Time History Analysis
FEM-Design 21
FEM-Design
Time History Analysis
Ground Response Spectra
Level Response Spectra
Structural Response Analysis
Forced Vibration Analysis
25
xg
x
20
Se [m/s^2]
m
k
15
10
5
0
ag
0
1
2
Tn [s]
version 1.1
2021
1
3
4
Time History Analysis
FEM-Design 21
StruSoft AB
Visit the StruSoft website for company and FEM-Design information at
www.strusoft.com
Time History Analysis
Copyright © 2021 by StruSoft, all rights reserved.
Trademarks
FEM-Design is a registered trademark of StruSoft.
Edited by
Zoltán I. Bocskai, Ph.D.
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Contents
List of symbols...............................................................................................................................4
1 Basics of time-history analysis....................................................................................................5
1.1 The general dynamic differential equation system and its solution.......................................5
1.2 The mass matrix.....................................................................................................................6
1.3 The Rayleigh damping matrix................................................................................................6
1.4 Time step and initial conditions.............................................................................................8
1.5 Calculations and results.........................................................................................................9
1.6 Restrictions of the calculation..............................................................................................10
2 Ground acceleration calculation................................................................................................11
2.1 Acceleration response spectra calculation...........................................................................12
2.1.1 Elastic pseudo acceleration response spectra.................................................................13
2.1.2 Design pseudo acceleration response spectra................................................................15
2.2 Structural response analysis.................................................................................................17
2.3 Level acceleration response spectra calculation..................................................................20
3 Forced vibration with arbitrary functions..................................................................................22
4 Verification examples................................................................................................................26
4.1 Response spectra calculations..............................................................................................26
4.1.1 Response spectra calculation from harmonic accelerogram..........................................26
4.1.2 Response spectra calculation from El Centro earthquake accelerogram.......................30
4.2 Time-history calculation with harmonic ground acceleration on MDOF system................35
4.3 Level spectrum calculation with harmonic accelerogram on MDOF system......................44
4.4 Time-history calculation with excitation force....................................................................47
4.4.1. Excitation with constant force on undamped SDOF system.........................................47
4.4.2. Excitation with constant force on damped SDOF system.............................................51
4.4.3. Excitation with harmonic force on undamped system..................................................54
4.4.4. Excitation with harmonic force on damped system......................................................60
References.....................................................................................................................................65
Notes.............................................................................................................................................66
Download link to the example files:
Link to verification example files
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List of symbols
ag(t)
ground acceleration function
C
damping matrix
f0
eigenfrequency of the SDOF system
f 0i
i-th eigenfrequency of the MDOF system
K
stiffness matrix
M
diagonal (lumped) mass matrix
q(t)
excitation force vector as a function of time
Se
pseudo acceleration
t
time
T
time period
v0
initial velocity vector
umax
maximum translational displacement
x(t)
displacement vector as a function of time
ẋ ( t)
velocity vector as a function of time
ẍ ( t)
acceleration vector as a function of time
ẍ g (t)
ground acceleration function
x0
initial displacement vector
α
Rayleigh damping matrix coefficient
β
Rayleigh damping matrix coefficient
Δt
time step
ξ
critical damping ratio
ξi
critical damping ratio of the i-th eigenfrequency
ω0
angular natural frequency of the SDOF system
ω0 *
damped angular natural frequency of the SDOF system
ω0i
i-th angular natural frequency of the MDOF system
ω
angular frequency of a harmonic excitation force
LTHA
linear time-history analysis
SDOF
single degree of freedom
MDOF
multiple degrees of freedom
DAF
dynamic amplification factor
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1 Basics of time-history analysis
1.1 The general dynamic differential equation system and its solution
In FEM-Design version 20 the so-called linear time-history analysis (LTHA) was implemented,
therefore only the theory connected to this topic will be introduced. Calculation of the dynamic
response of the structure is a very complicated issue. The dynamic load and response of the
structure could come from forced vibration (e.g.: time dependant excitation force) or from
ground accelerations (e.g: seismic effect). These kind of excitations will cause dynamic
responses on the structure. The results of the structural responses primarily the dynamic
amplification factor (DAF), the dynamic displacements and accelerations of different points of
the structure.
The second order linear inhomogeneous differential equation system which describes the
behaviour of the structure in case of excitation force is the following, see Ref. [1]:
M ẍ (t )+C ẋ (t)+ K x(t)=q(t ) ,
(Eq. 1.1)
where M is the diagonal mass matrix from the masses and the mass converted loads on the
structure (see Chapter 1.2). C is the so-called Rayleigh damping matrix (see Chapter 1.3). K is
the linear global structural stiffness matrix. q(t) is the load vector as a function of time. x(t) is
the displacement vector as a function of time. ẋ (t) is the velocity vector as a function of time.
ẍ ( t) is the acceleration vector as a function of time. Eq. 1.1 describes the so-called multiple
degrees of freedom (MDOF) system. Fig. 1.1 shows a single degree of freedom (SDOF) system
where the mentioned matrices are scalars.
K
C
x(t)
M
q(t)
Figure 1.1 – SDOF dynamic system with damping
The solution of the differential equation of motion is achieved by the direct integration method.
The used integration rule is the so-called Newmark or Wilson-Θ method depending on the
selected settings. The two different calculation methods follow what you can find in Ref. [1-2]
about the direct integration techniques.
WARNING: By these numerical techniques – analyzing harmonic responses – amplitude decay,
periodic elongation and by Wilson-Θ method numerical damping in the system could appear.
For further information see Ref. [1-2].
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1.2 The mass matrix
Usually by these calculations the mass matrix M in Eq. 1.1 is time independent and only
contains the mass of the structure (from dead loads) and masses which comes from specific
parts of some variable actions (see e.g.: EN 1998-1-1 Eq. 3.17).
In FEM-Design these could be user defined mass point or points and masses from the relevant
converted loads similarly by the eigenfrequency calculation (see Fig. 1.2).
FEM-Design during the calculation considers the lumped (diagonal) mass matrix in the finite
element mass matrix formulation.
Figure 1.2 – The load case to mass conversion dialog in FEM-Design
1.3 The Rayleigh damping matrix
The so-called Rayleigh damping matrix in Eq. 1.1 contains a mass-proportional and a stiffnessproportional part.
C=α M + β K
(Eq. 1.2)
The first term (mass-proportional part) refers to external damping (as a point damping), while
the second term (stiffness-proportional part) refers to coupled internal damping (as a point-topoint relationship). For further information see Ref. [1] and Fig. 1.3.
α m3
m3
x3
α m2
α m1
m2
k3
x2
k2
m1
x1
k1
β k3
β k2
β k1
Figure 1.3 – Visualization of Rayleigh damping,
mass-proportional (left) and stiffness-proportional (right)
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The necessary α and β scalar parameters depend on the analyzed structure. The recommended
values according to Ref. [1-2] are as follows, see Fig. 1.4:
1 α
+β ω 0i
2 ω 0i
)
(Eq. 1.3)
α + β ω 20i =2 ω 0i ξ i
(Eq. 1.4)
ξ i=
(
We can rearrange this equation:
Based on this equation the α and β parameters can be calculated by specifying 2-2 angular
frequencies and damping ratios. In the expression ω0i is the i-th angular natural frequency and ξi
is the i-th damping ratio related to the i-th mode shape.
ξn
Rayleigh damping
ξ n=
1 α
+β ω n
2 ωn
(
)
ξ
Stiffness
proportional
Mass
proportional
1
ξ n= β ω n
2
1 α
ξ n= ω
2 n
ωi
ωj
ωn
Figure 1.4 – Interpretation of Rayleigh damping
For example:
ξ 1 =0.03 ; ω 01=4 rad /s
ξ 2=0.12 ; ω 02=17 rad /s
Using the former equation to the 2-2 angular frequencies and damping ratios:
α +16 β =0.24
α +289 β =4.08
Based on these two equations:
α =0.01498
1
; β =0.01405 s
s
According to Ref. [1] the variations of modal damping ratios with natural frequencies are not
consistent with experimental data that indicate roughly the same damping ratios for several
vibration modes of a structure. If both modes are assumed to have same damping ratio ξ, which
is reasonable based on experimental data, then the parameters can be calculated as follows:
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2 ω 0i ω 0j
2
and β =ξ ω +ω
α =ξ ω +
0i ω 0j
0i
0j
(Eq. 1.5)
Or expressed with the natural eigenfrequencies:
α =ξ
4 π f 0i f 0j
and
f 0i + f 0j
ξ
β=π
1
f 0i + f 0j
(Eq. 1.6)
SUGGESTION: It is advisable to calculate first the eigenfrequencies and select the two specific
eigenfrequencies (and the associated vibration shapes) based on this calculation. The most
obvious method is to choose the two eigenfrequencies which provide the highest effective
masses. In FEM-Design the effective masses after the eigenfrequency calculation can be found
in the setup menu of the seismic calculation.
WARNING: If someone would like to consider only the so-called equivalent Kelvin-Voigt
damping, in that case the Rayleigh damping parameters:
ξ
α =0 and β =
(Eq. 1.7)
π f 0i
1.4 Time step and initial conditions
By a discrete model and solving the differential equation system with the direct integration
method besides the Rayleigh damping parameters another key input data is the Δt time step.
The adequate time step depends on the function of the excitation and its variablility. For
example in case of ground acceleration usually the input accelerograms contain the acceleration
data to every 0.01 or 0.02 s, therefore the Δt time step in this case should be this value.
The initial conditions are necessary to solve the differential equation system (Eq. 1.1). In FEMDesign it is considered in the following way.
In case of excitation force (forced vibration) the initial displacements at t = 0 time is x(0) = x0 ,
where x0 is the static displacements of the structure calculated with the loads from the specific
dynamic load combination at t = 0. The initial velocity is ẋ (0)=v 0=0 . For an example see
Chapter 4.4.1.
