D 
sin t  sin t (5.295) 1  2 /2  If  is small relative to , maximum D is nearly unity; thus, the system is practically statically loaded. If  is very large compared with , D is very small; thus, the mass cannot follow the rapid fluctuations in load and remains practically stationary. Therefore, when  differs appreciably from , the effects of unbalanced rotating parts are not too serious. But if   , resonance occurs; D increases with time. Hence, to prevent structural damage, measures must be taken to correct the unbalanced parts to change , or to change the natural frequency of the vibrating mass, or damping must be provided. The response as given by Eq. (5.294) consists of two parts, the free vibration and the forced part. When damping is present, the free vibration is of the form of Eq. (5.289) and is rapidly damped out. Hence, the free part is called the transient response, and the forced part, the steady-state response. The maximum value of the dynamic load factor for the steady-state response Dm is called the dynamic magnification factor. It is given by

Application of wall membranes should he started at the bottom of one end of the wall and the strips of fabric or felt laid vertically. Preparation of the surfaces and laying of the membrane proceed much as they do with floor membranes. The surfaces to which the membrane is attached must be dry and smooth, which may require that the faces of masonry walls be leveled with a thin coat of grout or mortar. The plies of the wall membrane should be lapped into those of the floor membrane. If the outside method of application is used and the membrane is faced with masonry, the narrow space between the units and the membrane should be filled with mortar as the units are laid. The membrane may be terminated at the grade line by a return into the superstructure wall facing. Waterstops in joints in walls and floors containing a bituminous membrane should be the metal-bellows type. The membrane should be placed on the exposed face of the joint and it may project into the joint, following the general outline of

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