Chapter 9
CONSTITUTIVE RELATIONS
FOR LINEAR ELASTIC SOLIDS
Figure 9.1: Hooke memorial window, St. Helen’s, Bishopsgate, City of London
211
212
9.1
CHAPTER 9. CONSTITUTIVE RELATIONS FOR LINEAR ELASTIC SOLIDS
Mechanical Constitutive Equations
Recall that in Chapters 2, 3 and 8 we briefly introduced the concept of a constitutive equation,
which generally relates kinetic variables to kinematic variables in the application of interest. With
respect to the application of the analysis of mechanical deformations in solids, the kinetic variable
is the stress tensor, σ, whereas the kinematic variables are the displacements ux , uy , uz , and the
strain tensor, ε which includes derivatives (sometimes called gradients) of the displacements. Since
it is generally observed that rigid body displacements do not induce stresses, the displacement field
ux , uy , uz , will not enter into a mechanical constitutive equation. Thus, the constitutive equations
will in general relate stress, σ, to strains, ε, and temperature T . In 1660, Robert Hooke observed
that for a broad class of solid materials called linear elastic (or Hookean), this relationship may be
described by a linear relationship. Hooke originally considered the test of a uniaxial body with a
force (stress) applied only in one direction and measured the corresponding elongation (strain) to
obtain:
σyy
σyy
y
x
E
A=cross-sectional
area
yy
= Eε yy
ε yy
σyy
Figure 9.2: Stress-Strain Curve for Linear Elastic Material
For a general three-dimensional state of stress, there are 6 independent stresses and 6 independent
strains; therefore, the linear relationship between stress and strain can be written in matrix form as:
σ = [C] ε
(9.1)
where [C] is a 6 × 6 matrix of elastic constants that must be determined from experiments. In
expanded form, these 6 equations become:
σxx
= C11 εxx + C12 εyy + C13 εzz + C14 εyz + C15 εzx + C16 εxy
σyy
= C21 εxx + C22 εyy + C23 εzz + C24 εyz + C25 εzx + C26 εxy
σzz
= C31 εxx + C32 εyy + C33 εzz + C34 εyz + C35 εzx + C36 εxy
σyz
σzx
= C41 εxx + C42 εyy + C43 εzz + C44 εyz + C45 εzx + C46 εxy
σxy
= C61 εxx + C62 εyy + C63 εzz + C64 εyz + C65 εzx + C66 εxy
(9.2)
= C51 εxx + C52 εyy + C53 εzz + C54 εyz + C55 εzx + C56 εxy
It is interesting to note that Robert Hooke first proposed the above “law” publicly in an anagram
at Hampton Court (1676) given by the group of letters:
ceiiinosssttuv.
9.1. MECHANICAL CONSTITUTIVE EQUATIONS
213
In 1678 he explained the anagram to be
“Ut tensio sic vis,”
which is Latin meaning “as the tension so the displacement” or in English “the force is proportional
to the displacement.” Students may recall that during this time period, science and scientific writing
was criticized and hence Hooke thought it necessary to discretely disclose his scientific finding with
an anagram.
Note that in the previous chapters stress and strain were represented as (3 × 3) matrices. It
is convenient, however, here to represent them as (6 × 1) column vectors since they have only
6 independent components (stress due to conservation of angular momentum and strain by its
definition). We then write

σxx




σyy



σzz
{σ} =
σ

yz



σ

zx


σxy















,

εxx




εyy



εzz
{ε} =
ε

yz



ε

zx


εxy















By adopting this representation for σ and ε, their linear relationship (9.2) can be easily written
in matrix form:

σxx




σyy



σzz
σ

yz



σ

zx


σxy












=










C11
C21
C31
C41
C51
C61
C12
C22
C32
C42
C52
C62
C13
C23
C33
C43
C53
C63
C14
C24
C34
C44
C54
C64
C15
C25
C35
C45
C55
C65
C16
C26
C36
C46
C56
C66
