In case of ground acceleration the initial relative displacements at t = 0 time is x(0) = 0. The
initial relative velocity is ẋ ( 0)=v 0=0 . For further information see Chapter 2.
We can see from the algorithms (Newmark or Wilson-Θ) of the direct solution of the differential
equation in Ref. [1] that the control parameter is the Δt time step. The correctly chosen Δt is
essential to obtain sufficiently accurate results and to the stability of the numerical solution.
In case of excitation force as a first guess the Eq. 1.8 time step is okay but always need a
convergence analysis to see the reduction of the time step how affects the results:
Δ t=
T
,
20
8
(Eq. 1.8)
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where T is the first time period of the structure.
1.5 Calculations and results
The analysis setup and results will be different by the different time-history calculation types
and detailed in the following specific chapters.
Here we summarized the possible results, but these may be available by the relevant calculation
options. After the dynamic response calculation the following results could be available:
–
Static displacements with graph or colour palette view at the different time steps and the
maximum value. Detailed graphical function results are also available to a single node.
–
Dynamic displacements with graph or colour palette view at the different time steps and
the maximum value. Detailed graphical function results are also available to a single node.
–
Accelerations colour palette view at the different time steps and the maximum value.
Detailed graphical function results are also available to a single node.
–
Dynamic factor colour palette view at the different time steps and the maximum values.
Detailed graphical function results are also available to a single node.
–
Normalized dynamic factor colour palette view at the different time steps and the
maximum values. Detailed graphical function results are also available to a single node.
NOTE: Two kinds of dynamic factor are available by the excitation force results.
The regular dynamic amplification factor (so-called dynamic factor) at a given i-th node is
calculated as the ratio of the dynamic and static displacements vector caused by the excitation
force at the i-th node.
x i ,dyn
x i , stat
(Eq. 1.9)
The normalized dynamic factor will be calculated in the same way, but in the denominator we
consider the maximum static displacement of the structure instead of the actual static
displacement of the specific node.
x i , dyn
x j , stat , max
9
(Eq. 1.10)
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1.6 Restrictions of the calculation
Since this type of dynamic calculation is linear, all non-linearity effects will be neglected during
this calculations e.g.:
–
Uplift
–
Cracked section analysis
–
Plastic calculation
–
Construction stage calculation
–
Diaphragm
–
Non-linear soil
–
2nd order analysis
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2 Ground acceleration calculation
Fig. 2.1 shows the SDOF problem of ground acceleration excitation. In this case there is no
excitation force, therefore the right side of Eq. 1.1 becomes to 0, however the acceleration of the
mass point will be extended with the acceleration of the ground, see Eq. 2.1 and Fig. 2.1. The
remaining part of the general differential equation system is the same as Eq. 1.1.
M ( ẍ (t)+ ẍ g ( t))+C ẋ (t )+K x (t)=0
xg
Eq. (2.1)
x
m
k
ag
Figure 2.1 – SDOF model to represent the ground acceleration calculation
The variables in Eq. 2.1 above the formerly mentioned variables: ẍ g (t)=a g (t) is the ground
acceleration function.
NOTE: As you can see in Fig. 2.1 the
x (t) is the displacement vector relative to the ground.
Thus, we have seen how the general MDOF differential equation system changes when we have
a known ground acceleration as a function of time. Now we should rearrange the equation
system to solve the problem, see Eq. 2.2.
M ẍ (t )+C ẋ (t)+ K x(t)= −M ẍ g ( t)
Eq. (2.2)
Or if we use the most common notation to the ground acceleration function:
M ẍ (t )+C ẋ (t)+ K x(t)= −M a g (t)
Eq. (2.3)
With Eq. 2.3 we can solve the problem with the mentioned direct integration technique (see
Chapter 1) because on the right side we have now a force-a-like function, thus the shape of the
differential equation system is the same as Eq. 1.1.
NOTE: We need to emphasize that here the solution function x (t) is the relative
displacement of the nodal points relative to the moving ground and not the absolute
displacement.
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2.1 Acceleration response spectra calculation
The displacement response spectra shows what is the maximum relative to the ground
displacement of the SDOF system caused by a ground accelerogram (see Fig. 2.1) considering
the equivalent Kelvin-Voigt damping (see Chapter 1.3).
m ẍ (t)+c ẋ ( t)+k x (t)=−m a g (t)
m ẍ (t)+2 ξ √ k m ẋ (t )+k x (t)=−m a g (t)
Eq. (2.4)
Where c is the damping constant, k is the stiffness of the system, m the mass of the system and ξ
is the critical damping ratio.
By SDOF systems the angular frequency of the system:
ω 0=
√
k
m
Eq. (2.5)
Considering this and rearrange Eq. 2.4 we get the following equation:
ẍ (t)+2 ξ ω 0 ẋ(t)+ω 20 x (t)= −a g (t)
Eq. (2.6)
Considering the time period of the SDOF system:
ẍ (t)+2 ξ
ẍ (t)+ξ
2
( ) x(t)= −a (t)
2π
2π
ẋ (t)+
T
T
g
4π
4π 2
ẋ (t)+ 2 x (t)=−a g (t)
T
T
Eq. (2.7)
Eq. (2.8)
Based on Eq. 2.8 we can see that next to a specific critical damping ratio the structural response
only depends on the time period T .
We should calculate the umax maximum displacement to every necessary time periods (e.g. see
Fig 2.2). After we get these displacements or with other words the displacements response
spectra we should calculate the pseudo acceleration response spectra because during a
replacement static calculation such as modal analysis we would like to get the maximum
internal forces in the structure caused by the ground acceleration. The maximum forces in the
structure (internal forces) arise from the maximum displacements. Based on the internal force in
the structure the definition of the pseudo acceleration is the following.
F =k u max =m ω 20 u max =m S e ,
where the Se is the pseudo acceleration response:
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Eq. (2.9)
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2
( )u
2π
S e =ω u max=
T
2
0
max
.
Eq. (2.10)
This value depends on the time period of the SDOF system (see Eq. 2.10), thus it will be an
S e (T ) function regarding the umax maximum displacement belongs to a time period caused
by the ground accelerogram. A detailed calculation about a specific case will be shown in the
verification examples Chapter 4.1.
350
300
275 mm
De [mm]
250
200
137 mm
150
112 mm
100
50
0
0
1
2
T [s]
3
4
5
Figure 2.2 – A displacement response spectra
Available results
–
Horizontal and vertical pseudo acceleration response spectras to the adjusted ground
accelerograms one-by-one.
–
Average of the calculated acceleration response spectras.
–
The selected pseudo acceleration response spectra is exportable into the modal analysis
settings.
2.1.1 Elastic pseudo acceleration response spectra
With Eq. 2.10 the elastic response spectra can be performed. In FEM-Design first of all on the
Loads tab – Ground acceleration button – by the ground accelerations we should define the
accelerograms. New/Delete/Modify options are available and we can indicate whether the
adjusted accelerogram horizontal or vertical. The ground accelerograms can be pasted/copied
from excel or import/export from saved data (see Fig. 2.3).
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Figure 2.3 – The ground acceleration diagram setting in FEM-Design
Figure 2.4 – The acceleration spectra parameters in FEM-Design
After the insertions of the accelerograms we can calculate the acceleration response spectras
with the Response spectra button (see Fig. 2.3). Here we should set different calculation
parameters to the acceleration spectra (see Fig. 2.4). Namely these settings are:
–
The selection of the calculation integration scheme (Newmark β or Wilson θ).
–
The damping factor (Ksi [%]) to Eq. 2.8 solution.
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–
Delta ti [s] time step (Δt) to direct integration solution of Eq. 2.8 (see Chapter 1.4 as well).
–
Delta T [s] is the steps of the considered time period interval (see Eq. 2.8).
–
T end [s] is the end time period of the calculated acceleration spectra (see Fig. 2.5).
–
q behaviour factor, if it equals to 1.0 the elastic pseudo acceleration response spectra will
be calculated. If it differs from 1.0 Chapter 2.1.2 calculation will be performed.
After the settings with the Calculate button the response spectras will be calculated. These can
be checked on the Horizontal or Vertical spectra tabs. Besides the adjusted accelerograms
response spectras the Average spectra also can be found.
Fig. 2.5 shows an example elastic pseudo acceleration response spectra (see detailed about it in
the verification examples Chapter 4.1).
Figure 2.5 – The pseudo acceleration response spectra
With the Set spectrum button the selected spectra can be exported as an input unique spectra of
the Modal analysis/Static linear shape/Static mode shape seizmic calculation.
2.1.2 Design pseudo acceleration response spectra
In line with the standard (EN 1998-1-1) by the pseudo acceleration response spectra calculation
parameters the q behaviour factor can be adjusted (see Fig.2.4). If the adjusted parameter of the
q factor differs from 1.0 the following modification will be performed on the elastic pseudo
acceleration response spectra.
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According to EN-1998-1-1 by the standard spectras the minimum q behaviour factor is 1.5. At
the beginning point all design spectras divisioned by 1.5. On the plato and after the plato the
elastic response spectra values are divided with the behaviour factor. Between the starting point
and the plato the standard elastic spectras divided with non-constant behaviour factor.
According to these EN-1998-1-1 considerations in FEM-Design if the adjusted behaviour factor
differs from 1.0 at the beginning the divisor number is 1.5. After this we find the time period
where the maximum acceleration spectra value arise. From the beginning point until this time
period the divisor number varying linearly. After this time period the divisor number on the
elastic response spectra will be constant (see. Fig. 2.6).
q
qadjusted≠ 1.0
1.5
1.0
qadjusted= 1.0
T(Se max)
elastic response
T
Figure 2.6 – The interpretation of the q behavior factor as a divisor
number to perform the design spectra from the elastic spectra
Fig. 2.7 shows a design spectra in case of q = 3.0. Fig. 2.5 shows the elastic spectra with q = 1.0.
