Dynamic Example

S-FRAME DYNAMIC & SEISMIC

ANALYSIS EXAMPLES

© SOFTEK Services Ltd. 2010

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S-FRAME ANALYSIS & DESIGN EXAMPLES Job Ref.

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Dynamic Analysis Sheet no./rev.

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Objective The objective of the following examples is to illustrate and provide guidance on the use of the features available in S-FRAME for seismic/dynamic Time History analysis. While they are necessarily discussed, the intention is not to explain or advise on theories of dynamic analysis. Those seeking such information are recommended to consult the references. We can also highly recommend the relevent courses (most of which are offered via the internet) of the Structural Engineers Association of BC’s Certificate in Structural Engineering (CSE) program – see http://www.seabc.ca/courses.html for more details. Discussions on aspects and methods of modeling, assumptions, theories etc are kept to a minimum to aid clarity and simplicity. The intention is to outline, for competent and professionally qualified individuals, the use of S-FRAME and S-STEEL as tools in the dynamic analysis.

Disclaimer While the authors of this document have tried to be as accurate as possible, they cannot be held responsible for any errors and omissions in it or in the designs of others that might be based on it. This document is intended for the use of professional personnel competent to evaluate the significance and limitations of its contents and recommendations, and who will accept the responsibility for its application. Users of information from this publication assume all liability. The authors and SOFTEK Services Ltd. disclaim any and all responsibility for the applications of the stated principles and for the accuracy of any of the material contained herein.

Acknowledgements With grateful acknowledgement to Luis E. García and Mete A. Sozen for their kind permission to include in this document excerpts from their Purdue University CE571 course notes (Reference [1]). This acknowledgement does not imply any endorsement by Dr García or Professor Sozen of SOFTEK programs, nor any checking or validation by them of these programs’ operation or output.


http://www.seabc.ca/courses.html

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Dynamic Analysis Sheet no./rev.

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S-FRAME VERIFICATION EXAMPLES Job Ref.

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Time History Analysis: Blast Load Sheet no./rev.

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S-FRAME BLAST LOAD TIME HISTORY ANALYSIS


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CONTENTS

EXAMPLE BRIEF .............................................................................................................................................................................5

S-FRAME MODEL ............................................................................................................................................................................6

SECTIONS ...........................................................................................................................................................................................7 MATERIALS........................................................................................................................................................................................8 RIGID DIAPHRAGM FLOORS ...............................................................................................................................................................9

FLOOR NUMBERS .........................................................................................................................................................................10

VIBRATION ANALYSIS ................................................................................................................................................................11

VIBRATION RESULTS COMPARISON .................................................................................................................................................12

TIME HISTORY ANALYSIS .........................................................................................................................................................13

TIME HISTORY LOADS .....................................................................................................................................................................13

INPUTTING TIME HISTORY LOADS ............................................................................................................................................14 DAMPING & RAYLEIGH DAMPING COEFFICIENTS ............................................................................................................................16 CONSTANT TIME STEP SIZE ...............................................................................................................................................................19 ANALYSIS DURATION ......................................................................................................................................................................20 NEWMARK COEFFICIENTS ................................................................................................................................................................21 TIME HISTORY ANALYSIS SETTINGS ................................................................................................................................................22

RESULTS ..........................................................................................................................................................................................23

TIME HISTORY RESPONSE ................................................................................................................................................................23 DEFLECTIONS...................................................................................................................................................................................24 FLOOR FORCES.................................................................................................................................................................................25 SENSITIVITY TO RAYLEIGH DAMPING COEFFICIENTS.......................................................................................................................26

FURTHER OUTPUT PARAMETERS...........................................................................................................................................27

PRACTICAL DISCUSSION; PEAK RESPONSE AND OUTPUT REDUCTION....................................................................29

FURTHER DESIGN VALUES...............................................................................................................................................................32

REFERENCES..................................................................................................................................................................................33

SUGGESTED VALUES FOR DAMPING RATIO .......................................................................................................................34


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1 Example Brief

Further aspects of the building/analysis model either discussed within the body of the example or not stated explicitly are:

• Supports are fully fixed. • Floors act as rigid diaphragms. • Mass distribution is idealized as concentrated at the floors only – column and girders elements are massless. • Contribution to stiffness matrix of shear stiffness is not considered.


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2 S-FRAME Model The frame is input as per the Example details and dimensions. One analysis element per beam/column is used. All connections are rigid and supports are fully fixed (it was found that this is the case, though it is not explicitly stated in the Example). The X-axis is chosen as the axis of displacement in S-FRAME.


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2.1 Sections S-FRAME’s Tapered section tool is used to create the ‘Girder’ and ‘Column’ sections (sections do not have to be tapered). Sections thus created will render correctly and can be readily edited.

To give closer agreement with the example (which does not consider shear stiffness) the Shear Area (Av) of the sections is set = 0. There is no reason why this should be done in practice; while shear effects are often small enough to be neglected they nevertheless occur.


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2.2 Materials Since the building mass in the example is idealized as concentrated at the floors, again to give closer agreement, a zero density material is created for the frame members so these do not introduce distributed mass which would change the vibration characteristics to some extent. A second material with the specified force density (per unit thickness) is created for the diaphragm panels which are used to introduce the floor mass.

Material Modulus of Elasticity; E = 25 GPa =

25000 N/mm2; (we note that this is a typical value for reinforced concrete for short term loading)

Note a sensible value of G (shear modulus) is also entered as this is a component of the S-FRAME Beam Element stiffness matrix for both shear and torsional stiffness.

Slab material force density; ρ = gacc × 1000 kg/m3 = 0.0000098067 N/mm3

(gravitational acceleration; gacc = 9.807 m/s2;)

Since the ‘Slab Material’ is only applied to the Rigid Diaphragm panel for mass generation, the E and G values for this material are irrelevant.


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2.3 Rigid Diaphragm Floors Since a ‘rigid diaphragm effect’ is specified in the Example, Rigid Diaphragm Panels are used to model this behaviour. These panels also provide a convenient method of introducing the floor mass. The panels are assigned thickness = 1000mm and a material with appropriate force density (see above). S-FRAME automatically calculates the mass of the diaphragm and applies it as a lumped mass at the centroid of the panel area. For convenience, since engineers usually work in terms of weight rather than mass, S-FRAME describes mass in terms of force units.

Note - The E & G values of the material assigned to a Rigid Diaphragm Panel have no effect on its stiffness – the diaphragm is rigid in-plane and has zero stiffness out of plane

Slab material force density; ρ = gacc × 1000 kg/m3 = 9.807×10-6 N/mm3

Floor Volume; Vi = 10m × 6m × 1000mm = 60 m3

Floor Weight; Wi = ρ× Vi = 588.40 kN

Floor Mass; Mi = ρ× Vi /gacc = 60.00 tonne

Although it is not necessary for analysis, the lumped mass of the panel can be exposed by running the S-FRAME command EDIT/Mesh Generator/Generate Rigid Diaphragm Mass. Note that the diaphragm produces translational mass in both lateral axes, as well as Z-rotational mass. The example only actually requires X-Translational Mass since it only considers one direction of response.


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3 Floor Numbers Finally Floor Numbers are applied to joints to identify floors. Note that Floor Number = 1 is applied to the base support joints. This is required for S-FRAME to calculate storey drifts, storey shears and floor forces. This completes the modeling process.


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4 Vibration Analysis Before running Time History Analysis, the vibration characteristics of the structure are assessed and verified. The following Vibration settings are initially used; Jacobi Threshold Eigenvalue Extraction Method (since the model is relatively small and has few modes) and 10 Eigenvalues (mode shapes) requested.

With this 3D model, some modes are found which are not given in the Example. S-FRAME finds typical Y-direction and torsional modes in addition to the required X-direction modes, since there is Y-Translational and Z-Rotational Mass.


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Undesired modes can be removed by deleting the Y-Trans and Z-Rotational lumped masses to restrict the DOF’s to the X-axis. This gives the following first 4 modes (only 4 Eigenvalues requested) which are all X-direction modes only. Removal of undesired modes could also be achieved by preventing Y-translation and rotation with appropriate supports.

