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CIVL 7117 Finite Elements Methods in Structural Mechanics Page 244

Development of the Plane Stress and PlaneStrain Stiffness Equations

Introduction

In Chapters 2 through 5, we considered only line elements. Line elements are

connected only at common nodes, forming framed or articulated structures such

as trusses, frames, and grids. Line elements have geometric properties such as

cross-sectional area and moment of inertia associated with their cross sections.

However, only one local coordinate along the length of the element is required to

describe a position along the element (hence, they are called line elements).

Nodal compatibility is then enforced during the formulation of the nodal

equilibrium equations for a line element.

This chapter considers the two-dimensional finite element. Two-dimensional

(planar) elements are thin-plate elements such that two coordinates define a

position on the element surface.

The elements are connected at common nodes and/or along common edges

to form continuous structures. Nodal compatibility is then enforced during the

formulation of the nodal equilibrium equations for two-dimensional elements. If

proper displacement functions are chosen, compatibility along common edges is

also obtained.



CIVL 7117 Finite Elements Methods in Structural Mechanics 45

The two-dimensional element is extremely important for:

(1) Plane stress analysis, which includes problems such as plates with

holes, fillets, or other changes in geometry that are loaded in their planeresulting in local stress concentrations; and

(2) Plane strain analysis, which includes problems such as a long

underground box culvert subjected to a uniform load acting constantly

over its length or a long cylindrical control rod subjected to a load that

remains constant over the rod length (or depth).




Plane Stress Problems

Plane Strain Problems

We begin this chapter with the development of the stiffness matrix for a basic

two-dimensional or plane finite element, called the constant-strain triangular

element . The constant-strain triangle (CST) stiffness matrix derivation is the

simplest among the available two-dimensional elements.We will derive the CST stiffness matrix by using the principle of minimum

potential energy because the energy formulation is the most feasible for the

development of the equations for both two- and three-dimensional finite

elements.




General Steps in the Formulation of the Plane Triangular Element

Equations

We will now follow the steps described in Chapter 1 to formulate the governing

equations for a plane stress/plane strain triangular element. First, we willdescribe the concepts of plane stress and plane strain. Then we will provide a

brief description of the steps and basic equations pertaining to a plane triangular

element.

Plane Stress

Plane stress is defined to be a state of stress in which the normal stress and

the shear stresses directed perpendicular to the plane are assumed to be zero.

That is, the normal stress σz and the shear stresses τ xz and τyz are assumed tobe zero. Generally, members that are thin (those with a small z dimension

compared to the in-plane x and y dimensions) and whose loads act only in the x-

y plane can be considered to be under plane stress.

Plane Strain

Plane strain is defined to be a state of strain in which the strain normal to the

x-y plane εz and the shear strains γ xz and γyz are assumed to be zero. The

assumptions of plane strain are realistic for long bodies (say, in the z direction)

with constant cross-sectional area subjected to loads that act only in the x and/or

y directions and do not vary in the z direction.

Two-Dimensional State of Stress and Strain

The concept of two-dimensional state of stress and strain and the stress/strain

relationships for plane stress and plane strain are necessary to understand fully

the development and applicability of the stiffness matrix for the plane

stress/plane strain triangular element.

A two-dimensional state of stress is shown in the figure below. The

infinitesimal element with sides dx and dy has normal stresses σz and σy acting in

the x and y directions (here on the vertical and horizontal faces), respectively.

The shear stress τ xy acts on the x edge (vertical face) in the y direction. The

shear stress τyx acts on the y edge (horizontal face) in the x direction.




Since τ xy equals τyx , three independent stress exist:

T

x y xy σ σ σ τ =

Recall, the relationships for principal stresses in two-dimensions are:

σ σ σ σ σ τ σ + − = + + =

2

21 max

2 2 x y x y

xy

min

2

2

222

σ τ σ σ σ σ

σ =+

−−

+=

xy

y x y x

Also, θ p is the principal angle which defines the normal whose direction is

perpendicular to the plane on which the maximum or minimum principle stress

acts.

2tan2

xy

p

x y

τ θ

σ σ =

−




The general two-dimensional state of strain at a point is show below.

