Free vibration and stability of axially functionally ...downloads.hindawi.com/journals/sv/2011/591716.pdf · In this section, the present element is employed in free vibration and

Shock and Vibration 18 (2011) 683–696 683DOI 10.3233/SAV-2010-0589IOS Press

Free vibration and stability of axiallyfunctionally graded tapered Euler-Bernoullibeams

Ahmad Shahbaa,b, Reza Attarnejada,b,∗ and Shahin HajilaraaSchool of Civil Engineering, University College of Engineering, University of Tehran, Tehran, IranbCentre of Numerical Methods in Engineering, University of Tehran, Tehran, Iran

Received 6 March 2010

Revised 11 May 2010

Abstract. Structural analysis of axially functionally graded tapered Euler-Bernoulli beams is studied using finite element method.A beam element is proposed which takes advantage of the shape functions of homogeneous uniform beam elements. The effectsof varying cross-sectional dimensions and mechanical properties of the functionally graded material are included in the evaluationof structural matrices. This method could be used for beam elements with any distributions of mass density and modulus ofelasticity with arbitrarily varying cross-sectional area. Assuming polynomial distributions of modulus of elasticity and massdensity, the competency of the element is examined in stability analysis, free longitudinal vibration and free transverse vibrationof double tapered beams with different boundary conditions and the convergence rate of the element is then investigated.

Keywords: Axially functionally graded material, tapered beam element, free vibration, stability

1. Introduction

Functionally graded (FG) materials are one of the most advanced materials whose mechanical properties varygradually with respect to a desired spatial coordinate. In comparison with laminated composites, employing FGmaterials in structural systems leads to elimination of stress concentration and also improves the strength andtoughness of the structure. Most of the literature on FG beams deals with beams whose mechanical properties varythrough thickness [2,25,28,30,31]. There are relatively few works on axially FG beams whose mechanical propertiesvary along the axis of the beam where most of them concern the special case of uniform beams. Due to varyingcross-sectional area, modulus of elasticity and mass density along the beam axis, the governing differential equationsof axially FG tapered beams for transverse and longitudinal vibrations and buckling are differential equations withvariable coefficients for which closed-form solutions could be hardly found or even impossible to obtain; henceapplication of numerical techniques is essential. Development of the technical literature on axially FG beams couldbe mostly tracked in the works written by Elishakoff and his co-workers [3–21,26,29,32,33] who used semi-inversemethod for solution of the governing differential equations. Assuming a uniform beam with constant mass density,Aydogdu [27] used semi-inverse method to study the free transverse vibration and stability of axially FG simplysupported beams.

In this paper, a new element is proposed for analysis of tapered beams with an arbitrarily varying cross-sectionmade of axially functionally graded materials. The element takes advantage of the shape functions of uniformhomogeneous Euler-Bernoulli beam elements. The idea of formulating tapered beams in terms of uniform beams

∗Corresponding author. Tel.: +98 2161112225; Fax: +98 2166403808; E-mail: [email protected].

ISSN 1070-9622/11/$27.50 2011 – IOS Press and the authors. All rights reserved

684 A. Shahba et al. / Free vibration and stability of axially functionally graded tapered Euler-Bernoulli beams

Fig. 1. Positive sign convention for nodal degrees of freedom and forces.

could be observed in the works of Bazoune [1], Banerjee [23] and Singh and Li [24]. In this paper, the varyingcross-sectional area, moment of inertia, mass density and Young’s modulus of elasticity are exactly considered in thederivation of the structural matrices while in the classical beam element method, the original non-prismatic beam ismodeled as an assemblage of some uniform elements; that is, constant values are considered for the cross-sectionalarea and moment of inertia of the beam element and finally these constant values are considered in the evaluation ofthe structural matrices. Therefore, the present element presents a more realistic model of the original non-prismaticbeam element.

