Section 8.

8.1 Introduction to Plasticity

8.1.1 Introduction

The theory of linear elasticity is useful for modelling materials which undergo small
deformations and which return to their original configuration upon removal of load.
Almost all real materials will undergo some permanent deformation, which remains after
removal of load. With metals, significant permanent deformations will usually occur
when the stress reaches some critical value, called the yield stress, a material property.

Elastic deformations are termed reversible; the energy expended in deformation is stored
as elastic strain energy and is completely recovered upon load removal. Permanent
deformations involve the dissipation of energy; such processes are termed irreversible, in
the sense that the original state can be achieved only by the expenditure of more energy.

The classical theory of plasticity grew out of the study of metals in the late nineteenth
century. It is concerned with materials which initially deform elastically, but which
deform plastically upon reaching a yield stress. In metals and other crystalline materials
the occurrence of plastic deformations at the micro-scale level is due to the motion of
dislocations and the migration of grain boundaries on the micro-level. In sands and other
granular materials plastic flow is due both to the irreversible rearrangement of individual
particles and to the irreversible crushing of individual particles. Similarly, compression
of bone to high stress levels will lead to particle crushing. The deformation of micro-
voids and the development of micro-cracks is also an important cause of plastic
deformations in materials such as rocks.

A good part of the discussion in what follows is concerned with the plasticity of metals;
this is the ‘simplest’ type of plasticity and it serves as a good background and
introduction to the modelling of plasticity in other material-types. There are two broad
groups of metal plasticity problem which are of interest to the engineer and analyst. The
first involves relatively small plastic strains, often of the same order as the elastic strains
which occur. Analysis of problems involving small plastic strains allows one to design
structures optimally, so that they will not fail when in service, but at the same time are not
stronger than they really need to be. In this sense, plasticity is seen as a material failure 1.

The second type of problem involves very large strains and deformations, so large that the
elastic strains can be disregarded. These problems occur in the analysis of metals
manufacturing and forming processes, which can involve extrusion, drawing, forging,
rolling and so on. In these latter-type problems, a simplified model known as perfect
plasticity is usually employed (see below), and use is made of special limit theorems
which hold for such models.

Plastic deformations are normally rate independent, that is, the stresses induced are
independent of the rate of deformation (or rate of loading). This is in marked

1
two other types of failure, brittle fracture, due to dynamic crack growth, and the buckling of some
structural components, can be modelled reasonably accurately using elasticity theory (see, for example, Part
I, §6.1, Part II, §5.3)

Solid Mechanics Part II 241 Kelly

 Section 8.1

contrast to classical Newtonian fluids for example, where the stress levels are
governed by the rate of deformation through the viscosity of the fluid.

Materials commonly known as “plastics” are not plastic in the sense described here.
They, like other polymeric materials, exhibit viscoelastic behaviour where, as the
name suggests, the material response has both elastic and viscous components. Due
to their viscosity, their response is, unlike the plastic materials, rate-dependent.
Further, although the viscoelastic materials can suffer irrecoverable deformation,
they do not have any critical yield or threshold stress, which is the characteristic
property of plastic behaviour. When a material undergoes plastic deformations, i.e.
irrecoverable and at a critical yield stress, and these effects are rate dependent, the
material is referred to as being viscoplastic.

Plasticity theory began with Tresca in 1864, when he undertook an experimental program
into the extrusion of metals and published his famous yield criterion discussed later on.
Further advances with yield criteria and plastic flow rules were made in the years which
followed by Saint-Venant, Levy, Von Mises, Hencky and Prandtl. The 1940s saw the
advent of the classical theory; Prager, Hill, Drucker and Koiter amongst others brought
together many fundamental aspects of the theory into a single framework. The arrival of
powerful computers in the 1980s and 1990s provided the impetus to develop the theory
further, giving it a more rigorous foundation based on thermodynamics principles, and
brought with it the need to consider many numerical and computational aspects to the
plasticity problem.

8.1.2 Observations from Standard Tests

In this section, a number of phenomena observed in the material testing of metals will be
noted. Some of these phenomena are simplified or ignored in some of the standard
plasticity models discussed later on.

At issue here is the fact that any model of a component with complex geometry, loaded in
a complex way and undergoing plastic deformation, must involve material parameters
which can be obtained in a straight forward manner from simple laboratory tests, such as
the tension test described next.

The Tension Test

Consider the following key experiment, the tensile test, in which a small, usually
cylindrical, specimen is gripped and stretched, usually at some given rate of stretching
(see Part I, §5.2.1). The force required to hold the specimen at a given stretch is recorded,
Fig. 8.1.1. If the material is a metal, the deformation remains elastic up to a certain force
level, the yield point of the material. Beyond this point, permanent plastic deformations
are induced. On unloading only the elastic deformation is recovered and the specimen
will have undergone a permanent elongation (and consequent lateral contraction).

In the elastic range the force-displacement behaviour for most engineering materials
(metals, rocks, plastics, but not soils) is linear. After passing the elastic limit (point A in
Fig. 8.1.1), the material “gives” and is said to undergo plastic flow. Further increases in
load are usually required to maintain the plastic flow and an increase in displacement; this

Solid Mechanics Part II 242 Kelly

 Section 8.1

phenomenon is known as work-hardening or strain-hardening. In some cases, after an

initial plastic flow and hardening, the force-displacement curve decreases, as in some
soils; the material is said to be softening. If the specimen is unloaded from a plastic state
(B) it will return along the path BC shown, parallel to the original elastic line. This is
elastic recovery. The strain which remains upon unloading is the permanent plastic
deformation. If the material is now loaded again, the force-displacement curve will re-
trace the unloading path CB until it again reaches the plastic state. Further increases in
stress will cause the curve to follow BD.

Two important observations concerning the above tension test (on most metals) are the
following:
(1) after the onset of plastic deformation, the material will be seen to undergo negligible
volume change, that is, it is incompressible.
(2) the force-displacement curve is more or less the same regardless of the rate at which
the specimen is stretched (at least at moderate temperatures).

force hardening D
B

A
Yield point

elastic unload load

loading

0 C displacement
plastic elastic
deformation deformation

Figure 8.1.1: force/displacement curve for the tension test

Nominal and True Stress and Strain

There are two different ways of describing the force F which acts in a tension test. First,
normalising with respect to the original cross sectional area of the tension test specimen
A0 , one has the nominal stress or engineering stress,

F
n  (8.1.1)
A0

Alternatively, one can normalise with respect to the current cross-sectional area A,
leading to the true stress,

F
 (8.1.2)
A

Solid Mechanics Part II 243 Kelly

 Section 8.1

in which F and A are both changing with time. For very small elongations, within the
elastic range say, the cross-sectional area of the material undergoes negligible change and
both definitions of stress are more or less equivalent.

Similarly, one can describe the deformation in two alternative ways. Denoting the
original specimen length by l 0 and the current length by l, one has the engineering strain

l  l0
 (8.1.3)
l0

Alternatively, the true strain is based on the fact that the “original length” is continually
changing; a small change in length dl leads to a strain increment d  dl / l and the
total strain is defined as the accumulation of these increments:

l
dl l 
t    ln  (8.1.4)
l0
l  l0 

The true strain is also called the logarithmic strain or Hencky strain. Again, at small
deformations, the difference between these two strain measures is negligible. The true
strain and engineering strain are related through

 t  ln 1    (8.1.5)

Using the assumption of constant volume for plastic deformation and ignoring the very
small elastic volume changes, one has also {▲Problem 3}

l
 n . (8.1.6)
l0

The stress-strain diagram for a tension test can now be described using the true
stress/strain or nominal stress/strain definitions, as in Fig. 8.1.2. The shape of the
nominal stress/strain diagram, Fig. 8.1.2a, is of course the same as the graph of force
versus displacement (change in length) in Fig. 8.1.1. A here denotes the point at which
the maximum force the specimen can withstand has been reached. The nominal stress at
A is called the Ultimate Tensile Strength (UTS) of the material. After this point, the
specimen “necks”, with a very rapid reduction in cross-sectional area somewhere about
the centre of the specimen until the specimen ruptures, as indicated by the asterisk.