The general definitions of normal and shear strains are:

x

v

y

u

x

v

x

u xy y x ∂

∂+

∂∂

=∂∂

=∂∂

= γ ε ε

The strain may be written in matrix form as:

T

x y xy ε ε ε γ =




For plane stress, the stresses σz , τ xz , and τyz are assumed to be zero. The

stress-strain relationship is:

( )

−−

=

xy

y

x

xy

y

x

E

γ

ε

ε

ν

ν

ν

ν τ

σ

σ

15.000

01

01

12

where

( )

−−

=

ν

ν

ν

ν 15.000

01

01

1][

2

E D

is called the stress-strain matrix (or the constitutive matrix ), E is the modulus

of elasticity, and ν is Poisson’s ratio.

For plane strain, the strains εz , γ xz , and γyz are assumed to be zero. The


( )( )

−

−

−

−+=

xy

y

x

xy

y

x

E

γ

ε

ε

ν

ν ν

ν ν

ν ν τ

σ

σ

5.000

01

01

211

where

( )( )

−

−

−

−+=

ν

ν ν

ν ν

ν ν 5.000

01

01

211][

E D

The partial differential equations for plane stress are:

∂∂

∂−∂

∂+=∂

∂+∂

∂y x

v

y

u

y

u

x

u 2

2

2

2

2

2

2

2

1 ν

∂∂

∂−

∂∂+

=∂∂

+∂∂

y x

u

y

v

y

v

x

v 2

2

2

2

2

2

2

2

1 ν




Steps in the Formulation of the Element Stiffness Equations

Consider the problem of a thin plate subjected to a tensile load as shown in

the figure below:

Step 1 - Discretize and Select Element Types

Discretize the thin plate into a set of triangular elements. Each element is

define by nodes i , j , and m.

We use triangular elements because boundaries of irregularly shaped bodies can

be closely approximated, and because the expressions related to the triangular

element are comparatively simple. This discretization is called a coarse-mesh

generation if few large elements are used. Each node has two degrees offreedom: displacements in the x and y directions. We will let ui and v i represent

the node i displacement components in the x and y directions, respectively.

The nodal displacements for an element with nodes i , j , and m are:




=

m

j

i

d

d

d

d

where the nodes are ordered counterclockwise around the element, and

=

i

i

i v ud

Therefore:

=

m

m

j

j

i

i

v

u

v

u

v

u

d

Step 2 - Select Displacement Functions

The general displacement function is:

=Ψ),(

),(

y x v

y x ui

The functions u( x, y ) and v ( x, y ) must be compatible with the element type.

Step 3 - Define the Strain-Displacement and

Stress-Strain Relationships

The general definitions of normal and shear strains are:

x

v

y

u

x

v

x

u xy y x ∂

∂+

∂∂

=∂∂

=∂∂

= γ ε ε

For plane stress, the stresses σz , τ xz , and τyz are assumed to be zero. The


( )

−−

=

xy

y

x

xy

y

x

E

γ

ε

ε

ν

ν

ν

ν τ

σ

σ

15.000

01

01

1 2




For plane strain, the strains εz , γ xz , and γyz are assumed to be zero. The


( )( )

−

−

−

−+

=

xy

y

x

xy

y

x

E

γ

ε

ε

ν

ν ν

ν ν

ν ν τ

σ

σ

5.000

01

01

211

Step 4 - Derive the Element Stiffness Matrix and Equations

Using the principle of minimum potential energy, we can derive the element

stiffness matrix.

d k f ][=

This approach is better than the direct methods used for one-dimensional

elements.

Step 5 - Assemble the Element Equations and

Introduce Boundary Conditions

The final assembled or global equation written in matrix form is:

d K F ][=

where F is the equivalent global nodal loads obtained by lumping distributed

edge loads and element body forces at the nodes and [K ] is the global structure

stiffness matrix.

Step 6 - Solve for the Nodal Displacements

Once the element equations are assembled and modified to account for the

boundary conditions, a set of simultaneous algebraic equations that can be

written in expanded matrix form as:

Step 7 - Solve for the Element Forces (Stresses)

For the structural stress-analysis problem, important secondary quantities of

strain and stress (or moment and shear force) can be obtained in terms of the

displacements determined in Step 6.