Although several works have been carried out on axially FG beams; there is still a gap in analysis of axially FGbeams with arbitrarily varying cross-sectional area and mechanical properties along the beam axis. This could beexplained in detail by recalling that the majority of previous works have followed a semi-inverse procedure in whichthe displacement field and the mass density distribution are prescribed and afterwards the distribution of modulus ofelasticity is obtained by satisfying the governing differential equations. As a result, semi-inverse method encountersdifficulty in calculation of displacement field of an axially FG beam whose distributions of mass density and modulusof elasticity are known. Finite element method (FEM) could overcome this problem. It should be also rememberedthat semi-inverse procedure should be formulated for each problem separately since the prescribed displacement fieldshould satisfy the boundary conditions; while FEM does not necessitate the reformulation for each set of boundaryconditions. Moreover, it is instructive to bear in mind that semi-inverse method provides exact results while FEMdoes not essentially predict the deformations of the structural system accurately.

2. New beam element

Consider a general beam element with 3 degrees of freedom per node with varying cross-sectional dimensionsalong the element axis as shown in Fig. 1. Based on the concept of FEM which is mostly considered as a displacement-based structural analysis method, the displacement field could be interpolated within the element in terms of thenodal degrees of freedom using shape functions as

w(x) = Nw

⎧⎪⎪⎨⎪⎪⎩

w1

θ1

w2

θ2

⎫⎪⎪⎬⎪⎪⎭

(1)

u(x) = Nu

{u1

u2

}(2)

in which x is the coordinate along the element; u(x) and w(x) are the axial and lateral displacements of the beamelement respectively; Nu =

{Nu1(x) Nu2(x)

}and Nw =

{Nw1(x) Nw2(x) Nw3(x) Nw4(x)

}are respectively

the axial and bending shape functions. The accuracy of the results predicted by FEM considerably depends on howwell these shape functions are selected. In this paper the shape functions for homogeneous uniform Euler-Bernoullibeam elements [22] are used to formulate the displacement field given in Eqs (1–2).

A. Shahba et al. / Free vibration and stability of axially functionally graded tapered Euler-Bernoulli beams 685

Nu ={

1 − xl

xl

}(3)

Nw ={

1 − 3(

xl

)2 + 2(

xl

)3x

[1 − (

xl

)]2 3(

xl

)2 − 2(

xl

)3x

[(xl

)2 − xl

]}(4)

where l is the element length.

3. Structural matrices

In order to carry out any structural analyses, the elemental structural matrices should be evaluated. The strainenergy of the beam element could be written as [22]

U (e) =12

l∫0

E(x)A(x)(

∂u

∂x

)2

dx +12

l∫0

E(x)I(x)(

∂2w

∂x2

)2

dx +12

l∫0

N(x)(

∂w

∂x

)2

dx (5)

Here N(x) is the tensile axial load; A(x) and I(x) are respectively the cross-sectional area and moment of inertiaand E(x) is the modulus of elasticity. The first and second terms in Eq. (5) refer respectively to the internal energydue to stretching and bending. The third term appears in Eq. (5) due to considering the case of large deflections butsmall strains. Using Eqs (1–2), Eq. (5) could be rewritten in matrix form as

U (e) =12dT Kd (6)

where d ={

u1 w1 θ1 u2 w2 θ2

}T. K is the general stiffness matrix given by

K = Ks + Kb + Kg (7)

in whichStretching stiffness matrix:

Ks =

l∫0

N′TuE(x)A(x)N′udx (8)

Bending stiffness matrix:

Kb =

l∫0

N′′TwE(x)I(x)N′′wdx (9)

Geometric stiffness matrix:

Kg =

l∫0

N′TwN(x)N′wdx (10)

The primes designate differentiation with respect to x. The kinetic energy of the beam element reads as

T (e) =12

l∫0

ρ(x)A(x)(

∂u

∂t

)2

dx +12

l∫0

ρ(x)A(x)(

∂w

∂t

)2

dx (11)

in which t is time and ρ(x) is the mass density. Similarly, kinetic energy could also be rewritten in matrix from as

T (e) =12dTMd (12)

Here the dot (.) designates differentiation with respect to t and


M = Ms + Mb (13)

whereStretching mass matrix:

Ms =

l∫0

NTuρ(x)A(x)Nudx (14)

Bending mass matrix:

Mb =

l∫0

NTwρ(x)A(x)Nwdx (15)

As observed, the effects of varying cross-sectional area and being axially functionally graded are considered in theevaluation of structural matrices through considering A(x), I(x), ρ(x) and E(x). Considering a linear distributionof modulus and elasticity and a cubic distribution of mass density, the structural matrices are derived explicitly andgiven in Appendix A for a non-prismatic beam whose breadth and height taper linearly.

