Question 13.6: Determine the reactions and draw the bending moment diagram ...
Determine the reactions and draw the bending moment diagram for the two-span continuous beam shown in Fig. 13.10(a) by the method of consistent deformations. Select the bending moment at the interior support B to be the redundant.
This beam was analyzed in Example 13.3 by selecting the vertical reaction at support B as the redundant.
Primary Beam The primary beam is obtained by removing the restraint corresponding to the redundant bending moment M_B by inserting an internal hinge at B in the given indeterminate beam, as shown in Fig. 13.10(b). Next, the primary beam is subjected separately to the external loading and a unit value of the redundant M_B, as shown in Fig. 13.10(b) and (c), respectively.
Compatibility Equation See Fig. 13.10(b) and (c):
θ_{BO rel.} + f_{BB rel.}M_B = 0 (1)
Slopes of Primary Beam Each of the spans of the primary beam can be treated as a simply supported beam of constant flexural rigidity EI, so we can use the beam-deflection formulas given inside the front cover of the book for evaluating the changes of slopes θ_{BO rel.} and f_{BB rel.}. From Fig. 13.10(b), we can see that
θ_{BO rel.} = θ_{BL} + θ_{BR}in which θ_{BL} and θ_{BR} are the slopes at the ends B of the left and the right spans of the primary beam, respectively, due to the external loading. By using the deflection formulas, we obtain
θ_{BL} = \frac{15(10)^3}{24EI} = \frac{625 kN-m^2}{EI}θ_{BR} = \frac{15(10)^3}{24E(2I)} + \frac{60(10)^2}{16E(2I)} = \frac{500 kN-m^2}{EI}
Thus,
θ_{BO rel.} = \frac{625 }{EI} + \frac{500 }{EI} = \frac{1,125 kN-m^2}{EI}The flexibility coefficient f_{BB rel.} can be computed in a similar manner. From Fig. 13.10(c), we can see that
f_{BB rel.} = f_{BBL} + f_{BBR}in which
f_{BBL} = \frac{10}{3EI} = \frac{3.33 m}{EI} and f_{BBR} = \frac{10}{3E(2I)} = \frac{1.67 m}{EI}
Thus,
f_{BB rel.} = \frac{3.33 }{EI} + \frac{1.67 }{EI} = \frac{5 m}{EI}Magnitude of the Redundant By substituting the values of θ_{BO rel.} and f_{BB rel.} into the compatibility equation (Eq. (1)), we obtain
\frac{1,125 }{EI} + (\frac{5 }{EI})M_B = 0M_B =-225 kN-m
Reactions The forces at the ends of the members AB and BD of the continuous beam can now be determined by applying the equations of equilibrium to the free bodies of the members shown in Fig. 13.10(d). By considering the equilibrium of member AB, we obtain
A_y = (\frac{1}{2})(15)(10) – (\frac{225}{10})= 52.5 kN ↑B^{AB}_y = (\frac{1}{2})(15)(10) + (\frac{225}{10}) = 97.5 kN ↑
Similarly, for member BD,
B^{BD}_y = (\frac{1}{2})(15)(10) + (\frac{60}{2}) + (\frac{225}{10}) = 127.5 kN ↑D_y = (\frac{1}{2})(15)(10) + (\frac{60}{2}) – (\frac{225}{10}) = 82.5 kN ↑
By considering the equilibrium of joint B in the vertical direction, we obtain
B_y = B^{AB}_y + B^{BD}_y = 97.5 + 127.5 = 225 kN ↑Bending Moment Diagram The bending moment diagram for the continuous beam, constructed by simple-beam parts, is shown in Fig. 13.10(e). The two parts of the diagram due to the external loading and the member end moments may be superimposed, if so desired, to obtain the resultant bending moment diagram shown in Example 13.3.