Both spectra were calculated from the same ground accelerogram.
Figure 2.7 – A design pseudo acceleration response spectra
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2.2 Structural response analysis
The calculation background of the structural response from ground acceleration was indroduced
at the beginning of Chapter 2. Here the excitation of the structure comes from ground
accelerations.
Input
In FEM-Design on the Loads tab – Ground acceleration button – by the ground accelerations we
could define a specific ground acceleration (accelerogram) function in time in horizontal or
vertical direction (see Chapter 2.1.1 as well). These functions could be independent or
dependant from each other. If we copy them from an earthquake database, the database could
contain coherent accelerograms which were recorded in the same time at the same place in
global X, Y and Z directions. But the given accelerogram in one direction can be independent
from the others. In this dialog box next to the Diagram tab in the Combination tab we could
define the so-called ground acceleration combinations (see Fig. 2.8).
Figure 2.8 – Ground acceleration combination dialog
We can see six different columns here. The meanings of the different columns are as follows:
–
Column 1: Sequence number of the ground acceleration combination
–
Column 2: Arbitrary name of ground acceleration combinations
–
Column 3: The Δt (dt) time step in secundum to solve the dynamic differential equation
(see Chapter 1.4)
–
Column 4: The directions of the accelerograms which will be involved in the combination
–
Column 5: A multiplication factor to the specific accelerogram in the current direction
–
Column 6: The optional ground acceleration diagrams which were defined in Diagrams tab
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Every ground acceleration combinations may contain X, Y or Z directional accelerograms, but
in a specific direction the multiplier can be zero or simply the line can be left blank. With this
method we can specify different ground acceleration combinations which will be the excitation
on our 3D structure according to Eq. 2.3.
Calculation request and settings
Similarly to other FEM-Design calculation requests it can be found on the Analysis main tab
under the Calculation dialog. New calculation request is available with the name of “Time
history, ground acceleration” (see Fig. 2.9).
Figure 2.9 – Ground acceleration calculations request dialog
According to Chapter 1 some additonal setup parameters are necessary to run such calculations.
In the setup under the “Time history, ground acceleration” calculation (see Fig. 2.9 and 2.10) the
following options can be adjusted:
Right side of Fig. 2.10:
–
Method of the integration scheme. Regarding Chapter 1.1 we can choose between
Newmark or Wilson-Θ direct integration method. The default is the Newmark one.
–
Rayleigh damping matrix. Here we should define the damping parameters (Alpha and
Beta) according to Chapter 1.3 which will determine the damping behaviour of our 3D
structure. It is very important to set these parameters correctly. Chapter 1.3 can help to
choose these parameters based on the relevant eigenfrequencies of the structure and its
critical damping ratios.
–
WARNING: The damping factor here (Ksi) only considered during the Level acceleration
spectra calculation (see Chapter 2.3). Thus the damping of the structure should be adjusted
with the Rayleigh damping matrix parameters (see the former paragraph).
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Left side of Fig. 2.10:
–
WARNING: The level acceleration calculation option only relevant for the Level
acceleration spectra which will be introduced in Chapter 2.3.
–
Time history calculation can be selected here as a global structural response analysis
option.
–
n result – During the analysis (direct integration method) we should calculate the
displacements and accelerations of the nodes in every time step to get proper solution,
but to save calculation time and hard drive space at the end of the calculation as results
only every n-th time step result will be saved and available.
–
t end [s] – This time moment will be the last during the calculation. It is useful because
it could happen that one of the accelerogram function in the specific combination
shorter in one direction (X, Y or Z) but the calculation will performed further than this
time moment. If one function ends earlier in any direction than the others, the shorter
function will be neglected and obviously only the remaining ones will be involved into
the last time interval of the calculation.
Figure 2.10 – Ground acceleration calculations setup dialog
related to structral response calculation
Available results
–
Dynamic relative to the ground displacements (translations or rotations) in the selected
time step and their envelope absolute maximum.
–
Detailed result diagram of the selected nodal point about the nodal displacements
(translations or rotations) as a function of time.
–
Absolute accelerations in the selected time step and their envelope absolute maximum.
–
Detailed result diagram of the selected nodal point about the nodal accelerations as a
function of time.
–
The dynamic reactions in the selected time step and their envelope absolute maximum,
positive maximum and negative maximum.
–
The internal forces in the structural elements in the selected time step and their envelope
absolute maximum, positive maximum and negative maximum
19
Time History Analysis
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We will show a benchmark example to this calculation type as well (see Chapter 4.2).
2.3 Level acceleration response spectra calculation
In this chapter the level spectra calculation method will be introduced. Chapter 2.2 will be the
basics of this calculation. It means that some part of the structural response analysis results will
be used to calculate the so-called level acceleration response spectra.
Fig. 2.11 shows the meanings of the level spectra calculation. In case of the ground acceleration
(earthquake) under the structure we can calculate the response of the main structure (e.g.: nodal
displacements, nodal accelerations). During the analysis if we have some machine or equipment
on one of the structural floor (see Fig. 2.11) the base foundation (supporting structure) of the
equipment between the floor (level) and the equipment is usually not part of the structural
analysis, but in reality it has a stiffness property which affects the response of the equipment
(see Ref. [4-6]).
The level spectra calculation will show the maximum acceleration response of the equipment
during the ground acceleration (earthquake) as a function of time period of the equipment (and
its supporting structure) which depends on m mass, k stiffness and ξ damping, see Fig. 2.11.
equipment
supporting structure
of the equipment
m
amax(T)
k,ξ
ai(t)
a2(t)
a1(t)
ag
Figure 2.11 – The interpretation of the level spectra calculation
To calculate the level spectra it is necessary to define at least one storey (or more) in the FEMDesign model. During the calculation first of all a time-history structural response analysis will
be run in the background with the defined settings and ground accelerations what were shown in
the previous chapter. One of the results of this calculation is the nodal (X,Y and Z directional)
absolute accelerations. According to the defined storey(s) we will get the average values of this
nodal acceleration vectors storey by storey. It means that we collect all the nodal acceleration
vectors on one storey and calculate an average acceleration vector function storey by storey.
During the level spectra calculation these average storey absolute acceleration funtions will be
the basics of the level spectra calculation. Basically we will performe the same calculation as it
was stated in Chapter 2.1 but instead of using the original ground acceleration functions we will
use these average level acceleration response functions as excitations on the SDOF system.
As by the original response spectra calculation, the level spectra calculation also has X, Y and Z
component, because the level acceleration vectors have three different component.
After the level spectra calculation method, we will have the level spectra responses of the
20
Time History Analysis
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storeys which can help us during the practical engineering to design the proper supporting
structure of our equipments or machines where it is important.
In the verification examples of time-history calculation we will show a benchmark example to
this calculation type as well (see Chapter 4.3).
Input
The same input is necessary as it was stated in Chapter 2.2 because a complete structural
response analysis (only displacements and accelerations) is necessary to get the level responses.
Therefore ground accelerations and ground acceleration combinations what were introduced in
Chapter 2.2 are necessary to define.
Calculation request and settings
By the analysis tab under the calculation dialog the level spectra settings are at the same place as
the structural response spectra settings (see Fig. 2.9).
Figure 2.12– Ground acceleration calculations setup dialog
related to level spectra calculation
The level spectra setup dialog can be seen in Fig. 2.12. You can set the same parameters on the
right side such as in Chapter 2.2 because firstly a structural response calculation is necessary to
get the response spectras of the levels.
We made a WARNING in Chapter 2.2 that on the right side of this dialog the Damping factor
[Ksi] only related to the level spectra calculation. This Ksi parameter will be the considered
damping to calculate the Level response spectra (see Fig. 2.11 about the interpretation of this
Ksi damping). On the left side of Fig. 2.12 we can set the same parameters for the level spectra
calculations which were introduced in Chapter 2.1.1.
Available results
–
Level pseudo acceleration response spectras to the storeys based on the adjusted ground
accelerograms.
–
A level spectra result contains three different diagrams, the global X, Y and Z directional
results respectively.
21
Time History Analysis
FEM-Design 21
3 Forced vibration with arbitrary functions
The calculation background of the forced vibration (excitation force) was introduced in Chapter
1.1. Here the excitation of the structure comes from different excitation force functions in time.
Input
In FEM-Design on the Loads tab – Excitation force button – by the excitation force Diagrams
we could define specific force multiplier functions in time. These function can have unique
name and arbitrary time steps. If we have some table data about our multiplier functions we may
export/import or copy them (see Fig. 3.1).
Figure 3.1 – The force excitation diagram setting in FEM-Design
After we defined these multiplier functions we can combine them and adjust our dynamic
excitation force combinations on the Combination tab in this dialog (see Fig. 3.2).
We can see six different columns here. The meanings of the different columns are as follows:
–
Column 1: Sequence number of the excitation force combination
–
Column 2: Arbitrary name of the excitation force combinations
–
Column 3: The Δt (dt) time step in secundum to solve the dynamic differential equation
(see Chapter 1.4)
–
Column 4: A multiplication factor to the selected multiplier function
–
Column 5: The selected multiplier function to the specific load case involved in the
dynamic combination
–
Column 6: The selected load case on which the multiplier function with the multiplication
factor is applied
22
Time History Analysis
FEM-Design 21
Every excitation force combinations may contain arbitrary number of load cases with the
selected multiplier functions. During the dynamic calculation FEM-Design will apply in the
specific time step the dynamic load according to the multiplication of the defined multiplication
factors, the multiplier function values (at the actual time step) and the intensity of the selected
loads in the adjusted load cases (see benchmark calculations with examples in the verification
Chapter 4.4).