Derivation of the Constant-Strain Triangular Element Stiffness

Matrix and Equations

Consider the problem of a thin plate subjected to a tensile load as shown in

the figure below:

Step 1 - Discretize and Select Element Types

The basic triangular element, shown below, is define by nodes i , j , and m

labeled in a counterclockwise manner.

The nodal displacement matrix is:

=

m

m

j

j

i

i

v

u

v

u

v

u

d

Step 2 - Select Displacement Functions

We will select a linear displacement function for each triangular element,

defined as:

++

++=

=Ψy a x aa

y a x aa

y x v

y x ui

654

321

),(

),(




A linear function ensures that the displacements along each edge of the element

and the nodes shared by adjacent elements are equal.

=

++

++=Ψ

6

5

4

3

2

1

654

321

1000

0001

a

a

a

a

a

a

y x

y x

y a x aa

y a x aai

To obtain the values for the a’s substitute the coordinated of the nodal points into

the above equations:

i i i i i i y a x aav y a x aau

654321 ++=++=

j j j j j j y a x aav y a x aau

654321 ++=++=

mmmmmmy a x aav y a x aau

654321 ++=++=

Solving for the a’s and writing the results in matrix forms gives:

[ ] u x a

a

a

a

y x

y x

y x

u

u

u

mm

j j

i i

m

j

i

1

3

2

1

1

1

1−=⇒

=

(x j, y j)

(xi, yi) (xm, ym)

ui

u j

um

u

y

x

Linear representation of u(x, y)




The inverse of the [ x ] matrix is:

=−

m j i

m j i

m j i

A x

γ γ γ

β β β

α α α

2

1][ 1

where

mm

j j

i i

y x

y x

y x

A

1

1

1

2 =

is the determinant of [ x ].

( ) j i mi m j m j i y y x y y x y y x A −+−+−=2

where A is the area of the triangle and

j mi m j i m j m j i x x y y x y y x −=−=−= γ β α

mi j i m j mi mi j x x y y x y y x −=−=−= γ β α

i j m j i m j i j i m x x y y x y y x −=−=−= γ β α

The values of a may be written matrix form as:

=

m

j

i

m j i

m j i

m j i

u

u

u

Aa

a

a

γ γ γ

β β β

α α α

2

1

3

2

1

and

=

m

j

i

m j i

m j i

m j i

v

v

v

Aa

a

a

γ γ γ

β β β

α α α

2

1

6

5

4




We will now derive the displacement function in terms of the coordinates x and y .

[ ]

=

3

2

1

1

a

a

a

y x u

Substituting the values for a into the above equation gives:

[ ]1

12

i j m i

i j m j

i j m m

u

u x y u A

u

α α α

β β β

γ γ γ

=

Expanding the above equations

[ ]1

12

i i j j m m

i i j j m m

i i j j m m

u u u

u x y u u u Au u u

α α α

β β β γ γ γ

+ +

= + + + +

Multiplying the matrices in the above equations gives

( ) ( ) ( ) 1

( , )2

i i i i j j j j m m m mu x y x y u x y u x y u A

α β γ α β γ α β γ = + + + + + + + +

A similar expression can be obtained for the y displacement

( ) ( ) ( ) 1

( , )2

i i i i j j j j m m m mv x y x y v x y v x y v A

α β γ α β γ α β γ = + + + + + + + +

The displacements can be written in a more convenience form as:

mm j j i i mm j j i i v N v N v N y x v uN uN uN y x u ++=++= ),(),(

where

( ) ( ) ( )y x A

N y x A

N y x A

N mmmm j j j j i i i i

γ β α γ β α γ β α ++=++=++=21

21

21

The elemental displacements can be summarized as:

++

++=

=Ψmm j j i i

mm j j i i

i v N v N v N

uN uN uN

y x v

y x u

),(

),(




In another form the above equations are:

0 0 0

[ ] 0 0 0

i

i

i j m j

i j m j

m

m

u

v

N N N u

N d N N N v

u

v

ψ

Ψ = =

where

[ ]