4.1 Vibration Results Comparison Comparing S-FRAME’S results to the example there is excellent agreement.

S-FRAME

Reference


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5 Time History Analysis

5.1 Time History Loads The Example specifies the following linearly varying time dependent pressure vs time load which has; a total duration of 0.6s, a maximum of 5 kPa @ 0.1s and minimum of -1 kPa @ 0.4s. This must be ‘converted’ to force vs time functions which can be applied to discrete joints in the structure.

Since the example considers only the displacement of the floors, it is sensible to apply the load at the beam/column intersection joints. We calculate the tributary area of each joint and hence the maximum and minimum forces for the function at this joint.

Three functions are required;

• Interior Joints; F_Max = 75 kN, F_Min = -15 kN • Edge Joints; F_Max = 37.5 kN, F_Min = -7.5 kN • Corner Joints; F_Max = 18.75 kN, F_Min = -3.75 kN


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5.2 Inputting Time History Loads First the required Force vs Time functions are created and saved to a file. Once the data is input and the curve Added the function is plotted and can be verified graphically. E.g. the following data is input for the function for ‘Interior Joints’:

Time (s) Force (kN)

0 0 0.1 75 0.4 -15 0.6 0

The Nodal Excitation Type is set and the appropriate functions created and added to a Time History Data file (file type *.DTH) with the Function Tool.


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When all the required functions have been created the Time History Load file is saved. Note that this file is saved seperately from the model file – i.e. the time history load data is not held in the *.TEL model file.

5.2.1 Assigning Functions to Model The functions are then assigned to the appropriate joints in the appropriate direction (X Translation) in the following manner. Each function is assigned a number so that correct assignment can be graphically verified.

1. Select the Direction to apply in – X-Translation 2. Select the Function to apply – e.g. function#1 3. Click on joint(s) using mouse to apply


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5.3 Damping & Rayleigh Damping Coefficients S-FRAME employs the Newmark direct time integration method for Time History Analysis, not a modal combination method. For more detailed information on theory and solutions see S-FRAME’s Theory Manual, and References 1 and 2. For the direct Integration methods damping can not be explicitly set to a single value for all modes. Damping is a function of frequency and is introduced via the Rayleigh Damping Coefficients α and β which are used to form the damping matrix [C]. It is important to note that these coefficients are not themselves damping values. For non-zero values of both α and β a desired value of damping ζ can be set precisely only for two frequencies, ωr and ωs. The damping value ζ is expressed as a percentage (of critical damping) and the frequencies ω are angular frequencies in terms of radians/s, not Hz. The values of α and β which produce precisely the desired damping value at two frequencies only can then calculated (from the following equation (see References 1. and 2) and input into S-FRAME for analysis.

Damping for other modes is a function of their frequency and hence will ≠ ζ (the desired damping). See the plots on the following pages for examples. Ideally the damping should be reasonably close to the desired value for modes which contribute significantly to the response. Hence a prior frequency analysis is generally required to determine the model’s frequency characteristics and make a rationale choice of two frequencies for which to calculate α and β. If only one or two frequencies dominate the response then the choice is obvious. Otherwise two frequencies can be chosen which give a reasonable approximation to the desired damping for a range of frequencies in which the modes which contribute most to the response fall. Reference [1] states:

It is convenient to take ωr as the value of the fundamental frequency and ωs as the frequency corresponding to the last of the upper modes that significantly contribute to the response. This way the first mode and mode s will have exactly the same damping, and all modes in between will have somewhat smaller similar values and the modes with frequencies larger than ωs will have larger damping values thus reducing their contribution to response.

5.3.1 Damping Value ζ The Example does not discuss the rationale for using ζ = 2%. See the table at the end of the document (page 34), reproduced

from Reference [2], for some recommended values. Inferring from the Example’s material E value and section sizes that the building is concrete, the value could have been chosen for “well-reinforced concrete (only slight cracking)”.


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For the Example, we could first choose Mode 1 and Mode 4 for initial values of α and β from:

sr

sr

ωωωξω

α+

=2

and sr ωω

ξβ+

=2

Damping; ζ = 2% ; 1st mode; ωr = 10.736; 2nd mode; ωs = 86.926

α = 2×ζ×ωr×ωs/(ωr + ωs) = 0.382232

β = 2×ζ/(ωr + ωs) = 0.0004096

These values produce the following damping/frequency relationship:

Rayleigh Damping

1.709 13.835

1.25%

0.00%

0.50%

1.00%

1.50%

2.00%

2.50%

3.00%

3.50%

4.00%

4.50%

5.00%

0 2 4 6 8 10 12 14Frequency (Hz)

Dam

ping

Rat

io

Total Damping RatioMass DampingStiffness DampingMode 1Mode 2Min DampingDesired Damping

The damping for other frequencies can be determined preceisely from; 22

i

ii

βωωαζ +=

Mode 2; f2 = 5.46 Hz; ω2 = 34.306; ζ2 = α/(2×ω2) + (β×ω2)/2 = 1.260 %

Mode 3; f3 = 9.839 Hz; ω3 = 61.822; ζ3 = α/(2×ω3) + (β×ω3)/2 = 1.575 %


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Alternatively Mode 1 and Mode 2 could be used giving:

1st mode; ωr = 10.736; 2nd mode; ωs = 34.306

α = 2×ζ×ωr×ωs/(ωr + ωs) = 0.327081

β = 2×ζ/(ωr + ωs) = 0.0008881

These values produce the following damping/frequency relationship:

Rayleigh Damping

1.709 5.460

1.70%

0.00%

0.50%

1.00%

1.50%

2.00%

2.50%

3.00%

3.50%

4.00%

4.50%

5.00%

0 2 4 6 8 10 12 14Frequency (Hz)

Dam

ping

Rat

io


The damping for other frequencies > 2%;

Mode 3; f3 = 9.839 Hz; ω3 = 61.822; ζ3 = α/(2×ω3) + (β×ω3)/2 = 3.010 %

Mode 4; f4 = 13.035 Hz; ω4 = 86.926; ζ4 = α/(2×ω4) + (β×ω4)/2 = 4.048 %


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5.4 Constant time step size Next the analysis Time Step size ∆t is considered. The time step size should be sufficiently small for accurate analysis.

However, the smaller the time step size, the higher the ‘cost’ (in terms of computational time) of analysis for a given duration of analysis – e.g. the Example’s first 2.5s of response. Additionally the analysis duration is limited by the maximum number of allowable time steps which is 32,767 (a limit imposed by current software architecture). So ideally the time step size should be

no smaller than it need be. The S-FRAME Theory Manual pg 34 gives a rule of thumb for ∆t as follows:

Where Ω = the highest (angular) frequency component of the forcing function in rads/s and Ω∗ = 4Ω. Since we more

commonly think in terms of period the above expression can be conveniently re-formulated as follows:

Ω==

π21f

T 80420

ffcr

TTt =

×≤∆⇒

Where Tf = 1/Ω. However, what if Tf is unknown or not applicable? The rule of thumb derives from two principles, the first of

which is illustrated by the following figure


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Principles:

1. Refer to the figure above; for a given forcing (or input) frequency Ω (ωA in the figure) the response of the system for a given natural frequency ωi (ωo in the figure) generally reduces as the frequencies diverge. The divergence is conveniently expressed in terms of the ratio of the forcing frequency to the natural frequency Ω/ωi. The figure illustrates this phenomenon. Furthermore it can be seen that the response in modes with a small ratio Ω/ωi is essentially static and the response in modes with a large ratio Ω/ωi is negligible. Below Ω/ωi = 1/4 further increases in ωi do not produce significant change in response hence frequencies (of response) higher than 4Ω = Ω∗ need not be considered.

2. Around 20 equal time intervals are required to discretize both the input and response motions with sufficient accuracy.

These principles explain the origin of the values of 4 and 20 in the rule of thumb. Thus a rational choice of time step size is

based on a) deciding which is the highest frequency required to be discretized considering both input and response and b) dividing the resulting (lowest) period by 20.

