Question 16.6: Determine the member end moments and reactions for the three...
Determine the member end moments and reactions for the three-span continuous beam shown in Fig. 16.10(a) due to the uniformly distributed load and due to the support settlements of 15 mm at B, 36 mm at C, and 18 mm at D. Use the slope-deflection method.
Degrees of Freedom Although all four joints of the beam are free to rotate, we will eliminate the rotations of the simple supports at the ends A and D from the analysis by using the modified slope-deflection equations for members AB and CD, respectively. Thus, the analysis will involve only two unknown joint rotations, θ_B and θ_C.
Fixed-End Moments
FEM_{AB} = FEM_{BC} = FEM_{CD} = \frac{32(5)^2}{12} = 66.7 kN-m \circlearrowleft or +66.7 kN-m
FEM_{BA} = FEM_{CB} = FEM_{DC} = 66.7 kN-m \circlearrowright or -66.7 kN-m
Chord Rotations The specified support settlements are depicted in Fig. 16.10(b) using an exaggerated scale. The inclined dashed lines in this figure indicate the chords (not the elastic curves) of the members in the deformed positions. It can be seen from this figure that since support A does not settle but support B settles by 15 mm the relative settlement between the two ends of member AB is 15 mm = 0.015 m. Because the length of member AB is 5 m, the rotation of the chord of member AB is
ψ_{AB} = -\frac{0.015}{5} = -0.003in which the negative sign has been assigned to the value of ψ_{AB} to indicate that its direction is clockwise, as shown in Fig. 16.10(b). The chord rotation for member BC can be computed in a similar manner by using the settlement of supports B and C. From Fig. 16.10(b), we observe that the relative settlement between the ends of member BC is 36 mm -15 mm = 21 mm = 0.021 m, and so
ψ_{BC} = -\frac{0.021}{5} = -0.0042Similarly, the chord rotation for member CD is
ψ_{CD} = -\frac{0.036 – 0.018}{5} = 0.0036Slope-Deflection Equations
M_{AB} = 0
M_{BA} =\frac{3EI}{5}(θ_B + 0.003) – 100 = 0.6EIθ_B + 0.0018EI – 100 (1)
M_{BC} =\frac{2EI}{5}[2θ_B + θ_C – 3(-0.0042)] + 66.7
= 0.8EIθ_B + 0.4EIθ_C + 0.00504EI + 66.7 (2)
M_{CB} =\frac{2EI}{5}[2θ_C + θ_B – 3(-0.0042)] + 66.7
=0.4EIθ_B + 0.8EIθ_C + 0.00504EI – 66.7 (3)
M_{CD} =\frac{3EI}{5}(θ_C – 0.0036) + 100 = 0.6EIθ_C – 0.00216EI + 100 (4)
M_{DC} = 0Equilibrium Equations See Fig. 16.10(c).
M_{BA} + M_{BC} = 0 (5)
M_{CB} + M_{CD} = 0 (6)
Joint Rotations By substituting the slope-deflection equations (Eqs. (1) through (4)) into the equilibrium equations (Eqs. (5) and (6)), we obtain
1.4EIθ_B + 0.4EIθ_C = -0.00684EI + 33.30.4EIθ_B + 1.4EIθ_C = -0.00288EI – 33.3
Substituting EI = 341000 kN-m² into the right sides of the above equations yields
1.4EIθ_B + 0.4EIθ_C = -2291.1 (7)
0.4EIθ_B + 1.4EIθ_C = -1015.4 (8)
By solving Eqs. (7) and (8) simultaneously, we determine the values of EIθ_B and EIθ_C to be
EIθ_B = -1562.6 kN-m^2EIθ_C = -278.8 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 (4)) to obtain
M_{BA} = -423.71 kN-m or 423.71 kN-m \circlearrowright
M_{BC} = 423.71 kN-m \circlearrowleft
M_{CB} = 803.84 kN-m \circlearrowleft
M_{CD} = -803.84 kN-m or 803.84 kN-m \circlearrowright
Member End Shears and Support Reactions See Fig. 16.10(d) and (e).
Equilibrium Check The equilibrium equations check.
Theoretically, the slope-deflection method and the method of consistent deformations should yield identical results for a given structure.
