In this case, since the terminations at A and C are FIXED, it is necessary to assume additional spans AA' and CC'. They are considered to have zerolengths and no loads. They satisfy, however, the condition of restraint, which Fig. 4. Continuous Beam of Two Spans, Restrained at the Ends, and with a UniformlyDistributed Load on One Span Only is that the tangent to the elastic curve is horizontal over these supports. The beam at the ends A' and C' of these supplementary spans is considered to be simply supported. Therefore V'3 = V2 - wal2 1 -wal22 9.6 Substituting the values of the bending moments in the above expressions, and remembering that la la, = FIG. 5. Typical Continuous Beam of Three Spans with Distributed Loads 3. Solutions of Problems by the Theorem of Three Moments. The particular application of this theorem to the design of buildings is best illustrated by the following examples. Example 1. A continuous beam of three spans is considered, as shown in Fig. 5. This type often occurs in the construction of school-buildings, hotels, etc. It is assumed that the left support is a brick wall and that the bending moment Mi = 0. As it is also assumed that the right support is a substantial concrete column, monolithic with the beam, a negative bending moment is provided for at that end. There is a uniform dead load of 1 000 lb per lin ft of beam, in addition to which the two outside, long spans have each a live load of 800 lb per lin ft, and the shorter, middle span, a live load of 1000 lb per lin ft of beam. It is required to find the CRITICAL BENDING MOMENTS, VERTICAL SHEARS, and REACTIONS, assuming a constant moment of inertia for the crosssections of beam. By substituting these values of the BENDING MOMENTS in the formulas for VERTICAL SHEAR, as explained in the preceding paragraphs, The distance in each span from the support at the left of that span to the section of zero shear is, This is the MAXIMUM POSITIVE BENDING MOMENT in span 11. This is the MINIMUM NEGATIVE BENDING MOMENT in span l1⁄2, and indicates that there is no positive bending moment in this span under this condition of loading. This is the MAXIMUM POSITIVE BENDING MOMENT in span 13. Having thus determined the MAXIMUM VERTICAL SHEARS and BENDING MOMENTS, the beams are designed as explained in the previous chapters by using these computed moment-values instead of the moment-values for equal spans and uniformlydistributed loads, which are only approximately correct for conditions similar to those in this example. As these computations for the exact conditions are somewhat tedious, it is convenient to prepare and refer to PLOTTED DIAGRAMS interpreting the types of construction of most frequent occurrence in practice. By the use of these diagrams, which show the proper bending moments and vertical shears for typical cases, the moments and shears for any particular problem can be closely approximated, with often a considerable saving in time. The theorem of three moments can be used, also, for beams with fixed ends, as shown, in a general case, in the following example. Example 2. To further illustrate the method of procedure, a single =2000 lb P1 = 016 = 20'--0 VM, = 0 =25'-0"-. R2 Fig. 6. span is considered of the beam shown in Fig. 6. This beam is fixed at the left support, simply supported at the right support, and carries a single concentrated load as shown. It is required to determine the bending moment, M2 at the left support. + (2 000)(1 − 4%) = 592 lb |