Transient Response
7.9 Large Mass Method Examples
The large mass examples are:
Model Analysis Type Enforced Motion
bd07two Frequency Response Constant Acceleration
bd07bar1 Transient Response Ramp, Acceleration
bd07bar2 Transient Response Ramp, Displacement
bd07bar3 Transient Response Ramp, Displacement, Discard Rigid-Body Mode
These examples are described in the following sections.
Two-DOF Model
Consider the two-DOF model first introduced in“Real Eigenvalue Analysis”and shown below in Figure 7-2. For this example, apply a constant magnitude base acceleration of 1.0m/sec2over the frequency range of 2 to 10 Hz and run modal frequency response with 5% critical damping in all modes. The acceleration input is applied to the large mass (grid point 3). The input file for this model is shown inFigure 7-3.
Figure 7-2. Two-DOF Model with Large Mass
LABEL = ENFORCED CONSTANT ACCELERATION MAGNITUDE
$
$ LARGE MASS AT BASE GRID POINT
CONM2 999 3 1.0E7
$
$ LOAD DEFINITION (INCLUDES SCALE FACTORS FOR ENFORCED ACCELERATION)
$DLOAD SID S S1 RLOAD1
$TABLED4 TID X1 X2 X3 X4 +TAB4
TABLED4 901 0. 1. 0. 100. +TAB901
$+TAB4 A0 A1 A2 A3 A4 A5
+TAB901 1.0 ENDT
$
$ MODAL EXTRACTION
$EIGRL SID V1 V2 ND MSGLVL
EIGRL 10 -1. 30. 0
$ MODAL DAMPING OF 5% CRITICAL
$TABDMP1 TID TYPE
$+TAB1 F1 G1 F2 G2 ETC
TABDMP1 777 CRIT +TABD7
+TABD7 0. 0.05 100. 0.05 ENDT
$
... basic model ...
$ ENDDATA
Figure 7-3. Input File for Enforced Constant Acceleration
The large mass value is chosen as 1.0E7 kilograms and is input via the CONM2 entry. The scale factor for the load (1.0E7) is input on the DLOAD Bulk Data entry. The factor of 1.0E7 is approximately six orders of magnitude greater than the overall structural mass (10.1 kg). The TABLED4 entry defines the constant acceleration input. (One of the other TABLEDi entries can also be used, but the TABLED4 entry is chosen to show how to use it for enforced constant velocity and displacement later in this example.)
Figure 7-4shows the X-Y plots resulting from the input point (grid point 3) and an output point (grid point 1). The plots show acceleration and displacement magnitudes. Note that the acceleration input is not precisely 1.0m/sec2; there is a very slight variation between 0.9999 and 1.0000 due to the large mass approximation.
Figure 7-4. Displacements and Accelerations for the Two-DOF Model
This model was analyzed with several values of large mass. Table 7-1shows the results. Note that the model with the 106mass ratio is the model discussed earlier. Peak frequency response results are compared for each model, and the natural frequencies are compared to those of the constrained model in“Real Eigenvalue Analysis”. The table shows that a mass ratio of 106is a good value to use for this model.
Table 7-1. Models with Different Large Mass Ratios
Response Peaks (m/sec2) Ratio of
Large Mass to Structure
Natural Frequencies1
(Hz)
102
10-9 4.8011 5.3025
52.0552 6.5531 1.0335 0.9524
104
10-10 4.7877 5.2910
52.2823 6.7641 1.0003 0.9995
106
0.0 4.7876 5.2909
52.2836 6.7661 1.0000 0.9999
Table 7-1. Models with Different Large Mass Ratios
1Resonant frequencies for the constrained model are 4.7876 and 5.2909 Hz.
This model can also be changed to apply constant velocity or constant displacement at its base.
Figure 7-5is an abridged input file for the model, showing the Bulk Data entries required for enforced constant acceleration, enforced constant velocity, and enforced constant displacement.
Note that only one of these is usually applied to any model, but all three are shown here for comparison purposes.
$...2...3...4...5...6...7...8...9...10...$
$
$ ENTRIES FOR ENFORCED MOTION
$
$ LARGE MASS AT BASE GRID POINT CONM2 999 3 1.0E7
$ UNIQUE ENTRIES FOR ENFORCED CONSTANT ACCELERATION MAGNITUDE
$
$RLOAD1 SID DAREA TC RLOAD1 998 997 901
$TABLED4 TID X1 X2 X3 X4 +TAB4 TABLED4 901 0. 1. 0. 100. +TAB901
$+TAB4 A0 A1 A2 A3 A4 A5 +TAB901 1.0 ENDT
$
$...2...3...4...5...6...7...8...9...10...$
$
$ UNIQUE ENTRIES FOR ENFORCED CONSTANT VELOCITY MAGNITUDE
$
$RLOAD1 SID DAREA TD RLOAD1 998 997 902
$TABLED4 TID X1 X2 X3 X4 +TAB4 TABLED4 902 0. 1. 0. 100. +TAB902
$+TAB4 A0 A1 A2 A3 A4 A5 +TAB902 0.0 6.283185 ENDT
$
$...2...3...4...5...6...7...8...9...10...$ $
$
$ UNIQUE ENTRIES FOR ENFORCED CONSTANT DISPLACEMENT MAGNITUDE
$
$RLOAD1 SID DAREA TC RLOAD1 998 997 903
$TABLED4 TID X1 X2 X3 X4 +TAB4 TABLED4 903 0. 1. 0. 100. +TAB903
$+TAB4 A0 A1 A2 A3 A4 A5 +TAB903 0.0 0.0 -39.4784 ENDT
Figure 7-5. Bulk Data Entries for Enforced Constant Motion Each input utilizes the TABLED4 entry. The TABLED4 entry uses the algorithm
Equation 7-22.
where x is input to the table, Y is returned, and N is the degree of the power series. When x<X3 , X3 is used for x ; when x>X4 , X4 is used for x . This condition has the effect of placing bounds on the TABLED4 entry; note that there is no extrapolation outside of the table boundaries.
