Seneca Seismic Design Handbook - First Edition PDF [PDF]

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Seneca Seismic Analysis and Design Principles and Practice

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Seneca Seismic Analysis and Design Principles and Practice First Edition K. Dirk Bondy & Bryan Allred

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Copyright © 2017 K. Dirk Bondy & Bryan Allred All rights reserved. No part of this book may be reproduced, stored, or transmitted by any means—whether auditory, graphic, mechanical, or electronic—without written permission of both publisher and author, except in the case of brief excerpts used in critical articles and reviews. Unauthorized reproduction of any part of this work is illegal and is punishable by law.

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Table of Contents Preface ....................................................................................................................................................... vi  1)

The Brief History of Seismic Design in California From a Practicing Engineer's Perspective .........1

2)

Site Specific Seismic Design Values & Base Shear Using the Equivalent Lateral Load Procedure ....................................................................................................1

3)

Vertical Distribution and Diaphragm Design Forces Using the Equivalent Lateral Force Procedure .................................................................................13

4)

Base Shear and Vertical Distribution Using Modal Response Spectrum Analysis ..........................31 

5)

Rigid Diaphragm Lateral Load Distribution ....................................................................................39 

6)

Rigid Diaphragm Shear and Moment Diagrams ..............................................................................66 

7)

Foundation Loads and Analysis ....................................................... Error! Bookmark not defined. 

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Design Considerations for Good Practice ........................................ Error! Bookmark not defined. 

9)

Concrete Shearwall and Foundation Design Example ..................... Error! Bookmark not defined. 

10) Concrete Moment Frame and Foundation Design Example ............ Error! Bookmark not defined. 11) Non-Linear Analysis of Concrete Moment Frames........................................................................ ??? About the Authors......................................................................................................................................87 

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Preface This book is not intended to comprehensively cover all aspects of seismic analysis and design, and it is priced to reflect that. In fact, we really do not cover much element design at all. The intent of this book is two-fold: first to provide notes and examples for use at the university level in a capstone fashion to prepare students for work in a design office, and second to address analysis issues that we believe many practicing engineers perform incorrectly – specifically in the areas of diaphragm analysis and foundation loading. We have arrived at this opinion based upon the number of seismic peer reviews that we have performed; many for the California State University system. The reader may find this next part annoying, but after writing “Post-Tensioned Concrete Principles and Practice” we learned the hard way that referencing code sections and equations numbers is great, until those code sections and equation numbers are changed by the respective code or document. And this seems to happen much more often than you would think it would. Therefore, we purposely describe code sections and equations sometimes without giving a reference to the section or equation numbers with the hope that the reader will be able to locate the section or equation themselves in both the near and distant future. We’ll see how that goes.

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1 The Brief History of Seismic Design in California From a Practicing Engineer's Perspective "History" began for me in the 1980's. I was a student at Cal Poly, San Luis Obispo and Bryan was about to enter the University of California at Irvine. I am writing this part because I am older than Bryan by about 7 years. The building code at that time was the 1985 Edition of the Uniform Building Code, pictured here.

This code included the requirements for the design of basically everything; wood, steel, concrete, prestressed concrete, masonry, live loads, wind loads and seismic loads. The book itself was 5-1/2" x 8" and about 1 inch thick. The seismic design portion started on page 114 in Section "Earthquake Regulations" and ended on page 123. The seismic base shear was a service level force calculated using the following equation: V = ZIKCSW Z was the zone factor, and was typically 1.0 for structures in high seismic zones. I was the importance factor, and remains the same today. K was the ductility factor and was 1.0 for a building frame system using shearwalls (like one of the examples in this book) and was 0.67 for ductile moment space frames (like the other example that we will follow through this book). C was a function of the building's fundamental period, and S was a factor relating to the soil type at the site, with a default value of 1.5. For short period buildings, the product CS needed not exceed 0.14. 1

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Seneca Seismic Analysis and Design Principles and Practice

Therefore, the base shear for a typical concrete shearwall building in a building frame system with a short period would be 0.14W. In order to design the concrete shearwalls we factored the base shear to bring it to strength level with a load factor of 1.4. The resulting base shear used for strength design was 1.4(0.14)W = 0.20W. Believe it or not, the shearwall building example in this book on a site in Los Angeles will actually require a lower strength based seismic coefficient than what we would have used in the 1980's. We distributed the base shear vertically in a very similar fashion to what we do today, and we accounted for the higher mode effects in longer period buildings by placing a lumped portion of the base shear at the very top of the building using the following equation: Ft = 0.07TV, when T > 0.7s We will see in this book, after going through the modal response spectrum analysis section, that a lumped force at the top is a very rational approach. In the subsequent years and building codes it would appear that the development of seismic loads has changed dramatically. But it really has not. I will admit that, while the theory of modal response spectrum analysis has not changed at all in 30 years, the code has gotten clearer and more descriptive on this subject. In the 1985 Uniform Building Code, dynamic analysis was addressed with the statement "Nothing in Section 2312 shall be deemed to prohibit the submission of properly substantiated technical data for establishing the lateral forces and distribution by dynamic analyses. In such analyses, the dynamic characteristics of the structure must be considered." There was no limit on base shear, or number of modes to consider, or scaling of results. That would be addressed in future codes. In today's ASCE 7 Standard, we do get more direction regarding modal response spectrum analysis, but not a lot more. For as large as all the codes necessary for design have gotten today, it may be a little surprising that the modal response spectrum analysis section fits on less than one page. Here are the codes and standards that an engineer must have in order to design a concrete or steel building. For other materials, you would require even more codes. I've cheated a little by showing the 2010 CBC, since we go online now to view the current codes.

And for scale, here are the relative sizes of some past building codes. Note that in 1994 there were two volumes, and in 1997 there were three volumes.

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In our opinion the basic theory of seismic design, and the theory of design of most materials such as post-tensioned concrete has not changed in the past thirty years. What has changed is our reliance on computers to do the analysis and design for us. And we're not convinced that this progress is good progress. We wrote our first book "Post-Tensioned Concrete Principles and Practice" primarily for our use in teaching the fundamentals of post-tensioned concrete at the university level. We focused on twodimensional analyses that could be performed by hand, and once we establish a thorough understanding in two-dimensions we explain the extrapolation to the three-dimensional building. In our opinion, it is impossible to be an expert and have a thorough understanding of a three-dimensional design or structure without first having a complete understanding in two dimensions. We take that same approach in this book. Basically, every design example in this book is done by hand, including the modal response spectrum analyses and the rigid diaphragm analyses. This is the way that we were taught in the 1980's and 1990's, and we remain convinced that this is the best way to fully grasp the concepts of both seismic design and post-tensioned concrete design. Once these concepts are fully understood in two dimensions the student should be able to use and understand three-dimensional computer software. But three-dimensional analyses of either seismic or post-tensioned concrete have never been done by hand (to the best of our knowledge), so it is impossible to start with three-dimensional analyses and ever completely understand what the computer is doing.

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Seneca Seismic Analysis and Design Principles and Practice

2 Site Specific Seismic Design Values & Base Shear Calculation Using the Equivalent Lateral Force Procedure The structural system of most buildings is defined by the function of the structure. For instance, open space parking structures are usually long-span post-tensioned concrete beams with one-way slabs utilizing perimeter moment frames as the lateral system. Residential mid-rise structures are typically post-tensioned flat plate systems utilizing concrete shearwalls as the lateral system. Therefore, contractors familiar with these framing types will usually be able to estimate the construction cost of the superstructure fairly accurately. The primary variables in the overall cost of construction are the foundations and increased costs associated with very high seismic demands. This is why one of the very first items that we ask for at the beginning of any project is the geotechnical report. The most critical initial structural decision is determining the most appropriate foundation system for the structure. Typically, the most economical foundation system consists of conventional spread and continuous footings and a “floating” grade slab. However, this is only feasible on very stable soil sites without high ground water or liquefaction potential. When the upper levels of the soil are poor, a mat foundation or intermediate ground improvement method such as rammed-aggregate piers or stone columns may be the most appropriate and economical solution. When the poor soil extends for a significant depth or the building loads are extremely high, piles may be the only feasible solution. Once the foundation system is chosen, the next step is to determine the ground motion data for the site.

The Response Modification Coefficient, R This factor has gone by a few names and definitions over the course of our careers. In the 1980's it was referred to as "K", and then "Rw" and was defined simply as a "numerical coefficient." In the 1997 Uniform Building Code the "w" was dropped and the definition became a "numerical coefficient representative of the inherent overstrength and global ductility capacity of lateral-force-resisting systems." We called it the "ductility factor." It is currently going by the name "R" and resides comfortably in South Florida, and its definition is "response modification coefficient." We like to think of the coefficient, R, as the "Rocky Balboa" factor, because it is a measure of how much of a beating the seismic system can take during an earthquake and still remain standing.

The Risk Category & Seismic Importance Factor, Ie Table 1.5-2 of ASCE 7 relates the Seismic Importance Factor to the Risk Category. The Risk Category is defined in Table 1.5-1. Most of the structures that we design (and that most engineers design) fall within Risk Category II, which is defined as “All Buildings and other structures except those listed in Risk Categories I, III, and IV.” The Seismic Importance Factor, Ie, associated with Risk Category II, which is the actual factor used in the following equations, is 1.0. Risk Category I is defined as “Buildings and other structures that represent a low risk to human like in the event of failure.” While it could be argued that parking structures and some other minimally occupied structures fall within this category, it makes no practical difference because both Risk Categories I and II have a Seismic Importance Factor, Ie, of 1.0 associated with them.

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Risk Category III has led to some confusion for some engineers. This category has many descriptive lines, but the first one is the one that has caused misunderstanding in that it states “Buildings and other structures, the failure of which could pose a substantial risk to human life.” At first glance, that would seem to describe just about any building. However, the generally accepted interpretation of the entirety of Risk Category III is one in which failure of the building or structure would endanger human life beyond the boundaries of the building or structure itself. Structures that house hazardous materials or explosive substances are good examples. The Seismic Importance Factor, Ie, associated with Risk Category III is 1.25. Risk Category IV is defined as buildings or structures that are essential facilities. Though it is not specifically stated, this includes hospitals and many public schools and includes any building or structure that is intended to remain operational after a major seismic event. Facilities that house large quantities of dangerous materials are also included in this category. The Seismic Importance Factor, Ie, associated with Risk Category IV is 1.5.

Geotechnical Data, Ss & S1 values All of our projects have a geotechnical report, eventually. However in the early stages of a project the report may be unavailable. When this is the case we can obtain most of the site specific seismic information that we need ourselves by first obtaining the site longitudinal and latitude coordinates, then the United States Geological Survey website at: https://earthquake.usgs.gov/designmaps/us/application.php to obtain ground motion data for our site. The site soil classification at most of our sites has been the default value of Site Class D – “Stiff Soil”, so lacking better information we will start with this assumption and verify it later when the geotechnical report is available. The USGS application will print all of the relevant ground motion data values (SS, S1, SMS, SM1, SDS & SD1) for the site. However, the only values that we actually require are the Ss and S1 values. The other values can (and should for educational purposes) be calculated using these two values and the ASCE 7 standard. Ss is defined as the mapped Maximum Considered Earthquake (MCER) spectral response acceleration parameter at short periods. ASCE 7 Section 12.8.1.3 states that “For regular structures five stories or less above the base as defined in Section 11.2 and with a period, T, of 0.5s or less, Cs is permitted to be calculated using a value of 1.5 for Ss”. S1 is defined as the mapped Maximum Considered Earthquake (MCER) spectral response acceleration parameter at a period of 1 second.

Calculating the SMS & SM1 values SMs is defined as the mapped Maximum Considered Earthquake (MCER) spectral response acceleration parameter at short periods adjusted for Site Class effects, and is equal to: SMS = FaSs SM1 is defined as the mapped Maximum Considered Earthquake (MCER) spectral response acceleration parameter at a period of 1 second adjusted for Site Class effects, and is equal to:

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Seneca Seismic Analysis and Design Principles and Practice

SM1 = FvS1 Fa and Fv are site acceleration and velocity coefficients found in ASCE 7 Tables 11.4-1 and 11.4-2, respectively and are a function of the Site Class and the SS & S1 values. The values in these tables shall be interpolated using straight-line interpolation as necessary.

Calculating the SDS & SD1 values SDS is defined as the design earthquake spectral response acceleration parameter at short periods, and is equal to: SDS = (2/3)SMS It may be beneficial not to blindly use the SDS value printed by the USGS application. That application will not apply the limit for SS of 1.5 that is available for certain low-rise regular structures. SD1 is defined as the design earthquake spectral response acceleration parameter at 1 second period, and is equal to: SD1 = (2/3)SM1

Determining the Building Fundamental Period, T ASCE 7 provides plenty of latitude in determining the building fundamental period for each primary building direction. The designer is only required to use “the structural properties and deformational characteristics of the resisting elements in a properly substantiated analysis.” However, regardless of the method used the upper limit of the fundamental period is equal to the approximate period, Ta, calculated in ASCE 7 Section 12.8.2.1 multiplied by the coefficient, Cu, taken from Table 12.8-1.

Determine the Fundamental Period Using Dynamic Modal Properties The most accurate method of determining the fundamental period is to model the seismic system in dynamic analysis computer program. The fundamental translational mode shape in each orthogonal direction will have a corresponding frequency of vibration. The inverse of that frequency is the period of that mode shape. This method, while being the most accurate, requires more sophisticated and expensive software than is often necessary for many buildings and requires a greater modeling effort on the part of the designer.

