Lecture 4
Development of Constitutive Equations for Continuum, Beams
and Plates
Problem 4-1: Summation Convention
Hooke’s law, a constitutive equation for a linear, elastic material, can be written in general form
as:
ij
kk
ij
2
ij
where and
are Làme constants.
a)
Expand Hooke’s Law. How many independent equations are there?
b)
Express and
c)
Where does the symmetry of the stress strain tensor come from?
d)
Re-write the Hooke’s law in terms of E and .
e)
Specify the constitutive equations to the case of plane stress.
in terms of Young’s Modulus, E, and Poisson’s ratio, .
Problem 4-1 Solution:
a)
Hooke’s law
ij
kk
ij
2
ij
where
ij
1, i = j
0,i j
For i=1
(
11
11
12
2
12
13
2
13
22
33
) 2
11
1
For i=2
2
21
21
(
22
11
23
2
23
31
2
31
32
2
32
22
33
) 2
22
22
33
) 2
33
For i=3
(
33
b)
Express ,
11
in terms of E,
We need to solve for , . Let’s assume a plane stress state in a uniaxial test
0
0 0
0 0
0
0 0
11
ij
Note: Assume
11 =0
because it is a slender bar
Substitute the stress components into the inversed form of the constitutive equations
1
ij
E
ij
E
kk ij
(1)
2
We have
1
E
11
11
22
E
11
11
33
E
11
11
(2)
From the constitutive equations
ij
kk
2
ij
(3)
ij
We have
22
(
11
22
33
) 2
22
(4)
11
(
11
22
33
) 2
11
(5)
Combining equation (2) and (4), we have
22
(
11
11
11
) 2 (
11
) 0
2
1 2
(6)
(7)
Combining equation (2) and (5), we have
11
(
11
22
33
(1 2 ) 2
) 2
E
11
E
11
(8)
(9)
Substitute (7) into (9), solve for
E
2(
1)
G
(10)
3
Substitute (10) into (7), solve for
E
(1 2 )(1
c)
(11)
)
Symmetry of the stress tensor
Let’s take an element in 2D(a unit square)
We know the normal forces are equal in magnitude and opposite in direction at opposite faces, so
as the shear forces.
Now let’s prove F1
xx
xx
yy
yy
F1
F1
F2
F2
F2 , take moments about the center
M
F2
1
1
1
1
F2
F1
F1
2
2
2
2
F2
xy
0
F1
yx
4
d)
Rewrite Hook’s law in terms of E and
Substitute equation (10) and (11) in part b into Hook’s law, we have:
ij
kk
ij
2
E
(1 2 )(1
ij
kk
)
ij
2
E
2(
1)
ij
E
ij
e)
(1
kk
) 1 2
ij
ij
For Plane stress
E
11
1
2
11
22
2
22
11
E
22
1
E
12
1
12
5
Problem 4-2: Inverting constitutive equations
The original form of the constitutive equation is to express stress ij in terms of strain. Invert the
3D constitutive equation and the 2D (plane stress) constitutive equation, meaning that strain ij
will be expressed in terms of stresses. The starting point of this problem is Eq. 4.19 for the 3D
case in the printed lecture notes. For the 2D case you can use Eq. 4.32 as a starting point.
Problem 4-2 Solution:
(1)
E,
The 3D constitutive equations for isotropic linear elastic materials expressed in terms of
is
E
ij
By making contraction i
j
(1
) 1 2
kk
ij
ij
(1)
) 1 2
kk
3
kk
(2)
k , we have
E
kk
where
kk
Express
11
kk
22
in terms of
33
kk
(1
3
, E and
kk
1 2v
E
(3)
kk
Substitute equation (3) into equation (1), we have
E
ij
(1
) 1 2
1 2
E
kk
ij
ij
(4)
Straight forwardly, the above equation gives
1
ij
E
ij
E
kk ij
(5)
which is the inverted form of 3D Hook’s law
6
(2)
The 2D constitutive equations for plane stress in terms of E,
E
(1
2
1
)
is
(1)
Where
1,
=
(2)
0,
Since
11
11
2
(3)
making contraction of equation (1):
E
(1
Replacing
with
)
)
(1
2
(4)
, Straight forwardly,
1
(5)
E
Substitute (5) into (1), we have
E
1
2
(1
1
)
E
(6)
Rearrange (6), we have the inverted form
1
E
E
(7)
7
Problem 4-3: Stress and strain deviator
Defining the stress deviator sij
sij
ij
eij
ij
1
3
kk ij
and the strain deviator eij
1
3
kk ij
Convert the constitutive equation into two separate equations, one for the spherical part and other
for the distortional part. The spherical part gives a relation between the hydrostatic pressure and
the change of volume (Eq. 4.21). For the distortional part the equation was not given in the notes,
so we are asking you to find it.
Problem 4-3 Solution:
From the lecture notes, express
kk
in terms of
kk
kk
E
1 2
, we have
kk
(1)
kk ij
(2)
kk ij
(3)
kk ij
(4)
From definition of stress deviator
sij
ij
1
3
We have
ij
sij
eij
ij
1
3
From definition of strain deviator
1
3
8
We have
1
3
eij
ij
(5)
kk ij
Substitute equation (3) and (5) into 3D constitutive equation
E
ij
(1
) 1 2
kk
ij
(6)
ij
We have
sij
1
3
E
kk ij
(1
) 1 2
kk ij
eij
1
3
(7)
kk ij
Combining equation (1) with equation (7)
sij
1
E
3 1 2
E
kk
ij
(1
) 1 2
kk ij
eij
1
3
kk ij
(8)
Finally,
sij
E
1
eij
(9)
9
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2.080J / 1.573J Structural Mechanics
Fall 2013
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