Note that, during loading into the plastic region, the yield stress increases. For example,
if one unloads and re-loads (as in Fig. 8.1.1), the material stays elastic up until a stress
higher than the original yield stress Y. In this respect, the stress-strain curve can be
regarded as a yield stress versus strain curve.

Solid Mechanics Part II 244 Kelly

 Section 8.1

n A  A


Y Y

 t
(a ) (b)

Figure 8.1.2: typical stress/strain curves; (a) engineering stress and strain, (b) true
stress and strain

Compression Test

A compression test will lead to similar results as the tensile stress. The yield stress in
compression will be approximately the same as (the negative of) the yield stress in
tension. If one plots the true stress versus true strain curve for both tension and
compression (absolute values for the compression), the two curves will more or less
coincide. This would indicate that the behaviour of the material under compression is
broadly similar to that under tension. If one were to use the nominal stress and strain,
then the two curves would not coincide; this is one of a number of good reasons for using
the true definitions.

The Bauschinger Effect

If one takes a virgin sample and loads it in tension into the plastic range, and then unloads
it and continues on into compression, one finds that the yield stress in compression is not
the same as the yield strength in tension, as it would have been if the specimen had not
first been loaded in tension. In fact the yield point in this case will be significantly less
than the corresponding yield stress in tension. This reduction in yield stress is known as
the Bauschinger effect. The effect is illustrated in Fig. 8.1.3. The solid line depicts the
response of a real material. The dotted lines are two extreme cases which are used in
plasticity models; the first is the isotropic hardening model, in which the yield stress in
tension and compression are maintained equal, the second being kinematic hardening, in
which the total elastic range is maintained constant throughout the deformation.

Solid Mechanics Part II 245 Kelly

 Section 8.1

Y1
Y0 2Y0

t

kinematic hardening Y1

isotropic hardening

Figure 8.1.3: The Bauschinger effect

The presence of the Bauschinger effect complicates any plasticity theory. However, it is
not an issue provided there are no reversals of stress in the problem under study.

Hydrostatic Pressure

Careful experiments show that, for metals, the yield behaviour is independent of
hydrostatic pressure. That is, a stress state  xx   yy   zz   p has negligible effect on
the yield stress of a material, right up to very high pressures. Note however that this is
not true for soils or rocks.

8.1.3 Assumptions of Plasticity Theory

Regarding the above test results then, in formulating a basic plasticity theory with which
to begin, the following assumptions are usually made:

(1) the response is independent of rate effects

(2) the material is incompressible in the plastic range
(3) there is no Bauschinger effect
(4) the yield stress is independent of hydrostatic pressure
(5) the material is isotropic

The first two of these will usually be very good approximations, the other three may or
may not be, depending on the material and circumstances. For example, most metals can
be regarded as isotropic. After large plastic deformation however, for example in rolling,
the material will have become anisotropic: there will be distinct material directions and
asymmetries.

Together with these, assumptions can be made on the type of hardening and on whether
elastic deformations are significant. For example, consider the hierarchy of models
illustrated in Fig. 8.1.4 below, commonly used in theoretical analyses. In (a) both the
elastic and plastic curves are assumed linear. In (b) work-hardening is neglected and the

Solid Mechanics Part II 246 Kelly

 Section 8.1

yield stress is constant after initial yield. Such perfectly-plastic models are particularly
appropriate for studying processes where the metal is worked at a high temperature – such
as hot rolling – where work hardening is small. In many areas of applications the strains
involved are large, e.g. in metal working processes such as extrusion, rolling or drawing,
where up to 50% reduction ratios are common. In such cases the elastic strains can be
neglected altogether as in the two models (c) and (d). The rigid/perfectly-plastic model
(d) is the crudest of all – and hence in many ways the most useful. It is widely used in
analysing metal forming processes, in the design of steel and concrete structures and in
the analysis of soil and rock stability.

 

Y Y

0  0 

(a) Linear Elastic-Plastic (b) Elastic/Perfectly-Plastic

 

Y Y

0  0 
(c) Rigid/Linear Hardening (d) Rigid-Perfectly-Plastic

Figure 8.1.4: Simple models of elastic and plastic deformation

8.1.4 The Tangent and Plastic Modulus

Stress and strain are related through   E in the elastic region, E being the Young’s
modulus, Fig. 8.1.5. The tangent modulus K is the slope of the stress-strain curve in the
plastic region and will in general change during a deformation. At any instant of strain,
the increment in stress d is related to the increment in strain d through2

d  Kd (8.1.7)

2
the symbol  here represents the true strain (the subscript t has been dropped for clarity); as mentioned,
when the strains are small, it is not necessary to specify which strain is in use since all strain measures are
then equivalent

Solid Mechanics Part II 247 Kelly

 Section 8.1

K



d p d e
d
E


Figure 8.1.5: The tangent modulus

After yield, the strain increment consists of both elastic,  e , and plastic, d p , strains:

d  d e  d p (8.1.8)

The stress and plastic strain increments are related by the plastic modulus H:

d  H d p (8.1.9)

and it follows that {▲Problem 4}

1 1 1
  (8.1.10)
K E H

8.1.5 Friction Block Models

Some additional insight into the way plastic materials respond can be obtained from
friction block models. The rigid perfectly plastic model can be simulated by a Coulomb
friction block, Fig. 8.1.6. No strain occurs until  reaches the yield stress Y. Then there
is movement – although the amount of movement or plastic strain cannot be determined
without more information being available. The stress cannot exceed the yield stress in
this model:

 Y (8.1.11)

If unloaded, the block stops moving and the stress returns to zero, leaving a permanent
strain, Fig. 8.1.6b.

Solid Mechanics Part II 248 Kelly

 Section 8.1

unload


Y

permanent
deformation
(a) (b)

Figure 8.1.6: (a) Friction block model for the rigid perfectly plastic material, (b)
response of the rigid-perfectly plastic model

The linear elastic perfectly plastic model incorporates a free spring with modulus E in
series with a friction block, Fig. 8.1.7. The spring stretches when loaded and the block
also begins to move when the stress reaches Y, at which time the spring stops stretching,
the maximum possible stress again being Y. Upon unloading, the block stops moving and
the spring contracts.

E


Figure 8.1.7: Friction block model for the elastic perfectly plastic material

The linear elastic plastic model with linear strain hardening incorporates a second,
hardening, spring with stiffness H, in parallel with the friction block, Fig. 8.1.8. Once the
yield stress is reached, an ever increasing stress needs to be applied in order to keep the
block moving – and elastic strain continues to occur due to further elongation of the free
spring. The stress is then split into the yield stress, which is carried by the moving block,
and an overstress   Y carried by the hardening spring.

Upon unloading, the block “locks” – the stress in the hardening spring remains constant
whilst the free spring contracts. At zero stress, there is a negative stress taken up by the
friction block, equal and opposite to the stress in the hardening spring.

The slope of the elastic loading line is E. For the plastic hardening line,

  Y d EH
 e  p    K  (8.1.12)
E H d E  H

It can be seen that H is the plastic modulus.

Solid Mechanics Part II 249 Kelly

 Section 8.1

H E 
(a)
Y 


(b) 

e 


(c)

e 
p

(d)
 0
e  0 
p

Figure 8.1.8: Friction block model for a linear elastic-plastic material with linear
strain hardening; (a) stress-free, (b) elastic strain, (c) elastic and plastic strain, (d)
unloading

8.1.6 Problems
1. Give two differences between plastic and viscoelastic materials.

2. A test specimen of initial length 0.01 m is extended to length 0.0101 m. What is the
percentage difference between the engineering and true strains (relative to the
engineering strain)? What is this difference when the specimen is extended to length
0.015 m?

3. Derive the relation 8.1.6,  /  n  l / l 0 .

4. Derive Eqn. 8.1.10.

5. Which is larger, H or K? In the case of a perfectly-plastic material?

6. The Ramberg-Osgood model of plasticity is given by

n
  
     
e p

E b
where E is the Young’s modulus and b and n are model constants (material
parameters) obtained from a curve-fitting of the uniaxial stress-strain curve.