Determine the Fundamental Period Using the Rayleigh Period Equation For old-timers like me (Bryan is not as old as I am and has no gray hair as of this writing, so I will not include him as an old-timer) the now defunct Uniform Building Code (UBC) had an option for determining the fundamental period based upon an equation known as the Rayleigh Equation (called “Method B” in the old UBC), and was typically used for moment frame structures. It was quite accurate compared to a dynamic modal analysis, and only required a simple frame analysis program in order to use. The Rayleigh Period equation is shown here and will be demonstrated in an example later. 2

∑ ∑

 

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We have done some research into why this method was not included in the International Building Code. The rationale seems to be that it required known properties of the frame, as well as the forces and displacements at each level in order to use the equation. In other words, the engineer needed to know the answer before he or she started. However, we believe that the code writers have somewhat missed the point. The structural design of anything is always iterative, and relies on a good educated guess in order to begin. Experienced designers will have a fairly good idea of how much seismic system (how many bays of moment frames, braced frames, or length of shearwall and their approximate sizes) will be required for a specific building. That estimate may be modified as the design progresses, but often only slightly if the estimating engineer was an experienced designer. The other important point being missed here is that virtually every time for a normal building, the calculated fundamental period of a moment frame structure using either a dynamic modal analysis program or the Rayleigh method will be larger than the maximum period allowed by ASCE 7, and therefore will be limited to the maximum value. As a young engineer who designed many structural steel seismic systems I often used the Rayleigh Period to verify that I could determine my base shear calculations using the code upper limit for the fundamental period. Whether or not my period calculation was completely “accurate” didn’t really matter provided my calculated period was well above the maximum permitted period. And since I would need to model the frames in a structural analysis program anyway in order to design them, it was an easy thing to do and it took very little additional time. Therefore, I remain a proponent of the Rayleigh Period method. And if the truth be told, we always knew that we would be able to achieve the maximum allowed period, CuTa, so we would actually just assume that, do the entire frame design, and then verify the period using the Rayleigh method last.

Approximate the Fundamental Period ASCE 7 Section 12.8.2.1 allows the fundamental period, Ta, to be approximated by using the following equation: Where hn is the structural height in feet above the seismic base of the structure and the coefficients Ct and x are determined from Table 12.8-2. Additionally, ASCE 7 provides two more alternative methods for approximating the fundamental period; one for moment frame structures not exceeding 12 stories in height with story heights of at least 10 feet, and another for masonry or concrete shearwall structures. It has been our experience that these two alternate methods are rarely, if ever, used by designers. And in our case specifically, we originally included the shearwall alternative fundamental period calculation in our software as an option but eventually removed it after realizing that for our projects the equation never once resulted in a longer period than the equation printed above.

Determining the Base Shear, V The seismic base shear in a given direction is calculated using the following equation:

where Cs is the seismic response coefficient and W is the effective seismic weight of the structure.

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Seneca Seismic Analysis and Design Principles and Practice

There are realistically two equations in ASCE 7 used to calculate the value Cs. There is technically a third, however it only applies to buildings with long fundamental periods. Depending upon the location of the site, the definition of long period varies from a minimum of 4 seconds up to 16 seconds. In the Western United States, the value of “long period” is at least 6 seconds and is typically greater. Most of us will never be involved in the design of a building or structure with a fundamental period in that range. We certainly never have.

Cs Determined with ASCE 7 Equation (12.8-2) The following equation will control the design of shorter period structures:

12.8

2

SDS and Ie were previously defined; R is the response modification factor (also known as the ductility factor) found in ASCE 7 Table 12.2-1. The next equation will limit the value of Cs for longer period structures, and is valid for structures with fundamental periods less than TL, the long period determined for the site. ASCE 7 states that the value of Cs in (12.8-2) above need not exceed:





12.8

3

SD1 , T, Ie and R were previously defined. Again, it is so rare that a building or structure will ever have a fundamental period longer than TL for the site that we will not include the Cs equation for that condition. The SDS and SD1/T portions of the above equations are shown graphically in ASCE 7 Figure 11.4-1 "Design Response Spectrum." While many engineers may think that this diagram only applies to modal response spectrum analysis, it actually applies equally to the Equivalent Lateral Force procedure. However, only the first mode, or the fundamental mode, is considered in that case. The graph depicts the CS value before it has been divided by the term (R/Ie), which are the ductility and importance factors. The CS value before it has been divided by (R/Ie) is also known as the Spectral Acceleration, Sa in terms of gravity, g.

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ASCE 7 Figure 11.4-1 Design Response Spectrum

ASCE 7 includes two minimum Cs value equations. The first of which applies to all buildings and the second only applies to buildings on sites where the S1 value is equal to or greater than 0.6g. The following minimum Cs equation always applies: 0.044



0.01 12.8

5

For buildings or structures on site with an S1 value equal to or greater than 0.6g the following equation also applies: .

12.8

6

Once the controlling value of Cs is determined the seismic base shear, V, is calculated by multiplying Cs by W, the effective seismic weight of the structure.

Redundancy Factor,  Discussion ASCE 7 Section 12.3.4 addresses redundancy, and the redundancy factor . The redundancy factor is a factor applied to the seismic lateral forces for the strength design only of the vertical lateral load resisting elements (those listed in ASCE 7 Table 12.2-1). Its value is either 1.0 or 1.3, depending upon the likelihood that degradation of one of the elements in the vertical lateral load resisting system would lead to a significant loss (more than 33%) of total story strength, or an extreme torsional irregularity. ASCE 7 Table 12.3-3 lists the conditions required for various lateral force-resisting elements to use a  factor of 1.0. Where these cannot be met, a  factor of 1.3 is required. It is not our intent in this book to go through each lateral force-resisting element and discuss the redundancy requirements for each. That is more appropriate for a material design textbook. However, we believe that the most valuable two

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Seneca Seismic Analysis and Design Principles and Practice

pieces of information that we can discuss here are the practical applications and the proper application of the  factor. Based upon our experience with our own designs and the seismic reviews of other's work that we perform, for all practical purposes concrete shearwall seismic force-resisting systems can usually easily attain a  factor of 1.0, whereas effectively all structural steel buildings utilizing steel lateral forceresisting systems usually cannot achieve this and are designed using a  factor of 1.3. Concrete momentframe systems also usually qualify for a  factor of 1.0 since they are typically designed as multi-span moment frames. However, where single-span concrete moment frames are used it is likely that a  factor of 1.3 will be required. With all that stated, it is always conservative to use a  factor of 1.3, and some firms simply use that as the default without performing any calculations. The second valuable piece of information that we want to discuss is the proper application of the  factor when it is 1.3. It has been our experience that the application of this factor is sometimes incorrect. In our opinion, the  factor of 1.3 should not be considered a multiplier in the calculation of the base shear. While it would be very conservative to do that, it is not necessary and is not the intent of the redundancy factor. The only part of the seismic design that is affected by the  factor is the strength design of the lateral force-resisting system itself (the frames, shearwalls, etc.). The following structures, elements and calculations always use a  factor of 1.0, regardless of the  calculation results of Table 12.3-1:  

   

Structures assigned to Seismic Design Category B or C Drift calculations EXCEPT that Section 12.12.1.1 states that for “seismic forceresisting systems comprised solely of moment frames in structures assigned to Seismic Design Categories D, E or F, the design story drift shall not exceed a/ for any story.” In other words, the designer must include  in the drift calculations for moment frame structures in SDC D, E or F buildings. Diaphragm loads determined using Eqn. (12.10-1) Design of diaphragm collector elements, splices, and their connections for which the seismic load effects including overstenth factor of Section 12.4.3 are required for design Design of non-structural components Additional items listed in ASCE 7 Section 12.3.4.1

There is a debate in the engineering community about whether or not foundations should be designed including the  factor. The general consensus is that they must, but the debate continues regarding whether or not this includes the allowable soil bearing pressure calculations. Of course it is always conservative to use it, but a  factor of 1.3 will usually have a significant effect on the size of the foundation. Clarification from the local building official for a specific structure should be obtained before ignoring the  factor for seismic soil bearing calculations. Again, it is not necessary to apply a  factor of 1.3 to the seismic base shear and carry that base shear calculation through all of the seismic calculations, and it is technically incorrect to do so. Therefore, in this book we will not use the  factor in the base shear calculations.

Determining the Seismic Design Category, SDC (ASCE 7 Section 11.6) We will admit that this can be a confusing part (or one of many confusing parts) of the ASCE 7 standard, but it is extremely important that the designer correctly identify the SDC and understand the

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potential restrictions associated with that SDC. First, do not confuse the Site Class with the Seismic Design Category as they are completely different concepts. The SDC is determined based upon the site characteristic values of S1, SDS and SD1 as well as the Risk Category for the structure. It would take us many pages to describe everything that goes into determining the SDC, so the designer should become familiar with the entirety of ASCE 7 Section 11.6. However, based upon our experience there is one key item to be aware of immediately upon review of the site characteristics or the geotechnical report. ASCE 7 Tables 11.6-1 and 11.6-2 are used to determine the SDC when the value of S1 is less than 0.75. When this is the case the SDC will be somewhere in the A to D range, and for those of us designing in higher seismic zones it will almost always be D. There are no serious restrictions to the design when the SDC is D or less. However, if the value of S1 is greater than or equal to 0.75 the SDC becomes either E or F; both of which have additional restrictions placed on the design. For all practical purposes, it is absolutely critical that a SDC of E or F be identified at the very beginning of the project as this may have a dramatic effect on the layout of the seismic system and the architectural design. Most buildings or structures fall within the Risk Category of I, II or II. When S1 is equal or greater than 0.75 for these buildings they are assigned a SDC of E. If however, the structure is a Risk Category IV classification the SDC assignment is F.

Limitations for SDC E and F Structures The designer should be very familiar with ASCE 7 Section 12.3.3 “Limitations and Additional Requirements for Systems with Structural Irregularities.” Tables 12.3-1 and 12.3-2 discuss restrictions for all types of horizontal and vertical irregularities, most of which are well understood by designers in high seismic regions. However, there are specific restrictions that apply only to SDC E and F structures that, based upon our peer review experience, are sometimes overlooked by designers.

Irrgularities Forbidden for SDC E and F Structures 1. Horizontal Irregularity Type 1b. Extreme Torsional Irregularity 2. Vertical Irregularity Type 1b. Stiffness-Extreme Soft Story Irregularity 3. Vertical Irregularity Type 5a. Discontinuity in Lateral Strength – Weak Story Irregularity Most designers in higher seismic regions would not design a structure with a soft story irregularity or a weak story irregularity regardless of the SDC assignment. However, extreme torsional irregularity occurs when one edge of one of the diaphragms displaces more than 1.4 times the average diaphragm displacement (see ASCE 7 Table 12.3-1 Type 1b. for the exact verbiage). This irregularity may occur when there are architectural restrictions on the seismic system layout requiring one side of the building to be completely open. Architects who design buildings and have them entitled through a City prior to consulting a structural engineer about the Seismic Design Category at the site have caused great heartache for owners and structural engineers, and of course themselves. It is not unusual for a site on the West Coast to have an S1 value in excess of 0.75. Architects need to be trained to ask a structural engineer about the SDC for a site before designing the building, and structural engineers need to immediately verify the S1 value for each site before proceeding with just about any other part of the project.

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Seneca Seismic Analysis and Design Principles and Practice

Now that the base shear has been determined, the following chapters will demonstrate how to vertically distribute that base shear to the upper floors of the building, through the diaphragms and to the vertical seismic system.

K. Dirk Bondy & Bryan Allred

Example #1 – 5 Story Steel Office Building with Steel Moment Frames at Cal Poly, San Luis Obispo Given: Building Data Steel Special Moment Frames, R = 8 Risk Category: II Regular Structure (No Vertical or Horizontal Irregularities) Site Latitude: 35.29607, Site Longitude: -120.65306 Seismic Floor Weight Shown Below Total Building Seismic Weight = 15,033 kips Calculated Dynamic Fundamental Period = 1.45s Ground Motion Data from the Geotechnical Report Site Class: D “Stiff Soil” Ss = 1.124g (from USGS Design Maps Website) S1 = 0.429g (from USGS Design Maps Website)

Find: a)

The Seismic Design Category (SDC)

b)

Maximum Useable Building Period, T (ASCE 7 12.8.2)

c)

SDS and SD1 values

d)

Cs Value

e)

Design Base Shear, V

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a)

Seneca Seismic Analysis and Design Principles and Practice

The Seismic Design Category (SDC)

The SDC is a function of the value determined from the geotechnical site conditions. In this example, the S1 value is 0.429g and is less than 0.75g. Therefore, given that the Risk Factor is II, the SDC assigned value is D. b)

Maximum Useable Building Period, T (ASCE 7 12.8.2)

In this example, the fundamental period has been calculated using the dynamic properties of the structure, and was given in the problem statement. This calculation may have come from a dynamic modal analysis of the structure, or from the Rayleigh Period estimate. ASCE 7 Section 12.8.2 allows the use of any period calculated, provided it was "established using the structural properties and deformational characteristics of the resisting elements in a properly substantiated analysis." However, ASCE 7 then limits this value to a maximum of "the product of the coefficient for upper limit on calculated period (Cu) from Table 12.8-1 and the approximate fundamental period, Ta, determined in accordance with Section 12.8.2.1. The building lateral system is classified as Steel Special Moment Frame System (see ASCE 7 Table 12.2-1). The Ct and x values determined from Table 12.8-2 are 0.028 and 0.8, respectively. Therefore; 0.028 68

.

0.82 The "Coefficient for Upper Limit on Calculated Period", Cu, in Table 12.8-1 is 1.4 for any value of SD1 greater than 0.3, and it is conservative to use Cu as 1.4 for any value of SD1. SD1 will be calculated in Part c), but we will use Cu = 1.4 regardless. Therefore, upper limit for the fundamental period, T is: 1.4 0.82 = 1.15s In this example, the calculated fundamental period calculated using the dynamic properties of the structure was found to be 1.45s. Since this is greater than the maximum period allowed, the limiting value of 1.15s will be the period used for the base shear calculations.  the Maximum Useable Period, T c)

.

SDS and SD1 values

ASCE 7 Tables 11.4-1 and 11.4-2 are used to determine the values of Fa and Fv, respectively. For a Site Class of D and an Ss value of 1.124 the Fa value from Table 11.4-1 is interpolated between 1.1 (for Ss = 1.0) and 1.0 (for Ss = 1.1), and is 1.05. For a site class of D and an S1 value of 0.429 the Fv value from Table 11.4-2 is interpolated between 1.6 (for S1 = 0.4) and 1.5 (for S1 = 0.5), and is 1.57. SMS = FaSs = (1.05)(1.124) =1.18

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Now, SM1 SM1 = FvS1 = (1.57)(0.429) = 0.67 And finally, SDS = (2/3)SMS = (2/3)(1.18) = 0.79 SD1 = (2/3)SM1 = (2/3)(0.67) = 0.45 d)

Cs Value

Recall the ASCE 7 Figure 11.4-1, which describes the response spectrum to be used in lieu of a sitespecific response spectrum.

