# Free Vibration Analysis of a Thin-Walled Beam with Shear Sensitive Material

Authors: K. B. Bozdogan; D. Ozturk
Journal: Mathematical Problems in Engineering
 (2013)

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## Abstract

This paper presents a method for a free vibration analysis of a thin-walled beam of doubly asymmetric cross section filled with shear sensitive material. In the study, first of all, a dynamic transfer matrix method was obtained for planar shear flexure and torsional motion. Then, uncoupled angular frequencies were obtained by using dynamic element transfer matrices and boundary conditions. Coupled frequencies were obtained by the well-known two-dimensional approaches. At the end of the study, a sample taken from the literature was solved, and the results were evaluated in order to test the convenience of the method.


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## Body

## 1. Introduction

In the last two decades research on the dynamics of beams has grown enormously. There are numerous studies [1–29] on the bending-torsion coupled beam. In the beams, the elastic center and the center of mass are not coincident, so the translational and torsional modes are inherently coupled as a result of this offset. Rafezy and Howson [24] proposed an exact dynamic stiffness matrix approach for the three-dimensional, bimaterial beam of doubly asymmetric cross-section. The beam comprises a thin-walled outer layer that encloses and works compositely with its shear sensitive core material.A dynamic transfer matrix method for the free vibration analysis of a thin-walled beam of doubly asymmetric cross-section filled with shear sensitive materials is suggested in this study. The following assumptions are made in this study: the behaviour of the material is linear elastic, small displacement theory is valid, and the dynamic coupling effect of structure caused by the eccentricity between the center of shear rigidity and the flexural rigidity center is ignored in analysis.


## 2. Analysis

### 2.1. Physical Model

Figure1 shows a uniform, three dimensional beam of length L. It has a doubly asymmetric cross-section comprising a thin-walled outer layer that encloses shear sensitive material [24]. The outer layer may have the form of an open or closed section that is assumed to provide warping and Saint-Venant rigidity, while the core materials provide Saint-Venant and shear rigidity. These assumptions lead to a model in which a typical cross-section has independent centers of flexure, shear, and mass denoted by O, S, and C, respectively [24]. For convenience, the origin of the coordinate system is located at the centre of flexure. O gives the result that the axis of elastic flexure coincides with the z-axis of the member. The x- and y-axes are subsequently aligned with the principle axes of the cross-section. The locations of the points S and C in the coordinate system Oxy are given by S(xs,ys) and C(xc,yc), respectively. The resulting elastic shear and mass axes then run parallel to the z-axis through (xs,ys) and (xc,yc) respectively. When the elastic axis of the beam does not coincide, the lateral and torsional motion of the beam will always be coupled in one or more planes [24].Typical thin-walled  beam [24].
(a)
(b)


