2.4 Cauchy's equations of motion
We have previously presented the integral formulations of Euler's axioms in (2.44) and (2.55), respectively. In the current section we will present these laws as more suitable field equations or differential equations, valid in each point/particle in a continuum.
2.4.1 Derivation of Cauchy's equations of motion
On component form, the balance of linear momentum in (2.44) may be presented: $$ \begin{equation} \int_V \dot{v}_i \rho \, dV = \int_A t_i \, dA + \int_V b_i \, \rho \, dV \tag{2.87} \end{equation} $$ and we observe that in (2.87) there are two volume integrals and one surface integral. A common way to transform integral equations like (2.87) to a differential equation, is to put all terms in a common integral in some way, and then claim that the integrand must vanish. For (2.87), a common integral may be obtained by transforming the surface integral to a volume integral. From Cauchy's stress theorem (2.65) and the divergence theorem in (8.31) we have: $$ \begin{equation} \int_A t_i \; dA = \int_A T_{ik} n_k \; dA = \int_V T_{ik,k} \; dV \tag{2.88} \end{equation} $$ Note that for the latter equality in (2.88) to hold, we let \( a_k = T_{ik} \) in the divergence theorem, which may be done as \( T_{ik} \) are scalars. Cauchy's equations of motion are finally obtained by substitution of (2.88) into (2.87) and claiming that the integrand must vanish for an arbitrary \( V \):Cauchy's equations of motion $$ \begin{equation} \rho\, \dot{v}_i = T_{ik,k} + \rho b_i \tag{2.89} \end{equation} $$
(2.89) is the fundamental equation motion in continuum mechanics, valid for the motion of any continuum, both fluids and solids. In solid mechanics the equations are normally formulated in Lagrangian coordinates in which the material derivative (see (2.24) and (2.25)) on the left hand side will reduce to a normal time derivative. In fluid mechanics (2.89) is normally rendered in Eulerian coordinates and consequently convective terms will appear on the right hand side. If a system is in equilibrium, the left hand side will vanish are there is no acceleration and consequently no change in linear momentum with respect to time.
By means of the permutation symbol \( e_{ijk} \) and the representation of vector products in (8.12), the law of balance of angular momentum (2.55) has the following component form: $$ \begin{equation} e_{ijk} \left ( \int_A x_i \, t_j \; dA + \int_V x_i \, (b_j - \dot{v}_j) \, \rho \; dV \right ) = 0 \tag{2.90} \end{equation} $$ Proceeding in the same manner as for the balance of linear momentum, the surface integral may be transformed to a volume integral by employing the divergence theorem in (8.31), the chain rule, and the Cauchy's stress theorem (2.65): $$ \begin{equation} \int_A x_i \, t_j \; dA = \int_A x_i \, T_{jl} n_l \; dA = \int_V x_{i,l} \, T_{jl} + x_{i} \, T_{jl,l} \; dV = \int_V T_{ji} + x_{i} \, T_{jl,l} \; dV \tag{2.91} \end{equation} $$ where the latter identity follow by realizing that: \( x_{i,l} = \delta_{il} \) and $ x_{i,l} \, T_{jl}= \delta_{il} \, T_{jl} = T_{ji}$. Substitution of (2.91) into (2.90) yields: $$ \begin{equation} e_{ijk} \; \int_V x_i \left ( T_{jl,l} + \rho b_j - \rho \dot{v}_j \right ) dV + \int_V e_{ijk} \, T_{jk} \;dV = 0 \tag{2.92} \end{equation} $$ Now, the first integral vanish due to balance of linear momentum as expressed in Cauchy's equations of motion (2.89), and consequently the integrand of the latter integral must vanish too: $$ \begin{equation} e_{ijk} \, T_{ji} = 0 \Leftrightarrow T_{ji} - T_{ij} = 0 \Rightarrow T_{ji} = T_{ij} \tag{2.93} \end{equation} $$
The stress tensor is symmetric \\ The balance of angular momentum implies that the stress matrix is symmetric, \( \mathbf{T=T}^T \).
2.4.2 Example 4: The hydrostatic pressure distribution
Figure 10: Fluid at rest in a container.
Consider a cylinder with a homogenous liquid at rest of constant density \( \rho \) (Figure 10). The only force acting is the constant gravitational force \( g \) in the negative z-direction.
