AN EXACT ELASTODYNAMIC SOLUTION TO VIBRATION PROBLEMS OF A COMPOSITE STRUCTURE IN THE PLANE STRESS STATE

Journal of Sound and Vibration (1996) 196(1), 85–96

AN EXACT ELASTODYNAMIC SOLUTION TO VIBRATION PROBLEMS OF A COMPOSITE STRUCTURE IN THE PLANE STRESS STATE S. K Institute of Machine Design Fundamentals, Warsaw University of Technology, 02-524 Warszawa, ul. Narbutta 84, Poland (Received 29 September 1995, and in final form 12 February 1996) An elastodynamic, closed form, exact solution of the plane stress state problems of layered structures is presented. The solution has been derived by applying semi-inverse techniques and the method of superposition, within the linear theory of elastodynamics. As an example, the vibration problem of a layered cantilever beam with base excitation is considered. Several numerical results are given for a two-layer cantilever beam consisting of isotropic layers. It is also shown that the final form of the eigenvalue problem for the beam is a set of three coupled transcendental equations. 7 1996 Academic Press Limited

1. INTRODUCTION

Levinson [1] derived, within the linear theory of elastodynamics, an exact, three-dimensional solution for the free vibrations problems of a homogeneous, simply supported, rectangular plate. The solution was obtained by using the semi-inverse techniques. Karczmarzyk [2–4] applied the same techniques and the superposition method in order to solve, within the linear theory of elastodynamics, several problems for layered beams and bands† of rectangular and non-rectangular cross-section, with various boundary conditions at their ends. The mathematical models of the problems include, among other things, shear stresses, cross-sectional warping in each layer and continuity of displacements and stresses between adjoining layers. The rigorous approach to modeling of the mechanical behaviour of the structures implies new computational problems. For instance, it has been shown [4] that the final form of the eigenvalue problem for the clamped-clamped band or beam with clamped ends is a set of three coupled, homogeneous, transcendental equations. (A ‘‘clamped-clamped band’’ is a rectangular plate with two opposite edges clamped and being in the plane strain state.) Due to the approach it is possible to include, in the boundary problem statement, practical boundary conditions at the ends of the beams: e.g., a one-sided attachment as shown in Figure 1b). The two-dimensional elastodynamic approach to vibration problems of composite beams is well-founded. Firstly, due to including in the elastodynamic formulation both the shear stresses and the cross-sectional warping in each layer one can calculate accurate values of eigenfrequencies of the beams. It is well documented that eigenfrequencies of †The word band is used to denote a rectangular plate in the plane strain state.

85 0022–460X/96/360085 + 12 $18.00/0

7 1996 Academic Press Limited

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homogeneous beams predicted by the classical theory based on the kinematic assumption of plane cross-sections and neglecting the shear stresses are overestimated [5]. The author verified [3, 4] that Huang’s conclusions refer to layered beams as well. Secondly, the author’s elastodynamic solutions (including this one presented here) can be directly applied to calculate accurate values of eigenfrequencies, displacements and stresses for various composite beams: e.g., beams consisting of a number of stiffness-comparable layers, typical sandwich beams consisting of thin stiff faces and a thick compliant (soft) core, etc. This wide application is possible since no simplifications, for example kinematic assumptions of a special form, have been introduced in the problem statement. Besides, one does not need a shear coefficient as, for instance, in Timoshenko beam theory. Thirdly, the approach can now be easily extended to beams and plates consisting of orthotropic layers, and can be applied for laminated structures since a cross-ply laminate is a typical orthotropic material. This paper contains a concise summary and generalization of the author’s efforts to apply the elastodynamic approach to the composite beams problems. In section 3 there is introduced and discussed in detail a kinematic model which is more general than the one introduced by Levinson [1] and applied by Karczmarzyk [2–4]. In section 4 a general solution of equations of motion is derived by using the kinematic model from section 3. In section 5 a new, exact, closed form, elastodynamic solution for the vibration of a layered, cantilever beam with a base excitation is presented. In section 6 a new, final form of the eigenvalue problem for a layered, cantilever beam is given. A reader can verify that, identically as in the case of a beam having both ends clamped [4], in order to calculate eigenfrequencies of the cantilevers one has to solve a set of three coupled, homogeneous, transcendental equations. To demonstrate the usefulness of the approach presented in the paper several numerical results for both homogeneous and layered beams with base excitations are given in section 7. In the same section a procedure for calculating displacements of cantilevers within the classical Bernoulli-Euler theory is outlined. The procedure was used to calculate deflections of the free ends of the beams in order to compare them with the results obtained according to the elastodynamic solution presented in sections 3–5. Considerations in the paper have been restricted to the case of beams of rectangular cross-sections with isotropic layers. However, the approach can easily be extended to take into account both non-rectangular cross-sections [3] and orthotropy of the layers. The free vibrations problem of a ribbed plate consisting of an orthotropic band and isotropic beams has been solved in reference [4] according to the approach presented here.

