Question 16.11: Determine the member end moments and reactions for the frame...

Degrees of Freedom Degrees of freedom are θ_C, θ_D, and Δ.

Fixed-End Moments Since no external loads are applied to the members, the fixed-end moments are zero.

Chord Rotations From Fig. 16.18(b), we can see that

Slope-Deflection Equations

M_{AC} = \frac{2EI}{5}[θ_C – 3(-0.25Δ)] = 0.4EIθ_C + 0.3EIΔ                  (1)

M_{CA} =\frac{2EI}{5}[2θ_C – 3(-0.25Δ)] = 0.8EIθ_C + 0.3EIΔ                 (2)

M_{BD} =\frac{2EI}{4}[θ_D – 3(-0.25Δ)] = 0.5EIθ_D + 0.375EIΔ                    (3)

M_{DB} = \frac{2EI}{4}[2θ_D – 3(-0.25Δ)] = EIθ_D + 0.375EIΔ                        (4)

M_{CD} = \frac{2EI}{5}[2θ_C + θ_D – 3(0.15Δ)] = 0.8EIθ_C + 0.4EIθ_D – 0.18EIΔ                  (5)

M_{DC} = \frac{2EI}{5}[2θ_D + θ_C – 3(0.15Δ)] = 0.8EIθ_D + 0.4EIθ_C – 0.18EIΔ                     (6)

Equilibrium Equations By considering the moment equilibrium of joints C and D, we obtain the equilibrium equations

M_{CA} + M_{CD} = 0                           (7)

M_{DB} + M_{DC} = 0                           (8)

The third equilibrium equation is established by summing the moments of all the forces and couples acting on the free body of the entire frame about point O, which is located at the intersection of the longitudinal axes of the two columns, as shown in Fig. 16.18(c). Thus

+\circlearrowleft ∑M_O = 0              M_{AC} – S_{AC}(13.33) + M_{BD} – S_{BD}(10.67) + 120(6.67) = 0

in which the shears at the lower ends of the columns can be expressed in terms of column end moments as (see Fig. 16.18(d))

S_{AC} = \frac{M_{AC} + M_{CA}}{5}             and            S_{BD} = \frac{M_{BD} + M_{DB}}{4}

By substituting these expressions into the third equilibrium equation, we obtain

1.67M_{AC} + 2.67M_{CA} + 1.67M_{BD} + 2.67M_{DB} = 800                           (9)

Joint Displacements Substitution of the slope-deflection equations (Eqs. (1) through (6)) into the equilibrium equations (Eqs. (7) through (9)) yields

1.6EIθ_C + 0.4EIθ_D + 0.12EIΔ = 0                           (10)

0.4EIθ_C + 1.8EIθ_D + 0.195EIΔ = 0                         (11)

2.804EIθ_C + 3.505EIθ_D + 2.93EIΔ = 800                  (12)

By solving Eqs. (10) through (12) simultaneously, we determine

Member End Moments By substituting the numerical values of EIθ_C , EIθ_D, and EIΔ into the slope-deflection equations (Eqs. (1) through (6)), we obtain

M_{CD} = -85.1 kN-m           or         85.1 kN-m \circlearrowright

M_{DC} = -91 kN-m             or           91 kN-m \circlearrowright

Back substitution of the numerical values of member end moments into the equilibrium equations yields

M_{CA} + M_{CD} = 85.1 – 85.1 = 0                      Checks

M_{DB} + M_{DC} = 91 – 91 = 0                          Checks

= 801.5 ≈ 800                      Checks

Member End Shears and Axial Forces See Fig. 16.18(e).

Support Reactions See Fig. 16.18(f ).

Equilibrium Check The equilibrium equations check.