In order to determine the natural frequencies and critical load of the system, the following eigenvalue problemsneed to be solved

Free longitudinal vibration: Kgs φ = ω2

LMgs φ

Free transverse vibration: Kgbφ = ω2

TMgbφ

Stability analysis:(Kg

b + ηKgg)

φ = 0where ωL and ωT are respectively the longitudinal and transverse natural frequencies and φ is the mode shape. Instability analysis, η is the eigenvalue and the critical load is determined as Pcr = ηP where P is an assumed constantcompressive load. The superscript g designates the global structural matrix which is obtained through assemblingthe elemental matrices and imposing the boundary conditions.

4. Numerical results and discussion

In this section, the present element is employed in free vibration and stability analyses of axially FG beamswith different boundary conditions. Numerical results are carried out for an axially FG beam with rectangularcross-section whose height and breadth both vary linearly and distributions of modulus of elasticity and mass

density are assumed to polynomially vary as E (X) =m∑

i=0

biXi and ρ (X) =

r∑i=0

aiXi, respectively. Therefore,

cross-sectional area and moment of inertia could be expressed respectively as A(X) = A0

(1 − cb

XL

) (1 − ch

XL

)and I(X) = I0

(1 − cb

XL

) (1 − ch

XL

)3where the breadth and height taper ratios could vary in the range of

0 � cb, ch � 1. If cb = ch = 0, the beam would become a uniform one and if cb = ch = 1, the beam would taperto a point at x = L which is merely a theoretical limit.

In order to facilitate the presentation of results, the boundary conditions including clamped, simple, free andguided are abbreviated respectively as C, S, F and G. Where possible, the computed results are compared with theprevious published results in the literature.

4.1. Stability analysis

Critical load (Pcr) is calculated for a non-prismatic beam which is subjected to a point compressive load at itsboth ends. Firstly, the critical load is calculated for a unit length uniform beam, i.e. cb = ch = 0. The results arecompared with those of Elishakoff [7] in Table 1 for different boundary conditions. It is observed that the results arein good agreement with those of Elishakoff [7].

Dividing the beam into 20 equal-length elements, critical load is calculated for an axially FG tapered beam fordifferent combinations of taper ratios in Table 2. As expected, the critical load decreases as the height and/or breadthtaper ratios increases due to the softening effect caused by the decrease in moment of inertia. It is observed that theeffect of height taper ratio is more pronounced than that of breadth taper ratio since the power of height in momentof inertia is 3; while it is 1 for the breadth.


Table 1Critical load for an axially FG uniform beam

Boundary Condition Coefficients of E (X) Present (NE∗=10) Elishakoff [7]

S-S b0 = 1, b1 = 1, b2 = −1 12.0003 12

C-C b0 = 16, b1 = 1, b2 = −1 12.0014 12

S-C b0 = 916

, b1 = 34, b2 = −1 12.0006 12

∗NE stands for number of elements.

Table 2Non-dimensional critical load (λ = PcrL2/E0I0) for an ax-ially FG tapered beam with E (X) = E0

(1 + X

L

)cb 0 0.2 0.4 0.6 0.8

ch C-C0 57.3948 51.7863 45.7362 38.9921 30.8925

0.2 41.9174 37.6028 32.9643 27.8175 21.68060.4 28.1798 25.0894 21.7817 18.1336 13.82470.6 16.3416 14.3963 12.3271 10.0645 7.42820.8 6.6812 5.7847 4.8410 3.8240 2.6665

C-F0 3.1177 2.9497 2.7676 2.5652 2.3285

0.2 2.6225 2.4638 2.2915 2.0992 1.87250.4 2.1054 1.9585 1.7988 1.6200 1.40740.6 1.5522 1.4217 1.2798 1.1208 0.93090.8 0.9245 0.8217 0.7109 0.5883 0.4441

