The Pennsylvania State University. The Graduate School. College of Engineering EVALUATION OF LIMIT DESIGN FOR EARTHQUAKE-RESISTANT MASONRY WALLS

Transcription

1 The Pennsylvania State University The Graduate School College of Engineering EVALUATION OF LIMIT DESIGN FOR EARTHQUAKE-RESISTANT MASONRY WALLS A Thesis in Architectural Engineering by Bradley S. Frederick 2014 Bradley S. Frederick Submitted in Partial Fulfillment of the Requirements for the Degree of Master of Science August 2014

2 ii The thesis of Bradley S. Frederick was reviewed and approved* by the following: Andres Lepage Assistant Professor of Architectural Engineering Thesis Co-Advisor Chimay J. Anumba Professor or Architectural Engineering Head of the Department of Architectural Engineering Thesis Co-Advisor Ali M. Memari Professor of Architectural Engineering *Signatures are on file in The Graduate School

3 iii ABSTRACT A Limit Design methodology is presented and evaluated for the design of reinforced masonry walls subjected to in-plane seismic forces. The method was recently incorporated into the Building Code Requirements for Masonry Structures (TMS 402, 2013) as an alternative design option. Using the framework of conventional design methods, Limit Design applies concepts of displacement-based design to the controlling yield mechanism of a given wall configuration subjected to lateral seismic loading. Two design examples illustrate the application of Limit Design. The examples represent typical instances of building structures where the seismic force-resisting system consists of Special Reinforced Masonry Shear Walls as permitted in Minimum Design Loads for Buildings and Other Structures (ASCE/SEI 7, 2010). The selected examples involve configurations in which design decisions are not favorably addressed by the conventional design methods in TMS 402. The structures are analyzed and designed according to the Strength Design provisions of the TMS 402 (2013) code and the new Limit Design alternative. The design outcomes for both methods are compared and analyzed to illustrate the limitations of the Strength Design method and advantages of Limit Design.

4 iv TABLE OF CONTENTS List of Tables... vi List of Figures... vii Acknowledgments... ix Chapter 1: Introduction Statement of the Problem Objectives and Scope Organization... 2 Chapter 2: Limit Design Method Framework Code Provisions Design Steps Limitations... 7 Chapter 3: Design Example Description Linear-Elastic Analysis Strength Design Limit Design Nonlinear Static Analysis Summary Chapter 4: Design Example 2: Multistory Coupled Shear Walls Description Linear-Elastic Analysis Strength Design Limit Design Nonlinear Static Analysis Summary Chapter 5: Conclusions Appendix A: Spreadsheet Formulations for Strength Design A.1 Strength Design Calculations for Design Example A.2 Strength Design Calculations for Design Example

5 v Appendix B: Spreadsheet Formulations for Limit Design B.1 Limit Design Calculations for Design Example B.2 Limit Design Calculations for Design Example Appendix C: Practical Nonlinear Static Analysis of Masonry Walls C.1 Nonlinear Layer Model C.2 Nonlinear Analysis Results Appendix D: Tables Appendix E: Figures List of References Biographical Sketch

6 vi LIST OF TABLES Table 2.1 Limit Design Code and Commentary, Taken from TMS 402 (2013) Table 3.1 Story Weight above the Seismic Base and Vertical Distribution of Seismic Forces Table 3.2 Wall Reinforcement Schedule for Strength Design, S DS = Table 3.3 Wall Reinforcement Schedule for Limit Design, S DS = Table 3.4 Lateral Stiffness for Different Mesh Sizes Table 3.5 Material Properties for Nonlinear Static Analysis Table 4.1 Story Weights above the Seismic Base and Vertical Distribution of Seismic Forces Table 4.2 Wall Reinforcement Schedule for Strength Design, S DS = Table 4.3 Wall Reinforcement Schedule for Limit Design, S DS = Table 4.4 Lateral Stiffness for Different Mesh Sizes Table 4.5 Material Properties for Nonlinear Static Analysis

7 vii LIST OF FIGURES Figure 3.1 Building Description, Wall Elevation Figure 3.2 Shear Wall Reinforcement Layout Figure 3.3 Member Forces Due to Dead Load (1.0D) Figure 3.4 Member Forces Due to Live Load (1.0L) Figure 3.5 Member Forces Due to Earthquake Load (1.0E), S DS = Figure 3.6 Values of M/(Vd) based on 1.0E Figure 3.7 Member Forces Due to 1.2D + 0.5L + 1.0E, S DS = Figure 3.8 Member Forces Due to 0.9D + 1.0E, S DS = Figure 3.9 Masonry Model, Axial Direction Figure 3.10 Reinforcement Steel Model, Axial Direction Figure 3.11 Wall Shear vs. Roof Displacement, Wall Reinforcement per Limit Design (Table 3.3), Eastward Loading Figure 3.12 Wall Shear vs. Roof Displacement, Wall Reinforcement per Limit Design (Table 3.3), Westward Loading Figure 3.13 Wall Shear vs. Roof Displacement, Wall Reinforcement per Strength Design (Table 3.2), Eastward Loading Figure 3.14 Wall Shear vs. Roof Displacement, Wall Reinforcement per Strength Design (Table 3.2), Westward Loading Figure 4.1 Building Description: Floor Plan, Material Properties, Loads, and Seismic Design Parameters Figure 4.2 Building Description, Wall Elevation (East Line of Resistance) Figure 4.3 Shear Wall Reinforcement Layout Figure 4.4 Member Forces Due to Dead Load (1.0D) Figure 4.5 Member Forces Due to Earthquake Load (1.0E), S DS = Figure 4.6 Values of M/(Vd) based on 1.0E Figure 4.7 Member Forces Due to 1.2D + 1.0E, S DS = Figure 4.8 Member Forces Due to 0.9D + 1.0E, S DS = Figure 4.9 Wall Shear vs. Roof Displacement, Wall Reinforcement per Limit Design (Table 4.3), Northward Loading

8 viii Figure 4.10 Deformed Shape for Simplified Nonlinear Layer Model, Wall Reinforcement per Limit Design (Table 4.3) Figure 4.11 Shear in Beams vs. Roof Displacement Figure 4.12 Axial Force in Beams vs. Roof Displacement, Northward Loading Figure 4.13 Axial Force in Beams vs. Roof Displacement, Southward Loading Figure 4.14 Wall Shear vs. Roof Displacement, Wall Reinforcement per Strength Design (Table 4.2), Northward Loading Figure 4.15 Deformed Shape for Simplified Nonlinear Layer Model, Wall Reinforcement per Strength Design (Table 4.2) Figure C.1 Linear-Elastic Model with 6 in. by 4 in. Mesh, Design Example Figure C.2 Simplified Nonlinear Layer Model, Design Example Figure C.3 Masonry Model, Axial Direction Figure C.4 Reinforcement Steel Model, Axial Direction

9 ix ACKNOWLEDGMENTS Financial support provided by the NCMA Education and Research Foundation, and additional support from the Department of Architectural Engineering and the College of Engineering of The Pennsylvania State University (Penn State) made this study possible. The writer is grateful to his advisor, Dr. Andres Lepage, for the assistance, guidance, and continued support throughout this project. Special recognition is also due to Steve Dill from KPFF Consulting Engineers and Jason Thompson from the National Concrete Masonry Association (NCMA) for their valuable discussions, suggestions, and contributions to this project. Appreciation is also given to Dr. Chimay Anumba, the writer s co-advisor, and Dr. Ali Memari for their participation as members of the thesis committee. Finally, special appreciation is due to my mother, Stephanie, and to my brother, sister, and grandparents for their support throughout my undergraduate and graduate studies at Penn State.

10 1 CHAPTER 1: INTRODUCTION 1.1 Statement of the Problem Structural engineers are provided with several design methods in the Building Code Requirements for Masonry Structures (TMS 402, 2013) for the seismic design of reinforced masonry shear walls. Code provisions in TMS are included for Allowable Stress Design (ASD) and Strength Design (SD). The conventional methods supported by ASD and SD are wellsuited for cantilever walls but become troublesome when applied to cases of walls perforated by windows or other openings. A new design procedure, Limit Design, was introduced in Appendix C of TMS to further encourage the use of structural masonry in earthquakeresistant construction, especially to address perforated wall types commonly found in building structures. Trial designs led to the development of the Limit Design method, and eventually the design approach evolved into Appendix C of TMS 402 (2013). With the adopted Limit Design code provisions, together with changes in the ASD and SD provisions, there is a need for evaluating the Limit Design method and comparing its design outcome with solutions derived from the use of conventional methods. 1.2 Objectives and Scope There are two main objectives in this study. The first objective is to evaluate the Limit Design method in TMS Appendix C by comparing the solutions of two design examples with the designs obtained by following the conventional Strength Design provisions in the main body of the TMS code. Second, the study aims to utilize the design examples to produce a design guide for the Limit Design method. For this purpose, the study documents a detailed set of structural design calculations. The design guide will be available for structural engineers to use as a resource alongside the new Limit Design provisions introduced in the TMS

11 2 code. The information gathered from this study directly influences design professionals in the architecture, engineering, and construction industry. The evaluation of Limit Design includes the implementation of static nonlinear analysis for corroborating the proper selection of the controlling yield mechanism. Recommended computer modeling assumptions are presented for a simplified but effective modeling technique. The practical nonlinear modeling approach is expected to motivate designers to use nonlinear computer models as a complementary design tool. 1.3 Organization The main body of this thesis comprises five chapters; Chapter 1 contains the Introduction and Chapter 5 the Conclusions. The manuscript is extensively supported by five appendices (A to E) and a List of References. Chapter 2 introduces the Limit Design method for new construction of earthquakeresistant masonry walls. The chapter describes the underlying framework of the method and presents the code and commentary language that has been adopted by the TMS 402 (2013) code. An outline of essential design steps and considerations is presented along with a summary of the limitations associated with the method. Chapter 3 presents Design Example 1, a reinforced masonry shear wall that is part of the lateral force-resisting system of a single-story auto repair facility. An exterior perforated wall, built with fully-grouted concrete masonry units, is designed according to the provisions of the TMS 402 (2013) code following the Strength Design approach and the Limit Design alternative. Comparisons of both design outcomes are based on the reinforcement quantities and the expected performance inferred from nonlinear static analysis. The nonlinear computer models are based on the simplified Nonlinear Layer model developed for this thesis. Design calculations

12 3 are presented in Appendix A and B for Strength Design and Limit Design, respectively. The nonlinear modeling technique is described in Appendix C. Chapter 4 presents Design Example 2, a reinforced masonry shear wall that is part of the lateral force-resisting system of a multistory residential building. The exterior wall of the building, built with fully-grouted clay masonry units, consists of three vertical wall segments coupled with deep beams at each floor level. The central vertical wall segment is supported by a deep transfer girder that allows for a large opening in the bottom story. Linear and nonlinear static analyses of the wall configuration are performed along with a comparison of the designs obtained following the conventional Strength Design provisions and the Limit Design alternative of the TMS 402 (2013) code. Strength Design calculations are included in Appendix A while those for Limit Design are presented in Appendix B. The simplified computer modeling technique, supporting the static nonlinear analyses, is presented in Appendix C.

13 4 CHAPTER 2: LIMIT DESIGN METHOD This chapter introduces the Limit Design method for reinforced masonry walls in earthquake-resistant construction. The chapter describes the underlying framework of the method, presents the code and commentary supporting the use of the method in TMS 402 (2013), outlines essential design steps, and indicates limitations of the method. 2.1 Framework Several design methods are available in the Building Code Requirements for Masonry Structures (TMS 402, 2013) for the use of special reinforced masonry shear walls as part of the lateral force-resisting system. Conventional design procedures in the TMS 402 code are wellsuited for cantilever rectangular walls controlled by flexural yielding, but the code is often difficult to apply for cases of irregular wall configurations or where the design of wall segments are predominantly controlled by shear strength. Traditional design of earthquake-resistant reinforced masonry shear walls has focused on the response of wall components without careful consideration of the global system behavior that may severely affect local component demands. Under many circumstances, these design procedures lead to impractical solutions or deficient designs that are incapable of meeting the desired seismic performance. Limit Design was developed as an alternative to the provisions for maximum area of flexural tensile reinforcement and special boundary elements presented in the TMS 402 code. The Limit Design provisions in TMS are intended to apply to perforated walls or where the use of conventional methods severely limits the usable strength of wall segments dominated by shear. The Limit Design method is based on the use of moment redistribution in a shear wall system due to the generalized yielding of its wall components, leading to the development of a limiting structural mechanism. The method applies to a single line of

14 5 resistance that is part of the seismic force-resisting system. The desired outcome of Limit Design is a more economical design capable of achieving superior performance. Limit Design directs the designer to focus on the portions of the structure that are subjected to inelastic deformation demands due to earthquake loading. 2.2 Code Provisions Initial efforts on developing a method similar to Limit Design originated during the early 2000s from activities of committee TS5 of the Building Seismic Safety Council (BSSC). Later in 2007, the Executive Committee of TMS 402 formed the Ductility Task Group for the purpose, in part, of developing code and commentary for inclusion of Limit Design into the TMS 402 code. The code and commentary in Appendix C of TMS for the Limit Design method, presented in Table 2.1, are a slightly modified version of those presented by Lepage et al. (2011). The version adopted in TMS incorporates changes in response to ballot comments submitted by the Seismic Subcommittee of TMS 402. The Limit Design provisions in Appendix C of TMS use the framework of the Strength Design provisions in Chapter 9 of TMS Limit Design combines linear-elastic analysis with concepts from simple plastic theory and displacement-based design to determine the in-plane strength and deformation capacities that may be safely assigned to a masonry wall configuration. The codified deformation capacities assigned to wall components are supported by experimental data of masonry walls subjected to reversed cyclic lateral loading. The evaluation of the experimental data by Sanchez (2012) included wall specimens tested by three different groups of researchers: Shing et al., 1989; Voon and Ingham, 2006; and Shedid et al., Sanchez (2012) observed that Limit Design (for new construction) assigns a lower deformation capacity to masonry wall segments than what is allowed in ASCE/SEI 41 (2013) for the seismic evaluation of existing buildings.

15 6 2.3 Design Steps An objective when developing the Limit Design method (Lepage et al., 2011) was to minimize changes to the requirements in ASCE/SEI 7 (2010) and TMS 402 (2013). Limit Design utilizes conventional seismic analysis, as specified in ASCE 7, and makes only minor adjustments to the Strength Design provisions of TMS 402. When using Limit Design, the designer is expected to follow a design process similar to that used in common practice. At the expense of requiring special checks, Limit Design generally leads to lower amount of reinforcement than conventional Strength Design. The additional checks include determining the controlling yield mechanism and evaluating the deformation capacity of the yielding wall components. The extra effort is only justified when the use of conventional Strength Design becomes impracticable. When applying Limit Design, the steps to follow involve: 1. Perform conventional seismic analysis Conduct a linear-elastic analysis in compliance with the Seismic Design Requirements for Building Structures in ASCE/SEI 7 Use reduced stiffness properties to account for the effects of cracked sections Determine the roof displacement and story forces associated with the maximum base shear at the line of lateral resistance under consideration 2. Define reinforcement layout Select reinforcement size and spacing Satisfy minimum reinforcement requirements for special reinforced masonry shear walls specified in TMS Chapter 7 3. Determine the controlling yield mechanism Identify the potential plastic hinge regions Assign plastic hinge strengths based on nominal flexural strength (M n ) Determine the limiting mechanism (assume profile of story forces are proportional to those determined in Step 1)

16 7 4. Check for shear-controlled wall segments and adjust plastic hinge strengths If V n < 2V Mn then wall segment is shear controlled [V Mn is the shear associated with M n ] If V n 2V Mn then wall segment is not shear controlled Adjust the plastic hinge strength, M p, using If V n < 2V Mn then M p = M n (V n / 2V Mn ) [shear-controlled condition] else M p = M n 5. Check mechanism strength Compute the limiting base shear strength, V lim, for the controlling yield mechanism (Step 3) using adjusted plastic hinge strengths (Step 4) Check φv lim V ub (φ = 0.8) [V ub is the base shear demand at the line of lateral resistance under consideration] 6. Check deformation capacities Deformation capacity, δ cap, of a wall segment is defined using δ cap = 0.5 l w h w ε mu / c [See notation in TMS 402 (2013)] For shear-controlled wall segments, δ cap = h w / 400, except that h w / 200 applies where conditions in TMS Section C.3.2 are met Compute deformation demands by imposing the calculated design roof displacement to the controlling yield mechanism. Compare deformation capacities with deformation demands. Based on the required checks in Step 4 to Step 6, the design may require proceeding iteratively (including restarting from Step 1 if wall dimensions are changed) until an acceptable solution is obtained. 2.4 Limitations Several restrictions are imposed on the use of Limit Design to ensure that its design outcome has adequate deformation capacity to mobilize the yield mechanism. Accordingly,

17 8 Limit Design applies only to Special Reinforced Masonry Shear Walls (SRMSW). The TMS 402 (2013) code defines SRMSW as a masonry shear wall designed to resist lateral forces while considering stresses in reinforcement and to satisfy special reinforcement and connection requirements. The special requirements provide the highest level of ductility for masonry walls. For masonry structures, ASCE/SEI 7 (2010) assigns the highest R value to SRMSW. Where masonry wall systems are selected as the seismic force-resisting system in buildings assigned to the highest Seismic Design Categories (D, E, and F), ASCE/SEI 7 requires the use of SRMSW. However, the use of SRMSW is still permitted for any Seismic Design Category. Combined axial loads due to gravity and seismic effects are limited to compressive forces not exceeding 0.3 f m A g. This limitation is intended to allow only those designs where yielding of the reinforcement in tension occurs before crushing of the masonry in compression. For any yielding wall segment, in the event that the nominal shear strength (V n ) does not exceed two times the shear associated with the development of the nominal flexural strength (2 V Mn ), the wall is considered shear controlled. An additional limitation requiring V n V Mn needs to be imposed. This limitation is intended to favor the development of flexural yielding consistent with the controlling mechanism (Step 3 of Section 2.3).

18 9 CHAPTER 3: DESIGN EXAMPLE 1 The first design example evaluated in this study is a single-story masonry wall structure providing lateral force resistance for a small commercial building. The relatively simple wall is perforated by two openings of different sizes. The resulting geometry is a wall that can be considered as a deep horizontal wall segment supported by three vertical wall segments. In this example, the wall is designed according to the provisions of the TMS 402 (2013) code for Strength Design (SD) and Limit Design (LD). For a direct comparison, the same reinforcement layout is used in both designs and the required bar sizes are determined for each design method. A nonlinear static analysis is conducted to allow comparison of the controlling mechanism and its strength for each of the two design outcomes (SD vs. LD). 3.1 Description The building represented in Design Example 1 is a small auto repair facility. An exterior wall forms part of the gravity and seismic force-resisting system of the building. The line of lateral force resistance under consideration acts in the east-west direction of the building and consists of a special reinforced masonry shear wall as defined in ASCE/SEI 7 (2010). The wall geometry, material properties, and design parameters are described in Figure 3.1. The seismic force-resisting system perpendicular to this line of resistance is not evaluated as part of this example. The minimum wall reinforcement was determined based on the prescriptive requirements in Chapter 7 of TMS The maximum spacing of the reinforcement for vertical wall segments must be the minimum of one-third the width of the wall, one-third its height, or 24 in. In this case, the governing condition is the width of the wall, which limits the spacing of both the horizontal and vertical reinforcement. A spacing of 16 in. was chosen for the end walls and 8 in. for the central wall. Reinforcement spacing for other wall segments were determined similarly. The resulting layout of reinforcement is shown in Figure 3.2.

19 10 Two design options are considered, Strength Design and Limit Design, to determine the required bar sizes for the reinforcement layout of Figure 3.2. Design actions are defined based on the basic load combinations in ASCE/SEI 7-10 Section (which include effects due to vertical seismic forces). Strength Design predominantly follows the requirements in Chapter 9 of TMS and Limit Design follows Appendix C of TMS The wall examined in Design Example 1 is constructed of fully grouted 8-in. by 8-in. by 16-in. concrete masonry units. The two openings in the wall form three vertical wall segments connected by coupling beams, see Figure 3.1. The walls are labeled A, B, and C from left to right. Wall A is 4-ft long by 10-ft tall, Wall B is 2-ft long by 8-ft tall, and Wall C is 4-ft long by 8-ft tall. The coupling beam joining Walls A and B has a clear span of 10 ft and a depth of 8 ft while the coupling beam joining Walls B and C has a clear span of 4 ft and a depth of 10 ft. The roof diaphragm connects to the wall 16 ft above the base, and a 2-ft parapet extends above the roof level, making the total height of the wall 18 ft. Dead and live loads from the tributary roof are carried by the wall. The roof framing consists of open-web steel trusses with plywood sheathing and built-up roofing. The tributary gravity loads correspond to half the spacing of the steel trusses. In addition to carrying these loads, the wall supports its own weight. The seismic base shear along the plane of the wall is determined according to the ASCE/SEI 7 (2010) equivalent lateral force procedure for a building responding in the constant acceleration region of the design spectrum. Data for story weight, height, and force are presented in Table 3.1. Trial designs to satisfy the Strength Design and Limit Design provisions of TMS led to the reinforcement indicated in Tables 3.2 and 3.3 (for the layout shown in Figure 3.2). Further commentary on the calculations supporting these designs is presented below.