From the foregoing discussion we can derive a more general and practical rule of thumb. Let us call the highest frequency of

the response fr with corresponding period Tr.

• If Tf is known or guessed at to a reasonable degree then it is not necessary to know Tr (following Principle 1) and

from Principle 2. we set the time step ∆t = (Tf /4)/20 = Tf /80.

• If Tf is unknown or inapplicable (as in Example 4) we consider the likely highest significant response frequency fr =

1/Tr and set ∆t = Tr /20.

For the Example Tf is inapplicable and Tr is known, being the period of mode 4.

Tr = T4 = 0.072 s; ∆t ≤; Tr/20 = 0.0036 s

A value of ∆t = 0.00125 was chosen since it is < the maximum calculated above and conveniently gives a time step at t = 0.2875s which is close to t = 0.2873s for which the Example gives results.

5.5 Analysis Duration The Example’s analysis duration of 2.5s is used. As discussed previously, the duration has a maximum possible value

governed by the time step size: Duration Limit; TA_max = 32,767 ∆t For a constant time step, the analysis duration is not explicitly entered; the user inputs the time step size ∆t and total time steps

N from which the duration derives.

∆t = 0.00125s; Desired analysis time; TA = 2.5s; total time steps; N = TA/ ∆t = 2000

The peak response may occur during the free-vibration portion of response – i.e. after the exciting function has ended – and the time to peak response (or steady state in the case of a periodic forcing function) may not be known with any great certainty. Hence in practice it is generally sensible to continue analysis beyond the end of excitation where applicable and some experimentation may be required with analysis duration having viewed initial results.


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5.6 Newmark Coefficients These are generally one of two sets of values as follows. See S-FRAME Theory Manual and References for more information.

Alpha Beta

Zero Damping 0.2525 0.5050

Non-zero Damping 0.25 0.5

Since the example has a specified non-zero damping the appropriate values are input in S-FRAME.


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5.7 Time History Analysis Settings The preceding consideration leads to the following initial Linear Dynamic Time History analysis settings

The Rayleigh damping coefficient values are those calculated on pagThe other Constant time-step integration parameters are discussed l

© SOFTEK Service

∆t =

a

s

N =

α =
β =
e 17 above for Modes 1 and 4. ter (see pg 27) and are not considered at this stage.

Ltd. 2010

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6 RESULTS 6.1 Time History Response The time history response at a chosen joint can be readily assessed by right-clicking the joint and choosing ‘Response Time

History…’ from the context menu: S-FRAME plots the chosen result parameter – e.g. X-Displacement – vs time so the user can easily identify the maximum response and the approximate time at which this occurs. There is a Trace function to assist with this – using this it can be seen that the maximum -ve X-displacement response occurs at around 0.58s for example.

Note that the response calculated by S-FRAME is the total response for all modes. S-FRAME does not explicitly evaluate the response of each mode as per the Example’s method as this is not usually required, the total response generally being of primary interest.


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6.2 Deflections

More detailed results for each time step are also available. S-FRAME presents each time step as a discrete loadcase for which all the usual results (both Graphical and Numerical) are available as for a static analysis. We wish to compare the displacement results, so the displacement diagram is chosen, viewing of Dx (X-displacement) values is enabled and time step 230 @ time = 0.2875 s is selected.

The results are given (in meters) on page 33 of the Example for t = 0.2873s

6.2.1 Comparison Floor Displacement

(mm) Reference S-FRAME

4th Floor 29.373 29.333

3rd Floor 25.103 25.107

2nd Floor 17.455 17.496

1st Floor 7.479 7.509


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6.3 Floor Forces Floor Forces which can be compared with those in the Example are available in Numerical Results.

The Example gives results for t = 0.2873s

6.3.1 Comparison Floor Force (kN) Reference S-FRAME

4th Floor 175.97 173.39

3rd Floor 180.18 180.46

2nd Floor 145.05 148.31

1st Floor 83.75 85.68


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6.4 Sensitivity to Rayleigh Damping Coefficients

If the α and β values calculated for Mode’s 1 and 2 (see page 18) are used the following results are obtained (Damping 2).

While these generally give marginally better agreement, this serves to illustrate that results for this example are relatively

insensitive to the choice of the second frequency (ωs) for deriving the damping coefficients, since the response is dominated by

that of mode1. However, this may not always be the case. In practice some experimentation may be required with the choice of frequencies and hence damping coefficient values.

Floor Displacement (mm) Reference S-FRAME

Damping 1 S-FRAME Damping 2 % Change

4th Floor 29.373 29.333 29.352 0.06%

3rd Floor 25.103 25.107 25.104 -0.01%

2nd Floor 17.455 17.496 17.473 -0.13%

1st Floor 7.479 7.509 7.493 -0.21%

Floor Force (kN) Reference S-FRAME

Damping 1 S-FRAME Damping 2 % Change

4th Floor 175.97 173.39 174.61 0.70%

3rd Floor 180.18 180.46 180.62 0.09%

2nd Floor 145.05 148.31 146.2 -1.42%

1st Floor 83.75 85.68 85.00 -0.79%


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7 Further Output Parameters Further analysis options are also available for Constant time-step integration.

Initial time 18 sec

This parameter applies to the input record and is the time at which S-FRAME begins to read data from the input function for analysis. This might be employed to use only part of a long record.

The final two parameters can be used to minimize the amount of analysis output – i.e. the number of time step result cases. They can best be understood by viewing their effect on the response result plot of this example.

Output to file after 1000 time steps


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N and ∆t remain as before – the analysis solution is unchanged but output (to result files) only begins at t = 1000×∆t = 1.25s. In this example this results in the peak response, which occurs before t = 1.25s, being missed. This option might be used to limit output to a portion where steady state is achieved for a periodic input.

Output to file every 100 time steps

N and ∆t remain as before – the analysis solution is unchanged but results are only available at time intervals = 100×∆t = 0.125s. This produces a more crude record of the response which may not capture some peak values with sufficient accuracy.


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8 Practical Issues; Peak Response and Output Reduction The Example does not discuss Peak Response but this will usually be of primary interest. This can be estimated for certain values using the Response Plot disucussed previously. ‘Exact’ values can be found in the Numerical Results Spreadsheet using the Find Max/Min function when viewing results for All Load Cases (i.e. Time Steps)

Additionally, the Numerical Results Spreadsheets can be exported wholesale (via a simple copy/paste operation) to a dedicated spreadsheet application like Microsoft Excel where customized response plots can be reproduced of all results and maxima and minima easily found. This was done to produce the following plots.


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The following plots displaying the response for 2.5s and the max/minima were produced in this manner using data for every 1 time step for; Roof Displacement (X-direction), Base Shear and Maximum (end) Moment in a 1st floor beam (parallel to X-axis)

X-Deflection vs Time: Every 1 Time Step

0.879, 29.879

0.581, -31.951-40

-30

-20

-10

0

10

20

30

40

0 0.5 1 1.5 2 2.5

Time (s)

Def

lect

ion

(mm

)

Base Shear vs Time: Every 1 Time Step

0.595, 649.239

0.879, -610.903-800

-600

-400

-200

0

200

400

600

800

0 0.5 1 1.5 2 2.5

Time (s)

Bas

e Sh

ear

(kN

)

1st Floor Beam Moment vs Time: Every 1 Time Step

0.879, 237.907

0.592, -252.295-300

-200

-100

0

100

200

300

0 0.5 1 1.5 2 2.5

Time (s)

Mom

ent (

kNm

)

It is interesting to note that the peak positive displacement, positive moment and negative base shear occur during the free vibration portion of response – i.e. after the cessation of excitation. It can also be seen that the response in general is dominated by that of the 1st mode, the period of response peaks being around 0.6s. Finally, we see that the peak values, which would be used for maximum design values, all occur at t < 1s. Thus analysis duration could sensibly be reduced to 1s to minimize analysis cost (i.e. time for an analysis run) and result-processing demand.