The following equations are used in the Equivalent Lateral Load procedure, but we can see that the equations simply come from the Design Response Spectrum above, divided by the term (R/Ie); the redundancy factor for the chosen seismic system divided by the importance factor for the structure.

12.8

2 &







12.8

3

According to ASCE 7 Table 1.5-2, a structure with a Risk Category of II has a corresponding Seismic Importance Factor, Ie, of 1.00. Therefore, the Cs value calculated in Equation 12.8-2 is: 0.79 0.098 8 1.00 ASCE 7 states that the value of Cs in (12.8-2) above need not exceed:

16

Seneca Seismic Analysis and Design Principles and Practice







12.8

3

0.45

0.049 0.098  use 0.049 8 1.15 1.00 An alternate way to look at this is to create the final Design Response Spectrum using the site-specific parameters SDS and SD1, dividing the graph by (R/Ie) and determining CS by finding the graph intersection at the fundamental period.

Using the values of SDS, SD1 previously found, Ie as 0, and the ASCE 7 definition of TS = SD1/SDS, we can create the graph of CS in terms of the period, T and then find CS at a period of 1.15s.

Our point in presenting the problem this way is show that the Equivalent Lateral Force procedure is very similar to what we will do later with the modal response spectrum analysis. The two analyses are actually very similar. The major difference is that in the Equivalent Lateral Force procedure we will only consider the fundamental period in the analysis, and ASCE 7 will require a vertical force distribution of the base shear that attempts to incorporate the effects of the higher modes and associated periods not considered in the analysis. ASCE 7 Equation 12.8-5 establishes one of the minimum Cs values. 0.044



0.01

= 0.044(0.79)(1.00) = 0.035 < 0.049  use 0.049 Since S1 is less than 0.6g, it is not necessary to check Equation (12.8-6).  The controlling value of Cs is 0.049. e)

Design Base Shear, V 0.049 15,033 kips = 737 kips

The following contains the output from our software, Seneca Seismic Design Values.

K. Dirk Bondy & Bryan Allred

SENECA SEISMIC DESIGN VALUES Licensed to: Seneca Structural Engineering, Inc. Project Name: Example #2 - 5 Story Steel Office Building: 1 Grand Ave., San Luis Obispo, CA Project Engineer: K. Dirk Bondy

Date: 8/24/2017

The program conforms to the 2013 California Building Code (CBC) and ASCE 7-10

BUILDING DATA Number of Elevated Levels: 5 |Analysis Direction: X |Seismic System: Steel Special Moment Frames R Factor: 8 |Risk Category: II |Importance Factor: 1.00

GROUND MOTION DATA FROM GEOTECHNICAL REPORT S S : 1.12g | S 1 : 0.43g | Site Class: D | Seismic Design Category: D

CALCULATED GROUND MOTION DATA Fa: 1.05 | Fv: 1.57 (from Tables 11.4-1 & 11.4-2 of ASCE 7) SMS = FaS S = 1.18 SM1 = FvS 1 = 0.67

SDS = (2/3)S MS = 0.79

BUILDING PERIOD

S D1 = (2/3)S M1 = 0.45 ASCE 7 Eqn. (12.8-7)

Ct: 0.028 |x: 0.8 |hn: 68 ft. Ta = Ct*(hn)^x = 0.82s Dynamic Period (from User Input), T(Dynamic) = 1.45s Controlling Period, T = 1.15s

Cs VALUES (Eqn. 12.8-2) Cs = Sds/(R/Ie) with: Sds = 0.79| R = 8| Ie = 1 Cs = 0.098

(Eqn. 12.8-3) Cs = Sd1/(T(R/Ie)) with: Sd1 = 0.45| R = 8| Ie = 1| T =1.15s Cs = 0.049

(Eqn. 12.8-5) Cs(min) = 0.044SdsIe >= 0.01 with: Sds = 0.79| Ie = 1 Cs(min) = 0.035

Controlling Cs = 0.049

DESIGN BASE SHEAR (Eqn. 12.8-1): V=CsW with Total Building Weight = 15,033 kips

V DESIGN = 737 kips

-1-

Seneca Software Solutions

17

Example #2 – 5 Story Residential Flat Plate with Concrete Shearwalls in Los Angeles Given: Building Data Special Concrete Shearwalls in a Building Frame System, R = 6 Risk Category: II Regular Structure (No Vertical or Horizontal Irregularities) Site Latitude: 34.04790, Site Longitude: -118.25602 Seismic Floor Weight Shown Below Total Building Seismic Weight = 23,162 kips Ground Motion Data from the Geotechnical Report Site Class: D “Stiff Soil” Ss = 2.367g (from USGS Design Maps Website) S1 = 0.831g (from USGS Design Maps Website)

Find: a)

The Seismic Design Category (SDC)

b)

Building Period, Ta (ASCE 7 12.8-7)

c)

SDS and SD1 values.

d)

Cs Value

e)

Design Base Shear, V

19

20

a)

Seneca Seismic Analysis and Design Principles and Practice

The Seismic Design Category (SDC)

The SDC is a function of the value determined from the geotechnical site conditions. In this example the S1 value is 0.831g and is greater than 0.75g. Therefore, given that the Risk Factor is II, the SDC assigned value is E. This SDC value has restrictions on irregularities allowed; most importantly the structure must not have extreme torsional irregularity as defined in ASCE 7 Table 13.3-1.1b. b)

Building Period, Ta (ASCE 7 12.8-7)

The building lateral system is classified as Special Concrete Shearwall in a Building Frame System (see ASCE 7 Table 12.2-1). The Ct and x values determined from Table 12.8-2 (All other structural systems) are 0.02 and 0.75, respectively. Therefore; 0.02 60

.

. c)

SDS and SD1 values

ASCE 7 Tables 11.4-1 and 11.4-2 are used to determine the values of Fa and Fv, respectively. For a Site Class of D and an Ss value greater than 1.25 the Fa value is 1.0. For a site class of D and an S1 value greater than 0.5 the Fv value is 1.5. SMS = FaSs This building is a 5 story, regular building with a period less than 0.5s. Therefore, ASCE 7 Section 12.8.1.3 states that we are not required to use an Ss factor greater than 1.5 for this building on this site. Therefore, SMS = (1.0)(1.5) = 1.5 Now, SM1 SM1 = FvS1 = (1.5)(0.831) = 1.25 And finally, SDS = (2/3)SMS = (2/3)(1.5) = 1.00

SD1 = (2/3)SM1 = (2/3)(1.25) = 0.83

K. Dirk Bondy & Bryan Allred

d)

21

Cs Value

12.8

2

According to ASCE 7 Table 1.5-2, a structure with a Risk Category of II has a corresponding Seismic Importance Factor, Ie, of 1.00. Therefore, the Cs value calculated in Equation 12.8-2 is:

1.00 6 1.00

0.167

ASCE 7 states that the value of Cs in (12.8-2) above need not exceed:



0.83 0.43

6 1.00



12.8



0.321

3

0.167  use 0.167

An alternate way to look at this is to create the final Design Response Spectrum using the site-specific parameters SDS and SD1, dividing the graph by (R/Ie) and determining CS by finding the graph intersection at the fundamental period. Using the values of SDS, SD1 previously found, Ie as 0, and the ASCE 7 definition of TS = SD1/SDS, we can create the graph of CS in terms of the period, T and then find CS at a period of 0.43s.

Our point in presenting the problem this way is show that the Equivalent Lateral Force procedure is very similar to what we will do later with the modal response spectrum analysis. The two analyses are actually very similar. The major difference is that in the Equivalent Lateral Force procedure we will only consider the fundamental period in the analysis, and ASCE 7 will require a vertical force distribution of the base shear that attempts to incorporate the effects of the higher modes and associated periods not considered in the analysis.

22

Seneca Seismic Analysis and Design Principles and Practice

ASCE 7 Equation 12.8-5 establishes one of the minimum Cs values. 0.044



0.01

= 0.044(1.00)(1.00) = 0.044 < 0.167  use 0.167 For buildings or structures on a site with an S1 value equal to or greater than 0.6g, which is the case in this example, the following equation also applies: .

12.8

.

.



6

0.069 < 0.167  use 0.167

.

The controlling value of Cs is 0.167. e)

Design Base Shear, V

0.167 23,162 kips = 3,868 kips It is interesting to note that had we not taken advantage of ASCE 7 Section 12.8.1.3, the required base shear would have been 6,092 kips, or 57% higher than the one we calculated. The following page contains the output from our software, Seneca Seismic Design Values.

K. Dirk Bondy & Bryan Allred

SENECA SEISMIC DESIGN VALUES Licensed to: Seneca Structural Engineering, Inc. Project Name: Example #2 - 5 Story Residential Flat Plate in Los Angeles Project Engineer: Bryan Allred

Date: 9/4/2017

The program conforms to the 2013 California Building Code (CBC) and ASCE 7-10

BUILDING DATA Number of Elevated Levels: 5 |Analysis Direction: X |Seismic System: Special Concrete Shearwalls - Building Frame System R Factor: 6 |Risk Category: II |Importance Factor: 1.00

GROUND MOTION DATA FROM GEOTECHNICAL REPORT S S : 1.50g (Limited to 1.50g per ASCE 7 Section 12.8.1.3) | S1 : 0.83g | Site Class: D | Seismic Design Category: E

CALCULATED GROUND MOTION DATA Fa: 1.00 | Fv: 1.50 (from Tables 11.4-1 & 11.4-2 of ASCE 7) SMS = FaS S = 1.50 SM1 = FvS 1 = 1.25

SDS = (2/3)S MS = 1.00

S D1 = (2/3)S M1 = 0.83

BUILDING PERIOD ASCE 7 Eqn. (12.8-7) Ct: 0.02 |x: 0.75 |hn: 52.5 ft. Ta = Ct*(hn)^x = 0.39s

Cs VALUES (Eqn. 12.8-2) Cs = Sds/(R/Ie) with: Sds = 1.00| R = 6| Ie = 1 Cs = 0.167

(Eqn. 12.8-3) Cs = Sd1/(T(R/Ie)) with: Sd1 = 0.83| R = 6| Ie = 1| T =0.39s Cs = 0.355

(Eqn. 12.8-5) Cs(min) = 0.044SdsIe >= 0.01 with: Sds = 1.00| Ie = 1 Cs(min) = 0.044

(Eqn. 12.8-6) - Only applies when S1 >= 0.6g Cs(min) = 0.5S1/(R/Ie) with: S1 = 0.83| R = 6| Ie = 1 Cs(min) = 0.069

Controlling Cs = 0.167

DESIGN BASE SHEAR (Eqn. 12.8-1): V=CsW with Total Building Weight = 23,162 kips

V DESIGN = 3,868 kips

-1-

Seneca Software Solutions

23

24

Seneca Seismic Analysis and Design Principles and Practice

3 Vertical Distribution and Diaphragm Design Forces Using the Equivalent Lateral Force Procedure Vertical Distribution of the Base Shear The vertical distribution of the base shear is intended to create a load combination that will envelope the maximum shear at any level. It is very important for young engineers to understand that the forces and their vertical distribution in the following calculations only make sense as a complete set of forces. The individual forces at each level in this analysis do not have any particular meaning, except that they contribute to the overall shear envelope. This concept will become clearer when we explore dynamic modal analysis later in the book. For now, understand that a multi-story building is oscillating under the influence of multiple modes acting simultaneously, and the purpose of the following calculations is to determine a singular set of forces that when applied to the structure will envelope the maximum story shears that are expected to occur as all of the oscillating modes participate in the structure's response to the earthquake. ASCE 7 Section 12.8.3 "Vertical Distribution of Seismic Forces" states that the portion of the base shear, V, applied at each level of the structure be determined from the following equations:

12.8

11

12.8

12

and,

∑

where, Cvx = the vertical distribution factor V = the total design lateral force or shear at the base of the structure wi and wx = the portion of the total effective seismic weight of the structure (W) located or assigned to Level i or x hi and hx = the height in feet from the base to Level i or x k = an exponent related to the structure period as follows:   

for structures having a period of 0.5s or less, k = 1 for structures having a period of 2.5s or more, k = 2 for structures having a period between 0.5s and 2.5s, k shall be 2 or determined by linear interpolation between 1 and 2

K. Dirk Bondy & Bryan Allred

25

Diaphragm Design Forces The diaphragm, in our opinion, is probably the most important element in the lateral load path and is often not given the respect it deserves, particularly by the academic world. Because of that, many engineering students graduate and begin their design career with a minimal understanding of the role of the diaphragm, and also often a statically incorrect view. The diaphragm is best thought of as the membrane that ties all of the lateral force-resisting elements together, requiring that they work together as a unit in resisting the earthquake. But it does more than just that. The diaphragm is unique in that it not only transfers the seismic load to the seismic force-resisting system; it is also where those lateral loads (weights) are generated. The diaphragm, or the laterally continuous portion of the floor, supports the majority of the building weight since it is an integral part of the floor. In post-tensioned concrete flat plates, the floor and the diaphragm are exactly the same thing. Every inch of the plate or flat slab is used to support the vertical gravity loads, and every inch of the plate is also used as the diaphragm to distribute the lateral loads to the seismic resisting elements. This is one of the reasons that flat plate structures are so cost efficient, especially when they are post-tensioned. However, in a long-span post-tensioned structure constructed of a one-way slab spanning to beams, only the slab portion is considered the diaphragm as it is the only laterally continuous portion of the floor system. While the beams can be used to resist collector forces (drag forces for us old folks) or chord forces, they do not participate in the membrane action that ties all of the lateral force-resisting members together. In wood structures, the portion of the floor considered as the diaphragm is typically only the plywood, and in most structural steel buildings the diaphragm is considered the concrete filled metal deck portion of the floor. Many engineers like to say that the analysis of the diaphragm is the same as the analysis of a beam on its side, where the diaphragm is modeled as a beam and the supports for the beam are the lateral load resisting elements (the shearwalls or moment frames, etc.). In our opinion this analogy, while somewhat accurate, often leads to a statically incorrect analysis of the diaphragm shears and moments. We will go into this in great depth later in the book. As previously stated, the forces found in the previous section represent a set of forces that will envelope the maximum seismic shear at any level, including base level. However, these forces do not represent the largest story force that may occur at any particular level during an earthquake. It is necessary to perform different calculations to determine the maximum force at any particular level regardless of the overall set of forces acting at that instant. These forces will be the forces that the diaphragms must be designed to withstand. Again, this will become clearer as we explore dynamic modal analysis later in the book. But for now, understand that as the building is oscillating under the influence of multiple modes, the maximum forces generated at any particular level will be the statistical combination of the modal story forces, not shears, at that level. ASCE 7 estimates these diaphragm forces using the following method. ASCE 7 Section 12.10 "Diaphragms, Chords and Collectors" states that the diaphragms at each level must be designed for loading according to the following equation:

26

Seneca Seismic Analysis and Design Principles and Practice



∑ ∑





12.10

1

where, Fpx = the diaphragm design force at Level x Fi = the design force applied at Level i wi = the weight tributary to Level i wpx = the weight tributary to the diaphragm at Level x The minimum diaphragm force (or “shall not be less than” force) at any level is: Fpx = 0.2SDSIewpx

(12.10-2)

And the maximum required diaphragm force (or “need not exceed” force) at any level is: Fpx = 0.4SDSIewpx

(12.10-3)

K. Dirk Bondy & Bryan Allred

27

Example #1 – 5 Story Steel Office Building with Steel Moment Frames at Cal Poly, San Luis Obispo Given: Base Shear = 737 kips (from Example #2 of Chapter 2) Seismic weight per level shown below Fundamental Period = 1.15s SDS = 0.79 (from Example #2 of Chapter 2) Ie = 1.00 (from Example #2 of Chapter 2)

Find: a)

Vertical Distribution of the Base Shear using the Equivalent Lateral Force Procedure of ASCE 7

b)

The Required Diaphragm Design Force at Each Level

a)

Vertical Distribution of the Base Shear using the Equivalent Lateral Force Procedure of ASCE 7

12.8

11

12.8

12

and,

∑

28

Seneca Seismic Analysis and Design Principles and Practice

Determine the Exponent, k The fundamental period of the structure is 1.15s, which is greater than 0.5. Therefore, according to ASCE 7 Section 12.8.3: 1.15 0.5 2 1 1.32 2.5 0.5 Set up a table for the vertical distribution calculations. 1

Level

b)



Roof

w (kips) 2,505

h (ft.) 68.0

5th

3,132

55.0

4th

3,132

42.0

3rd

3,132

29.0

2nd

3,132

16.0

hk

whk

(68.0)1.32 = 265.9 (55.0)1.32 = 200.8 (42.0)1.32 = 140.5 (29.0)1.32 = 86.1 (16.0)1.32 = 39.2

666,088

Cvx

(666,088/2,127,453) = 0.313 628,885 (628,885/2,127,453) = 0.296 440,163 (440,163/2,127,453) = 0.207 269,640 (269,640/2,127,453) = 0.127 122,757 (122,757/2,127,453) = 0.058 =2,127,453 =1.0

The Required Diaphragm Design Force at Each Level

∑ ∑





12.10

1

Fpx(min) = 0.2SDSIewpx

(12.10-2)

Fpx(max) = 0.4SDSIewpx

(12.10-3)

F (kips) 0.313(737) = 231 0.296(737) = 218 0.207(737) = 152 0.127(737) = 93 0.058(737) = 43 = 737

K. Dirk Bondy & Bryan Allred

Set up a table for the diaphragm calculations. Level Roof

w (kips) 2,505

F (kips) 231

5th

3,132

218

4th

3,132

152

3rd

3,132

93

2nd

3,132

43

Fpx (kips) (231/2,505)(2,505) = 231 (449/5,637)(3,132) = 249 (601/8,769)(3,132) = 215 (694/11,901)(3,132) = 183 (737/15,033)(3,132) = 153

Fpx(min) (kips) 0.2(0.79)(1.0)(2,505) = 394 0.2(0.79)(1.0)(3,132) = 493 0.2(0.79)(1.0)(3,132) = 493 0.2(0.79)(1.0)(3,132) = 493 0.2(0.79)(1.0)(3,132) = 493

Fpx(max) (kips) 0.4(0.79)(1.0)(2,505) = 789 0.4(0.79)(1.0)(3,132) = 986 0.4(0.79)(1.0)(3,132) = 986 0.4(0.79)(1.0)(3,132) = 986 0.4(0.79)(1.0)(3,132) = 986

Fpx(design) (kips) 394 493 493 493 493

Notice that, unlike the concrete building in the previous example, the diaphragm design loads are governed in this steel moment frame building by the minimum required Fpx force at every level. There is also no sense in drawing a force and shear diagram. The diaphragm forces are the maximum anticipated at any level at some time during the earthquake, but they are not assumed to be acting simultaneously. The following contains the output from our software, Seneca Seismic Design Values.

29

30

Seneca Seismic Analysis and Design Principles and Practice

SENECA SEISMIC DESIGN VALUES Licensed to: Seneca Structural Engineering, Inc. Project Name: Example #2 - 5 Story Steel Office Building: 1 Grand Ave., San Luis Obispo, CA Project Engineer: K. Dirk Bondy

Date: 8/24/2017

VERTICAL DISTRIBUTION TABLE (12.8.3)

(k per 12.8.3 = 1.32)

Level

w (kips)

Roof 5th 4th 3rd 2nd

 

F (kips)

h (ft.)

h^k

wh^k

Cvx

2,505

68.00

265.87

666,008

0.313

3,132

55.00

200.79

628,885

0.296

3,132

42.00

140.54

440,163

0.207

3,132

29.00

86.09

269,640

0.127

3,132

16.00

39.19

122,757

0.058

231 218 152 93 43

Sum = 2,127,453

1.00

737

 

 

 

DIAPHRAGM DESIGN LOADS (12.10.1.1) Level

W (kips)

F (kips)

Fpx (kips)

Roof 5th 4th 3rd 2nd

2,505

231

231

394

789

3,132

218

249

493

986

3,132

152

215

493

986

3,132

93

183

493

986

3,132

43

153

493

986

-2-

0.2*Sds*I*Wpx (kips)

0.4*Sds*I*Wpx (kips)

F(diaphragm) (kips)

Seneca Software Solutions

394 493 493 493 493

K. Dirk Bondy & Bryan Allred

31

Example #2 – 5 Story Residential Flat Plate with Concrete Shearwalls in Los Angeles Given: Base Shear = 3,868 kips (from Example #1 of Chapter 2) Seismic weight per level shown below Fundamental Period = 0.43s SDS = 1.00 (from Example #1 of Chapter 2) Ie = 1.00 (from Example #1 of Chapter 2)

Find: a)

Vertical Distribution of the Base Shear using the Equivalent Lateral Force Procedure of ASCE 7.

b)

The Required Diaphragm Design Force at Each Level.

a)

Vertical Distribution of the Base Shear using the Equivalent Lateral Force Procedure of ASCE 7

12.8

11

12.8

12

and,

∑

Determine the Exponent, k The fundamental period of the structure is 0.43s, which is less than 0.5. Therefore, according to ASCE 7 Section 12.8.3, k = 1.

32

Seneca Seismic Analysis and Design Principles and Practice

Set up a table for the vertical distribution calculations.

Level Roof

w (kips) 3,922

h (ft.) 60.0

5th

4,810

49.5

4th

4,810

39.0

3rd

4,810

28.5

2nd

4,810

18.0

hk

whk

Cvx

(60.0)1.0 = 60.0 (49.5)1.0 = 49.5 (39.0)1.0 = 39.0 (28.5)1.0 = 28.5 (18.0)1.0 = 18.0

235,320

(235,320/884,670) = 0.266 (238,095/884,670) = 0.269 (187,590/884,670) = 0.212 (137,085/884,670) = 0.155 (86,580/884,670) = 0.098 =1.0

238,095 187,590 137,085 86,580 =884,670

b)

F (kips) 0.266(3,868) = 1,029 0.269(3,868) = 1,041 0.212(3,868) = 820 0.155(3,868) = 599 0.098(3,868) = 379 = 3,868

The Required Diaphragm Design Force at Each Level

∑ ∑





12.10

1

Fpx(min) = 0.2SDSIewpx

(12.10-2)

Fpx(max) = 0.4SDSIewpx

(12.10-3)

Set up a table for the diaphragm calculations.

Level Roof

w (kips) 3,922

F (kips) 1,029

5th

4,810

1,041

4th

4,810

820

3rd

4,810

599

2nd

4,810

379

Fpx (kips) (1,029/3,922)(3,922) = 1,029 (2,070/8,732)(4,810) = 1,140 (2,890/13,542)( 4,810) = 1,027 (3,489/18,352)(4,810) = 915 (3,868/23,162)(4,810) = 803

Fpx(min) (kips) 0.2(1.0)(1.0)(3,922) = 784 0.2(1.0)(1.0)(4,810) = 962 0.2(1.0)(1.0)(4,810) = 962 0.2(1.0)(1.0)(4,810) = 962 0.2(1.0)(1.0)(4,810) = 962

Fpx(max) (kips) 0.4(1.0)(1.0)(3,922) = 1,569 0.4(1.0)(1.0)(4,810) = 1,924 0.4(1.0)(1.0)(4,810) = 1,924 0.4(1.0)(1.0)(4,810) = 1,924 0.4(1.0)(1.0)(4,810) = 1,924

The following contains the output from our software, Seneca Seismic Design Values.

Fpx(design) (kips) 1,029 1,140 1,027 962 962

K. Dirk Bondy & Bryan Allred

SENECA SEISMIC DESIGN VALUES Licensed to: Seneca Structural Engineering, Inc. Project Name: Example #1 - 5 Story Residential Flat Plate in Los Angeles Project Engineer: K. Dirk Bondy

Date: 9/7/2017

VERTICAL DISTRIBUTION TABLE (12.8.3)

(k per 12.8.3 = 1.00)

Level

w (kips)

  GS2.6 GS2.5 GS2.4 GS2.3 GS2.2

 

h (ft.)

h^k

wh^k

Cvx

 

F (kips)

3,922

60.00

60.00

235,320

0.266

4,810

49.50

49.50

238,095

0.269

4,810

39.00

39.00

187,590

0.212

4,810

28.50

28.50

137,085

0.155

4,810

18.00

18.00

86,580

0.098

1,029 1,041 820 599 379

Sum = 884,670

1.00

3,868

 

 

 

DIAPHRAGM DESIGN LOADS (12.10.1.1) Level

W (kips)

F (kips)

Fpx (kips)

GS2.6 GS2.5 GS2.4 GS2.3 GS2.2

3,922

1,029

1,029

784

1,569

4,810

1,041

1,140

962

1,924

4,810

820

1,027

962

1,924

4,810

599

915

962

1,924

4,810

379

803

962

1,924

 

-2-

0.2*Sds*I*Wpx (kips)

0.4*Sds*I*Wpx (kips)

F(diaphragm) (kips)

Seneca Software Solutions

1,029 1,140 1,027 962 962

33

34

Seneca Seismic Analysis and Design Principles and Practice

4 Base Shear and Vertical Distribution Using Modal Response Spectrum Analysis In no way is this book intended to replace your dynamics book. Our intent is to have you take what you learned in your dynamics class and apply it practically to the seismic design of a building, as you would in a design office. We tell our own college students and younger school children when we are asked to speak at high schools and middle schools that engineers are not mathematicians, nor are we physicists. Parents typically believe that because their young son or daughter "just loves math and science" that they will naturally be great engineers. Parents seem to hate hearing that in our opinion, the boy in the garage taking apart the microwave to see how it works, or the girl who asks for a soldering iron for her birthday is more likely to be successful as an engineer than the kid who won the math competition. That probably explains why we have not been asked back to speak to the middle school or high school students in a while. Engineers leave the high-level math to the mathematicians and the high-level physics to the physicists. An engineer's job is to take the results of what the mathematicians and physicists give us and apply those results in a practical and hopefully economical solution to a societal "problem." In this case, the problem is how to safely and economically design a building to withstand an earthquake. In the following section, we will do a minor review of dynamic theory as it applies to mode shapes and frequencies of vibration, but for anything beyond that you will need to consult Clough and Penzien.

The Basics of Free Vibration in an Undamped System

The classic equation of motion for an undamped freely vibrating single-degree of freedom (SDOF) system, or a system with no forcing function is:

0 where;

m is the mass of the SDOF system; k is the stiffness of the system; x is the displacement of the mass; is the acceleration of the mass, or second derivative of the displacement; 0 represents free vibration, or no forcing function.

K. Dirk Bondy & Bryan Allred

35

In words, this equation states that in a linear spring-type system, a mass experiencing an acceleration (a force) is equal to the stiffness of the spring multiplied by the displacement of the spring (also a force) and these two terms (forces) must be equal and opposite. In order to analyze a multi-degree of freedom (MDOF) system the same equation is used, but in vector or matrix form. Instead of x representing a displacement, it will represent a shape vector that oscillates in a simple harmonic motion represented by a sine curve. The number of oscillations in a given period of time is the frequency of the system's vibration. Therefore, the displacement in the above equation can be written in terms of time as a shape vector multiplied by a sine function with a frequency of oscillation, ;

sin  The acceleration, or second derivative of the displacement vector is:

= -

sin 

Substituting both the displacement vector and the acceleration vector back into the original equation of motion gives:

-

sin 





Dividing all terms by sin(t) and rewriting gives;





This equation represents a classic eigenvalue problem, where;

is an eigenvector, but more familiar to us as a mode shape;

 is an eigenvalue, but more familiar to us as the square of the frequency of vibration corresponding to a particular mode shape, which can then be related to a period of vibration, T. There are as many solutions to this problem as there are mass degrees of freedom in a lumped mass system, and there is only one corresponding eigenvector (mode shape) for each eigenvalue (squared frequency) solution. In simple terms, each mode shape is unique and has only one corresponding frequency of vibration, or period of vibration, T (T = 2π/). For our purposes, a three-story building modelled with lumped masses at each level that are limited in a two-dimensional system to translational movement only, will have three mode shape solutions and three corresponding periods.