### 2.2. Element Transfer Matrices for Planar Motion

The governing equations forith element of uncoupled thin-walled shear sensitive beam can be written as
(1)ρi∂2Ui(zi,t)∂t2+EIxi∂4Ui(zi,t)∂zi4-(GA)xi∂2Ui(zi,t)∂zi2=0,ρi∂2Vi(zi,t)∂t2+EIyi∂4Vi(zi,t)∂zi4-(GA)yi∂2Vi(zi,t)∂zi2=0,ρirm2∂2Ψi(zi,t)∂t2+EIwi∂4Ψi(zi,t)∂zi4-(GJ)oi∂2Ψi(zi,t)∂zi2=0,
where GJoi=GtiJti+GJci.E
I
x
i and EIyi are the flexural rigidity of the ith segment in the x-z and y-z planes, respectively, and GtiJti and EIwi are the Saint-Venant and warping torsion rigidity of the ith segment about O, where Iw is the warping moment of inertia or warping constant. GAxi and GAyi are the effective shear rigidities of the core material of the ith segment in x and y directions, respectively, and GJoi is the Saint-Venant torsional rigidity of the core material about O. ρi are the mass per unit length of the ith segment, and rm is the polar mass radius of gyration of cross section [24].If a sinusoidal variation ofU, V, and ψ with circular frequency ω is assumed then
(2)Ui(zi,t)=uisin(ωxt),Vi(zi,t)=visin(ωyt),ψi(zi,t)=θisin(ωθt),
where ui, vi, and θi are the amplitudes of the sinusoidally varying displacement.Substituting (2) in (1) results are
(3)d4uidzi4-(GA)xi(EI)xid2uidzi2-pi(EI)xiωx2ui=0,d4vidzi4-(GA)yi(EI)yid2vidzi2-pi(EI)yiωy2vi=0,d4θidzi4-(GJ)oi(EI)wid2θidθi2-pirm2(EI)wiωθ2θi=0.
When (3) is solved with respect to zi, ui(zi), vi(zi), and θi(zi) can be obtained as follows:
(4)ui(zi)=c1cosh(axizi)+c2sinh(axizi)+c3cos(bxizi)+c4sin(bxizi),(5)vi(zi)=c5cosh(ayizi)+c6sinh(ayizi)+c7cos(byizi)+c8sin(byizi),(6)θi(zi)=c9cosh(aθizi)+c10sinh(aθizi)+c11cos(bθizi)+c12sin(bθizi),
where αxi, αyi, αθi, bxi, byi, and bθi can be calculated as follows:
(7)axi=sxi+pxi2,ayi=syi+pyi2,aθi=sθi+pθi2,bxi=-sxi+pxi2,byi=-syi+pyi2,bθi=-sθi+pθi2,pxi=(GAxiEIxi)2+4*piωx2(EI)xi,pyi=(GAyiEIyi)2+4*piωy2(EI)yi,pθi=(GJoiEIwi)2+4*piωθ2(EI)wi,sxi=(GA)xi(EI)xi,syi=(GA)yi(EI)yi,sθi=(GJ)oi(EI)wi.
By using (4), (5), and (6), the rotation angles in x and y directions (ui′,vi′), rate of twist (θi′), bending moments in x and y directions (Mxi,Myi) and bimoment (Mwi), shear forces in x and y directions (Vxi,Vyi), and torque (Mti) for ith element can be obtained as follows:
(8)dui(zi)dzi=c1axisinh(axizi)+c2axicosh(axizi)-c3bxisin(bxizi)+c4bxicos(bxizi),(9)dvi(zi)dzi=c5ayisinh(ayizi)+c6ayicosh(ayizi)-c7byisin(byizi)+c8byicos(byizi),(10)dθi(zi)dzi=c9aθisinh(aθizi)+c10aθicosh(aθizi)-c11bθisin(bθizi)+c12bθicos(bθizi),(11)Mxi(zi)=EIxid2ui(zi)dzi2=EIxi[c1axi2cosh(axizi)+c2axi2sinh(axizi)kkkkkkk-c3bxi2cos(bxizi)-c4bxi2sin(bxizi)],(12)Myi(zi)=EIyid2vi(zi)dzi2=EIyi[c5ayi2cosh(ayizi)+c6ayi2sinh(ayizi)kkkkkkk-c7byi2cos(byizi)-c8byi2sin(byizi)],(13)Mwi(zi)=EIwid2θi(zi)dzi2=EIwi[c9aθi2cosh(aθizi)+c10aθi2sinh(aθizi)kkkkkkk-c11bθi2cos(bθizi)-c12bθi2sin(bθizi)],(14)Vxi(zi)=EIxid3ui(zi)dzi3-(GA)xidui(zi)dzi=[EIxiaxi3sinh(axizi)-(GA)xiaxisinh(axizi)]c1+[EIxiaxi3cosh(axizi)kkkkk-(GA)xiaxicosh(axizi)axi3]c2+[(EI)xibxi3sin(bxizi)+(GA)xibxisin(bxizi)]c3+[-(EI)xibxi3cos(bxizi)kkkkk-(GA)xibxicos(bxizi)bxi3]c4,(15)Vyi(zi)=EIyid3vi(zi)dzi3-(GA)yidvi(zi)dzi=[EIyiayi3sinh(ayizi)-(GA)yiayisinh(ayizi)]c5+[EIyiayi3cosh(ayizi)kkkkk-(GA)yiayicosh(ayizi)]c6+[(EI)yibyi3sin(byizi)+(GA)yibyisin(byizi)]c7+[-(EI)yibyi3cos(byizi)kkkkk-(GA)yibyicos(byizi)byi3]c8,(16)Mti(zi)=EIwid3θi(zi)dzi3-(GJ)oidθi(zi)dzi=[EIθiaθi3sinh(aθizi)-(GJ)oiaθisinh(aθizi)]c9+[EIwiaθi3cosh(aθizi)kkkkk-(GJ)ioaθicosh(aθizi)aθi3]c10+[(EI)wibθi3sin(bθizi)+(GJ)oibθisin(bθizi)]c11+[-(EI)wibθi3cos(bθizi)kkkkk-(GJ)oibyicos(bθizi)bθi3]c12.