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8.1.1 Output Every __ Time Steps The following plots illustrate the effect of Ouput Every 10 and Every 100 time steps – clearly this gives an increasingly approximate record of the response.

X-Deflection: Every 10 Time Steps

0.875, 29.857

0.588, -31.895

-40

-30

-20

-10

0

10

20

30

40

0 0.25 0.5 0.75 1

Time (s)

Def

lect

ion

(mm

)

Base Shear: Every 10 Time Steps

0.600, 647.606

0.875, -609.806

-800

-600

-400

-200

0

200

400

600

800

0 0.25 0.5 0.75 1

Time (s)

Bas

e Sh

ear (

kN)

1st Floor Beam Moment: Every 10 Time Steps

0.875, 237.565

0.588, -251.694-300

-200

-100

0

100

200

300

0 0.25 0.5 0.75 1

Time (s)

Mom

ent (

kNm

)

X-Deflection: Every 100 Time Steps

0.875, 29.857

0.625, -29.209

-40

-30

-20

-100

10

20

30

40

0 0.25 0.5 0.75 1

Time (s)

Def

lect

ion

(mm

)

Base Shear: Every 100 Time Steps

0.625, 589.456

0.875, -609.806

-800

-600

-400

-200

0

200

400

600

800

0 0.25 0.5 0.75 1

Time (s)

Bas

e Sh

ear (

kN)

1st Floor Beam Moment: Every 100 Time Steps

0.875, 237.565

0.625, -230.665

-300

-200

-100

0

100

200

300

0 0.25 0.5 0.75 1

Time (s)

Mom

ent (

kNm

)

Comparing with the plots for every 1 step; for ouptut every 10 steps the change in peak values is insignificant, while every 100 steps significantly under estimates some;

Every 1 Every 10 % diff Every 100 % dif

+ve Base Shear (kN) 649.2 647.6 -0.25% 589.5 -9.20%

-ve Beam Moment (kNm) -252.3 -251.7 -0.24% -230.7 -8.56%

Output every 10 time steps could thus sensibly be used to reduce output and improve post-analysis operations.


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8.2 Further Design Values Using the options discussed above to minimize analysis cost and output, S-FRAME’s Envelope function can be efficiently used to produce envelopes of design forces for members for example – e.g. Moment:


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9 REFERENCES [1] Luis E. García & Mete A. Sozen, Multiple Degrees of Freedom Structural Dynamics, Purdue University CE571 – Earthquake Engineering, 2002

[2] Anil K. Chopra, Dynamics of Structures: Theory and Applications to Earthquake Engineering (2nd Edition), Prentice Hall, 2000


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10 Suggested Values for Damping Ratio The following Table is reproduced from Reference [2].

Stress Level Type and Condition

of Structure

Damping Ratio

(%)

Working stress,

no more than about

21 yield point

Welded steel, prestresses

concrete, well-reinforced

concrete (only slight cracking)

2-3

Reinforced concrete with

considerable cracking

3-5

Bolted and/or riveted steel,

Wood structures with nailed or bolted

joints

5-7

At or just below

yield point

Welded steel, prestressed

concrete (without complete loss in

prestress)

5-7

Prestressed concrete with no prestress left 7-10

Reinforced concrete 7-10


Wood structures with

bolted joints

10-15

Wood structures with nailed

joints

15-20


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Time History & Response Spectrum Analysis of Seismic Load Sheet no./rev.

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S-FRAME SEISMIC TIME HISTORY ANALYSIS


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CONTENTS EXAMPLE 5 .......................................................................................................................................................................................5

S-FRAME MODEL ............................................................................................................................................................................6

SECTIONS ...........................................................................................................................................................................................7 MATERIALS........................................................................................................................................................................................8

RIGID DIAPHRAGM FLOORS .....................................................................................................................................................9 BUILDING MASS...............................................................................................................................................................................10

FLOOR NUMBERS .........................................................................................................................................................................11

STATIC ANALYSIS ........................................................................................................................................................................11

VIBRATION ANALYSIS ................................................................................................................................................................13

VIBRATION RESULTS COMPARISON .................................................................................................................................................14

TIME HISTORY ANALYSIS .................................................................................................................................................15

TIME HISTORY LOAD .......................................................................................................................................................................15 DAMPING & RAYLEIGH DAMPING COEFFICIENTS ............................................................................................................................16 CONSTANT TIME STEP SIZE ...............................................................................................................................................................18 ACCELEROGRAMS & SAMPLING.......................................................................................................................................................20 FREQUENCY CONTENT .....................................................................................................................................................................21 ANALYSIS DURATION ......................................................................................................................................................................22 NEWMARK COEFFICIENTS ................................................................................................................................................................22 TIME HISTORY ANALYSIS SETTINGS ................................................................................................................................................23

RESULTS ..........................................................................................................................................................................................24

TIME HISTORY RESPONSE ................................................................................................................................................................24 DEFLECTIONS...................................................................................................................................................................................26 FLOOR FORCES.................................................................................................................................................................................27 BASE SHEAR & OVERTURNING MOMENT (OTM) ............................................................................................................................28 TIME-VARIATION OF BASE SHEAR ...................................................................................................................................................29

TIME STEP SIZE SENSITIVITY ..................................................................................................................................................30

FLOOR FORCES AND STOREY SHEARS .................................................................................................................................31

FLOOR FORCES FROM STOREY SHEARS............................................................................................................................................32

EXAMPLE 6: MODAL SPECTRAL ANALYSIS.........................................................................................................................33

PRODUCING A RESPONSE SPECTRA FROM TIME HISTORY ANALYSIS ...............................................................................................34


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S-FRAME RESPONSE SPECTRUM ANALYSIS ...........................................................................................................................35 RESPONSE SPECTRUM ANALYSIS SETTINGS .....................................................................................................................................36 MODAL RESPONSE RESULTS ............................................................................................................................................................37

EXAMPLE 7 – MODAL COMBINATION....................................................................................................................................44

STOREY SHEAR ENVELOPE & FLOOR FORCES..................................................................................................................................46

FINAL COMPARISON ...................................................................................................................................................................50

CONCLUSION .................................................................................................................................................................................50

REFERENCES..................................................................................................................................................................................51

SUGGESTED VALUES FOR DAMPING RATIO .......................................................................................................................52


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1 Example 5

Further assumptions of the building/analysis model either discussed within the body of the example or not stated explicitly are:

a) Supports are fully fixed. b) Floors act as rigid diaphragms. c) Mass distribution is idealized as concentrated at the floors only; hence column and girders elements

are massless. d) Shear stiffness of elements is not considered. e) Beam-column connections are rigid producing a Moment Frame SFRS


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2 S-FRAME Model The frame is input as per the Example details and dimensions. One analysis element per beam/column is used. The X-axis is chosen as the axis of displacement in S-FRAME and since the frame is regular it is convenient to use S-FRAME’s regular framework model generator with the following input:


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2.1 Sections S-FRAME’s Tapered section tool is used to create the ‘Girder’ and ‘Column’ sections (sections do not have to be tapered). Sections created in this manner will render correctly and can be readily edited.

As per assumption (d) the Shear Area (Av) of the sections is set = 0 in which case S-FRAME will not calculate shear effects. There is no reason why this should be done in practice; while shear effects are often small enough to be neglected they nevertheless occur.


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2.2 Materials Since the building mass in the example is idealized as concentrated at the floors a zero density material is created for application to the frame members so extra distributed mass is not introduced. Material Modulus of Elasticity; E = 25 GPa = 25000 N/mm2; (we note that this is a typical value for reinforced concrete for short term loading)

A realistic value of G (shear modulus) is also entered as this is a component of the S-FRAME Beam Element stiffness matrix for both shear and torsional stiffness and hence is generally required for stability of a 3D model.


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2.3 Rigid Diaphragm Floors As an alternative to using Panel Elements, S-FRAME’s Constraint Tool is used to model the specified Rigid Diaphragm Floors. – Brief explanation of constraint concept here -- Firstly Constraints are created for each floor. The Constraint Tool includes the options of applying different translational and rotational masses and eccentricities for each constraint, though this is not required for the Example – all these values are left as zero.

