a. Remove the span AB, and introduce the shear VB and moment M 5 that the load on AB
induces at J5, as shown in Fig. 44Z).
b. Remove the fixed support at D and add the span DE of zero length, with a hinged sup-
port at E.
For the interval BD, the transformed beam is then identical in every respect with the
actual beam.
- Apply the equation for the theorem of three moments
Consider span BC as span 1 and CD as span 2. For the 5-kip (22.2-kN) load, k 2 = 12/16 =
0.75; for the 10-kip (44.5-kN) load, Ic 2 = 8/16 - 0.5. Then -12(10) + 2MC(10 + 16) +
16MD = -tt(4)(10)^3 - 5(16)^2 (0.75 - 0.422) - 10(16)^2 (0.5 - 0.125). Simplifying gives
13MC + 4MD = -565.0, Eq. a. - Apply the moment equation again
Considering CD as span 1 and DE as span 2, apply the moment equation again. Or, for the
5-kip (22.2-kN) load, Jt 1 = 0.25; for the 10-kip (44.5-kN) load, Jt 1 = 0.5. Then 16MC +
2MD(16 + O) = -5(16)^2 (0.25 - 0.016) - 10(16)^2 (0.50 - 0.125). Simplifying yields Mc +
2MD = -78.7,Eq. b. - Solve the moment equations
Solving Eqs. a and b gives Mc = -37.1 ft-kips (-50.30 kN-m); MD = -20.8 ft-kips (-28.20
kN-m). - Determine the reactions by using a free-body diagram
Find the reactions by drawing a free-body diagram of each span and taking moments with
respect to each support. Thus RB = 20.5 kips (91.18 kN); Rc = 32.3 kips (143.67 kN);
RD= 5.2 kips (23.12 kN).
BENDING-MOMENT DETERMINATION BY
MOMENT DISTRIBUTIONUsing moment distribution, determine the bending moments at the supports of the mem-
ber in Fig. 45. The beams are rigidly joined at the supports and are composed of the same
material.
Calculation Procedure:
- Calculate the flexural stiffness of each span
Using K to denote the flexural stiffness, we see that K = HL if the far end remains fixed
during moment distribution; K= 0.75//L if the far end remains hinged during moment dis-
tribution. Then KAB = 270/18 = 15; KBC = 192/12 = 16; KCD = 0.75(240/20) = 9. Record
all the values on the drawing as they are obtained. - For each span, calculate the required fixed-end moments at
those supports that will be considered fixed
These are the external moments with respect to the span; a clockwise moment is consid-
ered positive. (For additional data, refer to cases 14 and 15 in the AISC Manual.) Then
MAB = -wL^2 /12 - -2(18)^2 /12 = -54.0 ft-kips (-73.2 kN-m); MBA = +54.0 ft-kips (73.22
kN-m). Similarly, MBC = -48.0 ft-kips (-65.1 kN-m); MCB = +48.0 ft-kips (65.1 kN-m);
MCD = -24(15)(5)(15 + 20)/[2(20)^2 ] = -78.8 ft-kips (-106.85 kN-m).