The following equation shows the matrix form of (4), (8), (11), and (14):
(17)[ui(zi)ui′(zi)Mxi(zi)Vxi(zi)]=Axi(zi)[c1c2c3c4].
For the y direction, the following shows the matrix form of (5), (9), (12), and (15):
(18)[vi(zi)vi′(zi)Myi(zi)Vyi(zi)]=Ayi(zi)[c5c6c7c8].
Similarly, torsional motion can be written:
(19)[θi(zi)θi′(zi)Mwi(zi)Mti(zi)]=Aθi(zi)[c9c10c11c12].
At the initial point of the ith element, (17), (18), and (19) can be written as follows:
(20)[ui(0)ui′(0)Mxi(0)Vxi(0)]=Axi(0)[c1c2c3c4],(21)[vi(0)vi′(0)Myi(0)Vyi(0)]=Ayi(0)[c5c6c7c8],(22)[θi(0)θi′(0)Mwi(0)Mti(0)]=Aθi(0)[c9c10c11c12].
When vector c is solved from (20) and is substituted in (17), the following is obtained:
(23)[ui(zi)ui′(zi)Mxi(zi)Vxi(zi)]=Axi(zi)Axi(0)-1[ui(0)ui′(0)Mxi(0)Vxi(0)].
For zi=li, (23) can be written as
(24)[ui(li)ui′(li)Mxi(li)Vxi(li)]=Txi[ui(0)ui′(0)Mxi(0)Vxi(0)],
where Txi is the element dynamic transfer matrix of the ith element.For they direction, (23) and (24) can be written as follows:
(25)[vi(zi)vi′(zi)Myi(zi)Vyi(zi)]=Ayi(zi)Ayi(0)-1[vi(0)vi′(0)Myi(0)Vyi(0)],(26)[vi(li)vi′(li)Myi(li)Vyi(li)]=Tyi[vi(0)vi′(0)Myi(0)Vyi(0)].
Similarly, rotation motion can be written in equations as follows:
(27)[θi(zi)θi′(zi)Mwi(zi)Mti(zi)]=Aθi(zi)Aθi(0)-1[θi(0)θi′(0)Mwi(0)Mti(0)],(28)[θi(li)θi′(li)Mwi(li)Mti(li)]=Tθi[θi(0)θi′(0)Mwi(0)Mti(0)].
If (24) is written successively, the displacements—internal forces relationship between the initial part and end of the beam—can be found as follows:
(29)[uenduend′MxendVxend]=TxnTx(n-1)⋯Tx2Tx1[uinitialuinitial′MxinitialVxinitial]=tx[uinitialuinitial′MxinitialVxinitial].
For y and rotation motion, (29) can be written as follows:
(30)[vendvend′MyendVyend]=TynTy(n-1)⋯Ty2Ty1[vinitialvinitial′MyinitialVyinitial]=ty[vinitialvinitial′MyinitialVyinitial],[θendθend′MwendMtend]=TθnTθ(n-1)⋯Tθ2Tθ1[θinitialθinitial′MwinitialMtinitial]=tθ[θinitialθinitial′MwinitialMtinitial].
The eigenvalue equation for a thin-walled beam filled with shear sensitive material can be established using (29), (30), and the specific boundary conditions are as follows.(1)
Clamped-Free:fx=tx(3,3)*tx(4,4)-tx(3,4)*tx(4,3)=0, fy=ty(3,3)*ty(4,4)-ty(3,4)*ty(4,3)=0, fθ=tθ(3,3)*tθ(4,4)-tθ(3,4)*tθ(4,3)=0.(2)
Clamped-Clamped:fx=tx(1,3)*tx(2,4)-tx(1,4)*tx(2,3)=0, fy=ty(1,3)*ty(2,4)-ty(1,4)*ty(2,3)=0, fθ=tθ(1,3)*tθ(2,4)-tθ(1,4)*tθ(2,3)=0.(3)
Simply-Simply:fx=tx(1,2)*tx(3,4)-tx(3,2)*tx(1,4)=0,fy=ty(1,2)*ty(3,4)-  ty(3,2)*ty(1,4)=0, fθ=tθ(1,2)*tθ(3,4)-tθ(3,2)*tθ(1,4)=0.(4)
Free-Free:fx=tx(3,1)*tx(4,2)-tx(3,2)*tx(4,1)=0, fy=ty(3,1)*ty(4,2)-ty(3,2)*ty(4,1)=0, fθ=tθ(3,1)*tθ(4,2)-tθ(3,2)*tθ(4,1)=0.(5)
Clamped-Simply:fx=tx(1,3)*tx(3,4)-tx(1,4)*tx(3,3)=0, fy=ty(1,3)*ty(3,4)-ty(1,4)*ty(3,3)=0, fθ=tθ(1,3)*tθ(3,4)-tθ(1,4)*tθ(3,3)=0.In frequency equations the values ofω, which set the determinant to zero, are the uncoupled angular frequencies.


