Kinetics of a Continuous Media

In this section we will study the laws of motion applicable to a continuous medium similar to Newton's laws of motion studied in particle mechanics. We first review Newton's laws of motion below.

1. Newton's first law of motion: a free particle continues in its state of rest or of uniform motion.

2. Newton's second law of motion: In an inertial frame, the rate of change of linear momentum of a particle equals the resultant force acting on the particle. That is

   $$\Sigma \mathbf{F} = \frac{d}{dt} (m\mathbf{v}) = m\mathbf{a}.$$ (4.1.1)

   
   

3. Newton's third law of motion: To every action there is an equal and opposite reaction.

Newton's first law of motion defines an inertial frame. That is, an inertial frame is one in which Newton's first law of motion holds. Usually, it is taken as a frame attached to the Sun. However, in most engineering problems, one can take the co-ordinate axes fixed to the earth as an inertial frame without introducing any appreciable error. Hereafter, we will take an inertial frame as the frame of reference.

To write the laws of motion for a continuum we note that the linear momentum of the material enclosed in an infinitesimal volume $dv$ is $(\rho\ dv)\mathbf{v}$ where $\rho$ is the mass density and $\mathbf{v}$ is the velocity. Hence the linear momentum of the shaded portion is $\int (\rho dv)\mathbf{v}$ in which the integration is over the shaded region. To find the resultant force acting on this region of interest, we observe that we have two kinds of forces.

Body forces are forces that act on all particles in a body as a result of some external body or effect not in direct contact with the body under consideration. An example of this is the gravitational force exerted on a body. This type of force is defined as a force intensity per unit mass or per unit volume at a point in the continuum. Thus, if the body force per unit mass is $g_i$, then the body force on the material enclosed in the shaded region will be $dm\ g_i$.

Surface forces are contact forces that act across a surface of the body, which may be internal or external. In non-polar continuum mechanics we assume that the action of that part of the body which is exterior to the shaded region on the body enclosed in the shaded region is equipollent to a system of forces acting on the bounding surface of the shaded region. The assumption that the contact force is of this kind is the cut principle of Cauchy: Within the shape of a body at any given time, conceive a smooth, closed diaphragm; then the action of the part of the body outside that diaphragm and adjacent to it on that inside is equipollent to that of a field of vectors defined on the diaphragm. Note that no point moments are assumed to be exerted by one part of the body on its adjacent part across the common surface. Thus in nonpolar continuum mechanics, moments are caused by the forces. The contact force at a point $P$ on a surface is usually given as a force $\mathbf{f}$ acting on a unit area surrounding $P$ and lying on the surface. Through a given point in the body, there are infinitely many surfaces. The intensity of the contact force at the point $P$ on each of these surfaces will, in general, be different. How does $\mathbf{f}$ at the point $P$ depend upon the surface through $P$? In the classical continuum mechanics, it is assumed that the intensity of the contact force on all surfaces with a common tangent plane at $P$ is the same. That is, $\mathbf{f}$ at $P$ is assumed to depend upon the surface through $P$ only through the oriented normal $\mathbf{n}$ of the surface at $P$.

$$\mathbf{f} = \mathbf{f} (x_i,\ n_j).$$ (4.1.2)

This is Cauchy's Postulate. A unit normal to the surface which points out of the body is taken as positive. Thus $\mathbf{f} (\mathbf{x},\mathbf{n})$ is the intensity of the contact force at $P$ which the unshaded portion of the body exerts on the shaded portion and $\mathbf{f} (\mathbf{x},\ -\mathbf{n})$ is the intesntiy of the contact force at $P$ which the shaded portion exerts on the unshaded one. The intensity of the contact force is also known as surface traction or traction or stress vector. Denoting the magnitude of the element of area on the surface by $da$, the total contact force on the shaded region is $\int f_1da$. Thus eqn. (4.1.1) for the material contained in the shaded region takes the form

$$\frac{d}{dt}\int dm\ v_i = \int f_i\ da + \int dm\ g_i.$$ (4.1.3)

Equation (4.1.3) is known as the conservation of linear momentum. A similar equation

$$\frac{d}{dt} \int\varepsilon_{ijk} x_j (dm\ v_k) = \int\varepsilon_{ijk} x_jf_k da + \int\varepsilon_{ijk}x_j(dm\ g_k)$$ (4.1.4)

for the moment of momentum is known as the conservation of moment of momentum. Equations (4.1.3) and (4.1.4) are the BASIC LAWS OF MOTION of Continuum Mechanics, as far as this course is concerned.

Since $\frac{d}{dt}(dm) = 0$, therefore, equations (4.1.3) and (4.1.4) can also be written as

$$\int dm\ a_i = \int f_i\ da + \int dm\ g_i,$$ (4.1.5)

$$\int\varepsilon_{ijk}x_j(dm\ a_k) = \int\varepsilon_{ijk}x_jf_kda + \int\varepsilon_{ijk}x_j(dm\ g_k).$$ (4.1.6)

We now study the dependence of $\mathbf{f}$ upon $\mathbf{n}$ in some detail. Consider a cylinder of radius $R$ and height $\varepsilon$ with the top and bottom faces perpendicular to $\mathbf{n}$ and the point $P$ lying on one of the end faces. We apply the balance of linear momentum (4.1.5) to the material

contained within this cylinder. Using the mean-value theorem of calculus, we obtain

$$(\pi R^2\varepsilon \rho)\bar a_i = \int\limits_{\begin{array}{c}... ...e}}\!\!\! f_i (\mathbf{x},\mathbf{n})da + (\pi R^2\varepsilon \rho )\bar g_i\ .$$ (4.1.7)

Here $\bar a_i$ and $\bar g_i$ denote the values of $a_i$ and $g_i$ evaluated at some point in the cylinder. Let the height $\varepsilon$ of the cylinder go to zero and assume that the fields $a_i$ and $g_i$ are bounded. Then in the limit, eqn. (4.1.7) becomes

$$0 = \int [f_i (\mathbf{x},\mathbf{n}) + f_i(\mathbf{x},-\mathbf{n})]da.$$

Since this equation has to hold for all values of $R$, therefore, the integrand must be zero. That is

$$f_i (\mathbf{x},-\mathbf{n}) = -f_i (\mathbf{x},\mathbf{n}).$$ (4.1.8)

This is known as Cauchy's Fundamental Lemma and states that $\mathbf{f}$ is an odd function of $\mathbf{n}$. We now show that $\mathbf{f}$ is in fact linear in $\mathbf{n}$. Consider a tetrahedron, three sides of which are mutually orthogonal, the fourth having outward unit normal $\mathbf{n}$.

Let the area of the inclined plane $ABC$ be $A$. Then the areas of planes $PBC$, $PAC$ and $PAB$ are $An_1$, $An_2$ and $An_3$ respectively. On applying eqn. (4.1.5) to the material contained in the tetrahedron, and using the mean-value theorem, we obtain

$$\rho V\bar a_i = \bar f_i (\mathbf{x},\mathbf{n})A + \bar f_i (\m... ...mathbf{e}_2)An_2 + \bar f_i (\mathbf{x},- \mathbf{e}_3)An_3 + \rho V\bar g_i\ .$$ (4.1.9)

In eqn. (4.1.9), the superimposed bars indicate quantities evaluated at some point in the tetrahedron or at some point on a plane bounding the tetrahedron. In eqn. (4.1.9), dividing throughout by $A$, taking the limit as the tetrahedron shrinks to the point $P$, and using (4.1.8) we arrive at

$$f_i(\mathbf{x},\mathbf{n}) = f_i (\mathbf{x},\mathbf{e}_1)n_1 + f_i (\mathbf{x},\mathbf{e}_2)n_2 + f_i (\mathbf{x},\mathbf{e}_3)n_3\ ,$$ (4.1.10)

where all stress vectors are evaluated at the point $P$. Setting $\mathbf{n} = \alpha\mathbf{p} + \beta\mathbf{q}$ where $\mathbf{p}$ and $\mathbf{q}$ are unit vectors, we obtain

$$f_i (\mathbf{x},\alpha\mathbf{p} + \beta\mathbf{q}) =\ \alpha [f_i(\mathbf{x},\mathbf{e}_1) p_1 + f_i (\mathbf{x},\mathbf{e}_2)p_2 + f_i (\mathbf{x},\mathbf{e}_3)p_3]$$

$$+\beta [f_i(\mathbf{x},\mathbf{e}_1)q_1 + f_i(\mathbf{x},\mathbf{e}_2)q_2 + f_i (\mathbf{x}, \mathbf{e}_3)q_3]\ ,$$

$$=\ \alpha f_i (\mathbf{x},\mathbf{p}) + \beta f_i (\mathbf{x},\mathbf{q})\ .$$ (4.1.11)

Thus $\mathbf{f}$ is a linear function of $\mathbf{n}$ and we can write

$$f_i(\mathbf{x},\mathbf{n}) = T_{ij}(\mathbf{x})n_j.$$ (4.1.12)

By comparing the right-hand sides of (4.1.10) and (4.1.12) we get

$$T_{ij}(\mathbf{x}) = f_i (\mathbf{x},\mathbf{e}_j).$$ (4.1.13)

Thus $T_{i1},\ T_{i2}$ and $T_{13}$ denote, respectively, the surface tractions on planes whose outer normal points in the positive $x_1,x_2$ and $x_3$-directions. $T_{ij}$ is called the stress tensor. The plane whose outer normal points in the positive $x_1$-direction is simply known as the $x_1$-plane. Thus $T_{i1},\ T_{i2}$, and $T_{i3}$ are, respectively, the surface tractions on the $x_1,x_2$ and $x_3$-planes. Since

$$T_{11}(\mathbf{x}) = f_1(\mathbf{x},\mathbf{e}_1),\ T_{21} = f_2(\mathbf{x},\mathbf{e}_1),\ T_{31} = f_3 (\mathbf{x}, \mathbf{e}_1),$$

$T_{11},\ T_{21}$ and $T_{31}$ are, respectively, the normal and shearing stresses on the $x_1$-plane. Note that the resultant shear stress on the $x_1$-plane is $\sqrt{T^2_{21} + T^2_{31}}$. Also the positive values of $T_{11},\ T_{21}$ and $T_{31}$ point in the positive direction of the axes on the positive $x_1$-plane. Because of (4.1.8), positive values of $T_{11},\ T_{21}$ and $T_{31}$ point in the negative direction of the axes on the negative $x_1$-plane.

From equation (4.1.13) it is clear that stress vectors on three mutually perpendicular planes at a point determine the stress tensor at that point. Because of eqn. (4.1.12) or (4.1.10), stress vectors on three mutually perpendicular planes at a point also determine the stress vector on any other plane. This proves Cauchy's fundamental theorem: From the stress vectors acting on three mutually perpendicular planes at a point, stress vectors on every plane through the point can be determined; they are given by (4.1.12) as linear functions of the stress tensor $T_{ij}$.

A glance at equations (4.1.5) and (4.1.6) reveals that one integration in each equation is over the surface area whereas others are over the region under consideration. We now transform this surface integral into the volume integral by using the divergence theorem. Note that

$$\int f_ida = \int T_{ij}n_jda = \int T_{ij,j} dv,$$

and

$$\int \varepsilon_{ijk}x_jf_kda=\ \int\varepsilon_{ijk}x_j T_{kp}n_pda = \int (\varepsilon_{ijk}x_jT_{kp}),_pdv\ ,$$

$$=\ \int\varepsilon_{ijk}(\delta_{jp}T_{kp} + x_jT_{kp,p})dv.$$

Thus equations (4.1.5) and (4.1.6) can be written as

$$\int (\rho a_i - T_{ij,j} - \rho g_i)dv = 0,$$

and

$$\int \varepsilon_{ijk}[x_j(\rho a_k - T_{kp,p} - \rho g_k) - T_{kj}]dv = 0.$$

Since both these equations must hold for every region in the body, therefore, if the integrand is continuous throughout the body, then it must vanish. This gives

$$\rho a_i = T_{ij,j} + \rho g_i,$$ (4.1.14)

$$\varepsilon_{ijk}T_{kj} = 0\ {\rm or}\ T_{ij} = T_{ji}.$$ (4.1.15)