=

m j i

m j i

N N N

N N N N

000

000

The linear triangular shape functions are illustrated below

Step 3 - Define the Strain-Displacement and Stress-Strain Relationships

Elemental Strains: The strains over a two-dimensional element are:

∂∂

+∂∂

∂∂∂∂

=

=

x

v

y

uy

v x u

xy

y

x

γ

ε

ε

ε

j

i m

1

N i

y

x

j

i m

1

N j

y

x

j

i m

1

N m

y

x




Substituting our approximation for the displacement gives:

( )mm j j i i x

uN uN uN x

u x

u++

∂∂

==∂∂

,

m x m j x j i x i x

uN uN uN u,,,,

++=

where the comma indicates differentiation with respect to that variable. The

derivatives of the interpolation functions are:

( ) A

N A

N A

y x x A

N m

x m

j

x j

i

i i i x i 2222

1,,,

β β β γ β α ===++

∂∂

=

Therefore:

( )mm j j i i uuu A x

u β β β ++=∂∂ 21

In a similar manner, the remaining strain terms are approximated as:

( )1

2i i j j m m

v v v v

y Aγ γ γ

∂= + +

∂

( )mmmm j j j j i i i i

v uv uv u A x

v

y

uγ β γ β γ β +++++=

∂

∂+

∂

∂

2

1

We can write the strains in matrix form as:

or

[ ]

=

m

j

i

m j i

d

d

d

BBBε

0 0 01

0 0 02

i

i

x i j m

j

y i j m

j

xy i i j j m m

m

m

uuv x

uv

v y A

uu v y x v

ε β β β

ε ε γ γ γ

γ γ β γ β γ β

∂

∂ ∂

== = = ∂ ∂ ∂+ ∂ ∂

0 0 01

0 0 02

i

i

x i j m

j

y i j m

j

xy i i j j m m

m

m

uuv x

uv

v y A

uu v y x v

ε β β β

ε ε γ γ γ

γ γ β γ β γ β

∂

∂ ∂

== = = ∂ ∂ ∂+ ∂ ∂




where

[ ] [ ] [ ]

=

=

=

mm

m

m

m

j j

j

j

j

i i

i

i

i A

B A

B A

B

β γ

γ

β

β γ

γ

β

β γ

γ

β

0

0

2

10

0

2

10

0

2

1

These equations can be written in matrix form as:

][ d B=ε

Stress-Strain Relationship: The in-plane stress-strain relationship is:

=

xy

y

x

xy

y

x

D

γ

ε

ε

τ

σ

σ

][

where [D] for plane stress is:

( )

−−

=

ν

ν

ν

ν 15.000

01

01

1][

2

E D

and [D] for plane strain is:

( )( )

−

−

−

−+=

ν

ν ν

ν ν

ν ν 5.000

01

01

211][

E D

In-plane stress can be related to displacements by:

]][[ d BD=σ




Step 4 - Derive the Element Stiffness Matrix and Equations

The total potential energy is defined as the sum of the internal strain energy U

and the potential energy of the external forces Ω:

s pb p U Ω+Ω+Ω+=π

where the strain energy is:

∫=V

T dV U 2

1σ ε

or

∫=

V

T dV DU ][2

1ε ε

The potential energy of the body force term is:

∫ Ψ−=ΩV

T

bdV X

where Ψ is the general displacement function, and X is the body weight per

unit volume.

The potential energy of the concentrated forces are:

P d T

p −=Ω

where P are the concentrated forces, and d are the nodal displacements.

The potential energy of the distributed loads are:

∫ Ψ−=ΩS

T

sdST

where Ψ is the general displacement function, and T are the surface tractions.