### 2.3. Coupled Frequencies

Ignoring the dynamic coupling effect of structure caused by the eccentricity between the center of shear rigidity and the geometric center the coupled frequencies of the shear torsional beam can be obtained by using uncoupled frequencies and the well-known equation as follows [28]:
(31)|ωj(i)2-ωx(i)20-ycωj(i)20ωj(i)2-ωy(i)2xcωj(i)2-ycωj(i)2xcωj(i)2rm2(ωj(i)2-ωθ(i)2)|=0kkkkkkkkkkkkk(j=1,2,3)(i=1,2,3…).


## 2.1. Physical Model

Figure1 shows a uniform, three dimensional beam of length L. It has a doubly asymmetric cross-section comprising a thin-walled outer layer that encloses shear sensitive material [24]. The outer layer may have the form of an open or closed section that is assumed to provide warping and Saint-Venant rigidity, while the core materials provide Saint-Venant and shear rigidity. These assumptions lead to a model in which a typical cross-section has independent centers of flexure, shear, and mass denoted by O, S, and C, respectively [24]. For convenience, the origin of the coordinate system is located at the centre of flexure. O gives the result that the axis of elastic flexure coincides with the z-axis of the member. The x- and y-axes are subsequently aligned with the principle axes of the cross-section. The locations of the points S and C in the coordinate system Oxy are given by S(xs,ys) and C(xc,yc), respectively. The resulting elastic shear and mass axes then run parallel to the z-axis through (xs,ys) and (xc,yc) respectively. When the elastic axis of the beam does not coincide, the lateral and torsional motion of the beam will always be coupled in one or more planes [24].Typical thin-walled  beam [24].
(a)
(b)