Solid Mechanics Part II 250 Kelly

 Section 8.1

(i) Find the tangent and plastic moduli in terms of plastic strain  p (and the
material constants). 8
x 10
(ii) A material with model 5.5
E  70GPa
parameters n  4 , 5

E  70 GPa and 4.5 b  800MPa

 4
b  800 MPa is strained n4
3.5
n3
in tension to  p  0.02 3
and is subsequently 2.5 n2
unloaded and put into 2
compression. Find the 1.5
stress at the initiation of 1 n 1
compressive yield 0.5
assuming isotropic 0
hardening. 0 0.02 0.04 0.06 0.08 0.1

0.12 0.14 0.16 0.18 0.2

[Note that the yield stress is actually zero in this model, although the plastic strain at
relatively low stress levels is small for larger values of n.]

7. Consider the plasticity model shown below.

(i) What is the elastic modulus?
(ii) What is the yield stress?
(iii) What are the tangent and plastic moduli?
Draw a typical loading and unloading curve.

E2


E1

8. Draw the stress-strain diagram for a cycle of loading and unloading to the rigid -
plastic model shown here. Take the maximum load reached to be  max  4Y1 and
Y2  2Y1 . What is the permanent deformation after complete removal of the load?
[Hint: split the cycle into the following regions: (a) 0    Y1 , (b) Y1    3Y1 , (c)
3Y1    4Y1 , then unload, (d) 4Y1    3Y1 , (e) 3Y1    2Y1 , (f) 2Y1    0 .]
E2 E1

Y2 Y1

Solid Mechanics Part II 251 Kelly

 Section 8.2

8.2 Stress Analysis for Plasticity

This section follows on from the analysis of three dimensional stress carried out in §7.2.
The plastic behaviour of materials is often independent of a hydrostatic stress and this
feature necessitates the study of the deviatoric stress.

8.2.1 Deviatoric Stress

Any state of stress can be decomposed into a hydrostatic (or mean) stress σ m I and a
deviatoric stress s, according to

⎡σ 11 σ 12 σ 13 ⎤ ⎡σ m 0 0 ⎤ ⎡ s11 s12 s13 ⎤

⎢σ σ 23 ⎥ = ⎢ 0 σ m 0 ⎥⎥ + ⎢⎢ s 21 s 22
⎥ ⎢ s 23 ⎥⎥
⎢ 21 σ 22 (8.2.1)
⎣⎢σ 31 σ 32 σ 33 ⎦⎥ ⎣⎢ 0 0 σ m ⎦⎥ ⎣⎢ s31 s32 s 33 ⎦⎥

where

σ 11 + σ 22 + σ 33
σm = (8.2.2)
3

and

⎡ s11 s12 s13 ⎤ ⎡ 13 (2σ 11 − σ 22 − σ 33 ) σ 12 σ 13 ⎤

⎢ s 21 s 22 s 23 ⎥ = ⎢ σ 12 1
(2σ 22 − σ 11 − σ 33 ) σ 23 ⎥
s 33 ⎥⎦ ⎢⎣ )⎥
3
⎢s σ 13 σ 23 (
3 2σ 33 − σ 11 − σ 22 ⎦
⎣ 31 s 32 1

(8.2.3)

In index notation,

σ ij = σ mδ ij + sij (8.2.4)

In a completely analogous manner to the derivation of the principal stresses and the
principal scalar invariants of the stress matrix, §7.2.4, one can determine the principal
stresses and principal scalar invariants of the deviatoric stress matrix. The former are
denoted s1 , s 2 , s 3 and the latter are denoted by J 1 , J 2 , J 3 . The characteristic equation
analogous to Eqn. 7.2.23 is

s 3 − J1s 2 − J 2 s − J 3 = 0 (8.2.5)

and the deviatoric invariants are (compare with 7.2.24, 7.2.26)1

1
unfortunately, there is a convention (adhered to by most authors) to write the characteristic equation for
stress with a + I 2σ term and that for deviatoric stress with a − J 2 s term; this means that the formulae for
J2 in Eqn. 8.2.5 are the negative of those for I 2 in Eqn. 7.2.24

Solid Mechanics Part II 252 Kelly

 Section 8.2

J 1 = s11 + s 22 + s33
= s1 + s 2 + s3
(
J 2 = − s11 s 22 + s 22 s33 + s33 s11 − s122 − s 23 2
− s31
2
)
= −(s1 s 2 + s 2 s3 + s3 s1 ) (8.2.6)
J 3 = s11 s 22 s33 − s11 s 23
2
− s 22 s31
2
− s33 s122 + 2s12 s 23 s31
= s1 s 2 s3

Since the hydrostatic stress remains unchanged with a change of coordinate system, the
principal directions of stress coincide with the principal directions of the deviatoric stress,
and the decomposition can be expressed with respect to the principal directions as

⎡σ 1 0 0 ⎤ ⎡σ m 0 0 ⎤ ⎡ s1 0 0⎤
⎢0 σ 0 ⎥=⎢ 0 σm 0 ⎥⎥ + ⎢⎢ 0 s2 0 ⎥⎥
⎢ 2 ⎥ ⎢ (8.2.7)
⎣⎢ 0 0 σ 3 ⎦⎥ ⎣⎢ 0 0 σ m ⎦⎥ ⎣⎢ 0 0 s3 ⎦⎥

Note that, from the definition Eqn. 8.2.3, the first invariant of the deviatoric stress, the
sum of the normal stresses, is zero:

J1 = 0 (8.2.8)

The second invariant can also be expressed in the useful forms {▲Problem 3}

J2 = 1
2 (s
2
1 + s 22 + s32 , ) (8.2.9)

and, in terms of the principal stresses, {▲Problem 4}

J2 =
1
6
[
(σ 1 − σ 2 )2 + (σ 2 − σ 3 )2 + (σ 3 − σ 1 )2 . ] (8.2.10)

Further, the deviatoric invariants are related to the stress tensor invariants through
{▲Problem 5}

J2 = 1
3 (I 1
2
)
− 3I 2 , J 3 = 1
27 (2I 3
1 − 9 I 1 I 2 + 27 I 3 ) (8.2.11)

A State of Pure Shear

The stress state at a point is one of pure shear if for any one coordinate axes through the
point one has only shear stress acting, i.e. the stress matrix is of the form

⎡ 0 σ 12 σ 13 ⎤
[σ ]
ij = ⎢σ 12 0 σ 23 ⎥ (8.2.12)
⎢σ 0 ⎥⎦
⎣ 13 σ 23

Solid Mechanics Part II 253 Kelly

 Section 8.2

Applying the stress transformation rule 7.2.16 to this stress matrix and using the fact that
the transformation matrix Q is orthogonal, i.e. QQ T = Q T Q = I , one finds that the first
invariant is zero, σ 11′ + σ 22 ′ = 0 . Hence the deviatoric stress is one of pure shear.
′ + σ 33

8.2.2 The Octahedral Stresses

Examine now a material element subjected to principal stresses σ 1 , σ 2 , σ 3 as shown in

Fig. 8.2.1. By definition, no shear stresses act on the planes shown.

3
σ3

σ1

σ2 σ2
2

σ1
σ3
1

Figure 8.2.1: stresses acting on a material element

Consider next the octahedral plane; this is the plane shown shaded in Fig. 8.2.2, whose
normal n a makes equal angles with the principal directions. It is so-called because it cuts
a cubic material element (with faces perpendicular to the principal directions) into a
triangular plane and eight of these triangles around the origin form an octahedron.

na
2

Figure 8.2.2: the octahedral plane

Next, a new Cartesian coordinate system is constructed with axes parallel and
perpendicular to the octahedral plane, Fig. 8.2.3. One axis runs along the unit normal n a ;

Solid Mechanics Part II 254 Kelly

 Section 8.2

( )
this normal has components 1 / 3 ,1 / 3 ,1 / 3 with respect to the principal axes. The
angle θ 0 the normal direction makes with the 1 direction can be obtained from
n a ⋅ e1 = cos θ 0 , where e1 = (1,0,0 ) is a unit vector in the 1 direction, Fig. 8.2.3. To
complete the new coordinate system, any two perpendicular unit vectors which lie in
(parallel to) the octahedral plane can be chosen. Choose one which is along the
projection of the 1 axis down onto the octahedral plane. The components of this vector
( )
are {▲Problem 6} n c = 2 / 3 , − 1 / 6 , − 1 / 6 . The final unit vector n b is chosen so
that it forms a right hand Cartesian coordinate system with n a and n c , i.e. n a × n b = n c .
In summary,

