3.2.1 Mechanical Strain

We consider a thin layer of thickness dx in the yz-plane. The displacement of the left point of position x is u. But the displacement of the right point originally located at x + dx is determined by $u+\frac{\mathrm{\partial }u}{\mathrm{\partial }x}dx$. Thus the strain in the x-direction is given by:

 (3.9)

A similar relationship can be found for y and z-directions.

 (3.10)

 (3.11)

The deformation of a solid can be described by the superposition of the following particular deformations:

• or global volume change into the orthogonal dimensions of space, i.e. strain in ɛ_ii in dimension $i=x,y,z$. (Figure 3.3a)
• deformation with angle ${\gamma }_{ij}=2{\epsilon }_{ij}/2={\epsilon }_{ij}$. (Figure 3.3b)
• or rigid body motion with rotation angle β_i with i being the normal vector to the plane of rotation. (Figure 3.3c)

With those considerations we get the relationship between the shear angle $\gamma /2$ and the rotation angle β to the mechanical displacement. Assuming very small angles so that $\tan \gamma =\tan \epsilon \approx \gamma$ this leads to

 (3.12)

 (3.13)

The difference of equations (3.12) and (3.13) gives

 (3.14)

and the sum of both equations leads to an expression for the deformation angle ${\gamma }_{xy}={\epsilon }_{xy}$

 (3.15)

A similar expression can be found for the other shear angles ɛ_yz and ɛ_zx

 (3.16)

 (3.17)

3.2.1.1 Mechanical Strain - Voigt Notation

The mechanical strain ɛ_ij in three-dimensional space is the relative displacement of component i along dimension j. This is given by a symmetrical tensor of second order

 (3.18)

We separate out the normal strain in the diagonal ɛ_ii that is defined as the extension into dimension i related to unit length. The shear strain is given be the shearing angle ${\gamma }_{ij}={\epsilon }_{ij}$. Thus we can define for $i,j=x,y,z$, ${\eta }_i=u,v,w$ and ${\mathrm{\partial }}_i=\mathrm{\partial }x,\mathrm{\partial }y,\mathrm{\partial }z$

 (3.19)

Due to its symmetry the tensor in equation 3.18 has only six independent components, so the tensor can be alternatively represented in theVoigt notation

 (3.20)

3.2.1.2 Dilatation dilation – Relative Change in Volume

The relative change $\mathrm{\Delta }V$ of a volume $V=\mathrm{\Delta }x\mathrm{\Delta }y\mathrm{\Delta }z$ can be found from the linear strains:

 (3.21)

3.2.2 Mechanical Stress

According to the introduction of this section every force vector can be separated into two components (Figure 3.4):

1. normal stress: A normal force F_n in the direction of the normal vector of the surface.
2. shear stress: A tangential force perpendicular to the normal vector.

We consider the surfaces of a cube as shown in Figure 3.5 with the axes parallel to the coordinate axes. The symbol for stress is σ; the first index denotes the direction normal vector, and the second index the direction of the force. Thus, σ_xx, σ_yy and σ_zz are the normal stresses. σ_ij with $i\ne j$ and $i,j=x,y,z$ are the shear stresses. The definition is

 (3.22)

Because of the moment of equilibrium the stresses of exchanged indices must be equal

 (3.23)

Finally σ_ij is a symmetric tensor of second order.

 (3.24)

So we have six independent components of the stress tensor, and we can apply the Voigt notation

 (3.25)

3.2.3 Material Laws

The relationship between the strain and stress tensors is given by the material laws called Hooke’s law for linear elastic media. These laws in their most common form are given by:

 (3.26)

 (3.27)

The first tensor is called the stiffness tensor or elasticity tensor and the second is named compliance tensor. They have 81 components that can be reduced to 36 because of symmetry. Using the Voigt notation the stress tensor can be written as a matrix

 (3.28)

Those complex notations and equations are fortunately not that often necessary. For most engineering applications two special cases are applied: (1) Isotropic material, for example (steel, rubber, aluminium) and (2) orthotropic material (carbon fibre reinforced plastic (CFRP), steel enforced concrete, wood, plywood, etc.)