20 Linear-Elastic Analysis To determine the design forces acting on each wall segment, a two-dimensional (2D) linear-elastic computer model of the structure was developed using program SAP2000 (CSI, 2011b). The model provides the required design forces and displacements for each of the design methods considered. To properly examine the necessary load combinations, the model accounts for dead, live, and seismic loads. Figures 3.3, 3.4, and 3.5, respectively show the unfactored forces (axial, P; shear, V; and moment, M) acting at the ends of the vertical wall segments. Values of M/Vd corresponding to seismic forces are displayed in Figure 3.6. Factored forces corresponding to the design load combinations are presented in Figures 3.7 and 3.8. The following assumptions and simplifications were used in the 2D linear-elastic model of Design Example 1: The structure, loads, and response are defined in one vertical plane. Structural response accounts for the effects of shear, axial, and flexural deformations. The wall segments are modeled using area elements with an 8-in. square mesh. This size matches the nominal 8-in. modular dimension of each masonry unit with nominal dimensions of in. (thickness height length). See Table 3.4 for the effects of the mesh size on the computed lateral stiffness of the wall. The foundation of the structure is assumed rigid; all nodes at the ground level are fixed. Dead, live, and seismic loads are uniformly distributed to the nodes at the roof diaphragm level. P- effects are neglected.

21 Strength Design Spreadsheet formulations for the design of Design Example 1 according to Strength Design provisions of the TMS 402 (2013) code are presented in Appendix A.1. The spreadsheet performs a series of checks for code provisions related to the effects of combined axial, flexural, and shear forces that apply to the design of special reinforced masonry shear walls. The programming of the spreadsheet allows automatic recalculations for changes in selected input. The input data include: geometry, reinforcement, material properties, seismic design parameters, modeling assumptions, gravity loading, and seismic forces and displacements. To support compliance with ASCE/SEI 7 (2010) and TMS 402 (2013), the spreadsheet reports: design forces for load combinations, combined axial and flexural design, shear design, and boundary element compliance. Design forces are obtained from the linear-elastic model described in Section 3.2. An arbitrary base shear of 100 kip was applied to the structure to determine internal forces (axial, shear, and moment) acting at the ends of each wall segment. Once the actual design base shear from the equivalent lateral force procedure is determined, the spreadsheet automatically scales the design forces for each wall segment based on the actual design base shear. For example, if the design base shear for the wall is 90 kip, the user enters that value into the spreadsheet, and a scale factor of 90/100 is applied to all of the forces obtained from the linear model. The design earthquake for Design Example 1 is characterized by a short-period spectral response acceleration value of S DS = 1.0 which gives a base shear demand of 41.1 kip (see Table 3.1). For a direct comparison of the two design methods (Strength Design vs. Limit Design), the reinforcement layout is driven by the maximum spacing requirements in TMS Section In addition to spacing requirements, bar sizes were limited to minimum bar sizes of #4 bars in the vertical direction and #3 bars in the horizontal direction. The vertical wall

22 13 segments were symmetrically reinforced but were not limited to having identical reinforcement. Shear design of the wall segments was controlled by the provisions in Section of TMS , where the design strength, φv n, needs to exceed the smaller of: (a) the shear associated with 1.25 M n, and (b) 2.0 V u. In general, the cap of 2.0 V u controlled the shear demand of the vertical wall segments, indicating that Walls A, B, and C are likely to reach their shear capacity before yielding in flexure. The horizontal wall segments connecting the wall piers were reinforced satisfying the spacing requirements of TMS Section The Strength Design spreadsheet indicates that, for the reinforcement described in Table 3.2, the structure is adequate. Thus, the vertical wall segments (Walls A, B, and C) reinforced with the selected bar sizes at maximum spacing is sufficient. The selected minimum bar sizes (#4 verticals and #3 horizontals) at maximum prescriptive spacing is adequate for Wall B. For Wall A, flexural reinforcement was upsized to #5 verticals while for Wall C, #6 verticals and #4 horizontals are sufficient to satisfy demands. Careful inspection of the spreadsheet shows that for the reinforcement described in Table 3.2, maximum demand-to-capacity ratios, M u /φm n and Vu/φV n, were 0.99 and 1.0 for Wall A, 0.81 and 0.45 for Wall B, and 0.85 and 0.96 for Wall C. The controlling load combination for the vertical reinforcement was usually 0.9D + 1.0E. In general, the horizontal shear reinforcement, as stated above, was controlled by TMS Section where V n need not exceed 2.5V u. All wall segments complied with the requirements of TMS Section to avoid the need of special boundary elements. All walls (A, B, and C) satisfied the rapid screening method (Section ) but none of the walls satisfied the provisions related to the maximum extreme fiber compressive stress (Section ) or the maximum area of flexural tensile reinforcement (Section ).

23 Limit Design Spreadsheet formulations for Design Example 1 are presented in Appendix B (Section B.1) incorporating the Limit Design provisions of TMS Appendix C. The spreadsheet performs a series of checks for the code provisions addressing the effects of combined axial, flexural, and shear forces that apply to the design of special reinforced masonry shear walls. The programming of the spreadsheet allows direct recalculations for updates in selected input. The input data in the worksheets are organized in various categories: geometry, wall reinforcement, material properties, seismic design parameters, modeling assumptions, gravity loading, and seismic forces and displacements. The spreadsheet calculations support compliance with ASCE/SEI 7-10 and TMS by reporting design forces for load combinations, axial-flexure (P-M) interaction diagrams, wall hinge strengths including shear strengths, limit mechanism, mechanism strength, and deformation capacities of the yielding wall segments. Proceeding similarly as in Strength Design (Section 3.3), wall forces and displacements are obtained from the linear-elastic model described in Section 3.2. To check the necessary load combinations specified in ASCE/SEI 7 (2010), the model considers dead, live, and seismic loads, including vertical earthquake effects. The equivalent lateral force procedure of ASCE/SEI 7 is used to calculate the lateral force applied to the structure at the roof diaphragm. Since this building has only a single story, a single lateral load at the roof level equals the base shear demand for the line of lateral resistance. The design earthquake is defined by a short-period spectral response acceleration parameter of S DS = 1.0, which gives a base shear of 41.1 kip at the base of the wall (see Table 3.1). The structure responds in the constant acceleration region of the design spectrum. The chosen reinforcement layout presented in Figure 3.2 is based on the maximum allowed spacing (TMS Section ). The minimum size of reinforcement is set to #3

24 15 bars except for the longitudinal reinforcement in the vertical wall segments where the minimum size is set to #4 bars, and each vertical wall segment is symmetrically reinforced. Additionally, longitudinal reinforcement in the vertical wall segments is extended to the top of the wall to provide adequate development length and reinforcement in the joints. This minimum reinforcement layout provided a base shear strength of 41.2 kip, just greater than the base shear demand of 41.1 kip for this line of lateral resistance. The plastic hinge strengths are calculated for the wall sections where yielding is expected to occur, and the assigned flexural strengths of the plastic hinges are adjusted so that at any hinge the shear associated with the hinge flexural strength does not exceed half the shear strength of the yielding wall section. The required adjustment follows Section C.1 item (d) of TMS (see code provisions in Table 2.1). Any wall segment with adjusted plastic hinge strength is considered shear-controlled. Shear-controlled wall segments are assigned reduced deformation capacities (TMS Section C.3.2) and generally limit the mechanism strength (base shear strength) of the wall. The plastic hinge strengths are assessed using the axial forces corresponding to 0.9D 0.2S DS in accordance with TMS Section C.1 item (c). For a wall segment dominated by seismic forces or with negligible gravity loading, the adjustment factor (φ vo ) may be determined using φ vo = V n /2V Mn 1.0, where V Mn is the shear associated with the flexural strength, M n, and V n is the calculated shear strength. Values of φ vo less than one identify the shear controlled wall segments. It is recommended to limit φ vo to values greater than 0.5 in order to develop the flexural strength, M n, before reaching the shear strength, V n. The outcome of Limit Design for Design Example 1 shows various deviations with respect to Strength Design. Although Wall B is designed with the same reinforcement for both methods, Walls A and C require significantly less reinforcing steel for Limit Design than for Strength Design. As the tables show, Table 3.2 for Strength Design and Table 3.3 for Limit Design, Strength Design requires #5 vertical bars for Wall A whereas Limit Design requires #4

25 16 vertical bars. Wall C requires #6 vertical bars and #4 horizontal bars according to Strength Design, but only #4 vertical bars and #3 horizontal bars for Limit Design. The above suggests that by using Limit Design, the designer reduces the total wall reinforcement by about 30%, a more economical design than Strength Design. The spreadsheet calculations show that using the minimum shear reinforcement in the vertical walls does not cause a shear-controlled condition in any of the wall segments. In other words, the shear strength at the base of each wall is more than two times greater than the shear associated with the flexural strength at the base of the walls. As a result, there is no adjustment factor (φ vo ) applied to the hinge strengths of the walls and the deformation capacities of the walls do not have to comply with the more stringent limits of shear-controlled wall segments. The spreadsheet indicates that for the chosen wall reinforcement (size and spacing), the design has sufficient capacity for both strength and deformation demands. 3.5 Nonlinear Static Analysis A nonlinear static analysis was conducted to corroborate the controlling yield mechanism used in the application of the Limit Design method. The computer modeling technique is based on the simplified Nonlinear Layer model described in Appendix C of this document. The model incorporates nonlinear material properties based on the specified compressive strength of masonry (f m ) and the specified yield strength of reinforcement (f y ), as described in Table 3.5 and illustrated in Figures 3.9 and The computer model representing the wall designed using Limit Design incorporates the reinforcement schedule of Table 3.3 for the layout shown in Figure 3.2. The lateral force of this one-story structure is applied at the roof level and, because the wall in this example is not symmetric, the computer model considers separate cases for the force acting eastward and westward. The computed output is shown in terms of base shear versus roof displacement in Figures 3.11 and 3.12 for eastward and westward loading,

26 17 respectively. The yield mechanism is assumed to occur at the point where the slope of the force-displacement relationship (base shear vs. roof displacement) is reduced to 5% of the initial slope. This point is identified in Figures 3.11 and For eastward loading, the plotted data suggest a base shear strength of 59.1 kip. For westward loading, the data indicate a base shear strength of 51.0 kip. Wall A has a clear height greater than that of Wall C; therefore when Wall C is in compression due to seismic overturning, a greater base shear is attained. The increased capacity for the case of eastward loading is a consequence of Wall C having an increased flexural strength due to axial compression and a reduced clear height, both of which help increase the induced shear force. The base shear strength of 51.0 kip derived from the nonlinear analysis was nearly identical to the mechanism strength of 51.5 kip obtained from Limit Design (as documented in Appendix B Section B.1). The value of 51.5 kip does not include the strength reduction factor of 0.8 in TMS Section C.2. Careful inspection of the nonlinear analysis output shows that on the onset of developing the yield mechanism, the shear demand in the walls did not exceed the calculated shear strength in any of the yielding wall segments. A similar nonlinear static analysis was also performed using the wall reinforcement derived from the Strength Design provisions. The computer model representing the Strength Design product incorporates the reinforcement schedule of Table 3.2 for the reinforcement layout in Figure 3.2. Both designs, Strength Design and Limit Design, used the same reinforcement layout with different bar sizes. The computed base shear versus roof displacement is shown for two loading cases in Figures 3.13 and For eastward loading, Figure 3.13, the base shear strength is 86.7 kip. For westward loading, Figure 3.14, the base shear strength is indicated as 77.0 kip. These results show an increase in base shear capacity of nearly 50% for Strength Design when compared with Limit Design, even though the design base shear was the same in both design methods. The

27 18 results from the nonlinear analysis show that for the wall designed using Strength Design there is greater system overstrength than for the wall design using Limit Design. 3.6 Summary A single-story wall with two large openings was designed as a special reinforced masonry shear wall part of the seismic force-resisting system of a commercial building. The design outcomes were compared for two design options. The first design was based on the conventional Strength Design provisions in the TMS code and the second design was based on the Limit Design provisions in TMS Appendix C. The two design options used the same reinforcement layout but the required bar sizes were determined based on the specific requirements for each option in TMS The amount of reinforcement required by Limit Design was nearly 30% less than the amount required by Strength Design. A nonlinear static analysis was conducted for the wall reinforced following the Limit Design code provisions. Results from the analysis corroborated the controlling yield mechanism used with the Limit Design method. The base shear strength obtained from the nonlinear computer model was in close agreement with the mechanism strength calculated using Limit Design. The outcome of Design Example 1 indicates that Limit Design offers significant savings in the amount of reinforcement required when compared with the conventional Strength Design approach in TMS 402. More efficient designs with Limit Design expand the possibilities of earthquake-resistant masonry. In addition, the direct assessment of the deformation capacities of the yielding wall segments provides the designer with a better understanding of the expected seismic performance of the structural system.

28 19 CHAPTER 4: DESIGN EXAMPLE 2: MULTISTORY COUPLED SHEAR WALLS In the second design example, a five-story special reinforced masonry wall is designed for a residential structure. The wall consists of three vertical wall segments joined by coupling beams at each floor level. The center wall segment is supported by a deep transfer girder at the second floor level to create a large opening at the bottom level of the structure (below the seismic base). For this example, two design methods are compared: Strength Design (SD) and Limit Design (LD). Both methods follow the provisions of the TMS 402 (2013) code more specifically, the SD method follows Chapter 9 of the code and the LD method is carried out according to Appendix C of the code. A linear-elastic analysis is performed to determine design forces and displacements for the structure and a nonlinear static analysis is conducted to directly compare the controlling yield mechanism for the two designs (SD vs. LD). 4.1 Description Design Example 2 represents a typical multistory residential building where the lateral force-resisting system consists of special reinforced masonry shear walls according to ASCE/SEI 7 (2010). The typical exterior masonry shear wall in the north-south direction is designed in compliance with the detailing and reinforcement requirements of TMS 402 (2013). Details of the building are presented in Figure 4.1 including a description of materials, loads, and seismic design parameters. Description and design of the lateral force-resisting system in the east-west direction is not presented for this evaluation. The structure being studied in Design Example 2 is the exterior wall constructed of 6-in. by 4-in. by 12-in. (thickness by height by length) hollow brick masonry (fully grouted). The wall system consists of three vertical wall segments joined by coupling beams at each floor level as shown in Figure 4.2. The typical floor-to-floor height is 9 ft, resulting in a total height of 36 ft above the seismic base at level 2. The walls are labeled A, B, and C from left to right. The outer

29 20 walls (A and C) are each 10-ft long and the center wall (B) is 6-ft long. The coupling beams span 7 ft with a depth of 4 ft and the transfer girder at level 2 spans 20 ft with a depth of 8 ft. The floor at level 2 is a post-tensioned concrete slab above a reinforced concrete parking structure. The floor slab at level 2 acts as a rigid diaphragm. Additional walls at the first-story parking structure provide sufficient lateral stiffness to consider level 2 as the seismic base for the upper four stories of residential construction. The typical floor framing above level 2 consists of 1-in. lightweight concrete on 9/16-in. metal deck supported on light-gage steel joists. Story heights, story weights, story forces, and base shear for the north-south direction of loading are presented in Table 4.1. Seismic story forces are determined based on the equivalent lateral force procedure in ASCE/SEI 7 (2010) for a structure responding in the constant acceleration region of the design spectrum. The exterior masonry wall resists self-weight and lateral loads. The connections between the floor framing and the exterior wall are designed such that vertical floor loads are not transmitted to the wall. The exterior masonry shear walls provide lateral support in the north-south direction of the building while an interior steel frame carries the gravity loads. The typical exterior masonry shear wall is reinforced using the layout shown in Figure 4.3. This reinforcement layout meets requirements of the TMS 402 (2013) code for minimum reinforcement ratios, maximum spacing allowances, and strength requirements based on the forces determined from linear-elastic analysis of the structure. The reinforcement also complies with the prescriptive reinforcement in Section of TMS Trial designs, using Strength Design and Limit Design provisions in TMS , led to the reinforcement sizes presented in Tables 4.2 and 4.3. Details of the supporting calculations are presented below.

30 Linear-Elastic Analysis A two-dimensional (2D) linear-elastic model is developed for Design Example 2 using the program SAP2000 (CSI, 2011b). The linear-elastic model is used as a reference model to obtain design forces and displacements needed for the application of the Strength Design and Limit Design approaches, as described below in Sections 4.3 and 4.4. It is assumed that the 2D models in this study include the interaction with other lines of resistance and torsional effects, as required in ASCE/SEI 7 (2010). Figures 4.4 and 4.5 show forces (axial, P; shear, V; and moment, M) obtained from the linear-elastic model for gravity dead loads and seismic loads acting on the exterior wall. The P, V, and M values are given at the bottom and top sections of the vertical wall segments and at the left and right sections of the horizontal wall segments. Figure 4.6 displays the calculated M/(V d) ratios based on seismic forces at the interface of wall segments. These values are presented as a measure of the action (shear or flexure) dominating the behavior of the wall sections, higher values of M/(V d) lead to lower shear strength. Figures 4.7 and 4.8 show the section forces corresponding to the load combinations considered. The following general assumptions and simplifications were made in developing the 2D linear-elastic model for Design Example 2: The structure, loads, and response are defined in one vertical plane. Structural response accounts for the effects of shear, axial, and flexural deformations. The wall segments are modeled using area elements with a 6 4-in. mesh. This size corresponds to half the length of the two-core clay masonry unit with nominal dimensions of in. (thickness height length). See Table 4.4 for the effects of the mesh size on the computed lateral stiffness of the wall. A horizontal spring with a stiffness of 10,000 kip/in. is provided at level 2, which acts as the seismic base. The translational spring represents the combined lateral stiffness of the additional walls below level 2. The structure is assumed fixed at level 1 (ground level).

31 22 A rigid diaphragm constraint is assigned to nodes at level 2. Constraints are not assigned to nodes at other floor levels. Gravity loads due to self-weight are assigned to each 6 4-in. area element. No other gravity loads (dead or live) are assigned to the wall. Seismic story forces are applied at each floor level. The story force is uniformly distributed to the floor nodes. The vertical force distribution is in accordance to the values in Table 4.1. P- effects are neglected. 4.3 Strength Design The spreadsheet formulations presented in Appendix A (Section A.2) were developed for Design Example 2. The formulations incorporate the Strength Design provisions of the TMS 402 (2013) code. The spreadsheet performs a series of checks for code provisions related to the effects of combined axial, flexural, and shear forces that apply to the design of special reinforced masonry shear walls. The programming of the spreadsheet allows automatic recalculations for changes in selected input. The input data include: geometry, wall reinforcement, material properties, seismic design parameters, modeling assumptions, gravity loading, and seismic forces and displacements. To support compliance with ASCE/SEI 7 (2010) and TMS 402 (2013), the spreadsheet reports: design forces for load combinations, combined axial and flexural design, shear design, and boundary element compliance. Wall forces are obtained from the linear-elastic model described in Section 4.2. To check the necessary load combinations prescribed in ASCE/SEI 7 (2010), the model considers dead, live, and seismic loads, including vertical earthquake effects. However, since the wall is designed such that gravity loads from each floor are not transferred to the wall, there are no live loads applied to the wall. Therefore, the forces obtained from the model include only dead

32 23 (due to self-weight) and seismic loads. The seismic loads are applied to the structure as equivalent lateral forces at each floor level and are determined in accordance with the equivalent lateral force procedure of ASCE/SEI 7 for a structure responding in the constant acceleration region of the design spectrum. An arbitrary base shear of 100 kip is applied to the model with a vertical force distribution according to Table 4.1. When the actual base shear of the structure is calculated, based on the seismic design parameters (see Figure 4.1), the seismic forces in the spreadsheet are scaled based on the applied 100-kip base shear. For example, if the design base shear to assign to the line of lateral resistance is calculated to be 90 kip, all seismic forces obtained from the linear-elastic model are multiplied by a scale factor of 90/100. The spreadsheet automatically accounts for the scale factor derived after the user enters the design base shear. In this design example, the design earthquake is characterized by a shortperiod spectral response acceleration parameter of S DS = 1.0, which leads to a base shear of 86.6 kip at the seismic base (see Table 4.1). For easy comparison of the two design methods (Strength Design vs. Limit Design), the chosen reinforcement layout (see Figure 4.3) is based on the maximum allowed spacing (TMS Section ) and the smallest size of reinforcement for which the structure will pass all checks. The minimum size of reinforcement was set to #3 bars except for the longitudinal reinforcement in the vertical wall segments where the minimum size was set to #4 bars. The wall is symmetrically reinforced with coupling beams (from level 3 to roof) identically reinforced. The reinforcement layout in the vertical wall segments (walls A=C and wall B) remains constant from base to roof but the center wall (B) does not need to have the same bar sizes as those used in the end walls A and C. The coupling beams were exempted from the deep beams provisions (Section of TMS ) because the effective shear span ratio of the coupling beams is 1.75 (they are 4-ft deep and span 7 ft). Coupling beams subjected to seismic loading experience nearly constant shear throughout their clear span, a situation that differs from typical deep beams subjected to gravity loading. The deep beam provisions in the TMS 402 (2013) code are intended for

33 24 continuous beams with a shear span ratio not greater than 1.5. The transfer girder, supporting wall B, was considered a deep beam and therefore the spacing of the horizontal and transverse reinforcement was not permitted to exceed one-fifth of the beam depth nor 16 in. (TMS Sections and ). Shear design of the wall segments was controlled by the provisions in Section of TMS where the design strength, φv n, needs to exceed the smaller of: (a) the shear associated with 1.25 M n, and (b) 2.0 V u. With the exception of the coupling beams, the cap of 2.0 V u controlled, an indication that most wall segments are susceptible to reaching their shear capacity before yielding in flexure. The transfer girder at level 2 was checked using the overstrength factor, Ω 0, according to Section of ASCE/SEI The Strength Design spreadsheet indicates that, for the reinforcement described in Table 4.2, the structure is adequate. Thus, the vertical wall segments (walls A, B, and C) reinforced with the selected minimum bar sizes (#4 verticals and #3 horizontals) at maximum spacing is sufficient. The flexural and shear reinforcement in the coupling beams was upsized from #3 to #4 bars. For the transfer girder, only the flexural reinforcement was upsized from #3 to #4 bars. Careful inspection of the spreadsheet shows that for the reinforcement described in Table 4.2, the wall segments with the greater demand-to-capacity ratio (M u /φm n or V u /φv n ) were wall B (V u /φv n = 0.97, between levels 2 and 3) and the coupling beams (M u /φm n = 0.89, at level 3). The controlling load combination leading to the larger demand-to-capacity ratios was 1.2D + 0.5L + 1.0E. This load combination generally requires more flexural reinforcement in beams and more shear reinforcement in vertical wall segments when designing for the shear associated with 1.25M n. All wall segments comply with the requirements of TMS Section to avoid the need of special boundary element at the edges of shear walls. Additionally, all wall segments satisfy checks for the rapid screening method (Section ) and the maximum extreme fiber compressive stress (Section ).