1 ⎡1⎤ 1 ⎡0⎤ 1 ⎡2⎤

na = ⎢1⎥, nb = ⎢− 1⎥, nc = ⎢− 1⎥ (8.2.13)
3 ⎢⎣1⎥⎦ 2 ⎢⎣ 1 ⎥⎦ 6 ⎢⎣− 1⎥⎦

3
b a
θ0
e3
nb
na θ0

e2 2
nc
θ0
e1
c
1

Figure 8.2.3: a new Cartesian coordinate system

To express the stress state in terms of components in the a, b, c directions, construct the
stress transformation matrix:

⎡ e1 ⋅ n a e1 ⋅ n b e1 ⋅ n c ⎤ ⎡1 / 3 0 2/ 6 ⎤
Q = ⎢e 2 ⋅ n a e2 ⋅ nb ⎥ ⎢
e2 ⋅ nc = 1/ 3 − 1/ 2 − 1/ 6 ⎥ (8.2.14)
⎢e ⋅ n ⎢ ⎥
⎣ 3 a e3 ⋅ nb e 3 ⋅ n c ⎥⎦ ⎢1 / 3 1 / 2 − 1 / 6 ⎦⎥
⎣

and the new stress components are

⎡σ aa σ ab σ ac ⎤ ⎡σ 1 0 0⎤
⎢σ ba σ bb σ bc ⎥ = Q T ⎢ 0 σ 2 0 ⎥Q
⎢σ σ cb σ cc ⎥⎦ ⎢0 0 σ 3 ⎥⎦
⎣ ca ⎣
⎡ 13 (σ 1 + σ 2 + σ 3 ) − 16 (σ 2 − σ 3 ) 1
(2σ 1 − σ 2 − σ 3 )⎤
⎢
(σ 2 − σ 3 ) ⎥⎥
3 2
= ⎢ − 16 (σ 2 − σ 3 ) 1
2
(σ 2 + σ 3 ) 1
2 3
⎢ 3 1 2 (2σ 1 − σ 2 − σ 3 ) 2 1 3 (σ 2 − σ 3 ) (
6 4σ 1 + σ 2 + σ 3 ⎦
1
)⎥
⎣
(8.2.15)

Solid Mechanics Part II 255 Kelly

 Section 8.2

Now consider the stress components acting on the octahedral plane, σ aa , σ ab , σ ac ,

Fig. 8.2.4. Recall from Cauchy’s law, Eqn. 7.2.9, that these are the components of
the traction vector t (n a ) acting on the octahedral plane, with respect to the (a,b,c)
axes:

t (n a ) = σ aa n a + σ ab n b + σ ac n c (8.2.16)

3
•

t (na )
σ ab
σ aa = σ oct
τ oct
2
σ ac

Figure 8.2.4: the stress vector σ and its components

The magnitudes of the normal and shear stresses acting on the octahedral plane are called
the octahedral normal stress σ oct and the octahedral shear stress τ oct . Referring to
Fig. 8.2.4, these can be expressed as {▲Problem 7}

1
σ oct = σ aa = (σ 1 + σ 2 + σ 3 ) = 1 I1
3 3
τ oct = σ ab2 + σ ac2 (8.2.17)
1 2J 2
= (σ 1 − σ 2 )2 + (σ 2 − σ 3 )2 + (σ 3 − σ 1 )2 =
3 3

The octahedral normal and shear stresses on all 8 octahedral planes around the origin are
the same.

Note that the octahedral normal stress is simply the hydrostatic stress. This implies that
the deviatoric stress has no normal component in the direction n a and only contributes to
shearing on the octahedral plane. Indeed, from Eqn. 8.2.15,

⎡ s aa s ab s ac ⎤ ⎡ 0 − 16 (σ 2 − σ 3 ) 1
(2σ 1 − σ 2 − σ 3 )⎤
⎢
(σ 2 − σ 3 ) ⎥⎥
3 2
⎢ s ba sbb s bc ⎥ = ⎢ − 16 (σ 2 − σ 3 ) − 16 (2σ 1 − σ 2 − σ 3 ) 1
⎢s ⎥
s cc ⎦ ⎢ 1 (2σ 1 − σ 2 − σ 3 )
2 3
⎣ ca s cb
⎣3 2
1
2 3
(σ 2 − σ 3 ) 1
(
6 2σ 1 − σ 2 − σ 3 ⎦
)⎥
(8.2.18)

Solid Mechanics Part II 256 Kelly

 Section 8.2

The σ ’s on the right here can be replaced with s ’s since σ i − σ j = si − s j .

8.2.3 Problems

1. What are the hydrostatic and deviatoric stresses for the uniaxial stress σ 11 = σ 0 ?
What are the hydrostatic and deviatoric stresses for the state of pure shear σ 12 = τ ?
In both cases, verify that the first invariant of the deviatoric stress is zero: J 1 = 0 .

⎡σ 11 σ 12 σ 13 ⎤ ⎡1 2 4⎤
2. For the stress state ⎢σ 21 σ 22 σ 23 ⎥ = ⎢2 2 1⎥ , calculate
⎢σ σ 33 ⎥⎦ ⎢⎣4 1 3⎥⎦
⎣ 31 σ 32
(a) the hydrostatic stress
(b) the deviatoric stresses
(c) the deviatoric invariants

3. The second invariant of the deviatoric stress is given by Eqn. 8.2.6,

J 2 = −(s1 s 2 + s 2 s3 + s3 s1 )
By squaring the relation J 1 = s1 + s 2 + s3 = 0 , derive Eqn. 8.2.9,
J2 = 1
2 (s
2
1 + s 22 + s32 )
4. Use Eqns. 8.2.9 (and your work from Problem 3) and the fact that σ 1 − σ 2 = s1 − s 2 ,
etc. to derive 8.2.10,
[
J 2 = 16 (σ 1 − σ 2 ) + (σ 2 − σ 3 ) + (σ 3 − σ 1 )
2 2 2
]
5. Use the fact that J 1 = s1 + s 2 + s3 = 0 to show that
I 1 = 3σ m
I 2 = ( s1 s 2 + s 2 s 3 + s 3 s1 ) + 3σ m2
I 3 = s1 s 2 s3 + σ m ( s1 s 2 + s 2 s3 + s 3 s1 ) + σ m3
Hence derive Eqns. 8.2.11,
( ) (
J 2 = 13 I 12 − 3I 2 , J 3 = 271 2 I 13 − 9 I 1 I 2 + 27 I 3 )
6. Show that a unit normal n c in the octahedral plane in the direction of the projection
of the 1 axis down onto the octahedral plane has coordinates ( 2
3 ,− 1
6
,− 1
6
), Fig.
8.2.3. To do this, note the geometry shown below and the fact that when the 1 axis is
projected down, it remains at equal angles to the 2 and 3 axes.
na
x1
cos θ 0 = 1
3 θ0 project
down
nc

Octahedral plane

Solid Mechanics Part II 257 Kelly

 Section 8.2

7. Use Eqns. 8.2.15 to derive Eqns. 8.2.17.

8. For the stress state of problem 2, calculate the octahedral normal stress and the
octahedral shear stress

Solid Mechanics Part II 258 Kelly

 Section 8.3

8.3 Yield Criteria in Three Dimensional Plasticity

The question now arises: a material yields at a stress level Y in a uniaxial tension test, but
when does it yield when subjected to a complex three-dimensional stress state?

Let us begin with a very general case: an anisotropic material with different yield
strengths in different directions. For example, consider the material shown in Fig. 8.3.1.
This is a composite material with long fibres along the x1 direction, giving it extra
strength in that direction – it will yield at a higher tension when pulled in the x1 direction
than when pulled in other directions.

2 1
x2 x2

x3

x1 fibre binder x1
bundles
(a ) (b)

Figure 8.3.1: an anisotropic material; (a) microstructural detail, (b) continuum

model

We can assume that yield will occur at a particle when some combination of the stress
components reaches some critical value, say when

F ( 11 ,  12 ,  13 ,  22 ,  23 ,  33 )  k . (8.3.1)

Here, F is some function of the 6 independent components of the stress tensor and k is
some material property which can be determined experimentally1.

Alternatively, it is very convenient to express yield criteria in terms of principal stresses.

