Question 13.3: Determine the reactions and draw the shear and bending momen...
Determine the reactions and draw the shear and bending moment diagrams for the two-span continuous beam shown in Fig. 13.5(a) using the method of consistent deformations.
Degree of Indeterminacy The beam is supported by four reactions, so its degree of indeterminacy is equal to 4 – 3 = 1.
Primary Beam The vertical reaction B_y at the roller support B is selected to be the redundant, and the primary beam is obtained by removing the roller support B from the given indeterminate beam, as shown in Fig. 13.5(b). Next, the primary beam is subjected separately to the external loading and a unit value of the unknown redundant B_y, as shown in Fig. 13.5(b) and (c), respectively. As shown in these figures, \Delta_{BO} denotes the deflection at B due to the external loading, whereas f_{BB} denotes the flexibility coefficient representing the deflection at B due to the unit value of the redundant B_y.
Compatibility Equation Because the deflection at support B of the actual indeterminate beam is zero, the algebraic sum of the deflections of the primary beam at B due to the external loading and the redundant B_y must also be zero. Thus, the compatibility equation can be written as
\Delta_{BO} + f_{BB}B_y = 0 (1)
Deflections of Primary Beam The flexural rigidity EI of the primary beam is not constant (since the moment of inertia of the right half of the beam, BD, is twice the moment of inertia of the left half, AB), so we cannot use the formulas given inside the front cover of the book for computing deflections. Therefore, we will use the conjugate-beam method, discussed in Chapter 6, for determining the deflections of the primary beam.
To determine the deflection \Delta_{BO} due to the external loading, we draw the conjugate beams for the 15-kN/m uniformly distributed load and the 60-kN concentrated load, as shown in Fig. 13.5(d) and (e), respectively. Recalling that the deflection at a point on a real beam is equal to the bending moment at that point in the corresponding conjugate beam, we determine the deflection \Delta_{BO} due to the combined effect of the distributed and concentrated loads as
EI\Delta_{BO} = [-4,218.75(10) +(\frac{2}{3})(10)(750)(\frac{30}{8})] + [-718.75(10) +(\frac{1}{2})(10)(150)(\frac{10}{3})]\Delta_{BO} = -\frac{28,125 kN-m^3}{EI}
in which the negative sign indicates that the deflection occurs in the downward direction. Note that although the numerical values of E and I are given, it is usually convenient to carry out the analysis in terms of EI. The flexibility coefficient f_{BB} can be computed similarly by using the conjugate beam shown in Fig. 13.5(f ). Thus
EIf_{BB} = 20.833(10) -(\frac{1}{2})(10)(5)(\frac{10}{3}) = 125 kN-m^3/kNf_{BB} = \frac{125 kN – m^3/kN}{EI}
Magnitude of the Redundant By substituting the values of \Delta_{BO} and f_{BB} into the compatibility equation (Eq. (1)), we obtain
-\frac{28,125}{EI} + (\frac{125}{EI})B_y = 0 B_y = 225 kN \uparrow Ans.
Reactions To determine the remaining reactions of the indeterminate beam, we apply the equilibrium equations (Fig. 13.5(g)):
+\longrightarrow \sum{F_x}=0 A_x = 0 Ans.
+\circlearrowleft \sum{M_D}=0 -A_y(20) – 225(10) + 15(20)(10) + 60(5) = 0
A_y = 52.5 kN \uparrow Ans.
+\uparrow \sum{F_y}=0 52.5 + 225 – 15(20) – 60 + D_y = 0
D_y = 82.5 kN \uparrow Ans.
Shear and Bending Moment Diagrams See Fig. 13.5(h). Ans.
