end moments become negligible. To reduce the number of cycles, the unlocking of joints should start with those having the greatest unbalanced moments. Suppose the end moments are to be found for the prismatic continuous beam ABCD in Fig. 5.75. The I /L values for all spans are equal; therefore, the relative fixed-end stiffness for all members is unity. However, since A is a hinged end, the computation can be shortened by using the actual relative stiffness, which is 3/4. Relative stiffnesses for all members are shown in the circle on each member. The distribution factors are shown in boxes at each joint. The computation starts with determination of fixed-end moments for each member (Art. 5.11.4). These are assumed to have been found and are given on the first line in Fig. 5.75. The greatest unbalanced moment is found from inspection to be at hinged end A; so this joint is unlocked first. Since there are no other members at the joint, the full unlocking moment of 400 is distributed to AB at A and onehalf of this is carried over to B. The unbalance at B now is 400  480 plus the carry-over of 200 from A, or a total of 120. Hence, a moment of 120 must be applied and distributed to the members at B by multiplying by the distribution factors in the corresponding boxes. The net moment at B could be found now by adding the entries for each member at the joint. However, it generally is more convenient to delay the summation until the last cycle of distribution has been completed. The moment distributed to BA need not be carried over to A, because the carryover factor toward the hinged end is zero. However, half the moment distributed to BC is carried over to C. Similarly, joint C is unlocked and half the distributed moments carried over to B and D, respectively. Joint D should not be unlocked, since it actually is a fixed end. Thus, the first cycle of moment distribution has been completed. The second cycle is carried out in the same manner. Joint B is released, and the distributed moment in BC is carried over to C. Finally, C is unlocked, to complete the cycle. Adding the entries for the end of each member yields the final moments. 5.11.7 Maximum Moments in Continuous Frames In design of continuous frames, one objective is to find the maximum end moments and interior moments produced by the worst combination of loading. For maximum moment at the end of a beam, live load should be placed on that beam and on the FIGURE 5.76 Bending moments in a continuous frame obtained by moment distribution. beam adjoining the end for which the moment is to be computed. Spans adjoining these two should be assumed to be carrying only dead load. For maximum midspan moments, the beam under consideration should be fully loaded, but adjoining spans should be assumed to be carrying only dead load. The work involved in distributing moments due to dead and live loads in continuous frames in buildings can be greatly simplified by isolating each floor. The tops of the upper columns and the bottoms of the lower columns can be assumed fixed. Furthermore, the computations can be condensed considerably by following the procedure recommended in Continuity in Concrete Building Frames. EB033D, Portland Cement Association, Skokie, IL 60077, and indicated in Fig. Figure 5.74 presents the complete calculation for maximum end and midspan moments in four floor beams AB, BC, CD, and DE. Building columns are assumed to be fixed at the story above and below. None of the beam or column sections is known to begin with; so as a start, all members will be assumed to have a fixedend stiffness of unity, as indicated on the first line of the calculation. On the second line, the distribution factors for each end of the beams are shown, calculated from the stiffnesses (Arts. 5.11.3 and 5.11.4). Column stiffnesses are not shown, because column moments will not be computed until moment distribution to the beams has been completed. Then the sum of the column moments at each joint may be easily computed, since they are the moments needed to make the sum of the end moments at the joint equal to zero. The sum of the column moments at each joint can then be distributed to each column there in proportion to its stiffness. In this example, each column will get one-half the sum of the column moments. Fixed-end moments at each beam end for dead load are shown on the third line, just above the heavy line, and fixed-end moments for live plus dead load on the fourth line. Corresponding midspan moments for the fixed-end condition also are shown on the fourth line and, like the end moments, will be corrected to yield actual midspan moments. For maximum end moment at A, beam AB must be fully loaded, but BC should carry dead load only. Holding A fixed, we first unlock joint B, which has a totalload fixed-end moment of 172 in BA and a dead-load fixed-end moment of 37 in BC. The releasing moment required, therefore, is (172  37), or  135. When B is released, a moment of 135
