Question 16.12: Determine the member end moments, the support reactions, and...
Determine the member end moments, the support reactions, and the horizontal deflection of joint F of the two-story frame shown in Fig. 16.19(a) by the slope-deflection method.
Degrees of Freedom From Fig. 16.19(a), we can see that the joints C, D, E, and F of the frame are free to rotate, and translate in the horizontal direction. As shown in Fig. 16.19(b), the horizontal displacement of the first-story joints C and D is designated as Δ_1, whereas the horizontal displacement of the second-story joints E and F is expressed as Δ_1 + Δ_2, with Δ_2 representing the displacement of the second-story joints relative to the first-story joints. Thus, the frame has six degrees of freedom—that is, θ_C , θ_D , θ_E , θ_F , Δ_1, and Δ_2.
Fixed-End Moments The nonzero fixed-end moments are
FEM_{CD} = FEM_{EF} = 200 kN-mFEM_{CD} = FEM_{EF} = 200 kN-m
Chord Rotations See Fig. 16.19(b).
ψ_{AC} = ψ_{BD} = -\frac{Δ_1}{5}ψ_{CE} = ψ_{DF} = -\frac{Δ_2}{5}
ψ_{CD} = ψ_{EF} = 0
Slope-Deflection Equations Using I_{column} = I and I_{girder} = 2I , we write
M_{AC} = 0.4EIθ_C + 0.24EIΔ_1 (1)
M_{CA} = 0.8EIθ_C + 0.24EIΔ_1 (2)
M_{BD} = 0.4EIθ_D + 0.24EIΔ_1 (3)
M_{DB} = 0.8EIθ_D + 0.24EIΔ_1 (4)
M_{CE} = 0.8EIθ_C + 0.4EIθ_E + 0.24EIΔ_2 (5)
M_{EC} = 0.8EIθ_E + 0.4EIθ_C + 0.24EIΔ_2 (6)
M_{DF} = 0.8EIθ_D + 0.4EIθ_F + 0.24EIΔ_2 (7)
M_{FD} = 0.8EIθ_F + 0.4EIθ_D + 0.24EIΔ_2 (8)
M_{CD} = 0.8EIθ_C + 0.4EIθ_D + 200 (9)
M_{DC} = 0.8EIθ_D + 0.4EIθ_C – 200 (10)
M_{EF} = 0.8EIθ_E + 0.4EIθ_F + 200 (11)
M_{FE} = 0.8EIθ_F + 0.4EIθ_E – 200 (12)
Equilibrium Equations By considering the moment equilibrium of joints C, D, E, and F, we obtain
M_{CA} + M_{CD} + M_{CE} = 0 (13)
M_{DB} + M_{DC} + M_{DF} = 0 (14)
M_{EC} + M_{EF} = 0 (15)
M_{FD} + M_{FE} = 0 (16)
To establish the remaining two equilibrium equations, we successively pass a horizontal section just above the lower ends of the columns of each story of the frame and apply the equation of horizontal equilibrium (∑F_X = 0) to the free body of the portion of the frame above the section. The free-body diagrams thus obtained are shown in Fig. 16.19(c) and (d). By applying the equilibrium equation ∑F_X = 0 to the top story of the frame (Fig. 16.19(c)), we obtain
S_{CE} + S_{DF} = 40Similarly, by applying ∑F_X = 0 to the entire frame (Fig. 16.19(d)), we write
S_{AC} + S_{BD} = 120By expressing column end shears in terms of column end moments as
S_{AC} = \frac{M_{AC} + M_{CA}}{5} S_{BD} = \frac{M_{BD} + M_{DB}}{5}
S_{CE} = \frac{M_{CE} + M_{EC}}{5} S_{DF} = \frac{M_{DF} + M_{FD}}{5}
and by substituting these expressions into the force equilibrium equations, we obtain
M_{CE} + M_{EC} + M_{DF} + M_{FD} = 200 (17)
M_{AC} + M_{CA} + M_{BD} + M_{DB} = 600 (18)
Joint Displacements Substitution of the slope-deflection equations (Eqs. (1) through (12)) into the equilibrium equations (Eqs. (13) through (18)) yields
2.4EIθ_C + 0.4EIθ_D + 0.4EIθ_E + 0.24EIΔ_1 + 0.24EIΔ_2 = -200 (19)
0.4EIθ_C + 2.4EIθ_D + 0.4EIθ_F + 0.24EIΔ_1 + 0.24EIΔ_2 = 200 (20)
0.4EIθ_C + 1.6EIθ_E + 0.4EIθ_F + 0.24EIΔ_2 = -200 (21)
0.4EIθ_D + 0.4EIθ_E + 1.6EIθ_F + 0.24EIΔ_2 = 200 (22)
1.2EIθ_C + 1.2EIθ_D + 1.2EIθ_E + 1.2EIθ_F + 0.96EIΔ_2 = 200 (23)
0.4EIθ_C + 0.4EIθ_D + 0.32EIΔ_1 = 200 (24)
By solving Eqs. (19) through (24) by the Gauss-Jordan elimination method (Appendix B), we determine
EIθ_C = -203.25 kN-m^2EIθ_D = -60.389 kN-m^2
EIθ_E = -197.4 kN-m^2
EIθ_F = 88.31 kN-m^2
EIΔ_1 = 954.55 kN-m^3 or Δ_1 = 18.95 mm →
EIΔ_2 = 674.24 kN-m^3 or Δ_2 = 13.4 mm→
Thus, the horizontal deflection of joint F of the frame is as follows:
Δ_F = Δ_1 + Δ_2 = 18.95 + 13.4 = 32.35 mm →Member End Moments By substituting the numerical values of the joint displacements into the slope-deflection equations (Eqs. (1) through (12)), we obtain
M_{AC} = 147.8 kN-m \circlearrowleftM_{CA} = 66.5 kN-m \circlearrowleft
M_{BD} = 204.9 kN-m \circlearrowleft
M_{DB} = 180.8 kN-m \circlearrowleft
M_{CE} = -79.7 kN-m or 79.7 kN-m \circlearrowright
M_{EC} = -77.4 kN-m or 77.4 kN-m \circlearrowright
M_{DF} = 148.8 kN-m \circlearrowleft
M_{FD} = 208.3 kN-m \circlearrowleft
M_{CD} = 13.2 kN-m \circlearrowleft
M_{DC} = -329.6 kN-m or 329.6 kN-m \circlearrowright
M_{EF} = 77.4 kN-m \circlearrowleft
M_{FE} = -208.3 kN-m or 208.3 kN-m \circlearrowright
Back substitution of the numerical values of member end moments into the equilibrium equations yields
M_{CA} + M_{CD} + M_{CE} = 66.5 + 13.2 – 79.7 = 0 Checks
M_{DB} + M_{DC} + M_{DF} = 180.8 – 329.6 + 148.8 = 0 Checks
M_{EC} + M_{EF} = -77.4 + 77.4 = 0 Checks
M_{FD} + M_{FE} = 208.3 – 208.3 = 0 Checks
M_{CE} + M_{EC} + M_{DF} + M_{FD} = -79.7 – 77.4 + 148.8 + 208.3 = 200 Checks
M_{AC} + M_{CA} + M_{BD} + M_{DB} = 147.8 + 66.5 + 204.9 + 180.8 = 600 Checks
Member End Shears and Axial Forces See Fig. 16.19(e).
Support Reactions See Fig. 16.19(f ).
Equilibrium Check The equilibrium equations check.
