1.8 Localization of the Balance Laws

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An important notion in the classical theory of continuum mechanics is the localization

of the balance laws. In the previous section, the balance laws were formulated for finite

bodies. The localization principle postulates that the balance laws are valid for any body,

no matter how small. This strong assumption leads to a differential form of the balance

laws. Nonlocal theories exist (Eringen 1976), which do not make this assumption but rather

assume a sphere of influence for every point.

1.8.1 Conservation of mass

Since dv = J dV , one can write Equation (1.197) as

D

Dt

__

V0

ρJ dV

_ = 0. (1.218)

V0 is the volume of the mass at time t = t0 and as such not dependent on time. Hence,

Equation (1.218) is equivalent to

_

V0

D

Dt

(ρJ ) dV = 0. (1.219)

Since this equation must be satisfied for any volume, the balance of mass yields

D

Dt

(ρJ ) = 0 (1.220)

which can also be written as (Equation (1.178))

ρ∇ · v + Dρ

Dt

= 0 (1.221)

or

∂ρ

∂t

+∇ρ · v + ρ∇ · v = 0 (1.222)

which is equivalent to

∂ρ

∂t

+∇ · (ρv) = 0. (1.223)

DISPLACEMENTS, STRAIN, STRESS AND ENERGY 29

1.8.2 Conservation of momentum

Equation (1.201) can be written as

_

V0

D

Dt

(ρJ v) dV = _

A

t (n) da + _

V0

ρJf dV +_Fc. (1.224)

Before localization can be applied to Equation (1.224) the surface integral in the righthand

side has to be converted to a volume integral. To this end, the original conservation

of momentum Equation (1.201) is applied to the volume in Figure 1.5. In addition, the

mean value theorem is used, stating that for a continuous function φ in a domain _ a point

x

∗ ∈ _ exists such that

_

_

φ(x) d_ = φ(x

∗

) _

_

d_. (1.225)

Hence

D

Dt

(ρ

∗

v

∗

_v) = t (n)_a + t (−nk )_ak + ρ

∗

f

∗

_v (1.226)

assuming there are no point loads in the volume _v. Newton’s third law (action = reaction)

dictates that

t (−nk )

= −t (nk ) (1.227)

x1

x2

x3

g1

g2

g3

n

t (n)

t (−n1)

t (−n2)

t (−n3)

−n1

−n2

−n3

Figure 1.5 Equilibrium of an infinitesimal mass element

30 DISPLACEMENTS, STRAIN, STRESS AND ENERGY

and Equation (1.226) reduces to

D

Dt

(ρ

∗

v

∗

_v) = t (n)_a − t (nk )_ak + ρ

∗

f

∗

_v. (1.228)

Notice that n1, n2 and n3 are positive in the direction of g1, g2 and g3 respectively. Since

in the limit _v →0

lim

_v→0

_v

_a

= 0 (1.229)

for an infinitesimal volume, Equation (1.226) reduces to

t (n)_a = t (nk )_ak. (1.230)

Since

_ak = nk_a (1.231)

where nk are the components of the one-form n, that is, n = nkgk, one finds

t (n) = t (nk )nk. (1.232)

Notice that the normal to a surface is a one-form since the inner product with a length

vector produces a scalar volume. Denoting the traction vector on a surface with unit normal

nk by t k, Equation (1.232) reads

t (n) = t knk. (1.233)

Accordingly, the stress on a surface with normal n is a linear combination of the

stresses on surfaces perpendicular to the coordinate axes. Substituting Equation (1.233)

into Equation (1.224) and applying Cauchy’s theorem , which reads

_

A

t knk da = _

V

t k

;k dv (1.234)

one finds after localization

D

Dt

(ρJ v) = t k

;kJ + ρf J (1.235)

at locations without concentrated forces. Applying the balance of mass yields

ρ

Dv

Dt

= t k

;k

+ ρf (1.236)

or

ρ

_∂v

∂t

+ (v ⊗∇) · v_ = t k

;k

+ ρf . (1.237)

DISPLACEMENTS, STRAIN, STRESS AND ENERGY 31

1.8.3 Conservation of angular momentum

Localization of Equation (1.208) at points without concentrated forces nor moments, taken

Equation (1.232) into account, yields

D

Dt

(ρJ x × v) = J(x × t k),k + ρJx × f (1.238)

or

D

Dt

(ρJ )x × v + J x × ρ

Dv

Dt

= J x,k × t k + J x × t k

,k

+ J x × ρf . (1.239)

Using the balance of mass, Equation (1.220), and the balance of momentum, Equation (1.236),

yields

gk

× t k = 0 (1.240)

since x,k = gk. The meaning of Equation (1.240) will become clear in Section 1.9.

1.8.4 Conservation of energy

Similar operations as in the previous section convert Equation (1.213) into

D

Dt

_

ρJε + 1

2

ρJv · v_ = J(v · t k);k −∇ · qJ + ρJf · v + ρJh (1.241)

or

ρJ

Dε

Dt

+ J v · ρ

Dv

Dt

= J v · tk;

k

+ J v;k · t k − J∇ · q + J v · ρf + Jρh. (1.242)

Application of the balance of momentum finally leads to

ρ

Dε

Dt

= v;k · t k −∇ · q + ρh. (1.243)

Equation (1.243) shows that the change of the internal energy per unit of time is balanced

by the stress power (v;k · tk), the heat influx (−∇ · q) and the heat source power (ρh).

1.8.5 Entropy inequality

Along the same lines Equation (1.217) is reduced to

ρ

Dη

Dt

≥ ρb +∇ · s. (1.244)

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