 2001, W. E. Haisler
Chapter 9: Constitutive Relations
1
Constitutive Relations (Chapter 9)
“ceiiinosssttuv”  an anagram by Robert Hooke (1676)
ut tensio sic vis  the explanation by Robert Hooke (1678)
“as the tension so the displacement”
It is physically observed that the deformation and motion
of the particles in a continuous body are in some way related
to the forces applied to the body. Alternately we can say
that the strain (a measure of deformation per unit length) is
related to the stress (force per unit area) applied to the body.
This relationship essentially provides specific information
about the characteristics of the specific material that the
body is made of.
 2001, W. E. Haisler
Chapter 9: Constitutive Relations
2
In general, the 9 components of the stress tensor [] can be be
related to the 9 components of the strain tensor [E] through 81
material parameters. Due to the symmetry of both tensors
( yx   xy , etc), there are only 6 independent stresses and strains
and the number of material parameters is reduced to 36. For a
linear relationship between stress and strain, one can thus write:
 xx  C11 xx  C12 yy  C13 zz  C14 xy  C15 xz  C16 yz
 yy  C21 xx  C22 yy  C23 zz  C24 xy  C25 xz  C26 yz
.....
 xz  C61 xx  C62 yy  C63 zz  C64 xy  C65 xz  C66 yz
or
 2001, W. E. Haisler
3
Chapter 9: Constitutive Relations
 xx   C11
 yy  C
   21
 zz  C31
   C
 xy   41
 xz  C51
 yz  C61
C12
C22
C32
C42
C52
C62
C13 C14
C23 C24
C33 C34
C43 C44
C53 C54
C63
C64
C15
C25
C35
C45
C55
C65
C16   xx 
C26   yy 


C36   zz 
 
C46   xy
 
C56   xz 
C66  
 yz 
In general, Cij are functions of position x,y,z.
For a homogeneous material, Cij are independent of position
x,y,z.
For an isotropic material, Cij are independent of the orientation
of the coordinate axes (i.e., properties are same in all directions).
 2001, W. E. Haisler
Chapter 9: Constitutive Relations
4
For an orthotropic material, Cij are different in each of the 3
coordinate directions.
For the simplest solid material (linear, isotropic), one can deduce
from physical observation that there are only two independent
material constants that relate all stress and strain components.
These are
E Young’s modulus (slope of uniaxial stress-strain curve)
 Poisson’s ratio (ratio of contraction to extension strains)
These properties are typically measured in a uniaxial tensile test.
 2001, W. E. Haisler
5
Chapter 9: Constitutive Relations
Consider a test specimen with crosssectional area A and applied load P in
the x direction as shown below.
Assume a gauge length of L.
P
A
x
During the tensile test, we observe
that the length changes from L to L*
z
and the width decreases from W to
W*. The axial stress and strain in the
axial (x) direction are defined to be
 xx  P / A
 xx  L / L  ( L *  L) / L
and
L*
L
y
W*
W
t
P
A=W*t
The strain in the transverse (y) direction (due to the axial load) is
 yy  W / W  (W * W ) / W
 2001, W. E. Haisler
6
Chapter 9: Constitutive Relations
If we plot axial stress vs. axial strain and transverse strain vs.
axial strain, we obtain the following:
 yy  W / W
 xx  P / A
   yy /  xx   slope
E  slope
1
 xx  L / L
 xx  L / L
 xx  E xx ,where E is a material constant
 xx  L / L  ( L *  L) / L and  yy  W / W  (W * W ) / W

 yy   xx
or
 yy /  xx  
(a material constant)
 2001, W. E. Haisler
Chapter 9: Constitutive Relations
7
Similarly, the strain in the z direction is:  zz   xx
Combining the relation defining Poisson's ratio and Hooke's
Law, we can write the transverse strain  yy in terms of axial
stress  xx :
 yy   xx  ( / E ) xx
The strain in the z direction during the application of load P
(which causes stress  xx ) is also seen to be
 zz   xx  ( / E ) xx
E is called Young's modulus of elasticity, and
 is called Poisson's ratio.
 2001, W. E. Haisler
8
Chapter 9: Constitutive Relations
Consider a test where we apply normal tractions (stresses) in
the x, y and z directions simultaneously. For a linear
material, we can think of this as three separate problems:
 yy
 yy
 zz
 xx
 xx  xx
=
 zz
 yy
 xx
 zz
+
+
 yy
 zz
xx = normal strain in x direction due to  xx
+ normal strain in x direction due to  yy
+ normal strain in x direction due to  zz
From Hooke's Law, the strain in the x direction for each case is:
 2001, W. E. Haisler
9
Chapter 9: Constitutive Relations
 xx   xx due to  xx   xx due to  yy   xx due to  zz

