1.3.4.4 Reaction Forces and Moments on Continuous Beams
A continuous beam is one with three or more supports. Such a beam is statically
indeterminate and deflection equations must be applied to find the support reactions.
The three-moment equation is such an equation.
1.3.4.5 Application of the Three Moment Equation to Solving for the Reactions on
Continuous Beams
Figure 1-35(a) shows a uniform beam that is simply supported at three colinear points,
A, B, and C. In order to obtain the reactions, the beam is broken into two simply
supported sections with no end moments, as shown in Figure 1-35(b). The moment
diagrams are then found for these sections and the area A and centroid C of these
diagrams are found as shown in Figure 1-35(c). The quantities found may now be
substituted into the three moment equation:
MAL1+2MB(L1+L2)+MCL2=−6A1a¯1L1−6A2b¯2L2MAL1+2MB(L1+L2)+MCL2=−6A1a¯1L1
−6A2b¯2L2
(1-41)
If MA and MC are known, this equation may be solved for the moment at B, MB. Knowing
this moment, the support reactions at A, B, and C may be found by applying the
equations of statics.
The terms to the right of Equation (1-41) may be found for various simple loadings by
use of Table 1-10.
If a beam has a number of concentrated loads as shown in Figure 1-36, Equation (1-42)
becomes
MAL1+2MB(L1+L2)+McL2=−∑P1a1L1(L21−a21)−∑P2b2L2(L22−b22)MAL1+2MB(L1+L
2)+McL2=−∑P1a1L1(L12−a12)−∑P2b2L2(L22−b22)
(1-42)
where P1 denotes any one of several concentrated loads which may act on the left span
at a distance a1 from support A. Similarly, P2 denotes any load in the right span at a
distance from support C.
If a beam is simply supported at more than three points, the three-moment equation
may be written for each intermediate support. The equations may then be solved
simultaneously to obtain the moments at each support. This procedure is illustrated by
the sample problem in Section 1.3.4.6.
1.3.4.6 Sample Problem - Reactions on Continuous Beams by the Three Moment Equation
Given: The continuous beam shown in Figure 1-37.
Find: The support reactions.
Solution: The three-moment equation may be written for spans 1 and 2. Since only
concentrated loads are present, the special case given by Equation (1-42) may be used.
Thus,
M1L1+2M2(L1+L2)+M3L3=−P1a1L1(L21−a21)−P2b2L2(L22−b22)M1L1+2M2(L1+L2)+M
3L3=−P1a1L1(L12−a12)−P2b2L2(L22−b22)
Inserting numerical values gives
0(10)+2M2(10+15)+M3(20)=−500(5)10(102−52)−300(5)15(152−52)0(10)+2M2(
10+15)+M3(20)=−500(5)10(102−52)−300(5)15(152−52)
Simplifying gives 5M2 + 2M3 = 3875.
The more general form of the three-moment equation given by Equation (1-41) may
now be written for spans 2 and 3 with the aid of cases one and three of Table 1-10.
M2(15)+2M3(15+20)=−300(10)15(152+102)−10(20)34M2(15)+2M3(15+20)=−300(
10)15(152+102)−10(20)34
Simplifying gives 3M2 + 14M3 = -15,000.
The two equations in M2 and M3 that were just obtained may be solved simultaneously
to find that M2 = -376 and M3 = -990.
The equations of statics may now be applied as illustrated in Figure 1-38 to find the
reaction forces.
The beam may now be drawn as in Figure 1-39.