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

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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

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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.

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TABLE OF CONTENTS

List of Tables ................................................................................................................................................ vi

List of Figures .............................................................................................................................................. vii

Acknowledgments ........................................................................................................................................ ix

Chapter 1: Introduction ................................................................................................................................ 1

1.1 Statement of the Problem ............................................................................................................ 1

1.2 Objectives and Scope .................................................................................................................... 1

1.3 Organization .................................................................................................................................. 2

Chapter 2: Limit Design Method ................................................................................................................... 4

2.1 Framework .................................................................................................................................... 4

2.2 Code Provisions ............................................................................................................................. 5

2.3 Design Steps .................................................................................................................................. 6

2.4 Limitations..................................................................................................................................... 7

Chapter 3: Design Example 1 ........................................................................................................................ 9

3.1 Description .................................................................................................................................... 9

3.2 Linear-Elastic Analysis ................................................................................................................. 11

3.3 Strength Design ........................................................................................................................... 12

3.4 Limit Design ................................................................................................................................. 14

3.5 Nonlinear Static Analysis ............................................................................................................. 16

3.6 Summary ..................................................................................................................................... 18

Chapter 4: Design Example 2: Multistory Coupled Shear Walls ................................................................. 19

4.1 Description .................................................................................................................................. 19

4.2 Linear-Elastic Analysis ................................................................................................................. 21

4.3 Strength Design ........................................................................................................................... 22

4.4 Limit Design ................................................................................................................................. 25

4.5 Nonlinear Static Analysis ............................................................................................................. 28

4.6 Summary ..................................................................................................................................... 30

Chapter 5: Conclusions ............................................................................................................................... 31

Appendix A: Spreadsheet Formulations for Strength Design ..................................................................... 33

A.1 Strength Design Calculations for Design Example 1 ................................................................... 34

A.2 Strength Design Calculations for Design Example 2 ................................................................... 58

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Appendix B: Spreadsheet Formulations for Limit Design ........................................................................... 84

B.1 Limit Design Calculations for Design Example 1 ......................................................................... 85

B.2 Limit Design Calculations for Design Example 2 ......................................................................... 98

Appendix C: Practical Nonlinear Static Analysis of Masonry Walls .......................................................... 115

C.1 Nonlinear Layer Model ............................................................................................................. 117

C.2 Nonlinear Analysis Results ........................................................................................................ 119

Appendix D: Tables ................................................................................................................................... 121

Appendix E: Figures ................................................................................................................................... 130

List of References ...................................................................................................................................... 154

Biographical Sketch ................................................................................................................................... 156

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LIST OF TABLES

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

Table 3.1 – Story Weight above the Seismic Base and Vertical Distribution of Seismic Forces ............... 124

Table 3.2 – Wall Reinforcement Schedule for Strength Design, SDS = 1.0 ................................................ 124

Table 3.3 – Wall Reinforcement Schedule for Limit Design, SDS = 1.0 ...................................................... 125

Table 3.4 – Lateral Stiffness for Different Mesh Sizes .............................................................................. 125

Table 3.5 – Material Properties for Nonlinear Static Analysis .................................................................. 126

Table 4.1 – Story Weights above the Seismic Base and Vertical Distribution of Seismic Forces ............. 127

Table 4.2 – Wall Reinforcement Schedule for Strength Design, SDS = 1.0 ................................................ 127

Table 4.3 – Wall Reinforcement Schedule for Limit Design, SDS = 1.0 ...................................................... 128

Table 4.4 – Lateral Stiffness for Different Mesh Sizes .............................................................................. 128

Table 4.5 – Material Properties for Nonlinear Static Analysis .................................................................. 129

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LIST OF FIGURES

Figure 3.1 – Building Description, Wall Elevation ..................................................................................... 131

Figure 3.2 – Shear Wall Reinforcement Layout ........................................................................................ 131

Figure 3.3 – Member Forces Due to Dead Load (1.0D) ............................................................................ 132

Figure 3.4 – Member Forces Due to Live Load (1.0L) ............................................................................... 132

Figure 3.5 – Member Forces Due to Earthquake Load (1.0E), SDS = 1.0 ................................................... 133

Figure 3.6 – Values of M/(Vd) based on 1.0E ........................................................................................... 133

Figure 3.7 – Member Forces Due to 1.2D + 0.5L + 1.0E, SDS = 1.0 ............................................................ 134

Figure 3.8 – Member Forces Due to 0.9D + 1.0E, SDS = 1.0 ....................................................................... 134

Figure 3.9 – Masonry Model, Axial Direction ........................................................................................... 135

Figure 3.10 – Reinforcement Steel Model, Axial Direction ...................................................................... 135

Figure 3.11 – Wall Shear vs. Roof Displacement, Wall Reinforcement per Limit Design (Table 3.3),

Eastward Loading ................................................................................................................ 136

Figure 3.12 – Wall Shear vs. Roof Displacement, Wall Reinforcement per Limit Design (Table 3.3),

Westward Loading .............................................................................................................. 136

Figure 3.13 – Wall Shear vs. Roof Displacement, Wall Reinforcement per Strength Design (Table 3.2),

Eastward Loading ................................................................................................................ 137

Figure 3.14 – Wall Shear vs. Roof Displacement, Wall Reinforcement per Strength Design (Table 3.2),

Westward Loading .............................................................................................................. 137

Figure 4.1 – Building Description: Floor Plan, Material Properties, Loads, and Seismic Design

Parameters ......................................................................................................................... 138

Figure 4.2 – Building Description, Wall Elevation (East Line of Resistance) ............................................. 139

Figure 4.3 – Shear Wall Reinforcement Layout ........................................................................................ 140

Figure 4.4 – Member Forces Due to Dead Load (1.0D) ............................................................................ 141

Figure 4.5 – Member Forces Due to Earthquake Load (1.0E), SDS = 1.0 ................................................... 142

Figure 4.6 – Values of M/(Vd) based on 1.0E ........................................................................................... 143

Figure 4.7 – Member Forces Due to 1.2D + 1.0E, SDS = 1.0 ....................................................................... 144

Figure 4.8 – Member Forces Due to 0.9D + 1.0E, SDS = 1.0 ....................................................................... 145

Figure 4.9 – Wall Shear vs. Roof Displacement, Wall Reinforcement per Limit Design (Table 4.3),

Northward Loading ............................................................................................................. 146

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Figure 4.10 – Deformed Shape for Simplified Nonlinear Layer Model, Wall Reinforcement per Limit

Design (Table 4.3) ............................................................................................................... 147

Figure 4.11 – Shear in Beams vs. Roof Displacement ............................................................................... 148

Figure 4.12 – Axial Force in Beams vs. Roof Displacement, Northward Loading ..................................... 148

Figure 4.13 – Axial Force in Beams vs. Roof Displacement, Southward Loading ..................................... 149

Figure 4.14 – Wall Shear vs. Roof Displacement, Wall Reinforcement per Strength Design (Table 4.2),

Northward Loading ............................................................................................................. 149

Figure 4.15 – Deformed Shape for Simplified Nonlinear Layer Model, Wall Reinforcement per Strength

Design (Table 4.2) ............................................................................................................... 150

Figure C.1 – Linear-Elastic Model with 6 in. by 4 in. Mesh, Design Example 2 ......................................... 151

Figure C.2 – Simplified Nonlinear Layer Model, Design Example 2 .......................................................... 152

Figure C.3 – Masonry Model, Axial Direction ........................................................................................... 153

Figure C.4 – Reinforcement Steel Model, Axial Direction ........................................................................ 153

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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.

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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 402-13 are included for Allowable Stress Design

(ASD) and Strength Design (SD). The conventional methods supported by ASD and SD are well-

suited 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 402-13 to further encourage the use of structural masonry in earthquake-

resistant 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 402-13 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 402-13 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 402-13

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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 earthquake-

resistant 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

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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.

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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 well-

suited 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 402-13 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

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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 402-13 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 402-13 incorporates changes in response to ballot

comments submitted by the Seismic Subcommittee of TMS 402.

The Limit Design provisions in Appendix C of TMS 402-13 use the framework of the

Strength Design provisions in Chapter 9 of TMS 402-13. 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.,

2008. 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.

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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 402-13 Chapter 7

3. Determine the controlling yield mechanism

− Identify the potential plastic hinge regions

− Assign plastic hinge strengths based on nominal flexural strength (Mn)

− Determine the limiting mechanism (assume profile of story forces are

proportional to those determined in Step 1)

7

4. Check for shear-controlled wall segments and adjust plastic hinge strengths

− If Vn < 2VMn then wall segment is shear controlled

[VMn is the shear associated with Mn]

− If Vn ≥ 2VMn then wall segment is not shear controlled

− Adjust the plastic hinge strength, Mp, using

If Vn < 2VMn then Mp = Mn (Vn / 2VMn) [shear-controlled condition]

else Mp = Mn

5. Check mechanism strength

− Compute the limiting base shear strength, Vlim, for the controlling yield

mechanism (Step 3) using adjusted plastic hinge strengths (Step 4)

− Check φVlim ≥ Vub (φ = 0.8)

[Vub is the base shear demand at the line of lateral resistance under consideration]

6. Check deformation capacities

− Deformation capacity, δcap, of a wall segment is defined using

δcap = 0.5 lw hw εmu / c [See notation in TMS 402 (2013)]

For shear-controlled wall segments, δcap = hw / 400, except that hw / 200 applies

where conditions in TMS 402-13 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,

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 Ag. 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 (Vn) does

not exceed two times the shear associated with the development of the nominal flexural

strength (2 VMn), the wall is considered shear controlled. An additional limitation requiring Vn ≥ VMn

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).

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 402-13. 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.

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 12.4.2.3 (which include effects due to

vertical seismic forces). Strength Design predominantly follows the requirements in Chapter 9

of TMS 402-13 and Limit Design follows Appendix C of TMS 402-13.

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 402-13 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.

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3.2 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 8×8×16 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.

12

3.3 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 SDS = 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 402-13

Section 7.3.2.6. 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

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 7.3.2.6.1.1

of TMS 402-13, where the design strength, φVn, needs to exceed the smaller of: (a) the shear

associated with 1.25 Mn, and (b) 2.0 Vu. In general, the cap of 2.0 Vu 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 402-13 Section 7.3.2.6.

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,

Mu/φMn and Vu/φVn, 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

402-13 Section 7.3.2.6.1.1 where Vn need not exceed 2.5Vu.

All wall segments complied with the requirements of TMS 402-13 Section 9.3.6.5 to

avoid the need of special boundary elements. All walls (A, B, and C) satisfied the rapid screening

method (Section 9.3.6.5.1) but none of the walls satisfied the provisions related to the

maximum extreme fiber compressive stress (Section 9.3.6.5.4) or the maximum area of flexural

tensile reinforcement (Section 9.3.3.5).

14

3.4 Limit Design

Spreadsheet formulations for Design Example 1 are presented in Appendix B (Section

B.1) incorporating the Limit Design provisions of TMS 402-13 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 402-13 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 SDS = 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 402-13 Section 7.3.2.6). The minimum size of reinforcement is set to #3

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 402-13 (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 402-13 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 – 0.2SDS in accordance with TMS 402-13 Section C.1 item (c).

For a wall segment dominated by seismic forces or with negligible gravity loading, the

adjustment factor (φvo) may be determined using φvo = Vn/2VMn ≤ 1.0, where VMn is the shear

associated with the flexural strength, Mn, and Vn is the calculated shear strength. Values of φvo

less than one identify the shear controlled wall segments. It is recommended to limit φvo to

values greater than 0.5 in order to develop the flexural strength, Mn, before reaching the shear

strength, Vn.

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

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 (φvo) 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 (fy), as

described in Table 3.5 and illustrated in Figures 3.9 and 3.10. 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,

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 3.12. 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 402-13 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 3.14. 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

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 402-13 code and the second design was

based on the Limit Design provisions in TMS 402-13 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 402-13. 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.

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

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 7.3.2.6 of TMS 402-13. Trial designs, using

Strength Design and Limit Design provisions in TMS 402-13, led to the reinforcement sizes

presented in Tables 4.2 and 4.3. Details of the supporting calculations are presented below.

21

4.2 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 6×4×12 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).

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

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 short-

period spectral response acceleration parameter of SDS = 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

402-13 Section 7.3.2.6) 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 5.2.2.4 of

TMS 402-13) 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

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 402-13

Sections 5.2.2.3 and 5.2.2.4).

Shear design of the wall segments was controlled by the provisions in Section 7.3.2.6.1.1

of TMS 402-13 where the design strength, φVn, needs to exceed the smaller of: (a) the shear

associated with 1.25 Mn, and (b) 2.0 Vu. With the exception of the coupling beams, the cap of

2.0 Vu 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, Ω0, according to Section 12.3.3.3 of ASCE/SEI 7-10.

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 (Mu/φMn or Vu/φVn) were wall B (Vu/φVn =

0.97, between levels 2 and 3) and the coupling beams (Mu/φMn = 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 1.25Mn.

All wall segments comply with the requirements of TMS 402-13 Section 9.3.6.5 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 9.3.6.5.1) and the maximum

extreme fiber compressive stress (Section 9.3.6.5.4).

25

4.4 Limit Design

The spreadsheet formulations presented in Appendix B (Section B.2) incorporate the

Limit Design provisions of TMS 402-13 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 402-13 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

SDS = 1.0, which leads to a base shear of 86.6 kip at the seismic base (see Table 4.1).

26

The chosen reinforcement layout (see Figure 4.3) is based on the maximum allowed

spacing (TMS 402-13 Section 7.3.2.6). 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 5.2.2.4 of

TMS 402-13) 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 402-13

Sections 5.2.2.3 and 5.2.2.4).

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 402-13 (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 402-13 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 – 0.2SDS in accordance with TMS 402-13 Section C.1 item (c).

For a wall segment dominated by seismic forces or with negligible gravity loading, the

adjustment factor (φvo) may be determined using φvo = Vn/2VMn ≤ 1.0, where VMn is the shear

27

associated with the flexural strength, Mn, and Vn is the calculated shear strength. Values of φvo

less than one identify the shear controlled wall segments. It is recommended to limit φvo to

values greater than 0.5 in order to develop flexural strength, Mn, before reaching the shear

strength, Vn.

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 φvo = 1.50/2 = 0.75 for walls A and C, and φvo = 1.68/2 = 0.84 for wall B, see TMS

402-13 Section C.1 item (d). The deformation capacities of the yielding wall segments were also

limited, see TMS 402-13 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.

28

4.5 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 (fy), 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 4.10. 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 402-13 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

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 4.13.

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 4.5𝑓′𝑚 (psi). This value is nearly 1.5 times greater

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

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 402-

13 code and the second design was based on the Limit Design method in TMS 402-13 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 402-13 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.

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

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 earthquake-

resistant masonry.

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.

34

A.1 Strength Design Calculations for Design Example 1

Design Example 1 Geometry

Concrete masonry units, 16×8×8 in., fully grouted

Reinforcement

(Assumed sufficient for all non-seismic load combinations)

Wall A Flexural Reinforcement As,1 = (1) #5 = 0.31 in.2 @ d1 = 4 in.

As,2 = (1) #5 = 0.31 in.2 @ d2 = 20 in. As,3 = (1) #5 = 0.31 in.2 @ d3 = 28 in. As,4 = (1) #5 = 0.31 in.2 @ d4 = 44 in.

Wall A Shear Reinforcement Av = (1) #3 = 0.11 in.2 @ 16 in. o.c.

(Av/s) = 0.0069 in.2/in.

Wall B Flexural Reinforcement As,1 = (1) #4 = 0.2 in.2 @ d1 = 4 in.

As,2 = (1) #4 = 0.2 in.2 @ d2 = 12 in. As,3 = (1) #4 = 0.2 in.2 @ d3 = 20 in.

Wall B Shear Reinforcement Av = (1) #3 = 0.11 in.2 @ 8 in. o.c.

(Av/s) = 0.0069 in.2/in.

35

Wall C Flexural Reinforcement As,1 = (1) #6 = 0.44 in.2 @ d1 = 4 in.

As,2 = (1) #6 = 0.44 in.2 @ d2 = 20 in. As,3 = (1) #6 = 0.44 in.2 @ d3 = 28 in. As,4 = (1) #6 = 0.44 in.2 @ d4 = 44 in.

Wall C Shear Reinforcement Av = (2) #4 = 0.4 in.2 @ 16 in. o.c.

(Av/s) = 0.0250 in.2/in.

Material Properties f'm = 1500 psi

fy = 60 ksi Es = 29000 ksi Em = 1350 ksi

(TMS 402-13 §4.2.2) εmu = 0.0025 in./in.

(TMS 402-13 §9.3.2(c))

εsy = 0.0021 in./in.

Seismic Design Parameters SDS = 1

R = 5

(ASCE/SEI 7-10 Table 12.2-1) Cd = 3.5

Ω0 = 2.5 Ev = 0.2 SDS D

(ASCE/SEI 7-10 Eq. 12.14-6)

ρ = 1.0

Modeling Assumptions

(Section properties based on 50% of gross section properties) E = Em/2 = 675 ksi

G = 270 ksi ν = 0.25

Poisson's ratio

Gravity Loading

(Determined from linear-elastic analysis using SAP2000)

DSelf Weight = 80 psf DTributary = 150 plf LTributary = 225 plf

Wall A PD(Top) = 7.7 kip

PD(Bot) = 10.9 kip VD = 0.7 kip MD(Top) = 4.1 kip-ft MD(Bot) = 3.0 kip-ft

36

PL = 2.2 kip VL = 0.2 kip ML(Top) = 1.1 kip-ft ML(Bot) = 0.8 kip-ft

Wall B PD(Top) = 5.6 kip

PD(Bot) = 6.9 kip VD = 0.0 kip MD(Top) = 0.1 kip-ft MD(Bot) = 0.0 kip-ft

PL = 1.5 kip VL = 0.0 kip ML(Top) = 0.0 kip-ft ML(Bot) = 0.0 kip-ft

Wall C PD(Top) = 7.2 kip

PD(Bot) = 9.8 kip VD = 0.7 kip MD(Top) = 3.5 kip-ft MD(Bot) = 2.2 kip-ft

PL = 1.8 kip VL = 0.2 kip ML(Top) = 1.0 kip-ft ML(Bot) = 0.6 kip-ft

Seismic Forces and Displacement

(Determined from linear-elastic analysis using SAP 2000)

Base Shear, Vb = 41 kip

(Demand on one line of resistance) Roof Displacement, δR = 0.115 in.