(a)

(b)

z

z x

x z*2

A

A

Layer 1 Layer 2 Figure 1. (a) side view (above) and cross-section (below) a two layered cantilever beam; (b) layered beam one-sidedly attached to the support A.

   

87

2. THE PLANE STATE OF STRESS

Consider a composite beam of rectangular cross-section consisting of p homogeneous, isotropic layers as shown (for p = 2) in Figure 1. The beam vibrates in the xz-plane. It is assumed that the plane stress state occurs within the structure and thus within each jth layer. The plane state of stress means that the stress vector vanishes in the xz-plane: i.e., the following equations are satisfied: (syy )j = (sxy )j = (syz )j = 0.

(1)

Here (skl )j denotes the stress tensor within the jth layer of the structure. Taking into account equations (1) one obtains, on the basis of Hooke’s law, (skl )j = 2mj (okl )j + dkl lj (orr )j ,

k, l, r, = 1, 2, 3,

(2)

the following relationships: (oyy )j = −[nj /(1 − nj )][(oxx )j + (ozz )j ],

(3)

(sxx )j = 2mj [1/(1 − nj )](oxx )j + 2mj [nj /(1 − nj )](ozz )j ,

(4)

(sxz )j = mj [1uxj /1z + 1uzj /1x],

(5)

(szz )j = 2mj [1/(1 − nj )](ozz )j + 2mj [nj /(1 − nj )](oxx )j .

(6)

Here (okl )j denotes the strain tensor within the jth layer of the beam, mj , lj and nj are the Lame´ parameters and Poisson’s ratio, respectively. It is obvious that the stresses depend (only) on derivatives of the displacements ux and uz . The equations of motion, skl,k = r1 2ul /1t 2,

(7)

applied to the jth layer of the beam are now in the form 1sxx /1x + 1sxz /1z = r1 2ux /1t 2,

1szx /1x + 1szz /1z = r1 2uz /1t 2,

(8)

or in the form, mj 92uxj + mj 5j (1 2uxj /1x 2 + 1 2uzj /1z1x) − rj 1 2uxj /1t 2 = 0, mj 92uzj + mj 5j (1 2uxj /1x1z + 1 2uzj /1z 2 ) − rj 1 2uzj /1t 2 = 0,

(9)

where 92 = 1 2/1x 2 + 1 2/1z 2. Equations (9) are valid either for the plane stress state or for the plane strain state depending on the value of the coefficient 5j . The coefficient is defined as 5j =

6

(1 + nj )/(1 − nj ) 1/(1 − 2nj )

7

for the plane stress state , for the plane strain state

(10)

where nj is the Poisson’s ratio of the jth layer. Equations (9) for the plane stress state can be transformed into equations of motion for the plane strain state by replacing the Poisson’s ratio nj with the coefficient n'j = nj /(1 − nj ). Note finally that mj (1 + nj )/(1 − nj ) = l'j + mj ,

mj /(1 − 2nj ) = lj + mj ,

(11)

l'j = 2mj nj /(1 − nj ) 0 lj (1 − 2nj )/(1 − nj ) 0 2mj cj ,

(12)

cj = nj /(1 − nj ).

(13)

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88

3. SOLUTIONS OF THE PLANE STRESS STATE EQUATIONS OF MOTION

It is assumed that the displacements ux , uz within the jth layer of the beam are defined as uxj = −gj (dfj /dy)(dW/dx) t(t),

uzj = +fj (dfj /dy)W(x)t(t),

(14)

where t(t) = exp (ivt), v denotes frequency, t time, and fj = fj (y), gj (z), fj (z) are unknown functions. The assumptions (14) are an extended version of those given by Levinson [1]. The extension is the introduction of the factor dfj /dy. It is assumed in further considerations that the function W(x) satisfies Helmholtz’s equation: 1 2W/1x 2 = q 2W.