Let us suppose that we know the principal stresses everywhere, ( 1 ,  2 ,  3 ) , Fig. 8.3.1.
Yield must depend somehow on the microstructure – on the orientation of the axes
x1 , x2 , x3 , but this information is not contained in the three numbers ( 1 ,  2 ,  3 ) . Thus we
express the yield criterion in terms of principal stresses in the form

F ( 1 ,  2 ,  3 , n i )  k (8.3.2)

1
F will no doubt also contain other parameters which need to be determined experimentally

Solid Mechanics Part II 259 Kelly

 Section 8.3

where n i represent the principal directions – these give the orientation of the principal
stresses relative to the material directions x1 , x2 , x3 .

If the material is isotropic, the response is independent of any material direction –

independent of any “direction” the stress acts in, and so the yield criterion can be
expressed in the simple form

F ( 1 ,  2 ,  3 )  k (8.3.3)

Further, since it should not matter which direction is labelled ‘1’, which ‘2’ and which
‘3’, F must be a symmetric function of the three principal stresses.

Alternatively, since the three principal invariants of stress are independent of material
orientation, one can write

F ( I1 , I 2 , I 3 )  k (8.3.4)

or, more usually,

F ( I1 , J 2 , J 3 )  k (8.3.5)

where J 2 , J 3 are the non-zero principal invariants of the deviatoric stress. With the
further restriction that the yield stress is independent of the hydrostatic stress, one has

F (J 2 , J 3 )  k (8.3.6)

8.3.1 The Tresca and Von Mises Yield Conditions

The two most commonly used and successful yield criteria for isotropic metallic materials
are the Tresca and Von Mises criteria.

The Tresca Yield Condition

The Tresca yield criterion states that a material will yield if the maximum shear stress
reaches some critical value, that is, Eqn. 8.3.3 takes the form

1 1 1 
max   1   2 ,  2   3 ,  3   1   k (8.3.7)
2 2 2 

The value of k can be obtained from a simple experiment. For example, in a tension test,
 1   0 ,  2   3  0 , and failure occurs when  0 reaches Y, the yield stress in tension.
It follows that

Y
k . (8.3.8)
2

Solid Mechanics Part II 260 Kelly

 Section 8.3

In a shear test,  1   ,  2  0,  3   , and failure occurs when  reaches  Y , the yield

stress of a material in pure shear, so that k   Y .

The Von Mises Yield Condition

The Von Mises criterion states that yield occurs when the principal stresses satisfy the
relation

 1   2 2   2   3 2   3   1 2
k (8.3.9)
6

Again, from a uniaxial tension test, one finds that the k in Eqn. 8.3.9 is

Y
k . (8.3.10)
3

Writing the Von Mises condition in terms of Y, one has

1
 1   2 2   2   3 2   3   1 2 Y (8.3.11)
2

The quantity on the left is called the Von Mises Stress, sometimes denoted by  VM .
When it reaches the yield stress in pure tension, the material begins to deform plastically.
In the shear test, one again finds that k   Y , the yield stress in pure shear.

Sometimes it is preferable to work with arbitrary stress components; for this purpose, the
Von Mises condition can be expressed as {▲Problem 2}

 11   22 2   22   33 2   33   11 2  6 122   232   312   6k 2 (8.3.12)

The piecewise linear nature of the Tresca yield condition is sometimes a theoretical
advantage over the quadratic Mises condition. However, the fact that in many problems
one often does not know which principal stress is the maximum and which is the
minimum causes difficulties when working with the Tresca criterion.

The Tresca and Von Mises Yield Criteria in terms of Invariants

From Eqn. 8.2.10 and 8.3.9, the Von Mises criterion can be expressed as

f (J 2 )  J 2  k 2  0 (8.3.13)

Note the relationship between J 2 and the octahedral shear stress, Eqn. 8.2.17; the Von
Mises criterion can be interpreted as predicting yield when the octahedral shear stress
reaches a critical value.

With  1   2   3 , the Tresca condition can be expressed as

Solid Mechanics Part II 261 Kelly

 Section 8.3

f ( J 2 , J 3 )  4 J 23  27 J 32  36k 2 J 22  96k 4 J 2  64k 6  0 (8.3.14)

but this expression is too cumbersome to be of much use.

Experiments of Taylor and Quinney

In order to test whether the Von Mises or Tresca criteria best modelled the real behaviour
of metals, G I Taylor & Quinney (1931), in a series of classic experiments, subjected a
number of thin-walled cylinders made of copper and steel to combined tension and
torsion, Fig. 8.3.2.

 


Figure 8.3.2: combined tension and torsion of a thin-walled tube

The cylinder wall is in a state of plane stress, with  11     12   and all other stress
components zero. The principal stresses corresponding to such a stress-state are (zero
and) {▲Problem 3}

1
2  1
4  2  2 (8.3.15)

and so Tresca's condition reduces to

2 2
    
 2  4 2  4k 2 or      1 (8.3.16)
Y  Y /2

The Mises condition reduces to {▲Problem 4}

2 2
    
  3  3k
2 2 2
or       1 (8.3.17)
Y  Y / 3 

Thus both models predict an elliptical yield locus in  ,  stress space, but with
different ratios of principal axes, Fig. 8.3.3. The origin in Fig. 8.3.3 corresponds to an
unstressed state. The horizontal axes refer to uniaxial tension in the absence of shear,
whereas the vertical axis refers to pure torsion in the absence of tension. When there is a
combination of  and  , one is off-axes. If the combination remains “inside” the yield
locus, the material remains elastic; if the combination is such that one reaches anywhere
along the locus, then plasticity ensues.

Solid Mechanics Part II 262 Kelly

 Section 8.3

Y  Y / 3 Mises
Y  Y / 2
Tresca

Y

Figure 8.3.3: the yield locus for a thin-walled tube in combined tension and torsion

Taylor and Quinney, by varying the amount of tension and torsion, found that their
measurements were closer to the Mises ellipse than the Tresca locus, a result which has
been repeatedly confirmed by other workers2.

2D Principal Stress Space

Fig. 8.3.3 gives a geometric interpretation of the Tresca and Von Mises yield criteria in
 ,  space. It is more usual to interpret yield criteria geometrically in a principal stress
space. The Taylor and Quinney tests are an example of plane stress, where one principal
stress is zero. Following the convention for plane stress, label now the two non-zero
principal stresses  1 and  2 , so that  3  0 (even if it is not the minimum principal
stress). The criteria can then be displayed in  1 ,  2  2D principal stress space. With
 3  0 , one has

Tresca: max  1   2 ,  2 ,  1   Y
(8.3.18)
Von Mises:    1 2    Y
2
1
2
2
2

These are plotted in Fig. 8.3.4. The Tresca criterion is a hexagon. The Von Mises
criterion is an ellipse with axes inclined at 45 0 to the principal axes, which can be seen
by expressing Eqn. 8.3.18b in the canonical form for an ellipse:

2
the maximum difference between the predicted stresses from the two criteria is about 15%. The two
criteria can therefore be made to agree to within  7.5% by choosing k to be half-way between Y / 2 and
Y/ 3

Solid Mechanics Part II 263 Kelly

 Section 8.3

 1 1 / 2   1 
 12   1 2   22   1  2  
 1 / 2 1   2 
1 / 2 1 / 2  1 / 2 0   1 / 2 1 / 2   1 
  1  2      
1 / 2 1 / 2   0 3 / 2   1 / 2 1 / 2   2 
2 2
1     2  3   2  1 
  1     Y2
2 2  2 2 
2 2
      2 
  1    1
 2Y   2 / 3Y 

where  1,  2 are coordinates along the new axes; the major axis is thus 2Y and the
minor axis is 2 / 3Y .

Some stress states are shown in the stress space: point A corresponds to a uniaxial tension,
B to a equi-biaxial tension and C to a pure shear  .

2
Y

B 

A 1
 2
Y
C  1
E 
 D

Figure 8.3.4: yield loci in 2D principal stress space

Again, points inside these loci represent an elastic stress state. Any combination of
principal stresses which push the point out to the yield loci results in plastic deformation.

8.3.2 Three Dimensional Principal Stress Space

The 2D principal stress space has limited use. For example, a stress state that might start
out two dimensional can develop into a fully three dimensional stress state as deformation
proceeds.