Wall A

PE = 19.7 kip VE = 15.2 kip ME(Top) = 58.7 kip-ft ME(Bot) = 93.1 kip-ft

37

Wall B PE = 5.3 kip

VE = 4.6 kip ME(Top) = 15.5 kip-ft ME(Bot) = 21.3 kip-ft

Wall C PE = 25.0 kip

VE = 21.2 kip ME(Top) = 59.6 kip-ft ME(Bot) = 110.2 kip-ft

Design Forces for Load Combination 1.2D + 0.5L + 1.0E

(Including Ev) Wall A

Pu(Top) = 31.6 kip Pu(Bot) = 36.1 kip Vu = 16.3 kip Mu(Top) = 65.0 kip-ft Mu(Bot) = 97.7 kip-ft

Wall B Pu(Top) = 13.9 kip

Pu(Bot) = 15.7 kip Vu = 4.6 kip Mu(Top) = 15.7 kip-ft Mu(Bot) = 21.3 kip-ft

Wall C Pu(Top) = 36.0 kip

Pu(Bot) = 39.6 kip Vu = 22.3 kip Mu(Top) = 65.0 kip-ft Mu(Bot) = 113.6 kip-ft

Design Forces for Load Combination 0.9D + 1.0E

(Including Ev) Wall A

Pu(Top) = -14.3 kip

(Pu < 0, tension) Pu(Bot) = -12.1 kip

Vu = 15.7 kip Mu(Top) = 61.6 kip-ft Mu(Bot) = 95.2 kip-ft

38

Wall B Pu(Top) = -1.4 kip

(Pu < 0, tension)

Pu(Bot) = -0.5 kip Vu = 4.6 kip Mu(Top) = 15.6 kip-ft Mu(Bot) = 21.3 kip-ft

Wall C Pu(Top) = -20.0 kip

(Pu < 0, tension)

Pu(Bot) = -18.2 kip Vu = 21.7 kip Mu(Top) = 62.1 kip-ft Mu(Bot) = 111.7 kip-ft

39

Combined Axial and Flexure Design Top of Wall A

Pu = 31.6 kip

(1.2D+0.5L+E, incl. Ev) Mu = 65.0 kip-ft

φMn = 155.0 kip-ft > Mu

OK

Pu = -14.3 kip

(0.9D+E, incl. Ev) Mu = 61.6 kip-ft

φMn = 93.2 kip-ft > Mu

OK

Bottom of Wall A Pu = 36.1 kip

(1.2D+0.5L+E, incl. Ev)

Mu = 97.7 kip-ft φMn = 159.2 kip-ft > Mu

OK

Pu = -12.1 kip

(0.9D+E, incl. Ev) Mu = 95.2 kip-ft

φMn = 96.8 kip-ft > Mu

OK

-100

-50

0

50

100

150

200

250

300

350

400

0 50 100 150 200 250 300

Axia

l For

ce (k

ip)

Moment (kip-ft)

Wall A

1.2D+0.5L+E

0.9D-E

φ = 1.0 φ = 0.9

40

Combined Axial and Flexure Design (cont.) Top of Wall B

Pu = 13.9 kip

(1.2D+0.5L+E, incl. Ev) Mu = 15.7 kip-ft

φMn = 36.3 kip-ft > Mu

OK

Pu = -1.4 kip

(0.9D+E, incl. Ev) Mu = 15.6 kip-ft

φMn = 25.6 kip-ft > Mu

OK

Bottom of Wall B Pu = 15.7 kip

(1.2D+0.5L+E, incl. Ev)

Mu = 21.3 kip-ft φMn = 37.5 kip-ft > Mu

OK

Pu = -0.5 kip

(0.9D+E, incl. Ev) Mu = 21.3 kip-ft

φMn = 26.3 kip-ft > Mu

OK

-50

0

50

100

150

200

0 10 20 30 40 50 60 70

Axia

l For

ce (k

ip)

Moment (kip-ft)

Wall B

1.2D+0.5L+E

0.9D-E

φ = 1.0 φ = 0.9

41

Combined Axial and Flexure Design (cont.) Top of Wall C

Pu = 36.0 kip

(1.2D+0.5L+E, incl. Ev) Mu = 65.0 kip-ft

φMn = 187.8 kip-ft > Mu

OK

Pu = -20.0 kip

(0.9D+E, incl. Ev) Mu = 62.1 kip-ft

φMn = 128.9 kip-ft > Mu

OK

Bottom of Wall C Pu = 39.6 kip

(1.2D+0.5L+E, incl. Ev)

Mu = 113.6 kip-ft φMn = 190.8 kip-ft > Mu

OK

Pu = -18.2 kip

(0.9D+E, incl. Ev) Mu = 111.7 kip-ft

φMn = 131.5 kip-ft > Mu

OK

-200

-100

0

100

200

300

400

500

0 50 100 150 200 250 300

Axia

l For

ce (k

ip)

Moment (kip-ft)

Wall C

1.2D+0.5L+E

0.9D-E

φ = 1.0 φ = 0.9

42

Shear Design Top of Wall A

(1.2D+0.5L+E, incl. Ev) tWall = 7.625 in.

dv = 48 in.

Pu = 31.6 kip Vu = 16.3 kip Mu = 65.0 kip-ft

Mu/Vudv = 1.00

Vn = Vnm + Vns ≤ Vn,Max

(TMS 402-13 §9.3.4.1.2)

Vnm = [4-1.75(Mu/Vudv)]An√(f'm) + 0.25Pu

(TMS 402-13 Eq. 9-24) Vnm = 39.8 kip

Vns = 0.5(Av/s)fydv

(TMS 402-13 Eq. 9-25)

(Av/s) = 0.0069 in.2/in. Vns = 9.9 kip

Vn,Max = 4.0 An√(f'm)

(TMS 402-13 Eq. 9-22 and Eq. 9-23) Vn,Max = 56.8 kip

Vn = 49.7 kip

(TMS 402-13 §9.3.4.1.2)

φ = 0.8 φVn = 39.8 kip > 16.3 kip OK

Check φVn ≥ min( 2.0VE+VuG, 1.25VMn+VuG)

(TMS 402-13 §7.3.2.6.1.1)

VE = 15.2 kip

(Eh) ME = 58.7 kip-ft

(Eh)

VuG = Vu - VE = 1.1 kip

(1.2D+0.5L+E, incl. Ev)

Mn = 168.9 kip-ft

(Mn for P = Pu, φ = 1.0) VMn = MnVE/ME = 43.7 kip

2.0VE+VuG = 31.4 kip

1.25VMn+VuG = 55.7 kip φVn = 39.8 kip > 31.4 kip OK

43

Shear Design (cont.) Top of Wall A

(0.9D+E, incl. Ev) tWall = 7.625 in.

dv = 48 in.

Pu = -14.3 kip Vu = 15.7 kip Mu = 61.6 kip-ft

Mu/Vudv = 0.98

Vn = Vnm + Vns ≤ Vn,Max

(TMS 402-13 9.3.4.1.2)

Vnm = [4-1.75(Mu/Vudv)]An√(f'm) + 0.25Pu

(TMS 402-13 Eq. 9-24) Vnm = 28.8 kip

Vns = 0.5(Av/s)fydv

(TMS 402-13 Eq. 9-25)

(Av/s) = 0.0069 in.2/in. Vns = 9.9 kip

Vn,Max = 4.0 An√(f'm)

(TMS 402-13 Eq. 9-22 and Eq. 9-23) Vn,Max = 57.4 kip

Vn = 38.7 kip

(TMS 402-13 §9.3.4.1.2)

φ = 0.8 φVn = 30.9 kip > 15.7 kip OK

Check φVn ≥ min( 2.0VE+VuG, 1.25VMn+VuG)

(TMS 402-13 §7.3.2.6.1.1)

VE = 15.2 kip

(Eh) ME = 58.7 kip-ft

(Eh)

VuG = Vu - VE = 0.5 kip

(0.9D+E, incl. Ev)

Mn = 106.1 kip-ft

(Mn for P = Pu, φ = 1.0) VMn = MnVE/ME = 27.4 kip

2.0VE+VuG = 30.9 kip

1.25VMn+VuG = 34.8 kip φVn = 30.9 kip > 30.9 kip OK

44

Shear Design (cont.) Bottom of Wall A

(1.2D+0.5L+E, incl. Ev) tWall = 7.625 in.

dv = 48 in.

Pu = 36.1 kip Vu = 16.3 kip Mu = 97.7 kip-ft

Mu/Vudv = 1.50

Vn = Vnm + Vns ≤ Vn,Max

(TMS 402-13 §9.3.4.1.2)

Vnm = [4-1.75(Mu/Vudv)]An√(f'm) + 0.25Pu

(TMS 402-13 Eq. 9-24) Vnm = 40.9 kip

Vns = 0.5(Av/s)fydv

(TMS 402-13 Eq. 9-25)

(Av/s) = 0.0069 in.2/in. Vns = 9.9 kip

Vn,Max = 4.0 An√(f'm)

(TMS 402-13 Eq. 9-22 and Eq. 9-23) Vn,Max = 56.7 kip

Vn = 50.8 kip

(TMS 402-13 §9.3.4.1.2)

φ = 0.8 φVn = 40.7 kip > 16.3 kip OK

Check φVn ≥ min( 2.0VE+VuG, 1.25VMn+VuG)

(TMS 402-13 §7.3.2.6.1.1)

VE = 15.2 kip

(Eh) ME = 93.1 kip-ft

(Eh)

VuG = Vu - VE = 1.1 kip

(1.2D+0.5L+E, incl. Ev)

Mn = 173.1 kip-ft

(Mn for P = Pu, φ = 1.0) VMn = MnVE/ME = 28.2 kip

2.0VE+VuG = 31.4 kip

1.25VMn+VuG = 36.4 kip φVn = 40.7 kip > 31.4 kip OK

45

Shear Design (cont.) Bottom of Wall A

(0.9D+E, incl. Ev) tWall = 7.625 in.

dv = 48 in.

Pu = -12.1 kip Vu = 15.7 kip Mu = 95.2 kip-ft

Mu/Vudv = 1.52

Vn = Vnm + Vns ≤ Vn,Max

(TMS 402-13 §9.3.4.1.2)

Vnm = [4-1.75(Mu/Vudv)]An√(f'm) + 0.25Pu

(TMS 402-13 Eq. 9-24) Vnm = 28.9 kip

Vns = 0.5(Av/s)fydv

(TMS 402-13 Eq. 9-25)

(Av/s) = 0.0069 in.2/in. Vns = 9.9 kip

Vn,Max = 4.0 An√(f'm)

(TMS 402-13 Eq. 9-22 and Eq. 9-23) Vn,Max = 56.7 kip

Vn = 38.8 kip

(TMS 402-13 §9.3.4.1.2)

φ = 0.8 φVn = 31.0 kip > 15.7 kip OK

Check φVn ≥ min( 2.0VE+VuG, 1.25VMn+VuG)

(TMS 402-13 §7.3.2.6.1.1)

VE = 15.2 kip

(Eh) ME = 93.1 kip-ft

(Eh)

VuG = Vu - VE = 0.5 kip

(0.9D+E, incl. Ev)

Mn = 109.7 kip-ft

(Mn for P = Pu, φ = 1.0) VMn = MnVE/ME = 17.9 kip

2.0VE+VuG = 30.9 kip

1.25VMn+VuG = 22.9 kip φVn = 31.0 kip > 22.9 kip OK

46

Shear Design (cont.) Top of Wall B

(1.2D+0.5L+E, incl. Ev) tWall = 7.625 in.

dv = 24 in.

Pu = 13.9 kip Vu = 4.6 kip Mu = 15.7 kip-ft

Mu/Vudv = 1.69

Vn = Vnm + Vns ≤ Vn,Max

(TMS 402-13 9.3.4.1.2)

Vnm = [4-1.75(Mu/Vudv)]An√(f'm) + 0.25Pu

(TMS 402-13 Eq. 9-24) Vnm = 19.4 kip

Vns = 0.5(Av/s)fydv

(TMS 402-13 Eq. 9-25)

(Av/s) = 0.014 in.2/in. Vns = 9.9 kip

Vn,Max = 4.0 An√(f'm)

(TMS 402-13 Eq. 9-22 and Eq. 9-23) Vn,Max = 28.4 kip

Vn = 28.4 kip

(TMS 402-13 §9.3.4.1.2)

φ = 0.8 φVn = 22.7 kip > 4.6 kip OK

Check φVn ≥ min( 2.0VE+VuG, 1.25VMn+VuG)

(TMS 402-13 §7.3.2.6.1.1)

VE = 4.6 kip

(Eh) ME = 15.5 kip-ft

(Eh)

VuG = Vu - VE = 0.0 kip

(1.2D+0.5L+E, incl. Ev)

Mn = 39.4 kip-ft

(Mn for P = Pu, φ = 1.0) VMn = MnVE/ME = 11.7 kip

2.0VE+VuG = 9.2 kip

1.25VMn+VuG = 14.6 kip φVn = 22.7 kip > 9.2 kip OK

47

Shear Design (cont.) Top of Wall B

(0.9D+E, incl. Ev) tWall = 7.625 in.

dv = 24 in.

Pu = -1.4 kip Vu = 4.6 kip Mu = 15.6 kip-ft

Mu/Vudv = 1.69

Vn = Vnm + Vns ≤ Vn,Max

(TMS 402-13 §9.3.4.1.2)

Vnm = [4-1.75(Mu/Vudv)]An√(f'm) + 0.25Pu

(TMS 402-13 Eq. 9-24) Vnm = 15.6 kip

Vns = 0.5(Av/s)fydv

(TMS 402-13 Eq. 9-25)

(Av/s) = 0.014 in.2/in. Vns = 9.9 kip

Vn,Max = 4.0 An√(f'm)

(TMS 402-13 Eq. 9-22 and Eq. 9-23) Vn,Max = 28.4 kip

Vn = 25.5 kip

(TMS 402-13 §9.3.4.1.2)

φ = 0.8 φVn = 20.4 kip > 4.6 kip OK

Check φVn ≥ min( 2.0VE+VuG, 1.25VMn+VuG)

(TMS 402-13 §7.3.2.6.1.1)

VE = 4.6 kip

(Eh) ME = 15.5 kip-ft

(Eh)

VuG = Vu - VE = 0.0 kip

(0.9D+E, incl. Ev)

Mn = 28.6 kip-ft

(Mn for P = Pu, φ = 1.0) VMn = MnVE/ME = 8.5 kip

2.0VE+VuG = 9.2 kip

1.25VMn+VuG = 10.6 kip φVn = 20.4 kip > 9.2 kip OK

48

Shear Design (cont.) Bottom of Wall B

(1.2D+0.5L+E, incl. Ev) tWall = 7.625 in.

dv = 24 in.

Pu = 15.7 kip Vu = 4.6 kip Mu = 21.3 kip-ft

Mu/Vudv = 2.31

Vn = Vnm + Vns ≤ Vn,Max

(TMS 402-13 §9.3.4.1.2)

Vnm = [4-1.75(Mu/Vudv)]An√(f'm) + 0.25Pu

(TMS 402-13 Eq. 9-24) Vnm = 19.9 kip

Vns = 0.5(Av/s)fydv

(TMS 402-13 Eq. 9-25)

(Av/s) = 0.014 in.2/in. Vns = 9.9 kip

Vn,Max = 4.0 An√(f'm)

(TMS 402-13 Eq. 9-22 and Eq. 9-23) Vn,Max = 28.4 kip

Vn = 28.4 kip

(TMS 402-13 §9.3.4.1.2)

φ = 0.8 φVn = 22.7 kip > 4.6 kip OK

Check φVn ≥ min( 2.0VE+VuG, 1.25VMn+VuG)

(TMS 402-13 §7.3.2.6.1.1)

VE = 4.6 kip

(Eh) ME = 21.3 kip-ft

(Eh)

VuG = Vu - VE = 0.0 kip

(1.2D+0.5L+E, incl. Ev)

Mn = 40.5 kip-ft

(Mn for P = Pu, φ = 1.0) VMn = MnVE/ME = 8.7 kip

2.0VE+VuG = 9.2 kip

1.25VMn+VuG = 10.9 kip φVn = 22.7 kip > 9.2 kip OK

49

Shear Design (cont.) Bottom of Wall B

(0.9D+E, incl. Ev) tWall = 7.625 in.

dv = 24 in.

Pu = -0.5 kip Vu = 4.6 kip Mu = 21.3 kip-ft

Mu/Vudv = 2.31

Vn = Vnm + Vns ≤ Vn,Max

(TMS 402-13 §9.3.4.1.2)

Vnm = [4-1.75(Mu/Vudv)]An√(f'm) + 0.25Pu

(TMS 402-13 Eq. 9-24) Vnm = 15.8 kip

Vns = 0.5(Av/s)fydv

(TMS 402-13 Eq. 9-25)

(Av/s) = 0.014 in.2/in. Vns = 9.9 kip

Vn,Max = 4.0 An√(f'm)

(TMS 402-13 Eq. 9-22 and Eq. 9-23) Vn,Max = 28.4 kip

Vn = 25.7 kip

(TMS 402-13 §9.3.4.1.2)

φ = 0.8 φVn = 20.6 kip > 4.6 kip OK

Check φVn ≥ min( 2.0VE+VuG, 1.25VMn+VuG)

(TMS 402-13 §7.3.2.6.1.1)

VE = 4.6 kip

(Eh) ME = 21.3 kip-ft

(Eh)

VuG = Vu - VE = 0.0 kip

(0.9D+E, incl. Ev)

Mn = 29.2 kip-ft

(Mn for P = Pu, φ = 1.0) VMn = MnVE/ME = 6.3 kip

2.0VE+VuG = 9.2 kip

1.25VMn+VuG = 7.9 kip φVn = 20.6 kip > 7.9 kip OK

50

Shear Design (cont.) Top of Wall C

(1.2D+0.5L+E, incl. Ev) tWall = 7.625 in.

dv = 48 in.

