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TRANSCRIPT
Two-Dimensional
Problems
By
S. Ziaei Rad
Review of the Basic Theory
In general, the stresses and strains in a structure consist of
six components:
and
Review of the Basic TheoryUnder certain conditions, the state of stresses and strains
can be simplified. A general 3-D structure analysis can,
therefore, be reduced to a 2-D analysis.
Plane (2-D) Problems
• Plane stress:
A thin planar structure with constant thickness and
loading within the plane of the structure (xy-plane).
Review of the Basic Theory• Plane strain:
A long structure with a uniform cross section and
transverse loading along its length (z-direction).
Stress-Strain-Temperature (Constitutive) Relations
For plane stress elastic and isotropic materials, we have,
or,
which means that there are only two
independent materials constants for
homogeneous and isotropic materials.
Stress-Strain-Temperature
RelationsWe can also express stresses in terms of strains by
solving the above equation,
or
,The above relations are valid for plane stress case.
For plane strain case, we need to replace the material
constants in the above equations in the following
fashion,
plane stress
Stress-Strain-Temperature Relations
For example, the stress is related to strain by
in the plane strain case.
plane strain
Initial strains due to temperature change (thermal loading)
is given by,
where a is the coefficient of thermal expansion, ∆∆∆∆T the
change of temperature. Note that if the structure is free to
deform under thermal loading, there will be no (elastic)
stresses in the structure.
Strain and Displacement Relations
For small strains and small rotations, we have,
In matrix form,
From this relation, we know that the strains (and thus
stresses) are one order lower than the displacements, if the
displacements are represented by polynomials.
Equilibrium EquationsIn elasticity theory, the stresses in the structure must
satisfy the following equilibrium equations,
where fx and fy are body forces (such as gravity forces)
per unit volume. In FEM, these equilibrium conditions are
satisfied in an approximate sense.
Boundary Conditions
The boundary S of the body can be divided into two
parts, Su and St. The boundary conditions (BC’s) are
described as,
in which tx and ty are traction forces (stresses on the
boundary) and the barred quantities are those with known
values.In FEM, all types of loads (distributed surface loads, body
forces, concentrated forces and moments, etc.) are
converted to point forces acting at the nodes.
Exact Elasticity SolutionThe exact solution (displacements, strains and stresses) of a
given problem must satisfy the equilibrium equations, the
given boundary conditions and compatibility conditions
(structures should deform in a continuous manner, no cracks and
overlaps in the obtained displacement fields).
Example 3.1A plate is supported and loaded with distributed force p as shown in the figure. The material constants are E and νννν.
Exact Elasticity SolutionThe exact solution for this simple problem can be found
easily as follows,
Displacement:
Strain:
Stress:
Exact (or analytical) solutions for simple problems are
numbered (suppose there is a hole in the plate!).
That is why we need FEM!
A General Formula for the Stiffness Matrix
Displacements (u, v) in a plane element are interpolated
from nodal displacements (ui, vi) using shape functions Ni as
follows,
where N is the shape function matrix, u the displacement vector
and d the nodal displacement vector. Here we have assumed
that u depends on the nodal values of u only, and v on nodal
values of v only. From strain-displacement relation, the strain
vector is,
(1)
Finite Elements for 2-D Problems
where B = DN is the strain-displacement matrix.
Consider the strain energy stored in an element,
From this, we obtain the general formula for the element stiffness matrix,
(2)
(3)
Finite Elements for 2-D Problems
Note that unlike the 1-D cases, E here is a matrix which is given
by the stress-strain relation (e.g., Equation for plane stress).
The stiffness matrix k defined by (3) is symmetric since E
is symmetric. Also note that given the material property, the
behavior of k depends on the B matrix only, which in turn on
the shape functions. Thus, the quality of finite elements in
representing the behavior of a structure is entirely determined by
the choice of shape functions.
Most commonly employed 2-D elements are linear or
quadratic triangles and quadrilaterals.
Constant Strain Triangle (CST or T3)
This is the simplest 2-D element, which is also called
linear triangular element.
Linear Triangular Element
Constant Strain Triangle (CST or T3)
For this element, we have three nodes at the vertices of the
triangle, which are numbered around the element in the
counterclockwise direction. Each node has two degrees of
freedom (can move in the x and y directions). The
displacements u and v are assumed to be linear functions within
the element, that is,
(4)
where bi (i = 1, 2, ..., 6) are constants. From these, the strains
are found to be,
(5)
which are constant throughout the element. Thus, we have the
name “constant strain triangle”(CST).
Constant Strain Triangle (CST or T3)
Displacements given by (4) should satisfy the following
six equations,
Solving these equations, we can find the coefficients b1, b2, ...,
and b6 in terms of nodal displacements and coordinates.
Substituting these coefficients into (4) and rearranging the
terms, we obtain,
(5)
(6)
Constant Strain Triangle (CST or T3)
where the shape functions (linear functions in x and y) are
and
A is the area of the triangle
(Prove this!).