εxx
εyy
εzz
εyz
εzx
εxy















(9.3)
• If a material is homogeneous then the constants (Cij , i = 1, ..., 6 and j = 1, ..., 6) are all
independent of x, y, z for any time, t.
• If a material is isotropic, then for a given material point, Cij are independent of the orientation
of the coordinate system (i.e., the material properties are the same in all directions).
• If a material is orthotropic, then for a given material point, Cij can be defined in terms of
properties in three orthogonal coordinate directions.
• If a material is anisotropic, then for a given material point, Cij are different for all orientations
of the coordinate system.
In order to determine the material constants in equation (9.3), consider a uniaxial tensile test
using a test specimen of linear elastic isotropic material with cross-sectional area A and subjected
to a uniaxilly applied load F in the axial (y) direction as shown below. The cross-section may be
any shape but generally a rectangular or cylindrical shape is chosen. For a rectangular specimen,
assume a width W and thickness t so that the cross-sectional area is A = W t. Assume a small gauge
length of L for which the axial deformation will be measured during the load application.
During the uniaxial tensile test, we observe that the gauge length changes from L to L∗
and the gauge width decreases from W to W ∗ . We also observe a decrease (contraction) in the z
dimension. We further observe no change in angular orientation of the vertical or horizontal elements
and conclude that for uniaxial loading, no shear strains are produced. This leads us to postulate
the following strain state: εxx , εyy , εzz = 0, εxy , εyz , εxz = 0. The axial stress and strain in the axial
(y) direction are defined to be
214
CHAPTER 9. CONSTITUTIVE RELATIONS FOR LINEAR ELASTIC SOLIDS
σyy
after deformation
F
y
L
x
z
A=cross-sectional
area
L*
t
F
σyy
A=Wt
W*
W
test specimen
Figure 9.3: Experimental Measurement of Axial and Transverse Deformation
σyy
=
εyy
=
F
A
∆L
(L∗ − L)
=
L
L
The strain the in the transverse (x) direction due to the axial load is
εxx =
∆W
(W ∗ − W )
=
W
W
If we plot axial stress vs. axial strain and transverse strain vs. axial strain, we obtain the
following two plots:
σ yy =
F
A
εxx =
E
σ yy = Eεyy
∆W
W
εxx = −νεyy
1
εyy =
∆L
L
1
−ν
εyy
Figure 9.4: Experimental Results for Axial Stress vs. Axial Strain & Transverse Strain vs. Axial
Strain
From these two plots, we can write σyy = Eεyy and εxx = −νεyy for the uniaxial tension test.
Consequently, we may define the following two material constants from this single uniaxial test:
• E = slope of the uniaxial σyy vs. εyy curve = a material constant called Young’s modulus
9.1. MECHANICAL CONSTITUTIVE EQUATIONS
215
• ν = − εεxx
= negative ratio of the strain normal to the direction of loading over the strain in
yy
the loading direction = a material constant called Poisson’s ratio
