tanα=uy(x1, y)
tanβ =uy(x2, y+δy) tanγ=ux(x, y1) tanδ=ux(x+δx, y2)
wherex1andx2are the values ofxbetweenxandx+δx, andy1andy2are the values ofy betweeny andy+δy. Substituting these values in (3.3.1), we obtain
T δx[uy(x2, y+δy)−uy(x1, y)] +T δy[ux(x+δx, y2)−ux(x, y1)]
=ρ δxδy utt. Division byρ δxδy yields
T ρ
uy(x2, y+δy)−uy(x1, y)
δy +ux(x+δx, y2)−ux(x, y1) δx
=utt. (3.3.2) In the limit asδxapproaches zero andδyapproaches zero, we obtain
utt=c2(uxx+uyy), (3.3.3) wherec2=T /ρ. This equation is called thetwo-dimensional wave equation.
If there is an external force f per unit area acting on the membrane.
Equation (3.3.3) takes the form
utt =c2(uxx+uyy) +F, (3.3.4) whereF =f /ρ.
3.4 Waves in an Elastic Medium
If a small disturbance is originated at a point in an elastic medium, neigh- boring particles are set into motion, and the medium is put under a state of strain. We consider such states of motion to extend in all directions. We assume that the displacements of the medium are small and that we are not concerned with translation or rotation of the medium as a whole.
Let the body under investigation be homogeneous and isotropic. LetδV be a differential volume of the body, and let the stresses acting on the faces of the volume be τxx, τyy, τzz, τxy, τxz, τyx, τyz, τzx, τzy. The first three stresses are called the normal stresses and the rest are called the shear stresses. (See Figure 3.4.1).
We shall assume that the stress tensor τij is symmetric describing the condition of the rotational equilibrium of the volume element, that is,
Figure 3.4.1Volume element of an elastic body.
τij =τji, i=j, i, j=x, y, z. (3.4.1) Neglecting the body forces, the sum of all the forces acting on the volume element in thex-direction is
(τxx)x+δx−(τxx)x!
δyδz+"
(τxy)y+δy−(τxy)y# δzδx + (τxz)z+δz−(τxz)z!
δxδy.
By Newton’s law of motion this resultant force is equal to the mass times the acceleration. Thus, we obtain
(τxx)x+δx−(τxx)x!
δyδz+"
(τxy)y+δy−(τxy)y# δzδx + (τxz)z+δz−(τxz)z!
δxδy=ρ δxδyδz utt (3.4.2) whereρis the density of the body anduis the displacement component in thex-direction. Hence, in the limit asδV approaches zero, we obtain
∂τxx
∂x +∂τxy
∂y +∂τxz
∂z =ρ∂2u
∂t2. (3.4.3)
Similarly, the following two equations corresponding toy and z directions are obtained:
3.4 Waves in an Elastic Medium 71
∂τyx
∂x +∂τyy
∂y +∂τyz
∂z =ρ∂2v
∂t2, (3.4.4)
∂τzx
∂x +∂τzy
∂y +∂τzz
∂z =ρ∂2w
∂t2, (3.4.5)
wherev andware the displacement components in the y andz directions respectively.
We may now define linear strains [see Sokolnikoff (1956)] as εxx= ∂u
∂x, εyz =1 2
∂w
∂y +∂v
∂z
, εyy = ∂v
∂y, εzx= 1 2
∂u
∂z +∂w
∂x
, (3.4.6)
εzz = ∂w
∂z, εxy= 1 2
∂v
∂x +∂u
∂y
,
in whichεxx,εyy,εzzrepresent unit elongations andεyz,εzx,εxyrepresent unit shearing strains.
In the case of an isotropic body, generalized Hooke’s law takes the form τxx=λθ+ 2µεxx, τyz = 2µεyz,
τyy =λθ+ 2µεyy, τzx= 2µεzx, (3.4.7) τzz =λθ+ 2µεzz, τxy= 2µεxy,
whereθ=εxx+εyy+εzz is called the dilatation, and λandµareLame’s constants.
Expressing stresses in terms of displacements, we obtain τxx=λθ+ 2µ∂u
∂x, τxy=µ
∂v
∂x +∂u
∂y
, (3.4.8)
τxz =µ ∂w
∂x +∂u
∂z
. By differentiating equations (3.4.8), we obtain
∂τxx
∂x =λ∂θ
∂x + 2µ∂2u
∂x2,
∂τxy
∂y =µ ∂2v
∂x∂y +µ∂2u
∂y2, (3.4.9)
∂τxz
∂z =µ∂2w
∂x∂z +µ∂2u
∂z2. Substituting equation (3.4.9) into equation (3.4.3) yields
λ∂θ
∂x+µ ∂2u
∂x2 + ∂2v
∂x∂y + ∂2w
∂x∂z
+µ ∂2u
∂x2 +∂2u
∂y2 +∂2u
∂z2
=ρ∂2u
∂t2. (3.4.10) We note that
∂2u
∂x2 + ∂2v
∂x∂y + ∂2w
∂x∂z = ∂
∂x ∂u
∂x+∂v
∂y +∂w
∂z
= ∂θ
∂x, and introduce the notation
△=∇2= ∂2
∂x2 + ∂2
∂y2+ ∂2
∂z2.