Solid Mechanics Part II 264 Kelly

 Section 8.3

In three dimensional principal stress space, one has a yield surface f  1 ,  2 ,  3   0 ,

Fig. 8.3.5 3. In this case, one can draw a line at equal angles to all three principal stress
axes, the space diagonal. Along the space diagonal  1   2   3 and so points on it are
in a state of hydrostatic stress.

Assume now, for the moment, that hydrostatic stress does not affect yield and consider
some arbitrary point A,  1 ,  2 ,  3   a, b, c  , on the yield surface, Fig. 8.3.5. A pure
hydrostatic stress  h can be superimposed on this stress state without affecting yield, so
any other point  1 ,  2 ,  3   a   h , b   h , c   h  will also be on the yield surface.
Examples of such points are shown at B, C and D, which are obtained from A by moving
along a line parallel to the space diagonal. The yield behaviour of the material is
therefore specified by a yield locus on a plane perpendicular to the space diagonal, and
the yield surface is generated by sliding this locus up and down the space diagonal.

3
hydrostatic stress

the π - plane
ρ
σ  A a, b, c 
 B
yield 2
locus s 
C

D
1 deviatoric stress

Figure 8.3.5: Yield locus/surface in three dimensional stress-space

The -plane

Any surface in stress space can be described by an equation of the form

f  1 ,  2 ,  3   const (8.3.19)

and a normal to this surface is the gradient vector

f f f
e1  e2  e3 (8.3.20)
 1  2  3

where e1 , e 2 , e 3 are unit vectors along the stress space axes. In particular, any plane
perpendicular to the space diagonal is described by the equation

3
as mentioned, one has a six dimensional stress space for an anisotropic material and this cannot be
visualised

Solid Mechanics Part II 265 Kelly

 Section 8.3

 1   2   3  const (8.3.21)

Without loss of generality, one can choose as a representative plane the π – plane, which
is defined by  1   2   3  0 . For example, the point  1 ,  2 ,  3   1,1,0  is on the π
– plane and, with yielding independent of hydrostatic stress, is equivalent to points in
principal stress space which differ by a hydrostatic stress, e.g. the points
2,0,1, 0,2,1 , etc.
The stress state at any point A represented by the vector σ   1 ,  2 ,  3  can be regarded
as the sum of the stress state at the corresponding point on the π – plane, D, represented
by the vector s  s1 , s 2 , s 3  together with a hydrostatic stress represented by the vector
ρ   m ,  m ,  m  :

 1 ,  2 ,  3    1   m ,  2   m ,  3   m    m ,  m ,  m  (8.3.22)

The components of the first term/vector on the right here sum to zero since it lies on the π
– plane, and this is the deviatoric stress, whilst the hydrostatic stress is
 m   1   2   3  / 3 .

Projected view of the -plane

Fig. 8.3.6a shows principal stress space and Fig. 8.3.6b shows the π – plane. The heavy
lines  1 ,  2 ,  3 in Fig. 8.3.6b represent the projections of the principal axes down onto
the   plane (so one is “looking down” the space diagonal). Some points, A, B, C in
stress space and their projections onto the   plane are also shown. Also shown is some
point D on the   plane. It should be kept in mind that the deviatoric stress vector s in
the projected view of Fig. 8.3.6b is in reality a three dimensional vector (see the
corresponding vector in Fig. 8.3.6a).

 3
3
D
space 
D diagonal s
A
s 
A

B
2 
  C
 1  2
B C

1
(a ) ( b)

Figure 8.3.6: Stress space; (a) principal stress space, (b) the π – plane

Solid Mechanics Part II 266 Kelly

 Section 8.3

Consider the more detailed Fig. 8.3.7 below. Point A here represents the stress state
2,1,0 , as indicated by the arrows in the figure. It can also be “reached” in different
ways, for example it represents 3,0,1 and 1,2,1 . These three stress states of course
differ by a hydrostatic stress. The actual   plane value for A is the one for which
 1   2   3  0 , i.e.  1 ,  2 ,  3   s1 , s 2 , s3    53 , 43 , 13  . Points B and C also
represent multiple stress states {▲Problem 7}.

 3

  2    1 
B


E
A

 1  2
 C D
bisector
  3 

Figure 8.3.7: the -plane

The bisectors of the principal plane projections, such as the dotted line in Fig. 8.3.7,
represent states of pure shear. For example, the   plane value for point D is 0,2,2  ,
corresponding to a pure shear in the  2   3 plane.

The dashed lines in Fig. 8.3.7 are helpful in that they allow us to plot and visualise stress
states easily. The distance between each dashed line along the directions of the projected
axes represents one unit of principal stress. Note, however, that these “units” are not
consistent with the actual magnitudes of the deviatoric vectors in the   plane. To
create a more complete picture, note first that a unit vector along the space diagonal is
 
n   13 , 13 , 13 , Fig. 8.3.8. The components of this normal are the direction cosines; for
example, a unit normal along the ‘1’ principal axis is e1  1,0,0 and so the angle  0
between the ‘1’ axis and the space diagonal is given by n   e1  cos  0  13 . From Fig.
8.3.8, the angle  between the ‘1’ axis and the   plane is given by cos   2
3
, and so
a length of  1 units gets projected down to a length s  s  2
3 1 .

For example, point E in Fig. 8.3.7 represents a pure shear  1 ,  2 ,  3   2,2,0  , which is
on the   plane. The length of the vector out to E in Fig. 8.3.7 is 2 3 “units”. To

Solid Mechanics Part II 267 Kelly

 Section 8.3

convert to actual magnitudes, multiply by 2

3
to get s  2 2 , which agrees with
s  s  s12  s 22  s32  2 2  2 2  2 2 .

space
diagonal
1
 1
n
n1

s 0


-plane
s 2
3 1

Figure 8.3.8: principal stress projected onto the -plane

Typical -plane Yield Loci

Consider next an arbitrary point ( a, b, c) on the   plane yield locus. If the material is
isotropic, the points (a, c, b) , (b, a, c) , (b, c, a ) , (c, a, b) and (c, b, a ) are also on the yield
locus. If one assumes the same yield behaviour in tension as in compression, e.g.
neglecting the Bauschinger effect, then so also are the points (  a, b, c) , ( a,c,b) ,
etc. Thus 1 point becomes 12 and one need only consider the yield locus in one 30o
sector of the   plane, the rest of the locus being generated through symmetry. One such
sector is shown in Fig. 8.3.9, the axes of symmetry being the three projected principal
axes and their (pure shear) bisectors.

yield locus

Figure 8.3.9: A typical sector of the yield locus

The Tresca and Von Mises Yield Loci in the -plane

The Tresca criterion, Eqn. 8.3.7, is a regular hexagon in the   plane as illustrated in Fig.
8.3.10. Which of the six sides of the locus is relevant depends on which of  1 ,  2 ,  3 is
the maximum and which is the minimum, and whether they are tensile or compressive.

Solid Mechanics Part II 268 Kelly

 Section 8.3

For example, yield at the pure shear  1  Y / 2,  2  0,  3  Y / 2 is indicated by point A

in the figure.

Point B represents yield under uniaxial tension,  1  Y . The distance oB , the

“magnitude” of the hexagon, is therefore 2
3
Y ; the corresponding point on the  – plane
is s1 , s 2 s3    23 Y , 13 Y , 13 Y  .

A criticism of the Tresca criterion is that there is a sudden change in the planes upon
which failure occurs upon a small change in stress at the sharp corners of the hexagon.

 3
3  2  Y  3  1  Y
  2    1 

1   2  Y o  2  1  Y

B
2
3 Y
 1  2
A
1   3  Y 2 3  Y

  3 

Figure 8.3.10: The Tresca criterion in the -plane

Consider now the Von Mises criterion. From Eqns. 8.3.10, 8.3.13, the criterion is
J 2  Y / 3 . From Eqn. 8.2.9, this can be re-written as

2
s12  s 22  s 32  Y (8.3.23)
3

Thus, the magnitude of the deviatoric stress vector is constant and one has a circular yield
locus with radius 23 Y  2k , which transcribes the Tresca hexagon, as illustrated in Fig.
8.3.11.