1/4 is distributed to BA One-half of this is carried over to A, or 135
1/4
1/2  17. This value is entered as the carryover at A on the fifth line in Fig. 5.76. Joint B is then relocked. At A, for which we are computing the maximum moment, we have a total-load fixed-end moment of 172 and a carry-over of 17, making the total 189, shown on the sixth line. To release A, a moment of 189 must be applied to the joint. Of this, 189
1/3, or 63, is distributed to AB, as indicated on the seventh line of the calculation. Finally, the maximum moment at A is found by adding lines 6 and 7: For maximum moment at B, both AB and BC must be fully loaded but CD should carry only dead load. We begin the determination of the moment at B by first releasing joints A and C, for which the corresponding carry-over moments at BA and BC are 29 and (78  70)
1/4
1/2  1, shown on the fifth line in Fig. 5.76. These bring the total fixed-end moments in BA and BC to 201 and 79, respectively. The releasing moment required is (201  79)  122. Multiplying this by the distribution factors for BA and BC when joint B is released, we find the distributed moments, 30, entered on line 7. The maximum end moments finally are obtained by adding lines 6 and 7: 171 at BA and 109 at BC. Maximum moments at C, D, and E are computed and entered in Fig. 5.76 in a similar manner. This procedure is equivalent to two cycles of moment distribution. The computation of maximum midspan moments in Fig. 5.76 is based on the assumption that in each beam the midspan moment is the sum of the simple-beam midspan moment and one-half the algebraic difference of the final end moments (the span carries full load but adjacent spans only dead load). Instead of starting with the simple-beam moment, however, we begin with the midspan moment for the fixed-end condition and apply two corrections. In each span, these corrections are equal to the carry-over moments entered on line 5 for the two ends of the beam multiplied by a factor. For beams with variable moment of inertia, the factor is 1/2[(1/CF )  D  1] where CF is the fixed-end-carry-over factor toward the end for which the correction factor is being computed and D is the distribution factor for that end. The plus sign is used for correcting the carry-over at the right end of a beam, and the minus sign for the carry-over at the left end. For prismatic beams, the correction factor becomes For example, to find the corrections to the midspan moment in AB, we first multiply the carry-over at A on line 5, 17, by 1/2(1  1/3). The correction, 11, is also entered on the fifth line. Then, we multiply the carry-over at B,  29, by 1/2(1  1/4) and enter the correction, 18, on line 6. The final midspan moment is the sum of lines 4, 5, and 6: 99  11  18  128. Other midspan moments in Fig. 5.74 are obtained in a similar manner. See also Arts. 5.11.9 and 5.11.10. 5.11.8 Moment-Influence Factors In certain types of framing, particularly those in which different types of loading conditions must be investigated, it may be convenient to find maximum end moments from a table of moment-influence factors. This table is made up by listing for the end of each member in the structure the moment induced in that end when a moment (for convenience, 1000) is applied to every joint successively. Once this table has been prepared, no additional moment distribution is necessary for computing the end moments due to any loading condition. For a specific loading pattern, the moment at any beam end MAB may be obtained from the moment-influence table by multiplying the entries under AB for the various TABLE 5.6 Moment-Influence Factors

Shear Reinforcement. When Vu exceeds Vc, shear reinforcement must be provided to resist the excess factored shear. The shear reinforcement may consist of stirrups making an angle of 45 to 90 with the longitudinal reinforcement, longitudinal bars bent at an angle of 30 or more, or a combination of stirrups and bent bars. The nominal shear strength provided by the shear reinforcement Vs must not exceed 8

b d. c w Spacing of required shear reinforcement placed perpendicular to the longitudinal reinforcement should not exceed 0.5d for nonprestressed concrete, 75% of the overall depth for prestressed concrete, or 24 in. Inclined stirrups and bent bars should be spaced so that at least one intersects every 45 line extending toward the supports from middepth of the member to the tension reinforcement. When Vs is greater than 4