Then the total potential energy expression becomes:

1

[ ] [ ][ ] [ ] [ ] 2

T T T TT T T

p

V V S

d B D B d dV d N X dV d P d N T dSπ = − − −∫ ∫ ∫




The nodal displacements d are independent of the general x-y coordinates,

therefore

1

[ ] [ ][ ] [ ] [ ] 2

T T T TT T T

p

V V S

d B D B dV d d N X dV d P d N T dSπ = − − −∫ ∫ ∫

We can define the last three terms as:

∫∫ ++=S

T

V

T dST N P dV X N f ][][

Therefore:

1

[ ] [ ][ ]2

T T T

p

V

d B D B dV d d f π = −∫

Minimization of πp with respect to each nodal displacement requires that:

[ ] [ ][ ] 0

p T

V

B D B dV d f d

π ∂= − =

∂ ∫

The above relationship requires:

[ ] [ ][ ]T

V

B D B dV d f =∫

The stiffness matrix can be defined as:

[ ] [ ] [ ][ ]T

V

k B D B dV = ∫

For an element of constant thickness, t , the above integral becomes:

∫= A

T dy dx BDBt k ]][[][][

The integrand in the above equation is not a function of x or y (global

coordinates); therefore, the integration reduces to:

∫= A

T dy dx BDBt k ]][[][][




or

]][[][][ BDBtAk T =

where A is the area of the triangular element. Expanding the stiffness relationship

gives:

=

][][][

][][][

][][][

][

mmmj mi

jm jj ji

imij ii

k k k

k k k

k k k

k

where each [k ii ] is a 2 x 2 matrix define as:

tABDBk tABDBk tABDBk m

T

i im j

T

i ij i

T

i ii ]][[][][]][[][][]][[][][ ===

Recall:

[ ] [ ] [ ]

=

=

=

mm

m

m

m

j j

j

j

j

i i

i

i

i A

B A

B A

B

β γ

γ

β

β γ

γ

β

β γ

γ

β

0

0

2

10

0

2

10

0

2

1

Step 5 - Assemble the Element Equations to Obtain the Global Equations

and Introduce the Boundary ConditionsThe global stiffness matrix can be found by the direct stiffness method.

∑=

=N

e

ek K 1

)( ][][

The global equivalent nodal load vector is obtained by lumping body forces

and distributed loads at the appropriate nodes as well as including any

concentrated loads.

∑=

=N

e

ef F 1

)(

The resulting global equations are:




][ d K F =

where d is the total structural displacement vector.

In the above formulation of the element stiffness matrix, the matrix has been

derived for a general orientation in global coordinates. Therefore, no

transformation form local to global coordinates is necessary. However, for

completeness, we will now describe the method to use if the local axes for the

constant-strain triangular element are not parallel to the global axes for the whole

structure.

To relate the local to global displacements, force, and stiffness matrices we

will use:

T k T k Tf f Td d T ˆˆˆ ===

The transformation matrix T for the triangular element is:

where C = cos θ and S = sin θ, and q is shown in the figure above.

0 0 0 0

0 0 0 0

0 0 0 0

0 0 0 0

0 0 0 0

0 0 0 0

C S

S C

C ST S C

C S

S C

−

= −

−




Step 6 - Solve for the Nodal Displacements

Step 7 - Solve for Element Forces and Stress

Having solved for the nodal displacements, we can obtain strains and

stresses in x and y directions in the elements by using:

[ ] [ ][ ] B d D B d ε σ = =

Plane Stress Example

Consider the structure shown in the figure below.

Assume plane stress conditions. All coordinates are shown on the figure. Let E =

30 x 106 psi, ν = 0.25, and t = 1 in. Assume the element nodal displacements

have been determined to be u1 = 0.0, v 1 = 0.0025 in., u2 = 0.0012 in., v 2 = 0.0, u3

= 0.0, and v 3 = 0.0025 in. Determine the element stiffness matrix and the element

stresses.