Pu = 36.0 kip Vu = 22.3 kip Mu = 65.0 kip-ft

Mu/Vudv = 0.73

Vn = Vnm + Vns ≤ Vn,Max

(TMS 402-13 §9.3.4.1.2)

Vnm = [4-1.75(Mu/Vudv)]An√(f'm) + 0.25Pu

(TMS 402-13 Eq. 9-24) Vnm = 47.6 kip

Vns = 0.5(Av/s)fydv

(TMS 402-13 Eq. 9-25)

(Av/s) = 0.0250 in.2/in. Vns = 36.0 kip

Vn,Max = 4.7 An√(f'm)

(TMS 402-13 Eq. 9-22 and Eq. 9-23) Vn,Max = 67.0 kip

Vn = 67.0 kip

(TMS 402-13 §9.3.4.1.2)

φ = 0.8 φVn = 53.6 kip > 22.3 kip OK

Check φVn ≥ min( 2.0VE+VuG, 1.25VMn+VuG)

(TMS 402-13 §7.3.2.6.1.1)

VE = 21.2 kip

(Eh) ME = 59.6 kip-ft

(Eh)

VuG = Vu - VE = 1.1 kip

(1.2D+0.5L+E, incl. Ev)

Mn = 205.3 kip-ft

(Mn for P = Pu, φ = 1.0) VMn = MnVE/ME = 73.1 kip

2.0VE+VuG = 43.5 kip

1.25VMn+VuG = 92.5 kip φVn = 53.6 kip > 43.5 kip OK

51

Shear Design (cont.) Top of Wall C

(0.9D+E, incl. Ev) tWall = 7.625 in.

dv = 48 in.

Pu = -20.0 kip Vu = 21.7 kip Mu = 62.1 kip-ft

Mu/Vudv = 0.71

Vn = Vnm + Vns ≤ Vn,Max

(TMS 402-13 §9.3.4.1.2)

Vnm = [4-1.75(Mu/Vudv)]An√(f'm) + 0.25Pu

(TMS 402-13 Eq. 9-24) Vnm = 34.0 kip

Vns = 0.5(Av/s)fydv

(TMS 402-13 Eq. 9-25)

(Av/s) = 0.0250 in.2/in. Vns = 36.0 kip

Vn,Max = 4.8 An√(f'm)

(TMS 402-13 Eq. 9-22 and Eq. 9-23) Vn,Max = 67.5 kip

Vn = 67.5 kip

(TMS 402-13 §9.3.4.1.2)

φ = 0.8 φVn = 54.0 kip > 21.7 kip OK

Check φVn ≥ min( 2.0VE+VuG, 1.25VMn+VuG)

(TMS 402-13 §7.3.2.6.1.1)

VE = 21.2 kip

(Eh) ME = 59.6 kip-ft

(Eh)

VuG = Vu - VE = 0.5 kip

(0.9D+E, incl. Ev)

Mn = 146.5 kip-ft

(Mn for P = Pu, φ = 1.0) VMn = MnVE/ME = 52.2 kip

2.0VE+VuG = 42.9 kip

1.25VMn+VuG = 65.7 kip φVn = 54.0 kip > 42.9 kip OK

52

Shear Design (cont.) Bottom of Wall C

(1.2D+0.5L+E, incl. Ev) tWall = 7.625 in.

dv = 48 in.

Pu = 39.6 kip Vu = 22.3 kip Mu = 113.6 kip-ft

Mu/Vudv = 1.27

Vn = Vnm + Vns ≤ Vn,Max

(TMS 402-13 §9.3.4.1.2)

Vnm = [4-1.75(Mu/Vudv)]An√(f'm) + 0.25Pu

(TMS 402-13 Eq. 9-24) Vnm = 41.8 kip

Vns = 0.5(Av/s)fydv

(TMS 402-13 Eq. 9-25)

(Av/s) = 0.0250 in.2/in. Vns = 36.0 kip

Vn,Max = 4.0 An√(f'm)

(TMS 402-13 Eq. 9-22 and Eq. 9-23) Vn,Max = 56.7 kip

Vn = 56.7 kip

(TMS 402-13 §9.3.4.1.2)

φ = 0.8 φVn = 45.4 kip > 22.3 kip OK

Check φVn ≥ min( 2.0VE+VuG, 1.25VMn+VuG)

(TMS 402-13 §7.3.2.6.1.1)

VE = 21.2 kip

(Eh) ME = 110.2 kip-ft

(Eh)

VuG = Vu - VE = 1.1 kip

(1.2D+0.5L+E, incl. Ev)

Mn = 208.4 kip-ft

(Mn for P = Pu, φ = 1.0) VMn = MnVE/ME = 40.1 kip

2.0VE+VuG = 43.5 kip

1.25VMn+VuG = 51.3 kip φVn = 45.4 kip > 43.5 kip OK

53

Shear Design (cont.) Bottom of Wall C

(0.9D+E, incl. Ev) tWall = 7.625 in.

dv = 48 in.

Pu = -18.2 kip Vu = 21.7 kip Mu = 111.7 kip-ft

Mu/Vudv = 1.29

Vn = Vnm + Vns ≤ Vn,Max

(TMS 402-13 §9.3.4.1.2)

Vnm = [4-1.75(Mu/Vudv)]An√(f'm) + 0.25Pu

(TMS 402-13 Eq. 9-24) Vnm = 27.4 kip

Vns = 0.5(Av/s)fydv

(TMS 402-13 Eq. 9-25)

(Av/s) = 0.0250 in.2/in. Vns = 36.0 kip

Vn,Max = 4.0 An√(f'm)

(TMS 402-13 Eq. 9-22 and Eq. 9-23) Vn,Max = 56.7 kip

Vn = 56.7 kip

(TMS 402-13 §9.3.4.1.2)

φ = 0.8 φVn = 45.4 kip > 21.7 kip OK

Check φVn ≥ min( 2.0VE+VuG, 1.25VMn+VuG)

(TMS 402-13 §7.3.2.6.1.1)

VE = 21.2 kip

(Eh) ME = 110.2 kip-ft

(Eh)

VuG = Vu - VE = 0.5 kip

(0.9D+E, incl. Ev)

Mn = 149.1 kip-ft

(Mn for P = Pu, φ = 1.0) VMn = MnVE/ME = 28.7 kip

2.0VE+VuG = 42.9 kip

1.25VMn+VuG = 36.4 kip φVn = 45.4 kip > 36.4 kip OK

54

Boundary Element Compliance

(TMS 402-13 §9.3.6.5) Wall A

Rapid Screening

(TMS 402-13 §9.3.6.5.1)

Vu = 16.3 kip

(1.2D+0.5L+E, incl. Ev) Mu = 97.7 kip-ft

dv = 48 in. Mu/(Vudv) = 1.50

Pu = 36.1 kip

0.10Agf'm = 54.9 kip Conditions 1 and 2:

NG

3An√(f'm) = 42.5 kip > Vu Condition 3:

OK

Extreme Fiber Compressive Stress

(TMS 402-13 §9.3.6.5.4)

fmax = Pu/Ag+Mu/S ≤ 0.2f'm 0.2f'm = 300.0 psi

Pu = 36.1 kip

(1.2D+0.5L+E, incl. Ev)

Mu = 97.7 kip-ft Ag = 366 in2 S = 2928 in3

fmax = 498.9 psi > 300.0 psi NG

Both Sections, 9.3.6.5.1 and 9.3.6.5.4, are checked for illustration purposes (only one needs to be satisfied)

55

Boundary Element Compliance (cont.)

(TMS 402-13 §9.3.6.5) Wall B

Rapid Screening

(TMS 402-13 §9.3.6.5.1)

Vu = 4.6 kip

(1.2D+0.5L+E, incl. Ev) Mu = 21.3 kip-ft

dv = 24 in. Mu/(Vudv) = 2.31

Pu = 15.7 kip

0.10Agf'm = 27.45 kip Conditions 1 and 2:

NG

3An√(f'm) = 21.3 kip > Vu Condition 3:

OK

Extreme Fiber Compressive Stress

(TMS 402-13 §9.3.6.5.4)

fmax = Pu/Ag+Mu/S ≤ 0.2f'm 0.2f'm = 300.0 psi

Pu = 15.7 kip

(1.2D+0.5L+E, incl. Ev)

Mu = 21.3 kip-ft Ag = 183 in2 S = 732 in3

fmax = 435.6 psi > 300.0 psi NG

Both Sections, 9.3.6.5.1 and 9.3.6.5.4, are checked for illustration purposes (only one needs to be satisfied)

56

Boundary Element Compliance (cont.)

(TMS 402-13 §9.3.6.5) Wall C

Rapid Screening

(TMS 402-13 §9.3.6.5.1)

Vu = 22.3 kip

(1.2D+0.5L+E, incl. Ev) Mu = 113.6 kip-ft

dv = 48 in. Mu/(Vudv) = 1.27

Pu = 39.6 kip

0.10Agf'm = 54.9 kip Conditions 1 and 2:

NG

3An√(f'm) = 42.5 kip > Vu Condition 3:

OK

Extreme Fiber Compressive Stress

(TMS 402-13 §9.3.6.5.4)

fmax = Pu/Ag+Mu/S ≤ 0.2f'm 0.2f'm = 300.0 psi

Pu = 39.6 kip

(1.2D+0.5L+E, incl. Ev)

Mu = 113.6 kip-ft Ag = 366 in2 S = 2928 in3

fmax = 573.5 psi > 300.0 psi NG

Both Sections, 9.3.6.5.1 and 9.3.6.5.4, are checked for illustration purposes (only one needs to be satisfied)

57

Maximum Area of Flexural Tensile Reinforcement (ρmax Check)

(TMS 402-13 §9.3.3.5)

According to §7.3.2.12(d), if §9.3.6.5 is met then there is no need to check §9.3.3.5.

Wall A Pu = 22.9 kip

(D+0.75L+0.525E, not incl. Ev)

Vu = 8.8 kip Mu = 52.5 kip-ft dv = 48 in.

Mu/Vudv = 1.49

εs,Max = 4.00 εsy εs,Max = 8.3E-03 in./in. Pn,Max = 19.0 kip < 22.9 kip NG

Wall B Pu = 10.8 kip

(D+0.75L+0.525E, not incl. Ev)

Vu = 2.4 kip Mu = 11.2 kip-ft dv = 24 in.

Mu/Vudv = 2.31

εs,Max = 4.00 εsy εs,Max = 8.3E-03 in./in. Pn,Max = 10.0 kip < 10.8 kip NG

Wall C Pu = 24.2 kip

(D+0.75L+0.525E, not incl. Ev)

Vu = 12.0 kip Mu = 60.5 kip-ft dv = 48 in.

Mu/Vudv = 1.26

εs,Max = 4.00 εsy εs,Max = 8.3E-03 in./in. Pn,Max = -4.4 kip < 24.2 kip NG

58

A.2 Strength Design Calculations for Design Example 2

Design Example 2 Geometry

1750

OK

0

Clay masonry units, 12×6×4 in., fully grouted

Roof 36 ft

Level 2 Seismic Base

Level 3 9 ft

Level 5 27 ft

Level 4 18 ft

59

Reinforcement

(Assumed sufficient for all non-seismic load combinations)

Wall A = Wall C Longitudinal Reinforcement As,1 = (1) #4 = 0.2 in.2 @ d1 = 3 in.

As,2 = (1) #4 = 0.2 in.2 @ d2 = 21 in. As,3 = (1) #4 = 0.2 in.2 @ d3 = 39 in. As,4 = (1) #4 = 0.2 in.2 @ d4 = 57 in. As,5 = (1) #4 = 0.2 in.2 @ d5 = 63 in. As,6 = (1) #4 = 0.2 in.2 @ d6 = 81 in. As,7 = (1) #4 = 0.2 in.2 @ d7 = 99 in. As,8 = (1) #4 = 0.2 in.2 @ d8 = 117 in.

Wall A = Wall C Shear Reinforcement Av = (1) #3 = 0.11 in.2 @ 20 in. o.c.

(Av/s) = 0.0055 in.2/in.

Wall B Longitudinal Reinforcement As,1 = (1) #4 = 0.2 in.2 @ d1 = 9 in.

As,2 = (1) #4 = 0.2 in.2 @ d2 = 27 in. As,3 = (1) #4 = 0.2 in.2 @ d3 = 45 in. As,4 = (1) #4 = 0.2 in.2 @ d4 = 63 in.

Wall B Shear Reinforcement Av = (1) #3 = 0.11 in2 @ 20 in. o.c.

(Av/s) = 0.0055 in.2/in.

Coupling Beam Longitudinal Reinforcement As,1 = (1) #4 = 0.2 in.2 @ d1 = 3 in.

As,2 = (1) #4 = 0.2 in.2 @ d2 = 19 in. As,3 = (1) #4 = 0.2 in.2 @ d3 = 29 in. As,4 = (1) #4 = 0.2 in.2 @ d4 = 45 in.

Coupling Beam Shear Reinforcement Av = (1) #4 = 0.2 in.2 @ 12 in. o.c.

(Av/s) = 0.0167 in.2/in.

Transfer Girder Longitudinal Reinforcement As,1 = (1) #4 = 0.2 in.2 @ d1 = 9 in.

As,2 = (1) #4 = 0.2 in.2 @ d2 = 25 in. As,3 = (1) #4 = 0.2 in.2 @ d3 = 41 in. As,4 = (1) #4 = 0.2 in.2 @ d4 = 55 in. As,5 = (1) #4 = 0.2 in.2 @ d5 = 71 in. As,6 = (1) #4 = 0.2 in.2 @ d6 = 87 in.

60

Transfer Girder Shear Reinforcement Av = (1) #3 = 0.11 in.2 @ 12 in. o.c.

(Av/s) = 0.0092 in.2/in.

Material Properties

f'm = 2500 psi fy = 60 ksi Es = 29000 ksi Em = 1750 ksi

(TMS 402-13 §4.2.2) εmu = 0.0035 in./in.

(TMS 402-13 §9.3.2(c))

εsy = 0.0021 in./in.

Seismic Design Parameters SDS = 1.0

R = 5.5

(ASCE/SEI 7-10 Table 12.2-1) Cd = 4.0

Ω0 = 2.5 Ev = 0.2 SDS D

(ASCE/SEI 7-10 Eq. 12.14-6)

ρ = 1.0

Modeling Assumptions

(Section properties based on 50% of gross section properties) E = Em/2 = 875 ksi

G = 350 ksi ν = 0.25

Poisson's ratio

Gravity Loading

(Determined from linear-elastic analysis using SAP2000) DSelf Weight = 60 psf

DTributary = 0 plf LTributary = 0 plf

Wall A = Wall C, Story 3 PD(Top) = 26.6 kip

PD(Bot) = 29.6 kip VD = 3.4 kip MD(Top) = 4.3 kip-ft MD(Bot) = 12.5 kip-ft

PL = 0.0 kip VL = 0.0 kip ML(Top) = 0.0 kip-ft ML(Bot) = 0.0 kip-ft

61

Wall B, Story 3 PD(Top) = 8.7 kip

PD(Bot) = 10.5 kip VD = 0.0 kip MD(Top) = 0.0 kip-ft MD(Bot) = 0.0 kip-ft

PL = 0.0 kip VL = 0.0 kip ML(Top) = 0.0 kip-ft ML(Bot) = 0.0 kip-ft

Coupling Beams PD = 0 kip

(Envelope of forces)

VD = 2.5 kip MD = 6.7 kip-ft

PL = 0 kip VL = 0 kip ML = 0 kip-ft

Transfer Girder PD= 0 kip

VD(Left) = 10.0 kip

(At Wall A) VD(Right) = 6.7 kip

(At Wall B)

MD(Left) = 20.7 kip-ft MD(Right) = 24.1 kip-ft

PL = 0 kip VL(Left) = 0 kip VL(Right) = 0 kip ML(Left) = 0 kip-ft ML(Right) = 0 kip-ft

62

Seismic Forces and Displacements

(Determined from linear-elastic analysis using SAP 2000)

Base Shear, Vb = 86.6 kip

(Demand on one line of resistance)

Story Displacements δR = 0.250 in.

δ5 = 0.202 in. δ4 = 0.140 in. δ3 = 0.074 in. δ2 = 0.017 in.

