Question 16.9: Determine the member end moments and reactions for the frame...
Determine the member end moments and reactions for the frame of Example 16.8 due to a settlement of 18 mm at support B. Use the slope-deflection method.
The frame is shown in Fig. 16.14(a).
Degrees of Freedom θ_C and θ_D are the degrees of freedom.
Chord Rotations Since the axial deformation of member BD is neglected, the 18-mm settlement of support B causes the joint D to displace downward by the same amount, as shown in Fig. 16.14(b). The inclined dashed lines in this figure represent the chords (not the elastic curves) of members CD and DE in the deformed positions. The rotation of the chord of member CD is
ψ_{CD} = -\frac{0.018}{6} = -0.003in which the negative sign has been assigned to the value of ψ_{CD} to indicate that its sense is clockwise. Similarly, for member DE,
ψ_{DE} = 0.003Slope-Deflection Equations
M_{AC} = 0.5EIθ_C (1)
M_{CA} = EIθ_C (2)
M_{BD} = 0.5EIθ_D (3)
M_{DB} = EIθ_D (4)
M_{CD} = \frac{2E(2I)}{6}[2θ_C + θ_D – 3(-0.003)]
= 1.33EIθ_C + 0.67EIθ_D + 0.006EI (5)
M_{DC} = \frac{2E(2I)}{6}[2θ_D + θ_C – 3(-0.003)]
= 0.67EIθ_C + 1.33EIθ_D + 0.006EI (6)
M_{DE} =\frac{3E(2I)}{6}(θ_D – 0.003) = EIθ_D – 0.003EI (7)
M_{ED} = 0Equilibrium Equations See Fig. 16.14(c).
M_{CA} + M_{CD} = 0 (8)
M_{DB} + M_{DC} + M_{DE} = 0 (9)
Joint Rotations By substituting the slope-deflection equations into the equilibrium equations, we obtain
2.33EIθ_C + 0.67EIθ_D = -0.006EI0.67EIθ_C + 3.33EIθ_D = -0.003EI
Substitution of EI = 6 × 10^4 kN-m^2 into the right sides of the preceding equations yields
2.33EIθ_C + 0.67EIθ_D = -360 (10)
0.67EIθ_C + 3.33EIθ_D = -180 (11)
Solving Eqs. (10) and (11) simultaneously, we obtain
EIθ_C = -147.5 kN-m^2EIθ_D = -24.38 kN-m^2
Member End Moments By substituting the numerical values of EIθ_C and EIθ_D into the slope-deflection equations, we obtain
M_{AC} = -73.75 kN-m or 73.75 kN-m \circlearrowright
M_{CA} = -147.5 kN-m or 147.5 kN-m \circlearrowright
M_{BD} = -12.19 kN-m or 12.19 kN-m \circlearrowright
M_{DB} = -24.38 kN-m or 24.38 kN-m \circlearrowright
M_{CD} = 147.5 kN-m\circlearrowleft
M_{DC} = 228.75 kN-m\circlearrowleft
M_{DE} = -204.38 kN-m or 204.38 kN-m\circlearrowright
Back substitution of the numerical values of member end moments into the equilibrium equations (Eqs. (8) and (9)) yields
M_{CA} + M_{CD} = -147.5 + 147.5 = 0 Checks
M_{DB} + M_{DC} + M_{DE} = -24.38 + 228.75 – 204.38 = 0 Checks
Member End Shears and Axial Forces See Fig. 16.14(d).
Support Reactions See Fig. 16.14(e).
Equilibrium Check The equilibrium equations check.
