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( 67 .)


Mxx ax + Mxy ay + Mxz az

= ⎢ Myx ax + Myy ay + Myz az ⎥ . (7.34)


Mzx ax + Mzy ay + Mzz az

Along with the earlier speci¬ed matrix M the transposed matrix M T is de¬ned
according to (taking a mirror image along the principal diagonal):
⎡ ¤
Mxx Myx Mzx
⎢ ⎥
⎢ ⎥
⎢ ⎥
M = ⎢ Mxy Myy Mzy ⎥ .
⎢ ⎥
⎣ ¦
Mxz Myz Mzz

The tensor MT is associated with the matrix M T . Notice that

b= Ma bT = a T M T ,
is equivalent to
∼ ∼ ∼ ∼

b=M·a b = a · MT .
is equivalent to

The inverse of the tensor M is denoted by M’1 . By de¬nition:

M · M’1 = I, (7.36)

with I the unit tensor, I = ex ex + ey ey + ez ez . The inverse of matrix M is denoted
with M ’1 . By de¬nition:

M M ’1 = I , (7.37)

with I the unit matrix.
129 7.7 Mathematical preliminaries on tensors

The trace of tensor M (associated matrix M) is denoted as tr( M) = tr( M) and
given by

tr( M) = tr( M) = Mxx + Myy + Mzz . (7.38)

For the determinant of the tensor M with matrix representation M it can be written:

det( M) = det( M) = Mxx ( Myy Mzz ’ Myz Mzy )
’ Mxy ( Myx Mzz ’ Myz Mzx )
+ Mxz ( Myx Mzy ’ Myy Mzx ) . (7.39)

The deviatoric part of the tensor M is denoted by Md and de¬ned by

Md = M ’ tr( M) I. (7.40)
In matrix notation this reads:
M d = M ’ tr( M) I. (7.41)
Let M be an arbitrary tensor. A non-zero vector n is said to be an eigenvector of
M if a scalar » exists such that

M · n = »n ( M ’ »I) · n = 0.
or (7.42)

A non-trivial solution n from Eq. (7.42) only exists if

det( M ’ »I) = 0. (7.43)

Using the components of M and Eq. (7.39) will lead, after some elaboration, to
the following equation:

»3 ’ I1 »2 + I2 » ’ I3 = 0, (7.44)


I1 = tr( M)
I2 = Mxx Myy + Mxx Mzz + Myy Mzz ’ Mxy ’ Myz ’ Mxz
2 2 2

I3 = det( M) . (7.45)

Eq. (7.44) is called the characteristic equation and the scalar coef¬cients I1 , I2
and I3 are called the invariants of tensor M. In tensor form the invariants can be
written as
Biological materials and continuum mechanics

I1 = tr( M)
I2 = ( tr( M) )2 ’tr( M · M)
I3 = det( M) . (7.46)


7.1 A drug is administered to the blood via an intravenous drug delivery sys-
tem. In the ¬gure a two-dimensional schematic is given of the blood vessel
(in a ¬xed xy-coordinate system). The stationary concentration ¬eld c( x, y)
of the drug in the blood is sketched by means of iso-concentration lines. It
may be assumed that the concentration is constant in the z-direction. The
coordinates x and y are de¬ned in the unit [mm], the concentration is in
[mMol per litre].



30 70
60 30
10 20

Give an estimate of the gradient ∇ c of the concentration ¬eld in the point

with coordinates x = 20 [mm], y = 5 [mm].
A cube ABCDEFGH (with edge a) rotates with angular velocity ω around
the straight line , coinciding with the body diagonal CE. The points C
and E, and consequently the rotation axis , have a ¬xed position in the
xyz-coordinate system.