One excitation force combination may contain different load cases with different multiplier
functions which provides a wide range of opportunities to analyze several dynamic problems.
WARNING: From the load cases which were involved into the excitation force combinations
every kinematic load parts (such as thermal loads, stress loads and support motion loads) will be
excluded during the dynamic analysis.
Figure 3.2 – Force excitation combination dialog
Calculation request and settings
Similarly to other FEM-Design calculation requests it can be found on the Analysis main tab
under the Calculation dialog. New calculation request is available with the name of “Time
history, excitation force” (see Fig. 3.3).
According to Chapter 1 some additonal setup parameters are necessary to run such calculations.
In the setup under the “Time history, excitation force” calculation (see Fig. 3.3 and 3.4) the
following options can be adjusted:
Right side of Fig. 3.4:
–
Method of the integration scheme. Regarding Chapter 1.1 we can choose between
Newmark or Wilson-Θ direct integration method. The default is the Newmark one.
–
Rayleigh damping matrix. Here we should define the damping parameters (Alpha and
Beta) according to Chapter 1.3 which will determine the damping behaviour of our 3D
23
Time History Analysis
FEM-Design 21
structure. It is very important to set these parameters correctly. Chapter 1.3 can help to
choose these parameters based on the relevant eigenfrequencies of the structure and its
critical damping ratios.
–
WARNING: The damping factor here (Ksi) only considered during the Level acceleration
spectra calculation (see Chapter 2.3). This parameter is unused in this calculation option.
Thus the damping of the structure should be adjusted with the Rayleigh damping matrix
parameters (see the former paragraph).
Left side of Fig. 3.4:
–
n result – During the analysis (direct integration method) we should calculate the
displacements and accelerations of the nodes in every time step to get proper solution, but
to save calculation time and hard drive space at the end of the calculation as results only
every n-th time step result will be saved and available.
–
t end [s] – This time moment will be the last during the calculation. It is useful because it
could happen that one of the involved dynamic force on the structure stops and others keep
continuing. Thus if we applied a shorter multiplier function compared to others and the
end time moment is longer than the shorter multiplier functions on the remaining time
interval of the calculation that specific load case will be involved with zero multiplier and
only the relevants will be considered.
Figure 3.3 – Force excitation calculations request dialog
24
Time History Analysis
FEM-Design 21
Figure 3.4 – Force excitation calculations setup dialog
Available results
–
Static displacements (translations or rotations) in the selected time step considering the
static loads from the selected excitation force function combination in the selected time
step. The envelope absolute maximum also available.
–
Dynamic displacements (translations or rotations) in the selected time step considering the
dynamic effects from the selected excitation force function combination. The envelope
absolute maximum also available.
–
Detailed result diagram of the selected nodal point about the nodal displacements
(translations or rotations) as a function of time. Both static and dynamic displacements.
–
Accelerations in the selected time step in the selected time step considering the static loads
from the selected excitation force function combination in the selected time step. The
envelope absolute maximum also available,
–
Detailed result diagram of the selected nodal point about the nodal accelerations as a
function of time.
–
The dynamic reactions in the selected time step and their envelope absolute maximum,
positive maximum and negative maximum.
–
The internal forces in the structural elements in the selected time step and their envelope
absolute maximum, positive maximum and negative maximum
–
Dynamic amplification factors in the selected time step and their envelope absolute
maximum.
–
Detailed result diagram of the selected nodal point about the nodal dynamic amplification
factors as a function of time.
–
Normalized dynamic amplification factors in the selected time step and their envelope
absolute maximum.
–
Detailed result diagram of the selected nodal point about the nodal normalized dynamic
amplification factors as a function of time.
25
Time History Analysis
FEM-Design 21
4 Verification examples
4.1 Response spectra calculations
4.1.1 Response spectra calculation from harmonic accelerogram
In this example we will calculate with analytical form the elastic pseudo acceleration response
spectra from harmonic ground acceleration and will compare the results with FEM-Design
values. The harmonic ground accelerograms are not lifelike but for these we can provide a hand
calculation as benchmark. We will consider two different angular frequencies and amplitudes as
two different accelerograms (see Fig, 4.1.1.1). The necessary inputs are in the following table.
Function of the accelerograms:
Between 0-25 sec
Accelerogram #1
ag1 = 0.14*9.81*sin(2π*t)
Accelerogram #2
ag2 = 0.08*9.81*sin(π*t)
The critical damping ratio to the response spectra
ξ1= 5 % = 0.05
ξ2= 10 % = 0.10
0,15
a_g [m/s^2]
0,10
0,05
0,00
-0,05
0
2
4
6
8
10
12
14
-0,10
-0,15
t [s]
Figure 4.1.1.1 – The two different harmonic ground acceleration functions (1 blue; 2 red)
The calculation method in case of general ground acceleration was introduced in the
acceleration response spectra calculation chapter in the time-history theory description. But in
case of harmonic accelerograms the values of the response spectra can easily calculated
according to the resonance dynamic considerations.
The two different harmonic ground accelerations are (see Fig. 4.1.1.1):
a g1(t )=a g max1 sin( ω 1 t )=0.14⋅9.81sin (2 π t)=1.373 sin(2 π t)
a g2(t)=a g max2 sin (ω 2 t)=0.08⋅9.81 sin( π t)=0.7848sin ( π t )
26
Time History Analysis
FEM-Design 21
According to the differential connection between displacements and accelerations considering
zero initial conditions:
a g (t )= ẍ g (t ) ;
Considering the mentioned zero initial conditions the ground displacements:
x g1(t)=−
x g2(t)=−
a g max1
ω1
2
a g max2
ω2
2
sin( ω 1 t)=−
1.373
sin (2 π t) and
2
(2 π )
sin( ω 2 t)=−
0.7848
sin (π t )
π2
In Ref. [1] it can be found that the dynamic amplification factor (DAF) in case of damped
harmonic excitation :
DAF steady state damped =
1
√[
2 2
=
( )] [
ω
1− ω
0
+ 2 ξ ωω0
]
2
1
√[
(
1−
ω
2 π /T
2 2
)] [
+ 2ξ
ω
2 π /T
2
]
The elastic pseudo acceleration response spectra in this case comes from this amplification
factor as a function of T time period of the SDOF system:
S e1 (T )=a g max1 DAF 1 (T )=1.373
1
√[
S e2 (T )=a g max2 DAF 2 (T )=0.7848
(
2π
1−
2 π /T
2 2
)] [
2π
+ 2⋅0.05
2 π /T
2
]
1
√[
(
1−
π
2 π /T
2 2
)] [
+ 2⋅0.10
π
2 π /T
2
]
This solution only contains the steady-state pseudo accelerations, thus the total result in FEMDesign could be larger by some time periods where the transient solution gives larger
displacements than the steady-state ones. Fig. 4.1.1.2 shows these elastic acceleration response
spectras.
14
12
S_e [m/s^2]
10
8
6
4
2
0
0
0,5
1
1,5
2
2,5
T [s]
3
3,5
4
Figure 4.1.1.2 – The elastic pseudo acceleration response spectras
based on the two different accelerograms and dampings
27
4,5
5
Time History Analysis
FEM-Design 21
The maximum values of these response spectras:
S e1max =13.73
m
s2
S e2 max=3.944
m
s2
In FEM-Design we defined the harmonic ground accelerations between 0-25 s time interval to
get the pseudo accelerations spectras. Fig. 4.1.1.3 shows the response spectras based on FEMDesign Newmark calculation method.
agmax1=1.373 m/s2 ; ξ1 = 0.05
agmax2=0.7848 m/s2 ; ξ2 = 0.10
Figure 4.1.1.3 – FEM-Design results about the response spectras
The maximum values of these two spectras from FEM-Design:
S e1max FEM =13.75
m
s2
S e2 max FEM =3.942
m
s2
28
Time History Analysis
FEM-Design 21
We can say that the peak pseudo acceleration values are identical in both calculations. We can
observe that mostly the second parts of the diagrams (after the peaks) are bit larger in FEMDesign results than in the hand calculation results (see Fig. 4.1.1.2-3). This comes from that
FEM-Design considers the total solution of the differential equations (transient and steady-state
as well) but in the hand calculation only the steady-state solutions weres involved.
You can find the download link to the example file on the Contents page.
29
Time History Analysis
FEM-Design 21
4.1.2 Response spectra calculation from El Centro earthquake accelerogram
In this example we will calculate the elastic pseudo acceleration response spectra from El
Centro earthquake (1940, May 18) ground accelerogram. By the calculation we will consider
different critical damping ratios. After the FEM-Design calculations we will compare the results
with the results given in Ref. [3]. Fig. 4.1.2.1 shows the considered accelerogram to the
response spectra calculation.
Figure 4.1.2.1 – The El Centro 1940 May 18 earthquake accelerogram from
http://www.vibrationdata.com/elcentro.htm
In FEM-Design there is a calculation option which can directly prepare the pseudo acceleration
response specra. We can easily check some values with SDOF time-history (ground acceleration
excitation) calculation on different time periods (eigenfrequencies) SDOF systems.
Figure 4.1.2.2 – The response displacement function on SDOF system with T1=1 s
umax1=112 mm
30
Time History Analysis
FEM-Design 21
In the following calculations the considered critical damping ratio was ξ = 5 % = 0.05
Fig. 4.1.2.2 shows the response displacement function of the SDOF mass point with T1= 1 s
caused by the El Centro ground accelerogram. In Fig. 4.1.2.2 the maximum displacement
response is umax 1= 112 mm. This value will be the basics of the final pseudo response
acceleration spectra. We should calculate all the maximum displacement responses of the SDOF
systems with different time periods. With these maximum displacements the definition of the
pseudo acceleration:
ω 0=
√
k
m
is the angular frequency of the undamped SDOF system.