S-S0 14.5113 13.1398 11.6969 10.1451 8.3957

0.2 10.6860 9.5971 8.4543 7.2285 5.84990.4 7.2831 6.4715 5.6228 4.7164 3.70190.6 4.3287 3.7892 3.2284 2.6338 1.97490.8 1.8559 1.5952 1.3159 1.0242 0.7078

4.2. Free longitudinal vibration

In order to investigate the convergence of the method in free longitudinal vibration, the fundamental longitudinalfrequency (ωL1) for two types axially FG uniform beams is determined for different numbers of elements andcompared with exact results from Candan and Elishakoff [29] in Fig. 2. It is well-observed that the computed resultsconverge rapidly to exact ones as the number of elements increases. It is also instructive to know that FEM is a typeof stiffness method which is formulated on the basis of variational calculus; therefore it is expected that the resultsprovided by FEM show an upper bound of the exact ones. This fact could be clearly verified in Fig. 2.

Dividing the beam into 40 elements, the first three non-dimensional longitudinal frequencies of an axially FGtapered beam for two different boundary conditions are reported in Table 3. It is worthy to note that cross-sectionalarea plays a key role in free longitudinal vibration of rods and it is obvious that the order of contribution of ch and cb

in cross-sectional area is equal; thus the natural longitudinal frequencies of a rod with ch = c1 and cb = c2 are notdifferent from those of a bar with ch = c2 and cb = c1. This point is considered in tabulating the results in Table 3. Itis observed that unlike lower modes, the higher modes are not considerably affected by taper such that higher modesof an axially FG tapered rod is very close to those of uniform rod. Moreover, it is observed that natural frequenciesof a C-F rod increase with taper ratio; while they show a decreasing trend for a clamped-clamped rod. The first threemode shapes of both C-F and C-C axially FG tapered rods with ch = 0 and cb = 0.8 are plotted in Fig. 3.

4.3. Free transverse vibration

Firstly the fundamental transverse frequency (ωT1) of a unit-length axially FG uniform beam is obtained fordifferent boundary conditions and distributions of mass density and modulus of elasticity. The results are comparedwith those of Elishakoff and his co-workers [14–16] in Table 4 where it is observed that the results are comparablewith exact ones.


Table 3Non-dimensional longitudinal frequencies (μL = ωL

√ρ0L2/E0) for an

axially FG tapered beam with ρ = ρ0

(1 + X

L+

(XL

)2)

and E (X) =

E0

(1 + X

L

)C-F C-C

ch cb μL1 μL2 μL3 μL1 μL2 μL3

0 0 1.1902 4.2575 7.1768 2.8767 5.7687 8.66560.2 1.2504 4.2709 7.1846 2.8639 5.7622 8.66120.4 1.3293 4.2928 7.1976 2.8422 5.7514 8.65400.6 1.4401 4.3358 7.2240 2.8031 5.7312 8.64050.8 1.6129 4.4468 7.3007 2.7202 5.6830 8.6068

0.2 0.2 1.3120 4.2867 7.1939 2.8547 5.7575 8.65810.4 1.3929 4.3116 7.2087 2.8377 5.7490 8.65240.6 1.5060 4.3584 7.2376 2.8050 5.7321 8.64110.8 1.6819 4.4751 7.3184 2.7321 5.6893 8.6112

0.4 0.4 1.4760 4.3402 7.2259 2.8268 5.7436 8.64880.6 1.5918 4.3922 7.2583 2.8024 5.7311 8.64050.8 1.7707 4.5163 7.3446 2.7426 5.6957 8.6157

0.6 0.6 1.7105 4.4512 7.2956 2.7895 5.7251 8.63660.8 1.8919 4.5857 7.3904 2.7479 5.7009 8.6198

0.8 0.8 2.0723 4.7353 7.5006 2.7354 5.6972 8.6184

Fig. 2. Convergece of the present element in determination of the fundamental longitudinal frequency of a unit length C-F inhomogeneous rod.Case 1: ρ (X) = 1, E (X) = 2 + 2X − X2; Case 2: ρ (X) = 1 + X, E (X) = 13

3+ 13

3X + 1

3X2 − X3. (Results in squares are exact

ones reported in Candan and Elishakoff [29]).

