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the reference state, is fully described by the Green Lagrange strain matrix
E, with respect to a Cartesian xyz-coordinate system, with
⎡ ¤
1⎢ ⎥
E = ⎣ 3 1 0 ¦.

Calculate the volume change factor J for this particle.
10.5 A tendon is stretched in a uniaxial stress test. The tendon behaves like
an incompressible material. The length axis of the tendon coincides with
185 Exercises

the x-axis. For this test the following (time-independent) deformation rate
matrix D is applicable (expressed in [s’1 ]):

0.02 0 0

D=⎣ 0 ’0.01 ¦.
0 0
At time t = 0 [s] the tendon has a length 0 equal to 3 [cm]. From this time
on the above given matrix D can be applied. Calculate the length of the
tendon as a function of the time t.
10.6 At some material point the local deformation process is described by means
of the deformation tensor as a function of time: F( t). Based on this defor-
mation tensor, the left Cauchy Green tensor B = F · FT can be derived and
subsequently the Finger tensor µ F = 1 ( B ’ I).

T ), with: L = F · F’1 the velocity

Prove that µ F = 2 ( L · B + B · L

gradient tensor.
11 Local balance of mass, momentum
and energy

11.1 Introduction

A coherent amount of material (a material body or possibly a distinguishable
material fraction) is considered to be a continuum with current volume V in three-
dimensional space. In Chapter 8 attention was focussed on the local stress state
(the internal interaction between neighbouring volume elements), in Chapters 9
and 10 on the local kinematics (shape and volume changes of material particles).
To determine the stresses and kinematic variables as a function of the position
in the three-dimensional space, we need a description of the material behaviour,
which will be the subject of subsequent chapters, and we need local balance laws.
In the present chapter the balance of mass (leading to the continuity equation) and
the balance of momentum (leading to the equations of motion) for a continuum
will be formulated. In addition the balance of mechanical power will be derived
based on the balance of momentum.

11.2 The local balance of mass

Let us focus our attention on an in¬nitesimally small rectangular material element
dV = dxdydz in the current state, see Fig. 11.1.
The mass in the current volume element dV equals the mass in the refer-
ence con¬guration of the associated volume element dV0 , while the volumes are
related by

dV = JdV0 with J = det( F) . (11.1)

It is implicitly assumed that during the transformation from the reference state to
the current deformed state no material of the considered type is created or lost. So
there is no mass exchange with certain other fractions, for example in the form of
a chemical reaction. Based on mass conservation, it can be stated:

ρ0 dV0 = ρdV = ρJdV0 , (11.2)
187 11.3 The local balance of momentum



Figure 11.1
In¬nitesimally small element dV = dxdydz in the current state.

with ρ0 and ρ the (mass) density in the reference and current con¬guration, respec-
tively. So, balance of mass leads to the statement, that the product ρJ is time
independent and that the material time derivative of the product equals zero:
ρJ + ρ J = 0.
™ (11.3)

Using Eq. (10.70), with D the deformation rate tensor we obtain:

ρ = ’ρ tr( D) .
™ (11.4)

In this ¬nal result the reference con¬guration is no longer represented: all
variables in this equation are related to the current con¬guration.

11.3 The local balance of momentum

In Chapter 8 the equilibrium equations have been derived by considering the force
equilibrium of a material cube. The equilibrium equations are a special form of
the balance of momentum equations, which are the subject of the present section.
This means that some overlap between both derivations can be recognized.
Again, the material volume element dV from Fig. 11.1 is considered. Using the
balance of mass allows us to write for the current momentum of the element:

vρdV = vρJdV0 , (11.5)

with v the velocity of the material. Using that ρJdV0 = ρ0 dV0 is constant, the
change per unit time of the momentum of the volume element equals
™ ™
vρJdV0 = vρdV = aρdV, (11.6)

with a the acceleration vector (see Section 9.2). In the following, the above given
momentum change will be postulated, according to Newton™s law, to be equal to
the total ˜external™ force acting on the volume element.
For the element in Fig. 11.1 the resulting force in the x-direction is considered.
This force is the sum of the forces working in the x-direction on the outer surfaces
Local balance of mass, momentum and energy