1
 xx
E

  yy
E

  zz
E
or
1
 xx  [  (   )]
yy
zz
E xx
The stress in the x direction increases the strain while the
transverse stresses causes a contraction (decrease).
Doing similar experiments in the y and z directions gives:
1
 yy  [  (   )]
xx
zz
E yy
1
 zz  [  (   )]
xx
yy
E zz
 2001, W. E. Haisler
Chapter 9: Constitutive Relations
10
Experiments with shear tractions will show that an xy shear
stress in the xy plane produces only xy shear strain in the xy
plane and NO extensional strain* (e.g., the shear strain is
uncoupled from the extensional strain). Thus, we obtain the
following experimental observations for the shear strains:
1 
1 
1 
,  xz  (
, and  yz  (
 xy  (
)
)
)
xy
xz
yz
E
E
E
The term E /(1   )  2G defines a shear modulus, G, relating
shear strain and shear stress (similar to Young’s modulus, E,
for extensional strain). G  E /[2(1   )].
------------------------------
* Keep in mind that if one calculates shear strains at some angle
 from the x-axis (Mohr’s circle), you will obtain normal strains!
 2001, W. E. Haisler
Chapter 9: Constitutive Relations
11
For a linear, elastic, isotropic material, we can superimpose
all of the six experiments to obtain the constitutive relations:
1
 xx  [  (   )]
yy
zz
E xx
1
 yy  [  (   )]
xx
zz
E yy
1
 zz  [  (   )]
xx
yy
E zz
1 
 xy  (
)
xy
E
1 
 xz  (
)
xz
E
1 
 yz  (
)
yz
E
 2001, W. E. Haisler
Chapter 9: Constitutive Relations
The above may be solved for the stresses in terms of the strains to obtain:
E
 
[(1  )     ]
xx (1   )(1  2 )
xx
yy
zz
E
 
[  (1  )   ]
yy (1   )(1  2 ) xx
yy
zz
E
 
[    (1  ) ]
zz (1   )(1  2 ) xx
yy
zz
E
 

xy (1   ) xy
E
 

xz (1   ) xz
E
 

yz (1   ) yz
12
 2001, W. E. Haisler
Chapter 9: Constitutive Relations
13
The constitutive equation for the linear elastic isotropic
material can also be written in matrix notation as:


 1 

0 0 0

1  2 1  2 1  2
 xx 
 xx 



 
 

1 



0 0 0  yy 
 yy 
1  2 1  2 1  2

 zz 
  zz 
E  


1 
  
 


0
0
0
xy 
 xy  (1   ) 1  2 1  2 1  2


 xz 
 0
0
0
1 0 0   xz 


 


0
0
0 1 0   yz 
 yz 
 0
 0
0
0
0 0 1 
Note that only 2 material constants (E and  ) are required.
The material matrix [C] is symmetric!
 2001, W. E. Haisler
14
Chapter 9: Constitutive Relations
Important Note: Hooke's Law in frequently written in terms
of the engineering shear strain  . Recall, that the
engineering shear strain is defined to be twice that of the
tensor shear strain; for example,  xy  2 xy . Hence the shear
stress in terms of engineering shear strain becomes:
E
E
E
 
 
( / 2) 
  G
xy (1   ) xy (1   ) xy
xy
2(1   ) xy
Similarly,

xz
 G
xz
, and 
where
E
G
 shear modulus
2(1   )
yz
 G
yz
 2001, W. E. Haisler
Chapter 9: Constitutive Relations
15
For a linear, elastic, homogeneous material we note that the
extensional strains and stresses are uncoupled from the shear
strains and stresses.
Thermal Strain. Experimentally we observe that a
temperature increase, T, produces a uniform expansion but
no shear and the expansion is proportional to a material
constant  (coefficient of thermal expansion). The
additional strain due to heating is thus  xx   yy   zz  T .
1
1 
 xx  [  (   )]  T
 xy  (
)
yy
zz
xy
E xx
E
1
1 
 yy  [  (   )]  T
 xz  (
)
xx
zz
xz
E yy
E
1
1 
 zz  [  (   )]  T
 yz  (
)
xx
yy
yz
E zz
E
 2001, W. E. Haisler
Chapter 9: Constitutive Relations
16
It should be noted that the first term in the extensional strain
terms above (the [ ] term) is due to elastic behavior of the
material (ie, it has Young’s modulus in it). The second part
is due to thermal strain. We can separate the total strain into
elastic and thermal strains components:
elastic   thermal
 total


xx
xx
elastic   thermal
 total


yy
yy
elastic   thermal
 total


zz
zz
1
elastic

 [  (   )]
xx
yy
zz . yy and zz similar.
where
E xx
 thermal  T
 2001, W. E. Haisler
Chapter 9: Constitutive Relations
17
Recall, in the above, that shear strains have no thermal
component.
The constitutive relations for linear elastic isotropic material may
be written in a compact matrix notation as follows:
[ ] 
E [ ]   (tr[ ])[ I ]

(1   ) 
1  2
where tr  trace of the matrix, [I] = identity matrix. Similarly,
for the strain-stress equations
[ ]  1 (1   )[]  (tr[])[ I ]
E