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The uniformly distributed factored load of the beam is wf = 1.2 dead load + 1.6 live load = 1.2 x 3.2 kN/m + 1.6 x 12 kN/m = 23.04 kN/m From Table 23.3, the end-reaction moments of the beams associated with the factored load are wL2 3r1A(2 - r1B) 23.04(9)2 3 x 0.406(2 - 0.406)

m}0) = -M;B) =---—-— =-—---(-0-) = 78.73 kN m

End-reaction moments of beam 2 due to the load are

M2£> = -M£>= WL!. 3r2B(2 - r2C) = 23.04(10)2 . 3 x 0.432(2 - 0.432) = 102.32 kNm 2B 2C 12 4 + r2Br2C 12 4 - 0.4322

End-reaction moments of beam 3 due to the load are wL2 3r3C(2 - r3D) 23.04(9)2 3 x 0.406(2 - 1)

The modified moment distribution is conducted as shown in Table 23.5 and the final moment distribution of the beam is shown in Figure 23.10. The maximum span moment of Beam 2 is 201.09 kN m, which is located 5.06 m from joint B. The maximum negative moment at joint D is 193.08 kN m. Since the beam is laterally braced and W410 x 39 is a compact section, the flexural strength of the beam based on LRFD [2] is

Mr = f bMp = 0.9ZxEy = 0.9 x 730 x 103 mm3 x 345 MPa x 10-6 = 226.67 kN m > 201.09 kN m

Therefore, the flexural strength of the beam is adequate.

To calculate the mid-span deflection of Beam 2 due to the service load w = 3.2 kN/m + 12.0 kN/m = 15.2 kN/m, and the connection rotational stiffness R = 1.15 x 1010 Nmm/rad, the corresponding end-fixity factors of the beam are r1A = r1B = r3C = 0.578, rœ = 1.0, and r2B = r2D = 0.603

Following the foregoing procedure, the final moments obtained by the modified moment distribution at joints B and C are M2B = 81.67 kN m and M2B = -73.04 kN m.

TABLE 23.5 Moment Distribution of Three-Span Semirigid Continuous Beam

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