b d, the maximum spacing of shear reinforcement should be reduced c w by one-half. (See Art. 9.109 for shear-strength design for prestressed concrete members.) The area required in the legs of a vertical stirrup, in2, is V s A
s (9.40a) v d y where s
spacing of stirrups, in and y
yield strength of stirrup steel, psi. For inclined stirrups, the leg area should be at least V s A
s (9.40b) v (sin   cos ) d y where 
angle of inclination with longitudinal axis of member. For a single bent bar or a single group of parallel bars all bent at an angle  with the longitudinal axis at the same distance from the support, the required area

There are two primary categories of feltbase sheets and ply sheets. Base sheets are heavier felts that are often used for the first layer of felt to be installed. If the felt is to be nailed, a base sheet is recommended because of its greater strength. Ventilating base sheets are intended to allow for the venting of moisturevapor pressure by lateral (horizontal) movement. However, if a ventilated base sheet is to be used, the designer should take into account the small driving force for horizontal moisture transport and the small amount of moisture that can be moved horizontally. Surfacings as applied to built-up membranes, are typically small pieces of aggregate or slag, liquid-applied coatings, or a cap sheet. Common coatings include cutbacks and emulsions, which are both cold-applied. Cutbacks are composed of asphalt and solvent and often include an aluminum pigment for reflectivity. Emulsions consist of clay and asphalt particles dispersed in water. Some emulsions include aluminum pigment or titanium dioxide for reflectivity. The cutback and emulsion coatings are available in fibrated or nonfibrated grades. Latex (acrylic) coatings are also available, but for built-up roofs, these coatings are not used as often as the other types of coatings. Cap sheets are heavy coated felts that are factory surfaced with mineral granules. Cold-process roof coverings (also known as cold-applied) are similar to hotapplied BUR, except that instead of hot bitumen, asphalt-based cutbacks or emulsions are typically used. They are applied by sprayer, brush, broom, or squeegee. 12.4.2 Liquid-Applied Roof Coverings Liquid-applied systems are supplied as either single or two-component elastomeric materials. They are applied by sprayer, brush, roller, or squeegee. Typically these systems are applied directly over concrete or wood sheathing. Deck joints and cracks normally require special preparation. See also cold-process roof coverings (Art. 12.4.1) and coatings on polyurethane foam roofs (Art. 12.4.6). 12.4.3 Metal Roof Coverings These are generally used for steep-slope roofs rather than for low-slope roofs. See Art. 12.5.3. However, some standing-seam structural panel systems can be used successfully in low-slope situations. These systems are considered hydrostatic, that is, they have the ability to resist water intrusion under some pressure. These panel systems generally incorporate a sealant in the seam, or an anti-capillary hem to provide the necessary protection from moisture infiltration through the seams. FIGURE 12.3 Modified bitumen roof with a base sheet overlaying two layers of preformed insulation board. (NRCA Roofing and Waterproofing Manual.) 12.4.4 Modified Bitumen Membranes These are typically composed of prefabricated sheets of polymer-modified asphalt with polyester or glass-fiber reinforcement or a combination of these. The polymers most used for asphalt modification are atactic polypropylene (APP) or styrenebutadiene- styrene (SBS). These prefabricated sheets are commonly installed over a base sheet (Art. 12.4.1), which may or may not also be composed of modified bitumen. Sometimes the assembly also includes a ply sheet (Art. 12.4.1). In the past, modified bitumen membranes were occasionally applied in a single layer. However, two or more layers are now the predominant system (Fig. 12.3). SBS sheets are generally set in a continuous layer of hot asphalt, but some sheets may be torch-applied or set in cold adhesive. Self-adhering styrene-ethylenepropylene- styrene (SEPS) sheets are also available. SBS and SEPS sheets need protection from ultraviolet light (UV). Protection is typically provided by factoryapplied mineral granules. They may also be surfaced with coatings (Art. 12.4.1). APP sheets are generally torch-applied (Fig. 12.4). When APP sheets were introduced in the United States in the late 1970s, they were generally used without surfacing, since UV protection was reportedly provided by the APP modifier. While some such APP membranes weathered very well, others did not. Hence, coatings (cutbacks, emulsions, or latex) or granules are now often used. The ASTM material standards for polymer-modified bitumen sheet products are