Open access peer-reviewed chapter

# Identification of Eigen-Frequencies and Mode-Shapes of Beams with Continuous Distribution of Mass and Elasticity and for Various Conditions at Supports

Written By

Triantafyllos K. Makarios

Submitted: 19 November 2019 Reviewed: 19 March 2020 Published: 04 May 2020

DOI: 10.5772/intechopen.92185

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## Number Theory and Its Applications

Edited by Cheon Seoung Ryoo

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## Abstract

In the present article, an equivalent three degrees of freedom (DoF) system of two different cases of inverted pendulums is presented for each separated case. The first case of inverted pendulum refers to an amphi-hinge pendulum that possesses distributed mass and stiffness along its height, while the second case of inverted pendulum refers to an inverted pendulum with distributed mass and stiffness along its height. These vertical pendulums have infinity number of degree of freedoms. Based on the free vibration of the above-mentioned pendulums according to partial differential equation, a mathematically equivalent three-degree of freedom system is given for each case, where its equivalent mass matrix is analytically formulated with reference on specific mass locations along the pendulum height. Using the three DoF model, the first three fundamental frequencies of the real pendulum can be identified with very good accuracy. Furthermore, taking account the 3 × 3 mass matrix, it is possible to estimate the possible pendulum damages using a known technique of identification mode-shapes via records of response accelerations. Moreover, the way of instrumentation with a local network by three accelerometers is given via the above-mentioned three degrees of freedom.

### Keywords

• equivalent masses of continuous amphi-hinge vertical pendulum
• identification of mode-shapes
• distributed mass and stiffness
• continuous systems
• modal analysis of the continuous beam
• inverted pendulum

## 1. Introduction

An ideal three degrees of freedom system that is equivalent with the modal behavior of an infinity number of degree of freedom of two cases of pendulums is proposed for each case. The first pendulum has hinges at the two ends, while the second pendulum is a cantilever (inverted pendulum). Both pendulums are presented analytically in the present article. This equivalent three DoF system can be used in instrumentation of such pendulums, where the concept of the concentrated masses is not existing, with a local network of three accelerometers. This issue is a main problem that appears very common during the instrumentation of inverted pendulums or bridge beams or steel stairs [1, 2, 3] or wind energy powers [4, 5] in order to identify the real vibration mode-shapes and the fundamental eigen-frequencies of the structure via records of response accelerograms at specific positions due to ambient excitation [6, 7].

## 2. First case: modal analysis of undamped amphi-hinge vertical pendulum with distributed mass and stiffness

According to the theory of continuous systems [8, 9], consider a straight amphi-hinge vertical pendulum that is loaded by an external continuous dynamic loading pzxt, with reference to a Cartesian three-dimensional reference system oxyz, (Figure 1).

The vertical pendulum possesses a distributed mass mx per unit height, which in the special case of uniform distribution is given as mx=m¯ in tons per meter (tn/m). Furthermore, according to Bernoulli Technical Bending Theory, the beam has section flexural stiffness ΕΙyx, where in the special case of an uniform distribution of the stiffness it is given as ΕΙyx=ΕΙy, where E is the material modulus of elasticity and Ιy is the section moment of inertia about y-axis (Figure 1). Next, we are examining a such amphi-hinge vertical pendulum that possesses constant value of distributed mass along its height, as well as constant value of distributed section flexural stiffness ΕΙy. Due to fact that the vertical pendulum mass is continuously distributed, this pendulum/beam has infinity number of degrees of freedom for vibration along the horizontal oz-axis. In order to formulate of the motion equation of this beam, we consider an infinitesimal part of the vertical beam, at location x from the origin o, that has isolated by two very nearest parallel sections. The infinitesimal length of this part is the dx. On this infinitesimal length, we notice the flexural moment Mxt, the shear force Qxt with their differential increments, while the axial force Νxt is ignored, because it does not affect the horizontal beam vibration along z-axis. Moreover, noted the resulting force Pzxt of the external dynamic loading. Therefore, we can write:

Pzxt=pzxt·dxE1

where the resulting force Pzxt acts at the total beam infinitesimal part.