(15)

The parameter q can either be positive or negative. Taking into account equations (10), (12), (14) and (15) one can transform equations (9) to a set of equations containing only two unknown functions gj and fj : namely, 2

−mj d2gj /dz 2 − [(l'j + 2mj )q 2 + rj vm2 ]gj + (l'j + mj ) dfj /dz = 0, (l'j + 2mj ) d2fj /dz 2 + (mj q 2 + rj vm2 )fj − q 2(l'j + mj ) dgj /dz = 0.

(16)

Equations (16) are a generalized version of those given by Karczmarzyk [2]. The functions gj and fj in equations (16) are in the form of sums of two terms: i.e., gj (z) = g1j (z, b 1j ) + g2j (z, b 2j ),

fj (z) = f1j (z, b 1j ) + f2j (z, b 2j ),

(17)

The parameters b 1j and b 2j are defined as b 1j2 = q 2 + rj v 2/mj ,

b 2j2 = q 2 + rj v 2/(l'j + 2mj ).

(18)

Besides, Karczmarzyk [2] has established the following relationships: f1j = −(q 2/b 1j2 ) dg1j /dz,

df1j /dz = +q 2g1j , 2 2j

f2j = −dg2j /dz,

df2j /dz = +b g2j .

(19) (20)

Equations (16) are satisfied by each of the pairs of functions (g1j , f1j ) and (g2j , f2j ). Upon the basis of equations (19) and (20), one obtains d2f1j /dz 2 = −b 1j2 f1j , 2

2

2 1j

d g1j /dz = −b g1j ,

d2f2j /dz 2 = −b 2j2 f2j , 2

2

2 2j

d g2j /dz = −b g2j .

(21) (22)

Equations (21) and (22), together with equations (4)–(6), enable one to derive the following expressions for the stresses (sxx )j , (szx )j and (szz )j : (sxx )j = −2mj [df1j /dz + {1/(1 − nj )}( − nj + q 2/b 2j2 ) df2j /dz](dfj /dy)W(x)t(t), (sxz )j = mj [(1 + b 1j2 /q 2 )f1j + 2f2j ] (dfj /dy)(dW/dx)t(t), (szz )j = +2mj [df1j /dz + {1/(1 − nj )}(1 − nj q 2/b 2j2 ) df2j /dz](dfj /dy)W(x)t(t).

(23)

Each of expressions (23) consists of two components, one dependent on the function f1j or its derivative and the other on the function f2j or its derivative. The property is valid for all stresses and thus one can write the following expressions: 1) 2) (sxx )j = (sxx )j + (sxx )j ,

(szz )j = (szz1) )j + (szz2) )j ,

1) 2) (szx )j = (sxz )j = (szx )j + (szx )j , 1) 2) (sxy )j = (syx )j = (sxy )j + (sxy )j ,

(syz )j = (szy )j = (syz1) )j + (syz2) )j ,

1) 2) (syy )j = (syy )j + (syy )j .

(24)

   

89

Expressions for components (s ) for r = 1, 2 and s 0 x, z are as follows: r sy j

1) )j = −mj g1j (d2fj /dy 2 )(dW/dx )t(t), (sxy

(25)

(syz1) )j = −mj (q 2/b 1j2 )(dg1j /dz)(d2fj /dy 2 )(dW/dx)t(t),

(26)

2) )j = −mj g2j [d2fj /dy 2 + {nj /(1 − nj )}(b 2j2 − q 2 )fj ](dW/dx)t(t), (sxy

(27)

(syz2) )j = −mj (dg2j /dz)[d2fj /dy 2 + {nj /(1 − nj )}(b 2j2 − q 2 )fj ](dW/dx)t(t).

(28)

Note that the parameter q 2 appearing in the above expressions can be either positive or negative. Taking into consideration equations (24)–(28) one can write the relationships 1) 2) )j = 0 and (sxy )j = 0] [(sxy

c

[(sxy )j = 0],

[(syz1) )j = 0 and (syz2) )j = 0]

c

[(syz )j = 0],

1) )j = 0 and (syz1) )j = 0] [(sxy 2) )j = 0 and (syz2) )j = 0] \ [(sxy

\

[d2fj /dy 2 = 0],

[d2fj /dy 2 + {nj /(1 − nj )}(b 2j2 − q 2 )fj = 0].