3.2.3.1 Isotropic Materials

Isotropy means, there is no dependency of the Hooke’s law on space dimension or orientation. For the normal stress in the x-direction (or any other) we have

 (3.29)

This behavior is valid if the solid may perform lateral contraction. That means the elongation of a rod leads to lateral contraction. If this contraction is not constrained by surrounding solid material the elongation is reduced by the Poisson ratio ν, the negative ratio of transverse motion to axial strain. The strain due to stress in one direction must be corrected by the strain from other normal directions and the Poisson number.

 (3.30)

 (3.31)

 (3.32)

We can also apply shear stresses. There is a similar relationship to Equation (3.29)

 (3.33)

Those three material constants are not independent, and they are linked by

 (3.34)

Summarising all properties we can write the compliance and stiffness matrices:

 (3.35)

 (3.36)

3.2.3.2 Ortotropic Solids

This class of material occurs for example in composites with fibres or layered microstructures. The properties change with the normal axes of the three-dimensional coordinate system. Thus, orthotropic materials have three orthogonal planes of symmetry. If they coincide with the coordinate system we get for the stiffness matrix a common notation.

 (3.37)

This matrix can be derived from the inversion of the compliance matrix. However, this leads to lengthy expressions that will not provide any additional insight, whereas the compliance matrix can be presented in a compact form

 (3.38)

where $E_i$ is the Young’s modulus along axis i, G_ij is the shear modulus in direction j on the plane whose normal is in direction i. ν_ij, is the Poisson’s ratio that corresponds to a contraction in direction j when an extension is applied in direction i.

3.3.1 The One-dimensional Wave Equation

The field quantity for solids is the displacement $\mathbf{d}={\left\{\begin{array}{c}u,v,w\end{array}\right\}}^T$. We start from the control volume as shown in Figure 3.5. If we assume a cut-free cube there is a stress balance at each cube surface: i.e. that the stresses are in balance at each interface and have opposite direction (Figure 3.6). From the stress balance over the cube surfaces the resultant force in the x-direction is:

 (3.39)

From Newton’s law $d{F}_x=dm\ddot{u}$ and with $dm={\rho }_{0}dxdydz$ we get after division with $dV=dxdydz$ and the series expansion for ${\sigma }_{xx}(x+dx)$

 (3.40)

the equation of motion in the x-direction

 (3.41)

Similar treatment leads to the two equations of motion in y and z-directions

 (3.43)

 (3.44)

Thus, for deriving the wave equation in displacement coordinates the stresses must be expressed in terms of the displacement making use of the stress–strain relationship of isotropic solids. We start with the sum of equations (3.30)–(3.32).

 (3.45)

With the definition of the dilatation $\mathrm{\nabla }\mathbf{d}$ (3.21) and (3.34) this gives:

 (3.46)

The goal is to find an expression for σ_xx depending only on displacements not linked to other stresses. We start with (3.30) solving for σ_xx and adding $\nu {\sigma }_{xx}$ on both sides.

 (3.47)

With (3.45) this leads to

 (3.48)

Eliminating E yields

 (3.49)

or

 (3.50)

We introduce the Lamé constants

 (3.51)

 (3.52)

λ is a measure for the extension perpendicular to the stress. Applying these constants we get from (3.50)

 (3.53)

The equations for the y and z-directions can be derived similarly:

 (3.54)

 (3.55)

From the definition of the isotropic shear modulus (3.33) and the shear strains (3.15) the equations for the other components are found:

 (3.56)

 (3.57)

 (3.58)

Using equations (3.53)–(3.58) we get after some algebraic modifications the wave equation for the one-dimensional case in the x-direction

 (3.59)

Similar and equivalent derivation can be done for the remaining two dimensions.

 (3.60)

 (3.61)

3.3.2 The Three-dimensional Wave Equation

Collecting the one-dimensional wave equations and putting them together for the displacement vector d leads to

 (3.62)

With the identity $\mathrm{\nabla }(\mathrm{\nabla }\mathbf{d})=\mathrm{\Delta }\mathbf{d}+\mathrm{\nabla }\times \mathrm{\nabla }\mathbf{d}$ we get the wave equation in a different form

 (3.63)

This wave equation can only be solved in closed form for simple geometries. In most cases numerical methods like the finite element method must be applied to solve this equation for realistic systems.