Wall A = Wall C, Story 3 PE = 52.1 kip

VE = 29.0 kip ME(Top) = 37.7 kip-ft ME(Bot) = 182.5 kip-ft

Wall B, Story 3 PE = 0.0 kip

VE = 28.7 kip ME(Top) = 40.3 kip-ft ME(Bot) = 103.0 kip-ft

Coupling Beams PE = 0.0 kip

(Envelope of forces)

VE = 17.2 kip ME = 61.4 kip-ft

Transfer Girder PE = 0.0 kip

VE(Left) = 16.4 kip

(At Wall A) VE(Right) = 16.4 kip

(At Wall B)

ME(Left) = 77.4 kip-ft ME(Right) = 52.4 kip-ft

63

Design Forces for Load Combination 1.2D + 0.5L + 1.0E

(Including Ev) Wall A = Wall C, Story 3

Pu(Top) = 89.3 kip Pu(Bot) = 93.5 kip Vu = 33.7 kip Mu(Top) = 43.7 kip-ft Mu(Bot) = 200.1 kip-ft

Wall B, Story 3 Pu(Top) = 12.2 kip

Pu(Bot) = 14.7 kip Vu = 28.7 kip Mu(Top) = 40.3 kip-ft Mu(Bot) = 103.0 kip-ft

Coupling Beams Pu = 0.0 kip

(Envelope of forces)

Vu = 20.8 kip Mu = 70.7 kip-ft

Transfer Girder Pu = 0.0 kip

Vu(Left) = 30.5 kip

(At Wall A) Vu(Right) = 25.8 kip

(At Wall B)

Mu(Left) = 106.3 kip-ft Mu(Right) = 86.2 kip-ft

Design Forces for Load Combination 0.9D + 1.0E

(Including Ev)

Wall A = Wall C, Story 3 Pu(Top) = -33.6 kip

(Pu < 0, tension)

Pu(Bot) = -31.5 kip Vu = 31.3 kip Mu(Top) = 40.7 kip-ft Mu(Bot) = 191.3 kip-ft

Wall B, Story 3 Pu(Top) = 6.1 kip

Pu(Bot) = 7.3 kip Vu = 28.7 kip Mu(Top) = 40.3 kip-ft Mu(Bot) = 103.0 kip-ft

64

Coupling Beams Pu = 0.0 kip

(Envelope of forces)

Vu = 19.0 kip Mu = 66.0 kip-ft

Transfer Girder Pu = 0.0 kip

Vu(Left) = 23.5 kip

(At Wall A) Vu(Right) = 21.1 kip

(At Wall B)

Mu(Left) = 91.8 kip-ft Mu(Right) = 69.3 kip-ft

65

Combined Axial and Flexural Design Top of Wall A = Wall C, Story 3 Pu = 89.3 kip

(1.2D+0.5L+E, incl. Ev)

Mu = 43.7 kip-ft φMn = 757.5 kip-ft > Mu

OK

Pu = -33.6 kip

(0.9D+E, incl. Ev) Mu = 40.7 kip-ft

φMn = 253.8 kip-ft > Mu

OK

Bottom of Wall A = Wall C, Story 3 Pu = 93.5 kip

(1.2D+0.5L+E, incl. Ev)

Mu = 200.1 kip-ft φMn = 772.8 kip-ft > Mu

OK

Pu = -31.5 kip

(0.9D+E, incl. Ev) Mu = 191.3 kip-ft

φMn = 263.6 kip-ft > Mu

OK

-200

0

200

400

600

800

1000

1200

0 500 1,000 1,500 2,000

Axia

l For

ce (k

ip)

Moment (kip-ft)

Wall A = Wall C, Story 3

1.2D+0.5L+E

0.9D+E

φ = 1.0 φ = 0.9

66

Combined Axial and Flexural Design (cont.) Top of Wall B, Story 3

Pu = 12.2 kip

(1.2D+0.5L+E, incl. Ev) Mu = 40.3 kip-ft

φMn = 152.0 kip-ft > Mu

OK

Pu = 6.1 kip

(0.9D+E, incl. Ev) Mu = 40.3 kip-ft

φMn = 137.0 kip-ft > Mu

OK

Bottom of Wall B, Story 3 Pu = 14.7 kip

(1.2D+0.5L+E, incl. Ev)

Mu = 103.0 kip-ft φMn = 158.2 kip-ft > Mu

OK

Pu = 7.3 kip

(0.9D+E, incl. Ev) Mu = 103.0 kip-ft

φMn = 140.1 kip-ft > Mu

OK

-100

0

100

200

300

400

500

600

700

0 100 200 300 400 500 600 700

Axia

l For

ce (k

ip)

Moment (kip-ft)

Wall B, Story 3

1.2D+0.5L+E

0.9D+E

φ = 1.0 φ = 0.9

67

Combined Axial and Flexural Design (cont.) Coupling Beams

Pu = 0.0 kip

(1.2D+0.5L+E, incl. Ev) Mu = 70.7 kip-ft

φMn = 79.3 kip-ft > Mu

OK

Pu = 0.0 kip

(0.9D+E, incl. Ev)

Mu = 66.0 kip-ft φMn = 79.3 kip-ft > Mu

OK

-100

-50

0

50

100

150

200

250

300

350

400

450

0 50 100 150 200 250 300

Axia

l For

ce (k

ip)

Moment (kip-ft)

Coupling Beams

1.2D+0.5L+E

0.9D+E

φ = 1.0 φ = 0.9

68

Combined Axial and Flexural Design (cont.) Transfer Girder, Left End

(At Wall A)

Pu = 0.0 kip

(1.2D+0.5L+Ω0E, incl. Ev) Mu = 222.3 kip-ft

Ω0 applies to Eh

φMn = 239.9 kip-ft > Mu

OK

Pu = 0.0 kip

(0.9D+Ω0E, incl. Ev) Mu = -178.9 kip-ft

(Mu < 0, tension at bottom)

φMn = 239.9 kip-ft > Mu

OK

Transfer Girder, Right End

(At Wall B) Pu = 0.0 kip

(1.2D+0.5L+Ω0E, incl. Ev)

Mu = 164.8 kip-ft

Ω0 applies to Eh φMn = 239.9 kip-ft > Mu

OK

Pu = 0.0 kip

(0.9D+Ω0E, incl. Ev) Mu = -114.1 kip-ft

(Mu < 0, tension at bottom)

φMn = 239.9 kip-ft > Mu

OK

-200

-100

0

100

200

300

400

500

600

700

800

900

0 200 400 600 800 1,000 1,200

Axia

l For

ce (k

ip)

Moment (kip-ft)

Transfer Girder

1.2D+0.5L+E

0.9D+E

φ = 1.0 φ = 0.9

69

Shear Design Top of Wall A = Wall C, Story 3

(1.2D+0.5L+E, incl. Ev)

tWall = 5.5 in. dv = 120 in.

Pu = 89.3 kip Vu = 33.7 kip Mu = 43.7 kip-ft

Mu/Vudv = 0.13

Vn = Vnm + Vns ≤ Vn,Max

(TMS 402-13 §9.3.4.1.2)

Vnm = [4-1.75(Mu/Vudv)]An√(f'm) + 0.25Pu

(TMS 402-13 Eq. 9-24) Vnm = 146.8 kip

Vns = 0.5(Av/s)fydv

(TMS 402-13 Eq. 9-25)

(Av/s) = 0.0055 in.2/in. Vns = 19.8 kip

Vn,Max = 6.0 An√(f'm)

(TMS 402-13 Eq. 9-22 and Eq. 9-23) Vn,Max = 198.0 kip

Vn = 166.6 kip

(TMS 402-13 §9.3.4.1.2)

φ = 0.8 φVn = 133.3 kip > 33.7 kip OK

Check φVn ≥ min( 2.0VE+VuG, 1.25VMn+VuG)

(TMS 402-13 §7.3.2.6.1.1)

VE = 29.0 kip

(Eh) ME = 37.7 kip-ft

(Eh)

VuG = Vu - VE = 4.7 kip

(1.2D+0.5L+E, incl. Ev)

Mn = 805.2 kip-ft

(Mn for P = Pu, φ = 1.0) VMn = MnVE/ME = 619.1 kip

2.0VE+VuG = 62.7 kip

1.25VMn+VuG = 778.6 kip φVn = 133.3 kip > 62.7 kip OK

70

Shear Design (cont.) Top of Wall A = Wall C, Story 3

(0.9D+E, incl. Ev)

tWall = 5.5 in. dv = 120 in.

Pu = -33.6 kip Vu = 31.3 kip Mu = 40.7 kip-ft

Mu/Vudv = 0.13

Vn = Vnm + Vns ≤ Vn,Max

(TMS 402-13 §9.3.4.1.2)

Vnm = [4-1.75(Mu/Vudv)]An√(f'm) + 0.25Pu

(TMS 402-13 Eq. 9-24) Vnm = 116.1 kip

Vns = 0.5(Av/s)fydv

(TMS 402-13 Eq. 9-25)

(Av/s) = 0.0055 in.2/in. Vns = 19.8 kip

Vn,Max = 6.0 An√(f'm)

(TMS 402-13 Eq. 9-22 and Eq. 9-23) Vn,Max = 198.0 kip

Vn = 135.9 kip

(TMS 402-13 §9.3.4.1.2)

φ = 0.8 φVn = 108.7 kip > 31.3 kip OK

Check φVn ≥ min( 2.0VE+VuG, 1.25VMn+VuG)

(TMS 402-13 §7.3.2.6.1.1)

VE = 29.0 kip

(Eh) ME = 37.7 kip-ft

(Eh)

VuG = Vu - VE = 2.4 kip

(0.9D+E, incl. Ev)

Mn = 299.4 kip-ft

(Mn for P = Pu, φ = 1.0) VMn = MnVE/ME = 230.2 kip

2.0VE+VuG = 60.3 kip

1.25VMn+VuG = 290.2 kip φVn = 108.7 kip > 60.3 kip OK

71

Shear Design (cont.) Bottom of Wall A = Wall C, Story 3

(1.2D+0.5L, incl. Ev)

tWall = 5.5 in. dv = 120 in.

Pu = 93.5 kip Vu = 33.7 kip Mu = 200.1 kip-ft

Mu/Vudv = 0.59

Vn = Vnm + Vns ≤ Vn,Max

(TMS 402-13 §9.3.4.1.2)

Vnm = [4-1.75(Mu/Vudv)]An√(f'm) + 0.25Pu

(TMS 402-13 Eq. 9-24) Vnm = 121.1 kip

Vns = 0.5(Av/s)fydv

(TMS 402-13 Eq. 9-25)

(Av/s) = 0.0055 in.2/in. Vns = 19.8 kip

Vn,Max = 5.1 An√(f'm)

(TMS 402-13 Eq. 9-22 and Eq. 9-23) Vn,Max = 167.7 kip

Vn = 140.9 kip

(TMS 402-13 §9.3.4.1.2)

φ = 0.8 φVn = 112.7 kip > 33.7 kip OK

Check φVn ≥ min( 2.0VE+VuG, 1.25VMn+VuG)

(TMS 402-13 §7.3.2.6.1.1)

VE = 29.0 kip-ft

(Eh) ME = 182.5 kip-ft

(Eh)

VuG = Vu - VE = 4.7 kip

(1.2D+0.5L+E, incl. Ev)

Mn = 820.6 kip-ft

(Mn for P = Pu, φ = 1.0) VMn = MnVE/ME = 130.2 kip

2.0VE+VuG = 62.7 kip

1.25VMn+VuG = 167.5 kip φVn = 112.7 kip > 62.7 kip OK

72

Shear Design (cont.) Bottom of Wall A = Wall C, Story 3

(0.9D+E, incl. Ev)

tWall = 5.5 in. dv = 120 in.

Pu = -31.5 kip Vu = 31.3 kip Mu = 191.3 kip-ft

Mu/Vudv = 0.61

Vn = Vnm + Vns ≤ Vn,Max

(TMS 402-13 §9.3.4.1.2)

Vnm = [4-1.75(Mu/Vudv)]An√(f'm) + 0.25Pu

(TMS 402-13 Eq. 9-24) Vnm = 88.9 kip

Vns = 0.5(Av/s)fydv

(TMS 402-13 Eq. 9-25)

(Av/s) = 0.0055 in.2/in. Vns = 19.8 kip

Vn,Max = 5.0 An√(f'm)

(TMS 402-13 Eq. 9-22 and Eq. 9-23) Vn,Max = 166.3 kip

Vn = 108.7 kip

(TMS 402-13 §9.3.4.1.2)

φ = 0.8 φVn = 86.9 kip > 31.3 kip OK

Check φVn ≥ min( 2.0VE+VuG, 1.25VMn+VuG)

(TMS 402-13 §7.3.2.6.1.1)

VE = 29.0 kip-ft

(Eh) ME = 182.5 kip-ft

(Eh)

VuG = Vu - VE = 2.4 kip

(0.9D+E, incl. Ev)

Mn = 309.2 kip-ft

(Mn for P = Pu, φ = 1.0) VMn = MnVE/ME = 49.1 kip

2.0VE+VuG = 60.3 kip

1.25VMn+VuG = 63.7 kip φVn = 86.9 kip > 60.3 kip OK

73

Shear Design (cont.) Top of Wall B, Story 3

(1.2D+0.5L+E, incl. Ev) tWall = 5.5 in.

dv = 72 in.

Pu = 12.2 kip Vu = 28.7 kip Mu = 40.3 kip-ft

Mu/Vudv = 0.23

Vn = Vnm + Vns ≤ Vn,Max

(TMS 402-13 §9.3.4.1.2)

Vnm = [4-1.75(Mu/Vudv)]An√(f'm) + 0.25Pu

(TMS 402-13 Eq. 9-24) Vnm = 74.1 kip

Vns = 0.5(Av/s)fydv

(TMS 402-13 Eq. 9-25)

(Av/s) = 0.0055 in.2/in. Vns = 11.9 kip

Vn,Max = 6.0 An√(f'm)

(TMS 402-13 Eq. 9-22 and Eq. 9-23) Vn,Max = 118.8 kip

Vn = 86.0 kip

(TMS 402-13 §9.3.4.1.2)

φ = 0.8 φVn = 68.8 kip > 28.7 kip OK

Check φVn ≥ min( 2.0VE+VuG, 1.25VMn+VuG)

(TMS 402-13 §7.3.2.6.1.1)

VE = 28.7 kip

(Eh) ME = 40.3 kip-ft

(Eh)

VuG = Vu - VE = 0.0 kip

(1.2D+0.5L+E, incl. Ev)

Mn = 165.5 kip-ft

(Mn for P = Pu, φ = 1.0) VMn = MnVE/ME = 117.7 kip

2.0VE+VuG = 57.3 kip

1.25VMn+VuG = 147.1 kip φVn = 68.8 kip > 57.3 kip OK

74

Shear Design (cont.) Top of Wall B, Story 3

(0.9D+E, incl. Ev) tWall = 5.5 in.

dv = 72 in.

Pu = 6.1 kip Vu = 28.7 kip Mu = 40.3 kip-ft

Mu/Vudv = 0.23

Vn = Vnm + Vns ≤ Vn,Max

(TMS 402-13 §9.3.4.1.2)

Vnm = [4-1.75(Mu/Vudv)]An√(f'm) + 0.25Pu

(TMS 402-13 Eq. 9-24) Vnm = 72.6 kip

Vns = 0.5(Av/s)fydv

(TMS 402-13 Eq. 9-25)

(Av/s) = 0.0055 in.2/in. Vns = 11.9 kip

Vn,Max = 6.0 An√(f'm)

(TMS 402-13 Eq. 9-22 and Eq. 9-23) Vn,Max = 118.8 kip

Vn = 84.5 kip

(TMS 402-13 §9.3.4.1.2)

φ = 0.8 φVn = 67.6 kip > 28.7 kip OK

Check φVn ≥ min( 2.0VE+VuG, 1.25VMn+VuG)

(TMS 402-13 §7.3.2.6.1.1)

VE = 28.7 kip

(Eh) ME = 40.3 kip-ft

(Eh)

VuG = Vu - VE = 0.0 kip

(0.9D+E, incl. Ev)

Mn = 150.5 kip-ft

(Mn for P = Pu, φ = 1.0) VMn = MnVE/ME = 107.0 kip

2.0VE+VuG = 57.3 kip

1.25VMn+VuG = 133.8 kip φVn = 67.6 kip > 57.3 kip OK

75

Shear Design (cont.) Bottom of Wall B, Story 3

(1.2D+0.5L+E, incl. Ev) tWall = 5.5 in.

dv = 72 in.

Pu = 14.7 kip Vu = 28.7 kip Mu = 103.0 kip-ft

Mu/Vudv = 0.60

Vn = Vnm + Vns ≤ Vn,Max

(TMS 402-13 §9.3.4.1.2)

Vnm = [4-1.75(Mu/Vudv)]An√(f'm) + 0.25Pu

(TMS 402-13 Eq. 9-24) Vnm = 62.1 kip

Vns = 0.5(Av/s)fydv

(TMS 402-13 Eq. 9-25)

(Av/s) = 0.0055 in.2/in. Vns = 11.9 kip

Vn,Max = 5.1 An√(f'm)

(TMS 402-13 Eq. 9-22 and Eq. 9-23) Vn,Max = 100.4 kip

Vn = 74.0 kip

(TMS 402-13 §9.3.4.1.2)

φ = 0.8 φVn = 59.2 kip > 28.7 kip OK

Check φVn ≥ min( 2.0VE+VuG, 1.25VMn+VuG)

(TMS 402-13 §7.3.2.6.1.1)

VE = 28.7 kip-ft

(Eh) ME = 103.0 kip-ft

(Eh)

VuG = Vu - VE = 0.0 kip

(1.2D+0.5L+E, incl. Ev)

Mn = 171.7 kip-ft

(Mn for P = Pu, φ = 1.0) VMn = MnVE/ME = 47.8 kip

2.0VE+VuG = 57.3 kip

1.25VMn+VuG = 59.7 kip φVn = 59.2 kip > 57.3 kip OK

76

Shear Design (cont.) Bottom of Wall B, Story 3

(0.9D+E, incl. Ev) tWall = 5.5 in.

dv = 72 in.