The Constraints are applied in turn to each floor – the option to ‘Auto Find (joints) in Z Plane’ is used in conjunction with ‘Find Closest Joint’. The latter option refers to the Constraint Master Joint (CMJ), since a joint already exists at this position (the middle of each floor). The Master Joint is located at the centre of mass of each floor. When a constraint is applied, S-FRAME confirms this graphically by; colouring the CMJ red, drawing dashed lines from the CMJ to constrained joints to represent the influence of the constraint and assigning the constraint number to joints.


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2.4 Building Mass The ‘Translational Mass’ option of the Constraint is not used since this automatically applies mass in both X and Y directions (to the Constraint Master Joint) while the Example only requires X-Mass. Instead, the Lumped Mass Tool is used to deliberately apply X-Translational mass only to the CMJ’s. This will simplify the Vibration modes of the model. For convenience, since engineers usually work in terms of weight rather than mass, S-FRAME describes mass in terms of force units.

2.4.1 Floor Mass Total mass per unit area; m = 1780 kg/m2

Floor area; Ai = 12m×12m = 144.000 m2

Floor Mass; Mi = m × Ai = 256.320 tonne; (Example rounds this to 256 Mg)

Floor Weight; Wi = 256×103kg×gacc = 2510.5024 kN


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3 Floor Numbers Finally Floor Numbers are applied to joints to identify floors. This enables S-FRAME to calculate storey drifts, shears and floor forces. This completes the modeling process.

4 Static Analysis Although not strictly necessary for the Example, in practice it is sensible to first assess results of an analysis of static loads. We are interested in lateral inertial forces – i.e. a portion of the building weight applied laterally - and the building mass is modeled using lumped mass, so this can conveniently be performed by specifying a Gravitational Factor in the X-axis direction. We could, for example, enter a value of 0.25 for a uniform acceleration of all the masses of 0.25g.


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Linear Static Analysis of this gives the following displacements, reactions and total base shear.

Check; Floor force; Fj = 0.25×Wi = 627.63 kN

Total force; ΣF = 6×wj = 3765.75 kN

It is interesting to note at this point that Reference [3] gives an approximate formula based on Rayleigh’s principle that “is suggested in many building codes” for the fundamental natural period from a static lateral analysis (where perhaps a direct vibration analysis is not available). In practice this could be a good idea to provide an initial idea of the expected building analytical (rather than empirical) frequency

∑∑

=

== N

j jsj

N

j jj

uFg

uwT

1

12

1 2π Eq. 12.23 Ref[3]

Where; wj is floor weight; uj is floor displacement; Fsj is floor force Any convenient set of lateral forces can be used such as code ESFP forces, though to derive these a first guess of the building period is required, the code’s empirical method usually being used. Using the above forces and results gives an approximate period T = 1.15s as follows

Floor u(mm) u (m) W (kN) Fs (N) w (N) Fs*u w*u2

6 103.58 0.10358 2510.5 627625.6 2510502.4 65009.46 26934.725 96.42 0.09642 2510.5 627625.6 2510502.4 60515.66 23339.684 84.05 0.08405 2510.5 627625.6 2510502.4 52751.93 17735.23 66.1 0.0661 2510.5 627625.6 2510502.4 41486.05 10968.912 42.99 0.04299 2510.5 627625.6 2510502.4 26981.62 4639.761 17.08 0.01708 2510.5 627625.6 2510502.4 10719.85 732.3798

Σ = 257464.6 84350.65

T = 1.15 s


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5 Vibration Analysis Before running Time History Analysis, the vibration characteristics of the structure are assessed and verified. The following Vibration settings are initially used; Jacobi Threshold Eigenvalue Extraction Method (since the model is relatively small and has few modes) and 6 Eigenvalues (mode shapes) requested.

Since the mass has been idealized S-FRAME finds only X-direction modes – these can be viewed in the Numerical Results Frequency Spreadsheet.


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5.1 Vibration Results Comparison Comparing S-FRAME’S results to the example there is excellent agreement. We note that the approximate fundamental period of 1.15s from a Static Analysis discussed in section 4 above was indeed a good approximation to the first mode’s period.

S-FRAME

Reference


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6 Time History Analysis 6.1 Time History Load The Example specifies the N-S component of the ‘El Centro’ 1940 record which can be found in the S-FRAME library of Accelerograms. A new Time History Load case is created and the records accessed as follows.

The Support Excitation Type and the chart button clicked to access the record library files.

The correct record (in the ‘Accsmall.dac’ file) is #14 “Imperial Valley Earthquake – El Centro”. There are two such records with this main title, the header description gives the direction – ‘270 Degrees’ equating to the N-S direction. To apply the record for analysis in the X-axis direction it is selected and OK’d then an X-Scale factor <>0 is entered. We note that sign (not the magnitude) of the results of the Example are consistent with –ve values of the Accelerogram data presumably due to the choice of sign convention for analysis. Hence to obtain the same signs as the Example results X-Scale = -1 is entered.


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6.2 Damping & Rayleigh Damping Coefficients S-FRAME employs a Newmark direct time integration method for Time History Analysis, not the modal combination method used in the Example. For more detailed information on Time History Analysis theory and solutions see S-FRAME’s Theory Manual, and References.

While the Example sets a constant damping value for all modes, this is not possible for the direct integration method employed by S-FRAME. Damping is a function of frequency and is introduced via the Rayleigh Damping Coefficients α and β which are used to form the damping matrix [C]. It is important to note that these coefficients are NOT themselves damping values. For non-zero values of both α and β a desired value of damping ζ can be set precisely only for two frequencies, ωr and ωs. Note that the damping value ζ is a percentage (of critical damping) and the frequencies are angular frequencies ω in terms of radians/s, not Hz. The values of α and β which produce the desired damping value at these two frequencies only can then calculated (from the following equation (see References 1. and 2) and input into S-FRAME for analysis.

Damping for other modes is a function of their frequency and hence will ≠ ζ (the desired damping). Ideally the damping should be reasonably close to the desired value for modes which contribute significantly to the response. Hence a prior frequency analysis such as has already been performed is generally required to determine the model’s frequency characteristics and make a rationale choice of two frequencies for which to calculate α and β. If only one or two frequencies dominate the response then the choice is obvious. Otherwise two frequencies can be chosen which give a reasonable approximation to the desired damping for a range of frequencies in which the modes which contribute most to the response fall. Reference [1] states:

It is convenient to take ωr as the value of the fundamental frequency and ωs as the frequency corresponding to the last of the upper modes that significantly contribute to the response. This way the first mode and mode s will have exactly the same damping, and all modes in between will have somewhat smaller similar values and the modes with frequencies larger than ωs will have larger damping values thus reducing their contribution to response.

Since the Example uses a modal time history approach method of solution, where damping is explicitly set to precisely 5% for all modes, and since this is not possible in S-FRAME due to the nature of the direct time integration solution, we do not expect precisely the same results.