of the element and the distributed load (force per unit mass) acting on the element
in the x-direction, so successively (also see Chapter 8):
’σxx dydz back plane
σxx + dx dydz frontal plane
’σxy dxdz left plane
σxy + dy dxdz right plane
’σxz dxdy bottom plane
σxz + dz dxdy top plane
qx ρdxdydz volume,
with resultant (in x-direction):
‚σxy ‚σxy
‚σxx ‚σxz ‚σxx ‚σxz
+ + + qx ρ dxdydz = + + + qx ρ dV.
‚x ‚y ‚z ‚x ‚y ‚z
Similarly, the resulting forces in y- and z-direction can be determined. All external
forces applied to the volume element are stored in a column. Using the gradient
operator ∇ introduced earlier and the symmetrical Cauchy stress matrix σ the

column with external forces can be written as
⎡ ‚σ ¤
+ ‚y + ‚σxz + qx ρ
⎢ ‚x ‚z

⎢ ⎥
⎢ ‚σxy ‚σyy ‚σyz ⎥ T
⎢ + ‚y + ‚z + qy ρ ⎥ dV = ∇ T σ + ρq dV (11.8)
⎢ ‚x ⎥ ∼ ∼
⎢ ⎥
⎣ ¦
‚σxz ‚σzz
‚x + ‚y + ‚z + qz ρ

with the column q for the distributed load de¬ned according to

⎡ ¤
⎢ ⎥
q = ⎣ qy ¦ .

In vector/tensor notation the following expression can be given for the resulting
force on the element dV:
∇ · σ + ρ q dV.
For the considered element the time derivative of the momentum equals the
external load, leading to
∇ · σ + ρ q = ρ a, (11.9)
189 11.4 The local balance of mechanical power

and also

( ∇ T σ )T +ρ q = ρ a. (11.10)
∼ ∼

This equation is called the local equation of motion. Again, all variables in this
equation refer to the current con¬guration; the de¬ned reference con¬guration (as
usual for a solid) is not relevant for this.

11.4 The local balance of mechanical power

It has to be stated explicitly that no new balance law is introduced here; only use
will be made of relations that were already introduced before. Again the element
is considered, that was introduced in Fig. 11.1 (volume in reference state dV0 ,
current volume dV = JdV0 ). The dot product of the local equation of motion
Eq. (11.9), with the velocity vector v is taken. The result is multiplied by the
current volume of the element, yielding

v · ∇ · σ dV + v · q ρ dV = v · a ρ dV. (11.11)

The terms in this equation will be interpreted separately.
• For the term on the right-hand side it can be written:

v · a ρ dV = dUkin , (11.12)

1 1 1
dUkin = v · v ρ dV = v · v ρ JdV0 = v · v ρ0 dV0 , (11.13)
2 2 2
where dUkin is the current kinetic energy of the considered volume element. With respect
to the material time derivative that is used in Eq. (11.12) it should be realized that
ρJdV0 = ρ0 dV0 is constant.
The term on the right-hand side of Eq. (11.11) can be interpreted as the change of the
kinetic energy per unit time.
• The second term on the left-hand side of Eq. (11.11) can be directly interpreted as the
mechanical power, externally applied to the volume element by the distributed load q. It
can be noted that

dPext = v · qρdV. (11.14)
• For the ¬rst term on the left-hand side of Eq. (11.11) some careful mathematical
elaboration, using the symmetry of the stress tensor σ , leads to

v· ∇ · σ dV = ∇ ·(σ · v) dV ’tr σ ·( ∇v) dV = ∇ ·(σ · v) dV ’tr( σ ·D) dV. (11.15)
Local balance of mass, momentum and energy

For this, the de¬nition of the deformation rate tensor D of Section 10.6 is used.
The ¬rst term on the right-hand side of Eq. (11.15) represents the resulting, externally
applied power to the volume element by the forces (originating from neighbouring
elements) acting on the outer surfaces of the element:

dPextσ = ∇·( σ · v) dV. (11.16)

This can be proven with a similar strategy as was used in the previous section, to
derive the resultant of the forces acting on the outer surfaces of the volume element.
Interpretation of the second term on the right-hand side,

’tr( σ · D) dV,

will be given below.

Summarizing, after reconsidering Eq. (11.11), and using the results of the above,
it can be stated:

dPextσ ’ tr( σ · D) dV + dPext = dUkin , (11.17)
and after some re-ordering:

tr( σ · D) dV + dUkin = dPext , (11.18)
dPext = dPextσ + dPext . (11.19)
So, it can be observed that the total externally applied mechanical power is partly
used for a change of the kinetic energy. The remaining part is stored internally, so:
dPint = tr( σ · D) dV. (11.20)
This internally stored mechanical power (increase of the internal mechanical


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