Pu = 7.3 kip Vu = 28.7 kip Mu = 103.0 kip-ft

Mu/Vudv = 0.60

Vn = Vnm + Vns ≤ Vn,Max

(TMS 402-13 §9.3.4.1.2)

Vnm = [4-1.75(Mu/Vudv)]An√(f'm) + 0.25Pu

(TMS 402-13 Eq. 9-24) Vnm = 60.3 kip

Vns = 0.5(Av/s)fydv

(TMS 402-13 Eq. 9-25)

(Av/s) = 0.0055 in.2/in. Vns = 11.9 kip

Vn,Max = 5.1 An√(f'm)

(TMS 402-13 Eq. 9-22 and Eq. 9-23) Vn,Max = 100.4 kip

Vn = 72.2 kip

(TMS 402-13 §9.3.4.1.2)

φ = 0.8 φVn = 57.7 kip > 28.7 kip OK

Check φVn ≥ min( 2.0VE+VuG, 1.25VMn+VuG)

(TMS 402-13 §7.3.2.6.1.1)

VE = 28.7 kip-ft

(Eh) ME = 103.0 kip-ft

(Eh)

VuG = Vu - VE = 0.0 kip

(0.9D+E, incl. Ev)

Mn = 153.7 kip-ft

(Mn for P = Pu, φ = 1.0) VMn = MnVE/ME = 42.8 kip

2.0VE+VuG = 57.3 kip

1.25VMn+VuG = 53.5 kip φVn = 57.7 kip > 53.5 kip OK

77

Shear Design (cont.) Coupling Beams

(1.2D+0.5L+E, incl. Ev) tWall = 5.5 in.

dv = 48 in.

Pu = 0.0 kip Vu = 20.8 kip Mu = 70.7 kip-ft

Mu/Vudv = 0.85

Vn = Vnm + Vns ≤ Vn,Max

(TMS 402-13 §9.3.4.1.2)

Vnm = [4-1.75(Mu/Vudv)]An√(f'm) + 0.25Pu

(TMS 402-13 Eq. 9-24) Vnm = 33.2 kip

Vns = 0.5(Av/s)fydv

(TMS 402-13 Eq. 9-25)

(Av/s) = 0.0167 in.2/in. Vns = 24.0 kip

Vn,Max = 4.4 An√(f'm)

(TMS 402-13 Eq. 9-22 and Eq. 9-23) Vn,Max = 58.1 kip

Vn = 57.2 kip

(TMS 402-13 §9.3.4.1.2)

φ = 0.8 φVn = 45.7 kip > 20.8 kip OK

Check φVn ≥ min( 2.0VE+VuG, 1.25VMn+VuG)

(TMS 402-13 §7.3.2.6.1.1)

VE = 17.2 kip-ft

(Eh) ME = 61.4 kip-ft

(Eh)

VuG = Vu - VE = 3.5 kip

(1.2D+0.5L+E, incl. Ev)

Mn = 88.1 kip-ft

(Mn for P = Pu, φ = 1.0) VMn = MnVE/ME = 24.7 kip

2.0VE+VuG = 38.0 kip

1.25VMn+VuG = 34.5 kip φVn = 45.7 kip > 34.5 kip OK

78

Shear Design (cont.) Transfer Girder, Left End

(At Wall A) tWall = 5.5 in.

dv = 96 in.

Pu = 0.0 kip

(1.2D+0.5L+Ω0E, incl. Ev) VD = 10.0 kip

VL = 0.0 kip VE = 16.4 kip Vu = 55.1 kip

Ω0 applies to Eh Mu = 222.3 kip-ft

Mu/Vudv = 0.50

Vn = Vnm + Vns ≤ Vn,Max

(TMS 402-13 §9.3.4.1.2)

Vnm = [4-1.75(Mu/Vudv)]An√(f'm) + 0.25Pu

(TMS 402-13 Eq. 9-24) Vnm = 82.3 kip

Vns = 0.5(Av/s)fydv

(TMS 402-13 Eq. 9-25)

(Av/s) = 0.0092 in.2/in. Vns = 26.4 kip

Vn,Max = 5.3 An√(f'm)

(TMS 402-13 Eq. 9-22 and Eq. 9-23) Vn,Max = 140.5 kip

Vn = 108.7 kip

(TMS 402-13 §9.3.4.1.2)

φ = 0.8 φVn = 87.0 kip > 55.1 kip OK

79

Shear Design (cont.) Transfer Girder, Right End

(At Wall B) tWall = 5.5 in.

dv = 96 in.

Pu = 0.0 kip

(1.2D+0.5L+Ω0E, incl. Ev) VD = 6.7 kip

VL = 0.0 kip VE = 16.4 kip Vu = 50.4 kip

Ω0 applies to Eh Mu = 164.8 kip-ft

Mu/Vudv = 0.41

Vn = Vnm + Vns ≤ Vn,Max

(TMS 402-13 §9.3.4.1.2)

Vnm = [4-1.75(Mu/Vudv)]An√(f'm) + 0.25Pu

(TMS 402-13 Eq. 9-24) Vnm = 86.7 kip

Vns = 0.5(Av/s)fydv

(TMS 402-13 Eq. 9-25)

(Av/s) = 0.0092 in.2/in. Vns = 26.4 kip

Vn,Max = 5.6 An√(f'm)

(TMS 402-13 Eq. 9-22 and Eq. 9-23) Vn,Max = 147.3 kip

Vn = 113.1 kip

(TMS 402-13 §9.3.4.1.2)

φ = 0.8 φVn = 90.5 kip > 50.4 kip OK

80

Boundary Element Compliance

(TMS 402-13 §9.3.6.5) Wall A = Wall C

Rapid Screening

(TMS 402-13 §9.3.6.5.1)

Vu = 33.7 kip

(1.2D+0.5L+E, incl. Ev) Mu = 200.1 kip-ft

Mu/(Vudv) = 0.59

Pu = 93.5 kip 0.10Agf'm = 165 kip > Pu

Conditions 1 and 2:

OK

3An√(f'm) = 99.0 kip > Vu Condition 3:

OK

Extreme Fiber Compressive Stress

(TMS 402-13 §9.3.6.5.4)

fmax = Pu/Ag+Mu/S ≤ 0.2f'm 0.2f'm = 500.0 psi

Pu = 93.5 kip

(1.2D+0.5L+E, incl. Ev)

Mu = 200.1 kip-ft Ag = 660 in.2 S = 13200 in.3

fmax = 323.6 psi < 0.2f'm

OK

Both Sections, 9.3.6.5.1 and 9.3.6.5.4, are checked for illustration purposes (only one needs to be satisfied)

81

Boundary Element Compliance (cont.)

(TMS 402-13 §9.3.6.5) Wall B

Rapid Screening

(TMS 402-13 §9.3.6.5.1)

Vu = 28.7 kip

(1.2D+0.5L+E, incl. Ev) Mu = 103.0 kip-ft

Mu/(Vudv) = 0.60

Pu = 14.7 kip 0.10Agf'm = 99 kip > Pu

Conditions 1 and 2:

OK

3An√(f'm) = 59.4 kip > Vu Condition 3:

OK

Extreme Fiber Compressive Stress

(TMS 402-13 §9.3.6.5.4)

fmax = Pu/Ag+Mu/S ≤ 0.2f'm 0.2f'm = 500.0 psi

Pu = 14.7 kip

(1.2D+0.5L+E, incl. Ev)

Mu = 103.0 kip-ft Ag = 396 in.2 S = 4752 in.3

fmax = 297.0 psi < 0.2f'm

OK

Both Sections, 9.3.6.5.1 and 9.3.6.5.4, are checked for illustration purposes (only one needs to be satisfied)

82

Boundary Element Compliance (cont.)

(TMS 402-13 §9.3.6.5) Coupling Beams

Rapid Screening

(TMS 402-13 §9.3.6.5.1)

Vu = 20.8 kip

(1.2D+0.5L+E, incl. Ev) Mu = 70.7 kip-ft

Mu/(Vudv) = 0.85

Pu = 0.0 kip 0.10Agf'm = 66 kip > Pu

Conditions 1 and 2:

OK

3An√(f'm) = 39.6 kip > Vu Condition 3:

OK

Extreme Fiber Compressive Stress

(TMS 402-13 §9.3.6.5.4)

fmax = Pu/Ag+Mu/S ≤ 0.2f'm 0.2f'm = 500.0 psi

Pu = 0.0 kip

(1.2D+0.5L+E, incl. Ev)

Mu = 70.7 kip-ft Ag = 264 in.2 S = 2112 in.3

fmax = 401.7 psi < 0.2f'm

OK

Both Sections, 9.3.6.5.1 and 9.3.6.5.4, are checked for illustration purposes (only one needs to be satisfied)

83

Boundary Element Compliance (cont.)

(TMS 402-13 §9.3.6.5) Transfer Girder

Rapid Screening

(TMS 402-13 §9.3.6.5.1)

Vu = 55.1 kip

(1.2D+0.5L+Ω0E, incl. Ev) Mu = 222.3 kip-ft

Ω0 applies to Eh

Mu/(Vudv) = 0.50

Pu = 0.0 kip 0.10Agf'm = 132 kip > Pu

Conditions 1 and 2:

OK

3An√(f'm) = 79.2 kip > Vu Condition 3:

OK

Extreme Fiber Compressive Stress

(TMS 402-13 §9.3.6.5.4)

fmax = Pu/Ag+Mu/S ≤ 0.2f'm 0.2f'm = 500.0 psi

Pu = 0.0 kip

(1.2D+0.5L+Ω0E, incl. Ev)

Mu = 222.3 kip-ft

Ω0 applies to Eh Ag = 528 in.2

S = 8448 in.3

fmax = 315.8 psi < 0.2f'm

OK

Both Sections, 9.3.6.5.1 and 9.3.6.5.4, are checked for illustration purposes (only one needs to be satisfied)

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.

85

B.1 Limit Design Calculations for Design Example 1

Design Example 1 Geometry

Concrete masonry units, 16×8×8 in., fully grouted

Reinforcement

(Assumed sufficient for all non-seismic load combinations)

Wall A Flexural Reinforcement As,1 = (1) #4 = 0.2 in.2 @ d1 = 4 in.

As,2 = (1) #4 = 0.2 in.2 @ d2 = 20 in. As,3 = (1) #4 = 0.2 in.2 @ d3 = 28 in. As,4 = (1) #4 = 0.2 in.2 @ d4 = 44 in.

Wall A Shear Reinforcement Av = (1) #3 = 0.11 in.2 @ 16 in. o.c.

(Av/s) = 0.0069 in.2/in.

Wall B Flexural Reinforcement As,1 = (1) #4 = 0.2 in.2 @ d1 = 4 in.

As,2 = (1) #4 = 0.2 in.2 @ d2 = 12 in. As,3 = (1) #4 = 0.2 in.2 @ d3 = 20 in.

Wall B Shear Reinforcement Av = (1) #3 = 0.11 in.2 @ 8 in. o.c.

(Av/s) = 0.014 in.2/in.

86

Wall C Flexural Reinforcement As,1 = (1) #4 = 0.2 in.2 @ d1 = 4 in.

As,2 = (1) #4 = 0.2 in.2 @ d2 = 20 in. As,3 = (1) #4 = 0.2 in.2 @ d3 = 28 in. As,4 = (1) #4 = 0.2 in.2 @ d4 = 44 in.

Wall C Shear Reinforcement Av = (1) #3 = 0.11 in.2 @ 16 in. o.c.

(Av/s) = 0.0069 in.2/in.

Material Properties f'm = 1500 psi

fy = 60 ksi Es = 29000 ksi Em = 1350 ksi

(TMS 402-13 §4.2.2) εmu = 0.0025 in./in.

(TMS 402-13 §9.3.2(c))

εsy = 0.0021 in./in.

Seismic Design Parameters SDS = 1

R = 5

(ASCE/SEI 7-10 Table 12.2-1) Cd = 3.5

Ω0 = 2.5 Ev = 0.2 SDS D

(ASCE/SEI 7-10 Eq. 12.14-6)

ρ = 1.0

Modeling Assumptions

(Section properties based on 50% of gross section properties) E = Em/2 = 675 ksi

G = 270 ksi ν = 0.25

Poisson's ratio

Gravity Loading

(Determined from linear-elastic analysis using SAP2000)

DSelf Weight = 80 psf DTributary = 150 plf LTributary = 225 plf

Wall A PD(Top) = 7.7 kip

PD(Bot) = 10.9 kip VD = 0.7 kip MD(Top) = 4.1 kip-ft MD(Bot) = 3.0 kip-ft

87

PL = 2.2 kip VL = 0.2 kip ML(Top) = 1.1 kip-ft ML(Bot) = 0.8 kip-ft

Wall B PD(Top) = 5.6 kip

PD(Bot) = 6.9 kip VD = 0.0 kip MD(Top) = 0.1 kip-ft MD(Bot) = 0.0 kip-ft

PL = 1.5 kip VL = 0.0 kip ML(Top) = 0.0 kip-ft ML(Bot) = 0.0 kip-ft

Wall C PD(Top) = 7.2 kip

PD(Bot) = 9.8 kip VD = 0.7 kip MD(Top) = 3.5 kip-ft MD(Bot) = 2.2 kip-ft

PL = 1.8 kip VL = 0.2 kip ML(Top) = 1.0 kip-ft ML(Bot) = 0.6 kip-ft

Seismic Forces and Displacement

(Determined from linear-elastic analysis using SAP 2000)

Base Shear, Vb = 41 kip

(Demand on one line of resistance)

Roof Displacement, δR = 0.115 in.

Wall A PE = 19.7 kip

VE = 15.2 kip ME(Top) = 58.7 kip-ft ME(Bot) = 93.1 kip-ft

88

Wall B PE = 5.3 kip

VE = 4.6 kip ME(Top) = 15.5 kip-ft ME(Bot) = 21.3 kip-ft

Wall C PE = 25.0 kip

VE = 21.2 kip ME(Top) = 59.6 kip-ft ME(Bot) = 110.2 kip-ft

Design Forces for Load Combination 1.2D + 0.5L + 1.0E

(Including Ev) Wall A

Pu(Top) = 31.6 kip Pu(Bot) = 36.1 kip Vu = 16.3 kip Mu(Top) = 65.0 kip-ft Mu(Bot) = 97.7 kip-ft

Wall B Pu(Top) = 13.9 kip

Pu(Bot) = 15.7 kip Vu = 4.6 kip Mu(Top) = 15.7 kip-ft Mu(Bot) = 21.3 kip-ft

Wall C Pu(Top) = 36.0 kip

Pu(Bot) = 39.6 kip Vu = 22.3 kip Mu(Top) = 65.0 kip-ft Mu(Bot) = 113.6 kip-ft

Design Forces for Load Combination 0.9D + 1.0E

(Including Ev) Wall A

Pu(Top) = -14.3 kip

(Pu < 0, tension) Pu(Bot) = -12.1 kip

Vu = 15.7 kip Mu(Top) = 61.6 kip-ft Mu(Bot) = 95.2 kip-ft

89

Wall B Pu(Top) = -1.4 kip

(Pu < 0, tension)

Pu(Bot) = -0.5 kip Vu = 4.6 kip Mu(Top) = 15.6 kip-ft Mu(Bot) = 21.3 kip-ft

Wall C Pu(Top) = -20.0 kip

(Pu < 0, tension)

Pu(Bot) = -18.2 kip Vu = 21.7 kip Mu(Top) = 62.1 kip-ft Mu(Bot) = 111.7 kip-ft

90

Interaction Diagrams

= 1.0

φ = 1.0

-100

-50

0

50

100

150

200

250

300

350

0 50 100 150 200 250

Axia

l For

ce (k

ip)

Moment (kip-ft)

Wall A

φ = 0.9 φ = 1.0

φ = 0.9

-50

0

50

100

150

200

0 10 20 30 40 50 60 70

Axia

l For

ce (k

ip)

Moment (kip-ft)

Wall B

φ = 0.9 φ = 1.0

91

Interaction Diagrams (cont.)

= 1.0 = 0.9

-100

-50

0

50

100

150

200

250

300

350

400

0 50 100 150 200 250

Axia

l For

ce (k

ip)

Moment (kip-ft)

Wall C

φ = 0.9 φ = 1.0

92

Wall A Hinge Strength Mn and Vn determined for gravity load of 0.9D – 0.2SDSD

(Not including Eh)

Shear Corresponding to Development of Flexural Hinge (VMn) Pu,Top = 5.4 kip

(0.9D – 0.2SDSD)

Mn,Top = 95.0 kip-ft Pu,Bot = 7.64 kip Mn,Bot = 98.6 kip-ft VMn = Shear in Wall A associated with development of Mn

ME = 93.1 kip-ft VE = 15.2 kip VMn = (Mn / ME) VE VMn = 16.1 kip

Shear Strength Provided (Vn,Prov) tWall = 7.625 in.

dv = 48 in. Mu/Vudv = 1.52

(Using 0.9D + 1.0E)

Pu = 5.4 kip

(0.9D – 0.2SDSD) (Av/s) = 0.0069 in.2/in.

Vnm = [4-1.75(Mu/Vudv)]An√(f'm) + 0.25Pu

(TMS 402-13 Eq. 9-24) Vnm = 33.2 kip

Vns = 0.5(Av/s)fydv

(TMS 402-13 Eq. 9-25) Vns = 9.9 kip

Vn(Max) = 4.0 An√(f'm)

(TMS 402-13 Eq. 9-22 and Eq. 9-23) Vn(Max) = 56.7 kip

Vn,Prov = 43.1 kip

(TMS 402-13 §9.3.4.1.2)

Shear Demand VuG = 0.5 kip

VE = 15.2 kip

Check if Shear Controlled φvo = Reduction factor applicable where Vn < 2VMn + VuG

φvo = max((Vn-VuG)/(2VMn),(Vn-VuG)/(RVE))≤1 φvo = 1.00

Not Shear Controlled

Hinge Strength at Top of Wall A Hinge Strength = φvoMn(Top) Hinge Strength = 95.0 kip-ft

Hinge Strength at Base of Wall A Hinge Strength = φvoMn(Bot) Hinge Strength = 98.6 kip-ft

93

Wall B Hinge Strength Mn and Vn determined for gravity load of 0.9D – 0.2SDSD

(Not including Eh)

Shear Corresponding to Development of Flexural Hinge (VMn) Pu,Top = 3.9 kip

(0.9D – 0.2SDSD)

Mn,Top = 32.4 kip-ft Pu,Bot = 4.8 kip Mn,Bot = 33.1 kip-ft VMn = Shear in Wall B associated with development of Mn

ME = 21.3 kip-ft VE = 4.6 kip VMn = (Mn / ME) VE VMn = 7.1 kip

Shear Strength Provided (Vn,Prov) tWall = 7.625 in.

dv = 24 in. Mu/Vudv = 2.31

(Using 0.9D + 1.0E)

Pu = 3.9 kip

(0.9D – 0.2SDSD) (Av/s) = 0.014 in.2/in.