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As discussed in the Example, modes 1 and 2 dominate the response hence the frequencies of these are used to derive α and β from the relationship given above for 5% damping:

sr

sr

ωωωξω

α+

=2

and sr ωω

ξβ+

=2

Damping; ζ = 5% ; 1st mode; ωr = 5.436; 2nd mode; ωs = 17.372

α = 2×ζ×ωr×ωs/(ωr + ωs) = 0.414040

β = 2×ζ/(ωr + ωs) = 0.0043844

These values produce the following damping vs frequency plot

Rayleigh Damping

0.865 2.765

4.26%

0.00%

2.00%

4.00%

6.00%

8.00%

10.00%

12.00%

14.00%

0 2 4 6 8 10Frequency (Hz)

Dam

ping

Rat

io


The damping for other frequencies ≠ 5%, as can be seen from the plot, and can be determined precisely from;

22i

ii

βωωαζ +=

Mode 3; f3 = 5.113 Hz; ω3 = 32.128; ζ2 = α/(2×ω3) + (β×ω3)/2 = 7.7 %

Mode 6; f6 = 13.42 Hz; ω6 = 84.322; ζ3 = α/(2×ω6) + (β×ω6)/2 = 18.7 %


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6.3 Constant time step size Next the analysis Time Step size ∆t is considered. The time step size should be sufficiently small for accurate analysis. However, the smaller the time step size, the higher the ‘cost’ (in terms of computational time) of analysis for a given duration of analysis – e.g. the first 15s of the input record. Additionally the analysis duration is limited by the maximum number of allowable time steps which is 32,767 (a limit imposed by S-FRAME’s current software architecture). So ideally the time step size should be no smaller than it need be. The S-FRAME Theory Manual pg 34 gives a rule of thumb for the time step as follows:

Where Ω = the highest (angular) frequency component of the forcing function in rads/s and Ω∗ = 4Ω. Since we

more commonly think in terms of period the above expression can be conveniently re-formulated as follows:

Ω==

π21f

T 80420

ffcr

TTt =

×≤∆⇒

Where Tf = 2π/Ω. However, what if Tf is unknown or not applicable? The rule of thumb derives from two main principles, the first of which is illustrated by the following figure


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Principles:

1. Refer to the figure above; for a given forcing (or input) frequency Ω (ωA in the figure) the response of the system for a given natural frequency ωi (ωo in the figure) generally reduces as these frequencies diverge. The divergence is conveniently expressed in terms of the ratio of the forcing frequency to the natural frequency Ω/ωi. The figure illustrates this phenomenon. Furthermore it can be seen that the response in modes with a small ratio Ω/ωi is essentially static and the response in modes with a large ratio Ω/ωi is negligible. Below Ω/ωi = 1/4 further increases in ωi do not produce significant change in response hence frequencies (of response) higher than 4Ω = Ω∗ need not be considered.

2. Around 20 equal time intervals are required to discretize both the input and response motions with sufficient accuracy.

These principles explain the origin of the values of 4 and 20 in the rule of thumb. Thus a rational choice of time step size is based on a) deciding which is the highest frequency to be discretized considering both input and response and b) dividing the resulting (lowest) period by 20

From this we can derive a more general rule. Let us call the highest frequency of the response fr with

corresponding period Tr.

• If Tf is known or guessed at to a reasonable degree then it is not necessary to know Tr (following Principle

1) and from Principle 2. we cant set the time step ∆t = (Tf /4)/20 = Tf /80.


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• If Tf is unknown or inapplicable (as in Example 4) we consider the likely highest significant response

frequency fr = 1/Tr and set ∆t = Tr /20.

6.4 Accelerograms & Sampling Accelerograms with discrete data points such as those found in the S-FRAME Library are essentially a digital record of an analog signal and thus the maximum possible frequency content is finite and furthermore is determined by the sampling rate or frequency according to the following theorem:

The Nyquist–Shannon sampling theorem states that for an infinite record perfect reconstruction of a signal is possible when the sampling frequency is greater than twice the maximum frequency of the signal being sampled.

Stated mathematically; fs > 2fmax

where: fs is the sampling frequency and

fmax is the max frequency of signal (in this case the earthquake) For the ‘El Centro’ record the data points are given every 0.02s and, moreover, the record is not infinite. Thus:

Sampling period; Ts = 0.02s; Sampling rate; fs = 1/Ts = 50.000 Hz Max frequency; fmax < fs/2 = 25 Hz; Min period; Tmin > 2×Ts = 0.04s

So, according to the theorem, the highest frequency component of the El Centro record is < 25 Hz The time step size rule of thumb discussed above derives from this value; Ω the highest frequency component (in radians/s) of the forcing function. For a given accelerogram Ω may not be known, but from the preceding discussion we know it cannot contain (i.e. describe) frequencies higher than half the sample rate. Thus a practical lower bound can be calculated for the time step size from the record sample rate (or period).

sT

tTT scrsf 40

2 ≤∆⇒>

Thus for our ‘El Centro’ accelerogram min time step; ∆tcr = Ts/40 = 0.0005 s


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6.5 Frequency Content Another rationale for Tf and hence time step size would be from the frequency content of the earthquake record. This is commonly assessed by performing a frequency analysis of the record – see Reference [3]. For example the program NONLIN will do this automatically – below is a screenshot from NONLIN of a Fourier Amplitude Spectrum for the same El Centro record as that used in the Example. We note that the maximum of the Frequency Range of the analysis is 24.99, which accords with the previous discussion.

Such an analysis may indicate, for example, that the periods of interest in the record – those which contain

significant energy – lie in the range 0.1s-10s. In this case we could take 0.1s as a likely value for Tf : Tf = 0.1s; ; ∆t ≤ Tf/80 = 0.0013 s


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A time step size ∆t = 0.001s was chosen as a reasonable initial value from the foregoing and also since this produces a step at t=3.08s for which the Example gives results.

It should not be forgotten that the initial value of ∆t derives from a rule of thumb. Thus it is not guaranteed to be the precise value which produces the best accuracy in all circumstances without exception. In practice some experimentation with time step size may be required; to test for sensitivity of results to this value and also from

making some educated guesses about the value of Tr from a review of the results of an initial analysis. Thus obtaining good results may be an iterative process. Additionally some compromise may be required depending on the desired analysis duration and acceptable analysis cost (i.e. time), since these are both determined by the time step size.

6.6 Analysis Duration As discussed this has an upper limit deriving from the max possible number of time steps and the time step size ∆t.

Duration Limit; Tmax = 32,767 ∆t

The Example’s analysis duration of 15s is used. For a constant time step, the analysis duration is not explicitly

entered; the user inputs the time step size ∆t and total time steps N from which the duration derives.

∆t = 0.001s; Desired analysis time; TA = 15s

Total time steps required; N = TA/ ∆t = 15000 The peak response may not occur during excitation and furthermore the time to peak response (or steady state in the case of a periodic forcing function) is generally not known. Hence in practice experimentation may be required with analysis duration having viewed initial results.

6.7 Newmark Coefficients These are generally one of two sets of values as follows. See S-FRAME Theory Manual and References for more information.

Alpha Beta

Zero Damping 0.2525 0.5050

Non-zero Damping 0.25 0.5 Since the example has specified damping the appropriate values are input in S-FRAME.


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6.8 Time History Analysis Settings The preceding discussion leads to the following initial analysis settings. S-FRAME’s “Linear Dynamic Time History” option is selected which is equivalent to the Example’s ‘step by step’method employing Newmark’s Beta method – some references term this Response History Analysis.

To limit the amount of output the option to Output to file (i.e. results) every 10 time steps is set – this in conjunction with the time step size will give a time step at t = 3.08s for comparison with the Example.


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7 RESULTS 7.1 Time History Response

The response at a chosen joint can be readily assessed by right-clicking the joint and choosing ‘Response Time

History…’ from the context menu: S-FRAME plots the chosen result parameter – e.g. X-Displacement – vs time so the user can easily assess; the nature of the response, the maximum response and the approximate time at which this occurs. There is a Trace function to assist with this – using this it can be seen that the maximum +ve X-displacement response occurs at around 6s for example.


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A joint in the Roof level (any, since they all have the same displacement due to the rigid diaphragm behaviour) is selected in this manner to produce the following X-displacement response

It can be seen immediately that the shape of the response compares very closely with that of the Example.

The precise maximums and minimums for a Joint can be found in S-FRAME from the Numerical Results Spreadsheet by:

1. Select a single joint (e.g. any Roof Joint) 2. Select all time steps for viewing:

3. Use the find function to find the maximum and minimum values and corresponding time step

This produces the following results;

Max Displacement = 146.405 mm @ t = 5.92s Min Displacement = -130.219 mm @ t = 3.04s


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7.2 Deflections

More detailed results for each time step are also available. S-FRAME presents the results of each time step as a load case for which all the usual results (both Graphical and Numerical) are available as for a static analysis. We wish to compare the displacement results, so the displacement diagram is chosen, viewing of Dx (X-displacement) values is enabled and load case 311 corresponding to t = 3.08 s is selected.