(29)

One need not consider the vanishing of (syy )j since in order to derive expression (3) the stress has been assumed as equal to zero. After comparing the differential equations appearing in relationships (29) one can conclude that one has to introduce two functions fj1) (y) and fj2) (y) instead of one function fj (y). This means that all the expressions including the function fj (y) or its derivative should be modified. Generally, for a beam without twisting, the function fj1 (y) = y or its derivative are factors in the components of stresses and displacements denoted by superscripts 1) whereas the function fj2) (y) = sin (5* j y),

2 (5* 2j2 − q 2 ) 0 rj v 2nj /2mj , j ) = {nj /(1 − nj )}(b

(30)

or its derivative are factors in the components of stresses and displacements denoted by superscripts 2). Note that the ratio width/length is much less than 1 for a real beam and so that one can introduce the simplifications fj2) = sin (5* j y) 3 5* j y,

dfj2) /dy = 5* j cos (5* j y) 3 5* j .

(31)

On the other hand one has dfj1) /dy = Cj1) , where Cj1) is any constant of one’s choice. In particular one can make Cj1) = 5* j and thus expressions (23) for the stresses (sxx )j , (szx )j and (szz )j can be modified by replacing the factor dfj /dy with the coefficient 5* j . Expressions for the displacements should be modified as follows: uxj = uxj1) + uxj2) , uxj1) = −g1j (dW/dx)t(t),

uyj 0 uyj2) ,

uzj = uzj1) + uzj2) ,

uxj2) = −g2j 5* j cos (5* j y)(dW/dx)t(t),

uyj 0 uyj2) = −cj (df2j /dz − q 2g2j ) sin (5* j y)W(x)t(t), uzj1) = +f1j W(x)t(t),

uzj2) = +f2j 5* j cos (5* j y)W(x)t(t).

(32)

After taking into consideration the simplifications (31) one transforms expressions (32) for uxj and uzj to the form given by Levinson [1] and Karczmarzyk [2–4]. Note that the displacements field functions satisfy the Saint-Venant continuity conditions okl,ps + ops,kl − okp,ls − ols,kp = 0.

(33)

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90

4. BOUNDARY CONDITIONS AND CONTINUITY CONDITIONS BETWEEN ADJOINING LAYERS

To complete formulation of a boundary problem of a beam one has to introduce the boundary and continuity conditions. The boundary conditions on planes z1 (x, y, zˆ1 ) and z2 (x, y, zˆ2 ) are as follows; (szz1) )j = 1 + (szz2) )j = 1 = sˆ 1 ,

(szz1) )j = p + (szz2) )j = p = sˆ 2 ,

1) 2) (szx )j = 1 + (szx )j = 1 = tˆ 1 ,

1) 2) (szx )j = p + (szx )j = p = tˆ 2 .

(34)

Here sˆ 1 and tˆ 1 are given stress functions on the plane z1 , sˆ 2 and tˆ 2 are given stress functions on the plane z2 ; also zˆ2 − zˆ1 is equal to the thickness of the beam. If sˆ 1 = sˆ 2 = tˆ 1 = tˆ 2 = 0 the boundary conditions lead to the boundary eigenvalue problem. The continuity conditions between adjoining layers are as follows: (szz )j = l = (szz )j = l + 1 , (uz )j = l = (uz )j = l + 1 ,

(szx )j = l = (szx )j = l + 1 ,

(ux )j = l = (ux )j = l + 1 ,

l = 1, 2, . . . , p − 1.

(35)

Here p denotes the number of layers. The continuity conditions are valid for the beam with perfectly bonded layers. It can be seen that in order to satisfy equations (34) and (35) one has to apply the simplifications (31). Boundary conditions at the ends of the beam depend on the kind of supports at the ends. Recently Karczmarzyk [2, 4] has shown how to apply the approach presented here for two important cases of the boundary conditions: i.e., for completely clamped, free vibrating band and cantilever band or beam caused to vibrate by a sinusoidally varying force. In both these cases the superposition method was applied in order to link stresses and displacements components derived from the above expressions, successively for q2 = −a 2 less than zero and q2 = g 2 greater than zero. Note that for a simply supported beam one can satisfy the end conditions by assuming q2 = −am2 , am = mp/L, where m = 1, 2, 3, . . . , and L denotes the length of the beam. In this case one obtains directly from equations (18) and (21) the differential equations for the functions f1j and f2j : d2f1j /dz 2 = [ − b 1j2 0 (am2 − rj v 2/mj )]f1j , d2f2j /dz 2 = [ − b 2j2 0 {am2 − rj v 2/(l'j + 2mj )}]f2j .