Dividing the beam into 20 elements, the first three non-dimensional transverse frequencies of an axially FG taperedbeam are given in Tables 5–7 for different combinations of height and breadth taper ratios with different boundaryconditions.

As observed in Eq. (4), the shape functions do not reflect the effects of varying moment of inertia; thus it couldbe expected that the error introduced by this element is greater for those problems where the rate of change ofmoment of inertia is high i.e. height and/or breadth taper ratios increase towards unity. It is also obvious that theeffect of height tapering on moment of inertia is greater than that of breadth tapering; therefore the effects of breadthtapering is negligibly small on the convergence rate of the element. Seeking brevity, the convergence rate of themethod is higher for those beam configurations where the height taper ratio is small. This point could be observedin Fig. 4 which depicts the convergence of the method in determination of the first three natural frequencies of a

clamped-clamped axially FG tapered beam with ρ = ρ0

(1 + X

L +(

XL

)2)

and E (X) = E0

(1 + X

L

)with respect


Table 4Fundamental transverse frequency of an axially FG uniform beam

Boundary Coefficients Coefficients of E (X) Present Present Exactcondition of ρ (X) (NE = 6) (NE = 10)

G-S a0 = 1b0 = 61, b1 = 0, b2 = −14b3 = 0, b4 = 1

18.9737 18.9737√

360∗

C-G a0 = 1b0 = 88

9, b1 = 16

3, b2 = 4

3b3 = −4, b4 = 1

18.9741 18.9737√

360∗∗

C-F a0 = 1b0 = 26, b1 = 16, b2 = 6b3 = −4, b4 = 1

18.9739 18.9737√

360∗∗∗

C-F a0 = 1, a1 = 1b0 = 324

5, b1 = 214

5, b2 = 104

5b3 = −6

5, b4 = −11

5, b5 = 1

22.4502 22.4500√

504∗∗∗

C-Fa0 = 1, a1 = 1a2 = 1

b0 = 5785

, b1 = 4065

, b2 = 45

b3 = 445

, b4 = 35, b5 = −2

b6 = 1

25.9233 25.9230√

672∗∗∗

∗ Results from Ref. [14], ∗∗ Results from Ref. [15], ∗∗∗ Results from Ref. [16].

Table 5Non-dimensional transverse frequencies (μT = ωT

√ρ0A0L4/E0I0) for

an axially FG tapered beam with ρ = ρ0

(1 + X

L+

(XL

)2)

and E (X) =

E0

(1 + X

L

); Boundary Conditions: C-C

cb 0 0.2 0.4 0.6 0.8

ch First Mode0 20.4721 20.4152 20.2883 20.0186 19.3845

0.2 18.2170 18.1996 18.1286 17.9437 17.45660.4 15.8282 15.8498 15.8350 15.7367 15.40250.6 13.2293 13.2896 13.3319 13.3238 13.15290.8 10.2235 10.3229 10.4255 10.5168 10.5339

Second Mode0 56.5491 56.4715 56.2981 55.9205 54.9713

0.2 50.4801 50.4565 50.3599 50.1017 49.37280.4 44.0246 44.0553 44.0370 43.9027 43.40660.6 36.9653 37.0509 37.1137 37.1104 36.86780.8 28.7492 28.8912 29.0409 29.1842 29.2402

Third Mode0 110.9467 110.8615 110.6709 110.2501 109.1416

0.2 99.1734 99.1474 99.0414 98.7543 97.90460.4 86.6272 86.6608 86.6414 86.4932 85.91760.6 72.8740 72.9681 73.0382 73.0375 72.76150.8 56.8109 56.9674 57.1341 57.2991 57.3787

Fig. 3. The first three mode shapes of an axially FG tapered rod with ch = 0 and cb = 0.8 for (a) C-F and (b) C-C boundary conditions.



