First Steps in Vibration Analysis Using ANSYS

Ulrike Zwiers

Spring 2007

Referring to the two-mass-spring system sketched below, the basic steps in performinga vibration analysis with ANSYS are outlined. To validate the results of the modalanalysis, the natural frequencies, amplitude ratios and the particular solution are alsocomputed analytically. Furthermore, ANSYS is used to determine the system responseresulting from an harmonic excitation.

1 Equations of Motion and Closed-Form Solution

Formulating the balance of linear momentum for the free-body diagrams shown inFigure 2 yields the following equations of motion:

m1x1 = −k1x1 + k2(x2 − x1)

m2x2 = −k3x2 − k2(x2 − x1)

The corresponding matrix equation reads as

Mx + Kx = 0 ,

with the mass matrix M and the stiffness matrix K being given by

M =

[

m1 00 m2

]

and K =

[

k1 + k2 −k2

−k2 k2 + k3

]

,

respectively.

m1 m2

k1 k2 k3

Figure 1: Two-mass-spring system

1

1 EQUATIONS OF MOTION AND CLOSED-FORM SOLUTION 2

k1x1 k2(x2 − x1) k2(x2 − x1) k3x2

m1 m2

x1 x2

Figure 2: Free-body diagrams

To simplify the further notation, the masses and stiffness coefficients are assumed inthe form

m2 = m1 = m , k3 = k1 = k , and k2 = k1/5 = k/5 .

Consequently, the mass and stiffness matrices may be written as

M = m

[

1 00 1

]

and K =k

5

[

6 −1−1 6

]

,

respectively.

As for the analytical solution, an ansatz function of the form x(t) = xest may be chosen.However, since the considered system is undamped, it is known in advance that the tworoots s1 and s2 will turn out to be purely imaginary. Thus, one may express the solutiondirectly in terms of the natural frequency ω as x(t) = xeiωt. Substituting this ansatzinto the matrix equation of motion yields

(K − ω2M )x = 0 .

To obtain solutions other than the trivial solution x = 0, the determinant of the matrixof the coefficients must vanish, i.e., det(K −ω2M ) = 0, which gives for the problem athand the characteristic equation

(

6

5k − ω2m

)2

−1

25k2 = 0 .

This equation is quadratic in ω2 and may therefore be solved, for example, by usingthe so-called pq-formula resulting in

ω2

1=

k

mand ω2

2=

7

5

k

m.

The amplitude vector x is not uniquely defined, but the ratios of the amplitude com-ponents corresponding to the frequencies ω1 and ω2 can be determined, namely,

x1

x2

= 1 andx1

x2

= −1 ,

2 MODAL ANALYSIS USING ANSYS 3

respectively. Thus, one mode represents an in-phase vibration, while the other onerepresents an out-of-phase vibration.

As an example for forced vibrations, the first mass m1 is now assumed to be harmoni-cally excited in horizontal direction by an external force of the form

F = F sin Ωt ,

where Ω represents the excitation frequency. The governing equations of motion maynow be expressed in matrix notation as

Mx + Kx = f sin Ωt ,

with

f =

[

F0

]

.

The steady-state response is given by the particular solution of this inhomogeneousdifferential equation for which an ansatz function of the form

x = x sin Ωt

is chosen. Substituting this ansatz into the equation of motion yields

(K − Ω2M )x = f .

For the simplifying assumptions introduced before (m2 = m1 = m, k3 = k1 = k,k2 = k1/5 = k/5), the amplitudes are computed as

x1 =

(

6

5k − Ω2m

)

F(

6

5k − Ω2m

)2

− k2

25

,

x2 =k

5F

(

6

5k − Ω2m

)2

− k2

25

.

2 Modal Analysis Using ANSYS

A modal analysis is performed to determine the vibration characteristics (i.e., the nat-ural frequencies and mode shapes) of a structure. In ANSYS, a modal analysis isalso the starting point for other, more detailed, dynamic analyses, such as a harmonicresponse or a transient analysis.