There are two differential operators in space on the LHS of Equation (3.63) indicating two wave types in solids. Any vector field can be separated into two independent components. The irrotational and the source-free or zero-divergence component

 (3.64)

An irrotational field is characterized by

 (3.65)

and a source free field by

 (3.66)

Thus

 (3.67)

 (3.68)

These relationships show that there are two independent wave types in solids, called longitudinal waves (L) and shear or transversal waves (S).

3.4.1 Longitudinal Waves

Longitudinal waves are rotation free, thus $\mathrm{\nabla }\times \mathbf{d}=\mathbf{0}$ and the wave equation (3.63) simplifies to

 (3.69)

c_L is called the longitudinal wave speed. In isotropic materials this is the wave with the highest speed. When we calculate c_L for typical materials like iron, aluminium, wood etc., it becomes obvious why the practical relevance in vibroacoustics is quite low. For example, for iron the wavelength at f = 1 kHz equals 5 m, extending by far beyond the typical dimension of bulk parts in a car. In order to illustrate the propagation of a longitudinal wave we calculate the solution for propagation into the x-direction. That means $v=w=0$ and $\frac{\mathrm{\partial }}{\mathrm{\partial }y}=\frac{\mathrm{\partial }}{\mathrm{\partial }z}=0$. The longitudinal wave equation reduces to

 (3.70)

The wave motion is characterized by particles oscillating in the direction of propagation. All shear stresses are zero, and a plane of constant dilatation propagates in the wave direction. Thus, those waves are also called dilatation or density waves. Equation (2.34) from chapter 2 can be applied, and the harmonic solution is

 (3.71)

In the upper part of Figure 3.7 the particle motion is shown for this wave type.

3.4.2 Shear waves

In contrast to the longitudinal wave the shear wave is source free, thus $\mathrm{\nabla }\mathbf{d}=0$. In that case the wave Equation (3.63) leads to

 (3.72)

The wave motion is iso-voluminous. It can be shown by the introduction of a rotation vector that the particle motion is orthogonal or transversal to the wave propagation (Graff, 1991). This is the reason why this wave type is also called transversal wave. A simple plane wave equation for w is

 (3.73)

A solution in the positive x-direction is for example

 (3.74)

In the lower part of Figure 3.7 the particle motion for one fixed time is shown. From (3.34) it follows that shear waves are slower than longitudinal waves.

3.5.1 Longitudinal Waves

Longitudinal waves are associated with normal stresses in the direction of the beam axis. We consider a free cut small layer of the beam and take the stress balance (3.32). There are no constraints for y and z surfaces and therefore ${\sigma }_{yy},{\sigma }_{zz},{\sigma }_{yz},{\sigma }_{xz}=0$. Thus, the stress strain relationship in the x-direction reads

 (3.75)

Using this we get with Newton’s law from Equation (3.41)

 (3.76)

or in harmonic form for $u(x,t)=\mathit{u}{e}^{j\omega t}$

 (3.77)

The solution of this one-dimensional wave equation is

 (3.78)

Due to the fact that the bar is stress free in the direction normal to the bar axis, the wave speed in bars c_LB is slower than the longitudinal wave speed in infinite solids. This lead to displacement in y and z that can be calculated from the first column of (3.36):

 (3.79)

 (3.80)

The only information related to the cross section that is required for the determination of the longitudinal wave speed is that the dimensions of the cross sections must be small compared to the wave length.