The state of stress in the fluid is given by (2.86), which only contains the pressure \( p \). The pressure may in general vary in space, i.e., as a function of the spatial coordinates \( x \), \( y \), and \( z \). In this situation the left hand side of the Cauchy equations (2.89) vanishes and the balance of linear momentum reduce to: $$ \begin{equation} 0 = (-p \delta_{ik})_{,k} + \rho \, b_i \tag{2.94} \end{equation} $$ which yields: $$ \begin{equation} -p_{,i} + \rho \, b_i = 0\qquad \Leftrightarrow \qquad -\nabla p + \rho \mathbf{b} = 0 \tag{2.95} \end{equation} $$ on component and vector form, respectively. In this example with a constant gravitational force \( g \) in the negative z-direction, we have: \( b_1=b_x=b_2=b_y=0 \) and \( b_3=b_z=-g \). Consequently, (2.95) reduce to: $$ \begin{equation} \partd{p}{x} = 0, \quad \partd{p}{y} = 0, \quad \partd{p}{z} = -\rho g \tag{2.96} \end{equation} $$ From the first two equations in (2.96) we realize that the pressure is a function of \( z \) only, i.e., \( p=p(z) \). Further, we orient the coordinate system so that \( z=0 \) at the cylinder bottom and \( z=h \) at the liquid surface, where \( p(z)=p_0 \), i.e., atmospheric pressure at the liquid surface. By integration of the last equation in (2.96) we get: $$ \begin{equation} p(z) = p_0 + \rho g (h-z) \tag{2.97} \end{equation} $$ (2.97) is a hydrostatic pressure distribution, which is valid for all fluids at rest of constant density \( \rho \), regardless of e.g., their viscous properties.
2.4.3 Example 5: Cauchy equations in cylindrical coordinates
The challenge to express the Cauchy equations (2.89) in cylindrical coordinates, is to present the divergence of the stress tensor \( \mathbf{T} \). In this example we will first find expressions for the unit base vectors in a cylindrical coordinate system and their derivatives before we apply the generic expression for the divergence of a second order tensor derived in the section 8.3.1 Gradient, divergence and rotation in general orthogonal coordinates.
Consider the position vector \( \mathbf{r} \) on a cylinder with radius \( r \), at an angle \( \theta \) with the \( x_1 \)-axis, and at height \( z \). In a Cartesian coordinate system the position \( \mathbf{r} \) has the representation: $$ \begin{equation} \mathbf{r} = r \cos\theta \mathbf{e}_1 + r \sin\theta \mathbf{e}_2 + z \mathbf{e}_3 \tag{2.98} \end{equation} $$ For convenience we introduce cylindrical coordinates \( (y_1,y_2,y_3)=(r,\theta,z) \) and from the equation (8.13) we find expressions or the base vectors and base unit vectors, corresponding to the cylindrical directions. For simplicity we drop the y-superscript for the base unit vectors (i.e., \( \mathbf{e}_1^y =\mathbf{e}_r \) and so on). In the \( r \)-direction we find: $$ \begin{equation} \mathbf{g}_r = \partd{\mathbf{r}}{r} = \cos\theta \mathbf{e}_1 + \sin\theta \mathbf{e}_2 = \mathbf{e}_r \quad \Leftarrow \quad h_r^2 = \mathbf{g}_r \cdot \mathbf{g}_r = 1 \tag{2.99} \end{equation} $$ and in the \( z \)-direction: $$ \begin{equation} \mathbf{g}_z = \partd{\mathbf{r}}{z} = \mathbf{e}_3 = \mathbf{e}_z , \quad \Rightarrow \quad h_z^2 = \mathbf{g}_z \cdot \mathbf{g}_z = 1 \tag{2.100} \end{equation} $$ Only in the \( \theta \)-direction we find a base vector with length greater