If the transverse strain were measured in the z direction, we would find the same ratio for transverse
zz
to axial strain: ν = − εεyy
.
Combining these equations, we can write the two transverse
entirely in terms of the axial
σ strains
σ
stress σyy : εxx = −νεyy = −ν Eyy and εzz = −νεyy = −ν Eyy .
In order to obtain a complete description of three-dimensional constitutive behavior, consider
a test where we apply normal tractions (stresses) in the x, y and z directions simultaneously and
measure the strain only in the x direction. For a linear material response, we may use the principle
of linear superposition and consider three separate cases as shown below:
σ yy
σ yy
σ zz
σ xx σ xx
σ xx
σ xx
=
σ zz
σ zz
+
+
σ yy
σ yy
σ zz
Figure 9.5: Experimental Test with all Components of Normal Stresses Applied
εxx
= normal strain in x direction due to σxx
+ normal strain in x direction due to σyy
+ normal strain in x direction due to σzz
1
ν
ν
σxx − σyy − σzz
=
E
E
E
or
εxx =
1
[σxx − ν(σyy + σzz )]
E
(9.4)
The stress in the x direction increases the strain in the x direction while the transverse stresses
causes a contraction (decrease in εxx ).
Doing similar experiments in the y and z directions gives:
εyy
=
εzz
=
1
[σyy − ν(σxx + σzz )]
E
1
[σzz − ν(σxx + σyy )]
E
(9.5)
Experiments with shear tractions will show that a shear stress σxy in the x-y plane produces only
shear strain εxy in the x-y plane for a state of pure shear loading (i.e., no normal strain is observed
so that the shear strain is uncoupled from the normal strain).1 Thus, we obtain the following
1 Keep in mind that even for the case of pure shear, if one calculates shear stresses (or strains) at some angle θ
from the x-axis (Mohr’s circle), one may obtain non-zero normal stresses (or strains) for the off-axis planes.
216
CHAPTER 9. CONSTITUTIVE RELATIONS FOR LINEAR ELASTIC SOLIDS
experimental observations for the shear strains:
εxy
=
εxz
=
εyz
=
1+ν
σxy
E
1+ν
σxz
E
1+ν
σyz
E
(9.6)
Combining equations (9.4), (9.5) and (9.6), Hooke’s law for a linear elastic isotropic solid with a
three-dimensional stress state becomes:
Hooke’s Law for a Linear Elastic Isotropic Solid
εxx
=
εyy
=
εzz
=
εxy
=
εxz
=
εyz
=
1
[σxx − ν(σyy + σzz )]
E
1
[σyy − ν(σxx + σzz )]
E
1
[σzz − ν(σxx + σyy )]
E
1+ν
σxy
E
1+ν
σxz
E
1+ν
σyz
E
(9.7)
where E = Young’s modulus and ν = Poisson’s ratio.
It should be noted that in materials that undergo permanent deformation, the above model is
not accurate (such as metals beyond their yield point, or polymers that flow). A typical uniaxial
stress-strain curve for a ductile metal is shown below:
σ xx
yield stress
E
ultimate stress
failure stress
εxx
Figure 9.6: Typical Stress-Strain Curve for Ductile Metal
An algebraic inversion of the strain-stress relationship (9.7) provides the following relationship
of stress in terms of strain:
9.1. MECHANICAL CONSTITUTIVE EQUATIONS