First, the type of analysis is to be specified by using the command line ANTYPE,MODAL.The modal analysis options may then be specified by the command MODOPT. The massand stiffness matrices arising in the modal analysis of elastic structures are usuallylarge but sparse. The corresponding eigenvalue problem is typically of order 105 − 106,

3 HARMONIC ANALYSIS USING ANSYS 4

and several hundred to thousand eigenvectors are often required as the frequency rangeof interest for the modal analysis increases. Several algorithms for solving such largeeigenvalue problems have been implemented in ANSYS. Of course, the modal analysisof simple problems like the two-mass-spring system considered here does not requireany special solution method, thus, the default setting (Block-Lanczos algorithm) maybe used, while for more complex problems, other algorithms might be more suitable.However, in any case, the number of modes to extract must be specified, which is alsodone by the command MODOPT.

The ANSYS-element type MASS21 models a point mass in space, while simple spring-damper-combinations are represented by the element type COMBIN14. The mass andstiffness parameters are to be input as Real Constants.

The spring lengths can be arbitrarily selected since they are used only to define thespring direction.

After successful computation, the results of a modal analysis may be reviewed in theGeneral Postprocessor (\post1). The natural frequencies are listed in the Results Sum-mary SET,LIST, where each set is associated with a natural mode. For a certain mode(selected by using the SET-command), the corresponding mode shape may be animated(via the ANMODE-command).

The unit of frequencies computed by ANSYS is Hz specifying the number of cycles persecond, whereas the frequencies determined in the previous section represent angularfrequencies specifying the number of radians per second. Since there are 2π radians inone revolution, the frequency fn and the angular frequency ωn are related through

fn =ωn

2π.

Thus, for validation of the results, it is essential to designate frequencies in the correctunit.

The enclosed APDL-input file modal schwingerkette.txt documents the modal analy-sis of the two-mass-spring system at hand.

3 Harmonic Analysis Using ANSYS

An harmonic response analysis (ANTYPE,HARMIC) yields solutions of time-dependentequations of motion associated with linear structures undergoing steady-state vibration.To this end, all loads and displacements are assumed to vary sinusoidally at the sameknown frequency.

As for the two-mass-spring system introduced above, the first mass m1 is now assumedto be harmonically excited in horizontal direction by an external force as discussed in thefirst section. The excitation frequency Ω is now supposed to vary within a certain fre-quency range (to be defined by the HARFRQ-command), for which an amplitude-versus-frequency plot should be generated using the Time-History Postprocessor (\post26).

4 REFERENCE RESULTS 5

To get an adequate response curve, solutions at specific intervals are to be determined,i.e., a suitable number of substeps is to be specified (using the NSUBST-command). Here,it is essential to specify a stepped loading (KBC,1), since otherwise, the load amplitudeis gradually increased with each substep. In addition, harmonic analysis options maybe specified by the command HROPT.

The enclosed APDL-input file harm schwingerkette.txt documents the harmonic re-sponse analysis of the two-mass-spring system at hand.

4 Reference Results

The results obtained from the modal analysis and the harmonic analysis performed inANSYS are validated by comparing them against the corresponding analytical solutionsderived in the first section. To this end, the following parameters are chosen

m = 10 kg , k = 2000 N/m and F = 100 N . (1)

It is pointed out that the units of these parameters can be arbitrarily chosen as long asthey are consistent.

Figure 3: Harmonic response of the two-mass-spring system

4 REFERENCE RESULTS 6

The two natural angular frequencies of the considered system are ω1 = 14.1421 [s−1]and ω2 = 16.7332 [s−1], which agrees perfectly with the natural frequencies computedby ANSYS: f1 = 2.2508 [Hz] and f2 = 2.6632 [Hz]. These values are also confirmed bythe amplitude-versus-frequency plot shown in Figure 3.

As a further validation, the amplitudes x1 and x2 at a certain excitation frequencyshould be compared. At Ω = 2.4 [Hz], for example, both the analytical solution andthe ANSYS computation yield concordantly x1 = −0.0875 [m] and x2 = −0.2776 [m].