## 2.2. Element Transfer Matrices for Planar Motion

The governing equations forith element of uncoupled thin-walled shear sensitive beam can be written as
(1)ρi∂2Ui(zi,t)∂t2+EIxi∂4Ui(zi,t)∂zi4-(GA)xi∂2Ui(zi,t)∂zi2=0,ρi∂2Vi(zi,t)∂t2+EIyi∂4Vi(zi,t)∂zi4-(GA)yi∂2Vi(zi,t)∂zi2=0,ρirm2∂2Ψi(zi,t)∂t2+EIwi∂4Ψi(zi,t)∂zi4-(GJ)oi∂2Ψi(zi,t)∂zi2=0,
where GJoi=GtiJti+GJci.E
I
x
i and EIyi are the flexural rigidity of the ith segment in the x-z and y-z planes, respectively, and GtiJti and EIwi are the Saint-Venant and warping torsion rigidity of the ith segment about O, where Iw is the warping moment of inertia or warping constant. GAxi and GAyi are the effective shear rigidities of the core material of the ith segment in x and y directions, respectively, and GJoi is the Saint-Venant torsional rigidity of the core material about O. ρi are the mass per unit length of the ith segment, and rm is the polar mass radius of gyration of cross section [24].If a sinusoidal variation ofU, V, and ψ with circular frequency ω is assumed then
(2)Ui(zi,t)=uisin(ωxt),Vi(zi,t)=visin(ωyt),ψi(zi,t)=θisin(ωθt),
where ui, vi, and θi are the amplitudes of the sinusoidally varying displacement.Substituting (2) in (1) results are
(3)d4uidzi4-(GA)xi(EI)xid2uidzi2-pi(EI)xiωx2ui=0,d4vidzi4-(GA)yi(EI)yid2vidzi2-pi(EI)yiωy2vi=0,d4θidzi4-(GJ)oi(EI)wid2θidθi2-pirm2(EI)wiωθ2θi=0.
When (3) is solved with respect to zi, ui(zi), vi(zi), and θi(zi) can be obtained as follows:
(4)ui(zi)=c1cosh(axizi)+c2sinh(axizi)+c3cos(bxizi)+c4sin(bxizi),(5)vi(zi)=c5cosh(ayizi)+c6sinh(ayizi)+c7cos(byizi)+c8sin(byizi),(6)θi(zi)=c9cosh(aθizi)+c10sinh(aθizi)+c11cos(bθizi)+c12sin(bθizi),
where αxi, αyi, αθi, bxi, byi, and bθi can be calculated as follows:
(7)axi=sxi+pxi2,ayi=syi+pyi2,aθi=sθi+pθi2,bxi=-sxi+pxi2,byi=-syi+pyi2,bθi=-sθi+pθi2,pxi=(GAxiEIxi)2+4*piωx2(EI)xi,pyi=(GAyiEIyi)2+4*piωy2(EI)yi,pθi=(GJoiEIwi)2+4*piωθ2(EI)wi,sxi=(GA)xi(EI)xi,syi=(GA)yi(EI)yi,sθi=(GJ)oi(EI)wi.
By using (4), (5), and (6), the rotation angles in x and y directions (ui′,vi′), rate of twist (θi′), bending moments in x and y directions (Mxi,Myi) and bimoment (Mwi), shear forces in x and y directions (Vxi,Vyi), and torque (Mti) for ith element can be obtained as follows:
(8)dui(zi)dzi=c1axisinh(axizi)+c2axicosh(axizi)-c3bxisin(bxizi)+c4bxicos(bxizi),(9)dvi(zi)dzi=c5ayisinh(ayizi)+c6ayicosh(ayizi)-c7byisin(byizi)+c8byicos(byizi),(10)dθi(zi)dzi=c9aθisinh(aθizi)+c10aθicosh(aθizi)-c11bθisin(bθizi)+c12bθicos(bθizi),(11)Mxi(zi)=EIxid2ui(zi)dzi2=EIxi[c1axi2cosh(axizi)+c2axi2sinh(axizi)kkkkkkk-c3bxi2cos(bxizi)-c4bxi2sin(bxizi)],(12)Myi(zi)=EIyid2vi(zi)dzi2=EIyi[c5ayi2cosh(ayizi)+c6ayi2sinh(ayizi)kkkkkkk-c7byi2cos(byizi)-c8byi2sin(byizi)],(13)Mwi(zi)=EIwid2θi(zi)dzi2=EIwi[c9aθi2cosh(aθizi)+c10aθi2sinh(aθizi)kkkkkkk-c11bθi2cos(bθizi)-c12bθi2sin(bθizi)],(14)Vxi(zi)=EIxid3ui(zi)dzi3-(GA)xidui(zi)dzi=[EIxiaxi3sinh(axizi)-(GA)xiaxisinh(axizi)]c1+[EIxiaxi3cosh(axizi)kkkkk-(GA)xiaxicosh(axizi)axi3]c2+[(EI)xibxi3sin(bxizi)+(GA)xibxisin(bxizi)]c3+[-(EI)xibxi3cos(bxizi)kkkkk-(GA)xibxicos(bxizi)bxi3]c4,(15)Vyi(zi)=EIyid3vi(zi)dzi3-(GA)yidvi(zi)dzi=[EIyiayi3sinh(ayizi)-(GA)yiayisinh(ayizi)]c5+[EIyiayi3cosh(ayizi)kkkkk-(GA)yiayicosh(ayizi)]c6+[(EI)yibyi3sin(byizi)+(GA)yibyisin(byizi)]c7+[-(EI)yibyi3cos(byizi)kkkkk-(GA)yibyicos(byizi)byi3]c8,(16)Mti(zi)=EIwid3θi(zi)dzi3-(GJ)oidθi(zi)dzi=[EIθiaθi3sinh(aθizi)-(GJ)oiaθisinh(aθizi)]c9+[EIwiaθi3cosh(aθizi)kkkkk-(GJ)ioaθicosh(aθizi)aθi3]c10+[(EI)wibθi3sin(bθizi)+(GJ)oibθisin(bθizi)]c11+[-(EI)wibθi3cos(bθizi)kkkkk-(GJ)oibyicos(bθizi)bθi3]c12.
