Question 16.4: Determine the member end moments and reactions for the conti...

Since the moment and shear at end C of the cantilever member CD of the beam can be computed directly by applying the equations of equilibrium (see Fig. 16.8(b)), it is not necessary to include this member in the analysis. Thus, only the indeterminate part AC of the beam, shown in Fig. 16.8(c), needs to be analyzed. Note that, as shown in this figure, the 120-kN·m moment and the 30-kN force exerted at joint C by the cantilever CD must be included in the analysis.

Degrees of Freedom From Fig. 16.8(c), we can see that joints B and C are free to rotate. Thus, the structure to be analyzed has two degrees of freedom, which are the unknown joint rotations θ_B and θ_C.

Fixed-End Moments

FEM_{BC} = \frac{10(9)^2}{12} = 67.5 kN-m \circlearrowleft          or          +67.5 kN-m

FEM_{CB} = 67.5 kN-m \circlearrowright         or      -67.5 kN-m

Slope-Deflection Equations By applying Eq. (16.9) to members AB and BC, we write the slope-deflection equations:

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

M_{AB} = \frac{2EI}{6}(θ_B) = 0.333EIθ_B                             (1)

M_{BA} =\frac{2EI}{6}(2θ_B) = 0.667EIθ_B                              (2)

M_{BC} =\frac{2EI}{9}(2θ_B + θ_C) + 67.5 = 0.444EIθ_B + 0.222EIθ_C + 67.5                     (3)

M_{CB} =\frac{2EI}{9}(2θ_C + θ_B) – 67.5 = 0.222EIθ_B + 0.444EIθ_C – 67.5                         (4)

Equilibrium Equations By considering the moment equilibrium of the free bodies of joints B and C (Fig. 16.8(d)), we obtain the equilibrium equations

M_{BA} + M_{BC} = 0                    (5)

M_{CB} + 120 = 0                             (6)

Joint Rotations Substitution of the slope-deflection equations (Eqs. (2) through (4)) into the equilibrium equations (Eqs. (5) and (6)) yields

1.111EIθ_B + 0.222EIθ_C = -67.5                         (7)

0.222EIθ_B + 0.444EIθ_C = -52.5                       (8)

By solving Eqs. (7) and (8) simultaneously, we determine the values of EIθ_B and EIθ_C to be

Member End Moments The member end moments can now be computed by substituting the numerical values of EIθ_B and EIθ_C into the slope-deflection equations (Eqs. (1) through (4)):

M_{AB} = 0.333(-41.25) = -13.7 kN-m          or         13.7 kN-m \circlearrowright

M_{BA} = 0.667(-41.25) = -27.5 kN-m           or         27.5 kN-m \circlearrowright

= -120 kN-m           or         120 kN-m \circlearrowright

Note that the numerical values of M_{BA} , M_{BC}, and M_{CB} do satisfy the equilibrium equations (Eqs. (5) and (6)).

Member End Shears and Support Reactions See Fig. 16.8(e) and (f ).

Equilibrium Check The equilibrium equations check.