st

sn

σ1 σ2 σ3

Figure 8.15

Mohr™s circles for the stress.

Stress in three-dimensional continuous media

150

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

u

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

1

σM = (σxx ’ σyy )2 +(σyy ’ σzz )2 +(σzz ’ σxx )2 + 3 σxy + σxz + σyz .

2 2 2

2

(8.76)

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:

1

σM = (σ1 ’ σ2 )2 + (σ2 ’ σ3 )2 + (σ3 ’ σ1 )2 . (8.77)

2

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

152

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.

Exercises

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.

20

5

5

10

10

5

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 ±.

20

5

5

10 10

5

y

y

5

20

x x

(a) (b)

n

y

±

x

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

with

σ2 = 0 n2 = ’ 4 ex + 3 ey

with 5 5

σ3 = 25 n3 = 5 ex + 5 ey .

3 4

with

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

154

normal and shear stresses are depicted (expressed in [MPa]) that act on the

side faces of the cube.

z

4

2

2

3

2

2

y

1

x

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):

z

D F

E

y

C

A

x

B

A B C D E F

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.

z

E H

F G

D

A

y