Solid Mechanics Part II 269 Kelly

 Section 8.3

 3

  2    1 
s 2
3 Y

Tresca

Von Mises

 1  2

  3 

Figure 8.3.11: The Von Mises criterion in the -plane

The yield surface is a circular cylinder with axis along the space diagonal, Fig. 8.3.12.
The Tresca surface is a similar hexagonal cylinder.

3 Tresca yield
surface

Von Mises
yield surface

2

plane stress
yield locus ( 3  0)
 - plane
(   2   3  0)
yield locus 1
1

Figure 8.3.12: The Von Mises and Tresca yield surfaces

8.3.3 Haigh-Westergaard Stress Space

Thus far, yield criteria have been described in terms of principal stresses ( 1 ,  2 ,  3 ) . It
is often convenient to work with  , s,  coordinates, Fig. 8.3.13; these cylindrical
coordinates are called Haigh-Westergaard coordinates. They are particularly useful for
describing and visualising geometrically pressure-dependent yield-criteria.

Solid Mechanics Part II 270 Kelly

 Section 8.3

The coordinates  , s  are simply the magnitudes of, respectively, the hydrostatic stress
vector ρ   m ,  m ,  m  and the deviatoric stress vector s  s1 , s 2 , s 3  . These are given
by ( s can be obtained from Eqn. 8.2.9)

  ρ  3 m  I 1 / 3 , s  s  2J 2 (8.3.24)

3  3

ρ
e s
2  1  2
 1 s  

1

(a ) ( b)

Figure 8.3.13: A point in stress space

 is measured from the  1 ( s1 ) axis in the   plane. To express  in terms of

invariants, consider a unit vector e in the   plane in the direction of the  1 axis; this is
the same vector n c considered in Fig. 8.2.4 in connection with the octahedral shear
stress, and it has coordinates  1 ,  2 ,  2    2
3 , 1
6
, 1
6
, Fig. 8.3.13. The angle  can
now be obtained from s  e  s cos  {▲Problem 9}:

3s1
cos   (8.3.25)
2 J2

Further manipulation leads to the relation {▲Problem 10}

3 3 J3
cos 3  (8.3.26)
2 J 23 / 2

Since J 2 and J 3 are invariant, it follows that cos 3 is also. Note that J 3 enters
through cos 3 , and does not appear in  or s; it is J 3 which makes the yield locus
in the  -plane non-circular.

From Eqn. 8.3.25 and Fig. 8.3.13b, the deviatoric stresses can be expressed in terms
of the Haigh-Westergaard coordinates through

Solid Mechanics Part II 271 Kelly

 Section 8.3

 s1   cos  
 s 2   2 J 2 cos2 / 3    (8.3.27)
s  3 cos2 / 3   
 3

The principal stresses and the Haigh-Westergaard coordinates can then be related
through {▲Problem 12}

 1     cos  
 2   1     2 s cos  2 / 3 (8.3.28)
  3   3 cos  2 / 3
 3    

In terms of the Haigh-Westergaard coordinates, the yield criteria are

Von Mises: f ( s )  12 s 2  k  0
(8.3.29)
Tresca f ( s,  )  2 s sin   
3
 Y  0

8.3.4 Pressure Dependent Yield Criteria

The Tresca and Von Mises criteria are independent of hydrostatic pressure and are
suitable for the modelling of plasticity in metals. For materials such as rock, soils and
concrete, however, there is a strong dependence on the hydrostatic pressure.

The Drucker-Prager Criteria

The Drucker-Prager criterion is a simple modification of the Von Mises criterion,

whereby the hydrostatic-dependent first invariant I 1 is introduced to the Von Mises Eqn.
8.3.13:

f ( I 1 , J 2 )  I 1  J 2  k  0 (8.3.30)

with  is a new material parameter. On the  – plane, I 1  0 , and so the yield locus
there is as for the Von Mises criterion, a circle of radius 2k , Fig. 8.3.14a. Off the  –
plane, the yield locus remains circular but the radius changes. When there is a state of
pure hydrostatic stress, the magnitude of the hydrostatic stress vector is {▲Problem 13}
  ρ  k / 3 , with s  s  0 . For large pressures,  1   2   3  0 , the I 1 term in
Eqn. 8.3.30 allows for large deviatoric stresses. This effect is shown in the meridian
plane in Fig. 8.3.14b, that is, the (  , s) plane which includes the  1 axis.

Solid Mechanics Part II 272 Kelly

 Section 8.3

s
 3
 - plane

s 2k
meridian
plane k 
 2 3
 1

(a ) ( b)

Figure 8.3.14: The Drucker-Prager criterion; (a) the -plane, (b) the Meridian
Plane

The Drucker-Prager surface is a right-circular cone with apex at   k / 3 , Fig. 8.3.15.

Note that the plane stress locus, where the cone intersects the  3  0 plane, is an ellipse,
but whose centre is off-axis, at some ( 1  0,  2  0) .

3

s
ρ

2

 1

Figure 8.3.15: The Drucker-Prager yield surface

In terms of the Haigh-Westergaard coordinates, the yield criterion is

f (  , s )  6  s  2k  0 (8.3.31)

The Mohr Coulomb Criteria

The Mohr-Coulomb criterion is based on Coulomb’s 1773 friction equation, which can be
expressed in the form

  c   n tan  (8.3.32)

Solid Mechanics Part II 273 Kelly

 Section 8.3

where c,  are material constants; c is called the cohesion4 and  is called the angle of
internal friction.  and  n are the shear and normal stresses acting on the plane where
failure occurs (through a shearing effect), Fig. 8.3.16, with tan  playing the role of a
coefficient of friction. The criterion states that the larger the pressure   n , the more
shear the material can sustain. Note that the Mohr-Coulomb criterion can be considered
to be a generalised version of the Tresca criterion, since it reduces to Tresca’s when
  0 with c  k .

 n

Figure 8.3.16: Coulomb friction over a plane

This criterion not only includes a hydrostatic pressure effect, but also allows for different
yield behaviours in tension and in compression. Maintaining isotropy, there will now be
three lines of symmetry in any deviatoric plane, and a typical sector of the yield locus is
as shown in Fig. 8.3.17 (compare with Fig. 8.3.9)

yield locus

Figure 8.3.17: A typical sector of the yield locus for an isotropic material with
different yield behaviour in tension and compression

Given values of c and  , one can draw the failure locus (lines) of the Mohr-Coulomb
criterion in ( n , ) stress space, with intercepts   c and slopes  tan  , Fig. 8.3.18.
Given some stress state  1   2   3 , a Mohr stress circle can be drawn also in ( n ,  )
space (see §7.2.6). When the stress state is such that this circle reaches out and touches
the failure lines, yield occurs.

4
c  0 corresponds to a cohesionless material such as sand or gravel, which has no strength in tension

Solid Mechanics Part II 274 Kelly

 Section 8.3


failure line


 
c
 3 
2 
1 c n


Figure 8.3.18: Mohr-Coulomb failure criterion

From Fig. 8.3.18, and noting that the large Mohr circle has centre  12  1   3 ,0 and
radius 1
2
 1   3  , one has
1   3
 cos 
2 (8.3.33)
1   3 1   3
n   sin 
2 2

Thus the Mohr-Coulomb criterion in terms of principal stresses is

 1   3   2c cos    1   3 sin  (8.3.34)

The strength of the Mohr-Coulomb material in uniaxial tension, f Yt , and in uniaxial

compression, f Yc , are thus

2c cos  2c cos 
f Yt  , f Yc  (8.3.35)
1  sin  1  sin 

In terms of the Haigh-Westergaard coordinates, the yield criterion is

f (  , s,  )  2  sin   3s sin   3   s cos  3 sin   6c cos   0 (8.3.36)

The Mohr-Coulomb yield surface in the  – plane and meridian plane are displayed
in Fig. 8.3.19. In the  – plane one has an irregular hexagon which can be
constructed from two lengths: the magnitude of the deviatoric stress in uniaxial
tension at yield, s t 0 , and the corresponding (larger) value in compression, s c 0 ; these
are given by:

6 f Yc 1  sin   6 f Yc 1  sin  
st 0  , sc0  (8.3.37)
3  sin  3  sin 

Solid Mechanics Part II 275 Kelly

 Section 8.3

In the meridian plane, the failure surface cuts the s  0 axis at   3c cot 
{▲Problem 14}.

s
 3
 - plane

st 0 st 0
meridian
plane 
sc 0 sc 0
 1  2 3c cot 

(a ) ( b)

Figure 8.3.19: The Mohr-Coulomb criterion; (a) the -plane, (b) the Meridian Plane

The Mohr-Coulomb surface is thus an irregular hexagonal pyramid, Fig. 8.3.20.