(36)

5. AN EXAMPLE: NEW CLOSED FORM EXACT SOLUTION FOR VIBRATION OF A LAYERED, CANTILEVER BEAM WITH BASE EXCITATION

As was stated in section 3, the considerations refer to the function W(x) satisfying the Helmholtz equation (15) whereas the parameter q 2 can be either positive or negative. In order to obtain the solution of the boundary problem for the layered, cantilever beam one first has to solve equations (16), assuming q2 (subsequently) both less and greater than zero. In the further text two solutions of equations (16) corresponding to positive and negative values of q 2 are given. 5.1.        For the function W(x) of the form W(x) = WI(x) = sin (ax) + d I cos (ax),

(37)

   

91

where a, d are unknown real constants, one has q = −a and the functions gj (z) and fj (z) resulting from equations (16) are 2

I

2

fj (z) = C1j cosh (zb1j ) + C2j sinh (zb1j ) + C3j cosh (zb2j ) + C4j sinh (zb2j ), gj (z) = X'1j cosh (zb1j ) + X'2j sinh (zb1j ) + X'3j cosh (zb2j ) + X'4j sinh (zb2j ), b = −b = a − rj v /mj , 2 1j

2 1j

X'1j = −C2j b1j /a,

2

2

b = −b = a − rj v /(l'j + 2mj ), 2 2j

2 2j

X'2j = −C1j b1j /a,

2

2

X'3j = −C4j /b2j , X'4j = −C3j /b2j .

(38) (39) (40)

The functions gj and fj for the jth layer are dependent on four unknown constants Ckj , k = 1–4. Relationships (17), (19), (20) can now be written as gj = g1j (b1j , z) + g2j (b2j , z), 2

fj = f1j (b1j , z) + f2j (b2j , z),

2 1j

f1j = −(a /b )(d/dz)g1j , g1j = −(1/a 2 ) (d/dz) f1j ,

(41)

f2j = −(d/dz)g2j ,

(42)

g2j = −(1/b2j2 ) (d/dz) f2j .

(43)

After substituting functions (38) into the constitutive equation (2) and taking into consideration simplifications (31) one obtains the expression for the stresses: I (sxx )j = {−2mj (d/dz)f1j − [2mj /(1 − nj )](−nj + a 2/b2j2 )(d/dz)f2j }W I(x) exp (ivt),

(szzI )j = {+2mj (d/dz)f1j + [2mj /(1 − nj )](1 − nj a 2/b2j2 )(d/dz)f2j }W I(x) exp (ivt), I (szx )j = mj [(1 + b1j2 /a 2 )f1j + 2f2j ](dW I/dx)exp (ivt),

(44)

By using expressions (14), (31), (37) and (38) one can easily obtain expressions for the displacement vector uIj , corresponding to the stress vector sIj . 5.2.        For the function W(x) of the form [6] W(x) = W II (x) = exp [ − g(L − x)] + d II exp (−gx),

(45)

where g, d II are unknown, real constants, and L denotes the length of the beam, one has q 2 = g 2 and the functions Gj (z) 0 gj (z),

Fj (z) 0 fj (z)

(46)

resulting from equations (16) are as follows: Fj (z) = D1j cos (zr1j ) + D2j sin (zr1j ) + D3j cos (zr2j ) + D4j sin (zr2j ), Gj (z) = Y'1j cos (zr1j ) + Y'2j sin (zr1j ) + Y'3j cos (zr2j ) + Y'4j sin (zr2j ), r1j2 = b 1j2 = g 2 + rj v 2/mj , Y'1j = D2j r1j /g 2,

r2j2 = b 2j2 = g 2 + rj v 2/(l'j + 2mj ),

Y'2j = −D1j r1j /g 2,

Y'3j = D4j /r2j ,

Y'4j = −D3j /r2j .

(47) (48) (49)

The functions Gj and Fj for the jth layer are dependent on four unknown constants Dkj , k = 1 − 4. Relationships (17), (19), (20) can now be written in the form Gj = G1j (r1j , z) + G2j (r2j , z),

Fj = F1j (r1j , z) + F2j (r2j , z),

F1j = −(g 2/r1j2 )(d/dz)G1j , G1j = +(1/g 2 )(d/dz)F1j ,

(50)

F2j = −(d/dz)G2j ,

(51)

G2j = +(1/r2j2 )(d/dz)F2j .

(52)

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92

After substituting functions (47) into the constitutive equation (2) and taking into account simplifications (31) one obtains the expressions for the stresses: II (sxx )j = {−2mj (d/dz)F1j − [2mj /(1 − nj )]( − nj + g 2/r2j2 )(d/dz)F2j }W II (x) exp (ivt),

(szzII )j = {+2mj (d/dz)F1j + [2mj /(1 − nj )](1 − nj g 2/r2j2 )(d/dz)F2j }W II (x) exp (ivt), II (szx )j = mj [(1 + r1j2 /g 2 )F1j + 2F2j ](dW II/dx) exp (ivt).