E
{σ} =
1+ν







1−ν
1−2ν
ν
1−2ν
ν
1−2ν
ν
1−2ν
1−ν
1−2ν
ν
1−2ν
ν
1−2ν
ν
1−2ν
1−ν
1−2ν
0
0
0
0
0
0
0
0
0
217

0 0 0 
εxx



0 0 0 
εyy



εzz
0 0 0 


ε
1 0 0 
yz


εzx

0 1 0 


εxy
0 0 1








(9.8)







or,
Hooke’s Law for a Linear Elastic Isotropic Solid

σxx




σyy



σzz
{σ} =
σ

xy



σ

xz


σyz
























=








E
(1+ν)(1−2ν)
E
(1+ν)(1−2ν)
E
(1+ν)(1−2ν)
[(1 − ν)εxx + νεyy + νεzz ]
[νεxx + (1 − ν)εyy − νεzz ]
[νεxx + νεyy + (1 − ν)εzz ]
E
1+ν εxy
E
1+ν εxz
E
1+ν εyz

















(9.9)
where E = Young’s modulus and ν = Poisson’s ratio.
E
The term (1+ν)
≡ 2G defines a shear modulus, G, relating shear strain and shear stress (similar
to Young’s modulus, E, for extensional strain). Thus, the shear modulus is given by:
G=
E
2(1 + ν)
(9.10)
Note that the shear strain εxy is related to engineering shear strain γxy by γxy = 2εxy =
σxy
2( 1+ν
E )σxy = G so that σxy = Gγxy = 2Gεxy .
σ xy
σ xy = Gγ xy = G2εxy
G = shear modulus
G=
E
2(1+ν)
γ xy
Figure 9.7: Experimental Results for Shear Stress vs. Engineering Shear Strain
Note that G is defined in terms of E and ν and consequently G is not a new material property.
Thus, for a homogeneous linear elastic isotropic solid, we conclude that only two material properties
(Young’s modulus, E, and Poisson’s ratio, ν) are required to completely define the three-dimensional
constitutive behavior.
218
CHAPTER 9. CONSTITUTIVE RELATIONS FOR LINEAR ELASTIC SOLIDS
The stress-strain equations may also be written in terms of shear modulus to obtain:
σxx
=
σyy
=
σzz
=
σyz
=
σzx
=
σxy
=
E
2G
[(1 − ν)εxx + νεyy + νεzz ] =
[(1 − ν) εxx + νεyy + νεzz ]
(1 + ν)(1 − 2ν)
1 − 2ν
E
2G
[νεxx + (1 − ν)εyy − νεzz ] =
[νεxx + (1 − ν) εyy + νεzz ]
(1 + ν)(1 − 2ν)
1 − 2ν
E
2G
[νεxx + νεyy + (1 − ν)εzz ] =
[νεxx + νεyy + (1 − ν) εzz ]
(1 + ν)(1 − 2ν)
1 − 2ν
E
εyz = 2Gεyz
(9.11)
1+ν
E
εzx = 2Gεzx
1+ν
E
εxy = 2Gεxy
1+ν
zz
Side Note: If the definition of εxx and εzz from ν = − εεxx
= − εεyy
is substituted into equation (9.9),
yy
we obtain for the uniaxial bar extension experiment described previously:
E
[(1 − ν)(−νεyy ) + νεyy + ν(−ν)εyy ] = 0
(1 + ν)(1 − 2ν)
2
E
=
−ν εyy + (1 − ν)εyy − ν 2 εyy = Eεyy
(1 + ν)(1 − 2ν)
=0
σxx =
σyy
σzz
σxy = σxz = σyz = 0
This result is consistent with all observations made regarding the nature of stress for the uniaxial
test with an applied stress of σyy .
9.2
Constitutive Equations with Thermal Strain
Experimentally, we observe for a linear isotropic metal that a temperature increase, ∆T , produces a
uniform expansion but no shear and the expansion is proportional to a material constant α (coefficient of thermal expansion). The additional strain due to heating is thus: εxx = εyy = εzz = α∆T .
Thus, the constitutive equation for a linear elastic isotropic solid (9.7) may be modified by the
addition of the thermal strain to the normal strain components:
εxx
=
εyy
=
εzz
=
εxy
=
εxz
=
εyz
=
1
[σxx − ν(σyy + σzz )] + α∆T
E
1
[σyy − ν(σxx + σzz )] + α∆T
E
1
[σzz − ν(σxx + σyy )] + α∆T
E
1+ν
(
)σxy
E
1+ν
(
)σxz
E
1+ν
(
)σyz
E
(9.12)
9.2. CONSTITUTIVE EQUATIONS WITH THERMAL STRAIN
219
These equations can be inverted to obtain stress in terms of strain:
σxx
=
σyy
=
σzz
=
σxy
=
σxz
=
σyz
=
E
[(1 − ν)εxx + νεyy + νεzz ] −
(1 + ν)(1 − 2ν)
E
[νεxx + (1 − ν)εyy + νεzz ] −
(1 + ν)(1 − 2ν)
E
[νεxx + νεyy + (1 − ν)εzz ] −
(1 + ν)(1 − 2ν)
E
εxy
2(1 + ν)