√ρ0A0L4/E0I0) for

an axially FG tapered beam with ρ = ρ0

(1 + X

L+

(XL

)2)

and E (X) =

E0

(1 + X

L

); Boundary Conditions: C-F

cb 0 0.2 0.4 0.6 0.8

ch First Mode0 2.4256 2.6054 2.8508 3.2137 3.8310

0.2 2.5051 2.6863 2.9336 3.2993 3.92190.4 2.6155 2.7987 3.0486 3.4181 4.04710.6 2.7835 2.9699 3.2237 3.5985 4.23550.8 3.0871 3.2794 3.5401 3.9232 4.5695

Second Mode0 18.6042 19.0042 19.5304 20.2959 21.6760

0.2 17.3802 17.7501 18.2379 18.9501 20.24320.4 16.0705 16.4092 16.8571 17.5139 18.71640.6 14.6508 14.9567 15.3627 15.9616 17.06940.8 13.1142 13.3850 13.7466 14.2848 15.2955

Third Mode0 55.1800 55.5337 56.0226 56.7996 58.4352

0.2 50.0491 50.3934 50.8645 51.6029 53.13320.4 44.6181 44.9504 45.4003 46.0957 47.51290.6 38.7446 39.0605 39.4844 40.1304 41.42360.8 32.1309 32.4229 32.8123 33.3986 34.5521


√ρ0A0L4/E0I0)

for an axially FG tapered beam with ρ = ρ0

(1 + X

L+

(XL

)2)

and

E (X) = E0

(1 + X

L

); Boundary Conditions: S-S

cb 0 0.2 0.4 0.6 0.8

ch First Mode0 9.0286 9.0599 9.0867 9.0994 9.0685

0.2 8.1341 8.1462 8.1498 8.1336 8.06460.4 7.1531 7.1455 7.1254 7.0794 6.97030.6 6.0357 6.0082 5.9638 5.8868 5.73510.8 4.6520 4.6046 4.5355 4.4264 4.2283

Second Mode0 36.3717 36.3418 36.3153 36.2969 36.2769

0.2 32.5236 32.5123 32.5079 32.5164 32.53260.4 28.4747 28.4822 28.5003 28.5370 28.59280.6 24.1101 24.1371 24.1791 24.2469 24.34970.8 19.1314 19.1803 19.2509 19.3590 19.5300

Third Mode0 81.7318 81.6850 81.6449 81.6236 81.6394

0.2 73.1138 73.0959 73.0903 73.1116 73.18550.4 63.9942 64.0054 64.0350 64.1007 64.23740.6 54.0921 54.1330 54.1992 54.3126 54.52070.8 42.6954 42.7677 42.8742 43.0436 43.3451

to different number of elements. It is observed that the convergence rate for the case of ch = 0.4 and cb = 0.8 ishigher than the other two cases.

5. Concluding remarks

A new element is proposed for matrix analysis of axially FG tapered beams by using the shape functions ofhomogeneous uniform beams. The effects of varying cross-sectional area and mechanical properties of the beam


Fig. 4. Cnvergence of the element in free transverse vibration of a C-C axially FG tapered beam.

element are taken into account in the evaluation of the structural matrices. Carrying out several numerical examples,it was shown that the element could provide acceptable results in free vibration and stability analyses. Since limitedstudies have been carried out on axially FG tapered Euler-Bernoulli beams, the results given in this paper could beused as a means of comparison for any future studies. Moreover, this idea could be extended to derivation of a newelement for shear deformable beams.

Acknowledgment

The financial support from University College of Engineering, University of Tehran (Grant No. 8102033/1/02) isdeeply appreciated.

Appendix A.

Consider a non-prismatic beam whose cross-sectional area and moment of inertia vary respectively as

A(X) = A0

(1 − cb

X

L

) (1 − ch

X

L

)and I(X) = I0

(1 − cb

X

L

) (1 − ch

X

L

)3

where X is the coordinate along the beam; A0 and I0 are respectively the cross-sectional area and moment of inertiaat X = 0; ch and cb are respectively the height and breadth taper ratios and L is the beam length. It is assumed thatρ (X) = a0 + a1X + a2X

2 and E (X) = b0 + b1X . The beam is divided into several elements of length l. Thestructural matrices are evaluated for the nth element.