3.5.2 Power, Energy, and Impedance

When exciting a half beam as shown in Figure 3.9 we create a wave propagating in the positive direction leaving out the e^−jkx from (3.78)

 (3.81)

The stress in x is with (3.75):

 (3.82)

So, there is a longitudinal characteristic impedance $z_{LB}={\rho }_0{c}_{LB}$ of the longitudinal wave in beams. Assuming a force F_x exciting the beam in the x-direction via a stiff plate at the beam cross section leads to the boundary condition at x = 0 with ${\mathit{F}}_z=-A{\mathit{\sigma }}_1(x)$

 (3.83)

The mechanical impedance follows from this

 (3.84)

The power introduced into the system is, when we assume without loss of generality ${\mathit{F}}_x={\hat{F}}_x$

 (3.85)

This power propagates into the beam and creates an energy density per length $E^{\mathrm{\prime }}=\mathrm{\Pi }/c_{LB}$, because the power per time unit is distributed with velocity c_LB over the related length unit. This can also be proven in a more complicated way by deriving the kinetic and potential energy density per length

 (3.86)

For the potential energy we have to integrate the stress over the strain

 (3.87)

with ${\sigma }_{xx}(x)=Re(-\frac{{\hat{F}}_x}{A})$ we get for the potential energy density per length

 (3.88)

So we find $E_{kin}^{\mathrm{\prime }}=E_{pot}^{\mathrm{\prime }}$. With (3.83) we get for the total energy expressed for the velocity

 (3.89)

Integrating over one wavelength λ gives the total length specific energy

 (3.90)

proving the above arguments regarding power and speed of sound.

3.5.3 Bending Waves

The bending stiffness depends on area and shape of the cross section. In Figure 3.11 the cross section, forces, and moments of a cut-free slice of the beam are shown. We assume that there is a neutral axis without any dilation and pure bending. That means that each cross section of the un-loaded straight beam remains orthogonal to the neutral axis as shown in Figure 3.12. This is called the Bernoulli assumption.

Geometrically this assumption means that all fibres stay orthogonal to the cross section. The stress is linear with the bending radius R as depicted in Figure 3.12. The stress in the direction of the neutral axis vanishes at the neutral axis¹. The strain out of the neutral axis can be derived from Figure 3.12.

and therefore

Due to Equation (3.29) it can be shown, that the stress is a linear function of the distance from the neutral axis z and the curvature $1/R$.

 (3.91)

The moment vector $\mathbf{M}=(0,M_y,M_y)$ is given by the integral over all small moment contributions at r with $d\mathbf{M}=\mathbf{r}\times {\sigma }_{xx}dA$. Thus:

 (3.92)

and due to the assumptions the stress is symmetrical to the z-plane

 (3.93)

We call

 (3.94)

the second moment of area, in most literature called moment of inertia². The moment

 (3.95)

that vanishes in our special case is called the product moment of area. There are axes of the cross section where this product moment is zero. They are called the main axes and are for example the symmetry axes of a symmetrical section. Using (3.94) we can write (3.92) as

 (3.96)

Using the following differential equation for the curvature $1/R$

 (3.97)

We assume small displacements $w\ll 1$, and we get approximately

 (3.98)

and with that

 (3.99)

For the z-axis we get accordingly

 (3.100)

The force in the z-direction can be derived from the moment equilibrium at the position x + dx in Figure 3.11.

 (3.101)

Now, we set up the force balance at the control segment from Figure 3.11.

 (3.102)

Performing the limiting process for dx → 0 gives

 (3.103)

with m^ʹ being the mass per length or length specific density. The prime ${(}^{\mathrm{\prime }})$ specifies units per length in this section. Using the harmonic wave solution $w(x,t)=\mathit{w}{e}^{j(\omega t-k_{B}x)}$ we get the bending wave equation in the frequency domain

 (3.104a)

 (3.104b)

with the beam bending stiffness $B_y=E{I}_{yy}$ and $B_z=E{I}_{zz}$ for the principle y and z-axes of the cross section. Neglecting the time dependence and performing the spatial derivatives we get for the bending wavenumber

 (3.105)

with two real and two complex solutions

 (3.106)

Thus the solution includes two propagating waves and two decaying

 (3.107)

We neglect the complex wave numbers leading to exponential decaying waves and calculate the wave speed from the fourth root of (3.105)

 (3.108)

Here, we have the effect of dispersion, and the wave speed depends on the frequency. The main effect is that the propagation of energy is different from the propagation of constant phase in the wave. This was not the case for fluids where the speed of sound is constant over frequency. The consequences of this will be discussed in section 3.8.