34 Limit Design The spreadsheet formulations presented in Appendix B (Section B.2) incorporate the Limit Design provisions of TMS Appendix C. The spreadsheet performs a series of checks for the code provisions addressing the effects of combined axial, flexural, and shear forces that apply to the design of special reinforced masonry shear walls. The programming of the spreadsheet allows direct recalculations for updates in selected input. The input data in the worksheets are organized in various categories: geometry, wall reinforcement, material properties, seismic design parameters, modeling assumptions, gravity loading, and seismic forces and displacements. The spreadsheet calculations support compliance with ASCE/SEI 7-10 and TMS by reporting: design forces for load combinations, axial-flexure (P-M) interaction diagrams, wall hinge strengths including shear strengths, limit mechanism, mechanism strength, and deformation capacities of the yielding wall segments. In the same way as for Strength Design (Section 4.3), wall forces and displacements are obtained from the linear-elastic model described in Section 4.2. To check the necessary load combinations specified in ASCE/SEI 7 (2010), the model considers dead, live, and seismic loads, including vertical earthquake effects. Since the wall is designed such that gravity loads from each floor are not transferred to the wall, there are no live loads applied. The forces obtained from the model include only dead (due to self-weight) and seismic loads. The seismic loads are applied to the structure as equivalent lateral forces at each floor level and are determined in accordance with the equivalent lateral force procedure of ASCE/SEI 7. The seismic response of the structure is assumed to occur in the constant acceleration region of the design spectrum. The design earthquake is defined by a short-period spectral response acceleration parameter of S DS = 1.0, which leads to a base shear of 86.6 kip at the seismic base (see Table 4.1).

35 26 The chosen reinforcement layout (see Figure 4.3) is based on the maximum allowed spacing (TMS Section ). The minimum size of reinforcement was set to #3 bars except for the longitudinal reinforcement in the vertical wall segments where the minimum size was set to #4 bars. The wall is symmetrically reinforced. This minimum reinforcement layout led to a base shear strength of 90.2 kip, greater than the base shear demand of 86.6 kip at the exterior wall. These values are documented in Appendix B (Section B.2), see also Table 4.1. The coupling beams were exempted from the deep beams provisions (Section of TMS ) because the effective shear span ratio of the coupling beams is 1.75 (they are 4-ft deep and span 7 ft). Coupling beams subjected to seismic loading experience nearly constant shear throughout their clear span, a situation that differs from typical deep beams subjected to gravity loading. The deep beam provisions in the TMS 402 (2013) code are intended for continuous beams with a shear span ratio not greater than 1.5. The transfer girder, supporting wall B, was taken as a deep beam and therefore the spacing of the horizontal and transverse reinforcement was not permitted to exceed one-fifth of the beam depth nor 16 in. (TMS Sections and ). The plastic hinge strengths are calculated for the wall sections where yielding is expected and the assigned flexural strengths of the plastic hinges are adjusted so that at any hinge, the shear associated with the hinge flexural strength does not exceed half the shear strength of the yielding wall section. The required adjustment follows Section C.1 item (d) of TMS (see code provisions in Table 2.1). Any wall segment with adjusted plastic hinge strength is considered shear-controlled. Shear controlled wall segments are assigned reduced deformation capacities (refer to TMS Section C.3.2) and generally limit the mechanism strength (base shear strength). The plastic hinge strengths are assessed using the axial forces corresponding to 0.9D 0.2S DS in accordance with TMS Section C.1 item (c). For a wall segment dominated by seismic forces or with negligible gravity loading, the adjustment factor (φ vo ) may be determined using φ vo = V n /2V Mn 1.0, where V Mn is the shear

36 27 associated with the flexural strength, M n, and V n is the calculated shear strength. Values of φ vo less than one identify the shear controlled wall segments. It is recommended to limit φ vo to values greater than 0.5 in order to develop flexural strength, M n, before reaching the shear strength, V n. The outcome of Limit Design for Design Example 2 is very similar to that of Strength Design. However, using the reinforcement layout shown in Figure 4.3, the two designs led to different amounts of flexural and shear reinforcement in the coupling beams. Table 4.2 shows that Strength Design requires #4 bars for the longitudinal and transverse reinforcement of the coupling beams while Limit Design requires #3 bars. Thus, by using Limit Design, the designer reduces the coupling beam reinforcement by approximately 50%. This translates to a more economical design when using Limit Design versus Strength Design, with a reduction of about 25% in the overall wall reinforcement. The spreadsheet calculations show that using the minimum shear reinforcement in the vertical wall segments leads to a shear-controlled condition in all three walls (A, B, and C). The shear strength at the base of walls A and C was 1.50 times greater than the shear associated with the flexural strength at the base of the walls. At the base of wall B, the shear strength was 1.68 times greater than the shear associated with its flexural strength. Because 1.50 and 1.68 are smaller than 2, the walls are considered shear controlled and the hinge strength was reduced by φ vo = 1.50/2 = 0.75 for walls A and C, and φ vo = 1.68/2 = 0.84 for wall B, see TMS Section C.1 item (d). The deformation capacities of the yielding wall segments were also limited, see TMS Section C.3.2. The spreadsheet calculations in Appendix B (Section B.2) indicate that the design is still sufficient and does not require increasing the amount of flexural or shear reinforcement.

37 Nonlinear Static Analysis A nonlinear static analysis was performed to support the expected controlling yield mechanism used in the application of the Limit Design method. The computer modeling technique is based on the simplified Nonlinear Layer model described in Appendix C of this document. The model incorporates nonlinear material properties based on the specified compressive strength of masonry (f m ) and the specified yield strength of reinforcement (f y ), as described in Table 4.5 and illustrated in Figures C.3 and C.4. The computer model representing the wall designed using Limit Design incorporates the reinforcement schedule of Table 4.3 for the layout shown in Figure 4.3. Lateral forces were applied to the computer model following the vertical force distribution indicated in Table 4.1. The computed output is shown in terms of base shear versus roof displacement in Figure 4.9. The deformed shape associated with the development of the yield mechanism is shown in Figure It is reasonable to assume that the yield mechanism occurs at the point where the slope of the force-displacement relationship (base shear vs. roof displacement) is reduced to 5% of the initial slope. This point is identified in Figure 4.9, for which the base shear equals 156 kip. The deformed shape in Figure 4.10 clearly shows that all coupling beams above the seismic base yielded at their ends and walls A, B, and C yielded at the seismic base. The apparent mechanism in Figure 4.10 coincides with the mechanism used in the application of the Limit Design method. The mechanism strength calculated using the Limit Design method was 113 kip (without the strength reduction factor of 0.8 specified in TMS Section C.2) while the nonlinear analysis gave 156 kip, indicating that the Limit Design method is safe. The overstrength in the nonlinear computer model (in relation to the calculated mechanism strength) was mainly due to induced axial forces in the coupling beams causing an increase in their flexural strength. The computer model showed that, with increased roof displacement, there was an increase in the fraction of base shear

38 29 taken by the extreme compression wall (wall C for northward loading). The flow of lateral forces, from the tension walls to the compression wall, made the coupling beams act like drag struts. Figure 4.11 shows the variation in the coupling beam shears versus the roof displacement. The maximum shear acting in the coupling beams at any level is limited by the yield moment which, in turn, depends on the axial load. Figures 4.12 and 4.13 show the variation in the coupling beam axial forces versus the roof displacement. The shapes of the curves in Figure 4.11 are clearly related to the shapes of the curves in Figures 4.12 and Careful inspection of the nonlinear analysis output shows that on the onset of developing the yield mechanism, the shear demand did not exceed the calculated shear strength in any of the yielding wall segments. This inspection accounts for the axial forces induced by the lateral loads and their effects on the calculated shear strength. A similar nonlinear static analysis was also performed using the wall reinforcement derived from the Strength Design provisions. The computer model representing the Strength Design product incorporates the reinforcement schedule of Table 4.2 for the same reinforcement layout (Figure 4.3) that was used in Limit Design. The computed base shear versus roof displacement, for the wall designed using Strength Design, is shown in Figure 4.14 with a base shear strength of 195 kip, which is about 25% greater than the base shear strength for the wall designed using Limit Design. However, the deformed shape in Figure 4.15 indicates that the wall designed using Strength Design experiences tension yielding across the full section of the extreme tension wall and the coupling beams do not yield at all floor levels. By the time the wall develops its base shear strength, the shear stress acting on the compression wall was approximately 4.5 f (psi). This value is nearly 1.5 times greater m than the maximum shear stress reached in the wall designed using the Limit Design method. Considering that the vertical wall segments in both design outcomes (Strength Design and Limit Design) have the same reinforcement, these shear stresses suggest that Strength Design led to

39 30 vertical wall segments with a reduced margin of safety against a shear failure, a direct consequence of having stronger coupling beams in the wall designed using Strength Design. 4.6 Summary A multistory coupled wall was designed as a reinforced masonry shear wall part of the seismic force-resisting system of a residential building. The wall is a five-story structure with its seismic base at the top of a first-story podium. Shear wall design outcomes are compared for two design options. The first design is based on conventional Strength Design provisions in the TMS code and the second design was based on the Limit Design method in TMS Appendix C. The two design options were based on the use of the same reinforcement layout but the required bar sizes were determined following the TMS code provisions. The design outcomes differed only on the reinforcement of the coupling beams, with Strength Design requiring #4 bars for flexural and shear reinforcement instead of #3 bars required by Limit Design. This difference translated to Limit Design providing 25% savings in the total amount of wall reinforcement. The transfer girder at the podium level and all vertical wall segments required the same amount of reinforcement in both designs. A nonlinear static analysis was conducted on the wall reinforced according to the Limit Design code provisions. Results from the analysis confirmed the proper selection of the controlling yield mechanism used in the application of the Limit Design method. The controlling mechanism involved yielding of the coupling beams at all levels and yielding of the supporting vertical wall segments (walls A, B, and C) just above the seismic base. These results were more favorable than those obtained from the nonlinear analysis of the wall designed using conventional Strength Design. The Strength Design option led to a wall where a reduced number of plastic hinges developed in the coupling beams and the supporting walls yielded below the seismic base. The improved performance of the Limit Design product, in addition to the savings in the amount of reinforcement, clearly demonstrates the merits of Limit Design.

40 31 CHAPTER 5: CONCLUSIONS This study aimed at evaluating the Limit Design method presented in Appendix C of the TMS 402 (2013) code for earthquake-resistant masonry walls. Two design examples were analyzed using Strength Design and Limit Design. Detailed calculations were presented following both design approaches. Nonlinear static analyses were conducted to investigate the yielding mechanism and base shear strength of the design products. In Design Example 1, a single-story structure, both design methods used identical materials, seismic design parameters, reinforcement layout, and wall geometry. The amount of reinforcement required by the conventional Limit Design alternative was nearly 30% smaller than the amount required by the Strength Design approach. In Design Example 2, a multistory structure, both design methods were also conducted using identical materials, seismic design parameters, reinforcement layout, and wall geometry. For this more complicated wall configuration, the Limit Design method led to approximately 25% reduction in wall reinforcement when compared with the Strength Design solution. Results from nonlinear static analyses indicated that the yielding mechanism of the wall designed after Limit Design is characterized by flexural yielding of the coupling beams throughout the height of the structure in contrast with the wall designed after Strength Design, where yielding concentrated at the base of the wall. For a design method to be useful, it should be user-friendly and lead to efficient solutions. As the spreadsheet formulations for the methods show, Limit Design follows the framework of existing Strength Design. As such, it is relatively easy for designers to use Limit Design. The method allows for a direct assessment of the deformation capacity of individual wall components and by using limit analysis based on concepts of virtual work, designers can readily determine the base shear strength of the structure. The study also presented a practical

41 32 nonlinear modeling approach to facilitate the corroboration of the controlling yield mechanism and its strength. For the perforated wall configurations studied, Limit Design led to more rational and economical solutions. Compared with conventional Strength Design, Limit Design requires a smaller amount of reinforcement for a given base shear demand. In wall segments where flexural yielding is expected, the reduced amount of flexural reinforcement induces smaller shear and axial stresses, which improves deformation capacity. With the direct assessment of the deformation capacity of yielding wall segments, Limit Design provides a better understanding of the expected seismic performance of structural wall systems. With more efficient designs, Limit Design expands the possibilities of earthquakeresistant masonry.

42 33 APPENDIX A: SPREADSHEET FORMULATIONS FOR STRENGTH DESIGN This appendix contains detailed calculations for Design Examples 1 and 2 based on the Strength Design provisions of TMS 402 (2013). The Strength Design worksheet presents a general description of the wall; force and displacement demands; and capacities. The description of the structure includes geometry, material properties, and reinforcement (size and spacing). The demands are defined based on seismic design parameters, modeling assumptions, gravity loading, and load combinations. The capacities are determined using axial-flexure (P-M) interaction diagrams, shear strength calculations, and boundary element compliance.

43 34 A.1 Strength Design Calculations for Design Example 1 Design Example 1 Geometry Reinforcement Wall A Flexural Reinforcement Concrete masonry units, in., fully grouted A s,1 = (1) #5 = 0.31 in. d 1 = 4 in. A s,2 = (1) #5 = 0.31 in. d 2 = 20 in. A s,3 = (1) #5 = 0.31 in. d 3 = 28 in. A s,4 = (1) #5 = 0.31 in. d 4 = 44 in. (Assumed sufficient for all nonseismic load combinations) Wall A Shear Reinforcement A v = (1) #3 = 0.11 in. 16 in. o.c. (A v /s) = in. 2 /in. Wall B Flexural Reinforcement A s,1 = (1) #4 = 0.2 in. d 1 = 4 in. A s,2 = (1) #4 = 0.2 in. d 2 = 12 in. A s,3 = (1) #4 = 0.2 in. d 3 = 20 in. Wall B Shear Reinforcement A v = (1) #3 = 0.11 in. 8 in. o.c. (A v /s) = in. 2 /in.

44 35 Wall C Flexural Reinforcement A s,1 = (1) #6 = 0.44 in. d 1 = 4 in. A s,2 = (1) #6 = 0.44 in. d 2 = 20 in. A s,3 = (1) #6 = 0.44 in. d 3 = 28 in. A s,4 = (1) #6 = 0.44 in. d 4 = 44 in. Wall C Shear Reinforcement A v = (2) #4 = 0.4 in. 16 in. o.c. (A v /s) = in. 2 /in. Material Properties f' m = 1500 psi f y = 60 ksi E s = ksi E m = 1350 ksi (TMS ) ε mu = in./in. (TMS (c)) ε sy = in./in. Seismic Design Parameters S DS = 1 R = 5 (ASCE/SEI 7-10 Table ) C d = 3.5 Ω 0 = 2.5 E v = 0.2 S DS D (ASCE/SEI 7-10 Eq ) ρ = 1.0 Modeling Assumptions (Section properties based on 50% E = E m /2 = 675 ksi of gross section properties) G = 270 ksi ν = 0.25 Poisson's ratio Gravity Loading D Self Weight = 80 psf D Tributary = 150 plf L Tributary = 225 plf (Determined from linear-elastic analysis using SAP2000) Wall A P D(Top) = 7.7 kip P D(Bot) = 10.9 kip V D = 0.7 kip M D(Top) = 4.1 kip-ft M D(Bot) = 3.0 kip-ft

45 36 P L = 2.2 kip V L = 0.2 kip M L(Top) = 1.1 kip-ft M L(Bot) = 0.8 kip-ft Wall B P D(Top) = 5.6 kip P D(Bot) = 6.9 kip V D = 0.0 kip M D(Top) = 0.1 kip-ft M D(Bot) = 0.0 kip-ft P L = 1.5 kip V L = 0.0 kip M L(Top) = 0.0 kip-ft M L(Bot) = 0.0 kip-ft Wall C P D(Top) = 7.2 kip P D(Bot) = 9.8 kip V D = 0.7 kip M D(Top) = 3.5 kip-ft M D(Bot) = 2.2 kip-ft P L = 1.8 kip V L = 0.2 kip M L(Top) = 1.0 kip-ft M L(Bot) = 0.6 kip-ft Seismic Forces and Displacement (Determined from linear-elastic analysis using SAP 2000) Base Shear, V b = 41 kip (Demand on one line of Roof Displacement, δ R = in. resistance) Wall A P E = 19.7 kip V E = 15.2 kip M E(Top) = 58.7 kip-ft M E(Bot) = 93.1 kip-ft

46 37 Wall B P E = 5.3 kip V E = 4.6 kip M E(Top) = 15.5 kip-ft M E(Bot) = 21.3 kip-ft Wall C P E = 25.0 kip V E = 21.2 kip M E(Top) = 59.6 kip-ft M E(Bot) = kip-ft Design Forces for Load Combination 1.2D + 0.5L + 1.0E (Including E v ) Wall A P u(top) = 31.6 kip P u(bot) = 36.1 kip V u = 16.3 kip M u(top) = 65.0 kip-ft M u(bot) = 97.7 kip-ft Wall B P u(top) = 13.9 kip P u(bot) = 15.7 kip V u = 4.6 kip M u(top) = 15.7 kip-ft M u(bot) = 21.3 kip-ft Wall C P u(top) = 36.0 kip P u(bot) = 39.6 kip V u = 22.3 kip M u(top) = 65.0 kip-ft M u(bot) = kip-ft Design Forces for Load Combination 0.9D + 1.0E (Including E v ) Wall A P u(top) = kip (P u < 0, tension) P u(bot) = kip V u = 15.7 kip M u(top) = 61.6 kip-ft M u(bot) = 95.2 kip-ft

47 38 Wall B P u(top) = -1.4 kip (P u < 0, tension) P u(bot) = -0.5 kip V u = 4.6 kip M u(top) = 15.6 kip-ft M u(bot) = 21.3 kip-ft Wall C P u(top) = kip (P u < 0, tension) P u(bot) = kip V u = 21.7 kip M u(top) = 62.1 kip-ft M u(bot) = kip-ft

48 39 Combined Axial and Flexure Design Top of Wall A P u = 31.6 kip (1.2D+0.5L+E, incl. E v ) M u = 65.0 kip-ft φm n = kip-ft > M u OK P u = kip (0.9D+E, incl. E v ) M u = 61.6 kip-ft φm n = 93.2 kip-ft > M u OK Bottom of Wall A P u = 36.1 kip (1.2D+0.5L+E, incl. E v ) M u = 97.7 kip-ft φm n = kip-ft > M u OK P u = kip (0.9D+E, incl. E v ) M u = 95.2 kip-ft φm n = 96.8 kip-ft > M u OK 400 Wall A Axial Force (kip) φ = 0.9 φ = D+0.5L+E D-E Moment (kip-ft)

49 40 Combined Axial and Flexure Design (cont.) Top of Wall B P u = 13.9 kip (1.2D+0.5L+E, incl. E v ) M u = 15.7 kip-ft φm n = 36.3 kip-ft > M u OK P u = -1.4 kip (0.9D+E, incl. E v ) M u = 15.6 kip-ft φm n = 25.6 kip-ft > M u OK Bottom of Wall B P u = 15.7 kip (1.2D+0.5L+E, incl. E v ) M u = 21.3 kip-ft φm n = 37.5 kip-ft > M u OK P u = -0.5 kip (0.9D+E, incl. E v ) M u = 21.3 kip-ft φm n = 26.3 kip-ft > M u OK 200 Wall B 150 Axial Force (kip) φ = 0.9 φ = D+0.5L+E 0.9D-E Moment (kip-ft)

50 41 Combined Axial and Flexure Design (cont.) Top of Wall C P u = 36.0 kip (1.2D+0.5L+E, incl. E v ) M u = 65.0 kip-ft φm n = kip-ft > M u OK P u = kip (0.9D+E, incl. E v ) M u = 62.1 kip-ft φm n = kip-ft > M u OK Bottom of Wall C P u = 39.6 kip (1.2D+0.5L+E, incl. E v ) M u = kip-ft φm n = kip-ft > M u OK P u = kip (0.9D+E, incl. E v ) M u = kip-ft φm n = kip-ft > M u OK 500 Wall C Axial Force (kip) φ = 0.9 φ = D+0.5L+E 0.9D-E Moment (kip-ft)