E
εxz
2(1 + ν)
E
εyz
2(1 + ν)
Eα∆T
(1 − 2ν)
Eα∆T
(1 − 2ν)
Eα∆T
(1 − 2ν)
(9.13)
In the above, ∆T = ∆T ( x, y, z ) and represents the increase in temperature from a “reference”
temperature where the thermal strain is zero. It should be noted that the first term in the extensional
strain terms above (the [ ] term) is due to elastic behavior of the material (i.e., 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:
εtotal
xx
εtotal
yy
εtotal
zz
= εelastic
+ εthermal
xx
= εelastic
+ εthermal
yy
=
εelastic
zz
+ε
(9.14)
thermal
The elastic (also called mechanical) and thermal terms are given by:
εelastic
xx
εelastic
yy
εelastic
zz
εthermal
1
[σxx − ν(σyy + σzz )]
E
1
[σyy − ν(σxx + σzz )]
=
E
1
=
[σzz − ν(σxx + σyy )]
E
= α∆T
=
(9.15)
total
total
The terms εtotal
represent the total strain as measured or observed, and are thus equal
xx , εyy , εzz
to their deformation gradient definitions, i.e., for small strain,
∂ux
∂x
∂uy
=
∂y
∂uz
=
∂z
εtotal
xx
= εxx =
εtotal
yy
= εyy
εtotal
zz
= εzz
(9.16)
We state once again that shear strains have no thermal component for an isotropic material. Examples of problems involving thermal strain will be considered in Chapter 10.
Some typical values of material properties for isotropic metals are provided in the table below.
Note that the values of E (Young’s modulus) are typically in the million psi or GPa range for
engineering materials, while the values of ν are between zero and 0.5 (0 < ν < 0.5). The yield
strength represents the stress level at which the metal yields (becomes inelastic). For ductile metals,
220
CHAPTER 9. CONSTITUTIVE RELATIONS FOR LINEAR ELASTIC SOLIDS
the ultimate tensile strength is typically 10 to 50% higher than the yield strength. It should be
noted that material properties for commonly used metals must satisfy specifications established by
regulatory agencies. Values in Tables 9.1 and 9.2 that are not provided are unknown for purposes
of presentation herein and they should not to be interpreted as zero. The reader may with to consult
other sources for the omitted values.
For non-metals, properties may vary significantly depending upon many variables (for example,
wood has a Young’s modulus varying from 0.1×106 psi to 2×106 psi depending upon the tree species
and direction of wood grain; the modulus of concrete will depend on the concrete/aggregate ratio
and curing process). Examples of non-metals commonly used are concrete (ultimate compressive
strength of 5 ksi but zero tensile strength; with an elastic modulus in compression of 3 × 106 psi)
and Douglas fir (parallel to grain, ultimate compressive strength of 7 ksi; with an elastic modulus
of 1.6 × 106 psi).
Material
Density
Young’s
lb Modulus
in3
(106 psi)
Poisson’s
Ratio
Yield
Strength
(ksi)
Ultimate
Tensile
Strength
(ksi)
Coefficient of
Thermal
Expansion
62
42
12.9 (200 ◦ F)
13.0 (70-200 ◦ F)
65
55
240
6.5
6.8 (70-200 ◦ F)
7.1 (80-800 ◦ F)
10−6
◦F
Tension Shear
Aluminum
2024-T4
6061-T6
Steel
Structural (A36)
AISI 1025
5Cr-Mo-V
Copper
G3-Heat Treated
Titanium
Ti-5Al-2.5Sn
Ti-6M-4V
0.100
0.098
10.5
9.9
0.33
0.33
40
36
0.284
0.284
0.281
30.0
29.0
30.0
0.29
0.32
0.36
36
36
200
0.272
16.0
60
110
9.0 (70-570 ◦ F)
0.162
0.160
15.5
16.0
110
120
115
130
5.2 (200-400 ◦ F)
4.6 (200-400 ◦ F)
0.34
21
21
72
Table 9.1: Structural Material Properties for Selected Metals (US Customary Units)
Example 9-1
Given:

z
−y M
Izz

[σ] =
0
0
0
0
0

0
0 
0
where Mz and Izz are constants
Required :
(a) Verify that the stress tensor satisfies the Conservation of Linear Momentum.
(b) Determine the components of the infinitesimal strain tensor, ε.
(c) Determine the components of the displacement, ux , uy , and uz .
(d) Describe the displacement and physical problem described by these equations. Use as reference
the figure below.
9.2. CONSTITUTIVE EQUATIONS WITH THERMAL STRAIN
Material
Density
Young’s
Mg
Modulus
m3
(GPa)
Poisson’s
Ratio
221
Yield
Strength
(MPa)
Ultimate
Tensile
Strength
(MPa)
Coefficient of
Thermal
Expansion
469
290
23
24
445
517
12
17
655
17
1,000
895
9.4
8.3
10−6
◦C
Tension Shear
Aluminum
2024-T6
6061-T6
Steel
Structural A36
Stainless 304
Copper Alloy
Bronze C86100
Titanium
Ti-6Al-4V
Ti-6M-4V
2.79
2.71
73.1
68.9
0.35
0.33
414
245
7.85
7.86
207
193
0.29
0.27
248
207
8.83
103
0.34
345
4.43
4.34
120
110
0.36
0.34
924
827
145
145
495
Table 9.2: Structural Material Properties for Selected Metals (SI Units)
y
y
x
z
h
L
t
Figure 9.8:
(a) x-component of linear momentum:
x→
∂σxx
∂σxy
∂σxz
+
+
+ ρgx = 0
∂x
∂y
∂z
z
∂ −y M
Izz
∂x
=0
- Stress tensor satisfies the Conservation of Linear Momentum
(b) The strains are given by:
εxx =
σxx
,
E
εyy = εzz = −
εxx = −y
Mz
,
Izz E
ν
σxx ,
E
εyy =
νyMz
,
EIzz
εxy = εxz = εyz = 0
εzz =
νyMz
EIzz
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CHAPTER 9. CONSTITUTIVE RELATIONS FOR LINEAR ELASTIC SOLIDS
(c) Integrate the displacement equations and apply boundary conditions:
Mz
∂ux
= −y
∂x
EIzz
→
εyy =
→
εzz
∂uy
νyMz
=
∂y
EIzz
νyMz
∂uz
=
=
∂z
EIzz
→
εxx =
Mz
ux = −y
x + C1
EIzz
2
ν y2 Mz
uy =
+ C2
EIzz
νyMz
uz =
z + C3
EIzz
ux |x=0 = 0,
uy |y=0 = 0,
C1 = 0
uz |z=0 = 0,
C3 = 0
C2 = 0
Mz
Izz
νv 2 Mz
y
2E Izz
ν Mz
yz
E Izz
ux
= −yx
uy
=
uz
=
(d) The displacement in the x-direction is negative (shortening) when y is positive due to the
negative ux term. If y is negative the displacement in the x-direction is positive (expanding).
In the y-direction and z-direction the displacement is expanding when y is greater than zero and
vice versa. Displacement in the y-direction changes with respect to y 2 , and in the z-direction
it changes with respect to y.
9.2. CONSTITUTIVE EQUATIONS WITH THERMAL STRAIN
Deep Thought
Ut tensio sic vis!
Ut tensio sic vis!!
Ut tensio sic vis!!!
223
224
9.3
CHAPTER 9. CONSTITUTIVE RELATIONS FOR LINEAR ELASTIC SOLIDS
Questions
9.1 Which conservation laws are especially useful for describing stresses and strains? How are
stress and strain related?
9.2 Write the equations that result from an inversion of the stress-strain relationship.
9.3 Describe in your own words the meanings of the state of plane stress and the state of plane
strain?
9.4 Describe the two types of problems which when solved using the theory of plane elasticity
provide exact solutions.
9.5 Consider small shear strain for a moment. It is often given in terms of an angle. Explain why
this is done.
9.6 What is a constitutive relation? Write down the general constitutive relation in terms of
Cauchy stress and strain.
9.7 For an elastic, isotropic solid material, how many constants are required to define the constitutive relations? Name these and define their meaning.
9.4
Problems
9.8 Structural steel is subjected to the deformation defined by ux ( x, y, z ) = 0.002x, uy ( x, y, z ) =
0, uz ( x, y, z ) = 0 (displacements in inches). Determine the following in US units:
a) Infinitesimal strain tensor.
b) Stress tensor.
c) Draw Mohr’s Circle for the given state of stress.
d) Principal Stresses and Strains.
9.9 Repeat steps a and b in 9.8 for ux ( x, y, z ) = 0.002x2 + 0.001x, uy ( x, y, z ) = 0.002xy,
uz ( x, y, z ) = 0.001z 2 .
9.10 GIVEN : A Hookean material with E = 10 × 106 psi and ν = 0.5 experiences the following
deformation: ux ( x, y, z ) = 0, uy ( x, y, z ) = 0.004x, uz ( x, y, z ) = 0
REQUIRED:
a) Sketch ux versus x, uy versus y, and uz versus z, and calculate
∂ux ∂uy ∂uz
∂x , ∂y , ∂z .
b) Calculate the infinitesimal strain tensor.
c) Calculate the stress tensor.
9.11 GIVEN : ν = 0.25 and E = 2.0 × 1010 Pa, and strain tensors as follows