(53)

By using expressions (14), (31), (45) and (47) one can easily obtain expressions for the displacement vector uIIj , corresponding to the stress vector sIIj . 5.3.     The displacement and stress vectors, uj and sj , j = 1, 2, . . . , p, within the beam are composed of two components; sj = sIIj + sIj .

uj = uIj + uIIj ,

(54)

Each of the components must satisfy the boundary and continuity conditions (34) and (35). The boundary conditions at the ends of the beams as shown in Figure 1(b) are taken to be of the form ivt (uz (0, z* 2 ))j = U0 e ,

j=p

s

j=1

g

sj + 1

(fzx (0, z* 2 ))j = 0,

j=p

(szx (L, z))j dz = 0,

s

j=1

sj

g

(55)

sj + 1

(szz (L, z))j z dz = 0,

sj

=sj + 1 − sj = = tj ,

(56)

where sj , sj + 1 denote co-ordinates of the surfaces of the jth layer, tj is the thickness of the jth layer and fxz is the slope in the x-z plane. U0 and v are the amplitude and frequency of the base excitation, respectively. The conditions (55) can be replaced by the set 1 j=p s h j=1

g

sj + 1

sj

(uz (0, z))j dz = U0 eivt,

1 j=p s h j=1

g

sj + 1

(fzx (0, z))j dz = 0,

(57)

sj

where h denotes the thickness of the beam. Conditions (57) refer to the case shown in Figure 1(a). It is noticed that instead of equating to zero the normal stress sxx (L, z) and the shear stress szx (L, z), the simplified boundary conditions (56) have been applied. This is the only simplification introduced in this paper since the integral equations (57) can be but need not be, used instead of the conditions (55). According to the Saint-Venant principle the simplification seems to be without any influence on the accuracy of the solution. 5.4.       By using the functions gj , fj given in section 5.1. one can transform the boundary and continuity conditions (34) and (35) to the form AIX = 0,

(58)

X = [X1 = C11 , X2 = C21 , X3 = C31 , X4 = C41 , X5 = C12 , X6 = C22 , . . . . . . . . , X(4p − 3) = C1p , X(4p − 2) = C2p , X(4p − 1) = C3p , X(4p) = C4p ]T.

(59)

   

93

For this set of algebraic equations one obtains a non-zero vector X provided that the following equation is satisfied: det (AI) 0 F I(a) = 0.

(60)

By using the functions Gj and Fj given in section 5.2. one can transform the boundary and continuity conditions (34) and (35) to the form AIIY = 0,

(61)

Y = [Y1 = D11 , Y2 = D21 , Y3 = D31 , Y4 = D41 , Y5 = D12 , Y6 = D22 , . . . . . . . . , Y(4p − 3) = D1p , Y(4p − 2) = D2p , Y(4p − 1) = D3p , Y(4p) = D4p ]T.

(62)

As in the previous case one has the equation det (AII) 0 F II (g) = 0.

(63)

Matrices AI and AII are of dimensions 4p × 4p where p is the number of layers within the beam. When viscoelastic material is assumed the matrices are complex. After taking into account 4p − 1 equations of the set (58) one can determine 4p − 1 elements of the eigenvector X which are dependent on one unknown element Xm . Analogously one can determine 4p − 1 elements of the vector Y as dependent on one unknown element Ym . Boundary conditions at the ends of the beam are as follows: II ivt (uzI (0, z* 2 ))j + (uz (0, z* 2 ))j = U0 e , j=p

s

j=1

j=p

s

j=1

g

sj + 1

g

sj + 1

I II (fzx (0, z* 2 ))j + (fzx (0, z* 2 ))j = 0,

(64)

[(szzI (L, z))j + (szzII (L, z))j ]zdz = 0,

(65)

I II [(szx (L, z))j + (szx (L, z))j ] dz = 0.

(66)

sj

sj

The final form of the end conditions can then be written as AIIIR = BIII,

R = [Xm , d IXm , Ym , d IIYm ]T,

BIII = [U0 , 0, 0, 0]T.