If we calculate the bending wave speed for a typical system we see that we have a wave speed of similar order of magnitude as in air. Let us consider a rectangular cross section as shown in Figure 3.13. The moments are

 (3.109)

Assuming aluminium with isotropic material properties $E=71\times {10}^9$ Pa, ${\rho }_0=2700$ kg/m³ and ν = 0.34 the rectangular cross section with $b=3cm$ and $h=2cm$ yields the following wave speed for bending in the z and y-directions (Figure 3.14).

3.5.4 Power, Energy, and Impedance

The frequency dependence of the wave speed leads to a different propagation speed, point of constant phase, and energy stored in the wave. For the power considerations we summarize the following relationships from the above calculations

 (3.110)

One power component in beams is the force–velocity product as in (1.47). In addition the product of moment M_y and the rotational speed ${\dot{\beta }}_y$ must be taken into consideration

 (3.111)

with

 (3.112)

 (3.113)

We keep the oscillating part of the power in order to show a characteristic effect for wave propagation in beams. We will use the positively propagating part of the harmonic solution (3.107) and neglect the evanescent wave that is not relevant for power transport. For further simplifications of the expressions we assume a real displacement amplitude ${\mathit{w}}_2={\hat{w}}_2$ and we get for the force and the velocity

 (3.114)

 (3.115)

 (3.116)

Force and velocity are in phase and the same is true for moment and rotational velocity, but with a phase shift of $\pi /2$

 (3.117)

 (3.118)

 (3.119)

Thus, both contributions sum up to the constant value

 (3.120)

The same must be true for the energy. If we multiply the kinetic energy density e_kin by the beam area we get for the translational energy per length $E^{\prime }=Ae$

 (3.121)

The rotational component can be neglected, as we can see from

 (3.122)

The averaged ratio between both kinetic energies is

 (3.123)

The so called radius of gyration $\sqrt{I_{yy}/A}$ is the perpendicular position of a point mass which gives the same rotational inertia. As we stated in the beginning, one main assumption was that the cross sectional dimensions must be small compared to the wavelength λ_B. Consequently we can neglect the rotational part of the energy, too.

Due to the Bernoulli assumption only in-plane stress and strain must be considered. Here the stress and strain in the x-direction. Therefore, the potential energy per length is given by

 (3.124)

Both kinds of energy oscillate between 0 and the maximum value similar to the power. Using the dispersion relation (3.105) the sum of both gives

 (3.125)

So, we have a constant energy distribution in the wave. There is no energy oscillation as was the case for fluid waves. If we relate the power transport (3.120) to the above energy $E_{\text{tot}}^{\prime }$ we get

 (3.126)

The power transportation is given by twice the phase speed c_B, which is fine in this case, because the uniform distribution of power and energy allows this effect. This is not an error and comes from the dispersion of the bending waves as we will see in section 3.8. The power transport in dispersive media is not given by the phase velocity but by the group velocity $c_{B,\text{gr}}=2c_B$ for bending waves in beams.

A force exciting the beam radiates waves into the beam. In order to describe the source performance we come back to our impedance concept. As the force shall excite waves in the direction the bending occurs we apply a harmonic force $F_z(\omega )={\mathit{F}}_z{e}^{j\omega t}$ in the z-direction.

We start with a half infinite beam that is excited at x = 0, Figure 3.16. From the boundary condition it follows that $M_y=0$ as it is an open end. From the origin only waves that propagate in the positive x-direction occur. Thus, we use the following part of Equation (3.107)

 (3.127)

So we get for for M_y and F_z

 (3.128)

 (3.129)

At x = 0 the first condition leads to $w_2=w_{2D}$ and with this we get for the second equation balancing the force $F_z(x=0)$:

 (3.130)

and hence

 (3.131)

For the mechanical impedance at x = 0 we get now:

 (3.132)

For the infinite beam we assume a mirrored wave at x = 0 propagating away from the point of excitation, but now the moment does not need to be zero. For symmetry reasons the the rotation β_y must vanish at the exciting force position. Thus, with Equation (3.16) used in the y-direction and vanishing displacement the condition is

 (3.133)

The force is now driving two half beams, thus

 (3.134)

and

 (3.135)

This gives for the impedance of the full beam:

 (3.136)

The full beam impedance is four times higher than the impedance of the half infinite beam. With growing frequency the impedance decreases, but we must keep in mind that for higher frequencies the Bernoulli beam assumptions are no longer valid. The impedance is complex with a phase of $\pi /4$. The driving force experiences a reactive and a resistive part. In Figure 3.17 three time or phase slots of an infinite and half infinite beam excited by a force are shown in combination with the solution without the evanescent part. We see that this part contributes only to the near field at the point of excitation and therefore contributes only to the impedance.