It is at this point that we hand this problem over to the mathematicians to solve (or a computer) and go to lunch. Once we have the solutions (the mode shapes and periods) then we can go back to work as engineers and use this information in the seismic design of our building.

36

Seneca Seismic Analysis and Design Principles and Practice

The Participation Factor, n and the Participation Function, nn The Participation Factor, n, is defined as a scalar for intended to measure the relative contribution of mode "n" to the total system state, but we will see that this number by itself has very little meaning because it depends upon the arbitrary numbers used to define the mode shapes. The participation factor, 



∑ ∑

 

where; i = the level n = the mode number N = the total number of levels (lumped masses) mi = the mass at level i in = the mode shape value at level i in mode n.

The Participation Factor for a particular mode, when multiplied by the mode shape itself, provides the Participation Function, and this is the vector that is critical to the future calculations. It is important to note that regardless of the arbitrary original numbers used to describe the mode shape, the Participation Function will always result in the same vector because the Participation Factor will change based upon the numbers used to describe the shape. It appears that Bryan and I are the only remaining living people who use the term Participation Function. If you Google the term, you will probably find that nothing comes up, and we could not find the term in any of the newer dynamics books. In our opinion, dynamic analysis is now presented in a much more complicated fashion than it was back in the good old days, though nothing has really changed since I learned it in the 1980’s (and Bryan in the 1990’s). We like the way we learned it, and therefore we will present it that way. While we will analyze each mode in a MDOF system as a SDOF (single mass) response, it is the Participation Function vector that will provide the conversion that determines the response of each mass in the mode. Again, the Participation Function for mode n is the participation factor for that mode multiplied by the mode shape vector itself. Participation Function = nn. In some old books this is referred to as n. An interesting and useful fun fact about the Participation Function vectors is that when they are all added together a unity vector is created. In mathematical terms,

  where in this equation N equals the number of modes, and in our cases, will also equal the number of levels. In words, this means that when all modes are considered, 100% of the total response has been accounted for. While this calculation in itself will not aid in determining the modal forces, shears, etc. it is a very good check to determine if you have made any math mistakes up to this point.

K. Dirk Bondy & Bryan Allred

37

Modal Mass Participation Each mode will have a corresponding effective mass associated with it, which is a percentage of the total mass of the structure. The sum of all modal effective masses equals the total mass of the structure. The effective mass participating in Mode n is calculated as, ∑ ∑

 



This is the same equation as the Participation Factor, but with the numerator squared. ASCE 7 Section 12.9.1 states that the “analysis shall include a sufficient number of modes to obtain a combined modal mass participation of at least 90 percent of the actual mass in each of the orthogonal horizontal directions of response considered by the model.” This does not mean that 10% of the weight of the building can be ignored; it is simply the Code’s way of limiting the number of modes that are necessary in the analysis to only the significant modes. In other words, if a mode has a very small mass associated with it then its participation in the response to the earthquake is assumed to be negligible.

Determine the Forces and Story Shears for Each Individual Mode Each mode and its corresponding period of vibration will be analyzed individually as a SDOF system using a response spectrum. The forces will be applied to the masses in each mode based upon the following equation:

 

Where; = the force at level i for mode n is the spectral acceleration of a SDOF system with a period corresponding to mode n mi = the mass at level i

  is the participation function for mode n The Response Spectrum A response spectrum is a graph generated for a specific site by a geotechnical engineer that plots the maximum accelerations experienced by a SDOF mass for a spectrum of periods of vibration based upon that site’s specific characteristics. When a site specific response spectrum has not been generated by the geotechnical engineer for the site, the ASCE Standard provides a method for generating a design response spectrum for modal response spectrum analyses. This is described in ASCE 7 Section 11.4.5 and shown in Figure 11.4-1. That figure is replicated on the next page. Note that we have only included the graph up to a period of TL. In the Western United States, the value of “long period” is at least 6 seconds and is typically greater. Neither of us has ever been involved with the design of a structure with a period that exceeded the TL value of ASCE 7, and in all likelihood you will never either. Therefore, we are going to simplify the graph by eliminating the long period acceleration portion.

38

Seneca Seismic Analysis and Design Principles and Practice

ASCE 7 Figure 11.4-1 Design Response Spectrum Where;

0.2

& SDS, SD1 were described previously. Once the forces and story shears are calculated for each significant mode, the story shears must be combined using some rational method. In this book we will combine the story shears using the square root of the sum of the squares (SRSS) method, though engineers will sometimes use the complete quadratic combination (CQC) method. The CQC method claims to provide more accuracy for combining high frequency modes (short period modes) that have frequencies close to each other. This is rarely the case in most of the buildings that are built. Most buildings constructed are in the mid-rise range or shorter and will satisfy the 90% mass requirement within the first two or three translational modes. The mass participation in the higher modes is typically negligible. Therefore, how those higher modes are combined, if they are used at all, in the overall combination of modes makes little difference. ASCE 7 Section 12.9.2 states that the forces calculated from the Design Response Spectrum shall be divided by the quantity (R/Ie) in order to produce the design level forces.

Scaling the Dynamic Base Shear The base shear that is found using the modal response spectrum analysis will typically be less than the base shear calculated using the Equivalent Lateral Force (ELF) Procedure of ASCE 7 Section 12.8, and it will typically be less than 85% of the ELF base shear. When the base shear calculated using the dynamic analysis is less than 85% of the base shear calculated using the ELF procedure, ASCE 7 Section 12.9.4.1 requires that the dynamic base shear must be scaled up to 85% of ELF procedure base shear.

K. Dirk Bondy & Bryan Allred

39

Example #1 – 5 Story Steel Office Building with Steel Moment Frames at Cal Poly, San Luis Obispo Given:  Base shear using the ELF procedure = 737 kips (from Example #2 of Chapter 2)  Seismic weight per level shown below  Mode shapes and periods from a frame analysis program shown below (equal in both orthogonal directions)  Frame types shown below

40

Seneca Seismic Analysis and Design Principles and Practice

K. Dirk Bondy & Bryan Allred

41

Find: a)

Normalize the mode shapes such that the largest value for each mode is +1.0. Maintain three decimal places for each value.

b)

Using a two-dimensional frame analysis computer program, model all of the frames in one direction in series connected by rigid links with hinged connections. Approximate the first mode period using the Rayleigh Method and compare that to the Mode 1 period determined by dynamic analysis.

c)

Determine the Participation Function for each mode, and then sum the Participation Functions for all modes to verify that a unity vector is created. For the first mode only, recalculate the Participation Function using the original mode shape values given (not the normalized values) and show that the same Participation Function is calculated either way.

d)

Determine the mass (weight) participating in each mode. Sum the individual modal masses to verify that 100% of the total mass is accounted for by using all modes. Determine which modes only are necessary to satisfy the requirement that 90% of the modal mass be included in the modal response spectrum analysis.

e)

Create a response spectrum for the site based upon the site geotechnical properties given previously. Plot the graph for periods up to 2.0 seconds.

f)

Calculate the modal forces and story shears for all of the modes. Combine the story shears using the SRSS method and extract the vertical force distribution that creates that story shear envelope. Scale all of the story forces and base shear by dividing the forces obtained using the Design Response Spectrum by (R/Ie) and then scale the base shear and corresponding story forces back up to 85% of the ELF procedure base shear.

g)

Repeat f), but only use the modes that are required to account for 90% of the modal mass. Compare the forces and shears in this analysis side by side with the analysis using all modes.

h)

Combine the story forces (not shears) using the SRSS method using only the modes necessary to capture 90% of the mass to obtain maximum diaphragm forces. Scale the results by dividing each force by (R/Ie) and then multiplying each force by a scale factor equal to the ELF base shear divided by the base shear obtained in g) before it was scaled to 85% of the ELF base shear. Compare these results side by side to the diaphragm results obtained using ASCE 7 Equation (12.10-1) found previously.

42

a)

Seneca Seismic Analysis and Design Principles and Practice

Normalize the mode shapes such that the largest value for each mode is +1.0. Maintain three decimal places for each value.

This is entirely unnecessary numerically, but for some reason it is standard practice (or used to be in the good old days, at least) to “normalize” each shape to make all the values relative to 1.0. But we want to do it here for a different reason, which will become clear in Part c). b)

Using a two-dimensional frame analysis computer program, model all of the frames in one direction in series connected by rigid links with hinged connections. Approximate the first mode period using the Rayleigh Method and compare that to the Mode 1 period determined by dynamic analysis.

2

∑ ∑

 





The Rayleigh Period method previously existed in the Uniform Building Code as the “Method B” period. It was a completely valid and rational estimate of the building’s fundamental period, and was very useful to us young engineers, particularly when designing steel moment frame buildings. The reason that it was useful was that we are able to, without much additional work and using the frames that we had already modelled anyway, show that we could use the maximum Code allowed fundamental period when calculating the base shear of the building. The only thing that we were actually doing was demonstrating that the Rayleigh Period (Method “B” Period) was longer than the maximum allowed fundamental period. Any base shear may be applied to the building, but it should be applied using the same vertical distribution used for the Equivalent Lateral Load procedure. We could have used the base shear previously calculated, but it is common to use a nice round number, such as 1,000 kips for the analysis. The following demonstrates how we determine the Rayleigh Period using a 1,000 kip base shear.

K. Dirk Bondy & Bryan Allred

43

The results of this analysis are shown below.

Level

wi (kips) 2,505 3,132 3,132 3,132 3,132

Roof 5th 4th 3rd 2nd

2

∑ ∑

fi (kips) 313 296 207 127 58

i (in.) 2.366 2.036 1.545 1.095 0.608

i2

(in.)2 5.598 4.145 2.387 1.199 0.370

wii2 (kips-in2) 14,023 12,983 7,476 3,755 1,158 =39,395

fii (kip-in.) 740.6 602.7 319.8 139.1 35.3 =1,837.4

 

with g = 386.4 in/s2

2

39,395 / 386.4 1,837.4

1.48

This compares extremely well with the dynamic first mode period of 1.45s. The most important point is that we could have proven that we may use the maximum allowed value of CuTa = 1.15s as the fundamental period (see Chapter 2) when calculating the base shear without actually performing a dynamic analysis.

44

c)

Seneca Seismic Analysis and Design Principles and Practice

Determine the Participation Function for each mode, and then sum the Participation Functions for all modes to verify that a unity vector is created. For the first mode only, recalculate the Participation Function using the original mode shape values given (not the normalized values) and show that the same Participation Function is calculated either way. Participation Function = nn

We must first determine the participation factor. Remember, this factor is really meaningless by itself, as it depends entirely on the values that we used to describe the mode shape. 1st Mode Participation Factor

The participation factor, 

Level Roof 5th 4th 3rd 2nd





mi (kips) 2,505 3,132 3,132 3,132 3,132



 

∑

ϕi 2 (unitless) 1.0 0.778 0.473 0.251 0.082

ϕi (unitless) 1.0 0.882 0.688 0.501 0.286

9,887.1 7,466.3

∑

mi ϕ i (kips) 2,505.0 2,762.4 2,154.8 1,569.1 895.8 =9,887.1

mi ϕi2 (kips) 2,505.0 2,436.5 1,482.5 786.1 256.2 =7,466.3

1.324

1st Mode Participation Function = 11

1.0 0.882 1.324 0.688 0.501 0.286



. . . . .



This next step is not necessary, but just for fun we will redo the above calculations using the originally given mode shape numbers. The point of this exercise is to demonstrate that while the numbers used to describe the mode shape and the corresponding participation factor can vary, demonstrating that their values are meaningless in and of themselves, the Participation Function values are constant for a particular mode and are truly a distinct property of the mode itself. This is the money vector as it contains the numbers that truly matter in the modal force calculations.

K. Dirk Bondy & Bryan Allred



Level

mi (kips)

Roof 5th 4th 3rd 2nd

2,505 3,132 3,132 3,132 3,132



Original ϕi (unitless) 0.619 0.546 0.426 0.310 0.177

6,120.2 2,860.5

Original ϕi 2 (unitless) 0.383 0.298 0.182 0.096 0.031

mi ϕi (kips)

mi ϕi2 (kips)

1,550.6 1,710.1 1,334.2 970.9 554.4 =6,120.2

959.4 933.3 570.0 300.7 97.1 =2,860.5

45

2.139

1st Mode Participation Function = 11 0.619 0.546 2.139 0.426 0.310 0.177



. . . . .



This is, of course, the same Participation Function as previously calculated using the normalized mode shape. 2nd Mode Participation Factor The participation factor, 

Level Roof 5th 4th 3rd 2nd





mi (kips) 2,505 3,132 3,132 3,132 3,132

3,242.1 7,302.7



ϕi (unitless) 1.0 0.243 -0.594 -0.839 -0.645

∑ ∑

 

ϕi 2 (unitless) 1.0 0.059 0.353 0.704 0.416

0.444

2nd Mode Participation Function = 22

mi ϕi (kips) 2505.0 761.1 -1860.4 -2627.7 -2020.1 =-3,242.1

mi ϕi2 (kips) 2505.0 184.9 1105.1 2204.7 1303.0 =7,302.7

46

Seneca Seismic Analysis and Design Principles and Practice

1.0 0.243 0.444 0.594 0.839 0.645



. . .

. .