(67–69)

Equations (67) enable one to calculate Xm , d I, Ym and dII. The transcendental equations (60) and (63), and the sets of algebraic equations (58), (61) and (67) determine the new, closed form solution of the boundary problem for the composite cantilever beam. 6. FINAL FORM OF THE EIGENVALUE PROBLEM FOR THE BEAM

When the eigenvalue problem of the beam is considered, the frequency v appearing in the functions gj , fj , Gj and Fj , and thus in the expressions for displacements and stresses, is unknown. Therefore in this case elements of matrices AI, AII and AIII contain together three variables: i.e., a, g, v. On the other hand, in this case the vector BIII is equivalent to zero and one has to solve the set of three coupled, transcendental equations: namely, det (AI) 0 F I(a, v) = 0,

det (AII) 0 F II (g, v) = 0,

det (AIII) 0 F III (a, g, v) = 0. (70)

This set of equations is more general than that derived in reference [4] for a beam having both ends clamped. It can easily be proved that for the clamped-clamped beam the third of equations (70) contains only two unknowns, a and g.

. 

94

T 1 Numerical results for the homogeneous cantilever beam of rectangular cross-section for a range of length L; amplitude of the base excitation U0 = 0·2 (mm), excitation frequency v = 300(rad/s) L(m) aL (rad) gL (rad) dI d II u˜z (L) (mm) uBE (mm)

0·10

0·12

0·14

0·16

0·18

0·20

0·45397 0·45382 1·97215 1·52682 0·2011 0·2011

0·54476 0·54458 1·55569 1·63473 0·2022 0·2011

0·63556 0·63535 1·24250 1·73497 0·2041 0·2041

0·72635 0·72611 0·99280 1·82292 0·2071 0·2071

0·81715 0·81687 0·78419 1·89394 0·2115 0·2115

0·90794 0·90764 0·60291 1·94379 0·2179 0·2180

7. NUMERICAL RESULTS

Numerical results have been obtained for homogeneous and two-layer cantilever elastic beams with sinusoidal base excitations. Displacements at the free ends of the beams have been calculated after solving the set of equations (60), (63) and (67). The beams were considered to be one-sidedly fastened to a stiff support as it is shown in Figure 1(b). In the case of the two-layer beam the steel layer adheres to the support. Material and geometrical parameters of the beams are given in Appendices 1 and 2. The numerical results are presented in Tables 1–3. u˜z (L) denotes the maximum dynamic displacements at the free ends of the beams. Displacement values u˜z (L) are compared with results obtained from the simple Bernoulli-Euler theory of bending. uBE denotes the maximum deflection at the free end resulting from the theory. The author’s procedure for calculating uBE was as follows. To obtain uBE the set of equations which must first be solved is [7] y(x = 0, t) = U0 eivt, 1y/1x =x = 0 = 0,

Q(x = L, t) = 0,

M(x = L, t) = 0,

(71)

where y(x, t) is the deflection of the beam, U0 and v are defined in section 5, Q(x, t) denotes the shear force and M(x, t) is the bending moment. Formulas for the bending moment and the shear force for a layered beam, within the Bernoulli-Euler theory, are

$

%

p

M(x, t) = − s bj Ej (z2j3 − z1j3 )/3 (1 2y/1x 2 ) eivt, j=1

Q = 1M/1x.

(72)

T 2 Numerical results for the homogeneous cantilever beam of rectangular cross-section for a range of frequency v; amplitude of the base excitation U0 = 0·2 (mm), length L = 0·2(m) v (rad/s)

300

400

500

600

700

800

aL (rad) gL (rad) dI d II u˜z (L) (mm) uBE (mm)

0·90794 0·90764 0·60291 1·94379 0·2179 0·2180

1·04846 1·04799 0·35596 1·97208 0·2334 0·2335

1·17228 1·17162 0·15632 1·94276 0·2556 0·2558

1·28423 1·28338 −0·02015 1·87233 0·2871 0·2873

1·38721 1·38613 −0·18642 1·77344 0·3320 0·3324

1·48307 1·48175 −0·35111 1·65564 0·3986 0·3991

   

95

T 3 Numerical results for the two-layer cantilever beam of rectangular cross-section for a range of length L; amplitude of the base excitation U0 = 0·5 (mm), excitation frequency v = 500(rad/s) L (m) aL (rad) gL (rad) dI d II u˜z (L) (mm) uBE (mm)

0·15

0·20

0·25

0·30

0·35

0·40

0·65771 0·65526 1·18735 1·76907 0·5110 0·5118

0·87695 0·87367 0·67082 1·94592 0·5371 0·5385

1·09619 1·09209 0·28495 1·98659 0·5991 0·6015

1·31542 1·31051 −0·06327 1·86539 0·7412 0·7453

1·53446 1·52893 −0·43984 1·60035 1·1176 1·1240

1·75389 1·74735 −0·94085 1·23793 2·9494 2·9680

Following Baumgarten and Pearce [8] one can write the formulas for z1j , z2j for the two-layer beam (j = 1, 2) as z11 = zf ,

z21 = zf + t1 ,

z12 = zf − t2 ,

z22 = zf ,

(73)

zf = 0·5t2 [1 − (t1 /t2 )2(E1 /E2 )]/[1 + (t1 /t2 )(E1 /E2 )].