Vnm = [4-1.75(Mu/Vudv)]An√(f'm) + 0.25Pu

(TMS 402-13 Eq. 9-24) Vnm = 16.9 kip

Vns = 0.5(Av/s)fydv

(TMS 402-13 Eq. 9-25) Vns = 9.9 kip

Vn(Max) = 4 An√(f'm)

(TMS 402-13 Eq. 9-22 and Eq. 9-23) Vn(Max) = 28.4 kip

Vn,Prov = 26.8 kip

(TMS 402-13 §9.3.4.1.2)

Shear Demand VuG = 0.0 kip

VE = 4.6 kip

Check if Shear Controlled φvo = Reduction factor applicable where Vn < 2VMn + VuG

φvo = max((Vn-VuG)/(2VMn),(Vn-VuG)/(RVE))≤1 φvo = 1.00

Not Shear Controlled

Hinge Strength at Top of Wall B Hinge Strength = φvoMn(Top) Hinge Strength = 32.4 kip-ft

Hinge Strength at Base of Wall B Hinge Strength = φvoMn(Bot) Hinge Strength = 33.1 kip-ft

94

Wall C Hinge Strength Mn and Vn determined for gravity load of 0.9D – 0.2SDSD

(Not including Eh)

Shear Corresponding to Development of Flexural Hinge (VMn) Pu,Top = 5.1 kip

(0.9D – 0.2SDSD)

Mn,Top = 94.4 kip-ft Pu,Bot = 6.8 kip Mn,Bot = 97.3 kip-ft VMn = Shear in Wall C associated with development of Mn

ME = 110.2 kip-ft VE = 21.2 kip VMn = (Mn / ME) VE VMn = 18.7 kip

Shear Strength Provided (Vn,Prov) tWall = 7.625 in.

dv = 48 in. Mu/Vudv = 1.29

(Using 0.9D + 1.0E)

Pu = 5.1 kip

(0.9D – 0.2SDSD) (Av/s) = 0.0069 in.2/in.

Vnm = [4-1.75(Mu/Vudv)]An√(f'm) + 0.25Pu

(TMS 402-13 Eq. 9-24) Vnm = 33.2 kip

Vns = 0.5(Av/s)fydv

(TMS 402-13 Eq. 9-25) Vns = 9.9 kip

Vn(Max) = 4 An√(f'm)

(TMS 402-13 Eq. 9-22 and Eq. 9-23) Vn(Max) = 56.7 kip

Vn,Prov = 43.1 kip

(TMS 402-13 §9.3.4.1.2)

Shear Demand VuG = 0.5 kip

VE = 21.2 kip

Check if Shear Controlled φvo = Reduction factor applicable where Vn < 2VMN + VuG

φvo = max((Vn-VuG)/(2VMn),(Vn-VuG)/(RVE))≤1 φvo = 1.00

Not Shear Controlled

Hinge Strength at Top of Wall C Hinge Strength = φvoMn(Top) Hinge Strength = 94.4 kip-ft

Hinge Strength at Base of Wall C Hinge Strength = φvoMn(Bot) Hinge Strength = 97.3 kip-ft

95

Limit Mechanism

Virtual External Work = Virtual Internal Work Virtual External Work = VLim∆

Virtual Internal Work = (ΣMn/hw)Wall A + (ΣMn/hw)Wall B + (ΣMn/hw)Wall C

VLim = 51.5 kip (VLim = 51.5 kip, if not Shear Controlled) φLim = 0.8

φVLim = 41.2 kip

Base Shear, E = 41 kip < φVLim

OK

95.0 kip-ft32.4 kip-ft 94.4 kip-ft

Not Shear ControlledNot Shear Controlled Not Shear Controlled

33.1 kip-ft98.6 kip-ft 97.3 kip-ft

∆∆

8'

6'

10'

∆

96

Wall A Deformation Capacity Check

(1.2D+0.5L+E, incl. Ev)

Displacement Demand δR = 0.11 in.

Cd = 3.5 ∆m = δRCd = 0.40 in.

Displacement Capacity

Not Shear Controlled

Base Shear, E = 41 kip For Pu = 36.1 kip c = 9.85 in.

Neutral axis depth for Pu 0.5 hw Lw εmu/c = 0.73 in.

∆cap = 0.73 in. > 0.40 in. OK

Wall B Deformation Capacity Check

(1.2D+0.5L+E, incl. Ev)

Displacement Demand δR = 0.11 in.

Cd = 3.5 ∆m = δRCd = 0.40 in.

Displacement Capacity

Not Shear Controlled

Base Shear, E = 41 kip For Pu = 15.7 kip c = 5.42 in.

Neutral axis depth for Pu 0.5 hw Lw εmu/c = 0.53 in.

∆cap = 0.53 in. > 0.40 in. OK

97

Wall C Deformation Capacity Check

(1.2D+0.5L+E, incl. Ev)

Displacement Demand δR = 0.11 in.

Cd = 3.5 ∆m = δRCd = 0.40 in.

Displacement Capacity

Not Shear Controlled

Base Shear, E = 41 kip For Pu = 39.6 kip c = 10.33 in.

Neutral axis depth for Pu 0.5 hw Lw εmu/c = 0.56 in.

∆cap = 0.56 in. > 0.40 in. OK

98

B.2 Limit Design Calculations for Design Example 2

Design Example 2 Geometry

1750

OK

0

Clay masonry units, 12×6×4 in., fully grouted

Roof 36 ft

Level 2 Seismic Base

Level 3 9 ft

Level 5 27 ft

Level 4 18 ft

99

Reinforcement

(Assumed sufficient for all non-seismic load combinations)

Wall A = Wall C Longitudinal Reinforcement As,1 = (1) #4 = 0.2 in.2 @ d1 = 3 in.

As,2 = (1) #4 = 0.2 in.2 @ d2 = 21 in. As,3 = (1) #4 = 0.2 in.2 @ d3 = 39 in. As,4 = (1) #4 = 0.2 in.2 @ d4 = 57 in. As,5 = (1) #4 = 0.2 in.2 @ d5 = 63 in. As,6 = (1) #4 = 0.2 in.2 @ d6 = 81 in. As,7 = (1) #4 = 0.2 in.2 @ d7 = 99 in. As,8 = (1) #4 = 0.2 in.2 @ d8 = 117 in.

Wall A = Wall C Shear Reinforcement Av = (1) #3 = 0.11 in.2 @ 20 in. o.c.

(Av/s) = 0.0055 in.2/in.

Wall B Longitudinal Reinforcement As,1 = (1) #4 = 0.2 in.2 @ d1 = 9 in.

As,2 = (1) #4 = 0.2 in.2 @ d2 = 27 in. As,3 = (1) #4 = 0.2 in.2 @ d3 = 45 in. As,4 = (1) #4 = 0.2 in.2 @ d4 = 63 in.

Wall B Shear Reinforcement Av = (1) #3 = 0.11 in2 @ 20 in. o.c.

(Av/s) = 0.0055 in.2/in.

Coupling Beam Longitudinal Reinforcement As,1 = (1) #3 = 0.11 in.2 @ d1 = 3 in.

As,2 = (1) #3 = 0.11 in.2 @ d2 = 19 in. As,3 = (1) #3 = 0.11 in.2 @ d3 = 29 in. As,4 = (1) #3 = 0.11 in.2 @ d4 = 45 in.

Coupling Beam Shear Reinforcement Av = (1) #3 = 0.11 in.2 @ 12 in. o.c.

(Av/s) = 0.0092 in.2/in.

Transfer Girder Longitudinal Reinforcement As,1 = (1) #4 = 0.2 in.2 @ d1 = 9 in.

As,2 = (1) #4 = 0.2 in.2 @ d2 = 25 in. As,3 = (1) #4 = 0.2 in.2 @ d3 = 41 in. As,4 = (1) #4 = 0.2 in.2 @ d4 = 55 in. As,5 = (1) #4 = 0.2 in.2 @ d5 = 71 in. As,6 = (1) #4 = 0.2 in.2 @ d6 = 87 in.

100

Transfer Girder Shear Reinforcement Av = (1) #3 = 0.11 in.2 @ 12 in. o.c.

(Av/s) = 0.0092 in.2/in.

Material Properties f'm = 2500 psi

fy = 60 ksi Es = 29000 ksi Em = 1750 ksi

(TMS 402-13 §4.2.2) εmu = 0.0035 in./in.

(TMS 402-13 §9.3.2(c))

εsy = 0.0021 in./in.

Seismic Design Parameters SDS = 1.0

R = 5.5

(ASCE/SEI 7-10 Table 12.2-1) Cd = 4.0

Ω0 = 2.5 Ev = 0.2 SDS D

(ASCE/SEI 7-10 Eq. 12.14-6)

ρ = 1.0

Modeling Assumptions

(Section properties based on 50% of gross section

properties) E = Em/2 = 875 ksi G = 350 ksi ν = 0.25

Poisson's ratio

Gravity Loading

(Determined from linear-elastic analysis using SAP2000) DSelf Weight = 60 psf

DTributary = 0 plf LTributary = 0 plf

Wall A = Wall C, Story 3 PD(Top) = 26.6 kip

PD(Bot) = 29.6 kip VD = 3.4 kip MD(Top) = 4.3 kip-ft MD(Bot) = 12.5 kip-ft

PL = 0.0 kip VL = 0.0 kip ML(Top) = 0.0 kip-ft ML(Bot) = 0.0 kip-ft

101

Wall B, Story 3 PD(Top) = 8.7 kip

PD(Bot) = 10.5 kip VD = 0.0 kip MD(Top) = 0.0 kip-ft MD(Bot) = 0.0 kip-ft

PL = 0.0 kip VL = 0.0 kip ML(Top) = 0.0 kip-ft ML(Bot) = 0.0 kip-ft

Coupling Beams PD = 0 kip

(Envelope of forces)

VD = 2.5 kip MD = 6.7 kip-ft

PL = 0 kip VL = 0 kip ML = 0 kip-ft

Transfer Girder PD= 0 kip

VD(Left) = 10.0 kip

(At Wall A) VD(Right) = 6.7 kip

(At Wall B)

MD(Left) = 20.7 kip-ft MD(Right) = 24.1 kip-ft

PL = 0 kip VL(Left) = 0 kip VL(Right) = 0 kip ML(Left) = 0 kip-ft ML(Right) = 0 kip-ft

102

Seismic Forces and Displacements

(Determined from linear-elastic analysis using SAP 2000)

Base Shear, Vb = 86.6 kip

(Demand on one line of resistance)

Story Displacements δR = 0.250 in.

δ5 = 0.202 in. δ4 = 0.140 in. δ3 = 0.074 in. δ2 = 0.017 in.

Wall A = Wall C, Story 3 PE = 52.1 kip

VE = 29.0 kip ME(Top) = 37.7 kip-ft ME(Bot) = 182.5 kip-ft

Wall B, Story 3 PE = 0.0 kip

VE = 28.7 kip ME(Top) = 40.3 kip-ft ME(Bot) = 103.0 kip-ft

Coupling Beams PE = 0.0 kip

(Envelope of forces)

VE = 17.2 kip ME = 61.4 kip-ft

Transfer Girder PE = 0.0 kip

VE(Left) = 16.4 kip

(At Wall A) VE(Right) = 16.4 kip

(At Wall B)

ME(Left) = 77.4 kip-ft ME(Right) = 52.4 kip-ft

103

Design Forces for Load Combination 1.2D + 0.5L + 1.0E

(Including Ev) Wall A = Wall C, Story 3

Pu(Top) = 89.3 kip Pu(Bot) = 93.5 kip Vu = 33.7 kip Mu(Top) = 43.7 kip-ft Mu(Bot) = 200.1 kip-ft

Wall B, Story 3 Pu(Top) = 12.2 kip

Pu(Bot) = 14.7 kip Vu = 28.7 kip Mu(Top) = 40.3 kip-ft Mu(Bot) = 103.0 kip-ft

Coupling Beams Pu = 0.0 kip

(Envelope of forces)

Vu = 20.8 kip Mu = 70.7 kip-ft

Transfer Girder Pu = 0.0 kip

Vu(Left) = 30.5 kip

(At Wall A) Vu(Right) = 25.8 kip

(At Wall B)

Mu(Left) = 106.3 kip-ft Mu(Right) = 86.2 kip-ft

Design Forces for Load Combination 0.9D + 1.0E

(Including Ev) Wall A = Wall C, Story 3

Pu(Top) = -33.6 kip

(Pu < 0, tension) Pu(Bot) = -31.5 kip

Vu = 31.3 kip Mu(Top) = 40.7 kip-ft Mu(Bot) = 191.3 kip-ft

Wall B, Story 3 Pu(Top) = 6.1 kip

Pu(Bot) = 7.3 kip Vu = 28.7 kip Mu(Top) = 40.3 kip-ft Mu(Bot) = 103.0 kip-ft

104

Coupling Beams Pu = 0.0 kip

(Envelope of forces)

Vu = 19.0 kip Mu = 66.0 kip-ft

Transfer Girder Pu = 0.0 kip

Vu(Left) = 23.5 kip

(At Wall A) Vu(Right) = 21.1 kip

(At Wall B)

Mu(Left) = 91.8 kip-ft Mu(Right) = 69.3 kip-ft

105

Interaction Diagrams

-200

0

200

400

600

800

1000

1200

0 500 1,000 1,500 2,000

Axia

l For

ce (k

ip)

Moment (kip-ft)

Wall A = Wall C, Story 3

-100

0

100

200

300

400

500

600

700

0 100 200 300 400 500 600 700

Axia

l For

ce (k

ip)

Moment (kip-ft)

Wall B, Story 3

φ = 1.0

φ = 1.0

φ = 0.9

φ = 0.9

106

-200

-100

0

100

200

300

400

500

600

700

800

900

0 200 400 600 800 1,000 1,200

Axia

l For

ce (k

ip)

Moment (kip-ft)

Transfer Girder

φ = 1.0 φ = 0.9

-50

0

50

100

150

200

250

300

350

400

0 50 100 150 200 250 300

Axia

l For

ce (k

ip)

Moment (kip-ft)

Coupling Beams

φ = 1.0 φ = 0.9

107

Hinge Strength at Base of Wall A = Wall C, Story 3 Mn and Vn determined for gravity load of 0.9D – 0.2SDSD

(Not including Eh)

Shear Corresponding to Development of Flexural Hinge (VMn) Pu = 20.7

(0.9D – 0.2SDSD)

Mn = 538.9 kip-ft VMn = Shear in Wall A associated with development of Mn

ME = 182.5 kip-ft VE = 29.0 kip VMn = (Mn / ME) VE VMn = 85.5 kip

Shear Strength Provided (Vn,Prov) tWall = 5.5 in.

dv = 120 in. Mu/Vudv = 0.61

(Using 0.9D + 1.0E)

Pu = 20.7 kip

(0.9D – 0.2SDSD) (Av/s) = 0.0055 in.2/in.

Vnm = [4-1.75(Mu/Vudv)]An√(f'm) + 0.25Pu

(TMS 402-13 Eq. 9-24) Vnm = 101.9 kip

Vns = 0.5(Av/s)fydv

(TMS 402-13 Eq. 9-25) Vns = 19.8 kip

Vn(Max) = 5.0 An√(f'm)

(TMS 402-13 Eq. 9-22 and Eq. 9-23) Vn(Max) = 166.3 kip

Vn,Prov = 121.7 kip

(TMS 402-13 §9.3.4.1.2)

Shear Demand VuG = 2.4 kip

VE = 29.0 kip

Check if Shear Controlled φvo = Reduction factor applicable where Vn < 2VMN + VuG

φvo = max((Vn-VuG)/(2VMn),(Vn-VuG)/(RVE))≤1 φvo = 0.75

Shear Controlled

Hinge Strength Hinge Strength = φvoMn

Hinge Strength = 403.6 kip-ft

108

Hinge Strength at Base of Wall B, Story 3 Mn and Vn determined for gravity load of 0.9D – 0.2SDSD

(Not including Eh)

Shear Corresponding to Development of Flexural Hinge (VMn) Pu = 7.3

(0.9D – 0.2SDSD)

Mn = 153.7 kip-ft VMn = Shear in Wall B associated with development of Mn

ME = 103.0 kip-ft VE = 28.7 kip VMn = (Mn / ME) VE VMn = 42.8 kip

Shear Strength Provided (Vn,Prov) tWall = 5.5 in.

dv = 72 in. Mu/Vudv = 0.60

(Using 0.9D + 1.0E)

Pu = 7.3 kip

(0.9D – 0.2SDSD) (Av/s) = 0.0055 in.2/in.

