THREE-MOMENT EQUATIONThe three-moment equation, provides a convenient tool for analyzing continuous beams. The three-moment equation represents, in a general form, the compatibility condition that the slope of the elastic curve be continuous at an interior support of the continuous beam. Since the equation involves three moments—the bending moments at the support under consideration and at the two adjacent supports—it commonly is referred to as the three-moment equation. When using this method, the bending moments at the interior (and any fixed) supports of the continuous beam are treated as the redundants. The three-moment equation is then applied at the location of each redundant to obtain a set of compatibility equations which can be solved for the unknown redundant moments.

Derivation of Three-Moment Equation

Consider an arbitrary continuous beam subjected to external loads and support settlements as shown in Fig. (a). This beam can actually be analyzed by the method of consistent deformations by treating the bending moments at the interior supports to be the redundants. From Fig. (a), we can see that the slope of the elastic curve of the indeterminate beam is continuous at the interior supports. When the restraints corresponding to the redundant bending moments are removed by inserting internal hinges at the interior support points, the primary structure thus obtained consists of a series of simply supported beams. As shown in Figs.(b) and (c), respectively, when this primary structure is subjected to the known external loading and support settlements, discontinuities develop in the slope of the elastic curve at the locations of the interior supports. Since the redundant bending moments provide continuity of the slope of the elastic curve, these unknown moments are applied as loads on the primary structure as shown in Fig. (d), and their magnitudes are determined by solving the compatibility equations based on the condition that, at each interior support of the primary structure, the slope of the elastic curve, due to the combined effect of the external loading, support settlements, and unknown redundants, must be continuous.

The three-moment equation uses the foregoing compatibility condition of slope continuity at an interior support to provide a general relationship between the unknown bending moments at the support where compatibility is being considered and at the adjacent supports to the left and to the right, in terms of the loads on the intermediate spans and any settlements of the three supports.

To derive the three-moment equation, we focus our attention on the compatibility equation at an interior support c of the continuous beam, with prismatic spans and a constant modulus of elasticity, shown in Fig. (a). As indicated in this figure, the adjacent supports to the left and to the right of c are identified as l and r, respectively; the subscripts l and r are used to refer to the loads and properties of the left span, lc, and the right span, cr, respectively; and the settlements of supports l; c, and r are denoted by Δl; Δc, and Δr, respectively. The support settlements are considered positive when in the downward direction, as shown in the figure.

From Fig. (a), we can see that the slope of the elastic curve of the indeterminate beam is continuous at c. In other words, there is no change of slope of the tangents to the elastic curve at just to the left of c and just to the right of c; that is, the angle between the tangents is zero. However, when the primary structure, obtained by inserting internal hinges at the interior support points, is subjected to external loads, as shown in Fig. (b), a discontinuity develops in the slope of the elastic curve at c, in the sense that the tangent to the elastic curve at just to the left of c rotates relative to the tangent at just to the right of c. The

change of slope (or the angle) between the two tangents due to external loads is denoted by θ1 and can be expressed as (see Fig. (b))

(1)

in which θl1 and θr1 denote, respectively, the slopes at the ends c of the spans to the left and to the right of the support c, due to external loads.

Similarly, the slope discontinuity at c in the primary structure, due to support settlements (Fig. (c)), can be written as

(2)

in which θl2 and θr2 represent, respectively, the slopes of the spans to the left and to the right of c, due to support settlements. Finally, when the primary structure is loaded with the redundant support bending moments, as shown in Fig. (d), the slope discontinuity at c can be expressed as

(3)

in which θl3 and θr3 denote, respectively, the slopes at end c of the spans to the left and to the right of the support c, due to unknown redundant moments.