51 42 Shear Design Top of Wall A t Wall = in. d v = 48 in. (1.2D+0.5L+E, incl. E v ) P u = 31.6 kip V u = 16.3 kip M u = 65.0 kip-ft M u /V u d v = 1.00 V n = V nm + V ns V n,max (TMS ) V nm = [4-1.75(M u /V u d v )]A n (f' m ) P u (TMS Eq. 9-24) V nm = 39.8 kip V ns = 0.5(A v /s)f y d v (TMS Eq. 9-25) (A v /s) = in. 2 /in. V ns = 9.9 kip V n,max = 4.0 A n (f' m ) (TMS Eq and Eq. 9-23) V n,max = 56.8 kip V n = 49.7 kip (TMS ) φ = 0.8 φv n = 39.8 kip > 16.3 kip OK Check φv n min( 2.0V E +V ug, 1.25V Mn +V ug ) (TMS ) V E = 15.2 kip (E h ) M E = 58.7 kip-ft (E h ) V ug = V u - V E = 1.1 kip (1.2D+0.5L+E, incl. E v ) M n = kip-ft (M n for P = P u, φ = 1.0) V Mn = M n V E /M E = 43.7 kip 2.0V E +V ug = 31.4 kip 1.25V Mn +V ug = 55.7 kip φv n = 39.8 kip > 31.4 kip OK

52 43 Shear Design (cont.) Top of Wall A t Wall = in. d v = 48 in. (0.9D+E, incl. E v ) P u = kip V u = 15.7 kip M u = 61.6 kip-ft M u /V u d v = 0.98 V n = V nm + V ns V n,max (TMS ) V nm = [4-1.75(M u /V u d v )]A n (f' m ) P u (TMS Eq. 9-24) V nm = 28.8 kip V ns = 0.5(A v /s)f y d v (TMS Eq. 9-25) (A v /s) = in. 2 /in. V ns = 9.9 kip V n,max = 4.0 A n (f' m ) (TMS Eq and Eq. 9-23) V n,max = 57.4 kip V n = 38.7 kip (TMS ) φ = 0.8 φv n = 30.9 kip > 15.7 kip OK Check φv n min( 2.0V E +V ug, 1.25V Mn +V ug ) (TMS ) V E = 15.2 kip (E h ) M E = 58.7 kip-ft (E h ) V ug = V u - V E = 0.5 kip (0.9D+E, incl. E v ) M n = kip-ft (M n for P = P u, φ = 1.0) V Mn = M n V E /M E = 27.4 kip 2.0V E +V ug = 30.9 kip 1.25V Mn +V ug = 34.8 kip φv n = 30.9 kip > 30.9 kip OK

53 44 Shear Design (cont.) Bottom of Wall A t Wall = in. d v = 48 in. (1.2D+0.5L+E, incl. E v ) P u = 36.1 kip V u = 16.3 kip M u = 97.7 kip-ft M u /V u d v = 1.50 V n = V nm + V ns V n,max (TMS ) V nm = [4-1.75(M u /V u d v )]A n (f' m ) P u (TMS Eq. 9-24) V nm = 40.9 kip V ns = 0.5(A v /s)f y d v (TMS Eq. 9-25) (A v /s) = in. 2 /in. V ns = 9.9 kip V n,max = 4.0 A n (f' m ) (TMS Eq and Eq. 9-23) V n,max = 56.7 kip V n = 50.8 kip (TMS ) φ = 0.8 φv n = 40.7 kip > 16.3 kip OK Check φv n min( 2.0V E +V ug, 1.25V Mn +V ug ) (TMS ) V E = 15.2 kip (E h ) M E = 93.1 kip-ft (E h ) V ug = V u - V E = 1.1 kip (1.2D+0.5L+E, incl. E v ) M n = kip-ft (M n for P = P u, φ = 1.0) V Mn = M n V E /M E = 28.2 kip 2.0V E +V ug = 31.4 kip 1.25V Mn +V ug = 36.4 kip φv n = 40.7 kip > 31.4 kip OK

54 45 Shear Design (cont.) Bottom of Wall A t Wall = in. d v = 48 in. (0.9D+E, incl. E v ) P u = kip V u = 15.7 kip M u = 95.2 kip-ft M u /V u d v = 1.52 V n = V nm + V ns V n,max (TMS ) V nm = [4-1.75(M u /V u d v )]A n (f' m ) P u (TMS Eq. 9-24) V nm = 28.9 kip V ns = 0.5(A v /s)f y d v (TMS Eq. 9-25) (A v /s) = in. 2 /in. V ns = 9.9 kip V n,max = 4.0 A n (f' m ) (TMS Eq and Eq. 9-23) V n,max = 56.7 kip V n = 38.8 kip (TMS ) φ = 0.8 φv n = 31.0 kip > 15.7 kip OK Check φv n min( 2.0V E +V ug, 1.25V Mn +V ug ) (TMS ) V E = 15.2 kip (E h ) M E = 93.1 kip-ft (E h ) V ug = V u - V E = 0.5 kip (0.9D+E, incl. E v ) M n = kip-ft (M n for P = P u, φ = 1.0) V Mn = M n V E /M E = 17.9 kip 2.0V E +V ug = 30.9 kip 1.25V Mn +V ug = 22.9 kip φv n = 31.0 kip > 22.9 kip OK

55 46 Shear Design (cont.) Top of Wall B t Wall = in. d v = 24 in. (1.2D+0.5L+E, incl. E v ) P u = 13.9 kip V u = 4.6 kip M u = 15.7 kip-ft M u /V u d v = 1.69 V n = V nm + V ns V n,max (TMS ) V nm = [4-1.75(M u /V u d v )]A n (f' m ) P u (TMS Eq. 9-24) V nm = 19.4 kip V ns = 0.5(A v /s)f y d v (TMS Eq. 9-25) (A v /s) = in. 2 /in. V ns = 9.9 kip V n,max = 4.0 A n (f' m ) (TMS Eq and Eq. 9-23) V n,max = 28.4 kip V n = 28.4 kip (TMS ) φ = 0.8 φv n = 22.7 kip > 4.6 kip OK Check φv n min( 2.0V E +V ug, 1.25V Mn +V ug ) (TMS ) V E = 4.6 kip (E h ) M E = 15.5 kip-ft (E h ) V ug = V u - V E = 0.0 kip (1.2D+0.5L+E, incl. E v ) M n = 39.4 kip-ft (M n for P = P u, φ = 1.0) V Mn = M n V E /M E = 11.7 kip 2.0V E +V ug = 9.2 kip 1.25V Mn +V ug = 14.6 kip φv n = 22.7 kip > 9.2 kip OK

56 47 Shear Design (cont.) Top of Wall B t Wall = in. d v = 24 in. (0.9D+E, incl. E v ) P u = -1.4 kip V u = 4.6 kip M u = 15.6 kip-ft M u /V u d v = 1.69 V n = V nm + V ns V n,max (TMS ) V nm = [4-1.75(M u /V u d v )]A n (f' m ) P u (TMS Eq. 9-24) V nm = 15.6 kip V ns = 0.5(A v /s)f y d v (TMS Eq. 9-25) (A v /s) = in. 2 /in. V ns = 9.9 kip V n,max = 4.0 A n (f' m ) (TMS Eq and Eq. 9-23) V n,max = 28.4 kip V n = 25.5 kip (TMS ) φ = 0.8 φv n = 20.4 kip > 4.6 kip OK Check φv n min( 2.0V E +V ug, 1.25V Mn +V ug ) (TMS ) V E = 4.6 kip (E h ) M E = 15.5 kip-ft (E h ) V ug = V u - V E = 0.0 kip (0.9D+E, incl. E v ) M n = 28.6 kip-ft (M n for P = P u, φ = 1.0) V Mn = M n V E /M E = 8.5 kip 2.0V E +V ug = 9.2 kip 1.25V Mn +V ug = 10.6 kip φv n = 20.4 kip > 9.2 kip OK

57 48 Shear Design (cont.) Bottom of Wall B t Wall = in. d v = 24 in. (1.2D+0.5L+E, incl. E v ) P u = 15.7 kip V u = 4.6 kip M u = 21.3 kip-ft M u /V u d v = 2.31 V n = V nm + V ns V n,max (TMS ) V nm = [4-1.75(M u /V u d v )]A n (f' m ) P u (TMS Eq. 9-24) V nm = 19.9 kip V ns = 0.5(A v /s)f y d v (TMS Eq. 9-25) (A v /s) = in. 2 /in. V ns = 9.9 kip V n,max = 4.0 A n (f' m ) (TMS Eq and Eq. 9-23) V n,max = 28.4 kip V n = 28.4 kip (TMS ) φ = 0.8 φv n = 22.7 kip > 4.6 kip OK Check φv n min( 2.0V E +V ug, 1.25V Mn +V ug ) (TMS ) V E = 4.6 kip (E h ) M E = 21.3 kip-ft (E h ) V ug = V u - V E = 0.0 kip (1.2D+0.5L+E, incl. E v ) M n = 40.5 kip-ft (M n for P = P u, φ = 1.0) V Mn = M n V E /M E = 8.7 kip 2.0V E +V ug = 9.2 kip 1.25V Mn +V ug = 10.9 kip φv n = 22.7 kip > 9.2 kip OK

58 49 Shear Design (cont.) Bottom of Wall B t Wall = in. d v = 24 in. (0.9D+E, incl. E v ) P u = -0.5 kip V u = 4.6 kip M u = 21.3 kip-ft M u /V u d v = 2.31 V n = V nm + V ns V n,max (TMS ) V nm = [4-1.75(M u /V u d v )]A n (f' m ) P u (TMS Eq. 9-24) V nm = 15.8 kip V ns = 0.5(A v /s)f y d v (TMS Eq. 9-25) (A v /s) = in. 2 /in. V ns = 9.9 kip V n,max = 4.0 A n (f' m ) (TMS Eq and Eq. 9-23) V n,max = 28.4 kip V n = 25.7 kip (TMS ) φ = 0.8 φv n = 20.6 kip > 4.6 kip OK Check φv n min( 2.0V E +V ug, 1.25V Mn +V ug ) (TMS ) V E = 4.6 kip (E h ) M E = 21.3 kip-ft (E h ) V ug = V u - V E = 0.0 kip (0.9D+E, incl. E v ) M n = 29.2 kip-ft (M n for P = P u, φ = 1.0) V Mn = M n V E /M E = 6.3 kip 2.0V E +V ug = 9.2 kip 1.25V Mn +V ug = 7.9 kip φv n = 20.6 kip > 7.9 kip OK

59 50 Shear Design (cont.) Top of Wall C t Wall = in. d v = 48 in. (1.2D+0.5L+E, incl. E v ) P u = 36.0 kip V u = 22.3 kip M u = 65.0 kip-ft M u /V u d v = 0.73 V n = V nm + V ns V n,max (TMS ) V nm = [4-1.75(M u /V u d v )]A n (f' m ) P u (TMS Eq. 9-24) V nm = 47.6 kip V ns = 0.5(A v /s)f y d v (TMS Eq. 9-25) (A v /s) = in. 2 /in. V ns = 36.0 kip V n,max = 4.7 A n (f' m ) (TMS Eq and Eq. 9-23) V n,max = 67.0 kip V n = 67.0 kip (TMS ) φ = 0.8 φv n = 53.6 kip > 22.3 kip OK Check φv n min( 2.0V E +V ug, 1.25V Mn +V ug ) (TMS ) V E = 21.2 kip (E h ) M E = 59.6 kip-ft (E h ) V ug = V u - V E = 1.1 kip (1.2D+0.5L+E, incl. E v ) M n = kip-ft (M n for P = P u, φ = 1.0) V Mn = M n V E /M E = 73.1 kip 2.0V E +V ug = 43.5 kip 1.25V Mn +V ug = 92.5 kip φv n = 53.6 kip > 43.5 kip OK

60 51 Shear Design (cont.) Top of Wall C t Wall = in. d v = 48 in. (0.9D+E, incl. E v ) P u = kip V u = 21.7 kip M u = 62.1 kip-ft M u /V u d v = 0.71 V n = V nm + V ns V n,max (TMS ) V nm = [4-1.75(M u /V u d v )]A n (f' m ) P u (TMS Eq. 9-24) V nm = 34.0 kip V ns = 0.5(A v /s)f y d v (TMS Eq. 9-25) (A v /s) = in. 2 /in. V ns = 36.0 kip V n,max = 4.8 A n (f' m ) (TMS Eq and Eq. 9-23) V n,max = 67.5 kip V n = 67.5 kip (TMS ) φ = 0.8 φv n = 54.0 kip > 21.7 kip OK Check φv n min( 2.0V E +V ug, 1.25V Mn +V ug ) (TMS ) V E = 21.2 kip (E h ) M E = 59.6 kip-ft (E h ) V ug = V u - V E = 0.5 kip (0.9D+E, incl. E v ) M n = kip-ft (M n for P = P u, φ = 1.0) V Mn = M n V E /M E = 52.2 kip 2.0V E +V ug = 42.9 kip 1.25V Mn +V ug = 65.7 kip φv n = 54.0 kip > 42.9 kip OK

61 52 Shear Design (cont.) Bottom of Wall C t Wall = in. d v = 48 in. (1.2D+0.5L+E, incl. E v ) P u = 39.6 kip V u = 22.3 kip M u = kip-ft M u /V u d v = 1.27 V n = V nm + V ns V n,max (TMS ) V nm = [4-1.75(M u /V u d v )]A n (f' m ) P u (TMS Eq. 9-24) V nm = 41.8 kip V ns = 0.5(A v /s)f y d v (TMS Eq. 9-25) (A v /s) = in. 2 /in. V ns = 36.0 kip V n,max = 4.0 A n (f' m ) (TMS Eq and Eq. 9-23) V n,max = 56.7 kip V n = 56.7 kip (TMS ) φ = 0.8 φv n = 45.4 kip > 22.3 kip OK Check φv n min( 2.0V E +V ug, 1.25V Mn +V ug ) (TMS ) V E = 21.2 kip (E h ) M E = kip-ft (E h ) V ug = V u - V E = 1.1 kip (1.2D+0.5L+E, incl. E v ) M n = kip-ft (M n for P = P u, φ = 1.0) V Mn = M n V E /M E = 40.1 kip 2.0V E +V ug = 43.5 kip 1.25V Mn +V ug = 51.3 kip φv n = 45.4 kip > 43.5 kip OK

62 53 Shear Design (cont.) Bottom of Wall C t Wall = in. d v = 48 in. (0.9D+E, incl. E v ) P u = kip V u = 21.7 kip M u = kip-ft M u /V u d v = 1.29 V n = V nm + V ns V n,max (TMS ) V nm = [4-1.75(M u /V u d v )]A n (f' m ) P u (TMS Eq. 9-24) V nm = 27.4 kip V ns = 0.5(A v /s)f y d v (TMS Eq. 9-25) (A v /s) = in. 2 /in. V ns = 36.0 kip V n,max = 4.0 A n (f' m ) (TMS Eq and Eq. 9-23) V n,max = 56.7 kip V n = 56.7 kip (TMS ) φ = 0.8 φv n = 45.4 kip > 21.7 kip OK Check φv n min( 2.0V E +V ug, 1.25V Mn +V ug ) (TMS ) V E = 21.2 kip (E h ) M E = kip-ft (E h ) V ug = V u - V E = 0.5 kip (0.9D+E, incl. E v ) M n = kip-ft (M n for P = P u, φ = 1.0) V Mn = M n V E /M E = 28.7 kip 2.0V E +V ug = 42.9 kip 1.25V Mn +V ug = 36.4 kip φv n = 45.4 kip > 36.4 kip OK

63 54 Boundary Element Compliance (TMS ) Wall A Rapid Screening (TMS ) V u = 16.3 kip (1.2D+0.5L+E, incl. E v ) M u = 97.7 kip-ft d v = 48 in. M u /(V u d v ) = 1.50 P u = 36.1 kip 0.10A g f' m = 54.9 kip Conditions 1 and 2: NG 3A n (f' m ) = 42.5 kip > V u Condition 3: OK Extreme Fiber Compressive Stress (TMS ) f max = P u /A g +M u /S 0.2f' m 0.2f' m = psi P u = 36.1 kip (1.2D+0.5L+E, incl. E v ) M u = 97.7 kip-ft A g = 366 in 2 S = 2928 in 3 f max = psi > psi NG Both Sections, and , are checked for illustration purposes (only one needs to be satisfied)

64 55 Boundary Element Compliance (cont.) (TMS ) Wall B Rapid Screening (TMS ) V u = 4.6 kip (1.2D+0.5L+E, incl. E v ) M u = 21.3 kip-ft d v = 24 in. M u /(V u d v ) = 2.31 P u = 15.7 kip 0.10A g f' m = kip Conditions 1 and 2: NG 3A n (f' m ) = 21.3 kip > V u Condition 3: OK Extreme Fiber Compressive Stress (TMS ) f max = P u /A g +M u /S 0.2f' m 0.2f' m = psi P u = 15.7 kip (1.2D+0.5L+E, incl. E v ) M u = 21.3 kip-ft A g = 183 in 2 S = 732 in 3 f max = psi > psi NG Both Sections, and , are checked for illustration purposes (only one needs to be satisfied)

65 56 Boundary Element Compliance (cont.) (TMS ) Wall C Rapid Screening (TMS ) V u = 22.3 kip (1.2D+0.5L+E, incl. E v ) M u = kip-ft d v = 48 in. M u /(V u d v ) = 1.27 P u = 39.6 kip 0.10A g f' m = 54.9 kip Conditions 1 and 2: NG 3A n (f' m ) = 42.5 kip > V u Condition 3: OK Extreme Fiber Compressive Stress (TMS ) f max = P u /A g +M u /S 0.2f' m 0.2f' m = psi P u = 39.6 kip (1.2D+0.5L+E, incl. E v ) M u = kip-ft A g = 366 in 2 S = 2928 in 3 f max = psi > psi NG Both Sections, and , are checked for illustration purposes (only one needs to be satisfied)

66 57 Maximum Area of Flexural Tensile Reinforcement (ρ max Check) (TMS ) According to (d), if is met then there is no need to check Wall A P u = 22.9 kip (D+0.75L+0.525E, not incl. E v ) V u = 8.8 kip M u = 52.5 kip-ft d v = 48 in. M u /V u d v = 1.49 ε s,max = 4.00 ε sy ε s,max = 8.3E-03 in./in. P n,max = 19.0 kip < 22.9 kip NG Wall B P u = 10.8 kip (D+0.75L+0.525E, not incl. E v ) V u = 2.4 kip M u = 11.2 kip-ft d v = 24 in. M u /V u d v = 2.31 ε s,max = 4.00 ε sy ε s,max = 8.3E-03 in./in. P n,max = 10.0 kip < 10.8 kip NG Wall C P u = 24.2 kip (D+0.75L+0.525E, not incl. E v ) V u = 12.0 kip M u = 60.5 kip-ft d v = 48 in. M u /V u d v = 1.26 ε s,max = 4.00 ε sy ε s,max = 8.3E-03 in./in. P n,max = -4.4 kip < 24.2 kip NG

67 58 A.2 Strength Design Calculations for Design Example 2 Design Example 2 Geometry Roof 36 ft 1750 Level 5 27 ft Level 4 18 ft Level 3 9 ft Level 2 Seismic Base OK 0 Clay masonry units, in., fully grouted

68 59 Reinforcement (Assumed sufficient for all nonseismic load combinations) Wall A = Wall C Longitudinal Reinforcement A s,1 = (1) #4 = 0.2 in. d 1 = 3 in. A s,2 = (1) #4 = 0.2 in. d 2 = 21 in. A s,3 = (1) #4 = 0.2 in. d 3 = 39 in. A s,4 = (1) #4 = 0.2 in. d 4 = 57 in. A s,5 = (1) #4 = 0.2 in. d 5 = 63 in. A s,6 = (1) #4 = 0.2 in. d 6 = 81 in. A s,7 = (1) #4 = 0.2 in. d 7 = 99 in. A s,8 = (1) #4 = 0.2 in. d 8 = 117 in. Wall A = Wall C Shear Reinforcement A v = (1) #3 = 0.11 in. 20 in. o.c. (A v /s) = in. 2 /in. Wall B Longitudinal Reinforcement A s,1 = (1) #4 = 0.2 in. d 1 = 9 in. A s,2 = (1) #4 = 0.2 in. d 2 = 27 in. A s,3 = (1) #4 = 0.2 in. d 3 = 45 in. A s,4 = (1) #4 = 0.2 in. d 4 = 63 in. Wall B Shear Reinforcement A v = (1) #3 = 0.11 in 20 in. o.c. (A v /s) = in. 2 /in. Coupling Beam Longitudinal Reinforcement A s,1 = (1) #4 = 0.2 in. d 1 = 3 in. A s,2 = (1) #4 = 0.2 in. d 2 = 19 in. A s,3 = (1) #4 = 0.2 in. d 3 = 29 in. A s,4 = (1) #4 = 0.2 in. d 4 = 45 in. Coupling Beam Shear Reinforcement A v = (1) #4 = 0.2 in. 12 in. o.c. (A v /s) = in. 2 /in. Transfer Girder Longitudinal Reinforcement A s,1 = (1) #4 = 0.2 in. d 1 = 9 in. A s,2 = (1) #4 = 0.2 in. d 2 = 25 in. A s,3 = (1) #4 = 0.2 in. d 3 = 41 in. A s,4 = (1) #4 = 0.2 in. d 4 = 55 in. A s,5 = (1) #4 = 0.2 in. d 5 = 71 in. A s,6 = (1) #4 = 0.2 in. d 6 = 87 in.