First, we calculate the element β’s and γ’s as:

220110 −=−=−=−=−=−= j mi m j i

x x y y γ β

0002)1(0 =−=−==−−=−=mi j i m j

x x y y γ β

202101 =−=−=−=−−=−=i j m j i m

x x y y γ β




Therefore, the [B] matrix is:

[ ]

0 0 0 1 0 2 0 1 01 1

0 0 0 0 2 0 0 0 22 2(2)

2 1 0 2 2 1

i j m

i j m

i i j j m m

B A

β β β

γ γ γ

γ β γ β γ β

− − = = −

− − −

For plane stress conditions, the [D] matrix is:

−×

=

375.000

0125.0

025.01

)25.0(1

1030][

2

6

D

Substitute the above expressions for [D] and [B] into the general equations for

the stiffness matrix:

]][[][][ BDBtAk T =

6

1 0 2

0 2 11 0.25 0 1 0 2 0 1 0

2 0 1(2)30 10 1

0.25 1 0 0 2 0 0 0 24(0.9375) 2(2)2 0 20 0 0.375 2 1 0 2 2 1

1 0 2

0 2 1

k

− − − − − − ×

= − − − − −

−

Performing the matrix triple product gives:

inlbk

−−−

−−−−

−−−

−−−

−−−−

−−−

×=

375.425.175.01625.325.0

25.15.25.1225.05.0

75.05.15.1075.05.1

120412

625.325.075.01375.425.1

25.05.05.1225.15.2

104 6

The in-plane stress can be related to displacements by:




]][[ d BD=σ

6

0.0

0.00251 0.25 0 1 0 2 0 1 0

0.001230 10 1

0.25 1 0 0 2 0 0 0 20.9375 2(2) 0.00 0 0.375 2 1 0 2 2 1

0.0

0.0025

x

y

xy

in

in

in

σ

σ τ

− − ×

= − − − −

The stresses are:

−

=

psi

psi

psi

xy

y

x

000,15

800,4

200,19

τ

σ

σ

Recall, the relationships for principal stresses and principal angle in two-

dimensions are:

max

2

2

122

σ τ σ σ σ σ

σ =+

−+

+=

xy

y x y x

min

2

2

222

σ τ σ σ σ σ σ =+

−−+=

xy

y x y x

−= −

y x

xy

pσ σ

τ θ

2tan

2

1 1

Therefore:

( ) psi 639,28000,152

800,4200,19

2

800,4200,19 2

2

1 =−+

−+

+=σ

( ) psi 639,4000,152

800,4200,19

2

800,4200,19 2

2

1 −=−+

−−

+=σ

o

p3.32

800,4200,19

)000,15(2tan

2

1 1 −=

−

−= −θ




Treatment of Body and Surface Forces

The general force vector is defined as:

∫∫ ++=S

T

V

T dST N P dV X N f ][][

Body Force

Let’s consider the first term of the above equation.

[ ] = ∫ T

b

V

f N X dV

where

=

b

b

X X

Y

where X b and Y b are the weight densities in the x and y directions, respectively.

The force may reflect the effects of gravity, angular velocities, or dynamic inertial

forces.

The integration of the f b is simplified if the origin of the coordinate system is

chosen at the centroid of the element, as shown in the figure below. With the

origin placed at the centroid, we can use the definition of a centroid.

For a given thickness, t , the body force term becomes:

[ ] [ ] T T

b

V A

f N X dV t N X dA= =∫ ∫

0=∫ A

x dA

0=∫ A

y dA




Recall the interpolation functions for a place stress/strain triangle:

( ) ( ) ( )y x A

N y x A

N y x A

N mmmm j j j j i i i i

γ β α γ β α γ β α ++=++=++=2

1

2

1

2

1

Therefore the terms in the integrand are:

0 β γ = =∫ ∫i i

A A

x dA y dA

and

2

3i j m

Aα α α = = =

The body force at node i is given as:

3

=

b

bi

b

X tAf

Y

The general body force vector is:

3

bix b

biy b

bjx b

b

bjy b

bmx b

bmy b

f X

f Y

f X tAf

f Y f X

f Y

= =

Surface Force

The third term in the general force vector is defined as:

[ ] = ∫ T

s

S

f N T dS




Let’s consider the example of a uniform stress p acting between nodes 1 and 3

on the edge of element 1 as shown in figure below.