3.7.1 Strain–displacement Relations

Using the fourth assumption the tangential displacements are of the form

 (3.140)

 (3.141)

u₀ and v₀ are the tangential displacements of the middle plane. Entering these equations into the related strain–displacement relationships (3.16) and (3.17) and using assumption three yields

 (3.142)

 (3.143)

Therefore.

 (3.144)

This result is illustrated in Figure 3.19 for a pure bending configuration.

The strain displacement relation (3.19) leads to

 (3.145)

 (3.146)

 (3.147)

We consider an infinitesimally and cut-free small plate element as shown in figure 3.20. For the plate dynamics it is convenient to integrate over the plate thickness to get quantities defined per length instead of area. The integration over the normal stresses reads:

 (3.148)

3.7.2 In-plane Wave Equation

From Figure 3.19 it becomes obvious that the bending is decoupled from the in-plane motion of shear and stress in case of homogeneous thin plates. Bending leads to symmetric stress and strain, and the integrals in Equation (3.148) are thus decoupled from the in-plane motion.

From assumption five it follows that the strains ɛ_x and ɛ_y can be given according to equations (3.30) and (3.31)

 (3.149)

 (3.150)

With the above equations and component xy in Voigt notation from (3.33) corresponding to the xy index in (3.19) we get

 (3.151)

 (3.152)

 (3.153)

Entering equations (3.151)–(3.153) into (3.148) provides

 (3.154)

because the last term in Equation (3.145) vanishes due to symmetry. We get for $N_y^{\prime }$ and $N_{xy}^{\prime }$ accordingly

 (3.155)

 (3.156)

In Figure 3.21 the force balance of a plate element is shown. The balance in the x-direction is

 (3.157)

Division by area dxdy gives

 (3.158)

and accordingly for the y-direction

 (3.159)

With (3.154)–(3.156) we get finally, setting $u_0=u$ and $v_0=v$:

 (3.160)

Equalising the force per area to the external pressure plus inertia forces and using $m^{″}={\rho }_{0}h$

 (3.161)

leads to the wave equation for in-plane waves. Together with the following abbreviations

 (3.162a)

 (3.162b)

we get

 (3.163)

and in a similar way for the y-direction:

 (3.164)

The motion in the in-plane direction is coupled: a displacement in u leads to motion in v and vice versa. The solution of both equations leads to two waves analogical to the longitudinal and shear waves of the solid material in section 3.4.

3.7.3 Longitudinal Waves

In longitudinal waves the wave motion is in the direction of propagation. Without loss of generality we choose the x-direction as propagation direction. Thus, there is no change of u over y and v is zero.

 (3.165)

Using these assumptions in (3.163) we get

 (3.166)

and the wavenumber yields

 (3.167)

We see that this third longitudinal wave speed for plates is valued between the solution for the longitudinal waves in solids and beams.

The more constrained the bulk material is the stiffer it becomes due to the lateral contraction. The strain in thickness can be calculated from (3.32).

 (3.168)

Integrating the strain over the thickness delivers the shape of the wave motion.

3.7.4 Shear Waves

Shear waves have displacement perpendicular to their direction of propagation. For shear wave propagation in the x-direction we assume:

 (3.169)

Using (3.164) with these conditions gives the shear wave equation in the x-direction for plates:

 (3.170)

 (3.171)

The shear wave speed in plates or solid material is similar, because this wave type is free of dilatation due to condition 3.169 and thus not dependent on geometrical constraints.