69 60 Transfer Girder Shear Reinforcement A v = (1) #3 = 0.11 in. 12 in. o.c. (A v /s) = in. 2 /in. Material Properties f' m = 2500 psi f y = 60 ksi E s = ksi E m = 1750 ksi (TMS ) ε mu = in./in. (TMS (c)) ε sy = in./in. Seismic Design Parameters S DS = 1.0 R = 5.5 (ASCE/SEI 7-10 Table ) C d = 4.0 Ω 0 = 2.5 E v = 0.2 S DS D (ASCE/SEI 7-10 Eq ) ρ = 1.0 Modeling Assumptions (Section properties based on 50% E = E m /2 = 875 ksi of gross section properties) G = 350 ksi ν = 0.25 Poisson's ratio Gravity Loading D Self Weight = 60 psf D Tributary = 0 plf L Tributary = 0 plf (Determined from linear-elastic analysis using SAP2000) Wall A = Wall C, Story 3 P D(Top) = 26.6 kip P D(Bot) = 29.6 kip V D = 3.4 kip M D(Top) = 4.3 kip-ft M D(Bot) = 12.5 kip-ft P L = 0.0 kip V L = 0.0 kip M L(Top) = 0.0 kip-ft M L(Bot) = 0.0 kip-ft

70 61 Wall B, Story 3 P D(Top) = 8.7 kip P D(Bot) = 10.5 kip V D = 0.0 kip M D(Top) = 0.0 kip-ft M D(Bot) = 0.0 kip-ft P L = 0.0 kip V L = 0.0 kip M L(Top) = 0.0 kip-ft M L(Bot) = 0.0 kip-ft Coupling Beams P D = 0 kip (Envelope of forces) V D = 2.5 kip M D = 6.7 kip-ft P L = 0 kip V L = 0 kip M L = 0 kip-ft Transfer Girder P D = 0 kip V D(Left) = 10.0 kip (At Wall A) V D(Right) = 6.7 kip (At Wall B) M D(Left) = 20.7 kip-ft M D(Right) = 24.1 kip-ft P L = 0 kip V L(Left) = 0 kip V L(Right) = 0 kip M L(Left) = 0 kip-ft M L(Right) = 0 kip-ft

71 62 Seismic Forces and Displacements (Determined from linear-elastic analysis using SAP 2000) Base Shear, V b = 86.6 kip (Demand on one line of resistance) Story Displacements δ R = in. δ 5 = in. δ 4 = in. δ 3 = in. δ 2 = in. Wall A = Wall C, Story 3 P E = 52.1 kip V E = 29.0 kip M E(Top) = 37.7 kip-ft M E(Bot) = kip-ft Wall B, Story 3 P E = 0.0 kip V E = 28.7 kip M E(Top) = 40.3 kip-ft M E(Bot) = kip-ft Coupling Beams P E = 0.0 kip (Envelope of forces) V E = 17.2 kip M E = 61.4 kip-ft Transfer Girder P E = 0.0 kip V E(Left) = 16.4 kip (At Wall A) V E(Right) = 16.4 kip (At Wall B) M E(Left) = 77.4 kip-ft M E(Right) = 52.4 kip-ft

72 63 Design Forces for Load Combination 1.2D + 0.5L + 1.0E (Including E v ) Wall A = Wall C, Story 3 P u(top) = 89.3 kip P u(bot) = 93.5 kip V u = 33.7 kip M u(top) = 43.7 kip-ft M u(bot) = kip-ft Wall B, Story 3 P u(top) = 12.2 kip P u(bot) = 14.7 kip V u = 28.7 kip M u(top) = 40.3 kip-ft M u(bot) = kip-ft Coupling Beams P u = 0.0 kip (Envelope of forces) V u = 20.8 kip M u = 70.7 kip-ft Transfer Girder P u = 0.0 kip V u(left) = 30.5 kip (At Wall A) V u(right) = 25.8 kip (At Wall B) M u(left) = kip-ft M u(right) = 86.2 kip-ft Design Forces for Load Combination 0.9D + 1.0E (Including E v ) Wall A = Wall C, Story 3 P u(top) = kip (P u < 0, tension) P u(bot) = kip V u = 31.3 kip M u(top) = 40.7 kip-ft M u(bot) = kip-ft Wall B, Story 3 P u(top) = 6.1 kip P u(bot) = 7.3 kip V u = 28.7 kip M u(top) = 40.3 kip-ft M u(bot) = kip-ft

73 64 Coupling Beams P u = 0.0 kip (Envelope of forces) V u = 19.0 kip M u = 66.0 kip-ft Transfer Girder P u = 0.0 kip V u(left) = 23.5 kip (At Wall A) V u(right) = 21.1 kip (At Wall B) M u(left) = 91.8 kip-ft M u(right) = 69.3 kip-ft

74 65 Combined Axial and Flexural Design Top of Wall A = Wall C, Story 3 P u = 89.3 kip (1.2D+0.5L+E, incl. E v ) M u = 43.7 kip-ft φm n = kip-ft > M u OK P u = kip (0.9D+E, incl. E v ) M u = 40.7 kip-ft φm n = kip-ft > M u OK Bottom of Wall A = Wall C, Story 3 P u = 93.5 kip (1.2D+0.5L+E, incl. E v ) M u = kip-ft φm n = kip-ft > M u OK P u = kip (0.9D+E, incl. E v ) M u = kip-ft φm n = kip-ft > M u OK 1200 Wall A = Wall C, Story Axial Force (kip) φ = 0.9 φ = D+0.5L+E 0.9D+E ,000 1,500 2,000 Moment (kip-ft)

75 66 Combined Axial and Flexural Design (cont.) Top of Wall B, Story 3 P u = 12.2 kip (1.2D+0.5L+E, incl. E v ) M u = 40.3 kip-ft φm n = kip-ft > M u OK P u = 6.1 kip (0.9D+E, incl. E v ) M u = 40.3 kip-ft φm n = kip-ft > M u OK Bottom of Wall B, Story 3 P u = 14.7 kip (1.2D+0.5L+E, incl. E v ) M u = kip-ft φm n = kip-ft > M u OK P u = 7.3 kip (0.9D+E, incl. E v ) M u = kip-ft φm n = kip-ft > M u OK 700 Wall B, Story Axial Force (kip) φ = 0.9 φ = D+0.5L+E 0 0.9D+E Moment (kip-ft)

76 67 Combined Axial and Flexural Design (cont.) Coupling Beams P u = 0.0 kip (1.2D+0.5L+E, incl. E v ) M u = 70.7 kip-ft φm n = 79.3 kip-ft > M u OK P u = 0.0 kip (0.9D+E, incl. E v ) M u = 66.0 kip-ft φm n = 79.3 kip-ft > M u OK Coupling Beams φ = 0.9 φ = 1.0 Axial Force (kip) D+0.5L+E 0.9D+E Moment (kip-ft)

77 68 Combined Axial and Flexural Design (cont.) Transfer Girder, Left End (At Wall A) P u = 0.0 kip (1.2D+0.5L+Ω 0 E, incl. E v ) M u = kip-ft Ω 0 applies to E h φm n = kip-ft > M u OK P u = 0.0 kip (0.9D+Ω 0 E, incl. E v ) M u = kip-ft (M u < 0, tension at bottom) φm n = kip-ft > M u OK Transfer Girder, Right End (At Wall B) P u = 0.0 kip (1.2D+0.5L+Ω 0 E, incl. E v ) M u = kip-ft Ω 0 applies to E h φm n = kip-ft > M u OK P u = 0.0 kip (0.9D+Ω 0 E, incl. E v ) M u = kip-ft (M u < 0, tension at bottom) φm n = kip-ft > M u OK Transfer Girder φ = 0.9 φ = 1.0 Axial Force (kip) D+0.5L+E 0.9D+E ,000 1,200 Moment (kip-ft)

78 69 Shear Design Top of Wall A = Wall C, Story 3 (1.2D+0.5L+E, incl. E v ) t Wall = 5.5 in. d v = 120 in. P u = 89.3 kip V u = 33.7 kip M u = 43.7 kip-ft M u /V u d v = 0.13 V n = V nm + V ns V n,max (TMS ) V nm = [4-1.75(M u /V u d v )]A n (f' m ) P u (TMS Eq. 9-24) V nm = kip V ns = 0.5(A v /s)f y d v (TMS Eq. 9-25) (A v /s) = in. 2 /in. V ns = 19.8 kip V n,max = 6.0 A n (f' m ) (TMS Eq and Eq. 9-23) V n,max = kip V n = kip (TMS ) φ = 0.8 φv n = kip > 33.7 kip OK Check φv n min( 2.0V E +V ug, 1.25V Mn +V ug ) (TMS ) V E = 29.0 kip (E h ) M E = 37.7 kip-ft (E h ) V ug = V u - V E = 4.7 kip (1.2D+0.5L+E, incl. E v ) M n = kip-ft (M n for P = P u, φ = 1.0) V Mn = M n V E /M E = kip 2.0V E +V ug = 62.7 kip 1.25V Mn +V ug = kip φv n = kip > 62.7 kip OK

79 70 Shear Design (cont.) Top of Wall A = Wall C, Story 3 (0.9D+E, incl. E v ) t Wall = 5.5 in. d v = 120 in. P u = kip V u = 31.3 kip M u = 40.7 kip-ft M u /V u d v = 0.13 V n = V nm + V ns V n,max (TMS ) V nm = [4-1.75(M u /V u d v )]A n (f' m ) P u (TMS Eq. 9-24) V nm = kip V ns = 0.5(A v /s)f y d v (TMS Eq. 9-25) (A v /s) = in. 2 /in. V ns = 19.8 kip V n,max = 6.0 A n (f' m ) (TMS Eq and Eq. 9-23) V n,max = kip V n = kip (TMS ) φ = 0.8 φv n = kip > 31.3 kip OK Check φv n min( 2.0V E +V ug, 1.25V Mn +V ug ) (TMS ) V E = 29.0 kip (E h ) M E = 37.7 kip-ft (E h ) V ug = V u - V E = 2.4 kip (0.9D+E, incl. E v ) M n = kip-ft (M n for P = P u, φ = 1.0) V Mn = M n V E /M E = kip 2.0V E +V ug = 60.3 kip 1.25V Mn +V ug = kip φv n = kip > 60.3 kip OK

80 71 Shear Design (cont.) Bottom of Wall A = Wall C, Story 3 (1.2D+0.5L, incl. E v ) t Wall = 5.5 in. d v = 120 in. P u = 93.5 kip V u = 33.7 kip M u = kip-ft M u /V u d v = 0.59 V n = V nm + V ns V n,max (TMS ) V nm = [4-1.75(M u /V u d v )]A n (f' m ) P u (TMS Eq. 9-24) V nm = kip V ns = 0.5(A v /s)f y d v (TMS Eq. 9-25) (A v /s) = in. 2 /in. V ns = 19.8 kip V n,max = 5.1 A n (f' m ) (TMS Eq and Eq. 9-23) V n,max = kip V n = kip (TMS ) φ = 0.8 φv n = kip > 33.7 kip OK Check φv n min( 2.0V E +V ug, 1.25V Mn +V ug ) (TMS ) V E = 29.0 kip-ft (E h ) M E = kip-ft (E h ) V ug = V u - V E = 4.7 kip (1.2D+0.5L+E, incl. E v ) M n = kip-ft (M n for P = P u, φ = 1.0) V Mn = M n V E /M E = kip 2.0V E +V ug = 62.7 kip 1.25V Mn +V ug = kip φv n = kip > 62.7 kip OK

81 72 Shear Design (cont.) Bottom of Wall A = Wall C, Story 3 (0.9D+E, incl. E v ) t Wall = 5.5 in. d v = 120 in. P u = kip V u = 31.3 kip M u = kip-ft M u /V u d v = 0.61 V n = V nm + V ns V n,max (TMS ) V nm = [4-1.75(M u /V u d v )]A n (f' m ) P u (TMS Eq. 9-24) V nm = 88.9 kip V ns = 0.5(A v /s)f y d v (TMS Eq. 9-25) (A v /s) = in. 2 /in. V ns = 19.8 kip V n,max = 5.0 A n (f' m ) (TMS Eq and Eq. 9-23) V n,max = kip V n = kip (TMS ) φ = 0.8 φv n = 86.9 kip > 31.3 kip OK Check φv n min( 2.0V E +V ug, 1.25V Mn +V ug ) (TMS ) V E = 29.0 kip-ft (E h ) M E = kip-ft (E h ) V ug = V u - V E = 2.4 kip (0.9D+E, incl. E v ) M n = kip-ft (M n for P = P u, φ = 1.0) V Mn = M n V E /M E = 49.1 kip 2.0V E +V ug = 60.3 kip 1.25V Mn +V ug = 63.7 kip φv n = 86.9 kip > 60.3 kip OK

82 73 Shear Design (cont.) Top of Wall B, Story 3 (1.2D+0.5L+E, incl. E v ) t Wall = 5.5 in. d v = 72 in. P u = 12.2 kip V u = 28.7 kip M u = 40.3 kip-ft M u /V u d v = 0.23 V n = V nm + V ns V n,max (TMS ) V nm = [4-1.75(M u /V u d v )]A n (f' m ) P u (TMS Eq. 9-24) V nm = 74.1 kip V ns = 0.5(A v /s)f y d v (TMS Eq. 9-25) (A v /s) = in. 2 /in. V ns = 11.9 kip V n,max = 6.0 A n (f' m ) (TMS Eq and Eq. 9-23) V n,max = kip V n = 86.0 kip (TMS ) φ = 0.8 φv n = 68.8 kip > 28.7 kip OK Check φv n min( 2.0V E +V ug, 1.25V Mn +V ug ) (TMS ) V E = 28.7 kip (E h ) M E = 40.3 kip-ft (E h ) V ug = V u - V E = 0.0 kip (1.2D+0.5L+E, incl. E v ) M n = kip-ft (M n for P = P u, φ = 1.0) V Mn = M n V E /M E = kip 2.0V E +V ug = 57.3 kip 1.25V Mn +V ug = kip φv n = 68.8 kip > 57.3 kip OK

83 74 Shear Design (cont.) Top of Wall B, Story 3 (0.9D+E, incl. E v ) t Wall = 5.5 in. d v = 72 in. P u = 6.1 kip V u = 28.7 kip M u = 40.3 kip-ft M u /V u d v = 0.23 V n = V nm + V ns V n,max (TMS ) V nm = [4-1.75(M u /V u d v )]A n (f' m ) P u (TMS Eq. 9-24) V nm = 72.6 kip V ns = 0.5(A v /s)f y d v (TMS Eq. 9-25) (A v /s) = in. 2 /in. V ns = 11.9 kip V n,max = 6.0 A n (f' m ) (TMS Eq and Eq. 9-23) V n,max = kip V n = 84.5 kip (TMS ) φ = 0.8 φv n = 67.6 kip > 28.7 kip OK Check φv n min( 2.0V E +V ug, 1.25V Mn +V ug ) (TMS ) V E = 28.7 kip (E h ) M E = 40.3 kip-ft (E h ) V ug = V u - V E = 0.0 kip (0.9D+E, incl. E v ) M n = kip-ft (M n for P = P u, φ = 1.0) V Mn = M n V E /M E = kip 2.0V E +V ug = 57.3 kip 1.25V Mn +V ug = kip φv n = 67.6 kip > 57.3 kip OK

84 75 Shear Design (cont.) Bottom of Wall B, Story 3 (1.2D+0.5L+E, incl. E v ) t Wall = 5.5 in. d v = 72 in. P u = 14.7 kip V u = 28.7 kip M u = kip-ft M u /V u d v = 0.60 V n = V nm + V ns V n,max (TMS ) V nm = [4-1.75(M u /V u d v )]A n (f' m ) P u (TMS Eq. 9-24) V nm = 62.1 kip V ns = 0.5(A v /s)f y d v (TMS Eq. 9-25) (A v /s) = in. 2 /in. V ns = 11.9 kip V n,max = 5.1 A n (f' m ) (TMS Eq and Eq. 9-23) V n,max = kip V n = 74.0 kip (TMS ) φ = 0.8 φv n = 59.2 kip > 28.7 kip OK Check φv n min( 2.0V E +V ug, 1.25V Mn +V ug ) (TMS ) V E = 28.7 kip-ft (E h ) M E = kip-ft (E h ) V ug = V u - V E = 0.0 kip (1.2D+0.5L+E, incl. E v ) M n = kip-ft (M n for P = P u, φ = 1.0) V Mn = M n V E /M E = 47.8 kip 2.0V E +V ug = 57.3 kip 1.25V Mn +V ug = 59.7 kip φv n = 59.2 kip > 57.3 kip OK

85 76 Shear Design (cont.) Bottom of Wall B, Story 3 (0.9D+E, incl. E v ) t Wall = 5.5 in. d v = 72 in. P u = 7.3 kip V u = 28.7 kip M u = kip-ft M u /V u d v = 0.60 V n = V nm + V ns V n,max (TMS ) V nm = [4-1.75(M u /V u d v )]A n (f' m ) P u (TMS Eq. 9-24) V nm = 60.3 kip V ns = 0.5(A v /s)f y d v (TMS Eq. 9-25) (A v /s) = in. 2 /in. V ns = 11.9 kip V n,max = 5.1 A n (f' m ) (TMS Eq and Eq. 9-23) V n,max = kip V n = 72.2 kip (TMS ) φ = 0.8 φv n = 57.7 kip > 28.7 kip OK Check φv n min( 2.0V E +V ug, 1.25V Mn +V ug ) (TMS ) V E = 28.7 kip-ft (E h ) M E = kip-ft (E h ) V ug = V u - V E = 0.0 kip (0.9D+E, incl. E v ) M n = kip-ft (M n for P = P u, φ = 1.0) V Mn = M n V E /M E = 42.8 kip 2.0V E +V ug = 57.3 kip 1.25V Mn +V ug = 53.5 kip φv n = 57.7 kip > 53.5 kip OK

86 77 Shear Design (cont.) Coupling Beams (1.2D+0.5L+E, incl. E v ) t Wall = 5.5 in. d v = 48 in. P u = 0.0 kip V u = 20.8 kip M u = 70.7 kip-ft M u /V u d v = 0.85 V n = V nm + V ns V n,max (TMS ) V nm = [4-1.75(M u /V u d v )]A n (f' m ) P u (TMS Eq. 9-24) V nm = 33.2 kip V ns = 0.5(A v /s)f y d v (TMS Eq. 9-25) (A v /s) = in. 2 /in. V ns = 24.0 kip V n,max = 4.4 A n (f' m ) (TMS Eq and Eq. 9-23) V n,max = 58.1 kip V n = 57.2 kip (TMS ) φ = 0.8 φv n = 45.7 kip > 20.8 kip OK Check φv n min( 2.0V E +V ug, 1.25V Mn +V ug ) (TMS ) V E = 17.2 kip-ft (E h ) M E = 61.4 kip-ft (E h ) V ug = V u - V E = 3.5 kip (1.2D+0.5L+E, incl. E v ) M n = 88.1 kip-ft (M n for P = P u, φ = 1.0) V Mn = M n V E /M E = 24.7 kip 2.0V E +V ug = 38.0 kip 1.25V Mn +V ug = 34.5 kip φv n = 45.7 kip > 34.5 kip OK

87 78 Shear Design (cont.) Transfer Girder, Left End t Wall = 5.5 in. d v = 96 in. (At Wall A) P u = 0.0 kip (1.2D+0.5L+Ω 0 E, incl. E v ) V D = 10.0 kip V L = 0.0 kip V E = 16.4 kip V u = 55.1 kip Ω 0 applies to E h M u = kip-ft M u /V u d v = 0.50 V n = V nm + V ns V n,max (TMS ) V nm = [4-1.75(M u /V u d v )]A n (f' m ) P u (TMS Eq. 9-24) V nm = 82.3 kip V ns = 0.5(A v /s)f y d v (TMS Eq. 9-25) (A v /s) = in. 2 /in. V ns = 26.4 kip V n,max = 5.3 A n (f' m ) (TMS Eq and Eq. 9-23) V n,max = kip V n = kip (TMS ) φ = 0.8 φv n = 87.0 kip > 55.1 kip OK

88 79 Shear Design (cont.) Transfer Girder, Right End t Wall = 5.5 in. d v = 96 in. (At Wall B) P u = 0.0 kip (1.2D+0.5L+Ω 0 E, incl. E v ) V D = 6.7 kip V L = 0.0 kip V E = 16.4 kip V u = 50.4 kip Ω 0 applies to E h M u = kip-ft M u /V u d v = 0.41 V n = V nm + V ns V n,max (TMS ) V nm = [4-1.75(M u /V u d v )]A n (f' m ) P u (TMS Eq. 9-24) V nm = 86.7 kip V ns = 0.5(A v /s)f y d v (TMS Eq. 9-25) (A v /s) = in. 2 /in. V ns = 26.4 kip V n,max = 5.6 A n (f' m ) (TMS Eq and Eq. 9-23) V n,max = kip V n = kip (TMS ) φ = 0.8 φv n = 90.5 kip > 50.4 kip OK

89 80 Boundary Element Compliance (TMS ) Wall A = Wall C Rapid Screening (TMS ) V u = 33.7 kip (1.2D+0.5L+E, incl. E v ) M u = kip-ft M u /(V u d v ) = 0.59 P u = 93.5 kip 0.10A g f' m = 165 kip > P u Conditions 1 and 2: OK 3A n (f' m ) = 99.0 kip > V u Condition 3: OK Extreme Fiber Compressive Stress (TMS ) f max = P u /A g +M u /S 0.2f' m 0.2f' m = psi P u = 93.5 kip (1.2D+0.5L+E, incl. E v ) M u = kip-ft A g = 660 in. 2 S = in. 3 f max = psi < 0.2f' m OK Both Sections, and , are checked for illustration purposes (only one needs to be satisfied)