3rd Mode Participation Factor The participation factor, 

Level Roof 5th 4th 3rd 2nd





mi (kips) 2,505 3,132 3,132 3,132 3,132



 

∑

ϕi 2 (unitless) 0.759 0.542 0.607 0.162 1.000

ϕi (unitless) 0.871 -0.736 -0.779 0.402 1.000

1,828.0 9,135.7

∑

mi ϕi (kips) 2181.9 -2305.2 -2439.8 1259.1 3132.0 =1,828.0

mi ϕi2 (kips) 1900.4 1696.6 1900.6 506.1 3132.0 =9135.7

mi ϕi (kips) -1427.9 3132.0 -2067.1 -1196.4 2421.0 =861.6

mi ϕi2 (kips) 813.9 3132.0 1364.3 457.0 1871.5 =7,638.7

0.200

3rd Mode Participation Function = 33 0.871 0.736 0.200 0.779 0.402 1.0

. . .

. .



4th Mode Participation Factor The participation factor, 

Level Roof 5th 4th 3rd 2nd

mi (kips) 2,505 3,132 3,132 3,132 3,132

ϕi (unitless) -0.570 1.000 -0.660 -0.382 0.773



∑ ∑

 

ϕi 2 (unitless) 0.325 1.000 0.436 0.146 0.598

K. Dirk Bondy & Bryan Allred





861.6 7,638.7

47

0.113

4th Mode Participation Function = 44

0.570 1.0 0.113 0.660 0.382 0.773 th 5 Mode Participation Factor

.



.

.

The participation factor, 





∑



∑

mi

ϕi

ϕi 2

mi ϕ i

mi ϕi2

(kips)

(unitless)

(unitless)

(kips)

(kips)

Roof

2,505

-0.126

0.016

-315.6

39.8

5th

3,132

0.335

0.112

1049.2

351.5

4th

3,132

-0.744

0.554

-2330.2

1733.7

3rd

3,132

1.000

1.000

3132.0

3132.0

nd

3,132

-0.647

0.419

-2026.4

1311.1

=-491.0

=6,568.1

Level

2





. .



491.0 6,568.1

0.075

5th Mode Participation Function = 55

0.126 . 0.335 . 0.075 0.744 . 1.0 . 0.647 . Show that the Sum of All the Participation Function Vectors Equals a Unity Vector 1.324 0.444 0.174 0.064 0.009 1.168 0.108 0.147 0.113 0.025 0.911 + 0.264 + 0.156 + 0.075 + 0.056 = 0.663 0.373 0.080 0.043 0.075 0.379 0.286 0.200 0.087 0.049 d)

. . . . .



Determine the mass (weight) participating in each mode. Sum the individual modal masses to verify that 100% of the total mass is accounted for by using all modes. Determine which modes only are necessary to satisfy the requirement that 90% of the modal mass be included in the modal response spectrum analysis.

48

Seneca Seismic Analysis and Design Principles and Practice

The effective mass participating in an individual mode is calculated using the following formula:



∑



∑

equal to weight when gravity is constant

But notice that this is the same equation as the Participation Factor, but with the numerator squared. Therefore all of the numbers have already been calculated. 1st Mode Effective Mass



∑



∑



9,887.1 7,466.3



,





13,093 15,033

100



,





1,440 15,033

100

. %

. %

2nd Mode Effective Mass



∑



∑



3,242.1 7,302.7



3rd Mode Effective Mass

∑

 

∑



1,828.0 9,135.7







366 15,033

100

. %

4th Mode Effective Mass

∑

 

∑



861.6 7,638.7







97 15,033

100

. %

5th Mode Effective Mass

∑ ∑

 

491.0 6,568.1







37 15,033

100

. %

K. Dirk Bondy & Bryan Allred

49

Verify that the Sum of All the Effective Masses Equals the Mass (Weight) of the Structure

13,093

1440

366

97

37

,





Determine How Many Modes are Necessary to Account for 90% of the Total Mass It is clear that the first two modes account for 96.7% of the mass of the structure. Therefore, ASCE 7 Section 12.9.1 only requires that Modes #1 and #2 be included in the modal response spectrum analysis. e)

Create a response spectrum for the site based upon the site geotechnical properties given previously. Plot the graph for periods up to 2.0 seconds.

The parameters that are required to construct the response spectrum of ASCE 7 Figure 11.4-1 are the SDS and SD1 values previously calculated. Those were calculated for this site as 0.79g and 0.45g, respectively.

0.2 0.2

0.45 0.79

0.45 0.79

0.11

0.57

Therefore, the ASCE 7 Design Response Spectrum for the steel building site in San Luis Obispo, CA is generated as the following:

50

f)

Seneca Seismic Analysis and Design Principles and Practice

Calculate the modal forces and story shears for all of the modes. Combine the story shears using the SRSS method and extract the vertical force distribution that creates that story shear envelope. Scale all of the story forces and base shear by dividing the forces obtained using the Design Response Spectrum by (R/Ie) and then scale the base shear and corresponding story forces back up to 85% of the ELF procedure base shear.

The force at level i in Mode n is given by the following equation.

 



an0 is determined from the response spectrum at the period corresponding to Mode n.

Determine the Story Forces and Shears for Mode 1 The period of Mode 1 is 1.45s, which is greater than Ts of 0.57s. Therefore, the spectral acceleration, Sa is equal to SD1/T.

0.45 0.31 1.45 Or, by using the graph itself;



 





Story Mass Part. Story Force Story Shear (kips/g) Function (kips) (kips) , . 2,505 1.324 , . , . 3,132 1.168 , . , . 0.31 3,132 . 0.911 , . 3,132 0.663 . , . 3,132 0.379 .

K. Dirk Bondy & Bryan Allred

51

Determine the Story Forces and Shears for Mode 2 The period of Mode 2 is 0.53s, which is less than Ts of 0.57s. Therefore, the spectral acceleration, Sa is equal to SDS = 0.79g. Or, by using the graph itself;



 





Story Mass (kips/g) 2,505 3,132 0.79 3,132 3,132 3,132

Part. Function 0.444 0.108 0.264 0.373 0.286

Story Force Story Shear (kips) (kips) . . , . . . . . . , . .

52

Seneca Seismic Analysis and Design Principles and Practice

Determine the Story Forces and Shears for Mode 3 The period of Mode 3 is 0.30s, which is less than Ts of 0.57s. Therefore, the spectral acceleration, Sa is equal to SDS = 0.79g.







  Story Mass (kips/g) 2,505 3,132 0.79 3,132 3,132 3,132

Part. Function 0.174 0.147 0.156 0.080 0.200

Story Force Story Shear (kips) (kips) . . . . . . . . . .

Determine the Story Forces and Shears for Mode 4 The period of Mode 4 is 0.22s, which is less than Ts of 0.57s. Therefore, the spectral acceleration, Sa is equal to SDS = 0.79g.

K. Dirk Bondy & Bryan Allred



53

 



Story Mass Part. (kips/g) Function 2,505 0.064 3,132 0.113 0.79 3,132 0.075 3,132 0.043 3,132 0.087



Story Force Story Shear (kips) (kips) . . . . . . . . . .

Determine the Story Forces and Shears for Mode 5 The period of Mode 5 is 0.16s, which is less than Ts of 0.57s. Therefore, the spectral acceleration, Sa is equal to SDS = 0.79g.



 





Story Mass (kips/g) 2,505 3,132 0.79 3,132 3,132 3,132

0.009 0.025 0.056 0.075 0.049

Story Force Story Shear (kips) (kips) . . . . . . . . . .

54

Seneca Seismic Analysis and Design Principles and Practice

Combine the Story Shears Using the SRSS Method and Extract the Story Forces V5th = 1,028.2 V4th = 2,162.2 V3rd = 3,046.7 V2nd = 3,690.4 VBase = 4,058.4

878.7 1,145.9 492.7 430.2 1,137.8

344.3 19.4 405.4 207.5 287.4

126.7 152.9 32.7 139.1 76.2

17.8 , . 44.1 , . 94.5 , . 91.1 , . 30.1 , .

Extract the Level Forces from the Story Shears. Show as a proportion of the base shear. FRoof F5th = 2,452.3 - 1,401.5 kips F4th = 3,114.4 - 2,452.3 F3rd = 3,724.9 - 3,114.4 F2nd = 4,225.5 - 3,724.9

= 1,401.5 kips = 1,050.8 kips = 662.1 kips = 610.5 kips = 500.6 kips

(1,401.5 / 4,225.5 = 0.332V) (1,050.8 / 4,225.5 = 0.249V) (662.1 / 4,225.5 = 0.157V) (610.5 / 4,225.5 = 0.144V) (500.6 / 4,225.5 = 0.118V)

K. Dirk Bondy & Bryan Allred

55

ASCE 7 Section 12.9.2 states that the forces calculated from the Design Response Spectrum shall be divided by the quantity (R/Ie) in order to produce the design level seismic forces. However, ASCE 7 Section 12.9.4.1 also states that the final design base shear must not be less than 85% of the base shear determined using the ELF procedure. The dynamic base shear divided by (R/Ie) = 4,225.5 kips / [8/1.0] = 528.2 kips. However, 85% of the ELF procedure base shear is 0.85(737 kips) = 627 kips and this will control the design. Therefore, the final design forces using the Modal Response Spectrum Analysis considering all modes are: FRoof = 0.332(627 kips) = 208.2 kips F5th = 0.249(627 kips) = 156.1 kips F4th = 0.157(627 kips) = 98.4 kips F3rd = 0.144(627 kips) = 90.3 kips F2nd = 0.118(627 kips) = 74.0 kips  = 627.0 kips g)

Repeat f), but only use the modes that are required to account for 90% of the modal mass. Compare the forces and shears in this analysis side by side with the analysis using all modes.

We demonstrated previously that the first two modes account for well over 90% of the seismic mass. Therefore, we will repeat the modal combination using the SRSS method, but only with the first and second modes. Combine Only the First and Second Story Shears Using the SRSS Method and Extract the Story Forces V5th = 1,028.2 V4th = 2,162.2 V3rd = 3,046.7 V2nd = 3,690.4 VBase = 4,058.4

878.7 , . 1,145.9 , . 492.7 , . 430.2 , . 1,137.8 , .

Extract the Level Forces from the Story Shears. Show as a proportion of the base shear. FRoof F5th = 2,447.1 - 1,352.5 kips F4th = 3,086.3 - 2,447.1 F3rd = 3,715.4 - 3,086.3 F2nd = 4,214.9 - 3,715.4

= 1,352.5 kips = 1,094.6 kips = 639.2 kips = 629.1 kips = 499.5 kips

(1,352.5 / 4,214.9 = 0.321V) (1,094.6 / 4,214.9 = 0.260V) (639.2 / 4,214.9 = 0.152V) (629.1 / 4,214.9 = 0.149V) (499.5 / 4,214.9 = 0.118V)

As before, ASCE 7 Section 12.9.2 states that the forces calculated from the Design Response Spectrum shall be divided by the quantity (R/Ie) in order to produce the design level seismic forces. ASCE 7 Section 12.9.4.1 also states that the final design base shear must not be less than 85% of the base shear determined using the ELF procedure. The dynamic base shear divided by (R/Ie) = 4,214.9 kips / [8/1.0] = 526.9 kips.

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Seneca Seismic Analysis and Design Principles and Practice

However, 85% of the ELF procedure base shear is 0.85(737 kips) = 627 kips and this will control the design just as it did when we considered all modes. Therefore, there is no difference in design base shear in this example (or typically ever) between an analysis that considers all modes, or only the modes necessary to account for 90% of the structure's dynamic mass. Therefore, the final design forces using the Modal Response Spectrum Analysis considering only the first and second modes are: FRoof F5th F4th F3rd F2nd

= 0.321(627 kips) = 201.3 kips = 0.260(627 kips) = 163.0 kips = 0.152(627 kips) = 95.3 kips = 0.149(627 kips) = 93.4 kips = 0.118(627 kips) = 74.0 kips  = 627.0 kips

The results of this exercise demonstrate that the use of the higher modes tends to move the resultant force higher up the building. In previous codes, for longer period buildings a portion of the base shear was applied as a single force, Ft, at the top level of the building to account for the higher mode effects in the Equivalent Lateral Force procedure. In the current code that effect is accounted for with the use of the "k" factor in the vertical distribution equation:

12.8

11

12.8

12

and,

∑

In the late 1980's and early 1990’s, using an IBM 386 machine, it would often take a number of minutes, not quite hours, but a noticeable amount of time to run a modal response spectrum analysis. Using only the first few modes would significantly reduce the computer run-time, so designers would often choose this option. Today, a computer can run a large dynamic problem using all modes in a matter of seconds. Therefore, it is common to simply have the computer use all modes. However, the wiser (read “older”) folks realize that it is beneficial to the design to only use the modes that account for 90% of the mass because that analysis will result in lower overturning moments, deflections and foundation loading.

K. Dirk Bondy & Bryan Allred

h)

57

Combine the story forces (not shears) using the SRSS method using only the modes necessary to capture 90% of the mass to obtain maximum diaphragm forces. Scale the results by dividing each force by (R/Ie) and then multiplying each force by a scale factor equal to the ELF base shear divided by the base shear obtained in g) before it was scaled to 85% of the ELF base shear. Compare these results side by side to the diaphragm results obtained using ASCE 7 Equation (12.10-1) found previously.

We previously found that Modes 1 & 2 were the only modes necessary to capture over 90% of the building mass. The story forces for each mode are shown below. 1,028.2 878.7 1,134.0 267.2 653.2 884.5 643.7 922.9 368.0 707.6 The SRSS of the forces at each level are: FRoof = 1,028.2 F5th = 1,134.0 F4rd = 884.5 F3rd = 643.7 F2nd = 368.0

878.7 , . 267.2 , . 653.2 , . 922.9 , . 707.6 .