The surface traction becomes:

0

= =

x

y

p pT

p

and [N ]T is:

1

1

2

2

3

3

0

0

0[ ]

0

0

0

=

T

N

N

N N

N

N

N

evaluated at x=a

Therefore, the traction force vector is:

1

1

2

20 0

3

3

0

0

0

0 0

0

0

=

∫ ∫

t L

s

N

N

N pf dy dz N

N

N

evaluated at x=a




After some simplification, the traction force vector is:

1

2

0

3

0

0

0

=

∫

L

s

N p

N p

f t dy

N p

The interpolation function for i = 1 is:

( )1

2α β γ = + +i i i i N x y

A

For convenience, let’s choose the coordinate system shown in the figure below.

Recall:

α = −i j m j m x y y x

with i = 1, j = 2, and m = 3, we get

1 2 3 2 3α = − x y y x

If we substitute the coordinates of the triangle show above in the above equation

we get:




1 0α =

Similarly, we can find:

1 10 β γ = = a

Therefore, the interpolation function, N 1 is:

12

=ay

N A

The remaining interpolation function, N 2 and N 2 are:

2 3

( )

2 2

− −= =

L a x Lx ay N N

A A

Substituting the interpolation function in the traction force vector expression

gives:

1

1

2

2

3

3

1

0

0

2 0

10

= =

s x

s y

s x

s

s y

s x

s y

f

f

f pLt f

f

f f

Explicit Expression for the Constant-Strain Triangle Stiffness Matrix

Usually the stiffness matrix is computed internally by computer programs, but

since we are not computers, we need to explicitly evaluate the stiffness matrix.

For a constant-strain triangular element, considering the plane strain case, recall

that:

[ ] [ ] [ ][ ]= T k tA B D B

where [D] for plane strain is:




( )( )

−

−

−

−+=

ν

ν ν

ν ν

ν ν 5.000

01

01

211][

E D

Substituting the appropriate definition into the above triple product gives:

0

01 0

0[ ] 1 0

04 (1 )(1 2 )0 0 0.5

0

0

β γ

γ β ν ν

β γ ν ν

γ β ν ν ν

β γ

γ β

− = − + − −

i i

i i

j j

j j

m m

m m

tE k

A

0 0 0

0 0 0

β β β

γ γ γ

γ β γ β γ β

×

i j m

i j m

i i j j m m

Therefore the global stiffness matrix is:




The stiffness matrix is a function of the global coordinates x and y , the material

properties, and the thickness and area of the element.

Finite Element Solution of a Plane Stress Problem

Consider the thin plate subjected to the surface traction shown in the figure

below.

Assume plane stress conditions. Let E = 30 x 106 psi, n = 0.30, and t = 1 in.

Determine the nodal displacements and the element stresses.

Discretization

Lets discretize the plate into two elements as shown below:

This level of discretization will probably not yield practical results for

displacement and stresses: however; it is useful example for a longhand solution.




The tensile traction forces can be converted into nodal forces as follows:

1

1

2

2

3

3

1 1 5,000

0 0 0

0 0 01,000 (1 )10

2 20 0 01 1 5,000

0 0 0

s x

s y

s x

s

s y

s x

s y

f lb

f

f pLt psi in inf

f

f lb

f

= = = =

The governing global matrix equations are:

[ ] =F K d

Expanding the above matrices gives:

1 11

1 11

2 22

2 22

3 3 3

3 3 3

4 4 4

4 4 4

0

0

0

0[ ] [ ]

5,000

0

5,000

0

= = =

x x x

y y y

x x x

y y y

x x x

y y y

x x x

y y y

F d R

F d R

F d R

F d R K K

F d d lb

F d d

F d d lb

F d d

where [K ] is an 8 x 8 matrix before deleting the rows and columns accounting for

the boundary supports.

Assemblage of the Stiffness Matrix

The global stiffness matrix is assembled by superposition of the individual

element stiffness matrices. The element stiffness matrix is:

[ ] [ ] [ ][ ]= T k tA B D B




For element 1: the coordinates are x i = 0, y i = 0, x j = 20, y j = 10, x m = 0, and y m =

10. The area of the triangle is:

The matrix [B] is:

0 0 01

[ ] 0 0 02

β β β

γ γ γ γ β γ β γ β

=

i j m

i j m

i i j j m m

B A

We need to calculate the element β’s and γ’s as:

10 10 0 0 20 20 β γ = − = − = = − = − = −i j m i m j y y x x

1 10 0 10 0 0 0 β γ = − = − = = − = − = j m j i my y x x

0 10 10 20 0 20 β γ = − = − = − = − = − =m i j m i j y y x x


0 0 10 0 10 01 1[ ] 0 20 0 0 0 20

20020 0 0 10 20 10

− = − − −

Bin


61 0.3 0

30 10[ ] 0.3 1 0

0.910 0 0.35

× =

D psi

2=

bh

A

2(20)(10)100 .