0.002 0.004 0
0
0.005
0
0 
(1)  0.004 0.003 0  ,
(2)  0.005 0.04
0
0
0
0
0
0.006
REQUIRED:
(1) Calculate the stress tensors;
(2) How much is the relative volume change (the dilatation) for this deformation, and compare
the results obtained by using both finite strain formula and the small strain formula.
9.4. PROBLEMS
225
9.12 GIVEN : ν = 0.33 and E = 15.0 × 103 MPa, and stress tensors as follows




10 MPa 4 MPa 0
20 MPa 50 MPa
0

0
0
(1)  4 MPa 30 MPa 0 
(2)  50 MPa
0
0
0
0
0
6 MPa
REQUIRED: Calculate the strain tensors.
9.13 GIVEN : ux = z10−4 , uz = x10−4 , and uy = 0, and material constants E = 2.6 × 1010 , and
ν = 0.3.
(1) Compute infinitesimal strain tensor.
(2) Compute the corresponding stress tensor.
(3) What are the principal stresses and principal strains?
(4) Are the principal stresses and strains acting in the same directions?
9.14 A steel plate lies flat in the x-y plane and has dimensions 20 cm × 40 cm. If the plate is
uniformly heated throughout at 1000 ◦ C and the thermal expansion coefficient is given by
α = 11 × 10−6 (m m
- ◦ C) , calculate the new dimensions of the plate due to thermal expansion.
9.15 A thin rectangular sheet of linearly elastic material has an x-y coordinate system located at
its lower left corner. The body extends 15 in. in the x direction and 8 in. in the y direction.
The material is isotropic with an E = 35, 000, 000 psi and ν = 0.33. A plane stress condition
has been created by forces acting along the edges of the body with a displacement field of:
1.44 × 10−8 x2 y
ux
=
uy
= −1.44 × 10−8 xy 2
Write expressions for the surface force normal to and for the tangential surface force along the
upper 15 in. boundary as functions of x. Write expressions for the surface force normal to
and for the tangential surface force along the right 8 in. boundary as functions of y. Draw the
distribution of normal surface force along these two boundaries on a sketch of the body.
9.16 Use web resources to determine the following material properties. Provide the URL (http
address) that you used.
(a) Yield strength in shear of 2024-T4 and 2014-T6 aluminum.
(b) Poisson’s ratio and yield strength in shear for Ti-5Al-2.5Sn.
(c) All of the table values as presented in Table 9.1 for 4130 heat treated alloy steel.
(d) All of the table values as presented in Table 9.1 for balsa wood.
9.17 GIVEN : The isothermal (no temperature gradient) uniaxial bar specimen of 2024-T4 Aluminum (isotropic) shown below:
The axial displacements are measured to be:
ux
uy
= −0.02x in
= 0.000125x − 0.0005 in
x in inches!!
REQUIRED:
1. Sketch the deformed configuration of the test section boundary (using the displacements
given above).
2. Calculate the infinitesimal strain tensor for the test section.
226
CHAPTER 9. CONSTITUTIVE RELATIONS FOR LINEAR ELASTIC SOLIDS
Problem 9.17
3. Calculate the stress tensor for the test section.
9.18 GIVEN : A linear isotropic ThermoElastic plate of Stainless 304 is subjected to a uniform
temperature change of ∆T and is assumed to be in a state of stress as shown below. At
equilibrium, the ∆T is known.