Furthermore, according to D’Alembert principle, the resulting inertia force Faxt is noted, where:

Faxt=m¯·dx·2uzxt/t2Faxt=m¯·dx·u¨zxtE2

Here, we agree that the time derivatives of the displacements are going to symbolize with full stops, while the spatial derivatives of the displacements are going to symbolize with accent. Next, the damping and the second order differential are ignored, so the force equilibrium on the infinitesimal part of the beam along z-axis gives:

Fz=0Q+PzxtQ+Qxdx+Faxt=0
Qx=pzxtm¯·u¨zxtE3

Moreover, the moment equilibrium with reference to weight center (w.c.) of the infinitesimal part of the beam (see Figure 1) gives:

My=0M+Q·dx2+Q+Qxdx·dx2M+Mxdx=0
Q=MxE4

According to Euler-Bernoulli Bending theory (where the shear deformations are ignored), it is well-known that the following basic equation is true:

Mxt=ΕΙy·2uzxtx2E5

Eqs. (4) and (5) are inserted into Eq. (3), so the motion equation without damping for the examined vertical beam is given:

2Mx2=pzxtm¯·u¨zxt
2x2ΕΙy·2uzxtx2=pzxtm¯·u¨zxt
m¯2uzxtt2+ΕΙy4uzxtx4=pzxt
m¯·u¨zxt+ΕΙy·uzxt=pzxtE6

Eq. (6) is a partial differential equation that describes the motion uzxt of the vertical beam that is loaded with the external dynamic loading pzxt. In order to arise a unique solution from Eq. (6), the support conditions must be used at the two beam ends. It is worthy to note that the classical case of a beam with distributed mass and section flexural stiffness, under external horizontal excitation (Figure 2) on the two supports is mathematically equivalent with the vibration that is described by Eq. (6). Indeed, in the case of Figure 2, the total displacement uztotxt of the beam at x-location is given:

uztotxt=ugt+uzxtE7

where ugt is the displacement at the base, same for the two supports.

But, it is known that the inertia forces of the beam are depended by the total displacement uztotxt, while the distributed dynamic loading is null, pzxt=0. Thus, Eq. (3) is transformed into:

Qx=pzxtm¯·2uztotxtt2
Qx=0m¯·2ugtt2m¯·2uzxtt2E8

Following, Eqs. (4) and (5) are inserting into Eq. (8), thus we are taken:

2Mx2=m¯·2ugtt2+2uzxtt2
m¯2uzxtt2+ΕΙy4uzxtx4=m¯·2ugtt2E9

By the comparison of Eqs. (6) and (9), we notice that the undamped beam vibration due to horizontal motion of the two supports is mathematically equivalent with the undamped vibration of the same beam where the two supports are fixed and the beam is loaded with the equivalent distributed dynamic loading peqxt:

peqxt=m¯·2ugtt2E10

In the case of the horizontal pendulum/beam free vibration without damping, we consider the first part of Eq. (9) that must be null:

m¯2uzxtt2+ΕΙy4uzxtx4=0E11

Furthermore, we ask the unknown spatial time-function uzxt, which is the solution of Eq. (11), must have the form of separated variants:

uzxt=φx·qtE12

where φx is an unknown spatial function and qt is an unknown time-function. Eq. (12) has been derived two times with reference to time-dimension t and more two times with reference to spatial-dimension x, so:

2uzxtt2=φx·q¨t,2uzxtx2=φx·qtE13

Eqs. (13) are inserted into Eq. (11), giving:

m¯·φx·q¨t+ΕΙy·φx·qt=0E14

and, next, divided with the number m¯·φx·qt, thus we are getting:

q¨tqt=ΕΙy·φxm¯·φxE15

The left part of Eq. (15) is a time-function, but the right part is a spatial-function. In order to true Eq. (15) for all time values as well as for all spatial positions, the two parts of Eq. (15) must be equal with a constant λ. Thus, Eq. (15) is separated at two following differential equations:

q¨tqt=λq¨t+λ·qt=0E16
ΕΙy·φxm¯·φx=λΕΙy·φxλ·m¯·φx=0E17

However, the time equation (16) indicates a free vibration of an ideal single degree of freedom system that has eigen-frequency =λ . Inserting the eigen-frequency ω into Eq. (17) gives:

ΕΙy·φxω2·m¯·φx=0φxω2·m¯ΕΙy·φx=0E18

Next, we set the positive parameter β such as to be equal:

β4=ω2·m¯ΕΙyE19

because the parameters ω2,m¯,ΕΙy are always positive. By the mathematic theory it is known that the general solution of Eq. (18) has the following form:

φx=C1sinβx+C2cosβx+C3sinhβx+C4coshβxE20

where the four unknown parameters C1,C2,C3,C4 must be calculated. In order to achieve this, four support conditions of the beam have to used. Indeed, for x=0andx=L the displacement uz0t of the amphi-hinge vertical pendulum as well as the flexural moment M0t, both are equal zero. The spatial function φx, which is the solution of Eq. (20), gives the modal elastic line of the beam. Having as known data that the following equation is true:

sinhβx=eβxeβx2,coshβx=eβx+eβx2E21

The spatial function of the modal elastic line for x=0 is:

φ0=C1sin0+C2cos0+C3sinh0+C4cosh0=0
C2+C4=0E22

and also for x=0, the function of the flexural moment due to examined modal elastic line of the beam is given by Eq. (5):

M0t=ΕΙy·2φ0x2=0ΕΙy·φ0=0E23

Eq. (20) has been derived two times with reference to spatial-dimension x, thus arise:

φx=C1·β·cosβx+C2·β·sinβx+C3·β·coshβx+C4·β·sinhβxE24

and

φx=C1β2·sinβx+C2β2·cosβx+C3β2·sinhβx+C4β2·coshβxE25

Therefore, Eq. (23) is transformed:

ΕΙy·β2·C4C2=0E26

By Eqs. (22) and (26) directly arise C2=0 and C4=0, thus the general solution of Eq. (20) is the following:

φx=C1sinβx+C3sinhβxE27

In addition, the parameters C1,C3 are calculated by the support conditions of the second support of the beam. Therefore, for x=L the vertical displacement uLt=0 be true. Thus, from Eq. (12) arises that φL=0 and Eq. (27) gives:

φL=C1sinβL+C3sinhβL=0E28

In continuous, Eq. (23) gives:

MLt=ΕΙy·2φLx2=0ΕΙy·φL=0E29

where φL is directly getting from Eq. (25) that is equivalent with zero:

φL=C1β2·sinβL+C3β2·sinhβL=0E30

However, when Eqs. (28) and (30) are re-written again, we get:

C1·sinβL+C3·sinhβL=0E31
C1·sinβL+C3·sinhβL=0E32

And added part to part these two above-mentioned equations arise:

2·C3·sinhβL=0E33

But, the term sinhβL is not equal with zero, because then vibration is not existing. Therefore, C3 has to equal with zero, so Eq. (28) is formed:

φL=C1·sinβL=0E34

Moreover, by Eq. (34) arise that either C1=0 that is mpossibility because φx0 by Eq. (27), either sinβL=0 that means the following equation must be true:

βL=n·πn=1,2,3,E35

However, Eq. (35) is transformed to Eq. (36):

βL=n·πβ2L2=n2·π2β2=n2·π2L2E36

By the definition of parameter β, we can calculate the eigen-frequency ω:

β4=ω2·m¯ΕΙyω2=ΕΙy·β4m¯ω=β2·ΕΙym¯E37

Thus, inserting Eq. (36) into Eq. (37), the eigen-frequency ωn directly arises for each n-value.

ωn=n2·π2L2·ΕΙym¯n=1,2,3,E38

Therefore, the vibration mode-shape of the examined vertical pendulum arises by Eq. (27)—since previous inserting Eq. (35)—thus:

φnx=C1sinβx=C1sinn·π·xLn=1,2,3,E39

The value of C1 is arbitrary, and we usually get it equal to unit. Thus, for each value of parameter n, a mode-shape with its eigen-frequency is resulted. The fundamental (first) mode-shape is resulted for n=1, which shows a half sinusoidal wave, the second mode-shape shows a full sinusoidal wave, etc. (Figure 3). The order of the eigen-frequencies is ω1, ω2=4ω1, ω3=9ω1, ω4=16ω1, etc.