The following equation shows the matrix form of (4), (8), (11), and (14):
(17)[ui(zi)ui′(zi)Mxi(zi)Vxi(zi)]=Axi(zi)[c1c2c3c4].
For the y direction, the following shows the matrix form of (5), (9), (12), and (15):
(18)[vi(zi)vi′(zi)Myi(zi)Vyi(zi)]=Ayi(zi)[c5c6c7c8].
Similarly, torsional motion can be written:
(19)[θi(zi)θi′(zi)Mwi(zi)Mti(zi)]=Aθi(zi)[c9c10c11c12].
At the initial point of the ith element, (17), (18), and (19) can be written as follows:
(20)[ui(0)ui′(0)Mxi(0)Vxi(0)]=Axi(0)[c1c2c3c4],(21)[vi(0)vi′(0)Myi(0)Vyi(0)]=Ayi(0)[c5c6c7c8],(22)[θi(0)θi′(0)Mwi(0)Mti(0)]=Aθi(0)[c9c10c11c12].
When vector c is solved from (20) and is substituted in (17), the following is obtained:
(23)[ui(zi)ui′(zi)Mxi(zi)Vxi(zi)]=Axi(zi)Axi(0)-1[ui(0)ui′(0)Mxi(0)Vxi(0)].
For zi=li, (23) can be written as
(24)[ui(li)ui′(li)Mxi(li)Vxi(li)]=Txi[ui(0)ui′(0)Mxi(0)Vxi(0)],
where Txi is the element dynamic transfer matrix of the ith element.For they direction, (23) and (24) can be written as follows:
(25)[vi(zi)vi′(zi)Myi(zi)Vyi(zi)]=Ayi(zi)Ayi(0)-1[vi(0)vi′(0)Myi(0)Vyi(0)],(26)[vi(li)vi′(li)Myi(li)Vyi(li)]=Tyi[vi(0)vi′(0)Myi(0)Vyi(0)].
Similarly, rotation motion can be written in equations as follows:
(27)[θi(zi)θi′(zi)Mwi(zi)Mti(zi)]=Aθi(zi)Aθi(0)-1[θi(0)θi′(0)Mwi(0)Mti(0)],(28)[θi(li)θi′(li)Mwi(li)Mti(li)]=Tθi[θi(0)θi′(0)Mwi(0)Mti(0)].
If (24) is written successively, the displacements—internal forces relationship between the initial part and end of the beam—can be found as follows:
(29)[uenduend′MxendVxend]=TxnTx(n-1)⋯Tx2Tx1[uinitialuinitial′MxinitialVxinitial]=tx[uinitialuinitial′MxinitialVxinitial].
For y and rotation motion, (29) can be written as follows:
(30)[vendvend′MyendVyend]=TynTy(n-1)⋯Ty2Ty1[vinitialvinitial′MyinitialVyinitial]=ty[vinitialvinitial′MyinitialVyinitial],[θendθend′MwendMtend]=TθnTθ(n-1)⋯Tθ2Tθ1[θinitialθinitial′MwinitialMtinitial]=tθ[θinitialθinitial′MwinitialMtinitial].
The eigenvalue equation for a thin-walled beam filled with shear sensitive material can be established using (29), (30), and the specific boundary conditions are as follows.(1)
Clamped-Free:fx=tx(3,3)*tx(4,4)-tx(3,4)*tx(4,3)=0, fy=ty(3,3)*ty(4,4)-ty(3,4)*ty(4,3)=0, fθ=tθ(3,3)*tθ(4,4)-tθ(3,4)*tθ(4,3)=0.(2)
Clamped-Clamped:fx=tx(1,3)*tx(2,4)-tx(1,4)*tx(2,3)=0, fy=ty(1,3)*ty(2,4)-ty(1,4)*ty(2,3)=0, fθ=tθ(1,3)*tθ(2,4)-tθ(1,4)*tθ(2,3)=0.(3)
Simply-Simply:fx=tx(1,2)*tx(3,4)-tx(3,2)*tx(1,4)=0,fy=ty(1,2)*ty(3,4)-  ty(3,2)*ty(1,4)=0, fθ=tθ(1,2)*tθ(3,4)-tθ(3,2)*tθ(1,4)=0.(4)
Free-Free:fx=tx(3,1)*tx(4,2)-tx(3,2)*tx(4,1)=0, fy=ty(3,1)*ty(4,2)-ty(3,2)*ty(4,1)=0, fθ=tθ(3,1)*tθ(4,2)-tθ(3,2)*tθ(4,1)=0.(5)
Clamped-Simply:fx=tx(1,3)*tx(3,4)-tx(1,4)*tx(3,3)=0, fy=ty(1,3)*ty(3,4)-ty(1,4)*ty(3,3)=0, fθ=tθ(1,3)*tθ(3,4)-tθ(1,4)*tθ(3,3)=0.In frequency equations the values ofω, which set the determinant to zero, are the uncoupled angular frequencies.