(6a)
Constant Strain Triangle (CST or T3)
Using the strain-displacement relation, we have,
where xij = xi - xj and yij = yi - yj (i, j = 1, 2, 3). Again, we see
constant strains within the element. From stress-strain relation,
we see that stresses obtained using the
CST element are also constant.
Constant Strain Triangle (CST or T3)
Applying formula (3), we obtain the element stiffness
matrix for the CST element,
(7)
in which t is the thickness of the element. Notice that k for CST
is a 6 by 6 symmetric matrix. The matrix multiplication in (7)
can be carried out by a computer program.
Both the expressions of the shape functions in (6a) and
their derivations are lengthy and offer little insight into the
behavior of the element. Therefore, we introduce the
concept of natural
coordinate:
The Natural Coordinates
We introduce the natural coordinates on the triangle, then
the shape functions can be represented simply by,
Notice that,
which ensures that the rigid body translation is represented by
the chosen shape functions. Also, as in the 1-D case,
(7a)
The Natural Coordinates
and varies linearly within the element. The plot for shape
function N1 is shown in the following figure. N2 and N3 have
similar features.
Shape Function N1 for CST
The Natural CoordinatesWe have two coordinate systems for the element: the global
coordinates (x, y) and the natural coordinates . The
relation between the two is given by
or,
where and (i, j = 1, 2, 3)
as defined earlier.
Displacement u or v on the element can be viewed as
functions of (x, y) or . Using the chain rule for
derivatives, we have,
(8)
The Natural Coordinates
where J is called the Jacobian matrix of the transformation.
From (8), we calculate,
(9)
where
Prove it!
(10)
The Natural CoordinatesFrom (10), (9), (7a) and (6) we get:
Similarly,
(11)
(12)
Using (11) and (12) and the relations
we obtain the strain-displacement matrix,
The Natural Coordinates
which is the same as we derived earlier
Applications of the CST Element:
•••• Use in areas where the strain gradient is small.
•••• Use in mesh transition areas (fine mesh to coarse mesh).
•••• Avoid using CST in stress concentration or other crucial
areas in the structure, such as edges of holes and corners.
•••• Recommended for quick and preliminary FE analysis of
2-D problems.
Example
P
Px
y7a
a
xσσσσ
x
Stress along the x axis in a beam modeled by CTSs and
loaded in pure bending xσσσσ
The CTS does not work well in high strain gradient. The natural axis of the
beam should be stress-free, while the FE model display a square wave pattern.
CST ELEMENT
ηξηξ −−=== 1, , 321 NNN1
(1,0,0)
2
(0,1,1)
3
(0,0,1)
11321 ,0, 0, 1)1( yyxxNNN ==⇒===⇒
22321 ,0, 1, 0)2( yyxxNNN ==⇒===⇒
33321 ,1, 0, 0)3( yyxxNNN ==⇒===⇒
32313
3231
321
3)()(
)1(
xxxx
xxxxxx
xxxx
++=
+−+−=
−−++=
ηξ
ηξ
ηξηξ
CST ELEMENT
P
(3.85,4.8)1
(1.5,2)2
(7,3.5)
3
(4,7)
332211
332211
yNyNyNy
xNxNxNx
++=
++=
321
321
75.32
475.1
NNNy
NNNx
++=
++=
)1(75.328.4
)1(475.185.3
ηξηξ
ηξηξ
−−++=
−−++=
2.0
3.0
=
=
η
ξ
5.0
2.0
3.0
3
2
1
=
=
=
N
N
N
332211
332211
vNvNvNv
uNuNuNu
++=
++=
CST ELEMENT
32313
32313
321 )1(
yyyy
xxxx
xxxx
++=
++=
−−++=
ηξ
ηξ
ηξηξ
2313
2313
,
,
yy
yy
xx
xx
=∂
∂=
∂
∂
=∂
∂=
∂
∂
ηξ
ηξ
−
−=⇒
= −
1323
13231
2313
2313
det
1][][
xx
yy
JJ
yy
xxJ
∂
∂+
∂
∂−
∂
∂−
∂
∂
=
∂
∂
∂
∂
ηξ
ηξ
ux
ux
uy
uy
J
y
u
x
u
1323
1323
det
1
∂
∂+
∂
∂−
∂
∂−
∂
∂
=
∂
∂
∂
∂
ηξ
ηξ
vx
vx
vy
vy
J
y
v
x
v
1323
1323
det
1
CST ELEMENT
−−−+−+−−
−+−−
−−−
=
∂
∂+
∂
∂
∂
∂
∂
∂
=
)()()()(
)()(
)()(
det
1}{
3213312332133123
32133123
32133123
vvyvvyuuxuux
vvxvvx
uuyuuy
J
x
v
y
u
y
v
x
u
ε
=
∂
∂+
∂
∂
∂
∂
∂
∂
=
3
3
2
2
1
1
122131133232
231323
123123
000
000
det
1}{
v
u
v
u
v
u
yxyxyx
xxx
yyy
J
x
v
y
u
y
v
x
u
ε