Question 16.10: Determine the member end moments and reactions for the frame...

Degrees of Freedom The degrees of freedom are θ_C, θ_D, and Δ (see Fig. 16.17(b)).

Fixed-End Moments By using the fixed-end moment expressions given inside the back cover of the book, we obtain

FEM_{CD} = \frac{40(3)(4)^2}{(7)^2} = 39.2 kN-m \circlearrowleft        or       +39.2 kN-m

FEM_{DC} = \frac{40(3)^2(4)}{(7)^2} = 29.4 kN-m\circlearrowright         or        -29.4 kN-m

Chord Rotations From Fig. 16.17(b), we can see that

ψ_{AC} = -\frac{Δ}{7}              ψ_{BD} = -\frac{Δ}{5}            ψ_{CD} = 0

Slope-Deflection Equations

M_{AC} =\frac{2EI}{7}[θ_C -3(-\frac{Δ}{7})] = 0.286EIθ_C + 0.122EIΔ                  (1)

M_{CA} =\frac{2EI}{7}[2θ_C -3(-\frac{Δ}{7})] = 0.571EIθ_C + 0.122EIΔ                  (2)

M_{BD} =\frac{2EI}{5}[θ_D -3(-\frac{Δ}{5})] = 0.4EIθ_D + 0.24EIΔ                  (3)

M_{DB} =\frac{2EI}{5}[2θ_D -3(-\frac{Δ}{5})] = 0.8EIθ_D + 0.24EIΔ                  (4)

M_{CD} = \frac{2EI}{7}(2θ_C + θ_D) + 39.2 = 0.571EIθ_C + 0.286EIθ_D + 39.2                (5)

M_{DC} = \frac{2EI}{7}(θ_C + 2θ_D) – 29.4 = 0.286EIθ_C + 0.571EIθ_D – 29.4                (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)

To establish the third equilibrium equation, we apply the force equilibrium equation ∑F_X=0 to the free body of the entire frame (Fig. 16.17(c)), to obtain

in which S_{AC} and S_{BD} represent the shears at the lower ends of columns AC and BD, respectively, as shown in Fig. 16.17(c). To express the column end shears in terms of column end moments, we draw the free-body diagrams of the two columns (Fig. 16.17(d)) and sum the moments about the top of each column:

S_{AC} = \frac{M_{AC} + M_{CA}}{7}            and          S_{BD} = \frac{M_{BD} + M_{DB}}{5}

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

which can be rewritten as

5(M_{AC} + M_{CA}) + 7(M_{BD} + M_{DB}) = 0                                     (9)

Joint Displacements To determine the unknown joint displacements θ_C, θ_D, and Δ, we substitute the slopedeflection equations (Eqs. (1) through (6)) into the equilibrium equations (Eqs. (7) through (9)) to obtain

1.142EIθ_C + 0.286EIθ_D + 0.122EIΔ = -39.2                (10)

0.286EIθ_C + 1.371EIθ_D + 0.24EIΔ = 29.4                    (11)

4.285EIθ_C + 8.4EIθ_D + 4.58EIΔ = 0                        (12)

Solving Eqs. (10) through (12) simultaneously yields

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_{AC} = -14.6 kN-m        or     14.6 kN-m \circlearrowright

M_{CA} = -26 kN-m        or        26kN-m \circlearrowright

M_{DC} = -21.3 kN-m           or          21.3 kN-m \circlearrowright

To check that the solution of the simultaneous equations (Eqs. (10) through (12)) has been carried out correctly, we substitute the numerical values of member end moments back into the equilibrium equations (Eqs. (7) through (9)):

M_{CA} + M_{CD} = -26 + 26 = 0                   Checks

M_{DB} + M_{DC} = 21.3 – 21.3 = 0                        Checks

5(M_{AC} + M_{CA}) + 7(M_{BD} + M_{DB}) = 5(-14.6 – 26) + 7(7.7 + 21.3) = 0                            Checks

Member End Shears The member end shears, obtained by considering the equilibrium of each member, are shown in Fig. 16.17(e).

Member Axial Forces With end shears known, member axial forces can now be evaluated by considering the equilibrium of joints C and D. The axial forces thus obtained are shown in Fig. 16.17(e).

Support Reactions See Fig. 16.17(f ).

Equilibrium Check The equilibrium equations check.