## 2.3. Coupled Frequencies

Ignoring the dynamic coupling effect of structure caused by the eccentricity between the center of shear rigidity and the geometric center the coupled frequencies of the shear torsional beam can be obtained by using uncoupled frequencies and the well-known equation as follows [28]:
(31)|ωj(i)2-ωx(i)20-ycωj(i)20ωj(i)2-ωy(i)2xcωj(i)2-ycωj(i)2xcωj(i)2rm2(ωj(i)2-ωθ(i)2)|=0kkkkkkkkkkkkk(j=1,2,3)(i=1,2,3…).


## 3. Procedure of Computation

A program that considers the method presented in this study as a basis has been prepared in MATLAB, and the operation stages are presented below.(1)
element dynamic Transfer matrices are calculated for each element by using (24), (26), and (28).(2)
System dynamic transfer matrices (see (29)–(30)) are obtained with the help of element transfer matrices.(3)
The angular frequencies of uncoupled vibrations are obtained by using the boundary conditions.(4)
The coupled angular frequencies are found by using (31).


## 4. A Numerical Example

In this part of the study two numerical examples were solved by a program written in MATLAB to validate the presented method. The results are compared with those given in the literature.


### 4.1. Numerical Example 1

The first example considers the beam studied by Tanaka and Bercin [11]. A typical uniform thin-walled beam has a length of 1.5 m with a doubly asymmetric cross section. The properties of the cross section are as follows:xc=0.02316,yc=0.02625,ρ=1.947kg/m,rm2=3.0303*10-3m2,EIx=73480Nm2,EIy=16680Nm2,EIw=23.64Nm4,andGJ0=10.81Nm2.The first three coupled natural frequencies of the beam are calculated by the presented method and compared with the results by Tanaka and Bercin [11] and Rafezy and Howson [24] in Table 1 for clamped-free (C-F) and simply-simply (S-S) boundary conditions.Table 1
Coupled natural frequencies for the beam of example 1.
Natural frequencies (Hz)
BC
Proposed method
Tanaka and Bercin [11]
Rafezy and Howson [24]
f
1
f
2
f
3
f
1
f
2
f
3
f
1
f
2
f
3
C-F
17.17
27.31
59.10
17.03
27.58
59.25
17.17
27.31
59.10
S-S
44.71
75.14
164.87
41.48
74.12
164.11
44.71
75.14
164.87


### 4.2. Numerical Example 2

A typical continuous beam with a doubly asymmetric cross section is considered in this example (Figure2).Figure 2
The doubly asymmetric, continuous channel section and the cross section of  beam of example 2 with warping allowed at B, C, and D but fully constrained at A.The beam comprises a thin-walled outer layer and a shear core with the following properties between support points A and B. The typical uniform thin-walled beam has a length of 1.5 m with a doubly asymmetric cross section. The properties of the cross section are as follows:xs=0.08,ys=0.03,xc=0.05,yc=0.02,ρ=20kg/m,rm2=0.008m2,EIx=2.16*106Nm2,EIy=1.73*106Nm2,GtJt=3200Nm2,EIw=1.4*103Nm4,GAx=600000N,GAy=600000N,andGJc=3800Nm2.The shear core is omitted between points B and D, where the cross-sectional properties remain unchanged, except thatGAx=GAy=GJc=0, and the small change in ρ has been ignored.The first three coupled natural frequencies of the beam are calculated by the presented method and compared with the results of Rafezy and Howson [24] in Table 2.Table 2
Coupled natural frequencies of the continuous beam of example 2.
Frequency number
This study
Rafezy and Howson[24]
Difference (%)
1
6.906
6.940
−0.49
2
19.763
19.796
−0.17
3
35.461
33.836
4.80The main source of error between the proposed method and Rafezy and Howson methods is the eccentricity between the center of shear stiffness and flexural stiffness which was not taken into account in the proposed method.