(74)

where

The symbols bj , tj , Ej in the above formulas denote the width, thickness and Young’s modulus of the jth layer, respectively. The deflection function y(x, t) resulting from the equation of motion of the beam is of the form y(x, t) = (C1 sin kx + C2 cos kx + C3 sinh kx + C4 cosh kx)eiwt,

(75)

where k is of the form p

>

p

k4 = v 2 s rj bj tj s bj Ej (z2j3 − z1j3 )/3, j=1

(76)

j=1

in which rj is the mass density of the jth layer. After solving equations (71) one can obtain C1–C4 and then calculate, from the relationship (75), uBE = y(x = L, t = 0·5p/v). 8. CONCLUSIONS

On the basis of the numerical results it is seen that maximum deflections u˜z at the tip depend on both the length of the beam and the excitation frequency. As expected, within the ranges of length of the beams one can observe that the deflections increase with the increases in both the length and the frequency. The deflections for the homogeneous beams resulting from the two-dimensional solution given in the paper are in fact the same as the results obtained from the Bernoulli-Euler theory. However, for the two-layer beam the procedure based on the Bernoulli-Euler theory predicts slightly higher values of the deflection than the two-dimensional solution. The maximum difference, for L = 0·4 m, does not exceed 0·7%. On the basis of the past investigations of the author [3, 4] it is clear that the two-dimensional elastodynamic approach is especially useful for solving the eigenvalue problem. Eigenfrequencies resulting from the two-dimensional solutions are accurate and smaller than the corresponding eigenfrequencies obtained from the classical theory based on the assumption of plane cross-sections and neglecting the shear stresses. As regards

. 

96

forced vibrations, the two-dimensional solution, unlike the classical one, enables one to obtain both normal and shear stresses for each layer of the beam; the displacements of the composite beam predicted by the solution are very close to those resulting from classical theory.

REFERENCES 1. M. L 1985 Journal of Sound and Vibration 98, 289–298. Free vibrations of a simply supported, rectangular plate: an exact elasticity solution. 2. S. K 1995 Me´canique Industrielle et Mate´riaux 48, 107–110. New exact elastodynamic solutions to forced and free vibrations problems of plane viscoelastic composite structures. 3. S. K 1993 Journal of Theoretical and Applied Mechanics 31, 89–103. Free vibrations of layered beam with non-rectangular cross-section composed of viscoelastic layers. 4. S. K 1992 Journal of Theoretical and Applied Mechanics 30, 663–682. A new linear elastodynamic solution to boundary eigenvalue problem of flexural vibration of viscoelastic layered and homogeneous bands. 5. T. C. H 1961 Journal of Applied Mechanics 28, 579–584. The effect of rotatory inertia and of shear deformation on the frequency and normal mode equations of uniform beams with simple end conditions. 6. J. R. G and N. O 1982 Journal of Sound and Vibration 84, 481–489. Improved numerical computation of uniform beam characteristic values and characteristic functions. 7. A. W. L 1989 Journal of Sound and Vibration 134, 435–453. Closed form exact solutions for the steady state vibrations of continuous systems, subjected to distributed exciting forces. 8. J. R. B and B. K. P 1971 Journal of Engineering for Industry 93, 645–650. The damping effects of viscoelastic materials—part 1. Transverse vibrations of beams with viscoelastic coatings.

APPENDIX I: MATERIAL AND GEOMETRICAL PARAMETERS OF THE HOMOGENEOUS BEAM

Material Modulus of elasticity (N/m2 ) Poisson ratio Mass density (kg/m3 ) Thickness and width of the beam (m)

steel 0·2 × 1012 0·25 7860 0·01

APPENDIX II: MATERIAL AND GEOMETRICAL PARAMETERS OF THE TWO-LAYER BEAM

Modulus of elasticity (N/m2 ) Poisson ratio Mass density (kg/m3 ) Thickness (m) Width of the beam

Layer 1 0·16 × 1011 0·30 1750 0·02

Layer 2 0.2 × 1012 0·25 7860 0·005 0·035