90 81 Boundary Element Compliance (cont.) (TMS ) Wall B Rapid Screening (TMS ) V u = 28.7 kip (1.2D+0.5L+E, incl. E v ) M u = kip-ft M u /(V u d v ) = 0.60 P u = 14.7 kip 0.10A g f' m = 99 kip > P u Conditions 1 and 2: OK 3A n (f' m ) = 59.4 kip > V u Condition 3: OK Extreme Fiber Compressive Stress (TMS ) f max = P u /A g +M u /S 0.2f' m 0.2f' m = psi P u = 14.7 kip (1.2D+0.5L+E, incl. E v ) M u = kip-ft A g = 396 in. 2 S = 4752 in. 3 f max = psi < 0.2f' m OK Both Sections, and , are checked for illustration purposes (only one needs to be satisfied)

91 82 Boundary Element Compliance (cont.) (TMS ) Coupling Beams Rapid Screening (TMS ) V u = 20.8 kip (1.2D+0.5L+E, incl. E v ) M u = 70.7 kip-ft M u /(V u d v ) = 0.85 P u = 0.0 kip 0.10A g f' m = 66 kip > P u Conditions 1 and 2: OK 3A n (f' m ) = 39.6 kip > V u Condition 3: OK Extreme Fiber Compressive Stress (TMS ) f max = P u /A g +M u /S 0.2f' m 0.2f' m = psi P u = 0.0 kip (1.2D+0.5L+E, incl. E v ) M u = 70.7 kip-ft A g = 264 in. 2 S = 2112 in. 3 f max = psi < 0.2f' m OK Both Sections, and , are checked for illustration purposes (only one needs to be satisfied)

92 83 Boundary Element Compliance (cont.) (TMS ) Transfer Girder Rapid Screening (TMS ) V u = 55.1 kip (1.2D+0.5L+Ω 0 E, incl. E v ) M u = kip-ft Ω 0 applies to E h M u /(V u d v ) = 0.50 P u = 0.0 kip 0.10A g f' m = 132 kip > P u Conditions 1 and 2: OK 3A n (f' m ) = 79.2 kip > V u Condition 3: OK Extreme Fiber Compressive Stress (TMS ) f max = P u /A g +M u /S 0.2f' m 0.2f' m = psi P u = 0.0 kip (1.2D+0.5L+Ω 0 E, incl. E v ) M u = kip-ft Ω 0 applies to E h A g = 528 in. 2 S = 8448 in. 3 f max = psi < 0.2f' m OK Both Sections, and , are checked for illustration purposes (only one needs to be satisfied)

93 84 APPENDIX B: SPREADSHEET FORMULATIONS FOR LIMIT DESIGN In this appendix, supporting calculations are presented for Design Examples 1 and 2 based on the Limit Design provisions of the TMS 402 (2013) code. Data presented in the worksheet for Limit Design include a general description of the wall; design forces and displacement demands; and design strengths and deformations capacities. The description of the structure includes geometry, material properties, and wall reinforcement (size and spacing). The demands are defined based on seismic design parameters, modeling assumptions, gravity loading, and load combinations. The capacities are determined based on axial-flexure (P-M) interaction diagrams to obtain the flexural strength at potential plastic hinges and the controlling yield mechanism. Shear strengths are computed to identify the shear controlled wall segments. Neutral axis depths associated with the derivation of P-M interaction diagrams are used to support the calculation of the deformation capacities of yielding wall segments.

94 85 B.1 Limit Design Calculations for Design Example 1 Design Example 1 Geometry Concrete masonry units, in., fully grouted Reinforcement Wall A Flexural Reinforcement A s,1 = (1) #4 = 0.2 in. d 1 = 4 in. A s,2 = (1) #4 = 0.2 in. d 2 = 20 in. A s,3 = (1) #4 = 0.2 in. d 3 = 28 in. A s,4 = (1) #4 = 0.2 in. d 4 = 44 in. (Assumed sufficient for all nonseismic load combinations) Wall A Shear Reinforcement A v = (1) #3 = 0.11 in. 16 in. o.c. (A v /s) = in. 2 /in. Wall B Flexural Reinforcement A s,1 = (1) #4 = 0.2 in. d 1 = 4 in. A s,2 = (1) #4 = 0.2 in. d 2 = 12 in. A s,3 = (1) #4 = 0.2 in. d 3 = 20 in. Wall B Shear Reinforcement A v = (1) #3 = 0.11 in. 8 in. o.c. (A v /s) = in. 2 /in.

95 86 Wall C Flexural Reinforcement A s,1 = (1) #4 = 0.2 in. d 1 = 4 in. A s,2 = (1) #4 = 0.2 in. d 2 = 20 in. A s,3 = (1) #4 = 0.2 in. d 3 = 28 in. A s,4 = (1) #4 = 0.2 in. d 4 = 44 in. Wall C Shear Reinforcement A v = (1) #3 = 0.11 in. 16 in. o.c. (A v /s) = in. 2 /in. Material Properties f' m = 1500 psi f y = 60 ksi E s = ksi E m = 1350 ksi (TMS ) ε mu = in./in. (TMS (c)) ε sy = in./in. Seismic Design Parameters S DS = 1 R = 5 (ASCE/SEI 7-10 Table ) C d = 3.5 Ω 0 = 2.5 E v = 0.2 S DS D (ASCE/SEI 7-10 Eq ) ρ = 1.0 Modeling Assumptions (Section properties based on 50% E = E m /2 = 675 ksi of gross section properties) G = 270 ksi ν = 0.25 Poisson's ratio Gravity Loading (Determined from linear-elastic analysis using SAP2000) D Self Weight = 80 psf D Tributary = 150 plf L Tributary = 225 plf Wall A P D(Top) = 7.7 kip P D(Bot) = 10.9 kip V D = 0.7 kip M D(Top) = 4.1 kip-ft M D(Bot) = 3.0 kip-ft

96 87 P L = 2.2 kip V L = 0.2 kip M L(Top) = 1.1 kip-ft M L(Bot) = 0.8 kip-ft Wall B P D(Top) = 5.6 kip P D(Bot) = 6.9 kip V D = 0.0 kip M D(Top) = 0.1 kip-ft M D(Bot) = 0.0 kip-ft P L = 1.5 kip V L = 0.0 kip M L(Top) = 0.0 kip-ft M L(Bot) = 0.0 kip-ft Wall C P D(Top) = 7.2 kip P D(Bot) = 9.8 kip V D = 0.7 kip M D(Top) = 3.5 kip-ft M D(Bot) = 2.2 kip-ft P L = 1.8 kip V L = 0.2 kip M L(Top) = 1.0 kip-ft M L(Bot) = 0.6 kip-ft Seismic Forces and Displacement (Determined from linear-elastic analysis using SAP 2000) Base Shear, V b = 41 kip (Demand on one line of resistance) Roof Displacement, δ R = in. Wall A P E = 19.7 kip V E = 15.2 kip M E(Top) = 58.7 kip-ft M E(Bot) = 93.1 kip-ft

97 88 Wall B P E = 5.3 kip V E = 4.6 kip M E(Top) = 15.5 kip-ft M E(Bot) = 21.3 kip-ft Wall C P E = 25.0 kip V E = 21.2 kip M E(Top) = 59.6 kip-ft M E(Bot) = kip-ft Design Forces for Load Combination 1.2D + 0.5L + 1.0E (Including E v ) Wall A P u(top) = 31.6 kip P u(bot) = 36.1 kip V u = 16.3 kip M u(top) = 65.0 kip-ft M u(bot) = 97.7 kip-ft Wall B P u(top) = 13.9 kip P u(bot) = 15.7 kip V u = 4.6 kip M u(top) = 15.7 kip-ft M u(bot) = 21.3 kip-ft Wall C P u(top) = 36.0 kip P u(bot) = 39.6 kip V u = 22.3 kip M u(top) = 65.0 kip-ft M u(bot) = kip-ft Design Forces for Load Combination 0.9D + 1.0E (Including E v ) Wall A P u(top) = kip (P u < 0, tension) P u(bot) = kip V u = 15.7 kip M u(top) = 61.6 kip-ft M u(bot) = 95.2 kip-ft

98 89 Wall B P u(top) = -1.4 kip (P u < 0, tension) P u(bot) = -0.5 kip V u = 4.6 kip M u(top) = 15.6 kip-ft M u(bot) = 21.3 kip-ft Wall C P u(top) = kip (P u < 0, tension) P u(bot) = kip V u = 21.7 kip M u(top) = 62.1 kip-ft M u(bot) = kip-ft

99 90 Interaction Diagrams 350 Wall A = 1.0 Axial Force (kip) φ = 0.9 φ = Moment (kip-ft) 200 Wall B 150 Axial Force (kip) φ = 0.9 φ = φ 1.0 = Moment (kip-ft)

100 91 Interaction Diagrams (cont.) 400 Wall C Axial Force (kip) = 0.9 = 1.0 φ = 0.9 φ = Moment (kip-ft)

101 92 Wall A Hinge Strength M n and V n determined for gravity load of 0.9D 0.2S DS D (Not including E h ) Shear Corresponding to Development of Flexural Hinge (V Mn ) P u,top = 5.4 kip (0.9D 0.2S DS D) M n,top = 95.0 kip-ft P u,bot = 7.64 kip M n,bot = 98.6 kip-ft V Mn = Shear in Wall A associated with development of M n M E = 93.1 kip-ft V E = 15.2 kip V Mn = (M n / M E ) V E V Mn = 16.1 kip Shear Strength Provided (V n,prov ) t Wall = in. d v = 48 in. M u /V u d v = 1.52 (Using 0.9D + 1.0E) P u = 5.4 kip (0.9D 0.2S DS D) (A v /s) = in. 2 /in. V nm = [4-1.75(M u /V u d v )]A n (f' m ) P u (TMS Eq. 9-24) V nm = 33.2 kip V ns = 0.5(A v /s)f y d v (TMS Eq. 9-25) V ns = 9.9 kip V n(max) = 4.0 A n (f' m ) (TMS Eq and Eq. 9-23) V n(max) = 56.7 kip V n,prov = 43.1 kip (TMS ) Shear Demand V ug = 0.5 kip V E = 15.2 kip Check if Shear Controlled φ vo = Reduction factor applicable where V n < 2V Mn + V ug φ vo = max((v n -V ug )/(2V Mn ),(V n -V ug )/(RV E )) 1 φ vo = 1.00 Not Shear Controlled Hinge Strength at Top of Wall A Hinge Strength = φ vo M n(top) Hinge Strength = 95.0 kip-ft Hinge Strength at Base of Wall A Hinge Strength = φ vo M n(bot) Hinge Strength = 98.6 kip-ft

102 93 Wall B Hinge Strength M n and V n determined for gravity load of 0.9D 0.2S DS D (Not including E h ) Shear Corresponding to Development of Flexural Hinge (V Mn ) P u,top = 3.9 kip (0.9D 0.2S DS D) M n,top = 32.4 kip-ft P u,bot = 4.8 kip M n,bot = 33.1 kip-ft V Mn = Shear in Wall B associated with development of M n M E = 21.3 kip-ft V E = 4.6 kip V Mn = (M n / M E ) V E V Mn = 7.1 kip Shear Strength Provided (V n,prov ) t Wall = in. d v = 24 in. M u /V u d v = 2.31 (Using 0.9D + 1.0E) P u = 3.9 kip (0.9D 0.2S DS D) (A v /s) = in. 2 /in. V nm = [4-1.75(M u /V u d v )]A n (f' m ) P u (TMS Eq. 9-24) V nm = 16.9 kip V ns = 0.5(A v /s)f y d v (TMS Eq. 9-25) V ns = 9.9 kip V n(max) = 4 A n (f' m ) (TMS Eq and Eq. 9-23) V n(max) = 28.4 kip V n,prov = 26.8 kip (TMS ) Shear Demand V ug = 0.0 kip V E = 4.6 kip Check if Shear Controlled φ vo = Reduction factor applicable where V n < 2V Mn + V ug φ vo = max((v n -V ug )/(2V Mn ),(V n -V ug )/(RV E )) 1 φ vo = 1.00 Not Shear Controlled Hinge Strength at Top of Wall B Hinge Strength = φ vo M n(top) Hinge Strength = 32.4 kip-ft Hinge Strength at Base of Wall B Hinge Strength = φ vo M n(bot) Hinge Strength = 33.1 kip-ft

103 94 Wall C Hinge Strength M n and V n determined for gravity load of 0.9D 0.2S DS D (Not including E h ) Shear Corresponding to Development of Flexural Hinge (V Mn ) P u,top = 5.1 kip (0.9D 0.2S DS D) M n,top = 94.4 kip-ft P u,bot = 6.8 kip M n,bot = 97.3 kip-ft V Mn = Shear in Wall C associated with development of M n M E = kip-ft V E = 21.2 kip V Mn = (M n / M E ) V E V Mn = 18.7 kip Shear Strength Provided (V n,prov ) t Wall = in. d v = 48 in. M u /V u d v = 1.29 (Using 0.9D + 1.0E) P u = 5.1 kip (0.9D 0.2S DS D) (A v /s) = in. 2 /in. V nm = [4-1.75(M u /V u d v )]A n (f' m ) P u (TMS Eq. 9-24) V nm = 33.2 kip V ns = 0.5(A v /s)f y d v (TMS Eq. 9-25) V ns = 9.9 kip V n(max) = 4 A n (f' m ) (TMS Eq and Eq. 9-23) V n(max) = 56.7 kip V n,prov = 43.1 kip (TMS ) Shear Demand V ug = 0.5 kip V E = 21.2 kip Check if Shear Controlled φ vo = Reduction factor applicable where V n < 2V MN + V ug φ vo = max((v n -V ug )/(2V Mn ),(V n -V ug )/(RV E )) 1 φ vo = 1.00 Not Shear Controlled Hinge Strength at Top of Wall C Hinge Strength = φ vo M n(top) Hinge Strength = 94.4 kip-ft Hinge Strength at Base of Wall C Hinge Strength = φ vo M n(bot) Hinge Strength = 97.3 kip-ft

104 95 Limit Mechanism Virtual External Work = Virtual External Work = Virtual Internal Work = Virtual Internal Work V Lim (ΣM n /h w ) Wall A + (ΣM n /h w ) Wall B + (ΣM n /h w ) Wall C V Lim = 51.5 kip (V Lim = 51.5 kip, if not Shear Controlled) φ Lim = 0.8 φv Lim = 41.2 kip Base Shear, E = 41 kip < φv Lim OK 6' 10' 95.0 kip-ft 32.4 kip-ft 94.4 kip-ft Not Shear Controlled Not Shear Controlled Not Shear Controlled 33.1 kip-ft 98.6 kip-ft 97.3 kip-ft 8'

105 96 Wall A Deformation Capacity Check (1.2D+0.5L+E, incl. E v ) Displacement Demand δ R = 0.11 in. C d = 3.5 m = δ R C d = 0.40 in. Displacement Capacity Base Shear, E = 41 kip For P u = 36.1 kip Not Shear Controlled c = 9.85 in. Neutral axis depth for P u 0.5 h w L w ε mu /c = 0.73 in. cap = 0.73 in. > 0.40 in. OK Wall B Deformation Capacity Check (1.2D+0.5L+E, incl. E v ) Displacement Demand δ R = 0.11 in. C d = 3.5 m = δ R C d = 0.40 in. Displacement Capacity Base Shear, E = 41 kip For P u = 15.7 kip Not Shear Controlled c = 5.42 in. Neutral axis depth for P u 0.5 h w L w ε mu /c = 0.53 in. cap = 0.53 in. > 0.40 in. OK

106 97 Wall C Deformation Capacity Check (1.2D+0.5L+E, incl. E v ) Displacement Demand δ R = 0.11 in. C d = 3.5 m = δ R C d = 0.40 in. Displacement Capacity Base Shear, E = 41 kip For P u = 39.6 kip Not Shear Controlled c = in. Neutral axis depth for P u 0.5 h w L w ε mu /c = 0.56 in. cap = 0.56 in. > 0.40 in. OK

107 98 B.2 Limit Design Calculations for Design Example 2 Design Example 2 Geometry Roof 36 ft 1750 Level 5 27 ft Level 4 18 ft Level 3 9 ft Level 2 Seismic Base OK 0 Clay masonry units, in., fully grouted

108 99 Reinforcement (Assumed sufficient for all nonseismic load combinations) Wall A = Wall C Longitudinal Reinforcement A s,1 = (1) #4 = 0.2 in. d 1 = 3 in. A s,2 = (1) #4 = 0.2 in. d 2 = 21 in. A s,3 = (1) #4 = 0.2 in. d 3 = 39 in. A s,4 = (1) #4 = 0.2 in. d 4 = 57 in. A s,5 = (1) #4 = 0.2 in. d 5 = 63 in. A s,6 = (1) #4 = 0.2 in. d 6 = 81 in. A s,7 = (1) #4 = 0.2 in. d 7 = 99 in. A s,8 = (1) #4 = 0.2 in. d 8 = 117 in. Wall A = Wall C Shear Reinforcement A v = (1) #3 = 0.11 in. 20 in. o.c. (A v /s) = in. 2 /in. Wall B Longitudinal Reinforcement A s,1 = (1) #4 = 0.2 in. d 1 = 9 in. A s,2 = (1) #4 = 0.2 in. d 2 = 27 in. A s,3 = (1) #4 = 0.2 in. d 3 = 45 in. A s,4 = (1) #4 = 0.2 in. d 4 = 63 in. Wall B Shear Reinforcement A v = (1) #3 = 0.11 in 20 in. o.c. (A v /s) = in. 2 /in. Coupling Beam Longitudinal Reinforcement A s,1 = (1) #3 = 0.11 in. d 1 = 3 in. A s,2 = (1) #3 = 0.11 in. d 2 = 19 in. A s,3 = (1) #3 = 0.11 in. d 3 = 29 in. A s,4 = (1) #3 = 0.11 in. d 4 = 45 in. Coupling Beam Shear Reinforcement A v = (1) #3 = 0.11 in. 12 in. o.c. (A v /s) = in. 2 /in. Transfer Girder Longitudinal Reinforcement A s,1 = (1) #4 = 0.2 in. d 1 = 9 in. A s,2 = (1) #4 = 0.2 in. d 2 = 25 in. A s,3 = (1) #4 = 0.2 in. d 3 = 41 in. A s,4 = (1) #4 = 0.2 in. d 4 = 55 in. A s,5 = (1) #4 = 0.2 in. d 5 = 71 in. A s,6 = (1) #4 = 0.2 in. d 6 = 87 in.

