Question 16.8: Determine the member end moments and reactions for the frame...

Degrees of Freedom The joints C, D, and E of the frame are free to rotate. However, we will eliminate the rotation of the simple support at end E by using the modified slope-deflection equations for member DE. Thus the analysis will involve only two unknown joint rotations, θ_C and θ_D.

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

FEM_{AC} =\frac{200 × 4}{8} = 100 kN-m \circlearrowleft          or        +100 kN-m

FEM_{CA} = 100 kN-m \circlearrowright         or        -100 kN-m

FEM_{CD} = FEM_{DE} = \frac{50(6)^2}{12} = 150 kN-m \circlearrowleft       or       +150 kN-m

FEM_{DC} = FEM_{ED} = 150 kN-m \circlearrowright         or       -150 kN-m

Slope-Deflection Equations As indicated in Fig. 16.13(a), the moments of inertia of the columns and the girders of the frame are 300 × 10^6 mm^4 and 600 × 10^6 mm^4, respectively. Using I = I_{column} = 300 × 10^6 mm^4 as the reference moment of inertia, we express I_{girder} in terms of I as

Next, we write the slope-deflection equations by applying Eq. (16.9) to members AC, BD, and CD, and Eqs. (16.15) to member DE. Thus

M_{nf} = \frac{2EI}{L}(2θ_n + θ_f – 3ψ) + FEM_{nf}                      (16.9)

M_{rh} = \frac{3EI}{L}(θ_r – ψ) + (FEM_{rh} – \frac{FEM_{hr}}{2})                (16.15a)

M_{hr} = 0                      (16.15b)

M_{AC} = \frac{2EI}{4}(θ_C) + 100 = 0.5EIθ_C + 100                 (1)

M_{CA} =\frac{2EI}{4}(2θ_C) – 100 = EIθ_C – 100                (2)

M_{BD} =\frac{2EI}{4}(θ_D) = 0.5EIθ_D                      (3)

M_{DB} =\frac{2EI}{4}(2θ_D) = EIθ_D                            (4)

M_{CD} =\frac{2E(2I)}{6}(2θ_C + θ_D) + 150 = 1.33EIθ_C + 0.67EIθ_D + 150                     (5)

M_{DC} = \frac{2E(2I)}{6}(2θ_D + θ_C) – 150 = 0.67EIθ_C + 1.33EIθ_D – 150                     (6)

M_{DE} = \frac{3E(2I)}{6}(θ_D) + (150 + \frac{150}{2}) = EIθ_D + 225                     (7)

Equilibrium Equations By applying the moment equilibrium equation, ∑M =0, to the free bodies of joints C and D (Fig. 16.13(b)), we obtain the equilibrium equations

M_{CA} + M_{CD} = 0                    (8)

M_{DB} + M_{DC} + M_{DE} = 0                        (9)

Joint Rotations Substitution of the slope-deflection equations into the equilibrium equations yields

2.33EIθ_C + 0.67EIθ_D = -50                         (10)

0.67EIθ_C + 3.33EIθ_D = -75                          (11)

By solving Eqs. (10) and (11) simultaneously, we determine the values of EIθ_C and EIθ_D to be

Member End Moments The member end moments can now be computed by substituting the numerical values of EIθ_C and EIθ_D into the slope-deflection equations (Eqs. (1) through (7)).

M_{CA} = -115.9 kN-m           or        115.9 kN-m \circlearrowright

M_{BD} = -9.7 kN-m          or        9.7 kN-m \circlearrowright

M_{DB} = -19.3 kN-m           or        19.3 kN-m \circlearrowright

M_{DC} = -186.4 kN-m           or          186.4 kN-m \circlearrowright

To check that the solution of the simultaneous equations (Eqs. (10) and (11)) has been carried out correctly, we substitute the numerical values of member end moments back into the equilibrium equations (Eqs. (8) and (9)) to obtain

M_{CA} + M_{CD} = -115.9 + 115.9 = 0                     Checks

M_{DB} + M_{DC} + M_{DE} = -19.3 – 186.4 + 205.7 = 0                   Checks

Member End Shears The member end shears, obtained by considering the equilibrium of each member, are shown in Fig. 16.13(c).

Member Axial Forces With end shears known, member axial forces can now be evaluated by considering the equilibrium of joints C and D in order. The axial forces thus obtained are shown in Fig. 16.13(c).

Support Reactions See Fig. 16.13(d).

Equilibrium Check The equilibrium equations check.