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σ1 σ2 σ3

Figure 8.15
Mohr™s circles for the stress.
Stress in three-dimensional continuous media

8.7 Hydrostatic pressure and deviatoric stress

The hydrostatic pressure p is de¬ned as the average of the normal stresses in
Fig. 8.8. It is common practice to give the pressure the opposite sign of the average
normal stresses, so:
p = ’( σxx + σyy + σzz ) /3 = ’tr( σ ) /3 = ’tr( σ ) /3. (8.69)
It can be shown that, if expressed in principal stresses, we ¬nd
p = ’( σ1 + σ2 + σ3 ) /3. (8.70)
The associated hydrostatic stress tensor is denoted with σ h and de¬ned accord-
ing to
σ h = ’p I. (8.71)
The difference of the (total) stress tensor σ and the hydrostatic stress tensor σ is
called the deviatoric stress tensor σ d . Thus
σ = σ h + σ d = ’p I + σ d and also σ = σ h + σ d = ’p I + σ d . (8.72)
Splitting the stress state in a hydrostatic part and a deviatoric part appears to be
useful for the description of the material behaviour. This will be the theme of
Chapter 12.

8.8 Equivalent stress

In general mechanical failure of materials is (among other items) determined by
the stresses that act on the material. In principle, this means that all components of
the stress tensor σ in one way or another may contribute to failure. It is common
practice, to attribute a scalar property to the stress tensor σ that re¬‚ects the gravity
of the stress state with respect to failure. Such a scalar property is normalized,
based on a consideration of a uniaxial stress state (related to the extension of a
bar) and is called an equivalent stress σ . The equivalent stress is a scalar function
of the stress tensor and thus of the components of the stress matrix
σ = σ ( σ ) and also σ = σ ( σ ) . (8.73)
The formal relationship given above has to be speci¬ed, based on physical under-
standings of the failure of the material. Only experimentally can it be assessed at
which stress combinations a certain material will reach the limits of its resistance
to failure. It is very well possible and obvious that one material fails by means
of a completely different mechanism than another material. For example, consid-
ering technical materials, it is known that for a metal the maximum shear strain
151 8.8 Equivalent stress

( σs )max is often normative, while a ceramic material might fail because of a too
high maximum extensional stress ( σn )max . Such knowledge is important for the
design of hip and knee prostheses or tooth implants, where both types of materials
are used. But also biological materials may have different failure mechanisms. A
bone for example will often fail as a result of the maximum compression stress,
but a tendon will usually fail because it is overstretched, i.e. due to the maximum
extensional stress or maximum shear stress. The functional relationship σ ( σ ) is
primarily determined by the (micro) structure of the considered material. Thus,
depending on the material, different speci¬cations of σ may be applied. We will
limit ourselves to a few examples.
According to the equivalent stress σ T ascribed to Tresca (sometimes also called
Coulomb, Mohr, Guest) the maximum shear stress is held responsible for failure.
The de¬nition is

σ T = 2(σs )max = σ3 ’ σ1 , (8.74)

which is normalized in such a way that for a uniaxially loaded bar, with an
extensional stress σax > 0, we ¬nd σ T = σax .
The equivalent stress σM according to von Mises (also H¨ ber, Hencky) is based
on the deviatoric stress tensor:
3 3
σM = tr(σ d · σ d ) = tr( σ d σ d ). (8.75)
2 2
This can be elaborated to
σM = (σxx ’ σyy )2 +(σyy ’ σzz )2 +(σzz ’ σxx )2 + 3 σxy + σxz + σyz .
2 2 2


Here also the speci¬cation is chosen in such a way that for a uniaxially loaded bar
with axial stress σax , the equivalent von Mises stress satis¬es σ M = σax . It can be
proven that in terms of principal stresses:

σM = (σ1 ’ σ2 )2 + (σ2 ’ σ3 )2 + (σ3 ’ σ1 )2 . (8.77)
In general (for arbitrary σ ) the difference between the equivalent stresses accord-
ing to Tresca and von Mises are relatively small.
The equivalent stress σ R , according to Rankine (also Galilei) expresses that, in
absolute sense, the maximum principal stress determines failure. This means that

σ R = |σ3 | if |σ3 | ≥ |σ1 |
σ R = |σ1 | if |σ3 | < |σ1 |. (8.78)
Stress in three-dimensional continuous media

And again for a uniaxially loaded bar σ R = σax . For an arbitrary stress state
the equivalent stress according to Rankine can be completely different from the
equivalent stress according to Tresca or von Mises.
A few remarks on this subject are opportune at this point. The understanding of
failure thresholds and mechanisms for biological materials involves much more
than identifying a suitable equivalent stress expression. Biological materials can
also fail because of a disturbance of the metabolic processes in the cells. The
mechanical state in biological materials is not only determined by stresses and
strains, but also by rather complicated transport processes of nutrients, oxygen and
waste products and very complex biochemistry. These processes can be disturbed
by mechanical deformation (for example occlusion of blood vessels, causing an
ischemic state of the tissue, resulting in lack of oxygen and nutrients and accu-
mulation of waste products). After some time this may result in cell death and
thus damaged tissues. How these processes evolve and lead to tissue remodelling
and/or damage is still the subject of research.


8.1 Consider a two-dimensional plane stress state, with stress components:
σxx = ax2 + by
σyy = bx2 + ay2 ’ cx.
Use the equilibrium equations to determine the shear stress component
σxy ( x, y), satisfying σxy ( 0, 0) = 2a.
8.2 On an in¬nitesimal area segment, stress components are working as given
in the ¬gure.





y 5

x 20
153 Exercises

(a) Determine the stress tensor.
(b) Determine the stress vector s acting on a plane with unit normal
√ √
vector n = 1 2ex + 1 2ey .
2 2
(c) Determine the components of s perpendicular and parallel to the
plane de¬ned in item (b).
8.3 On an in¬nitesimal area segment two sets of stress components are working
as shown in the ¬gures (a) and (b).
(a) Determine the stress tensor for both situations.
(b) Determine the stress vector on the plane with normal n =
ex cos(±) + ey sin(±). The angle ± represents the angle of the normal
vector with the x-axis.
(c) Determine for both situations the length of the stress vector as a
function of ±.


10 10
x x
(a) (b)



8.4 In a material a stress state is observed that is characterized by the principal
stresses (in [MPa]) and the principal stress directions (unit vectors, de¬ned
with respect to a ¬xed Cartesian coordinate system):
σ1 = 0 n1 = ez
σ2 = 0 n2 = ’ 4 ex + 3 ey
with 5 5
σ3 = 25 n3 = 5 ex + 5 ey .
3 4
Calculate the associated stress tensor σ with respect to the basis vectors ex ,
ey and ez .
8.5 Consider a cube of material, with the edges oriented in the direction of
the axes of a xyz-coordinate system, see the ¬gure. In this ¬gure also the
Stress in three-dimensional continuous media

normal and shear stresses are depicted (expressed in [MPa]) that act on the
side faces of the cube.




Determine the equivalent stress σ M according to von Mises.
8.6 A prismatic piece of material ABCDEF is given, see the ¬gure.
The coordinates of the corner nodes are speci¬ed in the following table
(in mm):





x 0 8 0 0 8 0
y 1 7 7 1 7 7
z 0 0 0 5 5 5

For the faces of the prism that are visible in the ¬gure, the stresses that act
upon these faces (in [MPa]) are known:

pABED = ’5ex + 2ey + 6ez
pBCFE = 10ex + 5ey
pDEF = 10ex .

Calculate the associated stress tensor σ under the assumption that the stress
state in the considered piece of material is homogeneous.
155 Exercises

8.7 Consider a material cube ABCDEFGH. The edges of the cube are oriented
in the direction of the axes of a Cartesian xyz-coordinate system, see
the ¬gure.





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