The compatibility equation is based on the requirement that the slope of the elastic curve of the actual indeterminate beam is continuous at c; that is, there is no change of slope from just to the left of c to just to the right of c. Therefore, the algebraic sum of the angles between the tangents at just to the left and at just to the right of c due to the external loading, support settlements and the redundant bending moments must be zero. Thus,

(4)

By substituting Eqs. (1) through (3) into Eq. (4), we obtain

(5)

Since each span of the primary structure can be treated as a simply supported beam, the slopes at the ends c of the left and the right spans, due to the external loads (Fig. (b)), can be conveniently determined either by the conjugate-beam method or by using the beam-deflection formulas given inside the front cover of the book. By using the deflection formulas, we obtain

(6a)

(6b)

in which the summation signs have been added to the first terms on the right sides of these equations, so that multiple concentrated loads can be applied to each span (instead of a single concentrated load as shown in Figs. (a) and (b) for simplicity). As continuous beams usually are loaded with uniformly distributed loads over entire spans and concentrated loads, the effects of only these two types of loadings generally are considered in the three-moment equation. However, the effects of other types of loads can be included simply by adding the expressions of slopes due to these loads to the right sides of Eqs. (6a) and (6b).

The slopes θl2 and θr2, of the left and the right spans, respectively, due to support settlements, can be obtained directly from the deformed positions of the spans depicted in Fig. (c). Since the settlements are assumed to be small, the slopes can be expressed as

(7)

The slopes at ends c of the left and the right spans, due to redundant support bending moments, (Fig. (d)), can be determined conveniently by using the beam-deflection formulas. Thus,

(8a)

(8b)

in which Ml, Mc and Mr denote the bending moments at supports l, c and r, respectively. As shown in Fig. (d), these redundant bending moments are considered to be positive in accordance with the beam convention—that is, when causing compression in the upper fibers and tension in the lower fibers of the beam.

By substituting Eqs. (6) through (8) into Eq. (5), we write the compatibility equation as

By simplifying the foregoing equation and rearranging it to separate the terms containing redundant moments from those involving loads and support settlements, we obtain the general form of the three-moment equation:

(9)

in which Mc = bending moment at support c where the compatibility is being considered; Ml, Mr = bending moments at the adjacent supports to the left and to the right of c, respectively; E = modulus of elasticity;Ll; Lr = lengths of the spans to the left and to the right of c, respectively; Il; Ir = moments of inertia of the spans to the left and to the right of c, respectively; Pl; Pr = concentrated loads acting on the left and the right spans, respectively; kl (or kr) = ratio of the distance of Pl (or Pr) from the left (or right) support to the span length; wl; wr = uniformly distributed loads applied to the left and the right spans, respectively; Δc ¼ settlement of the support c under consideration; and Δl; Δr = settlements of the adjacent supports to the left and to the right of c, respectively. As noted before, the support bending moments are considered to be positive in accordance with the beam convention—that is, when causing compression in the upper fibers and tension in the lower fibers of the beam. Furthermore, the external loads and support settlements are considered positive when in the downward direction, as shown

in Fig. (a). If the moments of inertia of two adjacent spans of a continuous beam are equal (i.e., Il = Ir = I ), then the three-moment equation simplifies to

(10)

If both the moments of inertia and the lengths of two adjacent spans are equal (i.e., Il = Ir = I and Ll = Lr = L), then the three-moment equation becomes

(11)

The foregoing three-moment equations are applicable to any three consecutive supports, l, c and r, of a continuous beam, provided that there are no discontinuities, such as internal hinges, in the beam between the left support l and the right support r.

Application of Three-Moment EquationThe following step-by-step procedure can be used for analyzing continuousbeams by the three-moment equation.

1. Select the unknown bending moments at all interior supports of the beam as the redundants.

2. By treating each interior support successively as the intermediate support c, write a three-moment equation. When writing these equations, it should be realized that bending moments at the simple end supports are known. For such a support with a cantilever overhang, the bending moment equals that due to the external loads acting on the cantilever portion about the end support. The total number of three-moment equations thus obtained must be equal to the number of redundant support bending moments, which must be the only unknowns in these equations.

3. Solve the system of three-moment equations for the unknown support bending moments.

4. Compute the span end shears. For each span of the beam, (a) draw a free-body diagram showing the external loads and end moments and (b) apply the equations of equilibrium to calculate the shear forces at the ends of the span.