## 3. The equivalent three degrees of freedom system for amphi-hinges pendulum

At vertical pendulums or beams where the fundamental horizontal mode-shape does not activate the 90% of the total beam mass, we ask to consider the three first mode-shapes. Thus, for this purpose, we must define an ideal equivalent three degrees of freedom beam, which is going to give the first three frequencies and the first three mode-shapes of the examined beam. Therefore, which is the ideal three degrees of freedom system, where its three mode-shapes coincide with the real first three frequencies of the vertical pendulum with distributed mass and flexural stiffness?

In order to answer the above-mentioned question, consider a weightless vertical pendulum with height L and constant section in elevation, where carry three concentrated masses that each one has the same mass-value meq, located per distance 0.25L, between one to one, and each one mass possesses an horizontal degree of freedom (Figure 4).

The pendulum displacement vector u of the three degrees of freedom, as well as the diagonal beam mass matrix m is written:

u=u1u2u3,m=meq000meq000meqE40

Furthermore, the pendulum flexibility matrix f can be calculated using a suitable method (Figure 4), and the inverse matrix gives the stiffness matrixk of the three degrees of freedom beam.

f=D1,1D1,2D1,3D2,1D2,2D2,3D3,1D3,2D3,3=L348EIy·10.68750.68750.68750.56250.43750.68750.43750.5625E41
k=k1,1k1,2k1,3k2,1k2,2k2,3k3,1k3,2k3,3=48EIyL3·18.28571412.57142912.57142912.57142913.1428575.14285712.5714295.14285713.142857E42

The equations of motion for the case of the free undamped vibration of the ideal beam is given as:

mu¨t+kut=0E43

The eigen-problem is written as:

kωn2mφn=0n=1,2,3.E44

where the eigen-frequencies ωn and the three mode-shapes φn are known by Eq. (38) and (39) and Figure 4. Therefore, the unique unknown parameter is the mass meq. Thus,

detkω12m=0E45
meq3+Α·meq2+Β·meq+C=0E46

where

A=k11+k22+k33ω12,B=k11k33+k11k22+k22k33k122k132k232ω14
C=k11k22k33+2k12k13k23k11k232k22k132k33k122ω16

The numerical solution of Eq. (46) gives three roots for parameter meq, where only the first root is acceptable, because the other two values rejected since do not have natural meaning (appear values greater from the total pendulum mass m¯L). Thus, the only one acceptable root is given as:

meq=0.24984748·m¯LE47

Therefore, inserting the ideal equivalent mass meq by Eq. (47) at three degrees of freedom system of Figure 4, the three eigen-frequencies coincide with the real values of the initial vertical pendulum that has distributed mass and flexural stiffness.

## 4. Discussion about the amphi-hinge vertical pendulums

By the previous mathematic analysis, an ideal three degrees of freedom system that is equivalent with the modal behavior of the amphi-hinge vertical pendulum with distributed mass and flexural stiffness along its height has been presented. This ideal three degrees of freedom system can be used in instrumentation of a such vertical tower, which does not possess concentrated masses. In the framework of the identification of mode-shapes of an amphi-hinge pendulum, the equivalent mass by Eq. (47) permits to locate accelerometers per 0.25L (as shown at Figure 4) and there measure the response horizontal acceleration histories, in order to calculate the real first three mode-shapes of the vertical pendulum in order to avoid the instability and resonance-vibrations between the examined vertical pendulum and the supported rockets or space bus before the last launched.

## 5. Second case: modal analysis of undamped inverted pendulum with distributed mass and stiffness

According to the theory of continuous systems [8, 9], consider a straight vertical cantilever (inverted pendulum) that is loaded by an external continuous dynamic loading pzxt, with reference to a Cartesian three-dimensional reference system oxyz, (Figure 5).