The maximum force in the structure (spring) comes from the maximum displacement:
F =k u max =m ω 20 u max =m S e , where the Se is the pseudo response acceleration.
S e =ω 20 u max
The pseudo acceleration by the case in Fig. 4.1.2.2:
2
( )
S e 1=ω 02 u max 1=
2π
T1
2
( ) 0.112=4.422 ms
u max 1=
2π
1
2
Fig. 4.1.2.3-5 show the displacement responses to T2 = 2 s; T3 = 3 s; T4= 4 s time periods.
Based on these maximum displacement responses the pseudo accelerations:
u max1=112 mm ; u max2=137 mm ; u max3=275 mm ; u max4=257 mm ;
S e 1=4.422
m
m
m
m
; S e 2=1.352 2 ; S e 3=1.206 2 ; S e 4=0.6341 2 .
2
s
s
s
s
Figure 4.1.2.3 – The response displacement function on SDOF system with T2=2 s
umax2=137 mm ; ξ = 5 %
31
Time History Analysis
FEM-Design 21
Figure 4.1.2.4 – The response displacement function on SDOF system with T3=3 s
umax3=275 mm ; ξ = 5 %
Figure 4.1.2.5 – The response displacement function on SDOF system with T4=4 s
umax4=257 mm ; ξ = 5 %
350
300
275 mm
De [mm]
250
200
137 mm
150
112 mm
100
50
0
0
1
2
T [s]
3
4
Figure 4.1.2.6 – Displacement response spectra with 5% critical damping ratio
based on FEM-Design calculation
32
5
Time History Analysis
FEM-Design 21
Fig. 4.1.2.6 shows the so-called displacement response spectra. To prepare this figure it is
necessary to calculate every umax to every T periods. In Fig. 4.1.2.6 we indicated the formerly
calculated values.
Fig. 4.1.2.7 shows the pseudo acceleration response spectra made in FEM-Design based on the
formerly introduced method with ξ = 5 %.
Figure 4.1.2.7 – The response spectra with ξ= 5 % = 0.05
The elastic pseudo acceleration response spectra depends on the ground acceleration and the
critical damping ratio. Fig. 4.1.2.8 shows the response spectras based on the El Centro
earthquake and different critical damping ratios calculated with FEM-Design.
16
2%
5%
10%
20%
40%
60%
100%
14
12
Se [m/s^2]
10
8
6
4
2
0
0
1
2
3
4
T [s]
Figure 4.1.2.8 – FEM-Design response spectras with different critical damping ratios
33
5
Time History Analysis
FEM-Design 21
Compare the FEM-Design results with the results published in Ref. [3] we can say that the
results are identical to each other.
You can find the download link to the example file on the Contents page.
34
Time History Analysis
FEM-Design 21
4.2 Time-history calculation with harmonic ground acceleration on MDOF
system
Inputs:
Concrete elastic modulus
Ecm = 31000 MPa
Concrete shear modulus
G = 12917 MPa
Reduction on modulii due to cracking
fdyn = 0.5
Span length
L=6m
Wall height
H=3m
Wall width
w=4m
Wall thickness one by one (three walls exist)
t = 0.2 m
Mass conversion from distributed loads only
Level 1; q = 70 kN / m2
Level 2; q = 70 kN / m2
Level 3; q = 70 kN / m2
Critical damping ratio to time-history calculation
ξ = 0.05 (α =0; β =0.05181, see later)
Fig. 4.2.1 shows the analyzed 3 storey building with its geometry, the additional data indicated
in the previous table. The wall thicknesses were 0.2 m, the plate thicknesses were 0.5 m. We
would like to calculate by hand the structural response. Without numerical method it is only
possible if the accelerogram is a harmonic function, therefore here the accelerogram will be a
harmonic function.
Figure 4.2.1 – The analyzed three storey building
35
Time History Analysis
FEM-Design 21
The following accelerogam interpredeted in the global X direction paralell with the walls.
a g (t )=a g x ( t)=a g max sin( ω t )=0.14⋅9.81sin (15 t)=1.373 sin(15 t)
We would like to calculate on an approximated model the dynamic response of the different
levels (floors) and compare is with the FEM-Design time-history results.
First of all we need to interpret an approximated model for the problem. The storey number is
three, therefore it seems reasonable to get a 3 degrees of freedom model (see Fig. 4.2.2).
x3(t)
H
m3
x2(t)
H
m2
m1
x1(t)
H
EI
ρGA
Figure 4.2.2 – The analyzed MDOF three storey building in hand calculation
It was defined in the input table that there are distributed uniform loads on each floors which
will represent the considered mass in the dynamic analysis. The converted mass on each level
from the given distributed load:
m1=m 2=m3 =
q 2 L w 70 ⋅2 ⋅6 ⋅4
=
=342.5 ton
g
9.81
These mass points will represent the mass of the floors to our MDOF model (see Fig. 4.2.2).
According to the approximated model our mass matrix will be the following:
[
M=
342.5
0
0
0
342.5
0
0
0
342.5
]
ton
After we get the mass matrix we should calculate the stiffness matrix. In MDOF system in case
of hand calculation it is easier to calculate the compliance matrix first, then the stiffnes matrix
will be the inverse of the compliance matrix of the structure.
In this example, the accelerogram interpretation is parallel with the walls thus the walls will bear
the loads with its shear and bending rigidity (shear walls, see Fig. 4.2.2) in its plane.
We need to calculated the shear and bending stiffness of the walls to consider realistic
compliance matrix of the structure.
36
Time History Analysis
FEM-Design 21
The shear strain under unique load in the three walls (considering modulii reduction due to
cracking):
γ=
1
ρ f dyn GA
=
1
−8
=7.742 ⋅10 rad
0.8333 ⋅0.5 ⋅12917000 ⋅3 ⋅0.2 ⋅4
The bending stiffness of the three walls (considering modulii reduction):
EI = f dyn ⋅31000000
3 ⋅0.2 ⋅4 3
0.6 ⋅4 3
6
2
=0.5 ⋅31000000
=49.6 ⋅10 kNm
12
12
The compliance matrix (considering bending and shear deformation of the walls):
[
H3
+γ H
3 EI
3
3
W = H + H +γ H
3 EI 2 EI
H 3 H3
+ +γ H
3 EI EI
[
H3
H3
+
+γ H
3 EI 2 EI
8 H3
+γ 2 H
3 EI
8H3 2H3
+
+γ 2 H
3 EI
EI
]
]
H3 H3
+
+γ H
3 EI EI
8H3 2 H3
+
+γ 2 H =
3 EI
EI
27 H 3
+γ 3 H
3 EI
[
]
20.52 34.02 47.52
20.52 34.02 47.52
1
1
m
=
34.02 95.04 149.0 =
95.04 149.0 =
6 34.02
EI
kN
47.52 149.0 277.6 49.6⋅10 47.52 149.0 277.6
Based on the compliance matrix the inverse of it will be the stiffness matrix of the structure.
[
]
0.1321 −0.07486 0.01757 kN
K =W −1=49.6⋅10 6 −0.07486
0.1090 −0.04573
m
0.01757 −0.04573 0.02515
To calculate the angular frequencies of the system we should solve the following eigenvalue
problem.
( K−ω 0i2 M ) vi =0
;
where ω0i will be the i-th eigenfrequency of the MDOF system and vi will be the i-th
eigenvector. The eigenvectors should be normalized to the mass matrix to get a more simple
analytical solution to the storey responses.
The solution of the homogeneous equation system appears if the following determinant is zero:
∣K −ω 20i M ∣=0
37
Time History Analysis
FEM-Design 21
The previous determinant is zero in case of the following three angular frequencies:
ω 01=19.73
rad
rad
rad
; ω 02=90.73
; ω 03=173.0
;
s
s
s
The mass matrix normalized eigenvector which belongs to the first angular frequency:
ω 01=19.73
rad
; based on this the first eigenfrequency is:
s
[
]
0.008857
v 1= 0.02620
0.04642
f 01=3.140
1
s
is the first unitless mass normalized eigenvector.
The steady-state solution of a damped system under harmonic ground accelerations based on
Ref. [1]:
3
x ( t)=−∑
i=1
[ ]
a g max
1
T
v v M a g max sin (ω t−ϕ i )
2 i i
2
2 ω
0i
ω
a g max
+ 2ξ ω
0i
1
√(
( )) (
1− ωω
0i
2
)
where
ϕ i =arctg
2 ξ ωω
0i
(
2
1−( ωω0i )
)
is the phase shifting.
From previos analytical solution we can say that in the summation the ω0i has 1/ω0i2 effect on the
solution.
Here the second and third angular frequency were very far from the first one and there is no
resonance at the harmonics. It means that the first angular frequency has significant effect on the
steady-state result, thus during the solution we will approximate the result with only the first
angular frequency during the summation!
The steady state solution considering only the first eigenvalue and eigenvector:
[
][
0.008857 0.008857
1
x (t)≈−2.332
0.02620 0.02620
19.732 0.04642 0.04642
[
][
T
][ ]
342.5
0
0
1.373
0
342.5
0
1.373 sin (15 t−0.1783)
0
0
342.5 1.373
]
0.002033
x ( t)=− 0.006014 sin (15 t−0.1783) m .
0.010655
Fig. 4.2.3 shows the visualization of the solution displacement functions of the MDOF system.
38
Time History Analysis
FEM-Design 21
0,015
Level 1
Level 2
Level 3
x(t) [m]
0,010
0,005
0,000
-0,005
0
0,5
1
1,5
2
2,5
3
3,5
4
4,5
5
-0,010
-0,015
t [-]
Figure 4.2.3 – The steady-state displacement functions of the levels
With the former solution we get the relative displacement (relative to the ground) responses of
the floors (levels) from the given ground acceleration.
The absolute acceleration of the levels (floors) is made up of two parts, firstly the relative
acceleration as the second derivatives of the former relative displacements of the levels and
secondly the acceleration of the ground (based on the given ground acceleration).