109 100 Transfer Girder Shear Reinforcement A v = (1) #3 = 0.11 in. 12 in. o.c. (A v /s) = in. 2 /in. Material Properties f' m = 2500 psi f y = 60 ksi E s = ksi E m = 1750 ksi (TMS ) ε mu = in./in. (TMS (c)) ε sy = in./in. Seismic Design Parameters Modeling Assumptions S DS = 1.0 R = 5.5 (ASCE/SEI 7-10 Table ) C d = 4.0 Ω 0 = 2.5 E v = 0.2 S DS D (ASCE/SEI 7-10 Eq ) ρ = 1.0 E = E m /2 = 875 ksi G = 350 ksi (Section properties based on 50% of gross section properties) ν = 0.25 Poisson's ratio Gravity Loading D Self Weight = 60 psf D Tributary = 0 plf L Tributary = 0 plf (Determined from linear-elastic analysis using SAP2000) Wall A = Wall C, Story 3 P D(Top) = 26.6 kip P D(Bot) = 29.6 kip V D = 3.4 kip M D(Top) = 4.3 kip-ft M D(Bot) = 12.5 kip-ft P L = 0.0 kip V L = 0.0 kip M L(Top) = 0.0 kip-ft M L(Bot) = 0.0 kip-ft

110 101 Wall B, Story 3 P D(Top) = 8.7 kip P D(Bot) = 10.5 kip V D = 0.0 kip M D(Top) = 0.0 kip-ft M D(Bot) = 0.0 kip-ft P L = 0.0 kip V L = 0.0 kip M L(Top) = 0.0 kip-ft M L(Bot) = 0.0 kip-ft Coupling Beams P D = 0 kip (Envelope of forces) V D = 2.5 kip M D = 6.7 kip-ft P L = 0 kip V L = 0 kip M L = 0 kip-ft Transfer Girder P D = 0 kip V D(Left) = 10.0 kip (At Wall A) V D(Right) = 6.7 kip (At Wall B) M D(Left) = 20.7 kip-ft M D(Right) = 24.1 kip-ft P L = 0 kip V L(Left) = 0 kip V L(Right) = 0 kip M L(Left) = 0 kip-ft M L(Right) = 0 kip-ft

111 102 Seismic Forces and Displacements (Determined from linear-elastic analysis using SAP 2000) Base Shear, V b = 86.6 kip (Demand on one line of resistance) Story Displacements δ R = in. δ 5 = in. δ 4 = in. δ 3 = in. δ 2 = in. Wall A = Wall C, Story 3 P E = 52.1 kip V E = 29.0 kip M E(Top) = 37.7 kip-ft M E(Bot) = kip-ft Wall B, Story 3 P E = 0.0 kip V E = 28.7 kip M E(Top) = 40.3 kip-ft M E(Bot) = kip-ft Coupling Beams P E = 0.0 kip (Envelope of forces) V E = 17.2 kip M E = 61.4 kip-ft Transfer Girder P E = 0.0 kip V E(Left) = 16.4 kip (At Wall A) V E(Right) = 16.4 kip (At Wall B) M E(Left) = 77.4 kip-ft M E(Right) = 52.4 kip-ft

112 103 Design Forces for Load Combination 1.2D + 0.5L + 1.0E (Including E v ) Wall A = Wall C, Story 3 P u(top) = 89.3 kip P u(bot) = 93.5 kip V u = 33.7 kip M u(top) = 43.7 kip-ft M u(bot) = kip-ft Wall B, Story 3 P u(top) = 12.2 kip P u(bot) = 14.7 kip V u = 28.7 kip M u(top) = 40.3 kip-ft M u(bot) = kip-ft Coupling Beams P u = 0.0 kip (Envelope of forces) V u = 20.8 kip M u = 70.7 kip-ft Transfer Girder P u = 0.0 kip V u(left) = 30.5 kip (At Wall A) V u(right) = 25.8 kip (At Wall B) M u(left) = kip-ft M u(right) = 86.2 kip-ft Design Forces for Load Combination 0.9D + 1.0E (Including E v ) Wall A = Wall C, Story 3 P u(top) = kip (P u < 0, tension) P u(bot) = kip V u = 31.3 kip M u(top) = 40.7 kip-ft M u(bot) = kip-ft Wall B, Story 3 P u(top) = 6.1 kip P u(bot) = 7.3 kip V u = 28.7 kip M u(top) = 40.3 kip-ft M u(bot) = kip-ft

113 104 Coupling Beams P u = 0.0 kip (Envelope of forces) V u = 19.0 kip M u = 66.0 kip-ft Transfer Girder P u = 0.0 kip V u(left) = 23.5 kip (At Wall A) V u(right) = 21.1 kip (At Wall B) M u(left) = 91.8 kip-ft M u(right) = 69.3 kip-ft

114 105 Interaction Diagrams 1200 Wall A = Wall C, Story 3 Axial Force (kip) φ = 0.9 φ = ,000 1,500 2,000 Moment (kip-ft) 700 Wall B, Story 3 Axial Force (kip) φ = 0.9 φ = Moment (kip-ft)

115 Coupling Beams Axial Force (kip) φ = 0.9 φ = Moment (kip-ft) Axial Force (kip) Transfer Girder φ = 0.9 φ = ,000 1,200 Moment (kip-ft)

116 107 Hinge Strength at Base of Wall A = Wall C, Story 3 M n and V n determined for gravity load of 0.9D 0.2S DS D (Not including E h ) Shear Corresponding to Development of Flexural Hinge (V Mn ) P u = 20.7 (0.9D 0.2S DS D) M n = kip-ft V Mn = Shear in Wall A associated with development of M n M E = kip-ft V E = 29.0 kip V Mn = (M n / M E ) V E V Mn = 85.5 kip Shear Strength Provided (V n,prov ) t Wall = 5.5 in. d v = 120 in. M u /V u d v = 0.61 (Using 0.9D + 1.0E) P u = 20.7 kip (0.9D 0.2S DS D) (A v /s) = in. 2 /in. V nm = [4-1.75(M u /V u d v )]A n (f' m ) P u (TMS Eq. 9-24) V nm = kip V ns = 0.5(A v /s)f y d v (TMS Eq. 9-25) V ns = 19.8 kip V n(max) = 5.0 A n (f' m ) (TMS Eq and Eq. 9-23) V n(max) = kip V n,prov = kip (TMS ) Shear Demand V ug = 2.4 kip V E = 29.0 kip Check if Shear Controlled φ vo = Reduction factor applicable where V n < 2V MN + V ug φ vo = max((v n -V ug )/(2V Mn ),(V n -V ug )/(RV E )) 1 φ vo = 0.75 Shear Controlled Hinge Strength Hinge Strength = φ vo M n Hinge Strength = kip-ft

117 108 Hinge Strength at Base of Wall B, Story 3 M n and V n determined for gravity load of 0.9D 0.2S DS D (Not including E h ) Shear Corresponding to Development of Flexural Hinge (V Mn ) P u = 7.3 (0.9D 0.2S DS D) M n = kip-ft V Mn = Shear in Wall B associated with development of M n M E = kip-ft V E = 28.7 kip V Mn = (M n / M E ) V E V Mn = 42.8 kip Shear Strength Provided (V n,prov ) t Wall = 5.5 in. d v = 72 in. M u /V u d v = 0.60 (Using 0.9D + 1.0E) P u = 7.3 kip (0.9D 0.2S DS D) (A v /s) = in. 2 /in. V nm = [4-1.75(M u /V u d v )]A n (f' m ) P u (TMS Eq. 9-24) V nm = 60.3 kip V ns = 0.5(A v /s)f y d v (TMS Eq. 9-25) V ns = 11.9 kip V n(max) = 5.1 A n (f' m ) (TMS Eq and Eq. 9-23) V n(max) = kip V n,prov = 72.2 kip (TMS ) Shear Demand V ug = 0.0 kip V E = 28.7 kip Check if Shear Controlled φ vo = Reduction factor applicable where V n < 2V MN + V ug φ vo = max((v n -V ug )/(2V Mn ),(V n -V ug )/(RV E )) 1 φ vo = 0.84 Shear Controlled Hinge Strength Hinge Strength = φ vo M n Hinge Strength = kip-ft

118 109 Hinge Strength at Ends of Coupling Beams M n and V n determined for gravity load of 0.9D 0.2S DS D (Not including E h ) Shear Corresponding to Development of Flexural Hinge (V Mn ) P u = 0 kip (0.9D 0.2S DS D) M n = 49.9 kip-ft V Mn = Σ(M n )/L = 2 / 7 ft 49.9 kip-ft V Mn = 14.2 kip Shear Strength Provided (V n,prov ) t Wall = 5.5 in. d v = 48 in. M u /V u d v = 0.87 (Using 0.9D + 1.0E) P u = 0.0 kip (0.9D 0.2S DS D) (A v /s) = in. 2 /in. V nm = [4-1.75(M u /V u d v )]A n (f' m ) P u (TMS Eq. 9-24) V nm = 32.7 kip V ns = 0.5(A v /s)f y d v (TMS Eq. 9-25) V ns = 13.2 kip V n(max) = 4.4 A n (f' m ) (TMS Eq and Eq. 9-23) V n(max) = 57.4 kip V n,prov = 45.9 kip (TMS ) Shear Demand V ug = 1.8 kip V E = 17.2 kip Check if Shear Controlled φ vo = Reduction factor applicable where V n < 2V MN + V ug φ vo = max((v n -V ug )/(2V Mn ),(V n -V ug )/(RV E )) 1 φ vo = 1.00 Not Shear Controlled Hinge Strength Hinge Strength = φ vo M n Hinge Strength = 49.9 kip-ft

119 110 Limit Mechanism Virtual External Work = Virtual Internal Work Virtual External Work = Σ [f i i ] = Σ [(f i /V Lim )(h i /h R )] V Lim = V Lim Virtual Internal Work = kip-ft(2 /34 ft) kip-ft( /34 ft) kip-ft(16 /34 ft)(15 ft/7 ft) V Lim = kip (V Lim = kip, if not Shear Controlled) φ Lim = 0.8 φv Lim = 90.2 kip > E = 86.6 kip OK Not Shear Controlled Typ. for beams 49.9 kip-ft Typ. for beams 34' 25' 16' 7' Shear Controlled kip-ft kip-ft Shear Controlled Walls A and C Wall B

120 111 Deformation Capacity, Wall A = Wall C (1.2D+0.5L+E, incl. E v ) Displacement Demand δ R = 0.23 in. C d = 4 m = δ R C d = 0.93 in. Displacement Capacity Base Shear, E = 86.6 kip For P u = 93.5 kip Shear Controlled c = in. Neutral axis depth for P u 0.5 h w L w ε mu /c = 4.50 in. h/200 = 2.04 in. cap = 2.04 in. > 0.93 in. OK Deformation Capacity, Wall B (1.2D+0.5L+E, incl. E v ) Displacement Demand δ R = 0.23 in. C d = 4 m = δ R C d = 0.93 in. Displacement Capacity Base Shear, E = 86.6 kip For P u = 14.7 kip Shear Controlled c = 6.62 in. Neutral axis depth for P u 0.5 h w L w ε mu /c = 7.77 in. h/200 = 2.04 in. cap = 2.04 in. > 0.93 in. OK

121 112 Deformation Capacity, Coupling Beams (1.2D+0.5L+E, incl. E v ) Displacement Demand δ Beam = (δ/h) l δ = (δ R - δ 2 ) h = 34 ft l = 15 ft δ Beam = 0.10 in. C d = 4 m = δ Beam C d = 0.41 in. Displacement Capacity Base Shear, E = 86.6 kip For P u = 0.0 kip Not Shear Controlled c = 2.52 in. Neutral axis depth for P u 0.5 h w L w ε mu /c = 2.80 in. - - cap = 2.80 in. > 0.41 in. OK Design of Transfer Girder Flexural Design Left End P u = 0.0 kip (1.2D+0.5L+Ω 0 E, incl. E v ) M D = 20.7 kip-ft Ω 0 applies to E h M L = 0.0 kip-ft M E = 77.4 kip-ft M u = kip-ft φm n = kip-ft > M u OK Right End P u = 0.0 kip (1.2D+0.5L+Ω 0 E, incl. E v ) M D = 24.1 kip-ft Ω 0 applies to E h M L = 0.0 kip-ft M E = 52.4 kip-ft M u = kip-ft φm n = kip-ft > M u OK

122 113 Shear Design Left End t Wall = 5.5 in. d v = 96 in. (At Wall A) P u = 0.0 kip (1.2D+0.5L+Ω 0 E, incl. E v ) V D = 10.0 kip Ω 0 applies to E h V L = 0.0 kip V E = 16.4 kip V u = 55.1 kip M u = kip-ft M u /V u d v = 0.50 V n = V nm + V ns V n,max (TMS ) V nm = [4-1.75(M u /V u d v )]A n (f' m ) P u (TMS Eq. 9-24) V nm = 82.3 kip V ns = 0.5(A v /s)f y d v (TMS Eq. 9-25) (A v /s) = in. 2 /in. V ns = 26.4 kip V n,max = 5.3 A n (f' m ) (TMS Eq and Eq. 9-23) V n,max = kip V n = kip (TMS ) φ = 0.8 φv n = 87.0 kip > 55.1 kip OK

123 114 Right End t Wall = 5.5 in. d v = 96 in. (At Wall B) P u = 0.0 kip (1.2D+0.5L+Ω 0 E, incl. E v ) V D = 6.7 kip Ω 0 applies to E h V L = 0.0 kip V E = 16.4 kip V u = 50.4 kip M u = kip-ft M u /V u d v = 0.41 V n = V nm + V ns V n,max (TMS ) V nm = [4-1.75(M u /V u d v )]A n (f' m ) P u (TMS Eq. 9-24) V nm = 86.7 kip V ns = 0.5(A v /s)f y d v (TMS Eq. 9-25) (A v /s) = in. 2 /in. V ns = 26.4 kip V n,max = 5.6 A n (f' m ) (TMS Eq and Eq. 9-23) V n,max = kip V n = kip (TMS ) φ = 0.8 φv n = 90.5 kip > 50.4 kip OK

124 115 APPENDIX C: PRACTICAL NONLINEAR STATIC ANALYSIS OF MASONRY WALLS An analytical model is presented for performing practical nonlinear static analysis of masonry shear walls proportioned and detailed to resist strong ground motions. A simplified modeling technique was implemented to support the development and usage of the Limit Design provisions in TMS Appendix C (code and commentary are included in Table 2.1 of this document). The proposed computer model directly accounts for the effects of varying axial load caused by an increase in lateral forces. It facilitates finding the controlling yield mechanism and the associated limiting base shear strength for reinforced masonry wall configurations. The program SAP2000 (CSI, 2011) is used to implement a simplified modeling technique, representing a modified version of the modeling approach presented by Sanchez (2012). The model is based on the predominant use of linear-elastic area elements combined with a limited number of elements having nonlinear force-displacement relationships. Sanchez (2012) developed two modeling techniques; the Nonlinear Layer model and the Nonlinear Link model. This appendix describes a simplified version of the Nonlinear Layer model. The Nonlinear Layer model modifies the area elements at the potential plastic hinge regions with special layer definitions that account for material nonlinearity. To perform a nonlinear static analysis of a masonry shear wall configuration using nonlinear layers, the user must first develop a linear-elastic model. The linear-elastic model is used as a reference model to represent nonlinear response as permitted by modern building codes. This linear model is used to obtain the design roof displacement and to determine the axial forces due to the factored loads that are consistent with the design load combination producing the design roof displacement. These axial forces are used to calculate the deformation capacity of each wall segment based on simple rules described in the TMS Section C.3. Thus, the output of the linear model gives the necessary information to determine

125 116 deformation capacities and deformation demands. The nonlinear model is necessary in the event that the controlling yield mechanism cannot be easily determined by inspection after considering a handful of potential yield mechanisms. The simplified Nonlinear Layer model is described here through its application to Design Example 2 (presented in Chapter 4) which consists of a multistory coupled wall with openings as shown in Figure 4.2. The openings form a structure comprised of three multistory vertical wall segments joined by coupling beams at each floor level. The wall sits on top of a one-story parking structure where a deep transfer girder supports the center wall segment. The first elevated floor slab is laterally supported by additional walls providing ten times the lateral stiffness of any of the stories above and is therefore considered the seismic base. The residential structure above the seismic base is assumed to have flexible diaphragms at each floor level, which occurs at the center of the 4-ft deep coupling beams. The definition of material properties for modeling the nonlinear response are characterized by the specified material strengths, as shown in Table 4.5. To develop the nonlinear model, the area elements located at the interface of the wall segments are replaced with layered area elements. For the wall configuration in Design Example 2, the computer model for evaluating its linear-elastic response can be represented by Figure C.1. Figure C.2 shows the model with nonlinear wall segments. The linear-elastic model uses area elements with a 6 in. by 4 in. mesh. This level of discretization is sufficiently accurate considering that, for a unit load applied at the roof level, the resulting roof displacement is within 2% of the displacement calculated using a 2-in. square mesh, as reported in Table 4.4. The 6 in. by 4 in. mesh also allows a direct representation of the modular dimensions of clay masonry units (12-in. long, 4-in. high, and 5.5-in. thick) used in this wall. It is important to note that the modeling approach described here is not suitable for use in nonlinear dynamic analysis. Additional special definitions would be needed to properly

126 117 account for the cyclic behavior involving masonry cracking and reinforcement yielding, and their effects on stiffness and strength reductions. C.1 Nonlinear Layer Model The layered shell element, available in structural software SAP2000 (CSI, 2011b), is a special type of area element that may be defined with multiple layers in the thickness direction. Each layer may represent independent materials with user-defined nonlinear stress-strain relationships. A detailed description of the advanced features of the layered shell element is presented by CSI (2011a). The Nonlinear Layer model for reinforced masonry walls, as presented by Sanchez (2012), is based on the use of nonlinear area elements to represent the region at the interface of wall segments where yielding is likely to occur. For the wall in Design Example 2, the nonlinear model is shown in Figure C.2. The area elements outside the assumed yielding regions are modeled with linear-elastic area elements using full gross-section properties. For a planar wall configuration the area elements may be defined as membrane elements with layers assigned to materials with nonlinear behavior. Layers of masonry and steel flexural reinforcement are combined to represent reinforced masonry sections. For unreinforced masonry sections, a single masonry layer is typically used per area element. Because the nonlinear material properties are defined without bounds on maximum strains, the length of the region where the nonlinearity extends (away from wall interfaces) becomes unimportant as long as the user has other means to check deformations capacities, such as those provided in TMS Section C.3. Material stress-strain relationships are defined to represent the axial and shear behavior of the wall segments. The in-plane flexural behavior of the walls is controlled by the nonlinear axial response characteristics of the materials assigned to the layers. Independent materials are defined to represent the axial response of masonry and steel reinforcement. Masonry in

127 118 compression is assumed to have a bilinear stress-strain curve and is neglected in tension as shown in Figure C.3. Reinforcing steel is characterized by a bilinear and symmetrical stressstrain curve as shown in Figure C.4. The peak stress of masonry is taken as 0.8 times the specified compressive strength of masonry, f m, and the peak stress of the reinforcing steel is based on the specified yield strength, f y. Material property definitions neglect the strain hardening effects of steel and the expected overstrengths of steel and masonry. The Nonlinear Layer model implemented by Sanchez (2012) included the representation of the nonlinear shear response using a uniaxial material with a bilinear and symmetrical stressstrain curve. Sanchez defined the initial line segment of the stress-strain curve using the shear modulus, G m. For a uniaxial material, the shear stress values were entered as twice the actual values (CSI, 2011a). Therefore, Sanchez defined the input peak shear stress values as two times the calculated shear strength divided by the cross-sectional area of the wall. Because the shear strength of masonry walls depends on the ratio M u /(V u d v ) and on the axial load P u, different material definitions were required for the various wall segments involved. For this purpose, the values of M u, V u, and P u were based on values from the linear-response model used as a basis to create the nonlinear model. The nonlinear stress-strain idealization used for shear was meant to represent the combined effects of masonry and shear reinforcement. The modeling approach by Sanchez (2012), to represent nonlinear response in shear, was not intended to simulate realistic shear behavior but to help identify the wall segments that reach their codebased design shear strength before their flexural strength. For the wall in Design Example 2, this approach would have required, for each vertical wall segment, a different nonlinear material property definition to represent shear response due to changes in the axial load and M/Vd on each story. For this study, it was decided to adopt a simpler model. A simplified Nonlinear Layer model is proposed here where the shear response is modeled linearly. Instead of defining a nonlinear material to represent the shear response based on the M u /(V u d v ) ratio, axial load, masonry strength, and shear reinforcement, the simplified method assumes that the wall segments do not reach their shear strength. Using this

128 119 approach, the only nonlinear material layers needed for the model are those representing the axial response of the masonry and steel reinforcement as described above. The model output for the simplified modeling approach represents a realistic response as long as the shear strength of the wall components is not exceeded. To assemble the layers into an area section, the thickness of the layer representing masonry in compression or shear is set to be the actual wall thickness. The thickness of the layer representing the reinforcing steel, to resist flexure and axial loads, is defined by the steel area divided by the discretized length of area element represented. Thus, the program uses the area of steel reinforcement resulting from the product of the layer thickness and the length of the area element. The definition of a layer also requires assigning a material angle. For instance, an area element with nonlinear layers in Figure C.2 representing the reinforced masonry of the vertical wall segments, should incorporate a layer of masonry with nonlinear capabilities in the local 2-2 direction (or vertical direction) while linear-response is assigned to the local 1-1 direction (or horizontal direction). Similarly, it should incorporate a layer of steel with nonlinear capabilities in the local 2-2 direction. For more details, see CSI (2011a) and Lepage and Sanchez (2012). Before proceeding with the nonlinear static analysis for the lateral loads, the starting points on the stress-strain curves of each nonlinear layer need to be determined. This is typically accomplished by pre-loading the structure with a gravity load case. Based on the Limit Design provisions in TMS Appendix C, the gravity load case should be based on load combination 7 of Section of ASCE/SEI C.2 Nonlinear Analysis Results The nonlinear model for the structure representing Design Example 2, shown in Figure C.2, is analyzed for lateral loads in the north-south direction. The lateral load profile from base to roof follows the vertical force distribution obtained from the equivalent lateral force

129 120 procedure in ASCE/SEI 7 (2010) as indicated in Table 4.1. Output for global shear (total base shear) and local shear (individual wall shear) are monitored against the roof displacement in Figure 4.9. The simplified nonlinear static analysis has two main objectives: (1) identify where yielding occurs; and (2) determine the plastic base shear strength. The simplified model considers only nonlinear action due to flexure and axial loads and assumes linear response for shear forces. To identify the wall segments responding nonlinearly, the user needs to monitor the forces in the regions where nonlinear elements were assigned and check if the limiting strength of the nonlinear layers was reached. The plastic base shear strength, V p, of the wall configuration is determined using the base shear vs. roof displacement curves that result from the nonlinear static analysis as shown in Figure 4.9. On this figure, an open circle is used to identify the last point on the curve where the slope exceeds 5% of the slope referring to the initial stiffness. The initial stiffness was obtained from linear-elastic response using gross-section properties. The plastic base shear strength defined in this manner corresponds to the instance at which the structure has nearly developed a yield mechanism. Chapter 4 includes more details about the nonlinear response of the wall when subjected to lateral loads. For cases where the strength of the wall segments is controlled by flexural or axial yielding, the output of the proposed simplified Nonlinear Layer model will match the output obtained from the more elaborate Nonlinear Layer Model by Sanchez (2012). The caveat of using the simplified model is that it may require special processing of the output. Thus, after the nonlinear analysis of the simplified model is completed, the user should monitor the shear history that acted on any wall segment and compare it with the varying shear strength (V n ) associated with the concurrent axial force and M/Vd ratio. The limiting base shear (global shear) of the line of lateral resistance may then be adjusted using the value of base shear that corresponds to the instant where the shear in any wall segment (local shear) exceeded its calculated shear strength (V n ).

