the numerical values of EIθ_B and EIθ_C back into the slope-deflection equations (Eqs. (1) through (6)) to obtain
M_{AB} = 0.33(-32.08) + 32.4 = 21.6 kN-m\circlearrowleft
M_{BA} = 0.67(-32.08) – 48.6 = -70.2 kN-m or 70.2 kN-m \circlearrowright
M_{BC} = 0.67(-32.08) + 0.33(32.08) + 81 = 70.2 kN-m \circlearrowleft
M_{CB} = 0.33(-32.08) + 0.67(32.08) – 81
= -70.2 kN-m or 70.2 kN-m \circlearrowright
M_{CD} = 0.67(32.08) + 48.6 = 70.2 kN-m\circlearrowleft
M_{DC} = 0.33(32.08) – 32.4 = -21.6 kN-m or 21.6 kN-m \circlearrowright
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).
Equilibrium Check The equilibrium equations check.
Shear and Bending Moment Diagrams See Fig. 16.6(e) and (f ).