The symbol△or∇2is called theLaplace operator. Hence, equation (3.4.10) becomes
(λ+µ)∂θ
∂x+µ∇2u=ρ∂2u
∂t2. (3.4.11)
In a similar manner, we obtain the other two equations which are (λ+µ)∂θ
∂y+µ∇2v =ρ∂2v
∂t2. (3.4.12)
(λ+µ)∂θ
∂z+µ∇2w=ρ∂2w
∂t2 . (3.4.13)
The set of equations (3.4.11)–(3.4.13) is called theNavier equationsof mo- tion. In vector form, the Navier equations of motion assume the form
(λ+µ) grad divu+µ∇2u=ρutt, (3.4.14) whereu=ui+vj+wkand θ= divu.
(i) If divu= 0, the general equation becomes µ∇2u=ρutt, or
utt=c2T∇2u, (3.4.15) wherecT is called thetransverse wave velocity given by
cT = µ/ρ.
This is the case of an equivoluminal wave propagation, since the volume expansion θ is zero for waves moving with this velocity. Sometimes these waves are called waves of distortion because the velocity of propagation depends onµandρ; the shear modulusµcharacterizes the distortion and rotation of the volume element.
3.4 Waves in an Elastic Medium 73 (ii) When curlu= 0, the vector identity
curl curlu= grad divu− ∇2u, gives
grad divu=∇2u, Then the general equation becomes
(λ+ 2µ)∇2u=ρutt, or
utt=c2L∇2u, (3.4.16) wherecL is called thelongitudinal wave velocitygiven by
cL =
$ λ+ 2µ
ρ .
This is the case of irrotational or dilatational wave propagation, since curlu = 0 describes irrotational motion. Equations (3.4.15) and (3.4.16) are called thethree-dimensional wave equations.
In general, the wave equation may be written as
utt =c2∇2u, (3.4.17)
where the Laplace operator may be one, two, or three dimensional. The importance of the wave equation stems from the facts that this type of equation arises in many physical problems; for example, sound waves in space, electrical vibration in a conductor, torsional oscillation of a rod, shallow water waves, linearized supersonic flow in a gas, waves in an elec- tric transmission line, waves in magnetohydrodynamics, and longitudinal vibrations of a bar.
To give a more general method of decomposing elastic waves into trans- verse and longitudinal wave forms, we write the Navier equations of motion in the form
c2T∇2u+
c2L−c2T grad (divu) =utt. (3.4.18) We now decompose this equation into two vector equations by defining u=uT +uL, whereuT anduL satisfy the equations
divuT = 0 and curluL=0. (3.4.19ab) SinceuT is defined by (3.4.19a) that is divergenceless, it follows from vector analysis that there exists a rotation vectorψψψ such that
uT = curlψψψ, (3.4.20) whereψψψis called thevector potential.
On the other hand, uL is irrotational as given by (3.4.19b), so there exists a scalar functionφ(x, t), called thescalar potentialsuch that
uL = gradφ. (3.4.21)
Using (3.4.20) and (3.4.21), we can write
u= curlψψψ+ gradφ. (3.4.22) This means that the displacement vector field is decomposed into a diver- genceless vector and irrotational vector.
Insertingu=uT+uL into (3.4.18), taking the divergence of each term of the resulting equation, and then using (3.4.19a) gives
div c2L∇2uL−(uL)tt!
= 0. (3.4.23)
It is noted that the curl of the square bracket in (3.4.23) is also zero.
Clearly, any vector whose divergence and curl both vanish is identically a zero vector. Consequently,
c2L∇2uL= (uL)tt. (3.4.24) This shows thatuLsatisfies the vector wave equation with the wave velocity cL. SinceuL= gradφ, it is clear that the scalar potentialφalso satisfies the wave equation with the same wave speed. All solutions of (3.4.24) represent longitudinal waves that are irrotational (sinceψψψ=0).
Similarly, we substitute u=uL+uT into (3.4.18), take the curl of the resulting equation, and use the fact that curluL =0to obtain
curl c2T∇2uT −(uT)tt!
=0. (3.4.25)
Since the divergence of the expression inside the square bracket is also zero, it follows that
c2T∇2uT = (uT)tt. (3.4.26) This is a vector wave equation foruT whose solutions represent transverse waves that are irrotational but are accompanied by no change in volume (equivoluminal,transverse, rotational waves). These waves propagate with a wave velocitycT.
We close this section by seeking time-harmonic solutions of (3.4.18) in the form
u= Re U(x, y, z)eiωt!
. (3.4.27)