FIGURE 44- Take moments with respect to each support to find the
reactions
Span AB: SM 4 - 6(5) + 2(10)(5) + 80.2 - KXR 51 = O; Rm = 21.02 kips (93.496 kN);
SM 5 = IORA - 6(5) - 2(10)(5) + 80.2 = O; RA = 4.98 kips (22.151 kN).
Span BC: SM 5 = -80.2 + 10(9) + 3(15)(7.5) - 15RC = Q;RC = 23.15 kips (102.971
kN); 2MC = 15^ 52 - 80.2 - 10(6) - 3(15)(7.5) = O; R 82 = 31.85 kips (144.668 kN); RB =
21.02 + 31.85 = 52.87 kips (235.165 kN).
THEOREM OF THREE MOMENTS: BEAM
WITH OVERHANG AND FIXED END
Determine the reactions at the supports of the continuous beam in Fig. 440. Use the theo-
rem of three moments.
Calculation Procedure:- Transform the given beam to one amenable to analysis by the
theorem of three moments
Perform the following operations to transform the beam: