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3 Balance laws

In previous chapters we have studied forces acting on a continuum, the resulting deformation of the continuum, and the stresses that are caused by the deformation. The relation between forces, deformation and stress is governed by two types of equations. The first type, the balance laws, are mathematical formulations of universal physical principles: balance of mass, momentum, and moment of momentum. These laws will be discussed in this chapter.

3.1Balance of mass

3.1.1Lagrangian approach

We consider a body with a volume V0V_0, a surface area A0A_0 and a density ρ0\rho_0 in the reference state t0t_0. In the deformed state tt the body occupies a volume VV while having a surface area AA and a density ρ\rho.

A body in the initial state t_0 and the deformed state t, with arbitrary
subregion \bar{V}.

Figure 3.1:A body in the initial state t0t_0 and the deformed state tt, with arbitrary subregion Vˉ\bar{V}.

To derive the law of balance of mass[1], we concentrate on an arbitrary, fixed amount of material points in a subregion Vˉ0\bar{V}_0, i.e. we use a Langrangian approach. In the initial state these material points occupied a volume Vˉ0\bar{V}_0 while having a surface area Aˉ0\bar{A}_0 and a density ρ0\rho_0. According to the law of balance of mass, the mass of these material points must have remained constant during the deformation:

∫Vˉρ dV=∫Vˉ0ρ0 dV0\int\limits_{\bar{V}} \rho \,dV = \int\limits_{\bar{V}_0} \rho_0 \,dV_0

This equation is called the global formulation of the law of balance of mass. From the theory of kinematics, we know that the volumes before and after deformation are related as dVˉ=J  dVˉ0d\bar{V} = J\; d\bar{V}_0, where JJ represents the determinant of the deformation tensor. Thus we can transform the left-hand integral into an integral over the undeformed volume:

∫Vˉ0ρJ dV0=∫Vˉ0ρ0 dV0\int\limits_{\bar{V}_0} \rho J\,dV_0 = \int\limits_{\bar{V}_0} \rho_0 \,dV_0

which can also be written as:

∫Vˉ0(ρJ−ρ0) dV0=0\int\limits_{\bar{V}_0} ( \rho J - \rho_0 ) \,dV_0 = 0

Since this integral must hold for any arbitrary volume Vˉ0\bar{V}_0, it follows that the integrand must be zero:

ρJ=ρ0\rho J = \rho_0

This equation is called the local formulation of the law of balance of mass.

3.1.2Eulerian approach

The law of balance of mass can also be derived using an Eulerian approach. Here, we concentrate on a fixed arbitrary volume Vˉ\bar{V} in space, with surface area Aˉ\bar{A}. This volume is occupied by matter with a density ρ\rho. The total mass MM of the matter in this volume equals:

M=∫Vˉρ  dVM = \int\limits_{\bar{V}} \rho \; dV

The rate of change in time of the mass in the volume equals:

∂M∂t=∫Vˉ∂ρ∂t  dV\parder{M}{t} = \int\limits_{\bar{V}} \parder{\rho}{t} \; dV

According to the law of balance of mass, the change of mass within the volume VV can only be caused by mass transport across the boundary of the volume:

∫Vˉ∂ρ∂t  dV=−∫Aˉρv⃗⋅n⃗  dA\int\limits_{\bar{V}} \parder{\rho}{t} \; dV = - \int\limits_{\bar{A}} \rho \V{v} \inp \V{n} \; dA

Application of Gauss’s theorem to convert the surface integral into a volume integral yields:

∫Vˉ(∂ρ∂t+∇⃗⋅(ρv⃗))  dV=0\int\limits_{\bar{V}} \left( \parder{\rho}{t} + \grad \inp (\rho \V{v}) \right) \; dV = 0

Now again, because the volume Vˉ\bar{V} may be chosen arbitrarily, the integrand must be zero:

∂ρ∂t+∇⃗⋅(ρv⃗)=0\parder{\rho}{t} + \grad \inp (\rho \V{v}) = 0

or, in elaborated form:

∂ρ∂t+ρ∇⃗⋅v⃗+v⃗⋅∇⃗ρ=0\parder{\rho}{t} + \rho \grad \inp \V{v} + \V{v} \inp \grad \rho = 0

This local formulation of the equation of balance of mass is also called the continuity equation[2]. If we use the material time derivative of ρ\rho:

ρ˙=dρdt=∂ρ∂t+v⃗⋅∇⃗ρ\dot \rho = \tparder{\rho}{t} = \parder{\rho}{t} + \V{v} \inp \grad \rho

we obtain the following, slightly different form of the continuity equation:

ρ˙+ρ∇⃗⋅v⃗=0\dot \rho + \rho \grad \inp \V{v} = 0

The latter form is often used in fluid dynamics. These equations can also be derived directly by taking the material time derivative of the relation ρJ=ρ0\rho J = \rho_0,

3.2Balance of momentum

3.2.1Lagrangian approach

Again we concentrate on a certain, fixed amount of matter having a volume VV, a surface AA and a density ρ\rho. The matter is acted upon by body forces q⃗\V{q} and surface forces p⃗\V{p}.

Forces \V{q} and \V{p} acting on a subvolume \bar{V} of a body.

Figure 3.2:Forces q⃗\V{q} and p⃗\V{p} acting on a subvolume Vˉ\bar{V} of a body.

The total momentum[3] i⃗\V{i} of the matter equals:

i⃗=∫Vˉρv⃗ dV\V{i} = \int\limits_{\bar{V}} \rho \V{v} \,dV

The total resulting force f⃗\V{f}, acting on the matter, is the sum of all body forces and surface forces:

f⃗=∫Vˉρq⃗ dV+∫Aˉp⃗ dA=∫Vˉρq⃗ dV+∫Aˉσ⋅n⃗ dA=∫Vˉ(ρq⃗+∇⃗⋅σ) dV\begin{aligned} \V{f} &= \int\limits_{\bar{V}} \rho \V{q} \,dV + \int\limits_{\bar{A}} \V{p} \,dA \\ &= \int\limits_{\bar{V}} \rho \V{q} \,dV + \int\limits_{\bar{A}} \T{\sigma} \inp \V{n} \,dA \\ &= \int\limits_{\bar{V}} (\rho \V{q} + \grad \inp \T{\sigma} ) \,dV \end{aligned}

where we have applied Gauss’s theorem[4] to convert the surface integral into a volume integral.

According to the law of balance of momentum[5], this resulting force is equal to the rate of change of the momentum of the matter in Vˉ\bar{V}:

∫Vˉ(ρq⃗+∇⃗⋅σ) dV=ddt∫Vˉρv⃗ dV\int\limits_{\bar{V}} (\rho \V{q} + \grad \inp \T{\sigma} ) \,dV = \tparder{}{t} \int\limits_{\bar{V}} \rho \V{v} \,dV

To elaborate the differentiation with respect to time, we convert the right-hand volume integral to the undeformed situation:

ddt∫Vˉρv⃗ dV  =  ddt∫V0ρ0v⃗ dV0ˉ  =  ∫V0ρ0v⃗˙ dV0ˉ  =  ∫Vρv⃗˙ dVˉ\begin{aligned} \tparder{}{t} \int\limits_{\bar{V}} \rho \V{v} \,dV \;&=\; \tparder{}{t} \int\limits_{V_0} \rho_0 \V{v}\,d\bar{V_0} \\ \;&=\; \int\limits_{V_0} \rho_0 \dot{\V{v}}\,d\bar{V_0} \\ \;&=\; \int\limits_{V} \rho \dot{\V{v}} \,d\bar{V} \end{aligned}

where we used the local form of the law of balance of mass (ρdV=ρ0dV0\rho dV = \rho_0 dV_0). The global formulation of the law of balance of momentum can thus be written as:

∫Vˉ(ρq⃗+∇⃗⋅σ) dV  =  ∫Vˉρv⃗˙ dV\int\limits_{\bar{V}} (\rho \V{q} + \grad \inp \T{\sigma} ) \,dV \;=\; \int\limits_{\bar{V}} \rho \dot{\V{v}} \,dV

Now again, because the volume Vˉ\bar{V} may be chosen arbitrarily, the integrand must be zero, yielding the local formulation of the law of balance of momentum:

∇⃗⋅σ+ρq⃗  =  ρv⃗˙\grad \inp \T{\sigma} + \rho \V{q} \;=\; \rho \dot{\V{v}}

These equations are also called Cauchy’s equations of motion[6]. In case of static equilibrium, the acceleration v⃗˙\dot{\V{v}} is zero, and the equations of motion are reduced to the equilibrium equations[7]:

∇⃗⋅σ+ρq⃗=0⃗\grad \inp \T{\sigma} + \rho \V{q} = \V{0}

3.2.2Eulerian approach

Again we concentrate on a fixed volume VV in space, with surface area AA. The rate of change of momentum of the matter in this volume equals:

∂i⃗∂t=∂∂t∫Vˉρv⃗  dV+∫Aˉ(ρv⃗)v⃗⋅n⃗  dA.\parder{\V{i}}{t} = \parder{}{t} \int\limits_{\bar{V}} \rho \V{v} \;dV + \int\limits_{\bar{A}} (\rho \V{v}) \V{v} \inp \V{n} \;dA.

The first term on the right-hand side represents the change of momentum of the matter remaining within the volume, while the second term represents the change of momentum due to transport of matter across the boundary. According to the law of balance of momentum the rate of change of momentum is equal to the resulting force on the matter:

∫Vˉ(∂(ρv⃗)∂t+∇⃗⋅(ρv⃗v⃗)  dV=∫Vˉ(ρq⃗+∇⃗⋅σ) dV\int\limits_{\bar{V}} (\parder{(\rho \V{v})}{t} + \grad\inp(\rho \V{v} \V{v}) \;dV = \int\limits_{\bar{V}} (\rho \V{q} + \grad \inp \T{\sigma} ) \,dV

where Gauss’s theorem was applied for conversion to volume integrals. Using the local formulation of the law of balance of mass, and considering arbitrary volumes VV, we can derive the local formulation of the law of balance of momentum:

∇⃗⋅σ+ρq⃗=ρv⃗˙\grad \inp \T{\sigma} + \rho \V{q} = \rho \dot{\V{v}}

which is identical to the form, derived through the Lagrangian approach.

3.3Balance of moment of momentum

Forces \V{q} and \V{p}, acting on a subvolume \bar{V} of a body,
and position vector \V{x} with respect to the origin O, needed to calculate moments of
momentum.

Figure 3.3:Forces q⃗\V{q} and p⃗\V{p}, acting on a subvolume Vˉ\bar{V} of a body, and position vector x⃗\V{x} with respect to the origin OO, needed to calculate moments of momentum.

Using a Lagrangian approach, we concentrate on a certain, fixed amount of matter having a volume VV, a surface AA and a density ρ\rho. The matter is acted upon by body forces q⃗\V{q} and surface forces p⃗\V{p}. The total moment of momentum[8] L⃗O\V{L}_O of the material points with respect to the origin OO equals:

L⃗O=∫Vˉx⃗×ρv⃗ dV\V{L}_O = \int\limits_{\bar{V}} \V{x} \cross \rho \V{v} \,dV

The resulting moment M⃗O\V{M}_O of the forces acting on the body with respect to the origin OO, is given by:

M⃗O=∫Vˉx⃗×ρq⃗ dV+∫Aˉx⃗×p⃗ dA\V{M}_O = \int\limits_{\bar{V}} \V{x} \cross \rho \V{q} \,dV + \int\limits_{\bar{A}} \V{x} \cross \V{p} \,dA

Of course, the moment of momentum can be calculated with respect to any fixed point in space: for convenience we have chosen the origin. According to the law of balance of moment of momentum, the moment, exerted on the matter by the body and surface forces with respect to OO, is equal to the rate of change of the moment of momentum of the matter with respect to OO:

ddt∫Vˉx⃗×ρv⃗ dV=∫Vˉx⃗×ρq⃗ dV+∫Aˉx⃗×p⃗ dA\tparder{}{t} \int\limits_{\bar{V}} \V{x} \cross \rho \V{v} \,dV = \int\limits_{\bar{V}} \V{x} \cross \rho \V{q} \,dV + \int\limits_{\bar{A}} \V{x} \cross \V{p} \,dA

Through a laborious derivation, similar to that used for balance of mass and momentum, we can derive the following local formulation of the law of balance of moment of momentum:

σT=σ\T{\sigma}^T = \T{\sigma}

i.e., the Cauchy stress tensor is symmetric.

Footnotes
  1. massabalans

  2. continuïteitsvergelijking

  3. impuls

  4. divergentiestelling

  5. impulsbalans

  6. bewegingsvergelijkingen

  7. evenwichtsvergelijkingen

  8. impulsmoment