 (3.172)

3.7.5 Combination of Longitudinal and Shear Waves

The combined dynamics of in-plane wave motion is found by entering the ansatz

 (3.173)

with $i\in (S,L)$ being the index for both wave types. Entering this into (3.163) and (3.164) leads to the following system of equations

 (3.174)

We used ${\rho }_{0}h{\omega }^2=C{k}_L^2=S{k}_S^2$ following from (3.167) and (3.171). Equation (3.174) can be seen as a generalized eigenvalue problem with eigenvalue $\lambda =C{k}_L^2=S{k}_S^2$. Dividing the above equation by S leads to a form that is more practical for the next steps

 (3.175)

With the relationship

 (3.176)

This can be further modified to

 (3.177)

For non-trivial solutions of this matrix the determinant must vanish, and this leads to the equation

This is of the form $(a+b)(a+c)-bc=a(a+b+c)=0$. So the two solutions are

 (3.178)

 (3.179)

using relationship (3.176) again. Finally, we have derived the above wavenumbers for shear and longitudinal waves in a more formal way. Entering $k_S^2$ and $k_L^2$ for $k_x^2+k_y^2$ into (3.177) provides the solution for the shape of the wave propagation. With $k_{xS}^2+k_{yS}^2=k_S^2$ we get:

 (3.180)

So $k_{xS}{\mathit{u}}_S=-k_{yS}{\mathit{v}}_S$, and when we choose ${\mathit{u}}_S={\mathbf{\Psi }}_S{k}_{xS}$ we get ${\mathit{v}}_S=-{\mathbf{\Psi }}_S{k}_{yS}$; finally the displacement due to the shear wave motion is given by

 (3.181)

With $k_{xL}^2+k_{yL}^2=k_L^2$ the same equation reads as:

 (3.182)

providing the longitudinal wave motion

 (3.183)

The in-plane wave propagation is given by the superposition of both waves with:

 (3.184)

The displacement of longitudinal waves is parallel to the propagation direction given by ${\mathbf{k}}_L={\{k_{xL},k_{yL}\}}^T$, and the shear wave displacement is perpendicular to the propagation direction orientation k_S as shown in Figure 3.23

Similar to solids the wavelengths of longitudinal and transversal waves are very high. Thus, in-plane waves can be neglected in many technical acoustics problems at audible frequency, but they are required for a full understanding of the physics when plates are coupled.

3.7.6 Bending Wave Equation

For the derivation of the bending wave dynamics we define the moments per length M^ʹ parallel to the edges and as a torsional moment.

 (3.185)

The thickness integration over the shear stresses gives the shear forces per length:

 (3.186)

With those definitions and according to Figure 3.24 we can now set up the relationships for the required equilibrium of the total forces in the z-direction

 (3.187)

 (3.188)

and for the moments around the x-axis,

 (3.189)

and the y-axis accordingly:

 (3.190)

Neglecting terms of second order and with the moment equilibrium $M_{xy}^{\prime }=M_{yx}^{\prime }$ we get

 (3.191)

Entering $Q_x^{\prime }$ and $Q_y^{\prime }$ from (3.188) in (3.191) gives

 (3.192)

As mentioned above the bending motion is decoupled from the in-plane motion. So, we assume $u_0=v_0=0$ and get with (3.145) for the displacement in the thin layer an approximately linear dependency from z

 (3.193)

Entering (3.193) into (3.151)-(3.153) leads to the in-plane stresses as function of w and z

 (3.194)

 (3.195)

 (3.196)

This is applied in the definitions (3.185) and reads for $w(x,y)$

 (3.197)

 (3.198)

 (3.199)

with the bending stiffness B of the plate

 (3.200)

With equations (3.191) and (3.197)–(3.199) the shear forces can also be expressed as functions of w .

 (3.201)

 (3.202)

From (3.192) and with equations (3.197)–(3.199) we can derive the plate equation

 (3.203)

In order to get the wave equation for plates we set the force per area equal to external pressure plus the inertia force

 (3.204)

Hence.

 (3.205)

Now we are prepared to determine the wave speed for bending waves in plates using the solution in frequency domain $w(x,y,t)=\mathit{w}(x,y)e^{j\omega t}$

 (3.206)

Considering waves $\mathit{w}={\mathit{w}}_B{e}^{-jkx}$ in the x-direction without loss of generality the results for k_B are similar to the results of the beam bending waves

 (3.207)

and the wave speed

 (3.208)

Equation (3.206) can be written in Helmholtz form

 (3.209)

We get again two real and two complex solutions:

 (3.210)

Thus the solution includes four propagating waves. Using only the positive absolute values per definition with

we can write for the four waves in the x-direction

 (3.211)

The dispersion effects discussed in the beam section are also valid.