A.1. Stretching stiffness matrix

The stretching stiffness matrix is obtained as Ks = A0lL2

[ks

i,j

], i, j = 1, 2.

ks1,1 = α1

(n − 1

2

) (n2 − n +

12

)− α2

(n2 − n +

13

)+ α3

(n − 1

2

)+ b0L

2

ks2,2 = −ks

2,1 = ks1,1

where

α1 = chcbb1l3, α2 = l2 [b1L (ch + cb) − b0cbch] , α3 = lL [b1L − b0 (ch + cb)]

A.2. Bending stiffness matrix

The bending stiffness matrix is obtained as Kb = I0l3L4

[kb

i,j

], i, j = 1, ..., 4.

kb1,1 = 12

[β1

(n4 − 2n3 + 3n2 − 2n +

47

) (n − 1

2

)− β2

(n4 − 2n3 +

125

n2 − 75n +

1135

)+

β3

(n2 − n +

710

) (n − 1

2

)− β4

(n2 − n +

25

)+ β5

(n − 1

2

)+ b0L

4

]

kb2,1 = 6l

[β1

(n5 − 10

3n4 +

173

n3 − 5n2 +4721

n − 1742

)− β2

(n4 − 8

3n3 +

175

n2 − 2n +47105

)+

β3

(n3 − 2n2 − 17

10n − 1

2

)− β4

(n2 − 4

3n +

1730

)+ β5

(n − 2

3

)+ b0L

4

]

kb3,1 = −kb

1,1 = −kb3,3

kb4,1 = 6l

[β1

(n5 − 5

3n4 +

73n3 − 2n2 +

1921

n − 16

)− β2

(n4 − 4

3n3 +

75n2 − 4

5n +

19105

)+

β3

(n3 − n2 +

710

n − 15

)− β4

(n2 − 2

3n +

730

)+ β5

(n − 1

3

)+ b0L

4

]

kb2,2 = 4l2

[β1

(n5 − 15

4n4 +

193

n3 − 112

n2 +177

n − 73168

)− β2

(n4 − 3n3 +

195

n2 − 115

n +1735

)+

β3

(n3 − 9

4n2 +

1910

n − 1120

)− β4

(n2 − 3

2n +

1930

)+ β5

(n − 3

4

)+ b0L

4

]

kb3,2 = −kb

2,1

kb4,2 = 2l2

[β1

(n4 − 2n3 +

103

n2 − 73n +

2942

) (n − 1

2

)− β2

(n4 − 2n3 +

135

n2 − 85n +

1335

)+

β3

(n2 − n +

45

) (n − 1

2

)− β4

(n2 − n +

1330

)+ β5

(n − 1

2

)+ b0L

4

]

kb4,3 = −kb

4,1


kb4,4 = 4l2

[β1

(n5 − 5

4n4 +

43n3 − n2 +

37n − 13

168

)− β2

(n4 − n3 +

45n2 − 2

5n +

335

)+

β3

(n3 − 3

4n2 +

210

n − 110

)− β4

(n2 − 1

2n +

215

)+ β5

(n − 3

4

)+ b0L

4

]

where

β1 = b1l5c3

hcb, β2 = 3l4c2h

[ch

(13Lb1 − 1

3cbb0

)+ cbb1L

],

β3 = 3l3Lch

[−13

b0c2h + ch (b1L − cbb0) + Lcbb1

]

β4 = l2L2[−3c2

hb0 + ch (3b1L − 3cbb0) + b1cbL], β5 = lL3 (b1L − cbb0 − 3chb0)

A.3. Geometric stiffness matrix

Assuming N(x) = P , the geometric stiffness matrix is obtained as Kg = P[kg

i,j

], i, j = 1, ..., 4.

kg1,1 =

65l

, kg2,1 =

110

, kg3,1 = −kg

1,1 = −kg3,3, kg

4,1 = −kg3,2 = −kg

4,3 = kg2,1, kg

2,2 =215

l,

kg4,2 =

−130

l, kg4,4 = kg

2,2

A.4. Stretching mass matrix

The stretching mass matrix is obtained as Ma = A0l3L2

[Ma

i,j

], i, j = 1, 2.