The vertical inverted pendulum possesses a distributed mass mx per unit height, which in the special case of uniform distribution is given as mx=m¯ in tons per meter (tn/m). Furthermore, according to Bernoulli Technical Bending Theory, this cantilever has section flexural stiffness ΕΙyx, where in the special case of an uniform distribution of the stiffness it is given as ΕΙyx=ΕΙy, where E is the material modulus of elasticity and Ιy is the section moment of inertia about y-axis (Figure 1). Next, we are examining such an inverted pendulum that possesses constant value of distributed mass along its height, as well as constant value of distributed section flexural stiffness ΕΙy. Due to fact that the cantilever mass is continuously distributed, this inverted pendulum/beam has infinity number of degrees of freedom for vibration along the horizontal oz-axis. In order to formulate of the motion equation of this beam, we consider an infinitesimal part of the vertical beam, at location x from the origin o, that has isolated by two very nearest parallel sections.

The infinitesimal length of this part is the dx. On this infinitesimal length, we notice the flexural moment Mxt, the shear force Qxt with their differential increments, while the axial force Νxt is ignored, because it does not affect the horizontal cantilever vibration along z-axis. Moreover, noted the resulting force Pzxt of the external dynamic loading. Eqs. (1)(20) describe mathematically the modal analysis of the inverted pendulum. Afterward, the four unknown parameters C1,C2,C3,C4 of Eq. (20) must be calculated taking account the conditions of the cantilever. In order to achieve this, four support conditions of the cantilever have to be used. Indeed, for the end at x=0the displacement uz0t and the slope uz0t of displacement profile must be zero. So, Eq. (20) gives:

φ0=C1sin0+C2cos0+C3sinh0+C4cosh0=0C2+C4=0E48

and

φ0=βC1+C3=0C1+C3=0E49

Moreover, for the free end at x=L of the cantilever, the flexural moment MLt as well as the shear force QLt must be both zero. So, from Eqs. (20) and (5) and afterward from Eq. (48) we take:

MLt=ΕΙy·2φLx2=0ΕΙy·φL=0
C1sinβL+sinhβL+C2cosβL+coshβL=0E50

and

QLt=ΕΙy·3φLx3=0ΕΙy·φL=0
C1cosβL+coshβL+C2sinβL+sinhβL=0E51

However, re-writing Eqs. (50) and (51) again, we get the matrix form:

sinβL+sinhβLcosβL+coshβLcosβL+coshβLsinβL+sinhβLC1C2=00E52

Eq. (52) is a real eigenvalue problem. In order to calculate the eigenvalues, parameters C1 and C2 must not both equal zero. Thus, the determinant of the matrix by Eq. (52) must be zero, where it drives to the following frequency equation:

1+cosβL·coshβL=0E53

However, Eq. (54) can be solved numerically only, where we obtain the first four roots (n=1,2,3,4):

β1L=1.8751,β2L=4.6941,β3L=7.8548andβ4L=10.996E54

By the definition of parameter β, we can calculate the first four circular eigen-frequency ω using Eq. (36):

β4=ω2·m¯ΕΙyω2=ΕΙy·β4m¯ω=β2·ΕΙym¯E55

Thus, inserting Eq. (54) into Eq. (55), the first four eigen-frequency ωn is directly arise for each n-value.

ω1=3.516L2·ΕΙym¯,ω2=22.03L2·ΕΙym¯,
ω3=61.70L2·ΕΙym¯,ω4=120.9L2·ΕΙym¯E56

Therefore, the vibration mode-shape of the examined vertical inverted pendulum arises by Eq. (20)—since previous inserting Eq. (56)—thus:

φnx=C1coshβnxcosβnxcoshβnL+cosβnLsinhβnL+sinβnLsinhβnxsinβnxE57

The value of C1 is an arbitrary constant, and we usually get it equal to unit. Thus, for each value of parameter n, a mode-shape with its eigen-frequency is resulted. The fundamental (first) mode-shape is resulted for n=1, etc. (Figure 6).

## 6. The equivalent three degrees of freedom system of inverted pendulum

At inverted pendulums cantilevers, where the fundamental horizontal mode-shape does not activate the 90% of the total cantilever mass, we ask to consider the three first mode-shapes. Thus, for this purpose, we must define an ideal equivalent three degrees of freedom beam, which is going to give the three mode-shapes of the examined beam. Therefore, which is the ideal three degrees of freedom system, where its three eigen-frequencies coincide with the real first three frequencies of the inverted pendulum with distributed mass and flexural stiffness?

