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