5. Determine support reactions by considering the equilibrium of the support joints of the beam.

6. If so desired, draw shear and bending moment diagrams of the beam by using the beam sign convention.

Fixed SupportsThe three-moment equations, as given by Eqs. (9) through (11), were derived to satisfy the compatibility condition of slope continuity at the interior supports of continuous beams. These equations can, however, be used to satisfy the compatibility condition of zero slope at

the fixed end supports of beams. This can be achieved by replacing the fixed support by an imaginary interior roller support with an adjoining end span of zero length simply supported at its outer end, as shown in Fig. 2. The reaction moment at the actual fixed support is now treated as the redundant bending moment at the imaginary interior support, and the three-moment equation when applied to this imaginary support satisfies the compatibility condition of zero slope of the elastic curve at the actual fixed support. When analyzing a beam for support settlements, both imaginary supports—that is, the interior roller support and the outer simple end support—are considered to undergo the same settlement as the actual fixed support.

Example 1:

Determine the reactions and draw the shear and bending moment diagrams for the beam shown in Fig. (a) by using the three-moment equation.

Solution

Redundant The beam has one degree of indeterminacy. The bending moment MB, at the interior support B, is the redundant.

Three-Moment Equation at Joint B Considering the supports, A, B, and C as l, c, and r, respectively, and substituting Ll = 24 ft, Lr = 20 ft, Il = 2I , Ir = I , Pl1 = 30 k, kl1 = 1/3, Pl2 = 20 k, kl2 = 2/3, wr = 2.5 k/ft, and Pr = wl = Δl = Δc = Δr = 0, into Eq. (9), we obtain

Since A and C are simple end supports, we have by inspection

Thus, the three-moment equation becomes

from which we obtain the redundant bending moment to be

Span End Shears and Reactions The shears at the ends of the spans AB and BC of the continuous beam can now be determined by applying the equations of equilibrium to the free bodies of the spans shown in Fig. (b). Note that the negative bending moment MB is applied at the ends B of spans AB and BC so that it causes tension in

the upper fibers and compression in the lower fibers of the beam. By considering the equilibrium of span AB, we obtain

Similarly, for span BC,

By considering the equilibrium of joint B in the vertical direction, we obtain

The reactions are shown in Fig. (c).

Example 2:

Determine the reactions for the continuous beam shown in Fig. (a) due to the uniformly distributed load and due to the support settlements of 10 mm at A, 50 mm at B, 20 mm at C, and 40 mm at D. Use the three-moment equation.

SolutionRedundants The bending moments MB and MC, at the interior supports B and C, respectively, are the redundants.

Three-Moment Equation at Joint B By considering the supports A, B, and C as l, c, and r, respectively, and substituting L = 10 m, E = 200 GPa = 200(106) kN/m2, I = 700(106) mm4 = 700(10-6) m4, wl = wr = 30 kN/m, Δl = ΔA = 10 mm = 0.01 m, Δc = ΔB = 50 mm = 0.05 m, Δr = ΔC = 20 mm = 0.02 m and Pl = Pr = 0, into Eq. (11), we write

Since A is a simple end support, MA = 0. The foregoing equation thus simplifies to

(1)

Three-Moment Equation at Joint C Similarly, by considering the supports B, C, and D as l, c, and r, respectively, and by substituting the appropriate numerical values in Eq. (11), we obtain

Since D is a simple end support, MD = 0. Thus, the foregoing equation becomes

(2)

Support Bending Moments Solving Eqs. (1) and (2) simultaneously for MB and MC, we obtain

Span End Shears and Reactions

With the redundants MB and MC known, the span end shears and the support reactions can be determined by considering the equilibrium of the free bodies of the spans AB, BC, and CD, and joints B and C, as shown in Fig. (b). The reactions are shown in Fig. (c).

Example 3:

Determine the reactions for the continuous beam shown in Fig. (a) by the three-moment equation.

SolutionSince support A of the beam is fixed, we replace it with an imaginary interior roller support with an adjoining end span of zero length, as shown in Fig. (b).

Redundants From Fig. (b), we can see that the bending moments MA and MB at the supports A and B, respectively, are the redundants.

Three-Moment Equation at Joint A By using Eq. (10) for supports A’, A, and B, we obtain

Or Three-Moment Equation at Joint B Similarly, applying Eq. (10) for supports A, B, and C, we write

The bending moment at end C of the cantilever overhang CD is computed as

By substituting MC = -90 k-ft into the foregoing three-moment equation and simplifying, we obtain

Support Bending Moments Solving Eqs. (1) and (2), we obtain

.

Span End Shears and Reactions See Figs. 14.5(c) and (d).