Determining the Redundancy Factor

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Example 13

Determining the Redundancy Factor Section 12.3.4 describes the methodology for determination of the redundancy factor, r, which is used in several of the seismic load combinations that are specified in Section 12.4. This example demonstrates how the redundancy factor is determined for a variety of structural systems. The use of the redundancy factor in the context of the load combinations is demonstrated in Example 18 of this guide. The value of the redundancy factor is either 1.0 or 1.3, depending on the Seismic Design Category (SDC) and the structural configuration. Section 12.3.4 indicates that the redundancy factor may be different in the two orthogonal loading directions of the structure. For structures in SDCs B and C, r is taken as 1.0 in each direction. Section 12.3.4.1 provides a list of other situations in which the redundancy factor may be taken as 1.0. These situations (e.g., r = 1.0 in drift calculations) are applicable for all SDCs. Note, however, that Section 12.12.1.1 requires that drift limits be multiplied by 1 π r for moment frames in SDCs D, E, and F.

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For structures in SDCs D, E, and F, the redundancy factor is 1.3 unless it can be shown that certain conditions exist, as outlined in Section 12.3.4.2 and Table 12.3-3. Condition (b) in Section 12.3.4.2 provides a description of situations in which the redundancy factor may be taken as 1.0 in both directions. If these conditions are not met, then condition (a) must be evaluated, from which the redundancy factor is either 1.0 or 1.3 in the direction of interest. Condition (b) states that r may be taken as 1.0 where the structure has no plan irregularities and where there are two bays of perimeter seismic force resisting elements on each side of the building for each story of the building resisting more than 35 percent of the base shear. One or more collinear shear walls may be considered a bay only if the total plan length of the wall (or walls) is greater than the story height. Fig. G13–1 illustrates several cases for which condition (b) may be evaluated. The floor plans in the figure are applicable at levels for which the design shear is greater than 35 percent of the base shear. Only building A in the figure satisfies the condition (b) test. The walls on each side of this planregular building are long enough to be classified as a bay, and there are two perimeter bays on each side of the building. Building B violates the criteria because the two walls marked with asterisks are not on the perimeter. Additionally, the system has a nonparallel system irregularity because the walls are not symmetrically placed about the center of resistance. Building C, which is assumed to have no irregularities, does not satisfy the criteria in the Y direction because the plan length of the four walls marked with asterisks is insufficient to classify the walls as bays. Buildings D, E, and F cannot automatically be classified with r = 1.0 because each has a plan irregularity. In building F, the plan irregularity occurs because of an out-of-plane offset of the shear walls marked with asterisks. The walls at the upper level are on the interior of the building and transfer to the exterior at the lower levels.

Figure G13–1 Evaluation of the Redundancy Factor for Various Buildings.

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The fact that condition (b) has not been satisfied does not mean that the redundancy factor is 1.3. This situation would be the case only if condition (a) in Section 12.3.4.2 is also not met. Consider again building B of Fig. G13–1. In this shear wall system, each of the walls has a plan length greater than the height of the wall. Thus, the height-to-width ratio of the walls is less than 1.0, and the system defaults to “Other” lateral force resisting elements in Table 12.3-3. Presumably, therefore, this system can be assigned a redundancy factor of 1.0 in each direction because no requirements dictate otherwise. It appears that the same situation would occur even if the walls marked by asterisks in building B of Fig. G13–1 were removed entirely. In the opinion of the author, this situation violates the spirit of the redundancy factor concept, and a factor of 1.3 should be assigned in this case. In building C of Fig. G13–1, each of the walls marked with an asterisk has a length less than the story height. Removal of one of these walls does not cause an extreme torsional irregularity. It would appear at first glance that the removal of one wall would reduce the strength of the system by only 25 percent in the Y direction. However, this situation does not consider the effect of torsion. The reduction in strength must be based on the questions “How much lateral load can be applied in the Y direction for the system with one wall missing, and how does that compare to the strength of the system with the wall in place?” Two interpretations exist for evaluation of the strength of the system with elements removed. The first is based on elastic analysis, and the second is based on inelastic analysis. An important consideration of the use of an inelastic analysis is that the system must be sufficiently ductile to handle the continued application of loads after the lateral load resisting elements begin to yield. Consider, for example, the system shown in Fig. G13–2. This system has eight identical walls, marked A through H, each with a force–deformation relationship as shown in Fig. G13–3. The lateral load carrying capacity of each wall is 100 kips. The system is evaluated on the basis of the following situations: 1. 2. 3. 4.

elastic behavior with all walls in place elastic behavior with one wall removed inelastic behavior analysis with all walls in place inelastic behavior with one wall removed

For each situation, the lateral force V is applied in the Y direction at an eccentricity of 5 percent of the width (eccentricity = 0.05 × 125 ft = 6.25 ft) of the building in the X direction. The eccentricity of 5 percent of the plan width is consistent with the accidental torsion requirements of Section 12.8.4.2. The analysis was performed for this example by use of a computer program that can model inelastic structures. This analysis provided the curves

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Figure G13–2 System with One Wall (C) Removed.

Figure G13–3 Force–Deformation Relationship for Shear Wall. shown in Fig. G13–4. The upper curve represents the behavior of the system with all four walls in place, and the lower curve is for the system with wall C removed. The elastic analysis for each system is represented by the response up to first yield (the first change in slope of the curves), and the inelastic response is represented by the full curve. From the perspective of the elastic analysis, the structure with all walls in place can resist a lateral load V of 370 kips. At this load, walls C and D on the right side of the building carry 100 kips, and walls A and B carry 85 kips.

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Figure G13–4 Force–Deformation Plot for Structure with Three or Four Walls Using Inelastic Analysis. The ratio of the displacement d2 relative to the displacement at point d1 is 1.14, so according to Table 12.3-1, the structure does not have a torsional irregularity. When the inelastic response is considered, the structure can carry additional lateral load because walls A and B can each resist an additional 15 kips before they reach their 100-kips capacity. With four walls resisting 100 kips each, the total lateral capacity of the system is 400 kips. Recall that the force– displacement plot for this four-wall system is shown by the upper curve in Fig. G13–4. The displacement shown in the figure is the Y direction displacement d2. The first change of slope in the curve occurs when walls C and D yield, and the second change occurs when the walls A and B yield. When wall C is removed, the center of rigidity moves 9.375 ft to the left. When an elastic analysis is performed, the system can resist a lateral load of only 230 kips because at this point wall D reaches its 100-kips capacity. At this point, walls A and B resist 65 kips each. Additionally, the ratio of the displacement at point d2 with respect to the displacement at point d1 is 1.35. Hence, a torsional irregularity, but not an extreme irregularity, exists. The ratio of the resisting force of the three-wall system to that of the fourwall system is 230/370 or 0.621. On this basis, the system must be designed with r = 1.3 because it loses more than 33 percent of its strength. From the perspective of an inelastic analysis, the three-wall system can resist a total of 300 kips. This is shown by the force–displacement diagram (the lower curve) in Fig. G13–4. The ratio of the inelastic resisting force of the three-wall system to that of the four-wall system is 300/400 = 0.75. In this

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case, the system could theoretically be designed with r = 1.0 because it passes both the strength and the nonextreme torsion irregularity tests. The inelastic strength ratios for the system with and without one wall removed could be obtained a priori, by simply summing the wall strengths in the direction of loading. Thus, it is not necessary to obtain the full inelastic force–deformation responses shown in Fig. G13–4. It is necessary, however, to perform sufficient analysis to determine whether the removal of one wall causes an extreme torsional irregularity.

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