Ma1,1 = χ1

(n4 − 3n3 +

185

n2 − 2n +37

)− χ2

(n3 − 9

4n2 +

95n − 1

2

)+ χ3

(n2 − 3

2n +

35

)

+χ4

(n − 3

4

)+ a0L

2

Ma2,1 =

12

[χ1

(n4 − 2n3 +

95n2 − 4

5n +

17

)− χ2

(n2 − n − 2

5

) (n − 1

2

)+ χ3

(n2 − n +

310

)

+χ4

(n − 1

2

)+ a0L

2

]

Ma2,2 = χ1

(n4 − n3 +

35n2 − 1

5n +

135

)− χ2

(n3 − 3

4n2 +

310

n − 120

)+ χ3

(n2 − 1

2n +

110

)

+χ4

(n − 1

4

)+ a0L

2

where

χ1 = l4cbcha2, χ2 = l3 [La2 (ch + cb) − a1chcb] ,

χ3 = l2[a2L

2 − a1L (ch + cb) + a0chcb

], χ4 = lL [a1L − a0 (ch + cb)]


A.5. Bending mass matrix

The bending mass matrix is given by Mb = A0L2

[M b

i,j

], i, j = 1, ..., 4.

M b1,1 =

13l

35

[γ1

(n4 − 40

13n3 +

14539

n2 − 16178

n +63143

)− γ2

(n3 − 30

13n2 +

14578

n − 161312

)+

γ3

(n2 − 20

13n +

145234

)+ γ4

(n − 10

13

)+ a0L

2

]

M b2,1 =

11l2

210

[γ1

(n4 − 30

11n3 +

6522

n2 − 4933

n +35121

)− γ2

(n3 − 45

22n2 +

6544

n − 49132

)+

γ3

(n2 − 15

11n +

65132

)+ γ4

(n − 15

22

)+ a0L

2

]

M b3,1 =

9l

70

[γ1

(n4 − 2n3 +

4627

n2 − 1927

n +23198

)− γ2

(n2 − n +

1954

) (n − 1

2

)+

γ3

(n2 − n +

2381

)+ γ4

(n − 1

2

)+ a0L

2

]

M b4,1 =

−13l2

420

[γ1

(n4 − 28

13n3 +

2513

n2 − 3239

n +119858

)− γ2

(n3 − 21

13n2 +

2526

n − 839

)+

γ3

(n2 − 14

13n +

2578

)+ γ4

(n − 7

13

)+ a0L

2

]

M b2,2 =

l3

105

[γ1

(n4 − 5

2n3 +

52n2 − 7

6n +

733

)− γ2

(n3 − 15

8n2 +

54n − 7

24

)+

γ3

(n2 − 5

4n +

512

)+ γ4

(n − 5

8

)+ a0L

2

]

M b3,2 =

13l2

420

[γ1

(n4 − 24

13n3 +

1913

n2 − 2239

n +25286

)− γ2

(n3 − 18

13n2 +

1926

n − 1178

)+

γ3

(n2 − 12

13n +

1978

)+ γ4

(n − 6

13

)+ a0L

2

]

M b4,2 =

−l3

140

[γ1

(n4 − 2n3 +

53n2 − 2

3n +

766

)− γ2

(n2 − n +

13

) (n − 1

2

)+

γ3

(n2 − n +

518

)+ γ4

(n − 1

2

)+ a0L

2

]

M b3,3 =

13l

35

[γ1

(n4 − 12

13n3 +

1939

n2 − 1178

n +5

286

)− γ2

(n3 − 9

13n2 +

1978

n − 11312

)+

γ3

(n2 − 6

13n +

19234

)+ γ4

(n − 3

13

)+ a0L

2

]


M b4,3 =

−11l2

210

[γ1

(n4 − 14

11n3 +

1722

n2 − 833

n +23726

)− γ2

(n3 − 21

22n2 +

1744

n − 233

)+

γ3

(n2 − 7

11n +

17132

)+ γ4

(n − 7

22

)+ a0L

2

]

M b4,4 =

l3

105

[γ1

(n4 − 3

2n3 + n2 − 1

3n +

122

)− γ2

(n3 − 9

8n2 +

12n − 1

12

)+

γ3

(n2 − 3

4n +

16

)+ γ4

(n − 3

8

)+ a0L

2

]

where

γ1 = l4a2cbch, γ2 = l3 [a2L (ch + cb) − a1cbch] ,

γ3 = l2[a2L

2 − a1L (ch + cb) + a0chcb

], γ4 = lL [La1 − a0 (ch + cb)]

References

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