These are Cauchy's laws of motion. Equation (4.1.14) expresses the balance of linear momentum and eqn. (4.1.15) the balance of moment of momentum. We remark that these hold in an inertial frame. Thus on the assumption that the balance of linear momentum is satisfied, the balance of moment of momentum reduces to the requirement that the stress tensor be symmetric. In classical continuum mechanics, $T_{ij}$ is always taken to be symmetric so that the balance of moment of momentum is identically satisfied.

For static problems $a_i = 0$ and eqn. (4.1.14) gives

$$T_{ij,j} + \rho g_i = 0$$ (4.1.16)

as the three equations of equilibrium. If a given stress field satisfies eqn. (4.1.16), we may or may not be able to produce that stress field statically in a continuous body. However, if a given stress field does not satisfy even only one of the three equations of equilibrium, then it certainly cannot be produced statically in a continuous body.

Example: Show that the following stress field

$$T_{11} = x^2_2 + \nu \left(x^2_1 - x^2_2\right),\ T_{12} = -2\nu x_1x_2,$$

$$T_{22} = x^2_1 + \nu \left(x^2_2 - x^2_1\right),\ T_{23} = T_{13} = 0,$$

$$T_{33} = \nu \left(x^2_1 + x^2_2\right),$$

satisfies equations of equilibrium with zero body forces.

Solution

$$T_{11,1} = 2\nu x_1,\ T_{12,2} = -2\nu x_1,\ T_{13,3} = 0,$$

$$T_{21,1} = -2\nu x_2,\ T_{22,2} = 2\nu x_2,\ T_{23,3} = 0,$$

$$T_{31,1} = 0,\ T_{32,2} = 0,\ T_{33,3} = 0.$$

Thus

$$T_{11,1} + T_{12,2} + T_{13,3} = 2\nu x_1 - 2\nu x_1 + 0 = 0,$$

$$T_{21,1} + T_{22,2} + T_{23,3} = -2\nu x_2 + 2\nu x_2 + 0 = 0,$$

$$T_{31,1} + T_{32,2} + T_{33,3} = 0 + 0 + 0 = 0,$$

and the equations of equilibrium

$$T_{ij,j} = 0$$

with zero body force are satisfied.

Exercise: Suppose that the body force is $\mathbf{g} = -g\mathbf{e}_3$, where $g$ is a constant. Consider the following stress tensor

$$[\textbf{T}] = \alpha \left[\begin{array}{ccc} x_2 & -x_3 & 0\\ -x_3 & 0 & -x_2\\ 0 & -x_2 & T_{33}\end{array}\right]\ .$$

Find an expression for $T_{33}$ so that $T_{ij}$ satisfies the equations of equilibrium.

Exercise: Suppose that the stress distribution has the form (called plane stress)

$$[T_{ij}] = \left[\begin{array}{ccc} T_{11} (x_1,x_2) & T_{12} (x_... ... 0\\ T_{12} (x_1,x_2) & T_{22} (x_1,x_2) & 0\\ 0 & 0 & 0\end{array}\right]\ .$$

(a)

What are the equilibrium equations in this special case?

(b)

If we introduce a function $\phi (x_1,x_2)$ such that

$$T_{11} = \phi,_{22},\ \ T_{22} = \phi,_{11},\ \ T_{12} = -\phi,_{12},$$

will this stress distribution be in equilibrium with zero body force?