Vnm = [4-1.75(Mu/Vudv)]An√(f'm) + 0.25Pu

(TMS 402-13 Eq. 9-24) Vnm = 60.3 kip

Vns = 0.5(Av/s)fydv

(TMS 402-13 Eq. 9-25) Vns = 11.9 kip

Vn(Max) = 5.1 An√(f'm)

(TMS 402-13 Eq. 9-22 and Eq. 9-23) Vn(Max) = 100.4 kip

Vn,Prov = 72.2 kip

(TMS 402-13 §9.3.4.1.2)

Shear Demand VuG = 0.0 kip

VE = 28.7 kip

Check if Shear Controlled φvo = Reduction factor applicable where Vn < 2VMN + VuG

φvo = max((Vn-VuG)/(2VMn),(Vn-VuG)/(RVE))≤1 φvo = 0.84

Shear Controlled

Hinge Strength Hinge Strength = φvoMn

Hinge Strength = 129.7 kip-ft

109

Hinge Strength at Ends of Coupling Beams Mn and Vn determined for gravity load of 0.9D – 0.2SDSD

(Not including Eh)

Shear Corresponding to Development of Flexural Hinge (VMn) Pu = 0 kip

(0.9D – 0.2SDSD)

Mn = 49.9 kip-ft VMn = Σ(Mn)/L = 2 / 7 ft 49.9 kip-ft

VMn = 14.2 kip

Shear Strength Provided (Vn,Prov) tWall = 5.5 in.

dv = 48 in. Mu/Vudv = 0.87

(Using 0.9D + 1.0E)

Pu = 0.0 kip

(0.9D – 0.2SDSD) (Av/s) = 0.0092 in.2/in.

Vnm = [4-1.75(Mu/Vudv)]An√(f'm) + 0.25Pu

(TMS 402-13 Eq. 9-24) Vnm = 32.7 kip

Vns = 0.5(Av/s)fydv

(TMS 402-13 Eq. 9-25) Vns = 13.2 kip

Vn(Max) = 4.4 An√(f'm)

(TMS 402-13 Eq. 9-22 and Eq. 9-23) Vn(Max) = 57.4 kip

Vn,Prov = 45.9 kip

(TMS 402-13 §9.3.4.1.2)

Shear Demand VuG = 1.8 kip

VE = 17.2 kip

Check if Shear Controlled φvo = Reduction factor applicable where Vn < 2VMN + VuG

φvo = max((Vn-VuG)/(2VMn),(Vn-VuG)/(RVE))≤1 φvo = 1.00

Not Shear Controlled

Hinge Strength Hinge Strength = φvoMn

Hinge Strength = 49.9 kip-ft

110

Limit Mechanism

Virtual External Work = Virtual Internal Work Virtual External Work = Σ [fi ∆i] = Σ [(fi/VLim)(hi/hR)] VLim ∆ = 0.690 VLim∆

Virtual Internal Work = 403.6 kip-ft(2∆/34 ft) + 129.7 kip-ft(∆/34 ft) +

49.9 kip-ft(16∆/34 ft)(15 ft/7 ft)

VLim = 112.7 kip (VLim = 125.3 kip, if not Shear Controlled)

φLim = 0.8 φVLim = 90.2 kip > E = 86.6 kip OK

Not Shear Controlled 49.9 kip-ft

Typ. for beams Typ. for beams

Shear Controlled 403.6 kip-ft

129.7 kip-ft Shear Controlled

Walls A and C

Wall B

34'

16'

25'

7'

111

Deformation Capacity, Wall A = Wall C

(1.2D+0.5L+E, incl. Ev)

Displacement Demand δR = 0.23 in.

Cd = 4 ∆m = δRCd = 0.93 in.

Displacement Capacity

Shear Controlled

Base Shear, E = 86.6 kip For Pu = 93.5 kip c = 19.05 in.

Neutral axis depth for Pu 0.5 hw Lw εmu/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. Ev)

Displacement Demand δR = 0.23 in.

Cd = 4 ∆m = δRCd = 0.93 in.

Displacement Capacity

Shear Controlled

Base Shear, E = 86.6 kip For Pu = 14.7 kip c = 6.62 in.

Neutral axis depth for Pu 0.5 hw Lw εmu/c = 7.77 in.

h/200 = 2.04 in. ∆cap = 2.04 in. > 0.93 in. OK

112

Deformation Capacity, Coupling Beams

(1.2D+0.5L+E, incl. Ev)

Displacement Demand δBeam = (δ/h) l

δ = (δR - δ2) h = 34 ft

l = 15 ft δBeam = 0.10 in. Cd = 4

∆m = δBeamCd = 0.41 in.

Displacement Capacity

Not Shear Controlled Base Shear, E = 86.6 kip

For Pu = 0.0 kip c = 2.52 in.

Neutral axis depth for Pu 0.5 hw Lw εmu/c = 2.80 in.

- - ∆cap = 2.80 in. > 0.41 in. OK

Design of Transfer Girder

Flexural Design

Left End Pu = 0.0 kip

(1.2D+0.5L+Ω0E, incl. Ev)

MD = 20.7 kip-ft

Ω0 applies to Eh ML = 0.0 kip-ft

ME = 77.4 kip-ft Mu = 222.3 kip-ft

φMn = 239.9 kip-ft > Mu

OK

Right End Pu = 0.0 kip

(1.2D+0.5L+Ω0E, incl. Ev)

MD = 24.1 kip-ft

Ω0 applies to Eh ML = 0.0 kip-ft

ME = 52.4 kip-ft Mu = 164.8 kip-ft

φMn = 239.9 kip-ft > Mu

OK

113

Shear Design Left End

(At Wall A) tWall = 5.5 in.

dv = 96 in.

Pu = 0.0 kip

(1.2D+0.5L+Ω0E, incl. Ev) VD = 10.0 kip

Ω0 applies to Eh

VL = 0.0 kip VE = 16.4 kip Vu = 55.1 kip Mu = 222.3 kip-ft

Mu/Vudv = 0.50

Vn = Vnm + Vns ≤ Vn,Max

(TMS 402-13 §9.3.4.1.2)

Vnm = [4-1.75(Mu/Vudv)]An√(f'm) + 0.25Pu

(TMS 402-13 Eq. 9-24) Vnm = 82.3 kip

Vns = 0.5(Av/s)fydv

(TMS 402-13 Eq. 9-25)

(Av/s) = 0.0092 in.2/in. Vns = 26.4 kip

Vn,Max = 5.3 An√(f'm)

(TMS 402-13 Eq. 9-22 and Eq. 9-23) Vn,Max = 140.5 kip

Vn = 108.7 kip

(TMS 402-13 §9.3.4.1.2)

φ = 0.8 φVn = 87.0 kip > 55.1 kip OK

114

Right End

(At Wall B) tWall = 5.5 in.

dv = 96 in.

Pu = 0.0 kip

(1.2D+0.5L+Ω0E, incl. Ev) VD = 6.7 kip

Ω0 applies to Eh

VL = 0.0 kip VE = 16.4 kip Vu = 50.4 kip Mu = 164.8 kip-ft

Mu/Vudv = 0.41

Vn = Vnm + Vns ≤ Vn,Max

(TMS 402-13 §9.3.4.1.2)

Vnm = [4-1.75(Mu/Vudv)]An√(f'm) + 0.25Pu

(TMS 402-13 Eq. 9-24) Vnm = 86.7 kip

Vns = 0.5(Av/s)fydv

(TMS 402-13 Eq. 9-25)

(Av/s) = 0.0092 in.2/in. Vns = 26.4 kip

Vn,Max = 5.6 An√(f'm)

(TMS 402-13 Eq. 9-22 and Eq. 9-23) Vn,Max = 147.3 kip

Vn = 113.1 kip

(TMS 402-13 §9.3.4.1.2)

φ = 0.8 φVn = 90.5 kip > 50.4 kip OK

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 402-13 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 402-13

Section C.3. Thus, the output of the linear model gives the necessary information to determine

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

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 402-13 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

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 stress-

strain 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, fy. 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 stress-

strain curve. Sanchez defined the initial line segment of the stress-strain curve using the shear

modulus, Gm. 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 Mu/(Vu dv) and on the axial load Pu, different

material definitions were required for the various wall segments involved. For this purpose, the

values of Mu, Vu, and Pu 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 code-

based 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 Mu/(Vu dv) ratio, axial load, masonry strength, and shear reinforcement, the

simplified method assumes that the wall segments do not reach their shear strength. Using this

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 402-13 Appendix C, the gravity load case should be based on load

combination 7 of Section 2.3.2 of ASCE/SEI 7-10.

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

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, Vp, 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 (Vn)

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 (Vn).

121

APPENDIX D:

TABLES

122

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

123

Table 2.1 – Continued

124

Table 3.1 – Story Weight above the Seismic Base and Vertical Distribution of Seismic Forces

Floor Level

Floor Elevation

Floor Weighta

PartitionWeight

Tributary Floor Area

Wall Weight

Wall Areab

Story Weightc

Story Forced

hx (ft) (psf) (psf) (ft2) (psf) (ft2) wx (kip) Fx (kip)

R 16 31 5 4200 80 440 187 41.1 G 0 - - - - -

Σ = 187 41.1 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, Vb, acting on the wall (Figure 3.1) is calculated using Vb = 1.1 W SDS/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, SDS = 1.0

Wall Element Level Reinforcement

Vertical Horizontal

Strength Design

Wall A Ground to Roof #5 @ 16" #3 @ 16"

Wall B Ground to Roof #4 @ 8" #3 @ 8"

Wall C Ground to Roof #6 @ 16" #5 @ 16"

125

Table 3.3 – Wall Reinforcement Schedule for Limit Design, SDS = 1.0

Wall Element Level Reinforcement

Vertical Horizontal

Limit Design

Wall A Ground to Roof #4 @ 16" #3 @ 16"

Wall B Ground to Roof #4 @ 8" #3 @ 8"

Wall C Ground to Roof #4 @ 16" #3 @ 16"

Table 3.4 – Lateral Stiffness for Different Mesh Sizes

Mesh Size Lateral Force at Roofa

Roof Displacementb Stiffness Relative Error

(in.) (kip) (in.) (kip/in.) (%)

1x1 100 0.1444 693 0.0

2x2 100 0.1439 695 0.3

4x4 100 0.1428 700 1.0

8x8 100 0.1402 713 2.9

12x12 100 0.1373 728 5.1

a Applied at roof level (see Figure 3.1). b Displacements based on linear-elastic model using gross section properties.

126

Table 3.5 – Material Properties for Nonlinear Static Analysis

Material IDa Pointa Strainb Stressb

M2500

2 -8.89E-03 -1.20E+00 [k,in]

1 -8.89E-04 -1.20E+00

f 'm = 1.50 ksi 0 0 0 Em = 1350 ksi 1' 8.89E-03 1.20E-04 2' 8.89E-02 1.20E-04

R60

2 -2.07E-02 -6.00E+01 [k,in]

1 -2.07E-03 -6.00E+01

fy = 60 ksi 0 0 0 Es = 29000 ksi 1' 2.07E-03 6.00E+01 2' 2.07E-02 6.00E+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 fy = 60 ksi. b Modulus of Elasticity, Em, is taken as 700 f'm for clay masonry.

127

Table 4.1 – Story Weights above the Seismic Base and Vertical Distribution of Seismic Forces

Floor Level

Floor Elevation

Floor Weighta

PartitionWeight

Tributary N-S Width

Tributary E-W Width

Wall Weight

Wall Areab

Story Weightc

hx (ft) (psf) (psf) (ft) (ft) (psf) (ft2) wx (kip) wxhx Fx/Vbd

R 36 17 5 54 42 60 450 77 2768 0.30 5 27 27 10 54 42 60 580 119 3205 0.35 4 18 27 10 54 42 60 580 119 2137 0.23 3 9 27 10 54 42 60 580 119 1068 0.12 2 0 - - - - - - G -15 - - - - - -

Σ = 433 9179 1.00 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). Fx is the lateral seismic force at any level. The base shear, Vb, acting on the wall (Figure 4.2) is calculated using Vb = 1.1 W SDS / R = 86.6 kip; where W = 433 kip, SDS = 1.0, and R = 5.5. The factor 1.1 accounts for torsional effects.

Table 4.2 – Wall Reinforcement Schedule for Strength Design, SDS = 1.0

Wall Element Level Reinforcement

Vertical Horizontal

Strength Design

Wall A = Wall C Ground to Roof #4 @ 18" #3 @ 20"

Wall B Ground to Roof #4 @ 18" #3 @ 20"

Coupling Beams 3 to Roof #4 @ 12" #4 @ 16"

Transfer Girder 2 #3 @ 12" #4 @ 16"

128

Table 4.3 – Wall Reinforcement Schedule for Limit Design, SDS = 1.0

Wall Element Level Reinforcement

Vertical Horizontal

Limit Design

Wall A = Wall C Ground to Roof #4 @ 18" #3 @ 20"

Wall B Ground to Roof #4 @ 18" #3 @ 20"

Coupling Beams 3 to Roof #3 @ 12" #3 @ 16"

Transfer Girder 2 #3 @ 12" #4 @ 16"

Table 4.4 – Lateral Stiffness for Different Mesh Sizes

Mesh Size Lateral Force at Roofa

Roof Displacementb Stiffness Relative Errorc

(in.) (kip) (in.) (kip/in.) (%)

2x2 100 0.2073 482 0.0

3x2 100 0.2069 483 0.2

6x4 100 0.2046 489 1.5

6x6 100 0.2036 491 1.9

12x12 100 0.1970 508 5.4

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.

129

Table 4.5 – Material Properties for Nonlinear Static Analysis

Material IDa Pointa Strainb Stressb

M2500

2 -1.14E-02 -2.00E+00 [k,in]

1 -1.14E-03 -2.00E+00

f 'm = 2.50 ksi 0 0 0 Em = 1750 ksi 1' 1.14E-03 2.00E-04 2' 1.14E-02 2.00E-04

R60

2 -2.07E-02 -6.00E+01 [k,in]

1 -2.07E-03 -6.00E+01

fy = 60 ksi 0 0 0 Es = 29000 ksi 1' 2.07E-03 6.00E+01 2' 2.07E-02 6.00E+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 fy = 60 ksi. b Modulus of Elasticity, Em, is taken as 700 f'm for clay masonry.

130

APPENDIX E:

FIGURES

131

Materials: fy = 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: SDS = 1.0 SD1 = 0.6 R = 5.0 Cd = 3.5 ρ = 1.0 δe,roof = 0.28” for Vb = 100 kip (based on 50% gross section properties)

Figure 3.1 – Building Description, Wall Elevation

Figure 3.2 – Shear Wall Reinforcement Layout

132

Figure 3.3 – Member Forces Due to Dead Load (1.0D)

Figure 3.4 – Member Forces Due to Live Load (1.0L)

M= 4.1V= 0.70P= 7.7

M= 0.12 M= 3.5V= 0.01 V= 0.72P= 5.6 P= 7.2

M= 3.0 M= 0.01 M= 2.2V= 0.70 V= 0.01 V= 0.72P= 10.9 Units: kip, ft P= 6.9 P= 9.8

M= 1.1V= 0.20P= 2.2

M= 0.01 M= 0.96V= 0.00 V= 0.20P= 1.5 P= 1.8

M= 0.84 M= 0.02 M= 0.62V= 0.20 V= 0.00 V= 0.20P= 2.2 Units: kip, ft P= 1.5 P= 1.8

133

Figure 3.5 – Member Forces Due to Earthquake Load (1.0E), SDS = 1.0

Figure 3.6 – Values of M/(Vd) based on 1.0E

M= 58.7V= 15.2P= 19.7

M= 15.5 M= 59.6V= 4.6 V= 21.2P= 5.3 P= 25.0

M= 93.1 M= 21.3 M= 110.2V= 15.2 V= 4.6 V= 21.2P= 19.7 Units: kip, ft P= 5.3 P= 25.0

0.97

≥ 1.0 0.70

≥ 1.0 Units: kip, ft ≥ 1.0 ≥ 1.0

134

Figure 3.7 – Member Forces Due to 1.2D + 0.5L + 1.0E, SDS = 1.0

Figure 3.8 – Member Forces Due to 0.9D + 1.0E, SDS = 1.0

M u = 65.0V u = 16.3P u = 31.6

M u = 15.7 M u = 65.0V u = 4.6 V u = 22.3P u = 13.9 P u = 36.0

M u = 97.7 M u = 21.3 M u = 113.6V u = 16.3 V u = 4.6 V u = 22.3P u = 36.1 Units: kip, ft P u = 15.7 P u = 39.6

M u = 61.6V u = 15.7P u = -14.3

M u = 15.6 M u = 62.1V u = 4.6 V u = 21.7P u = -1.4 P u = -20.0

M u = 95.2 M u = 21.3 M u = 111.7V u = 15.7 V u = 4.6 V u = 21.7P u = -12.1 Units: kip, ft P u = -0.5 P u = -18.2

135

Figure 3.9 – Masonry Model, Axial Direction

Figure 3.10 – Reinforcement Steel Model, Axial Direction

-1.5

-1.0

-0.5

0.0

0.5

1.0

1.5

-0.01 -0.005 0 0.005 0.01

Stre

ss, k

si

Strain

0.8 f'm

Em = 1350 ksi

-80

-60

-40

-20

0

20

40

60

80

-0.01 -0.005 0 0.005 0.01

Stre

ss, k

si

Strain

fy

fy

Es = 29000 ksi

136

0.0 0.1 0.2 0.3 0.4 0.5

0

20

40

60

80

100

0

20

40

60

80

100

0.0 0.1 0.2 0.3 0.4 0.5

Shea

r, ki

p

Roof Displacement, in.

Base Shear

VA

VB VC

0.0 0.1 0.2 0.3 0.4 0.5

0

20

40

60

80

100

0

20

40

60

80

100

0.0 0.1 0.2 0.3 0.4 0.5

Shea

r, ki

p

Roof Displacement, in.

Base Shear

VA VB

VC

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

51.0

59.1

137

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

0.0 0.1 0.2 0.3 0.4 0.5

0

20

40

60

80

100

0

20

40

60

80

100

0.0 0.1 0.2 0.3 0.4 0.5

Shea

r, ki

p

Roof Displacement, in.