The relative acceleration of the mass points (the second derivatives of the previously calculated
x(t) function):
[
]
[ ]
0.002033
0.4574
m
a rel (t)=15 2 0.006014 sin(15 t−0.1783)= 1.353 sin(15 t−0.1783) 2
s
0.010655
2.397
To obtain the absolute accelerations we should add the ground acceleration function to the
previous function:
[ ]
[ ]
0.4574
1.373
a abs (t)=a rel (t )+a g (t )= 1.353 sin(15 t−0.1783)+ 1.373 sin (15 t)
2.397
1.373
m
s2
Fig. 4.2.4 shows the visualization of the absolute acceleration functions.
4
Level 1
Level 2
Level 3
3
a_rel(t) [m/s^2]
2
1
0
-1
0
0,5
1
1,5
2
2,5
3
3,5
4
4,5
-2
-3
-4
t [s]
Figure 4.2.4 – The steady-state absolute acceleration functions of the levels
39
5
Time History Analysis
FEM-Design 21
Based on Fig. 4.2.4 the maximum of the previous functions:
[ ]
1.826 m
a abs max = 2.716 2
3.758 s
Now we have the steady-state response (relative displacements and absolute accelerations) of
the structure (levels) from the given ground acceleration.
In FEM-Design we modelled the structure what was shown in Fig. 4.2.1 with the given input
data.
Based on the FEM-Design eigenfrequency calculation the relevant eigenfrequeny which firstly
contributes in the X directional response (the applied ground acceleration was global X
directional):
f 01FEM =3.072
1
The difference between this and the hand calculation is around 2%.
s
With this frequency we can assume the Rayleigh damping parameter to the time-history analysis
considering the equivalent Kelvin-Voigt damping:
α =0 and β =
ξ
0.05
=
=0.005181 .
π f 01FEM π 3.072
These values were indicated in the input table.
After the FEM-Design time-history ground acceleration calculation with the given input
function and the above calculated Rayleigh damping we will show the relative displacement and
absolute acceleration functions in the middle of each levels (floors).
Fig. 4.2.5 shows the nodal relative displacement and absolute acceleration function in time
interpreted in the middle node of the Level 1 floor.
Fig. 4.2.6 shows the nodal relative displacement and absolute acceleration function in time
interpreted in the middle node of the Level 2 floor.
Fig. 4.2.7 shows the nodal relative displacement and absolute acceleration function in time
interpreted in the middle node of the Level 3 floor.
We can see that the steady-state parts of the FEM-Design results (after the damping of the
transient response) equal to the hand calculation results. In hand calculation we only generated
the steady-state results. Compare Fig. 4.2.3-7 to each other.
The phase shift and the maximum steady-state displacements and accelerations are identical to
each other.
40
Time History Analysis
FEM-Design 21
Figure 4.2.5 – FEM-Design results of Level 1 middle node
displacement [mm] ; acceleration [m/s2]
41
Time History Analysis
FEM-Design 21
Figure 4.2.6 – FEM-Design results of Level 2 middle node
displacement [mm] ; acceleration [m/s2]
42
Time History Analysis
FEM-Design 21
Figure 4.2.7 – FEM-Design results of Level 3 middle node
displacement [mm] ; acceleration [m/s2]
You can find the download link to the example file on the Contents page.
43
Time History Analysis
FEM-Design 21
4.3 Level spectrum calculation with harmonic accelerogram on MDOF system
Based on the previous example (see Fig. 4.2.1 as well), we would like to form the level spectra
results of the three different levels due to the harmonic ground acceleration of the structure.
To the level spectra calculation the considered critical damping ratio was ξ = 0.10.
After we get the absolute acceleration values of the levels (see Fig. 4.2.4) from the ground
acceleration with hand calculation we should multiply these values with the dynamic
amplification factor to get the pseudo acceleration response spectra of the different levels.
However this DAF is depends on the time period (T) and the damping (ξ = 0.10) of the supports
between the equipment and the considered storey level:
DAF (T )=
1
√[
(
1−
ω
2 π /T
2 2
=
)] [
+ 2ξ
ω
2 π /T
2
]
1
√[
(
15
1−
2 π /T
2 2
)] [
+ 2⋅0.10
15
2π /T
2
]
NOTE: The critical damping ratio is not the same which was involved in the acceleration
calculation of the levels (Fig. 4.2.3-4). Here it represents the damping between the floor and the
supporting structure of the equipment on the floor.
The elastic pseudo acceleration level response spectra from the steady-state solution of the three
different levels (storey) as a function of time period of the supporting structure between the
floor and the equipment:
[ ][
√
1.826
state
S steady
(T
)=a
⋅DAF
(T
)=
2.716
e level
abs max
3.758
1
2
1−
]
225⋅T 2
9T2
+
4π 2
4π 2
NOTE: The angular frequency of the ground acceleration and the maximum absolute level
acceleration values were based on the previous example hand calculation results.
Fig. 4.3.1 shows the previously defined level spectra functions.
The maximum values in this level response spectras appear at the T = (2π) / ω = (2π) / 15 =
0.4189 s time period.
The maximum values are (see Fig. 4.3.1 as well):
[ ]
9.13 m
state
S steady
=
13.58 2 .
e level max
18.79 s
44
Time History Analysis
FEM-Design 21
The maximum response values (including the transient and steady-state solutions as well) based
on FEM-Design calculation:
[ ]
9.43 m
S e level max FEM = 14.21 2
19.75 s
Fig. 4.3.2-4 show the FEM-Design level response spectras.
20
18
Level 1
Level 2
Level 3
S_e level
DAF(T)
(T) [m/s^2]
[-]
16
14
12
10
8
6
4
2
0
0
0,2
0,4
0,6
0,8
1
1,2
1,4
1,6
1,8
2
T [s]
Figure 4.3.1 – The acceleration response spectra of the levels from the steady-state solution
Figure 4.3.2 – The Level 1 response spectra from FEM-Design
45
Time History Analysis
FEM-Design 21
Figure 4.3.3 – The Level 2 response spectra from FEM-Design
Figure 4.3.4 – The Level 3 response spectra from FEM-Design
The differences between the hand calculation and FEM-Design results are around 5%. But in the
hand calculation we considered a very pure model and only the steady-state responses. Thus we
can say that the results are indenctical to each other.
You can find the download link to the example file on the Contents page.
46
Time History Analysis
FEM-Design 21
4.4 Time-history calculation with excitation force
4.4.1. Excitation with constant force on undamped SDOF system
In Fig. 4.4.1.1 you can see the analyzed SDOF structure with constant excitation force and
neglecting damping. The input parameters can be found in the following table.
Inputs:
Section (bending around strong axis)
HEA400
Mass at the top
100 t
Structural steel
S235
Bending stiffness
EI = 94646 kNm2
Shear stiffness
ρGA = 338994 kN
Column length
h=6m
Excitation force
F = 10 kN (constant)
x(t)
F
m
k
Figure 4.4.1.1 – The SDOF system with constant excitation force (undamped)
In this SDOF case according to Ref. [1] the dynamic differential equation reduced to:
m ẍ (t)+k x (t)=F
On the right side the excitation force is constant, thus time independent.
To get the analytical solution first of all we should calculate some miscellaneous values. First of
all the k “spring” constant as stiffness should be determined what represents the SDOF system.
This can be very easily calculated based on strength of material basics considering that
according to Fig. 4.4.1.1 the relevant degree of freedom is the horizontal displacement of the
mass in the plane of the system.
Stiffness k (considering bending and shear deformation as well):
k=
1
3
h
h
+
3 EI ρ GA
=
1
3
6
6
+
3⋅94646 338994
=1285
kN
m
47
Time History Analysis
FEM-Design 21
Considering the solution of the differential equation above the angular frequency of the system
will have a great influence on the final result.
Angular frequency: ω 0=
√ √
k
1285
1
=
=3.585
m
100
s
During the solution of this kind of differential equation it is necessary to consider the initial
conditions. In this example we considered the followings:
Displacement: x 0=0 ; Velocity: v 0 =0 .
NOTE: Comparing the analytical solution with FEM-Design it is very important, because FEMDesign considers the initial conditions according to time-history description sub-chapter 1.5
which means that we should give the input excitation force according to Fig. 4.4.1.3.
The final solution of the differential equation with the given initial conditions:
x (t)=−
F
F F
10
cos ω 0 t+ = ( 1−cos ω 0 t )=
( 1−cos (3.585 t) ) .
k
k k
1285
Fig. 4.4.1.2 shows the solution function in time. This vibration arises from the homogeneous
part of the solution of the differential equation, because in this example the system was
undamped. We can see an oscillation around the static displacement (see Fig. 4.4.1.2). Later in
damped cases we will see that the homogeneous part (transient part) diasappers after short time.
The second derivative gives us the acceleration function of the mass point:
F ω 20
10⋅3.5852
a (t)=
cos ω 0 t=
cos 3.585 t .
k
1285
0,016
0,014
0,012
x(t) [m]
0,010
0,008
0,006
0,004
0,002
0,000
0
2
4
6
t [s]
8
10
12
14
Figure 4.4.1.2 – The solution function of the dynamic problem and the static displacement
The static displacement under the given force amplitude is (see also Fig. 4.4.1.2):
x stat =
F
10
=
=7.782 mm .
k 1285
48
Time History Analysis
FEM-Design 21
The maximum displacement amplitude of the dynamic problem is:
x dynmax =
10
( 1−(−1) )=15.56 mm .
1285
The maximum acceleration is:
a max =
10⋅3.5852
m
=0.100 2 .
1285
s
The dynamic displacement amplification factor is:
DAF =
x dynmax 15.56
=
=2 .
x stat
7.782
Fig. 4.4.1.3 shows the FEM-Design model and the adjusted dynamic excitation force function
multiplier.