In order to answer the above-mentioned question, consider a weightless vertical inverted pendulum with height L and constant section in elevation, where carry three concentrated masses that each one has the same mass-value meq, located per distance 0.333L, between one to one, and each one mass possesses an horizontal degree of freedom (Figure 7).

The inverted pendulum displacement vector u of the three degrees of freedom, as well as the diagonal beam mass matrix m are written:

u=u1u2u3,m=meq000meq000meqE58

Furthermore, the inverted pendulum flexibility matrix f can be calculated using a suitable method (Figure 7), and the inverse matrix gives the stiffness matrix k of the three degrees of freedom beam.

f=D1,1D1,2D1,3D2,1D2,2D2,3D3,1D3,2D3,3=L33EIy·10.51850.14810.51850.29630.09260.14810.09260.0370E59
k=k1,1k1,2k1,3k2,1k2,2k2,3k3,1k3,2k3,3=3EIyL3·14.5384633.2307724.9230833.2307791.3846295.5384624.9230895.53846166.15385E60

The equations of motion for the case of the free undamped vibration of the ideal beam is given as:

mu¨t+kut=0E61

The eigen-problem is written as:

kωn2mφn=0n=1,2,3.E62

where the eigen-frequencies ωn and the three mode-shapes φn are known by Eq. (56) and Figure 6. Therefore, the unique unknown parameter is the mass meq. Thus,

detkω12m=0E63
meq3+Α·meq2+Β·meq+C=0E64

where

A=k11+k22+k33ω12,B=k11k33+k11k22+k22k33k122k132k232ω14
C=k11k22k33+2k12k13k23k11k232k22k132k33k122ω16

The numerical solution of Eq. (64) gives three roots for parameter meq, where only the first root is acceptable, because the other two values rejected since do not have natural meaning (appear values greater from the total pendulum mass m¯L). Thus, the only one acceptable root is given:

meq=0.1868388·m¯LE65

Therefore, inserting the ideal equivalent mass meq by Eq. (65) at three degrees of freedom system of Figure 7, the three eigen-frequencies coincide with the real values of the initial vertical inverted pendulum that has distributed mass and flexural stiffness.

On the contrary to above-mentioned about the examined cantilever, it is worth noting that in the case where ask an ideal single degree of freedom system that has eigen-frequency equal to fundamental eigen-frequency of the real cantilever we write the following equations:

• Eigen-frequency of single degree of freedom system with k the cantilever lateral stiffness and meq,sdof the concentrated mass at the top of the cantilever:

ω2=kmeq,sdof=3EIy/L3meq,sdofE66

• Fundamental (first) eigen-frequency (see Eq. (56)) of the real cantilever with continuous distribution of mass and flexural stiffness:

ω=3.516L2·ΕΙym¯E67

• Thus, inserting Eq. (67) into Eq. (66) the equivalent concentrated mass meq,sdof at the top of the cantilever is given [9]:

meq,sdof=0.2426742·m¯LE68

## 7. Conclusions

The present article has presented a mathematic ideal three degrees of freedom system that is equivalent with the modal behavior of two cases of pendulums. First, an amphi-hinge vertical pendulum with distributed mass and flexural stiffness along its height has been examined. Second, an inverted pendulum that can be simulates a tower (or cantilever) has been examined too. For each case, an ideal three degrees of freedom system has been proposed that can be used in instrumentation of such a vertical tower, which does not possess concentrated masses. In the framework of the identification of mode-shapes of the above-mentioned pendulums, the equivalent mass by Eqs. (47) and (65) permits to locate accelerometers (as shown at Figures 4 and 7) and measures the response acceleration histories, in order to calculate the real first three frequencies.

## Acknowledgments

The author wishes to acknowledge the German Research Foundation (DFG) program on Initiation of International Collaboration entitled “Data-driven analysis models for slender structures using explainable artificial intelligence”, Project No. 417973400 for the period 2019–2021, Prof. Dr.-Ing. Kay Smarsly, Bauhaus University Weimar, project coordinator.

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Written By

Triantafyllos K. Makarios

Submitted: 19 November 2019 Reviewed: 19 March 2020 Published: 04 May 2020