Question 16.1: Determine the reactions and draw the shear and bending momen...

Degrees of Freedom From Fig. 16.5(a), we can see that only joint B of the beam is free to rotate. Thus, the structure has only one degree of freedom, which is the unknown joint rotation, θ_B.

Fixed-End Moments By using the fixed-end moment expressions given inside the back cover of the book, we evaluate the fixed-end moments due to the external loads for each member:

FEM_{AB} =\frac{Pab^2}{L^2} = \frac{90(2)(3)^2}{(5)^2} = 64.8 kN-m \circlearrowleft              or           +64.8 kN-m

FEM_{BA} =\frac{Pa^2b}{L^2} = \frac{90(2)^2(3)}{(5)^2} = 43.2 kN-m \circlearrowright              or         -43.2 kN-m

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

FEM_{CB} = 150 kN-m \circlearrowright                 or                           -150 kN-m

Note that in accordance with the slope-deflection sign convention, the counterclockwise fixed-end moments are considered as positive, whereas the clockwise fixed-end moments are considered to be negative.

Chord Rotations Since no support settlements occur, the chord rotations of both members are zero; that is, ψ_{AB} = ψ_{BC} = 0.

Slope-Deflection Equations To relate the member end moments to the unknown joint rotation, θ_B, we write the slope-deflection equations for the two members of the structure by applying Eq. (16.9). Note that since the supports A and C are fixed, the rotations θ_A = θ_C = 0. Thus the slope-deflection equations for member AB can be expressed as

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

M_{AB} = \frac{2EI}{5}(θ_B) + 64.8 = 0.4EIθ_B + 64.8                      (1)

M_{BA} = \frac{2EI}{5}(2θ_B) – 43.2 = 0.8EIθ_B – 43.2                      (2)

Similarly, by applying Eq. (16.9) for member BC, we obtain the slope-deflection equations

M_{BC} = \frac{2EI}{6}(2θ_B) + 150 = 0.67EIθ_B + 150                      (3)

M_{CB} = \frac{2EI}{6}(θ_B) – 150 = 0.33EIθ_B – 150                      (4)

Equilibrium Equation The free-body diagram of joint B is shown in Fig. 16.5(b). Note that the member end moments, which are assumed to be in a counterclockwise direction on the ends of the members, must be applied in the (opposite) clockwise direction on the free body of the joint, in accordance with Newton’s third law.

By applying the moment equilibrium equation ∑M_B=0 to the free body of joint B, we obtain the equilibrium equation

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

Joint Rotation To determine the unknown joint rotation, θ_B, we substitute the slope-deflection equations (Eqs. (2) and (3)) into the equilibrium equation (Eq. (5)) to obtain

or

from which

Member End Moments The member end moments can now be computed by substituting the numerical value of EIθ_B back into the slope-deflection equations (Eqs. (1) through (4)). Thus,

M_{BA} = 0.8(-72.65) – 43.2 = -101.5 kN-m          or          101.5 kN-m\circlearrowright

M_{CB} = 0.33(-72.65) – 150 = -174.3 kN-m        or        174.3 kN-m \circlearrowright

Note that a positive answer for an end moment indicates that its sense is counterclockwise, whereas a negative answer for an end moment implies a clockwise sense. Since the end moments M_{BA} and M_{BC} are equal in magnitude but opposite in sense, the equilibrium equation, M_{BA} + M_{BC} = 0, is indeed satisfied.

Member End Shears The member end shears, obtained by considering the equilibrium of each member, are shown in Fig. 16.5(c).

Support Reactions The reactions at the fixed supports A and C are equal to the forces and moments at the ends of the members connected to these joints. To determine the reaction at the roller support B, we consider the equilibrium of the free body of joint B in the vertical direction (see Fig. 16.5(c)), to obtain

The support reactions are shown in Fig. 16.5(d).

Equilibrium Check To check our calculations of member end shears and support reactions, we apply the equations of equilibrium to the free body of the entire structure. Thus (see Fig. 16.5(d)),

40.82 – 90 + 187.05 – 50 × 6 + 162.13 = 0                        Checks

35.6 – 40.82(11) + 90(9) – 187.05(6) + 50(6)(3) – 174.3 = -0.02 ≈ 0                             Checks

Shear and Bending Moment Diagrams The shear and bending moment diagrams can now be constructed by using the beam sign convention described in Section 5.1. These diagrams are shown in Fig. 16.5(e) and (f).