NOTE: At the beginning of the factor function we can see that instead of the 1.0 factor it starts
with 0 and after that at 0.001 sec it jumps up to 1.0. It is necessary because in FEM-Design the
initial conditions considered according to time-history description sub-chapter 1.5 and these
inputs represent the initial condition which was considered during the analytical solution.
Figure 4.4.1.3 – The FEM-Design model with the adjusted force multiplier function
Fig. 4.4.1.4 shows the FEM-Design result. We can say that the calculated function in Fig.
4.4.1.4 and the analytical function in Fig. 4.4.1.2 are indentical to each other.
49
Time History Analysis
FEM-Design 21
Figure 4.4.1.4 – The FEM-Design detailed result about the undamped displacement of the mass point
The maximum displacement amplitude of the dynamic problem in FEM-Design:
x dynmaxFEM =15.57 mm .
The maximum acceleration in FEM-Design:
a maxFEM =0.100
m
.
s2
The dynamic displacement amplification factor is:
DAF FEM =2 .
The analytical and FEM results are indentical.
You can find the download link to the example file on the Contents page.
50
Time History Analysis
FEM-Design 21
4.4.2. Excitation with constant force on damped SDOF system
This example is the same as the former one but the only different is that now we will consider
damping in the system. The inputs are as follows.
Inputs:
Section
HEA400
Mass
100 t
Structural steel
S235
Bending stiffness
EI = 94646 kNm2
Shear stiffness
ρGA = 338994 kN
Column length
h=6m
Excitation force
F = 10 kN
Critical damping ratio
ξ = 0.10 (α = 0 ; β = 0.05579)
x(t)
c
F
m
k
Figure 4.4.2.1 – The SDOF system with constant excitation force (damped)
In this SDOF damped case according to Ref. [1] the dynamic differential equation is:
m ẍ (t)+c ẋ (t)+k x (t)=F .
On the right side the excitation force is constant, thus time independent.
Some of the necessary parameters are identical with the former example values:
Stiffness k (considering bending and shear deformation as well), and angular frequency:
k =1285
kN
1
; ω 0=3.585
.
m
s
Based on these (considering the damping ) the damped angular frequency:
ω *0=ω 0 √ 1−ξ 2=3.585 √ 1−0.12=3.567
1
.
s
51
Time History Analysis
FEM-Design 21
The damping parameter with the critical damping ratio:
ξ=
c
kNs
thus c=ξ⋅2 √ k m=0.1⋅2 √1285⋅100=71.69
m
2 √k m
Initial conditions: x 0=0 ; v 0 =0
The final solution of the differential equation with the given initial conditions (see Ref. [1]):
x (t)=
(
c
)
c
−
t
−
t
F
c
1−e 2 m cos ω *0 t−e 2 m
sin ω *0 t =
*
k
2mω0
(
71.69
71.69
−
t
−
t
10
71.69
=
1−e 2⋅100 cos 3.567 t−e 2⋅100
sin 3.567 t
1285
2⋅100⋅3.567
)
Fig. 4.4.2.2 shows the solution function in time. Compare this with the result of the former
example we can say that this oscillation around the static displacement arises from the
homogeneous solution of the differential equation but in it disappers shortly in time and we get
the static displacement due to damping.
0,014
0,012
x(t) [m]
0,010
0,008
0,006
0,004
0,002
0,000
0
2
4
6
t [s]
8
10
12
14
Figure 4.4.2.2 – The solution function of the dynamic problem and the static displacement
The static displacement under the given force amplitude is (see also Fig. 4.4.2.2):
x stat =
F
10
=
=7.782 mm .
k 1285
The maximum displacement amplitude of the dynamic problem is (see also Fig. 4.4.2.2):
( ) (
c π
)
(
71.69
π
)
−
−
2m ω
F
10
. x dynmax =x π* =
1+e
=
1+e 2⋅100 3.567 =13.46 mm
k
1285
ω0
*
0
The dynamic displacement amplification factor is:
DAF =
x dynmax 13.46
=
=1.730 .
x stat
7.782
52
Time History Analysis
FEM-Design 21
NOTE: In FEM-Design calculation the Rayleigh damping factor in this case should be
considered according to time-history description sub-chapter 1.4 Eq. 7, namely, the equivalent
Kelvin-Voigt damping:
The Rayleigh damping parameters should be the followings:
α =0 and β =
ξ
=
π f0
0.1
=0.05579 , where f is the eigenfrequency.
3.585
π
2π
Fig. 4.4.2.3 shows the FEM-Design result. We can say that the calculated function in Fig.
4.4.2.3 and the analytical function in Fig. 4.4.2.2 are indentical to each other.
Figure 4.4.2.3 – The FEM-Design detailed result about the damped displacement of the mass point
The maximum displacement amplitude of the dynamic problem in FEM-Design:
x dynmaxFEM =13.46 mm .
The dynamic displacement amplification factor is:
DAF FEM =1.729 .
The analytical and FEM results are indentical.
You can find the download link to the example file on the Contents page.
53
Time History Analysis
FEM-Design 21
4.4.3. Excitation with harmonic force on undamped system
In this example we would like to calculate a clamped-clamped slab with concentrated masses at
its mid-span and harmonic excitation forces on the masses. By the calculation we will consider
three different angular frequencies of the excitation force to represent the different cases when
the vibration is in the same phase as the excitation force, when the vibration phase is in opposite
phase compared to the excitation force and we will show the resonance when the angular
frequency of the excitation force is very close to the angular frequency of the free vibration
system.
Inputs:
Slab thickness
t = 0.2 m
Span length
L = 14 m
Width
B=6m
Concrete (no reduction due to cracking)
C25/30 Ecm = 31 GPa, ν = 0.0
Mass points
see Fig. 4.4.3.1
Amplitude of excitation forces
see Fig. 4.4.3.1
Multiplier factor function of the excitation force
1. excitation
sin(ω1 t) = sin(2.892 t)
2. excitation
sin(ω2 t) = sin(14.46 t)
3. excitation
sin(ω3 t) = sin(72.29 t)
Fig. 4.4.3.1. shows the problem with the masses and the amplitude of the forces and the input
table contains the harmonic function multipliers for the three different cases.
Figure 4.4.3.1 – The clamped-clamped slab with the amplitude
of the harmonic excitation force and the mass points
During this calculation we would like to compare the FEM-Design results with some analytical
solution therefore we will make some simplification to get a proper closed form solution by
hand. Fig 4.4.3.2 shows a possible simplification of the problem with one summarized mass
point and SDOF system.
54
Time History Analysis
FEM-Design 21
Q sinωt
k
m=42 t
L = 14 m
Figure 4.4.3.2 – Simplification of the problem to get analytical solution
The dynamic differential equation reduced to:
m ẍ (t)+k x (t)=Q sin ω t
The total mass and the total amplitude of the given forces:
m=42 t ; Q=420 kN
Stiffness k (considering fixed-fixed beam analogy and elastic behaviour):
k=
384 EI 384 31000000⋅6⋅0.23 /12
kN
=
=8676
3
3
2 L
2
m
14
Considering the solution of the above differential equation the angular frequency of the system
has great influence on the final results.
Angular frequency: ω 0=
√ √
k
8676
1
=
=14.37
m
42
s
In this example we considered the following initial conditions:
Displacement: x 0=0 ; Velocity: v 0 =0 .
The final solution of the differential equation with the given initial conditions:
x (t)=
Q
k
1
ω
1− ω
0
2
( )
(sin ω t− ωω sin ω t )
0
0
And the functions with the given excitation angular frequencies:
x 1 (t)=
420
8676
x 2 (t )=
420
8676
x 3 (t)=
420
8676
1
2.892
1−
14.37
(
2
)
1
2
( )
1−
14.46
14.37
1
72.29
1−
14.37
(
2
)
2.892
sin 14.37 t ;
(sin 2.892t− 14.37
)
(
sin 14.46 t−
14.46
sin 14.37 t
14.37
(
sin 72.29 t−
72.29
sin 14.37 t
14.37
55
ω 1 2.892
ω 0 = 14.37 =0.2013
)
ω 14.46
=1.006
; ω2=
0
14.37
)
ω 72.29
=5.031
; ω3=
0
14.37
Time History Analysis
FEM-Design 21
x(t) [m]
Fig. 4.4.3.3 shows the solution functions in time.
0,06
0,05
0,04
0,03
0,02
0,01
0
-0,01 0
-0,02
-0,03
-0,04
-0,05
-0,06
0,5
1
1,5
2
2,5
3
3,5
4
4,5
5
3
3,5
4
4,5
5
3
3,5
4
4,5
5
t [s]
ω1 = 2.892 1/s
2
1,5
1
x(t) [m]
0,5
0
-0,5
0
0,5
1
1,5
2
2,5
-1
-1,5
-2
t [s]
ω2 = 14.46 1/s
x(t) [m]
0,02
0,01
-0,01 0
0,5
1
1,5
2
2,5
-0,02
t [s]
ω3 = 72.29 1/s
Figure 4.4.3.3 – The solution functions of the vibrations with different excitation frequencies
These differential equation results contain the homogeneous part (free vibration) and particular
part (from the external force) as well.
56
Time History Analysis
FEM-Design 21
NOTE: Generally if the angular frequency of the excitation force exactly equals to the angular
frequency of the free vibration of the system then the above shown solution for the problem is
not true. In reality damping always exists and two real numbers can not be exactly the same.