130 121 APPENDIX D: TABLES

131 Table 2.1 Limit Design Code and Commentary, Taken from TMS 402 (2013) 122

132 Table 2.1 Continued 123

133 124 Table 3.1 Story Weight above the Seismic Base and Vertical Distribution of Seismic Forces Floor Level Floor Elevation Floor Weight a Partition Weight Tributary Floor Area Wall Weight Wall Area b Story Weight c Story Force d h x (ft) (psf) (psf) (ft 2 ) (psf) (ft 2 ) w x (kip) F x (kip) R G Σ = a Roof weight: plywood sheathing, 3 psf; roofing, 5 psf; joists, 2 psf; MEP, 3psf; ceiling, 3 psf; misc. garage equipment, 15 psf b Same area of walls assumed for both directions of analysis c Based on the total seismic weight tributary to the line of lateral resistance under consideration (Figure 3.1) d The base shear, V b, acting on the wall (Figure 3.1) is calculated using V b = 1.1 W S DS /R = 41.1 kip; where W = 187 kip, and R = 5.0. The factor 1.1 accounts for torsional effects. Table 3.2 Wall Reinforcement Schedule for Strength Design, S DS = 1.0 Wall Element Reinforcement Level Vertical Horizontal Strength Design Wall A Ground to Roof 16" 16" Wall B Ground to Roof 8" 8" Wall C Ground to Roof 16" 16"

134 125 Table 3.3 Wall Reinforcement Schedule for Limit Design, S DS = 1.0 Wall Element Level Limit Design Reinforcement Vertical Horizontal Wall A Ground to Roof 16" 16" Wall B Ground to Roof 8" 8" Wall C Ground to Roof 16" 16" Table 3.4 Lateral Stiffness for Different Mesh Sizes Mesh Size Lateral Force at Roof a Roof Displacement b Stiffness Relative Error (in.) (kip) (in.) (kip/in.) (%) 1x x x x x a Applied at roof level (see Figure 3.1). b Displacements based on linear-elastic model using gross section properties.

135 126 Table 3.5 Material Properties for Nonlinear Static Analysis Material ID a Point a Strain b Stress b M E E+00 [k,in] E E+00 f ' m = 1.50 ksi E m = 1350 ksi 1' 8.89E E-04 2' 8.89E E-04 R E E+01 [k,in] E E+01 f y = 60 ksi E s = ksi 1' 2.07E E+01 2' 2.07E E+01 Notes: a M2500 represents the masonry (f' m = 2500 psi) subjected to axial forces and with negligible tensile strength. R60 represents axially loaded reinforcement with f y = 60 ksi. b Modulus of Elasticity, E m, is taken as 700 f' m for clay masonry.

136 127 Table 4.1 Story Weights above the Seismic Base and Vertical Distribution of Seismic Forces Floor Level Floor Elevation Floor Weight a Partition Weight Tributary N-S Width Tributary E-W Width Wall Weight Wall Area b Story Weight c h x (ft) (psf) (psf) (ft) (ft) (psf) (ft 2 ) w x (kip) w x h x F x /V b d R G Σ = a Roof weight: metal deck, 2 psf; roofing, 5 psf; joists, 2 psf; MEP, 3psf; ceiling and miscellaneous, 5 psf Floor weight: metal deck and topping, 14 psf; floor finish, 3 psf; joists, 2 psf; MEP, 3 psf; ceiling and miscellaneous, 5 psf b Tributary wall area (seismic and non-seismic). Same area of walls assumed for both directions of analysis c Based on the total seismic weight tributary to the line of lateral resistance under consideration (Figure 4.1) d Vertical force distribution based on the equivalent lateral force procedure in ASCE/SEI 7 (2010). F x is the lateral seismic force at any level. The base shear, V b, acting on the wall (Figure 4.2) is calculated using V b = 1.1 W S DS / R = 86.6 kip; where W = 433 kip, S DS = 1.0, and R = 5.5. The factor 1.1 accounts for torsional effects. Table 4.2 Wall Reinforcement Schedule for Strength Design, S DS = 1.0 Wall Element Reinforcement Level Vertical Horizontal Strength Design Wall A = Wall C Ground to Roof 18" 20" Wall B Ground to Roof 18" 20" Coupling Beams 3 to Roof 12" 16" Transfer Girder 2 12" 16"

137 128 Table 4.3 Wall Reinforcement Schedule for Limit Design, S DS = 1.0 Wall Element Level Limit Design Reinforcement Vertical Horizontal Wall A = Wall C Ground to Roof 18" 20" Wall B Ground to Roof 18" 20" Coupling Beams 3 to Roof 12" 16" Transfer Girder 2 12" 16" Table 4.4 Lateral Stiffness for Different Mesh Sizes Mesh Size Lateral Force at Roof a Roof Displacement b Stiffness Relative Error c (in.) (kip) (in.) (kip/in.) (%) 2x x x x x a Applied at centerline of Wall B (see Figure 4.2). b Measured at edge node of roof level. Displacements based on linear-elastic model using gross section properties. c All models used a translational spring at level 2 (with a stiffness of 10,000 kip/in.) to represent the combined lateral stiffness of the additional walls below level 2.

138 129 Table 4.5 Material Properties for Nonlinear Static Analysis Material ID a Point a Strain b Stress b M E E+00 [k,in] E E+00 f ' m = 2.50 ksi E m = 1750 ksi 1' 1.14E E-04 2' 1.14E E-04 R E E+01 [k,in] E E+01 f y = 60 ksi E s = ksi 1' 2.07E E+01 2' 2.07E E+01 Notes: a M2500 represents the masonry (f' m = 2500 psi) subjected to axial forces and with negligible tensile strength. R60 represents axially loaded reinforcement with f y = 60 ksi. b Modulus of Elasticity, E m, is taken as 700 f' m for clay masonry.

139 130 APPENDIX E: FIGURES

140 131 Materials: f y = 60 ksi f m = 1500 psi 8-in. concrete masonry units, fully grouted Loads: Self weight = 80 psf Trib. Dead = 150 plf Trib. Live = 225 plf Seismic Design Parameters: S DS = 1.0 S D1 = 0.6 R = 5.0 C d = 3.5 ρ = 1.0 δ e,roof = 0.28 for V b = 100 kip (based on 50% gross section properties) Figure 3.1 Building Description, Wall Elevation Figure 3.2 Shear Wall Reinforcement Layout

141 132 M= 4.1 V= 0.70 P= 7.7 M= 0.12 M= 3.5 V= 0.01 V= 0.72 P= 5.6 P= 7.2 M= 3.0 M= 0.01 M= 2.2 V= 0.70 V= 0.01 V= 0.72 P= 10.9 Units: kip, ft P= 6.9 P= 9.8 Figure 3.3 Member Forces Due to Dead Load (1.0D) M= 1.1 V= 0.20 P= 2.2 M= 0.01 M= 0.96 V= 0.00 V= 0.20 P= 1.5 P= 1.8 M= 0.84 M= 0.02 M= 0.62 V= 0.20 V= 0.00 V= 0.20 P= 2.2 Units: kip, ft P= 1.5 P= 1.8 Figure 3.4 Member Forces Due to Live Load (1.0L)

142 133 M= 58.7 V= 15.2 P= 19.7 M= 15.5 M= 59.6 V= 4.6 V= 21.2 P= 5.3 P= 25.0 M= 93.1 M= 21.3 M= V= 15.2 V= 4.6 V= 21.2 P= 19.7 Units: kip, ft P= 5.3 P= 25.0 Figure 3.5 Member Forces Due to Earthquake Load (1.0E), S DS = Units: kip, ft Figure 3.6 Values of M/(Vd) based on 1.0E

143 134 M u = 65.0 V u = 16.3 P u = 31.6 M u = 15.7 M u = 65.0 V u = 4.6 V u = 22.3 P u = 13.9 P u = 36.0 M u = 97.7 M u = 21.3 M u = V u = 16.3 V u = 4.6 V u = 22.3 P u = 36.1 Units: kip, ft P u = 15.7 P u = 39.6 Figure 3.7 Member Forces Due to 1.2D + 0.5L + 1.0E, S DS = 1.0 M u = 61.6 V u = 15.7 P u = M u = 15.6 M u = 62.1 V u = 4.6 V u = 21.7 P u = -1.4 P u = M u = 95.2 M u = 21.3 M u = V u = 15.7 V u = 4.6 V u = 21.7 P u = Units: kip, ft P u = -0.5 P u = Figure 3.8 Member Forces Due to 0.9D + 1.0E, S DS = 1.0

144 135 Stress, ksi Strain -0.5 E m = 1350 ksi 0.8 f' m Figure 3.9 Masonry Model, Axial Direction Stress, ksi f y E s = ksi Strain f y Figure 3.10 Reinforcement Steel Model, Axial Direction

145 Shear, kip Base Shear V C Roof Displacement, in. V A V B 20 Figure 3.11 Wall Shear vs. Roof Displacement, Wall Reinforcement per Limit Design (Table 3.3), Eastward Loading Shear, kip Base Shear V A V C V B Roof Displacement, in. Figure 3.12 Wall Shear vs. Roof Displacement, Wall Reinforcement per Limit Design (Table 3.3), Westward Loading

146 Base Shear V C Shear, kip V A 20 V B Roof Displacement, in. Figure 3.13 Wall Shear vs. Roof Displacement, Wall Reinforcement per Strength Design (Table 3.2), Eastward Loading Base Shear 80 Shear, kip V A V C 20 V B Roof Displacement, in. Figure 3.14 Wall Shear vs. Roof Displacement, Wall Reinforcement per Strength Design (Table 3.2), Westward Loading

147 138 Material Properties: fy = 60 ksi f m = 2500 psi Clay masonry units, 12 x 4 x 6 in., fully grouted Weights: Wall self-weight: 60 psf (in elevation) Roof dead: 77 k Floor dead: 119 k W = 433 k (for 42 x 54 building footprint) Notes: 1. Lateral force-resisting system in the E-W direction not shown. 2. Gravity loads not transferred to masonry shear walls. N Seismic Design Parameters (ASCE/SEI 7-10): SDS = 1.0 R = 5.5 (Special Reinforced Masonry Shear Walls) Cd = 4.0 Ie = 1.0 δe,roof = 0.29 for Vb = 100 k (per 40 wall) (Based on 50% of gross section properties) ΩO = 2.5 CT = 0.02 x = 0.75 hn = 36 ft Ta = 0.29 s Cu = 1.4 k = 1.0 ρ = 1.0 Cs = 0.18 for SDS = 1.0 Figure 4.1 Building Description: Floor Plan, Material Properties, Loads, and Seismic Design Parameters

148 139 5 x 7 Window (typ.) Roof 36 Level 5 27 Level 4 18 Level 3 9 Level 2 Seismic Base Ground Figure 4.2 Building Description, Wall Elevation (East Line of Resistance)

149 Figure 4.3 Shear Wall Reinforcement Layout 140

150 141 M= 2.4 M= 1.8 M= 1.8 M= 2.4 R V= 1.4 V= 0.2 V= 0.2 V= 1.4 M= 6.9 M= 0.0 M= 6.9 V= 1.4 V= 0.0 V= 1.4 P= 3.8 P= 1.9 P= 3.8 M= 0.1 M= 0.0 M= 0.1 V= 1.4 V= 0.0 V= 1.4 P= 6.8 P= 3.6 P= 6.8 M= 4.1 M= 3.0 M= 3.0 M= V= 1.8 V= 0.2 V= 0.2 V= 1.8 M= 7.3 M= 0.0 M= 7.3 V= 1.7 V= 0.0 V= 1.7 P= 11.1 P= 4.8 P= 11.1 M= 1.2 M= 0.0 M= 1.2 V= 1.7 V= 0.0 V= 1.7 P= 14.1 P= 6.6 P= 14.1 M= 5.2 M= 3.9 M= 3.9 M= V= 2.1 V= 0.5 V= 0.5 V= 2.1 M= 6.9 M= 0.0 M= 6.9 V= 2.2 V= 0.0 V= 2.2 P= 18.6 P= 7.2 P= 18.6 M= 3.9 M= 0.0 M= 3.9 V= 2.2 V= 0.0 V= 2.2 P= 21.6 P= 9.0 P= 21.6 M= 6.7 M= 5.2 M= 5.2 M= V= 2.5 V= 0.9 V= 0.9 V= 2.5 M= 4.3 M= 0.0 M= 4.3 V= 3.4 V= 0.0 V= 3.4 P= 26.6 P= 8.7 P= 26.6 M= 12.5 M= 0.0 M= 12.5 V= 3.4 V= 0.0 V= 3.4 P= 29.6 P= 10.5 P= M= 20.7 M= 24.1 M= 24.1 M= 20.7 V= 10.0 V= 6.7 V= 6.7 V= 10.0 M= 32.7 M= 32.7 V= 4.7 V= 4.7 P= 44.4 P= 44.4 M= 10.1 M= 10.1 V= 4.7 V= 4.7 G P= 49.8 Units: kip, ft P= 49.8 Figure 4.4 Member Forces Due to Dead Load (1.0D)

151 142 M= 23.2 M= 20.0 M= 20.0 M= 23.2 R V= 6.2 V= 6.2 V= 6.2 V= 6.2 M= 42.5 M= 48.4 M= 42.5 V= 5.8 V= 14.4 V= 5.8 P= 6.2 P= 0.0 P= 6.2 M= 13.4 M= 18.6 M= 13.4 V= 5.8 V= 14.4 V= 5.8 P= 6.2 P= 0.0 P= 6.2 M= 43.8 M= 42.1 M= 42.1 M= V= 12.3 V= 12.3 V= 12.3 V= 12.3 M= 72.9 M= 61.2 M= 72.9 V= 17.0 V= 22.2 V= 17.0 P= 18.4 P= 0.0 P= 18.4 M= 12.2 M= 49.9 M= 12.2 V= 17.0 V= 22.2 V= 17.0 P= 18.4 P= 0.0 P= 18.4 M= 58.8 M= 56.5 M= 56.5 M= V= 16.5 V= 16.5 V= 16.5 V= 16.5 M= 45.5 M= 63.6 M= 45.5 V= 24.7 V= 26.8 V= 24.7 P= 34.9 P= 0.0 P= 34.9 M= 77.9 M= 70.6 M= 77.9 V= 24.7 V= 26.8 V= 24.7 P= 34.9 P= 0.0 P= 34.9 M= 61.4 M= 59.3 M= 59.3 M= V= 17.2 V= 17.2 V= 17.2 V= 17.2 M= 37.7 M= 40.3 M= 37.7 V= 29.0 V= 28.7 V= 29.0 P= 52.1 P= 0.0 P= 52.1 M= M= M= V= 29.0 V= 28.7 V= 29.0 P= 52.1 P= 0.0 P= M= 77.4 M= 52.4 M= 52.4 M= 77.4 V= 16.4 V= 16.4 V= 16.4 V= 16.4 M= 72.7 M= 72.7 V= 0.2 V= 0.2 P= 68.6 P= 68.6 M= 70.6 M= 70.6 V= 0.2 V= 0.2 G P= 68.6 Units: kip, ft P= 68.6 Figure 4.5 Member Forces Due to Earthquake Load (1.0E), S DS = 1.0

152 143 R G 1.00 Units: kip, ft 1.00 Figure 4.6 Values of M/(Vd) based on 1.0E

153 144 M u = 26.5 M u = 22.6 M u = 22.6 M u = 26.5 R V u = 8.2 V u = 6.5 V u = 6.5 V u = 8.2 M u = 52.1 M u = 48.4 M u = 52.1 V u = 7.7 V u = 14.4 V u = 7.7 P u = 11.6 P u = 2.7 P u = 11.6 M u = 13.6 M u = 18.6 M u = 13.6 V u = 7.7 V u = 14.4 V u = 7.7 P u = 15.7 P u = 5.0 P u = M u = 49.5 M u = 46.3 M u = 46.3 M u = 49.5 V u = 14.9 V u = 12.5 V u = 12.5 V u = 14.9 M u = 83.1 M u = 61.2 M u = 83.1 V u = 19.4 V u = 22.2 V u = 19.4 P u = 34.0 P u = 6.8 P u = 34.0 M u = 14.0 M u = 49.9 M u = 14.0 V u = 19.4 V u = 22.2 V u = 19.4 P u = 38.2 P u = 9.3 P u = M u = 66.1 M u = 61.9 M u = 61.9 M u = 66.1 V u = 19.5 V u = 17.1 V u = 17.1 V u = 19.5 M u = 55.2 M u = 63.6 M u = 55.2 V u = 27.7 V u = 26.8 V u = 27.7 P u = 61.0 P u = 10.0 P u = 61.0 M u = 83.4 M u = 70.6 M u = 83.4 V u = 27.7 V u = 26.8 V u = 27.7 P u = 65.2 P u = 12.5 P u = M u = 70.7 M u = 66.5 M u = 66.5 M u = 70.7 V u = 20.8 V u = 18.4 V u = 18.4 V u = 20.8 M u = 43.7 M u = 40.3 M u = 43.7 V u = 33.7 V u = 28.7 V u = 33.7 P u = 89.3 P u = 12.2 P u = 89.3 M u = M u = M u = V u = 33.7 V u = 28.7 V u = 33.7 P u = 93.5 P u = 14.7 P u = M u = 106 M u = 86.2 M u = 86.2 M u = 106 V u = 30.5 V u = 25.8 V u = 25.8 V u = 30.5 M u = M u = V u = 6.9 V u = 6.9 P u = P u = M u = 84.7 M u = 84.7 V u = 6.9 V u = 6.9 G P u = Units: kip, ft P u = Figure 4.7 Member Forces Due to 1.2D + 1.0E, S DS = 1.0

154 145 M u = 24.9 M u = 21.3 M u = 21.3 M u = 24.9 R V u = 7.2 V u = 6.3 V u = 6.3 V u = 7.2 M u = 47.3 M u = 48.4 M u = 47.3 V u = 6.8 V u = 14.4 V u = 6.8 P u = -3.5 P u = 1.3 P u = -3.5 M u = 13.5 M u = 18.6 M u = 13.5 V u = 6.8 V u = 14.4 V u = 6.8 P u = -1.4 P u = 2.5 P u = M u = 46.7 M u = 44.2 M u = 44.2 M u = 46.7 V u = 13.6 V u = 12.4 V u = 12.4 V u = 13.6 M u = 78.0 M u = 61.2 M u = 78.0 V u = 18.2 V u = 22.2 V u = 18.2 P u = P u = 3.4 P u = M u = 13.1 M u = 49.9 M u = 13.1 V u = 18.2 V u = 22.2 V u = 18.2 P u = -8.6 P u = 4.6 P u = M u = 62.4 M u = 59.2 M u = 59.2 M u = 62.4 V u = 18.0 V u = 16.8 V u = 16.8 V u = 18.0 M u = 50.3 M u = 63.6 M u = 50.3 V u = 26.2 V u = 26.8 V u = 26.2 P u = P u = 5.0 P u = M u = 80.7 M u = 70.6 M u = 80.7 V u = 26.2 V u = 26.8 V u = 26.2 P u = P u = 6.3 P u = M u = 66.0 M u = 62.9 M u = 62.9 M u = 66.0 V u = 19.0 V u = 17.8 V u = 17.8 V u = 19.0 M u = 40.7 M u = 40.3 M u = 40.7 V u = 31.3 V u = 28.7 V u = 31.3 P u = P u = 6.1 P u = M u = M u = M u = V u = 31.3 V u = 28.7 V u = 31.3 P u = P u = 7.3 P u = M u = 91.8 M u = 69.3 M u = 69.3 M u = 91.8 V u = 23.5 V u = 21.1 V u = 21.1 V u = 23.5 M u = 95.6 M u = 95.6 V u = 3.6 V u = 3.6 P u = P u = M u = 77.7 M u = 77.7 V u = 3.6 V u = 3.6 G P u = Units: kip, ft P u = Figure 4.8 Member Forces Due to 0.9D + 1.0E, S DS = 1.0

155 Shear, kip Base Shear V C V B Roof Displacement, in. V A 0 Figure 4.9 Wall Shear vs. Roof Displacement, Wall Reinforcement per Limit Design (Table 4.3), Northward Loading

156 147 Beam Yielding Wall Yielding Seismic Base Figure 4.10 Deformed Shape for Simplified Nonlinear Layer Model, Wall Reinforcement per Limit Design (Table 4.3)

157 Beam Section at Wall A Interface Shear, kip 0-20 Roof 5th 3rd Southward Loading Northward Loading th Roof Displacement, in. Figure 4.11 Shear in Beams vs. Roof Displacement Beam Section at Wall A Interface Compression (+) 4th Axial Load, kip rd 5th Roof Roof Displacement, in. Figure 4.12 Axial Force in Beams vs. Roof Displacement, Northward Loading -60

158 149 Axial Load, kip Beam Section at Wall A Interface Compression (+) 3rd 4th 5th Roof Roof Displacement, in. Figure 4.13 Axial Force in Beams vs. Roof Displacement, Southward Loading Base Shear 195 V C Shear, kip V B 50 0 V A Roof Displacement, in. 0 Figure 4.14 Wall Shear vs. Roof Displacement, Wall Reinforcement per Strength Design (Table 4.2), Northward Loading

159 150 Seismic Base Wall Yielding Figure 4.15 Deformed Shape for Simplified Nonlinear Layer Model, Wall Reinforcement per Strength Design (Table 4.2)

160 151 Seismic base with springs Linear-elastic area elements Nodes fixed at ground Figure C.1 Linear-Elastic Model with 6 in. by 4 in. Mesh, Design Example 2

161 152 Seismic base with springs Area elements with nonlinear layers Nodes fixed at ground Linear-elastic area elements Figure C.2 Simplified Nonlinear Layer Model, Design Example 2