Divide these forces by R/Ie to obtain the response spectrum design diaphragm forces. FRoof = 1,352.5 kips / (8/1.0) = 169.1 kips F5th = 1,165.1 kips / (8/1.0) = 145.6 kips F4th = 1,099.6 kips / (8/1.0) = 137.5 kips F3rd = 1,125.2 kips / (8/1.0) = 140.7 kips F2nd = 797.6 kips / (8/1.0) = 99.7 kips In the story shear analysis we were required to scale the results to 85% of the ELF base shear. The base shear from the response spectrum analysis using only Modes 1 & 2 was found to be 526.9 kips. It follows that we could use the same scale factor for the diaphragm forces. However, to be conservative and to be able to compare “apples to apples” we will scale these forces to the full ELF base shear of 737 kips. Therefore, the scale factor will be 737 kips / 526.9 kips = 1.40. The modal response spectrum diaphragm forces that we will use to compare to the ELF diaphragm forces are: FRoof = 169.1 kips x 1.40 = 236.7 kips F5th = 145.6 kips x 1.40 = 203.8 kips F4th = 137.5 kips x 1.40 = 192.5 kips F3rd = 140.7 kips x 1.40 = 197.0 kips F2nd = 99.7 kips x 1.40 = 139.5 kips

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Seneca Seismic Analysis and Design Principles and Practice

The comparison of the ELF diaphragm forces and the modal response spectrum forces is shown below. Level Roof 5th 4th 3rd 2nd

Equivalent Lateral Force Procedure Diaphragm Fpx Values (kips) 231 249 215 183 153

Modal Response Spectrum Diaphragm Forces (kips) 236.7 203.8 192.5 197.0 139.5

This was really an academic exercise to demonstrate how the diaphragm force can be calculated using a modal response spectrum analysis, and how this compares to the ASCE 7 simplified diaphragm equation. If we included all of the modes we would see that the dynamic diaphragm forces increase slightly at every level. In actuality for this example however, the diaphragm forces at every level are governed by the minimum diaphragm force, just as they were in the ELF calculations: Fpx(min) = 0.2SDSIewpx

(12.10-2)

K. Dirk Bondy & Bryan Allred

59

5 Rigid Diaphragm Lateral Load Distribution In the previous two chapters we developed the vertical distribution of the seismic base shear using both the Equivalent Lateral Force and modal response spectrum procedures. Those methods were based upon two-dimensional analyses considering the earthquake in one of the building's two primary orthogonal directions. Those forces now must be placed on each respective diaphragm and distributed to the vertical lateral load resisting elements (the shearwalls, moment frames, etc.). In this book we are limiting our discussions to rigid diaphragms (concrete or concrete filled metal decks). Flexible diaphragms such as wood and metal deck without concrete fill are simpler to analyze and the loads that they deliver to the vertical seismic elements are typically based upon tributary seismic area to each vertical seismic resisting element.

The Diaphragm - What is it, Exactly? The diaphragm is best thought of as a membrane that is rigid in its own plane. It is similar to a horizontal concrete shearwall but it laterally connects to and ties all of the seismic and non-seismic elements together such that their spacing from each other in two horizontal dimensions remains constant during a seismic event. The theory is that all of the elements attached to the diaphragm will move with the diaphragm as it laterally translates and torsionally rotates as a rigid body.

This figure shows a very stiff in-plane concrete seismic shearwall, a flexible concrete column and a metal stud wall with no continuity or stiffness at all. All three elements will maintain their connection points to each diaphragm as the diaphragms translate and rotate during an earthquake, therefore maintaining the same horizontal distance from each other in all directions. The stiff shearwall will experience primarily shear deformations while the flexible concrete column will experience primarily bending deformations, and the metal stud wall will simply rack. In our seismic analyses, we will assume that only the seismic elements resist the lateral loads because the stiffness of the seismic system will be much greater than any other non-seismic element, and the stiffer the element the more load will be required to displace it. However, it is important to realize that every element attached to a diaphragm must be capable of experiencing all the same lateral movements that the seismic system is experiencing without collapsing. Once this becomes clear, most engineers

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Seneca Seismic Analysis and Design Principles and Practice

arrive at the conclusion that a seismic system that limits the lateral story drifts (the displacement between diaphragms from one story to the next) to very manageable amounts for all the elements within the structure is often the preferred seismic system for that structure.

Determining the Rigidity of the Lateral Load Resisting Elements The inertial seismic loads will be distributed to the lateral load resisting elements through the rigid diaphragm translation and rotation based upon the stiffness of the lateral load resisting elements. The first step is to determine the stiffness, or rigidity of those elements. The most accurate way to do this without modeling the building in three dimensions would be to link in series all of the seismic load resisting elements and apply a force distribution to the system matching the force distribution percentages previously calculated. The sum of the shears in the vertical elements of a frame or wall in a given story divided by the total shears for all frames or walls at that story would represent the relative rigidity of that frame or wall at that story.

This is somewhat more complicated than we want to demonstrate in this book, and typically unnecessary unless the building has dramatic stiffness differences between elements or an unusual configuration where not all elements connect to all levels. At that point, a three-dimensional model would be advisable. We are going to assume that the building is fairly regular in stiffness distribution with connectivity to all floors. We will define the rigidity of an element as the force required to displace the element a unit value at the level of the force application. It will be necessary to determine the rigidity of each element at each floor level in order to properly calculate the distribution of seismic forces to those elements.

K. Dirk Bondy & Bryan Allred

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The Rigidity of Concrete Shearwalls

The lateral displacement of a cantilevered concrete shearwall under a lateral load is comprised of two components; a shear deformation, and a bending deformation. The shear deformation component is calculated using the following equation, where G is the shear modulus of elasticity:





1.2

1.2 0.4





3

The bending deformation component is determined from the traditional "flagpole" equation:





3

Therefore, the total lateral displacement at any level due to a force, F applied at that level is:





3 3

Since the rigidity is the force per unit displacement, the rigidity of a concrete shearwall at a particular level a distance h above the base can be written as:



∆



1 3

3

We will primarily be interested in only the relative rigidities of the walls (relative to each other) as opposed to the actual rigidities, so if we can simplify this equation that will help simplify our calculations. When the modulus of elasticity, E, is constant for all walls, as it often is, the rigidity equation for a cantilevered wall simplifies to:



∆



1 3

3

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Seneca Seismic Analysis and Design Principles and Practice

The Rigidity of Moment Frames, Braced Frames, Etc.

The simplest way to determine the rigidity of a moment frame or braced frame is to model it in a frame analysis program and place a lateral force at the level of interest and allow the program to calculate the deflection at that level. The rigidity is simply the force divided by the lateral displacement at that level.



∆

Direct Translational Forces and Rigid Body Rotational Forces Once the rigidities of all of the seismic elements have been determined the diaphragms will be analyzed in two conditions; direct translation and rigid body rotation. In both cases the centroid of the diaphragm force will be applied at the center of gravity, or C.G. Direct Translational Forces Direct forces will be calculated with the diaphragm “locked” against rotation. Only direct translation will be allowed, and the direct forces will be distributed to the seismic system based solely on their relative rigidities.

K. Dirk Bondy & Bryan Allred

Direct Forces (All Forces Shown Acting On Diaphragm) Rotational Forces Rotational forces will be calculated by locking the diaphragm this time against translation, and only allowing rotation. The rotation must occur about the center of rigidity, C.R. The torsional moment is equal to the resultant diaphragm force multiplied by the eccentricity between the center of gravity and the center of rigidity. The resultant diaphragm force will act through the center of gravity, but will be resisted through the center of rigidity.

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Seneca Seismic Analysis and Design Principles and Practice

Calculation of Rotational Forces Consider the diaphragm rotating about the center of rigidity, with a rotation angle, .

Rotational Forces (All Forces Shown Acting On Diaphragm) The deflection of any wall, i is: i = di, where di is the perpendicular distance between the wall and the center of rigidity. The force in any wall, i due to rotation only is: Fi = Rii = Ridi By summing moments about the center of rigidity; Fe = MTORS = F1d1+ F2d2+ F3d3 + F4d4 Or in general terms; Fe = MTORS =Fidi Substituting Fi = Ridi Fe = MTORS = R1(d1)2 + R2(d2)2 + R3(d3)2 + R4(d4)2 Or in general terms; Fe = MTORS = [ Ri(di)2] But  is constant, and for any wall, i: ∆

∴



∆

K. Dirk Bondy & Bryan Allred

65

Rewriting in terms of the force, Fi;



∑

ASCE 7 Section 12.8.4.2 Accidental Torsion ASCE 7 requires that the total torsional moment applied to the diaphragm include the inherent torsional moment plus an accidental torsional moment resulting from the shift in the center of mass equal to 5 percent of the dimension of the structure perpendicular to the direction of applied forces. This accidental shift in the center of mass must be applied “in the direction that produces the greater effect.” Therefore, in order to capture the greatest forces resulting from torsion, it may be necessary to shift the center of mass in the positive and negative directions.

ASCE 7 Section 12.8.4.3 Amplification of Accidental Torsional Moment According to ASCE 7, structures assigned to Seismic Design Category C, D, E or F where Type 1a or 1b torsional irregularity exists as defined in Table 12.3-1 shall have the effects accounted for by multiplying the accidental torsional moment at each level by a torsional amplification factor (Ax), where;



 . 

(12.8-14)

Where; max = the maximum displacement at Level x computed assuming Ax = 1 avg = the average of the displacements at the extreme points of the structure at Level x computed assuming Ax = 1

Ax shall not be less than 1 and is not required to exceed 3.0. The simplest way to account for the Ax factor is to multiply it by the accidental eccentricity percent value. In other words, the analysis must include an accidental torsional moment resulting from a shift in the center of mass equal to Ax(5%).

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Seneca Seismic Analysis and Design Principles and Practice

Example #1 – 5 Story Steel Office Building with Steel Moment Frames at Cal Poly, San Luis Obispo Given:

Roof Level Force (Previously Found in Modal Analysis using 1st two modes): 201.3 kips Levels 5 Force (Previously Found in Modal Analysis using 1st two modes): 163.0 kips Recall that the Redundancy Factor, , will not be applied here. It would be applied, if it is equal to 1.3, when sizing the frame members for strength, and it is also used to decrease the allowable drift values in moment frame structures.

K. Dirk Bondy & Bryan Allred

67

Find: a)

Determine the frame rigidities for each frame at each level

b)

Determine the center of gravity and center of rigidity locations

c)

For loading in the North-South direction applied with a positive eccentricity shift, determine the required seismic loads to each frame at the Roof and 5th Levels using the forces found using the modal response spectrum analysis considering only Modes 1 & 2

d)

For the Roof Level, determine if torsional irregularity Type 1a or 1b exists according to ASCE 7 Table 12.3-1

a)

Determine the frame rigidities for each frame at each level

Model the two frame types and apply a 100 kip load at each level, n for frame i. Determine the displacement at that level and calculate the frame rigidity at that level for each frame based upon:

∆

Frame Type 1 Rigidity Loading and Deflections at the Roof and 5th Levels

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Seneca Seismic Analysis and Design Principles and Practice

Frame Type 2 Rigidity Loading and Deflections at the Roof and 5th Levels The following table contains the deflection at each level for each frame under a 100 kip lateral load at that level, as well as the frame rigidities and relative rigidities. The relative rigidities are calculated by setting Frame Type 1 to 1.0 and calculating the corresponding relative rigidity of Frame Type 2. This is only done for convenience as it will simplify the calculations. Level

Frame Type 1,  (in.)

Frame Type 2,  (in.)

Frame Type 1, R (kips/in.)

Frame Type 2, R (kips/in.)

Frame Type 1 Rel. Rigidity

Frame Type 2 Rel. Rigidity

Roof 5th 4th 3rd 2nd

2.539 1.767 1.180 0.770 0.395

1.479 0.957 0.662 0.425 0.226

39.39 56.59 84.75 129.87 253.16

67.61 104.49 151.06 235.29 442.48

1.0 1.0 1.0 1.0 1.0

1.716 1.846 1.782 1.812 1.748

It is important to note that the relative rigidities of the frames vary between levels.

b)

Determine the center of gravity and center of rigidity locations

All locations will be measured from the southwest corner (the bottom left corner) of the diaphragm. Roof Level The East-West, or X coordinate of the center of gravity at the Roof Level is:

X

. .

182 ft 122 ft 2

120 ft 122 ft 122 ft

60 ft

36,844ft

109.1 ft

The North-South, or Y coordinate of the center of gravity at the Roof Level is:

Y.

.

122 ft 182 ft 2

120 ft 122 ft

60 ft

122 ft 2

36,844ft

102.9 ft

The East-West, or X coordinate of the center of rigidity at the Roof Level is:

X

1 ft 1.0 . .

1 ft 1.0

61 ft 1.716 121 ft 1.716 2 1.0 3 1.716

241 ft 1.716

101.8 ft

The North-South, or Y coordinate of the center of rigidity at the Roof Level is:

Y.

1 ft 1.716 .

61 ft 1.716

121 ft 1.716 2 1.0 3 1.716

181 ft 1.0

181 ft 1.0

94.6 ft

K. Dirk Bondy & Bryan Allred

c)

For loading in the North-South direction applied with a positive eccentricity shift, determine the required seismic loads to each frame at the Roof and 5th Levels using the forces found using the modal response spectrum analysis considering only Modes 1 & 2

Roof Level Load Distribution for North-South Forces The moment frames have been labeled below, and the centers of gravity and rigidity are shown.

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Seneca Seismic Analysis and Design Principles and Practice

Direct Forces in the Y Direction

F

F

F

F

. .

F

.

.

.

201.3 kips

. .

.

.

.

.

.

. .

201.3 kips

201.3 kips . .

28.2 kips

201.3 kips

48.3 kips

Rotational Forces for Loading in the Y Direction The force in any wall due to the torsional moment (pure rotation of the diaphragm) includes all of the walls in the system and is found from:

F

M

Rd ∑R d

Where: MTORS = Torsional moment = Fp(XCR - XCG) Ri = Rigidity of wall “i” di = Perpendicular distance between the center of rigidity and wall “i” ASCE 7 requires that the total torsional moment applied to the diaphragm include the inherent torsional moment plus an accidental torsional moment resulting from the shift in the center of mass equal to 5 percent of the dimension of the structure perpendicular to the direction of applied forces. This accidental shift in the center of mass must be applied “in the direction that produces the greater effect.”

K. Dirk Bondy & Bryan Allred

71

The accidental eccentricity of 5 percent of the building dimension perpendicular to the applied forces is:

e

0.05 242 ft

12.1 ft

Therefore, the x dimension of the center of gravity with a positive shift is:

X

. .