The results are given (in meters) on page 43 of the Example for t = 3.08s

Floor Displacement (mm) Reference S-FRAME

Roof 125.33 124.92 5th Floor 117.51 117.04 4th Floor 103.31 102.88 3rd Floor 79.56 81.18 2nd Floor 48.55 52.31 1st Floor 18.23 20.28


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7.3 Floor Forces Floor forces, which can be compared with those in the Example, are available in Numerical Results.

The Example gives the following values in kN for time t = 3.08s

Floor Force (kN) Reference S-FRAME Roof -658.49 -658.17

5th Floor -636.50 -724.87 4th Floor -920.33 -883.99 3rd Floor -1041.9 -933.26 2nd Floor -663.43 -742.89 1st Floor -308.29 -407.15


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7.4 Base Shear & Overturning Moment (OTM)

The Base Shear, V, and Overturning Moment for each time step can also be readily viewed graphically:

Reference S-FRAME Total Base Shear (kN) -4229.0 4350.3

Total Overturning Moment (kNm) -46727 47406


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7.5 Time-Variation of Base Shear From the Example:

S-FRAME does not automatically produce such a plot. However, the Numerical Results Base Shear Spreadsheet for all time steps can be directly copied to Excel, for example, to produce such a plot. This was performed to produce the following which shows good agreement in both general shape and maximum and minimum

Base Shear vs Time

5.88, 4,384.13

3.07, 4,363.42-5000

-2500

0

2500

5000

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

Time (s)

Bas

e Sh

ear (

kN)


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8 Time Step Size Sensitivity Examining the sensitivity of results to time step size, it can be seen that using ∆t = 0.01s which is 10× larger than the initial value does not change results significantly. Since, as the Example discusses, the response is dominated by the first 2 modes this makes sense; even though the input contains higher frequency signals and the model exhibits higher frequency modes these do not contributed greatly to the response. As discussed, around 20 (equal time) intervals are required to discretize the response motions with sufficient accuracy. So for the highest significant response frequency (of mode 2) this gives ∆t ≤ 0.362/20 = 0.018s > 0.01s.

Floor Displacement at t = 3.08s (mm) Reference

S-FRAME ∆t = 0.001s


% Change

Roof 125.33 124.92 124.78 0.1 5th Floor 117.51 117.04 117.02 0.0 4th Floor 103.31 102.88 102.93 0.0 3rd Floor 79.56 81.18 81.19 0.0 2nd Floor 48.55 52.31 52.28 0.1 1st Floor 18.23 20.28 20.25 0.1

Reference S-FRAME

∆t = 0.001s S-FRAME ∆t = 0.01s

% Change Max Displacement

(mm) δ (mm) t (s) δ (mm) t (s) δ (mm) t (s) δ t

Maximum 148.729 -- 146.405 5.92 146.334 5.89 0.0 0.5 Minimum -128.367 -- -130.219 3.04 -130.226 3.01 0.0 1.0

Floor Force at t = 3.08s (kN) Reference



% Change

Roof -658.49 -658.17 -640.38 2.7 5th Floor -636.50 -724.87 -733.08 -1.1 4th Floor -920.33 -883.99 -900.31 -1.8 3rd Floor -1041.9 -933.26 -936.82 -0.4 2nd Floor -663.43 -742.89 -730.57 1.7 1st Floor -308.29 -407.15 -399.64 1.8

At t = 3.08s (kN) Reference S-FRAME

∆t = 0.001s S-FRAME ∆t = 0.01s

% Change

Total Base Shear (kN) -4229.0 4350.3 4340.79 0.2 Total Overturning Moment (kNm) -46727 47406 47340 0.1


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9 Floor Forces and Storey Shears Although not necessary, it can be shown that the Storey Shears can be obtained graphically using the Freebody Forces Tool by selecting only the elements which comprise the Freebody above a particular floor. S-FRAME calculates the resultant forces acting on the Freebody (from unselected elements) from equilibrium, the sum of which is the total storey shear and is reported in the legend.

Freebody

If the above is performed for successive storeys the building shear distribution can be developed as shown below (for a particular time step, in the case of Time History Analysis). These are the same as the Storey Shears (e.g. Shear X) reported in the Numerical Results Storey Forces Spreadsheet. The Floor Forces can be derived from these as shown below.


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9.1 Floor Forces from Storey Shears

Storey Shear Envelope t = 3.08s

-1383.05

-658.17

-4350.33

-2267.03

-3200.29

-3943.18

0

3

6

9

12

15

18

-6000.00 -5000.00 -4000.00 -3000.00 -2000.00 -1000.00 0.00

Shear (kN)

Floo

r h

(m)

Fx6

Vx6Fx5

Vx5Fx4

Vx4Fx3

Vx3Fx2

Vx2Fx1

Vx1

The Floor forces Fxi can be determined from the Storey Shears Vxi from:

)1()()( +−= ixixix VVF

For Example;

Shear storey 4; Vx4 = -2267.03 kN; Shear storey 5; Vx5 = -1383.05 kN

Force at Floor 4; Fx4 = Vx4 –Vx5 = -884.0 kN


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10 Example 6: Modal Spectral Analysis

Example 6 goes on to look at an alternative analysis of the same record using the Displacement Response Spectrum of the ground motion. This is the response to the ground motion record of a SDOF (single degree of freedom) system for a range periods.


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10.1 Producing a Response Spectra from Time History Analysis S-FRAME can automatically produce a response spectrum plot for any joint in the structure. In practice, this might be used for the design of a component attached to a part of a structure – e.g. the roof. The accelerations experienced by such a component can be very different to the ground motions – accelerations at points in a structure can be highly amplified depending on its dynamic characteristics. The generated Spectra (at a joint) are accessed in the same manner as the Time History Response; by right-clicking the desired joint and selecting the Response Time History option from the resulting context menu (see page 24) then choosing the Response

Spectra tab. To obtain the ground motion response spectrum this is performed for a support joint.

This is performed in S-FRAME to produce a plot similar to that of the example – note that the following are specified to produce a comparable plot;

• Relative = OFF (i.e. Absolute) • Damping value = 0.05 = 5% • Plot Option = Sd (displacement), Direction = X • period range End Period = 2s

Compare this to the plot in the Example (see above). The plot can be printed and also saved to an input file for a Response Spectrum Analysis (a *.DRS file) by clicking the Save As… button.


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10.2 S-FRAME Response Spectrum Analysis An input response spectrum file was produced from the preceding Time History Analysis as discussed above in order to perform a Response Spectrum Analysis in S-FRAME to compare results to Example 6 and 7. A Response Spectrum type load case is created and the generated spectra file selected. Similarly to Time History Analysis, a scale factor <>0 is required to specify the application of all or a portion of the curve in a particular direction.

• X-Scale = 1.0 is entered


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10.3 Response Spectrum Analysis Settings The Unstressed Response Spectrum analysis option is chosen. Note that the majority of the analysis settings are associated with a Vibration Analysis, since this is a precursor to Response Spectrum calculations. These settings are the same as those previously used in the investigation prior to the Time History Analysis requesting 6 Eigenvalues (mode shapes). The Scale options are not used, since the Examples do not discuss scaled or Design Code-based results.


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10.4 Modal Response Results Though ultimately we are interested in the combined response of all modes, the Example first looks at the Modal response – the response of each individual mode. The modal response is a single displacement vector of the structure for which it is in static equilibrium (an important point). By default S-FRAME initially presents the results of modal combination for the RSA Load case, since these are usually of primary interest and it is necessary to toggle on the Modal Response option – e.g. modal displacements can be viewed in S-FRAME by selecting the following options when viewing graphical results; Modal Response = ON, Direction = X, Mode Shape = 1, displacement values = Dx

These exhibit good agreement to the results of the Example; all mode shapes are displayed on the next page.


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S-FRAME Results; Modal displacements (mm)


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The Example gives the following detailed displacements for each mode.