3.7.6.1 Cylindrical Solution of Bending Wave Equation

In order to derive the mechanical impedance for a plate we must agree on a certain set of boundary conditions and assumptions that allow for a solution:

• The solution is rotationally symmetric.
• The point of excitation is rotation free.
• The solution fulfils the Sommerfeld condition: i.e. that the wave decays in large distances.

Starting with (3.206), using the wavenumber result from (3.207) and rewriting it to

 (3.212)

gives the following system of equations:

 (3.213a)

 (3.213b)

The first Equation (3.213a) is similar to the Helmholtz equation for fluids but for two dimensions. The solution is the Hankel function of second kind:

 (3.214)

With r being the distance from the point of excitation. For the far field properties and the behavior at the excitation point we use the asymptotic behavior for small and large values of x

The solution of the second Equation (3.213b) can be found by replacing k_Br with $-j{k}_{B}r$, hence

 (3.215)

For large kr we get

Thus, Equation (3.215) represents the exponentially decaying near field, as in (3.106). The total solution is

 (3.216)

With the assumption that the rotation

 (3.217)

must vanish at r = 0, so that

we finally get

 (3.218)

3.7.6.2 Power, Impedance, and Energy

The value at r = 0 follows from the asymptotic expression for small values:

 (3.219)

So C₁ equals the displacement amplitude at the excitation point.

 (3.220)

Some shapes at specific phases of the force are shown in Figure 3.26.

In order to derive the displacement amplitude ${\hat{\mathit{w}}}_0$ we consider a small circle around the the point of excitation and use equations (3.201)–(3.202) in cylindrical coordinates

 (3.221)

Entering this into (3.218) leads to:

 (3.222)

Using the approximation of the Hankel function for small arguments gives

 (3.223)

Multiplication of (3.223) with the perimeter of a small circle finally provides the force

 (3.224)

Entering this into (3.220) we get the shape of the plate due to force excitation

 (3.225)

With the velocity at the centre ${\mathit{v}}_z(0)=j\omega {\mathit{w}}_0$ we get a surprisingly simple expression for the mechanical impedance at r = 0:

 (3.226)

Regarding the energy in the plate, we rely on the considerations of bending waves in beams in section 3.5. The property characterising bending for beams was the Young’s modulus times the second moment of area, e.g. $B_y=E{I}_{yy}$. For plates this must be replaced by the bending stiffness that is defined per length as given in equation 3.200. Thus, after adjusting Equation (3.125) to this length reference we get area energy density $E^{″}=E/A$

 (3.227)

The equality of kinetic and potential energy, as far as the fact that the energy is not pulsating but constant over time, is also similar to beams. The expression for the power flow has to be adjusted to power per length or length intensity with

 (3.228)

Note that the energy is transported with twice the phase wave speed, which is not in contradiction to the energy conservation because of the fact that the energy is equally distributed over the phase cycle. For the point force excitation the introduced energy flows through the edge of a circle of radius r with the circumference $2\pi r$.

 (3.229)

The input power is known through the point impedance (3.226)

 (3.230)

Combining equations (3.228) to (3.229) we get an expression for the displacement amplitude derived from energy considerations:

 (3.231)

The same expression can be derived from Equation (3.225) when using the asymptotic expression for the Hankel function for $kr\gg 1$.

Damping

The damping of wave propagation is related to energy or the transport of energy. This complicates the treatment for dispersive waves. To overcome this issue we use the damping loss as the complex ratio of stiffness related quantities from (1.62). For plates this is the complex Young’s modulus E:

 (3.232)

impacting linearly the bending stiffness:

 (3.233)

When we introduce this into Equation (3.207)

and

 (3.234)

In contrast to Equation (2.59) this is one-fourth of the wavenumber. The reason for this is that due to dispersion of bending waves the energy propagates twice as fast as the phase.