Base Shear

VA

VB

VC

86.5

77.0

0.0 0.1 0.2 0.3 0.4 0.5

0

20

40

60

80

100

0

20

40

60

80

100

0.0 0.1 0.2 0.3 0.4 0.5

Shea

r, ki

p

Roof Displacement, in.

Base Shear

VA

VB

VC

138

Mat

eria

l Pro

pert

ies:

f y

= 60

ksi

f ’ m =

250

0 ps

i Cl

ay m

ason

ry u

nits

, 12

x 4

x 6

in.,

fully

gro

uted

W

eigh

ts:

Wal

l sel

f-wei

ght:

60 p

sf (i

n el

evat

ion)

Ro

of d

ead:

77

k Fl

oor d

ead:

119

k

W =

433

k (f

or 4

2’x

54’

bui

ldin

g fo

otpr

int)

Seis

mic

Des

ign

Para

met

ers (

ASCE

/SEI

7-1

0):

S DS =

1.0

R

= 5.

5 (S

peci

al R

einf

orce

d M

ason

ry S

hear

Wal

ls)

C d =

4.0

I e

= 1.

0 δ e

,roof

= 0

.29”

for V

b = 1

00 k

(per

40’

wal

l)

(Ba

sed

on 5

0% o

f gro

ss se

ctio

n pr

oper

ties)

Ω

O =

2.5

C T

= 0

.02

x =

0.75

h n

= 3

6 ft

T a =

0.2

9 s

C u =

1.4

k

= 1.

0 ρ

= 1.

0 C s

= 0

.18

for S

DS =

1.0

Figu

re 4

.1 –

Bui

ldin

g De

scrip

tion:

Flo

or P

lan,

Mat

eria

l Pro

pert

ies,

Load

s, an

d Se

ismic

Des

ign

Para

met

ers

N

Not

es:

1. L

ater

al fo

rce-

resis

ting

syst

em in

the

E-W

dire

ctio

n no

t sho

wn.

2.

Gra

vity

load

s not

tran

sfer

red

to m

ason

ry sh

ear w

alls.

139

Figure 4.2 – Building Description, Wall Elevation (East Line of Resistance)

Roof 36’

Level 5 27’

Level 4 18’

Level 3 9’

Level 2 Seismic Base

Ground

5’ x 7’ Window

(typ.)

140

Figure 4.3 – Shear Wall Reinforcement Layout

141

Figure 4.4 – Member Forces Due to Dead Load (1.0D)

M= 2.4 M= 1.8 M= 1.8 M= 2.4R V= 1.4 V= 0.2 V= 0.2 V= 1.4

M= 6.9 M= 0.0 M= 6.9V= 1.4 V= 0.0 V= 1.4P= 3.8 P= 1.9 P= 3.8

M= 0.1 M= 0.0 M= 0.1V= 1.4 V= 0.0 V= 1.4P= 6.8 P= 3.6 P= 6.8

M= 4.1 M= 3.0 M= 3.0 M= 4.15 V= 1.8 V= 0.2 V= 0.2 V= 1.8

M= 7.3 M= 0.0 M= 7.3V= 1.7 V= 0.0 V= 1.7P= 11.1 P= 4.8 P= 11.1

M= 1.2 M= 0.0 M= 1.2V= 1.7 V= 0.0 V= 1.7P= 14.1 P= 6.6 P= 14.1

M= 5.2 M= 3.9 M= 3.9 M= 5.24 V= 2.1 V= 0.5 V= 0.5 V= 2.1

M= 6.9 M= 0.0 M= 6.9V= 2.2 V= 0.0 V= 2.2P= 18.6 P= 7.2 P= 18.6

M= 3.9 M= 0.0 M= 3.9V= 2.2 V= 0.0 V= 2.2P= 21.6 P= 9.0 P= 21.6

M= 6.7 M= 5.2 M= 5.2 M= 6.73 V= 2.5 V= 0.9 V= 0.9 V= 2.5

M= 4.3 M= 0.0 M= 4.3V= 3.4 V= 0.0 V= 3.4P= 26.6 P= 8.7 P= 26.6

M= 12.5 M= 0.0 M= 12.5V= 3.4 V= 0.0 V= 3.4P= 29.6 P= 10.5 P= 29.6

2 M= 20.7 M= 24.1 M= 24.1 M= 20.7V= 10.0 V= 6.7 V= 6.7 V= 10.0

M= 32.7 M= 32.7V= 4.7 V= 4.7P= 44.4 P= 44.4

M= 10.1 M= 10.1V= 4.7 V= 4.7

G P= 49.8 Units: kip, ft P= 49.8

142

Figure 4.5 – Member Forces Due to Earthquake Load (1.0E), SDS = 1.0

M= 23.2 M= 20.0 M= 20.0 M= 23.2R V= 6.2 V= 6.2 V= 6.2 V= 6.2

M= 42.5 M= 48.4 M= 42.5V= 5.8 V= 14.4 V= 5.8P= 6.2 P= 0.0 P= 6.2

M= 13.4 M= 18.6 M= 13.4V= 5.8 V= 14.4 V= 5.8P= 6.2 P= 0.0 P= 6.2

M= 43.8 M= 42.1 M= 42.1 M= 43.85 V= 12.3 V= 12.3 V= 12.3 V= 12.3

M= 72.9 M= 61.2 M= 72.9V= 17.0 V= 22.2 V= 17.0P= 18.4 P= 0.0 P= 18.4

M= 12.2 M= 49.9 M= 12.2V= 17.0 V= 22.2 V= 17.0P= 18.4 P= 0.0 P= 18.4

M= 58.8 M= 56.5 M= 56.5 M= 58.84 V= 16.5 V= 16.5 V= 16.5 V= 16.5

M= 45.5 M= 63.6 M= 45.5V= 24.7 V= 26.8 V= 24.7P= 34.9 P= 0.0 P= 34.9

M= 77.9 M= 70.6 M= 77.9V= 24.7 V= 26.8 V= 24.7P= 34.9 P= 0.0 P= 34.9

M= 61.4 M= 59.3 M= 59.3 M= 61.43 V= 17.2 V= 17.2 V= 17.2 V= 17.2

M= 37.7 M= 40.3 M= 37.7V= 29.0 V= 28.7 V= 29.0P= 52.1 P= 0.0 P= 52.1

M= 182.5 M= 103.0 M= 182.5V= 29.0 V= 28.7 V= 29.0P= 52.1 P= 0.0 P= 52.1

2 M= 77.4 M= 52.4 M= 52.4 M= 77.4V= 16.4 V= 16.4 V= 16.4 V= 16.4

M= 72.7 M= 72.7V= 0.2 V= 0.2P= 68.6 P= 68.6

M= 70.6 M= 70.6V= 0.2 V= 0.2

G P= 68.6 Units: kip, ft P= 68.6

143

Figure 4.6 – Values of M/(Vd) based on 1.0E

0.94 0.81 0.81 0.94R

0.56 0.73

0.22 0.23

0.89 0.86 0.86 0.895

0.46 0.43

0.37 0.07

0.89 0.86 0.86 0.894

0.40 0.18

0.44 0.32

0.89 0.86 0.86 0.893

0.23 0.13

0.60 0.63

2 0.59 0.40 0.40 0.59

≥ 1.00 ≥ 1.00

G ≥ 1.00 Units: kip, ft ≥ 1.00

0.13

0.63

0.73

0.23

0.43

0.07

0.18

0.32

144

Figure 4.7 – Member Forces Due to 1.2D + 1.0E, SDS = 1.0

M u = 26.5 M u = 22.6 M u = 22.6 M u = 26.5R 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 = 52.1V u = 7.7 V u = 14.4 V u = 7.7P u = 11.6 P u = 2.7 P u = 11.6

M u = 13.6 M u = 18.6 M u = 13.6V u = 7.7 V u = 14.4 V u = 7.7P u = 15.7 P u = 5.0 P u = 15.7

5 M u = 49.5 M u = 46.3 M u = 46.3 M u = 49.5V 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 = 83.1V u = 19.4 V u = 22.2 V u = 19.4P u = 34.0 P u = 6.8 P u = 34.0

M u = 14.0 M u = 49.9 M u = 14.0V u = 19.4 V u = 22.2 V u = 19.4P u = 38.2 P u = 9.3 P u = 38.2

4 M u = 66.1 M u = 61.9 M u = 61.9 M u = 66.1V 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 = 55.2V u = 27.7 V u = 26.8 V u = 27.7P u = 61.0 P u = 10.0 P u = 61.0

M u = 83.4 M u = 70.6 M u = 83.4V u = 27.7 V u = 26.8 V u = 27.7P u = 65.2 P u = 12.5 P u = 65.2

3 M u = 70.7 M u = 66.5 M u = 66.5 M u = 70.7V 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 = 43.7V u = 33.7 V u = 28.7 V u = 33.7P u = 89.3 P u = 12.2 P u = 89.3

M u = 200.1 M u = 103.0 M u = 200.1V u = 33.7 V u = 28.7 V u = 33.7P u = 93.5 P u = 14.7 P u = 93.5

2 M u = 106 M u = 86.2 M u = 86.2 M u = 106V u = 30.5 V u = 25.8 V u = 25.8 V u = 30.5

M u = 118.5 M u = 118.5V u = 6.9 V u = 6.9P u = 130.7 P u = 130.7

M u = 84.7 M u = 84.7V u = 6.9 V u = 6.9

G P u = 138.3 Units: kip, ft P u = 138.3

145

Figure 4.8 – Member Forces Due to 0.9D + 1.0E, SDS = 1.0

M u = 24.9 M u = 21.3 M u = 21.3 M u = 24.9R 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 = 47.3V u = 6.8 V u = 14.4 V u = 6.8P u = -3.5 P u = 1.3 P u = -3.5

M u = 13.5 M u = 18.6 M u = 13.5V u = 6.8 V u = 14.4 V u = 6.8P u = -1.4 P u = 2.5 P u = -1.4

5 M u = 46.7 M u = 44.2 M u = 44.2 M u = 46.7V 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 = 78.0V u = 18.2 V u = 22.2 V u = 18.2P u = -10.7 P u = 3.4 P u = -10.7

M u = 13.1 M u = 49.9 M u = 13.1V u = 18.2 V u = 22.2 V u = 18.2P u = -8.6 P u = 4.6 P u = -8.6

4 M u = 62.4 M u = 59.2 M u = 59.2 M u = 62.4V 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 = 50.3V u = 26.2 V u = 26.8 V u = 26.2P u = -21.9 P u = 5.0 P u = -21.9

M u = 80.7 M u = 70.6 M u = 80.7V u = 26.2 V u = 26.8 V u = 26.2P u = -19.8 P u = 6.3 P u = -19.8

3 M u = 66.0 M u = 62.9 M u = 62.9 M u = 66.0V 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 = 40.7V u = 31.3 V u = 28.7 V u = 31.3P u = -33.6 P u = 6.1 P u = -33.6

M u = 191.3 M u = 103.0 M u = 191.3V u = 31.3 V u = 28.7 V u = 31.3P u = -31.5 P u = 7.3 P u = -31.5

2 M u = 91.8 M u = 69.3 M u = 69.3 M u = 91.8V u = 23.5 V u = 21.1 V u = 21.1 V u = 23.5

M u = 95.6 M u = 95.6V u = 3.6 V u = 3.6P u = -37.5 P u = -37.5

M u = 77.7 M u = 77.7V u = 3.6 V u = 3.6

G P u = -33.7 Units: kip, ft P u = -33.7

146

Figure 4.9 – Wall Shear vs. Roof Displacement, Wall Reinforcement per Limit Design (Table 4.3),

Northward Loading

0 1 2 3 4

0

50

100

150

200

250

0

50

100

150

200

250

0 1 2 3 4

Shea

r, ki

p

Roof Displacement, in.

Base Shear

VA VB

VC 156

147

Figure 4.10 – Deformed Shape for Simplified Nonlinear Layer Model, Wall Reinforcement per

Limit Design (Table 4.3)

Seismic Base

Wall Yielding

Beam Yielding

148

Figure 4.11 – Shear in Beams vs. Roof Displacement

Figure 4.12 – Axial Force in Beams vs. Roof Displacement, Northward Loading

0 1 2 3 4

-60

-40

-20

0

20

40

60

-60

-40

-20

0

20

40

60

0 1 2 3 4

Shea

r, ki

p

Roof Displacement, in.

Southward Loading Northward LoadingRoof

5th

4th

3rd

Beam Section at Wall A Interface

0 1 2 3 4

-60

-40

-20

0

20

40

60

-60

-40

-20

0

20

40

60

0 1 2 3 4

Axia

l Loa

d, k

ip

Roof Displacement, in.

Roof

5th

4th

3rd

Beam Section at Wall A InterfaceCompression (+)

149

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

0 1 2 3 4

-60

-40

-20

0

20

40

60

-60

-40

-20

0

20

40

60

0 1 2 3 4

Axia

l Loa

d, k

ip

Roof Displacement, in.

5th

4th

3rd

Roof

Beam Section at Wall A InterfaceCompression (+)

0 1 2 3 4

0

50

100

150

200

250

0

50

100

150

200

250

0 1 2 3 4

Shea

r, ki

p

Roof Displacement, in.

Base Shear

VA

VB

VC

195

150

Figure 4.15 – Deformed Shape for Simplified Nonlinear Layer Model, Wall Reinforcement per

Strength Design (Table 4.2)

Seismic Base

Wall Yielding

151

Figure C.1 – Linear-Elastic Model with 6 in. by 4 in. Mesh, Design Example 2

Linear-elastic area elements

Nodes fixed at ground

Seismic base with springs

152

Figure C.2 – Simplified Nonlinear Layer Model, Design Example 2

Linear-elastic area elements

Area elements with nonlinear layers

Nodes fixed at

ground

Seismic base with springs

153

Figure C.3 – Masonry Model, Axial Direction

Figure C.4 – Reinforcement Steel Model, Axial Direction

-2.5

-2.0

-1.5

-1.0

-0.5

0.0

0.5

1.0

1.5

2.0

2.5

-0.01 -0.005 0 0.005 0.01

Stre

ss, k

si

Strain

0.8 f'm

Em = 1750 ksi

-80

-60

-40

-20

0

20

40

60

80

-0.01 -0.005 0 0.005 0.01

Stre

ss, k

si

Strain

fy

fy

Es = 29000 ksi

154

LIST OF REFERENCES

ASCE/SEI 7-10 (2010). Minimum Design Loads for Buildings and Other Structures, ASCE

Standard ASCE/SEI 7-10, Structural Engineering Institute of the American Society of Civil

Engineers, Reston, Virginia.

ASCE/SEI 41-13 (2013). Seismic Evaluation and Retrofit of Existing Buildings, ASCE Standard

ASCE/SEI 41-13, Structural Engineering Institute of the American Society of Civil Engineers,

Reston, Virginia.

CSI (2011a). Analysis Reference Manual for SAP2000, ETABS, SAFE, and CSiBridge. Computers

and Structures Inc., Berkeley, California.

CSI (2011b). SAP2000: Static and Dynamic Finite Element Analysis of Structures, Nonlinear

Version 15.0. Computers and Structures Inc., Berkeley, California.

Lepage A., Dill S., Haapala M., and Sanchez R. (2011). “Seismic Design of Reinforced Masonry

Walls: Current Methods and Proposed Limit Design Alternative”, The 11th North American

Masonry Conference, Minneapolis, Minnesota.

Lepage A., and Sanchez, R. E. (2012). “Practical Nonlinear Analysis for Limit Design of Masonry

Walls”, The Open Civil Engineering Journal, Bentham Science Publishers, Vol. 6, pp. 107-118.

Park R. and Paulay T. (1975). Reinforced Concrete Structures, Wiley-Interscience.

Sanchez, R. E. (2012). “Limit Design of Reinforced Masonry Walls for Earthquake-Resistant

Construction”, M.S. Thesis, The Pennsylvania State University, University Park, Pennsylvania.

155

Shedid M.T., Drysdale R.G., and El-Dakhakhni W.W. (2008). “Behavior of Fully Grouted

Reinforced Concrete Masonry Shear Walls Failing in Flexure: Experimental Results”, Journal of

Structural Engineering, Vol. 134, No. 11, pp. 1754-1767.

Shing P.B., Noland J.L., Klamerus E., and Spaeh H. (1989). “Inelastic Behavior of Concrete

Masonry Shear Walls”, Journal of Structural Engineering, Vol. 115, No. 9, pp. 2204-2225.

TMS 402 (2013). Building Code Requirements for Masonry Structures (TMS 402-13). The

Masonry Society, Boulder, Colorado.

Voon K.C. and Ingham J.M. (2006). “Experimental In-Plane Shear Strength Investigation of

Reinforced Concrete Masonry Walls”, Journal of Structural Engineering, Vol. 132, No. 3, pp. 400-

408.

156

BIOGRAPHICAL SKETCH

Bradley S. Frederick

Brad Frederick grew up in the small town of Bradford, Pennsylvania. He began attending

The Pennsylvania State University in University Park, Pennsylvania, for an undergraduate

degree in Architectural Engineering in the fall of 2008 and entered the integrated Bachelor of

Architectural Engineering and Master of Science in Architectural Engineering degree program in

the fall of 2012.

While in pursuit of his degrees, Brad worked as an assistant in the Architectural

Engineering department laboratories on diverse projects, which included studies on concrete

members reinforced with ultrahigh strength steel and curtain walls subjected to simulated

seismic loading. He also worked for Dr. Andres Lepage as a grader for the undergraduate course

on concrete design for building structures. Brad worked as a Graduate Research Assistant from

spring of 2013 through spring of 2014.

Brad received his B.A.E. undergraduate degree in the fall of 2013 and is expected to

receive his M.S. degree in Architectural Engineering in the summer of 2014. Brad will begin

working as a design engineer for Atlantic Engineering Services, a structural consulting firm in

Pittsburgh, Pennsylvania, in the summer of 2014.