than unity: $$ \begin{equation} \tag{2.101} \mathbf{g}_{\theta} = \partd{\mathbf{r}}{\theta} = -r \sin \theta \mathbf{e}_1 + r \cos \theta \mathbf{e}_2 \quad \Rightarrow \quad h_{\theta}^2 = r^2 \end{equation} $$ and thus the unit base vector in the \( \theta \)-direction become: $$ \begin{equation} \tag{2.102} \mathbf{e}_{\theta} = \frac{\mathbf{g}_{\theta}}{h_{\theta}} = -\sin \theta \mathbf{e}_1 + \cos \theta \mathbf{e}_2 \end{equation} $$ Now, from equation (2.98) and (2.99) we find two equivalent expressions for the position vector \( \mathbf{r} \): $$ \begin{equation} \mathbf{r} = r \cos\theta \mathbf{e}_1 + r \sin\theta \mathbf{e}_2 + z \mathbf{e}_3 = r \mathbf{e}_r + z \mathbf{e}_z \tag{2.103} \end{equation} $$ Further, we also find from equation (2.99) and (2.102): $$ \begin{equation} \partd{\mathbf{e}_r}{\theta} = \mathbf{e}_{\theta} \quad \mathrm{and} \quad \partd{\mathbf{e}_{\theta}}{\theta} = -\mathbf{e}_r \tag{2.104} \end{equation} $$
With the relations in equation (2.104) we may use the expression derived in equation (8.29) to present the divergence of the stress tensor \( \mathbf{T} \) in cylindrical coordinates. Note that we for convenience use subscripts \( (r,\theta,z) \) for the components rather than \( (1,2,3) \). $$ \begin{align} \mathrm{div} \, \mathbf{T} & = \sum_i \left (\frac{1}{h_i}\partd{T_{ri}}{y_i} \, \mathbf{e}_r + \frac{1}{h_i}\partd{T_{\theta i}}{y_i} \, \mathbf{e}_{\theta} + \frac{1}{h_i}\partd{T_{z i}}{y_i} \, \mathbf{e}_z \right ) \tag{2.105}\\ & + \frac{1}{h_{\theta}} T_{r \theta} \, \partd{\mathbf{e}_r}{\theta} + \frac{1}{h_{\theta}} T_{\theta \theta} \partd{\mathbf{e}_{\theta}}{\theta} \tag{2.106}\\ & + \frac{1}{h_{\theta}} T_{kl} \, \mathbf{e}_k \otimes \partd{\mathbf{e}_l}{\theta} \cdot \mathbf{e}_{\theta} \tag{2.107} \end{align} $$
The expression in equation (2.107) may be simplified by taking into account the relations in equation (2.104): $$ \begin{align} \mathrm{div} \, \mathbf{T} & = \sum_i \left (\frac{1}{h_i}\partd{T_{ri}}{y_i} \, \mathbf{e}_r + \frac{1}{h_i}\partd{T_{\theta i}}{y_i} \, \mathbf{e}_{\theta} + \frac{1}{h_i}\partd{T_{z i}}{y_i} \, \mathbf{e}_z \right ) \tag{2.108}\\ & + \frac{1}{h_{\theta}} T_{r \theta} \, \mathbf{e}_{\theta} - \frac{1}{h_{\theta}} T_{\theta \theta} \, \mathbf{e}_{r} \tag{2.109}\\ & + \frac{1}{h_{\theta}} T_{kr} \, \mathbf{e}_{k} \tag{2.110} \end{align} $$
By introducing the common engineering notation introduced in equation (2.60) and collecting terms in each of the cylindrical directions in equation (2.110), results in the following representation of the the Cauchy equations (2.89) in cylindrical coordinates: $$ \begin{align} \rho a_r &= \partd{\sigma_r}{r} + \frac{\sigma_r - \sigma_{\theta}}{r} + \frac{1}{r} \partd{\tau_{r\theta}}{\theta} + \partd{\tau_{rz}}{z} + \rho b_r \tag{2.111}\\ \rho a_{\theta}& = \partd{}{r} \left (r^2 \tau_{r\theta} \right ) + \frac{1}{r} \partd{\sigma_{\theta}}{\theta} + \partd{\tau_{\theta z}}{z} + \rho b_\theta \tag{2.112}\\ \rho a_z &= \frac{1}{r} \partd{}{r} \left ( r \tau_{zr} \right ) + \frac{1}{r} \partd{\tau_{z \theta}}{\theta} + \partd{\sigma_z}{z} + \rho b_z \tag{2.113} \end{align} $$
This formulation of the Cauchy equation will turn out to be useful e.g., when we later will study wave propagation phenomena in compliant vessels.