## 4.1. Numerical Example 1

The first example considers the beam studied by Tanaka and Bercin [11]. A typical uniform thin-walled beam has a length of 1.5 m with a doubly asymmetric cross section. The properties of the cross section are as follows:xc=0.02316,yc=0.02625,ρ=1.947kg/m,rm2=3.0303*10-3m2,EIx=73480Nm2,EIy=16680Nm2,EIw=23.64Nm4,andGJ0=10.81Nm2.The first three coupled natural frequencies of the beam are calculated by the presented method and compared with the results by Tanaka and Bercin [11] and Rafezy and Howson [24] in Table 1 for clamped-free (C-F) and simply-simply (S-S) boundary conditions.Table 1
Coupled natural frequencies for the beam of example 1.
Natural frequencies (Hz)
BC
Proposed method
Tanaka and Bercin [11]
Rafezy and Howson [24]
f
1
f
2
f
3
f
1
f
2
f
3
f
1
f
2
f
3
C-F
17.17
27.31
59.10
17.03
27.58
59.25
17.17
27.31
59.10
S-S
44.71
75.14
164.87
41.48
74.12
164.11
44.71
75.14
164.87


## 4.2. Numerical Example 2

A typical continuous beam with a doubly asymmetric cross section is considered in this example (Figure2).Figure 2
The doubly asymmetric, continuous channel section and the cross section of  beam of example 2 with warping allowed at B, C, and D but fully constrained at A.The beam comprises a thin-walled outer layer and a shear core with the following properties between support points A and B. The typical uniform thin-walled beam has a length of 1.5 m with a doubly asymmetric cross section. The properties of the cross section are as follows:xs=0.08,ys=0.03,xc=0.05,yc=0.02,ρ=20kg/m,rm2=0.008m2,EIx=2.16*106Nm2,EIy=1.73*106Nm2,GtJt=3200Nm2,EIw=1.4*103Nm4,GAx=600000N,GAy=600000N,andGJc=3800Nm2.The shear core is omitted between points B and D, where the cross-sectional properties remain unchanged, except thatGAx=GAy=GJc=0, and the small change in ρ has been ignored.The first three coupled natural frequencies of the beam are calculated by the presented method and compared with the results of Rafezy and Howson [24] in Table 2.Table 2
Coupled natural frequencies of the continuous beam of example 2.
Frequency number
This study
Rafezy and Howson[24]
Difference (%)
1
6.906
6.940
−0.49
2
19.763
19.796
−0.17
3
35.461
33.836
4.80The main source of error between the proposed method and Rafezy and Howson methods is the eccentricity between the center of shear stiffness and flexural stiffness which was not taken into account in the proposed method.


## 5. Conclusions

This paper presents a method for a free vibration analysis of a thin-walled beam of doubly asymmetric cross section filled with shear sensitive material. In the study, first of all, a dynamic transfer matrix method was obtained for planar shear flexure and torsional motion. Then, uncoupled angular frequencies were obtained by using dynamic element transfer matrices and boundary conditions. Coupled frequencies were obtained by the well-known two-dimensional approaches. It was observed from the sample taken from the literature that the presented method gave sufficient results. The error margin of the proposed method is shown to be less than 5%. The main source of error is the eccentricity between the center of shear stiffness and flexural stiffness which was not taken into account in the proposed method.The transfer matrix method is an efficient and computerized method which also provides a fast and practical solution since the dimension of the matrix for the elements and system never changes. Because of this the proposed method is simple and accurate enough to be used both at the concept design stage and for final analyses.



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*Source: 493905-2013-10-12.xml*