There are N+1 entries to this table.
Constant acceleration is the easiest to apply since the force is proportional to the mass for all frequencies. The power series for this case becomes
Equation 7-23.
where:
A0 = 1.0
X1 = 0.0
X2 = 1.0
Therefore, these terms define a constant (1.0 in this case).
Constant velocity involves a scale factor that is directly proportional to circular frequency (2πf ).
The power series for this case becomes
Equation 7-24.
where:
A0 = 0.0
A1 = 2π= 6.283185
X1 = 0.0
X2 = 1.0
Note that a phase change of 90 degrees is also required; this change is input using the TD field (field 7) of the RLOAD1 entry.
Constant displacement involves a scale factor that is proportional to the circular frequency squared (2πf)2with a sign change. The power series for this case becomes
Equation 7-25.
where:
A0 = 0.0 A1 = 0.0
A2 = —(2π)2= —39.4784 X1 = 0.0
X2 = 1.0
Table 7-2 summarizes the coefficients for the power series.
Table 7-2. Coefficients for the Power Series
Type of Excitation A0 A1 A2
Enforced u 0.0 0.0 —(2π)2
Enforced 0.0 2π –
Enforced 1.0 – –
Cantilever Beam Model
Consider the cantilever beam first introduced inFrequency Response Analysis and shown in Figure 7-6. In this case the planar model is analyzed for bending; therefore, only three DOFs per grid point are considered: T1 (x-translation), T2 (y-translation), and R3 (z-rotation). An acceleration ramp function in the y-direction is enforced at the base (grid point 1) by applying a large mass and a force. T1 and R3 are constrained at grid point 1 since the enforced motion is in only the T2-direction. Modal transient response analysis (SOL 112) is run with 5% critical damping used for all modes. Modes up to 3000 Hz are computed with the Lanczos method. Figure 7-7shows the idealized ramp function and the NX Nastran implementation. The excitation is not cut off abruptly; instead it is cut off over two time steps. A time step of 0.001 second is used, and the analysis is run for 1.0 second. Figure 7-8shows the abridged input file.
Figure 7-6. Beam Model with Large Mass
Figure 7-7. Idealized Ramp Function Versus NX Nastran Ramp Function
$ CONSTRAIN MASS IN 1,6 DIRECTIONS SPC 21 1 16
$
$ DYNAMIC LOADING
$DLOAD SID S S1 L1 DLOAD 22 1.0E9 0.102 23
$TLOAD1 SID DAREA DELAY TYPE TID TLOAD1 23 24 0 25
$DAREA SID P1 C1 A1
$ CONVERT WEIGHT TO MASS: MASS = (1/G)*WEIGHT
$ G = 9.81 m/sec**2 --> WTMASS = 1/G = 0.102
$ MODAL DAMPING OF 5% IN ALL MODES
$TABDMP1 TID TYPE +TABD
$+TABD F1 G1 F2 G2 ETC.
TABDMP1 25 CRIT +TABD
+TABD 0. 0.05 1000. 0.05 ENDT
$
... basic model ...
$ ENDDATA
Figure 7-8. Abridged Input File for Enforced Acceleration
A large mass of 1.0E9 kg is placed at grid point 1. This grid point is constrained in the T1- and R3-directions but is free in the T2-direction. The load is scaled to give a peak input acceleration of 0.15m/sec2. This scaling is performed by applying a scale factor of 1.0E9 in the S field (field 3) of the DLOAD entry, a scale factor of 0.102 in the S1 field (field 4) of the DLOAD entry, and a factor of 0.15 in the A1 field (field 5) of the DAREA entry. The applied load is scaled by 0.102 because the large mass is also scaled by 0.102 due to the PARAM,WTMASS entry (seeEquation 7-15). The time variation is specified with the TABLED1 entry. The TLOAD1 entry specifies the type of loading (field 5) as 0 (applied force); this gives the same answers if the type is specified as 3 (enforced acceleration).
Figure 7-9shows the displacement and acceleration response at grid points 1 (base) and 11 (tip).
Note that at the end of the acceleration pulse the base has a constant velocity, and therefore, a linearly increasing displacement.
Figure 7-9. Response for Enforced Acceleration
Next, consider the same model with a 0.015 meter displacement imposed instead of an
acceleration. The same ramp time history function is used (with a peak enforced displacement of 0.015 meter) so that the only change to the input file is to change the excitation type from 0 (applied force) to 1 (enforced displacement on field 5 of the TLOAD1 entry) and the amplitude in the DAREA entry from 0.15 to 0.015. Figure 7-10shows the idealized input displacement time history. Figure 7-11shows the displacement and acceleration response at grid points 1 and 11.
Figure 7-10. Response for Enforced Displacement (With the Rigid-Body Mode)
Figure 7-11. Response for Enforced Displacement (Without the Rigid-Body Mode) Now, consider a change to the enforced displacement run. In this case, remove the rigid-body mode’s contribution either by not computing the rigid-body mode (by setting V1 to a small positive value, such as 0.01 Hz) or by neglecting the rigid-body mode in the transient response (by setting PARAM,LFREQ to a small positive number, such as 0.01 Hz). Figure 7-12shows the resulting displacement and acceleration responses at grid points 1 and 11. Note that the responses are relative to the structure and are not absolute. The relative displacement of grid point 1 should be zero, and it is very close to zero (i.e., 10-10) as a result of the sufficiently large mass.
Figure 7-12. Response for Enforced Displacement (Without the Rigid-Body Mode)