Question 13.14: 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. 13.21(a) due to the uniformly distributed load and due to the support settlements of 15 mm at B, 37 mm at C, and 18 mm at D. Use the method of consistent deformations.
This beam was previously analyzed in Example 13.8 for the 30 kN/m uniformly distributed loading by selecting the vertical reactions at the interior supports B and C as the redundants. We will use the same primary beam as used previously.
Relative Settlements The specified support settlements are depicted in Fig. 13.21(b) using an exaggerated scale. It can be seen from this figure that the settlements of supports B and C relative to the chord of the primary beam (which is the line connecting the displaced positions of supports A and D) are
Δ_{BR} = -9 mm and Δ_{CR} = -25 mm
in which the negative signs for the magnitudes of Δ_{BR} and Δ_{CR} indicate that these settlements occur in the downward direction—that is, opposite to the upward direction assumed for the redundants B_y and C_y.
Compatibility Equations The compatibility equations for the beam remain the same as in Example 13.8, except that the right-hand sides of the equations must now be set equal to the settlements Δ_{BR} and Δ_{CR}. Thus
Δ_{BO} + f_{BB}B_y + f_{BC}C_y = Δ_{BR} (1)
Δ_{CO} + f_{CB}B_y + f_{CC}C_y = Δ_{CR} (2)
Deflections of Primary Beam In Example 13.8, the deflections and the flexibility coefficients of the beam were expressed in terms of EI. Since the right-hand sides of the compatibility equations were zero, the EI terms simply canceled out of the computations. In the present example, however, because of the presence of support settlements on the righthand sides of the compatibility equations, the EI terms cannot be canceled out; therefore, the actual numerical values of deflections and flexibility coefficients must be computed.
Δ_{BO} = Δ_{CO} = -\frac{35640 kN-m^3}{EI} = \frac{-35640}{(200(10^6)3000(10^{-6})} = -0.0594 mf_{BB} = f_{CC} = \frac{96 m^3}{EI} = \frac{96}{(200(10^6)3000(10^{-6})} = 0.00016 m
f_{CB} = f_{BC} = \frac{84 m^3}{EI} = \frac{84 }{(200(10^6)3000(10^{-6})} = 0.00014 m
Magnitudes of the Redundants By substituting the numerical values into the compatibility equations, we write
-0.0594 + 0.00016B_y + 0.00014C_y = -0.009 (1a)
-0.0594 + 0.00014B_y + 0.00016C_y = -0.025 (2a)
By solving Eqs. (1a) and (2a) simultaneously for B_y and C_y, we obtain
B_y = 541.3 kN↑ and C_y = -258.6 kN = 258.6 kN↓
Reactions and Shear and Bending Moment Diagrams The remaining reactions of the continuous beam can now be determined by equilibrium. The reactions and the shear and bending moment diagrams of the beam are shown in Fig. 13.21(c). A comparison of these results with those of Example 13.8 (without settlement) indicates that even small support settlements may have a significant effect on the reactions and the shear and bending moment diagrams of indeterminate structures.