−125
σ =  −50
0
−50
−100
0

0
0  MPa
0
REQUIRED:
a) Calculate the infinitesimal strain tensor when ∆T = 0 ◦ C. (review equations 9.12 and
9.13 in the notes)
b) Calculate the infinitesimal strain tensors for the two cases:
when ∆T = 100 ◦ C, and when ∆T = 25 ◦ C.
c) Find the temperature change ∆T necessary to produce zero strain.
9.19 GIVEN : Consider the state of stress called plane stress in which non-zero stresses exist in only
one plane.
REQUIRED:
a) For a state of plane stress in the x-y plane, show that the constitutive equations for
an elastic isotropic material (isothermal case) reduce to the following. Hint: start with
the constitutive equations for the general 3-D elastic, isotropic case and reduce to plane
stress; see equation 10.6): SHOW ALL STEPS.
σxx
=
σyy
=
σxy
=
E
[εxx + νεyy ]
(1 − ν 2 )
E
[νεxx + εyy ]
(1 − ν 2 )
E
εxy
(1 + ν)
9.4. PROBLEMS
227
b) Starting with the above relations, show that the strains for plane stress in the x-y plane
become those shown below (see equation 10.7). You must show all steps necessary to
obtain the relations below.
εxx
εyy
εxy
εzz
1
[σxx − νσyy ]
E
1
=
[σyy − νσxx ]
E
1+ν
= (
)σxy
E
ν
= − (σxx + σyy )
E
=
You may use Scientific Workplace to do the matrix algebra.
9.20 GIVEN : Given that for a general orthotropic elastic material there are 12 unique coeffecients
such that:





[D] = 



1
E11
− Eν21
22
− Eν31
33
− Eν12
11
0
0
0
0
0
0
1
E22
− Eν32
33
− Eν13
11
− Eν23
22
0
0
0
0
0
0
1
µ23
1
E33
0
0
0
0
0
0
1
µ31
0
0
0
0
0
0









1
µ12
The constitutive equation for this form would then be:
{ε} = [D] {σ}
where the stress have the following values

σxx = 5 ksi




σ

yy = 10 ksi


σzz = 20 ksi
{σ} =
 σyz = 0 ksi



 σzx = 0 ksi


σxy = 7.5 ksi















;

εxx




εyy



εzz
{ε} =
ε
 yz



εzx



εxy















REQUIRED:
a) Write the stress tensor in its more common form (i.e., as a tensor or matrix). Does this
constitute generalized plane stress? Why or why not?
Recall that generalized plane stress is a requirement for Mohr’s Circle
b) Suppose that the 12 material coefficients have the following values:
E11
=
106 psi
E22
=
=
3 × 107 psi
0.2 × 106 psi
E33
228
CHAPTER 9. CONSTITUTIVE RELATIONS FOR LINEAR ELASTIC SOLIDS
ν12
=
=
0.2
0.25
=
0.33
ν23
=
0.43
ν31
=
0.05
ν32
=
0.06
ν13
ν21
µ23
µ31
µ12
=
=
104 psi
2 × 104 psi
=
3 × 104 psi
Calculate the infinitesimal strain tensor.
c) Write the strain tensor in its more common form. Does this constitute generalized plane
strain? Why or why not?