Comparable results can be viewed in S-FRAME’s Numerical Results as follows by selecting a single joint in each floor – here only the CMJ’s are selected - and choosing Modal Response for All Mode Shapes


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10.4.1 Displacement Comparison Mode 1 Floor

Displacement (mm) Reference S-FRAME


10.4.2 Storey Drift

Comparable results can be viewed in S-FRAME’s Storey Drift spreadsheet:


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10.4.3 Maximum Modal Lateral Forces

Comparable results can be viewed in S-FRAME’s Numerical Results Floor Forces by choosing Modal Response for All Mode Shapes


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10.4.4 Modal Lateral Force Comparison

Reference S-FRAME Modal Lateral Floor Force (kN) Mode1 Mode2 Mode1 Mode2

Roof 1108.3 -748.9 1108.2 -721.6 5th Floor 1016.2 -264.8 1016.5 -255.8 4th Floor 860.2 331.8 863.5 314.1 3rd Floor 632.9 761.5 654.2 699.3 2nd Floor 369.4 765.9 405.7 692.2 1st Floor 135.1 363.0 152.5 329.0

10.4.5 Graphical Modal Lateral Forces The Modal Lateral Forces can be viewed graphically also as follows. S-FRAME displays the forces at the CMJ’s. In the following screenshot the Reactions Tool is activated and Modal results for Mode 1 are displayed so the Lateral forces and resulting Base Shear and OTM can be viewed simultaneously.

* note that since the model is viewed from the front (i.e. the X-Z plane) in this screen-shot the reactions at support joints are superimposed on top of each other. Thus there are three reactions with the value 418.61 (kN) for example. The reactions thus do in fact sum to the same as the total base shear displayed in the Legend:

Total Reaction; 3×(-418.61kN -562.94kN – 418.61kN) = -4200.48 kN


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10.4.6 Base Shear & Overturning Moment From Example:

Comparable results can be viewed in S-FRAME’s Numerical Results Base Shears Spreadsheet by choosing Modal Response for All Modes.

RSA Modal Base Shears (kN) Reference S-FRAME

Mode 1 4122.1 4200.5 Mode 2 1208.5 1057.2 Mode 3 444.6 419.8 Mode 4 257.9 250.9 Mode 5 106.1 94.8 Mode 6 29.1 26.2


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11 Example 7 – Modal Combination In S-FRAME modal combination of results does not require a further analysis – these results are available for a single Response Spectrum analysis run. When the load case is created, the Modal Combination Method is specified on the same page as the direction scale factor. The Example first looks at the SRSS combination method.

11.1.1 Maximum Credible Storey Displacements All the following results in S-FRAME are viewed with the Modal Result option OFF and are hence the Modal Combination values which result from applying the Modal Combination Method (MCM) to the modal results.

Example


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11.1.2 Modal Combination Storey Shears & Floor Forces The Example warns about the dangers of applying the modal combination method to the modal floor forces – this does not produce meaningful results. First the modal combination (maximum credible) Storey Shears are calculated

S-FRAME performs this operation automatically for all storeys by applying the MCM to the modal Storey shears:

Comparison

RSA SRSS Modal Combination Storey Shears

(kN) Reference S-FRAME

6th 1417.6 1393.9 5th 2369.8 2351.0 4th 3080.3 3080.0 3rd 3640.1 3663.8 2nd 4080.2 4116.7 1st 4327.6 4360.1


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11.2 Storey Shear Envelope & Floor Forces A plot of the Storey Shears Vxi produces the Storey Shear Envelope. The modal combination Floor Forces Fxi are derived from this envelope, not by applying the MCM to the modal Floor Forces.

RSA SRSS Storey Shear Envelope & Floor Forces

Floor Force, 243.37

Floor Force, 452.91

Floor Force, 583.85

Floor Force, 728.99

Floor Force, 957.09

4116.73

3663.82

3079.97

4360.11

1393.89

2350.98

Floor Force, 1393.89

0

3

6

9

12

15

18

0.00 1000.00 2000.00 3000.00 4000.00 5000.00

Shear, Force (kN)

Floo

r h (m

)

The Floor forces Fxi are calculated from Storey Shears Vxi from:

)1()()( +−= ixixix VVF

For Example;

Shear storey 4; Vx4 =3079.97 kN; Shear storey 5; Vx5 = 2350.98 kN

Force at Floor 4; Fx4 = Vx4 –Vx5 = 729.0 kN


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11.2.1 Base Shear From the Example (this is of course the same result as the shear for Storey 1)

A comparable result is directly available from the S-FRAME Base Shear results Spreadsheet.

11.2.2 Overturning Moments Storey: Example S-FRAME Storey Forces Spreadsheet

Base:

A comparable result is directly available from the S-FRAME Base Shears results Spreadsheet.

Comparison

RSA SRSS Modal Combination Base Results Reference S-FRAME

Base Shear (kN) 4327.6 4360.1 Overturning Moment (kNm) 53865.8 54374.2


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11.2.3 Static Equivalent Lateral Forces As noted these cannot be derived by applying the MCM to the modal Floor Forces Fxi . They are derived from the Modal Combination Storey Shears as disucussed on page 32 above.

Comparison

RSA SRSS Modal Combination Equivalent

Lateral Force (kN) Reference S-FRAME



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The Example notes that the Overturning Moment resulting from these forces does not equal that computed by applying the MCM to the modal Storey Overturning moments:

This result can be reproduced easily in S-FRAME using the option to Generate a Load case of Equivalent Lateral Forces (ELF’s) from the results of a Response Spectrum Analysis Load case. This is performed in the Loads view via the command Edit/Generate Equivalent Static Loads from RSA Case… The ELF’s are automatically applied to the CMJ’s. Analysis of this load case produces the following result; the Base shear is the same as the MCM base shear, but the OTM is not and is comparable to that computed from the ELF’s of the Example.


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12 Final Comparison From the Example

S-FRAME

Parameter Example 5

THA

Example 6 RSA

Absolute Combination*

Example 7 RSA

SRSS Combination

Example 7 RSA

CQC Combination

Roof lateral displacement (mm) 146.4 158 146.9 146.8

Base shear (kN) 4 380 6 049 4 360 4 372 Overturning

Moment (kNm) 54 588 57 498 54 374 54 368

*These Absolute Sum Combination values can be obtained directly from S-FRAME by choosing this as the MCM for the Response Spectrum Load case.

13 Conclusion Generally we see excellent agreement in results between the example and S-FRAME. Small differences are explained by for example: different method of time history solution (modal vs direct integration) involving different damping % for some modes; some rounding in the Examples ‘hand’ computations; the Example does not consider other phenomenon like elastic shortening of columns which are implicit in the S-FRAME analysis; possible differences in the accelerogram and/or the displacement response spectrum data used by the Example.

It is interesting to note that S-FRAME’s results for RSA SRSS Combination Base Shear and OTM agree almost exactly with those of the Examples step-by-step (time history). S-FRAME’s results for a CQC combination are also included for comparison, showing that there is negligible difference to those of SRSS combination in this case.


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14 References [1] Luis E. García & Mete A. Sozen, Multiple Degrees of Freedom Structural Dynamics, Purdue University CE571 – Earthquake Engineering, 2002

[2] Anil K. Chopra, Dynamics of Structures: Theory and Applications to Earthquake Engineering (2nd Edition), Prentice Hall, 2000 [3] Roberto Villaverde, Fundamental Concepts of Earthquake Engineering, CRC Press, 2009


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15 Suggested Values for Damping Ratio The following Table is reproduced from Reference [2].

Stress Level Type and Condition

of Structure

Damping Ratio

(%)

Working stress,

no more than about

21 yield point


concrete, well-reinforced

concrete (only slight cracking)

2-3

Reinforced concrete with

considerable cracking

3-5


Wood structures with nailed or bolted

joints

5-7

At or just below

yield point


concrete (without complete loss in

prestress)

5-7

Prestressed concrete with no prestress left 7-10

Reinforced concrete 7-10


Wood structures with

bolted joints

10-15

Wood structures with nailed

joints

15-20


Dynamic Example

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Transcript of Dynamic Example