The static displacement under the given force amplitude is:
x stat =
Q 420
=
=48.41 mm .
k 8676
The maximum displacement amplitude of the dynamic problem is:
x dynmax ω =51.83 mm ; x dynmax ω =1729 mm ; x dynmax ω =10.37 mm
1
2
3
We can see in Fig. 4.4.3.3 that the first and the third harmonic excitation give a periodic
response in the analyzed 0-5 sec interval with finite maximum amplitude. In the second case the
displacement of the mass point is constanly increasing, thus the above given value is the
maximum displacement at the end of the considered time period. This phenomenon appers if the
angular frequency of the excitation force very very close to the angular frequency of the free
vibration of the system.
The dynamic displacement amplification factors are (this DAF includes the complete solution
and not only the DAF regarding the steady-state vibration, see also Fig. 4.4.3.3, and Fig.
4.4.3.4):
DAF ω =
1
51.83
1729
10.37
=1.071 ; DAF ω =
=35.72 ; DAF ω =
=0.2142
48.41
48.41
48.41
2
3
Fig. 4.4.3.5 shows DAF considering only the steady-state solution, thus the above calculated
DAF values are bit greater than the ordinary DAF values. In reality there is damping which
causes that the transient part disappers shortly in time, we will see it in the next example.
1
ω
1− ω
0
∣
( )∣
2
35
30
25
DAF [-]
DAF steady state=
20
15
10
5
0
0
0,5
1
1,5
2
2,5
3
3,5
4
4,5
ω / ω_0 [-]
Figure 4.4.3.4 – The undamped DAF considering only the steady state solution
compared to the values based on the total solution
57
5
Time History Analysis
FEM-Design 21
Fig. 4.4.3.5 shows the FEM-Design results about the vibrations.
Figure 4.4.3.5 – The results from FEM-Design with different excitation frequencies
58
Time History Analysis
FEM-Design 21
The static displacement:
x statFEM =48.00 mm .
The maximum displacement amplitude of the dynamic problem is:
x dynmax ω
1
FEM
=52.51 mm ; x dynmax ω
2
FEM
=1715 mm ; x dynmax ω
3
FEM
=10.36 mm
The dynamic displacement amplification factors are (this DAF includes the complete solution
and not only the DAF regarding the steady state vibration, see also Fig. 4.4.3.3, and Fig.
4.4.3.4):
DAF ω
1
FEM
=1.094 ; DAF ω
2
FEM
=35.74 ; DAF ω
3
FEM
=0.2160
Basically the hand calculation and FEM-Design results are identical.
You can find the download link to the example file on the Contents page.
59
Time History Analysis
FEM-Design 21
4.4.4. Excitation with harmonic force on damped system
In this case we will calculate the same structure what was in the former example (see Fig.
4.4.3.1), but in here the damping will be considered (see the inputs as well).
Inputs:
Slab thickness
t = 0.2 m
Span length
L = 14 m
Width
B=6m
Concrete
C25/30 Ecm = 31 GPa, ν = 0.0
Mass points
see Fig. 4.4.3.1
Amplitude of excitation forces
see Fig. 4.4.3.1
Multiplier factor function of the excitation force
1. excitation
sin(ω1 t) = sin(2.892 t)
2. excitation
sin(ω2 t) = sin(14.46 t)
3. excitation
sin(ω3 t) = sin(72.29 t)
Critical damping ratio
ξ = 0.10 (α = 0 ; β = 0.01383)
Fig. 4.4.3.1. shows the problem with the masses and the amplitude of the forces and the input
table contains the harmonic function multipliers for the three different cases.
During this calculation we would like to compare the FEM-Design results with some analytical
solution therefore we will make some simplification to get a proper closed form solution by
hand. Fig 4.4.3.2 shows a possible simplification of the problem with one summerized mass
point and SDOF system.
The dynamic differential equation:
m ẍ (t)+c ẋ (t)+k x (t)=Q sin ω t
The total mass and the total amplitude of the given forces:
m=42 t ; Q=420 kN
Stiffness k (considering fixed-fixed beam analogy and elastic behaviour):
k=
384 EI 384 31000000⋅6⋅0.23 /12
kN
=
=8676
3
2 L3
2
m
14
Considering the solution of the differential equation above the angular frequency of the system
has great influence on the final results.
Angular frequency: ω 0=
√ √
k
8676
1
=
=14.37
m
42
s
60
Time History Analysis
FEM-Design 21
Based on these (considering the damping ) the damped angular frequency:
ω *0=ω 0 √ 1−ξ 2=14.37 √ 1−0.12=14.30
1
.
s
In this example we considered the following initial conditions:
Displacement: x 0=0 ; Velocity: v 0 =0 .
The final solution of the differential equation with the given initial conditions:
−ξ ω 0 t
x (t)=e
C=
( A cos ω *0 t+B sin ω *0 t )+C sin ω t+D cos ω t
ω
1− ω
0
2
( )
Q
k
2 2
[1−( ωω ) ] +[ 2ξ ωω ]
2
0
[
;
0
ω
−2 ξ ω
0
Q
D=
k
, where
2
]
ω 2 + 2ξ ω
1−( ω
ω0
0)
[
2
;
]
A=−D ;
B=
−ξ ω 0 D−ω C
ω *0
We can easily write the functions with the given angular frequencies:
ω 1 2.892
ω 2 14.46
ω 3 72.29
ω 0 = 14.37 =0.2013 ; ω 0 = 14.37 =1.006 ; ω 0 = 14.37 =5.031
Fig. 4.4.4.1 shows the solution functions in time. These differential equation results contain the
homogeneous part (damped vibration) and particular part (from the external force) as well.
The static displacement under the given force amplitude is:
x stat =
Q 420
=
=48.41 mm .
k 8676
The maximum displacements amplitude of the dynamic problem is (see Fig. 4.4.4.1):
x dynmax ω =50.40 mm ; x dynmax ω =240.0 mm ; x dynmax ω =9.541 mm
1
2
3
We can see in Fig. 4.4.4.1 that after a short disturbance at the beginning (what caused by the
damped homogeneous solution part) we get periodic vibration due to the particular solution.
The dynamic displacement amplification factors are (this DAF includes the complete solution
and not only the DAF regarding the steady-state vibration, see also Fig. 4.4.4.1, and Fig.
4.4.4.2):
DAF ω =
1
50.40
240
9.541
=1.041 ; DAF ω =
=4.958 ; DAF ω =
=0.1970
48.41
48.41
48.41
2
3
61
x(t) [m]
Time History Analysis
0,06
0,05
0,04
0,03
0,02
0,01
0
-0,01 0
-0,02
-0,03
-0,04
-0,05
-0,06
FEM-Design 21
0,5
1
1,5
2
2,5
3
3,5
4
4,5
5
3
3,5
4
4,5
5
3,5
4
4,5
5
t [s]
ω1 = 2.892 1/s
0,3
0,2
x(t) [m]
0,1
0
0
0,5
1
1,5
2
2,5
-0,1
-0,2
-0,3
t [s]
x(t) [m]
ω2 = 14.46 1/s
0,008
0,006
0,004
0,002
0,000
-0,002 0
-0,004
-0,006
-0,008
-0,010
0,5
1
1,5
2
2,5
3
t [s]
ω3 = 72.29 1/s
Figure 4.4.4.1 – The solution functions of the vibrations with different excitation frequencies
62
Time History Analysis
FEM-Design 21
Fig. 4.4.4.2 shows DAF considering only the steady-state solution, thus the above calculated
DAF values are bit greater than the ordinary DAF values.
NOTE: In reality damping always exists, but the dynamic amplification factor by a real damped
harmonic vibration could be larger than the DAF from the formula below!
The damped DAF considering only the steady-state part:
DAF steady state damped =
1
√[
2
]
ω 2 + 2ξ ω
1−( ω
ω0
0)
[
]
2
5
4,5
4
3,5
DAF [-]
3
2,5
2
1,5
1
0,5
0
0
0,5
1
1,5
2
2,5
3
3,5
4
4,5
5
ω / ω_0 [-]
Figure 4.4.4.2 – The undamped DAF considering only the steady-state solution
compared to the values based on the total solution
The static displacement with FEM-Design:
x statFEM =48.00 mm .
The maximum displacement amplitude of the dynamic problem is (see. Fig. 4.4.4.3):
x dynmax ω
1
FEM
=49.92 mm ; x dynmax ω
2
FEM
=238.1 mm ; x dynmax ω
3
FEM
=9.360 mm
The dynamic displacement amplification factors are (this DAF includes the complete solution
and not only the DAF regarding the steady state vibration, see also Fig. 4.4.4.2, and Fig.
4.4.4.3):
DAF ω
1
FEM
=1.040 ; DAF ω
2
FEM
=4.961 ; DAF ω
3
FEM
=0.1950
The hand calculation and FEM-Design results are identical.
You can find the download link to the example file on the Contents page.
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Time History Analysis
FEM-Design 21
Figure 4.4.4.3 – The results from FEM-Design with different excitation frequencies
64
Time History Analysis
FEM-Design 21
References
[1] Chopra A.K., Dynamics of Structures, Prentice Hall, Fourth edition, 2011.
[2] Bathe K.J., Finite element procedures, Second Edition, Prentice Hall, 2016.
[3] Dulácska E., Joó A., Kollár L.P., Structural design against earthquakes (in Hungarian),
Tartószerkezetek tervezése földrengési hatásokra, Akadémiai Kiadó, Budapest, 2008.
[4] Bourahla N., Bouriche F., Benghalia Y., Response spectrum transformation for seismic
qualification testing, International Scholarly and Scientific Research & Innovation, Vol. 5(11),
pp. 1199-1202., 2011.
[5] U.S. Nuclear Regulatory Commission, Regulatory Guide 1.122, Development of floor
design response spectra for seismic design of floor-supported equipment or components, pp. 14., 1976.
[6] Sullivan T.J., Nascimbene R., Towards improved floor spectra estimates for seismic design,
Earthquakes and Structures, Vol. 4., No. 1., pp. 109-132., 2013.
65
Time History Analysis
FEM-Design 21
Notes
66