109.1 ft

12.1 ft

121.2 ft

And the x dimension of the center of gravity with a negative shift is:

X

. .

109.1 ft

12.1 ft

97.0 ft

The accidental eccentricity shifts result in center of gravity locations to each side of the center of rigidity. Therefore, two rotational analyses will be required to capture the "greatest effects", or maximum possible forces to all the frames since the positive shift creates a counter-clockwise torsional moment, and the negative shift creates a clockwise torsional moment. In this example only the forces resulting from the positive shift will be shown. The torsional moment including a positive accidental eccentricity shift for loading in the North-South direction is: MTORS = 201.3k(121.2 ft - 101.8 ft) = 3,905 ft-kips, applied counter-clockwise on the diaphragm about the center of rigidity.

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Seneca Seismic Analysis and Design Principles and Practice

Wall Grid Y1 Y2 Y3 Y4 Y5 X1 X2 X3 X4 X5

Ri 1.0 1.0 1.716 1.716 1.716 1.716 1.716 1.716 1.0 1.0

di (ft.) 100.8 100.8 40.8 -19.2 -139.2 93.6 33.6 -26.4 -86.4 -86.4

Ridi 100.8 100.8 70.0 -32.9 -238.9 160.6 57.7 -45.3 -86.4 -86.4

Ridi2 10,161 10,161 2,857 633 33,250 15,034 1,937 1,196 7,465 7,465 =90,158

It is easy to get confused when trying to keep track of positive and negative signs while doing this analysis. However, if the applied torsional moment is drawn at the center of rigidity it is easier to visualize the rotated shape, and it becomes easy to see the direction that the shearwall forces act on the diaphragm. Then the correct signs can be applied after performing the calculations. Rotational Forces on Diaphragm without Convention Signs Applied: F

3,905 ft

kips

F

3,905 ft

kips

F

3,905 ft

kips

F

3,905 ft

kips

F

3,905 ft

kips

100.8 90,158 100.8 90,158 70.0 90,158 32.9 90,158 238.9 90,158

4.4 F 4.4 F 3.0 F 1.4 F 10.3 F

160.6 90,158 57.7 3,905 ft kips 90,158 45.3 3,905 ft kips 90,158 86.4 3,905 ft kips 90,158 86.4 3,905 ft kips 90,158 3,905 ft

kips

6.9 2.5 2.0 3.7 3.7

Apply these forces to the diaphragm in the direction they must act to resist torsion, then determine the appropriate sign and combine with the direct forces to determine the final forces.

K. Dirk Bondy & Bryan Allred

Final Total Forces on Diaphragm FY1 = -28.2k + 4.4k FY2 = -28.2k + 4.4k FY3 = -48.3k + 3.0k FY4 = -48.3k - 1.4k FY5 = -48.3k - 10.3k

= -23.8k = -23.8k = -45.3k = -49.7k = -58.6k

FX1 = -6.9k FX2 = -2.5k FX3 = +2.0k FX4 = +3.7k FX5 = +3.7k

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Seneca Seismic Analysis and Design Principles and Practice

Verify Statics on Diaphragm

FE-W = -6.9k - 2.5k + 2.0k + 3.7k + 3.7k = 0  FN-S = 201.3k - 23.8k - 23.8k - 45.3k - 49.7k - 58.6k  0  M0,0 = (2)(23.8)k(1 ft) + 45.3k(61 ft) + 49.7k(121 ft) + 58.6k(241 ft) – 6.9k(1 ft) - 2.5k(61 ft) + 2.0k(121 ft) + (2)(3.7k)(181 ft) - 201.3k(121.2 ft)  0  Calculate the forces at the 5th Level In this example, the center of gravity at the 5th Level is the same as it was at the Roof Level.

X.. Y. .

109.1 ft 102.9 ft

However, the center of rigidity will not be the same at the 5th Level as it was at the Roof Level.

1 ft 1.0

X

. .

Y.

.

1 ft 1.846

1 ft 1.0

61 ft 1.846 121 ft 1.846 2 1.0 3 1.846

61 ft 1.846

121 ft 1.846 2 1.0 3 1.846

181 ft 1.0

241 ft 1.846 181 ft 1.0

103.9 ft 92.8 ft

K. Dirk Bondy & Bryan Allred

75

The forces placed on the 5th Level diaphragm must be the total shear at that level, not the force. The loads from the frames above are assumed to be redistributed back into the diaphragm, and distributed again through the net center of gravity about the center of rigidity at this level. Since the center of gravity is the same at both levels, the centroid of the shear will be at the same location that it was at the Roof Level. The total shear applied at the 5th Level will be:

The results of this analysis are shown in the final shear diagram below:

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Seneca Seismic Analysis and Design Principles and Practice

For example, the Roof and 5th Level forces on Frame Y5 on Grid 9 are calculated as follows:

The remaining forces to each frame will be calculated in the same way. The centroid of the shear is applied at each level, distributed through each diaphragm based upon the rigidities of the seismic resisting elements at that level, and the forces are determined by subtracting the shear at the level of concern from the shear in that element at the level above. Also remember, the analysis shown is for a positive shift in the center of gravity, or a positive accidental eccentricity. Since the accidental eccentricity will cause a shift in the center of gravity to the opposite side of the center of rigidity, a second complete analysis is required. Frames in which the rotational forces subtracted from the direct forces will now be additive in the second, negative shift of the center of gravity analysis. Finally, a similar analysis must be done for forces in the East-West direction. This will make anybody a fan of a three-dimensional analysis program that does all of these calculations for the engineer.

K. Dirk Bondy & Bryan Allred

d)

77

For the Roof Level, determine if torsional irregularity Type 1a or 1b exists according to ASCE 7 Table 12.3-1

ASCE 7 Table 12.3-1 states that torsional irregularity Type 1a exists when "the maximum story drift, computed including accidental torsion with Ax = 1.0, at one end of the structure transverse to an axis is more than 1.2 times the average of the story drifts at the two ends of the structure." Torsional irregularity Type 1b "Extreme Torsional Irregularity" exists when the same ratio exceeds 1.4. According to ASCE 7, structures assigned to Seismic Design Category C, D, E or F where Type 1a or 1b torsional irregularity exists as defined in Table 12.3-1 shall have the effects accounted for by multiplying the accidental torsional moment at each level by a torsional amplification factor (Ax), where;



 . 

(12.8-14)

Where; max = the maximum displacement at Level x computed assuming Ax = 1 avg = the average of the displacements at the extreme points of the structure at Level x computed assuming Ax = 1

Ax shall not be less than 1 and is not required to exceed 3.0. Recall the final forces acting on the Roof Level diaphragm due to a positive shift in the center of gravity due to the required accidental eccentricity. The forces shown are acting on the diaphragm, and the force on the frames will be in the opposite direction.

The rotational calculations are relative values, so it is not required that the actual deflections be calculated. However, we will use the original rigidity values to calculate the diaphragm displacements

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Seneca Seismic Analysis and Design Principles and Practice

in this example, with the understanding that the actual deflection of the frames at the roof will ultimately be calculated with the application of all of the lateral loads at each level. Since the application of the centroid of load at the center of gravity causes a counter-clockwise rotation, the maximum displacement of the roof diaphragm will occur at the eastern-most edge of the diaphragm along Grid 9. Since Frame Y5 occurs on this gridline it can be used to calculate the maximum diaphragm displacement. The total final force, including rotational effects in Frame Y5 was found to be 58.6 kips, and this frame occurs at one of extreme edges of the diaphragm along Grid 9. Therefore, the deflection of this frame at the roof level under a concentrated roof force of 58.6 kips is:

∆

∆

58.6 kips kips 67.61 in



0.87 in

The total final force, including rotational effects in Frame Y1 was found to be 23.8 kips, and this frame occurs at one of extreme edges of the diaphragm along Grid 1. Therefore, the deflection of this frame at the roof level under a concentrated roof force of 23.8 kips is:

∆

∆





23.8 kips kips 39.39 in

0.60 in

The average displacement of the diaphragm is:





0.87 in

0.60 in 2

0.74 in

K. Dirk Bondy & Bryan Allred

79

And the ratio of the maximum displacement to the average displacement is: 

/



0.87 in / 0.74 in

1.18

Therefore, for this example since the ratio of the maximum displacement to the average displacement is less than 1.2 there is no torsional irregularity, and Ax will be taken as 1.0.

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Seneca Seismic Analysis and Design Principles and Practice

6 Rigid Diaphragm Shear and Moment Diagrams Until now the mass at each level has been considered a lumped mass at the center of gravity of the floor or roof. That was an appropriate assumption for determining the vertical force distribution of the seismic loads, either in the Equivalent Lateral Force procedure, or using dynamic analysis. This was also a statically valid assumption in the rigid diaphragm analysis for determining the design shears and forces to the lateral load resisting elements. However, for the design of the diaphragm itself, the lateral load must be applied proportionally to the mass distribution of the floor system. The lateral load that will be used to determine the diaphragm shear and moment diagrams was generated previously using the Fpx values previously calculated. That Fpx value must be distributed to the diaphragm based proportionally on the seismic mass distribution on the floor. But this brings up an issue that is often debated among structural engineers; how is accidental eccentricity addressed?

To Include Accidental Eccentricity or Not Include Accidental Eccentricity? That is the question. The short answer is that accidental eccentricity should not be included in the diaphragm design because it is statically incorrect to do so. A simple statics example will demonstrate why moving the resultant force from the centroid of the load for any reason will result in a statically incorrect analysis. In the following example, the reactions at Grids A & B are found by applying the resultant 100 kip load at the centroid of the distributed load, and summing moments and forces based upon this resultant load and location. With those reactions the shear and moment diagrams are generated.

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Now, using the same example, arbitrarily relocate the centroid of the load one foot to the right and find the new reactions at Grids A & B as we would if we included an accidental eccentricity in order to maximize the reactions at the seismic resisting systems, but leave the distributed load as it exists.

The shear diagram will close because the sum of the reactions equals the total applied load, but neither the reactions nor the shear diagram correctly reflect the applied 10.0 klf distributed load. The moment diagram will not close and makes no sense at all. The analysis will never be statically correct if the resultant load location does not equal the actual resultant of the applied (distributed) loading. Therefore, the only analysis that will make static sense for diaphragms is one in which the location of the resultant load used to calculate the forces to the seismic system (the reactions) is equal to the resultant of the applied loading. In other words, we must use an analysis with no applied accidental eccentricity.

The Role of the Perpendicular Seismic System The perpendicular seismic system plays a significant role in the diaphragm shear and moment diagrams, and it is statically incorrect to ignore it. The following will demonstrate why. Consider the following example of a 100 foot by 50 foot diaphragm, first with a seismic layout only in the North-South direction. The walls have rigidities, R, of 1, 2 & 3 for simplicity. The diaphragm force is 1,000 kips, also for simplicity, but with greater mass on the left side.

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The shear and moment diagrams look like one might expect for a rigid beam on flexible supports. However, now consider the same layout with the same loading, but with a perpendicular seismic system along Grids A & B, each wall with a rigidity of 3.

K. Dirk Bondy & Bryan Allred

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Some engineers and reputable publications advocate "smoothing" the moment jump over the entire moment diagram because they are apparently bothered by the jump, and believe that is somehow does not represent reality. The problem with doing that is that it makes the problem mathematically and statically incorrect. The shear diagram will no longer be the derivative of the moment diagram. And likewise, the moment diagram will no longer be the integral of the shear diagram (taken from each end to the center of rigidity). Therefore, to correct this the engineer would need to modify the shear diagram to match the "new and improved" moment diagram. That, in turn, would technically change the loading diagram and the results of the rigid diaphragm analysis. In other words, to match the moment diagram that is more palatable, all of the initial design parameters would need to change. The final diaphragm reactions, shears and moments would no longer represent the diaphragm that is actually being designed. We recommend learning to live with the jump in the moment diagram. The following diagram was generated by our in-house software program for the Roof Level of the steel office building example, with the diaphragm force of 394 kips applied as line loads based upon the mass distribution of the diaphragm.

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About the Authors K. Dirk Bondy

Dirk Bondy earned his Bachelor’s degree from California Polytechnic State University at San Luis Obispo in 1988 and his Master’s degree from the University of California at Berkeley in 1989. He is the president of Seneca Structural Engineering, Inc. and The Great American Cable Company II, Inc., both located in Laguna Hills, California. While he has been the engineer of record on millions of square feet of post-tensioned concrete structures he is probably best known for being the son of Ken Bondy. His professional interest and expertise are in the area of post-tensioned and reinforced concrete, seismic design, seismic retrofit and vertical load retrofit. In addition to his structural design experience, Mr. Bondy currently teaches Prestressed Concrete Design at the University of California at Los Angeles in the Spring Quarter, and has been an instructor at the University of California, Irvine and California Polytechnic State University, Pomona where he taught courses on Prestressed Concrete Design, Reinforced Concrete Design, Steel Design, Structural Design and Seismic Design. He has been published in numerous journals and conference proceedings. He is a registered Civil and Structural Engineer in the states of California, Nevada, Hawaii and Arizona as well as a licensed C50 contractor in the state of California. He is a licensed private pilot (single-engine, multi-engine and instrument ratings) with over 800 hours total pilot-in-command time. He resides in Laguna Hills, California with his wife Kristen and their three children; Cameron, Ryan and Elizabeth. Bryan Allred

Bryan Allred has been a practicing engineer since 1993 and is the Vice-President of Seneca Structural Engineering Inc. in Laguna Hills California. He received his Bachelors and Master’s degree in Civil Engineering from the University of California at Irvine and is a licensed civil and structural engineer in the state of California. Bryan specializes in the design of concrete buildings utilizing post-tensioned floor systems, post-tensioned slab on ground foundations and retrofits of existing building using external post-tensioning. He is a member of the Structural Engineers Association of California, American Concrete Institute and is a Fellow of the Post-Tensioning Institute where he is also a member of the Building Design and Education Committees. He has written several magazine articles relating to posttensioned construction and engineering and has also given numerous post-tensioning educational seminars and webinars across the country. Bryan and his wife Marisa have three children, two smart beautiful girls and one ultra-cool son.

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