Question 13.8: Determine the reactions and draw the shear and bending momen...
Determine the reactions and draw the shear and bending moment diagrams for the three-span continuous beam shown in Fig. 13.13(a) using the method of consistent deformations.
Degree of Indeterminacy i = 2.
Primary Beam The vertical reactions B_y and C_y at the interior supports B and C, respectively, are selected as the redundants. The roller supports at B and C are then removed to obtain the primary beam shown in Fig. 13.13(b). Next, the primary beam is subjected separately to the 30 kN/m external load and the unit values of the redundants B_y and C_y, as shown in Fig. 13.13(b), (c), and (d), respectively.
Compatibility Equations Since the deflections of the actual indeterminate beam at supports B and C are zero, we set equal to zero the algebraic sum of the deflections at points B and C, respectively, of the primary beam due to the 30 kN/m external load and each of the redundants to obtain the compatibility equations:
Δ_{BO} + f_{BB} B_y + f_{BC} C_y = 0 (1)
Δ_{CO} + f_{CB} B_y + f_{CC} C_y = 0 (2)
Deflections of the Primary Beam By using the beam-deflection formulas, we obtain
Δ_{BO} = Δ_{CO} = -\frac{35640 kN-m^3}{EI}f_{BB} = f_{CC} = \frac{96 m^3}{EI}
f_{CB} = \frac{84 m^3}{EI}
By applying Maxwell’s law,
f_{CB} = \frac{84 m^3}{EI}Magnitudes of the Redundants By substituting the values of the deflections and flexibility coefficients of the primary beam just computed into the compatibility equations (Eqs. (1) and (2)), we obtain
-35640 + 96B_y + 84C_y = 0-35640 + 84B_y + 96C_y = 0
or
96B_y + 84C_y = 35640 (1a)
84B_y + 96C_y = 35640 (2a)
Solving Eqs. (1a) and (2a) simultaneously for B_y and C_y, we obtain
B_y = C_y = 198 kN ↑Reactions The remaining reactions can now be determined by applying the three equations of equilibrium to the free body of the continuous beam as follows (Fig. 13.13(e)):
+→∑F_x = 0 A_x = 0
+\circlearrowleft ∑M_D = 0 -A_y(18) + 30(18)(9) – 198(6 + 12) = 0
A_y = 72 kN ↑
+↑∑F_y = 0 72- 30(18) + 198 + 198 + D_y = 0
D_y = 72 kN ↑
Shear and Bending Moment Diagrams The shear and bending moment diagrams of the beam are shown in Fig. 13.13(f ).
The shapes of the shear and bending moment diagrams for continuous beams, in general, are similar to those for the three-span continuous beam shown in Fig. 13.13(f ). As shown in this figure, negative bending moments generally develop at the interior supports of continuous beams, whereas the bending moment diagram is usually positive over the middle portions of the spans. The bending moment at a hinged support at an end of the beam must be zero, and it is generally negative at a fixed end support. Also, the shape of the bending moment diagram is parabolic for the spans subjected to uniformly distributed loads, and it consists of linear segments for spans subjected to concentrated loads. The actual values of the bending moments, of course, depend on the magnitude of the loading as well as on the lengths and flexural rigidities of the spans of the continuous beam.
