Seismic design handbook pdf

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.

Plot the graph for periods up to 2. Combine the story shears using the SRSS method and extract the vertical force distribution that creates that story shear envelope. Compare the forces and shears in this analysis side by side with the analysis using all modes.

But we want to do it here for a different reason, which will become clear in Part c. 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.

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, kips for the analysis.

The following demonstrates how we determine the Rayleigh Period using a 1, kip base shear. Roof 2, 2. This compares extremely well with the dynamic first mode period of 1. 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.

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. This is, of course, the same Participation Function as previously calculated using the normalized mode shape. Therefore all of the numbers have already been calculated. Those were calculated for this site as 0. The force at level i in Mode n is given by the following equation.

Or, by using the graph itself;. Extract the Level Forces from the Story Shears. Show as a proportion of the base shear. Therefore, we will repeat the modal combination using the SRSS method, but only with the first and second modes.

Therefore, the final design forces using the Modal Response Spectrum Analysis considering only the first and second modes are:. 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:. 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. The story forces for each mode are shown below. It follows that we could use the same scale factor for the diaphragm forces. The modal response spectrum diaphragm forces that we will use to compare to the ELF diaphragm forces are:.

The comparison of the ELF diaphragm forces and the modal response spectrum forces is shown below. 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:. 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.

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.

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

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

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:. 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. Only direct translation will be allowed, and the direct forces will be distributed to the seismic system based solely on their relative rigidities. 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. 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. The simplest way to account for the Ax factor is to multiply it by the accidental eccentricity percent value. It would be applied, if it is equal to 1. Determine the displacement at that level and calculate the frame rigidity at that level for each frame based upon:.

The following table contains the deflection at each level for each frame under a 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.

This is only done for convenience as it will simplify the calculations. Rigidity Rigidity Roof 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.

Direct Forces in the Y Direction. F F 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:.

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. The accidental eccentricity of 5 percent of the building dimension perpendicular to the applied forces is: e 0.

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. Wall Grid Ri di ft. Ridi Ridi2 Y1 1. 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. 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.

However, the center of rigidity will not be the same at the 5th Level as it was at the Roof Level. 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 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. Torsional irregularity Type 1b "Extreme Torsional Irregularity" exists when the same ratio exceeds 1. 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. 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 Therefore, the deflection of this frame at the roof level under a concentrated roof force of The average displacement of the diaphragm is: 0.

Therefore, for this example since the ratio of the maximum displacement to the average displacement is less than 1. 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?

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. With those reactions the shear and moment diagrams are generated.

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 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 foot by 50 foot diaphragm, first with a seismic layout only in the North-South direction.

The diaphragm force is 1, kips, also for simplicity, but with greater mass on the left side. The shear and moment diagrams look like one might expect for a rigid beam on flexible supports. 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 kips applied as line loads based upon the mass distribution of the diaphragm.

He is the president of Seneca Structural Engineering, Inc. 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. 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 hours total pilot-in-command time. 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 has written several magazine articles relating to post- tensioned construction and engineering and has also given numerous post-tensioning educational seminars and webinars across the country.

Share this:. Yahoo Finance. Engineering News. Currently, there are 19 results released and the latest one is updated on 26 Dec The above search results can partly answer users' queries, however, there will be many other problems that users are interested in.

We list the most common ones below. So, the information you read is accurate and will surely meet your desired needs. Of course not, searching and using information are completely free for users.

Only when you have a need to use the service will we request information and make payment. When you are in need of home decor, this is where you can find the best ideas. We collect the latest and fastest ideas and tips. In addition, when you need to buy or need advice on decor, we will provide you with the most reputable websites to support and meet your needs.

Decorwiki is the place to help you find the most reputable services, tips, shops and decor websites matching your condition and preferences. We constantly update the most prestigious decor websites, the highest quality and the most reasonable prices. We have everything you need related to the field of decor. And searching on the site is completely free for all users.

Visit site. Antenna Theory Balanis book and solutions manual Antenna Theory Propagation. Computer system Architecture Third Edition by M. Purchasing Digital Textbooks on VitalSource.

His interests include constitutive modeling of engineering materials, soil and concrete plasticity, structural connections, and structural stability, and he has received several national engineering awards. In , he was elected to the U. National Academy of Engineering. Chen has authored and coauthored more than 20 engineering books and technical papers.

Lian Duan is a senior bridge engineer and structural steel committee chair with the California Department of Transportation Caltrans. He earned his diploma in civil engineering in , MS in structural engineering in from Taiyuan University of Technology, China, and PhD in structural engineering from Purdue University in His interests include inelastic behavior of reinforced concrete and steel structures, structural stability, seismic bridge analysis, and design.

Duan has authored and coauthored more than 70 papers, chapters, and reports, and is the coeditor of the Handbook of International Bridge Engineering C P. He has received several awards, including the prestigious Arthur M. Disclaimer : learncreative.