y A D
131 Exercises

At a certain time the edges AB, AD and AE are exactly oriented in the
directions of the x-, y- and z-axes, respectively (see ¬gure). Determine for
that time instant the velocity vector ∼B of the point B.
7.3 Bone mineral density is often measured, because this quantity can be
related to the strength and stiffness of bone. The measurement can also
be used as a diagnostic tool for osteoporosis. Consider a rectangular piece
of bone in the xy-plane of a Cartesian xyz-coordinate system, with varying
density ρ = ρ( x, y). For the corner points ABCD of the specimen the den-
sity is given (expressed in [kg dm’3 ), see ¬gure. Prove, that it is impossible
for the gradient ∇ρ of the density ¬eld in the ¬gure to be constant.
ρD = 1.02 ρC = 1.09

ρA = 1.00 ρB = 1.05

7.4 Consider a domain in the form of a cube in three-dimensional space, given
by ’ ¤ x ¤ 3 , ’2 ¤ y ¤ 2 and ’2 ¤ z ¤ 2 with a constant and
x, y and z the Cartesian coordinates. Within this domain a temperature ¬eld
exists with known gradient ∇T which is given by
∇T = θey with θ a constant.
In the domain a curve is given. The points on the curve satisfy
( x ’ )2 + y2 = and z = .

Calculate the coordinates of the point (or points) on the curve where the
derivative of the temperature in the direction of the curve is equal to zero.
7.5 In a two-dimensional xy-coordinate system (mutually perpendicular unit
vectors ex and ey along the axes) a two-dimensional stationary ¬‚uid ¬‚ow
is considered. The ¬‚uid ¬‚ow is caused by a ¬‚uid source in the origin. An
arbitrary point in the coordinate system is given by the vector x = xex +yey .
At some distance from the origin (with |x| > ») the velocity ¬eld v( x) can
be written as
v( x) = ± 2 .
The location of a point P is de¬ned by the position vector xP = ( ex + ey )
with > ». Determine the velocity gradient tensor L in point P.
8 Stress in three-dimensional
continuous media

In this chapter the concepts introduced in Chapter 6 for a one-dimensional con-
tinuous system are generalized to two-dimensional con¬gurations. Extension to
three-dimensional problems is brie¬‚y discussed. First the equilibrium conditions
in a two- or three-dimensional body are derived from force equilibrium of an
in¬nitesimally small volume element. Thereafter, the concept of a stress tensor, as
a sum of dyads, is introduced to compute the stress vector acting on an arbitrary
surface in a material point of the body.

8.1 Stress vector

Before examining the equilibrium conditions in a two-dimensional body, the
concept of a stress vector is introduced. For this purpose we consider an
in¬nitesimally small surface element having area A, see Fig. 8.1.
On this surface an in¬nitesimally small force vector F is applied with com-
ponents in the x-, y- and z-direction: F = Fx ex + Fy ey + Fz ez . Following
the de¬nition of stress, Eq. (6.8), three stresses may be de¬ned:
sx = lim , (8.1)

that acts in the x-direction, and
sy = lim , (8.2)

that acts in the y-direction, and
sz = lim , (8.3)

that acts in the z-direction. Hence, a stress vector may be de¬ned:

s = s x e x + sy e y + sz e z . (8.4)
133 8.2 From stress to force


ey ”Fxex


Figure 8.1
Force F on an in¬nitesimal surface element having area A.

8.2 From stress to force

Suppose, that a free body diagram is created by means of an imaginary cutting
plane through a body. The cutting plane is chosen in such a way that it coincides
with the xy-plane (Fig. 8.2). On the imaginary cutting plane a stress vector s is
given as a function of x and y. How can the total force vector acting on that plane
be computed based on this stress vector? The complete answer to this question is
somewhat beyond the scope of this course because it requires the integration of a
multi-variable function. However, the more simple case where the stress vector is
a function x (or y) only, while it acts on a rectangular plane at constant z is more
easy to answer and suf¬ciently general to be useful in the remainder of this chap-
ter. Therefore, suppose that the stress vector is a function of x such that it may


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