3

 2

 1

Figure 8.3.20: The Mohr-Coulomb yield surface

By adjusting the material parameters  , k , c,  , the Drucker-Prager cone can be

made to match the Mohr-Coulomb hexagon, either inscribing it at the minor vertices,
or circumscribing it at the major vertices, Fig. 8.3.21.

Solid Mechanics Part II 276 Kelly

 Section 8.3

 3

 1  2

Figure 8.3.21: The Mohr-Coulomb and Drucker-Prager criteria matched in the -

plane

Capped Yield Surfaces

The Mohr-Coulomb and Drucker-Prager surfaces are open in that a pure hydrostatic
pressure can be applied without affecting yield. For many geomaterials, however, for
example soils, a large enough hydrostatic pressure will induce permanent deformation. In
these cases, a closed (capped) yield surface is more appropriate, for example the one
illustrated in Fig. 8.3.22.

3

 2

 1

Figure 8.3.22: a capped yield surface

An example is the modified Cam-Clay criterion:


3 J 2   13 I 1 M 2 2 p c  13 I 1  or s   3 2 3 M 2 2 p c  1
3

 ,  0 (8.3.38)

with M and p c material constants. In terms of the standard geomechanics notation, it

reads

q 2  M 2 p 2 p c  p  (8.3.39)

where

Solid Mechanics Part II 277 Kelly

 Section 8.3

1 1 3
p   I1   , q  3J 2  s (8.3.40)
3 3 2

The modified Cam-Clay locus in the meridian plane is shown in Fig. 8.3.23. Since s
is constant for any given  , the locus in planes parallel to the  - plane are circles.
The material parameter p c is called the critical state pressure, and is the pressure
which carries the maximum deviatoric stress. M is the slope of the dotted line
shown in Fig. 8.3.23, known as the critical state line.
q
 - plane
Critical state line

pc M

pc p

2 pc

Figure 8.3.23: The modified Cam-Clay criterion in the Meridian Plane

8.3.5 Anisotropy

Many materials will display anisotropy. For example metals which have been processed
by rolling will have characteristic material directions, the tensile yield stress in the
direction of rolling being typically 15% greater than that in the transverse direction. The
form of anisotropy exhibited by rolled sheets is such that the material properties are
symmetric about three mutually orthogonal planes. The lines of intersection of these
planes form an orthogonal set of axes known as the principal axes of anisotropy. The
axes are (a) in the rolling direction, (b) normal to the sheet, (c) in the plane of the sheet
but normal to rolling direction. This form of anisotropy is called orthotropy (see Part I,
§6.2.2). Hill (1948) proposed a yield condition for such a material which is a natural
generalisation of the Mises condition:

f ( ij )  F  22   33   G  33   11   H  11   22 
2 2 2

(8.3.41)
 2 L 23
2
 2 M  312  2 N 122  1  0

where F, G, H, L, M, N are material constants. One needs to carry out 6 tests: uniaxial
tests in the three coordinate directions to find the uniaxial yield strengths
( Y ) x , ( Y ) y , ( Y ) z , and shear tests to find the shear strengths ( Y ) xy , ( Y ) yz , ( Y ) zx . For a
uniaxial test in the x direction, Eqn. 8.3.41 reduces to G  H  1/ ( Y ) 2x . By considering
the other simple uniaxial and shear tests, one can solve for the material parameters:

Solid Mechanics Part II 278 Kelly

 Section 8.3

1 1 1 1  1 1
F     L
2   Y  2  Y  2  Y  2  2  Y 2
 y z x  yz

1 1 1 1  1 1
G     M (8.3.42)
2   Y  2   Y  2  Y  2  2  Y 2
 z x y  zx

1 1 1 1  1 1
H     N
2   Y  2  Y  2  Y  2  2  Y 2
 x y z  xy

The criterion reduces to the Mises condition 8.3.12 when

L M N 1
F GH     2 (8.3.43)
3 3 3 6k

The 1, 2, 3 axes of reference in Eqn. 8.3.41 are the principal axes of anisotropy. The form
appropriate for a general choice of axes can be derived by using the usual stress
transformation formulae. It is complicated and involves cross-terms such as  11 23 , etc.

8.3.6 Problems
1. A material is to be loaded to a stress state
 50  30 0
 
 ij   30 90 0 MPa
 0 0 0
What should be the minimum uniaxial yield stress of the material so that it does not
fail, according to the
(a) Tresca criterian
(b) Von Mises criterion
What do the theories predict when the yield stress of the material is 80MPa?

2. Use Eqn. 8.2.6, J 2  s122  s 23

2
 s31
2
 s11 s 22  s 22 s33  s33 s11  to derive Eqn. 8.3.12,
 11   22 2   22   33 2   33   11 2  6 122   232   312   6k 2 , for the Von
Mises criterion.

2
 11   22     22 
3. Use the plane stress principal stress formula  1, 2    11    12
2

2  2 
to derive Eqn. 8.3.15 for the Taylor-Quinney tests.

4. Derive Eqn. 8.3.17 for the Taylor-Quinney tests.

5. Describe the states of stress represented by the points D and E in Fig. 8.3.4. (The
complete stress states can be visualised with the help of Mohr’s circles of stress, Fig.
7.2.17.)

Solid Mechanics Part II 279 Kelly

 Section 8.3

6. Suppose that, in the Taylor and Quinney tension-torsion tests, one has   Y / 2 and
  3Y / 4 . Plot this stress state in the 2D principal stress state, Fig. 8.3.4. (Use
Eqn. 8.3.15 to evaluate the principal stresses.) Keeping now the normal stress at
  Y / 2 , what value can the shear stress be increased to before the material yields,
according to the von Mises criterion?

7. What are the   plane principal stress values for the points B and C in Fig. 8.3.7?

8. Sketch on the   plane Fig. 8.3.7 a line corresponding to  1   2 and also a region
corresponding to  1   2  0   3
3s1
9. Using the relation s  e  s cos  and s 2  s 3   s1 , derive Eqn. 8.3.25, cos   .
2 J2

10. Using the trigonometric relation cos 3  4 cos 3   3 cos and Eqn. 8.3.25,

cos  
3s1
, show that cos 3 
3 3s1 2
 
s1  J 2 . Then using the relations 8.2.6,
2 J2 2 J 23 / 2
3 3 J3
J 2  s1 s 2  s 2 s 3  s 3 s1  , with J 1  0 , derive Eqn. 8.3.26, cos 3 
2 J 23 / 2

11. Consider the following stress states. For each one, evaluate the space coordinates
(  , s, ) and plot in the   plane (see Fig. 8.3.13b):
(a) triaxial tension:  1  T1   2   3  T2  0
(b) triaxial compression:  3   p1   1   2   p 2  0 (this is an important test for
geomaterials, which are dependent on the hydrostatic pressure)
(c) a pure shear  xy   :  1   ,  2  0,  3  
(d) a pure shear  xy   in the presence of hydrostatic pressure p:
 1   p   ,  2   p,  3   p   , i.e.  1   2   2   3

12. Use relations 8.3.24,   I 1 / 3 , s  2 J 2 and Eqns. 8.3.27 to derive Eqns. 8.3.28.

13. Show that the magnitude of the hydrostatic stress vector is   ρ  k / 3 for the
Drucker-Prager yield criterion when the deviatoric stress is zero

14. Show that the magnitude of the hydrostatic stress vector is   3c cot  for the
Mohr-Coulomb yield criterion when the deviatoric stress is zero

15. Show that, for a Mohr-Coulomb material, sin   (r  1) /(r  1) , where r  f Yc / f Yt is

the compressive to tensile strength ratio

16. A sample of concrete is subjected to a stress  11   22   p,  33   Ap where the

constant A  1 . Using the Mohr-Coulomb criterion and the result of Problem 15,
show that the material will not fail provided A  f Yc / p  r

Solid Mechanics Part II 280 Kelly