2= = A in




Therefore:

6

0 0 20

0 20 01 0.3 0

10 0 030(10 )

[ ] [ ] 0.3 1 0200(0.91) 0 0 100 0 0.35

10 0 20

0 20 10

− −

= −

−

T

B D

Simplifying the above expression gives:

6

0 0 7

6 20 0

10 3 030(10 )[ ] [ ]200(0.91) 0 0 3.5

10 3 7

6 20 3.5

T B D

− − −

= − −

−

The element stiffness matrix is:

[ ] [ ] [ ][ ]= T k tA B D B

therefore:

6

0 0 7

6 20 0

10 3 0(0.15)(10 )[ ] [ ][ ] 1(100)

0.91 0 0 3.5

10 3 7

6 20 3.5

T tA B D B

− − −

= − −

−

0 0 10 0 10 01

0 20 0 0 0 20200

20 0 0 10 20 10

− × − − −





1 1 3 3 2 2

140 0 0 70 140 70

0 400 60 0 60 400

0 60 100 0 100 6075,000[ ]

0.91 70 0 0 35 70 35

140 60 100 70 240 130

70 400 60 35 130 435

u v u v u v

k

− − − − −

− −=

− − − − −

− − −

For element 2: the coordinates are x i = 0, y i = 0, x j = 20, y j = 0, x m = 20, and y m =

10. The area of the triangle is:

We need to calculate the element β’s and γ’s as:

0 10 10 20 20 0 β γ = − = − = − = − = − =i j m i m j y y x x

1 10 0 10 0 20 20 β γ = − = − = = − = − = − j m j i my y x x

0 0 0 20 0 20 β γ = − = − = = − = − =m i j m i j y y x x


10 0 10 0 0 01 1[ ] 0 0 0 20 0 20

2000 10 20 10 20 0

− = − − −

Bin

2(20)(10)100 .

2= = A in





61 0.3 0

30 10[ ] 0.3 1 0

0.910 0 0.35

× =

D psi

Therefore:

6

10 0 0

0 0 101 0.3 0

10 0 2030(10 )[ ] [ ] 0.3 1 0

200(0.91) 0 20 100 0 0.35

0 0 20

0 20 0

− − − = −

T B D


6

10 3 0

0 0 3.5

10 3 730(10 )[ ] [ ]

200(0.91) 0 20 3.5

6 0 7

6 20 0

− − − −

= −

−

T B D

The element stiffness matrix is:

[ ] [ ] [ ][ ]= T k tA B D B

therefore:

6

10 3 0

0 0 3.510 3 7(0.15)(10 )

[ ] [ ][ ] 1(100)0.91 0 20 3.5

6 0 7

6 20 0

− −

− −

= −

−

T tA B D B




10 0 10 0 0 01

0 0 0 20 0 20200

0 10 20 10 20 0

− × − − −


1 1 4 4 3 3

100 0 100 60 0 60

0 35 70 35 70 0

100 70 240 130 140 6075,000[ ]

0.91 60 35 130 435 70 400

0 70 140 70 140 0

60 0 60 400 0 400

u v u v u v

k

− − − − − − −

= − − −

− −

− −

Element 1:

Element 2:

1 1 2 2 3 3 4 4

28 0 28 14 0 14 0 0

0 80 12 80 12 0 0 0

28 12 48 26 20 14 0 0

14 80 26 87 12 7 0 0375,000[ ]

0.91 0 12 20 12 20 0 0 0

14 0 14 7 0 7 0 00 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0

− −

− −

− − −

− − −=

− −

− −

u v u v u v u v

k

1 1 2 2 3 3 4 4

20 0 0 0 0 12 20 12

0 7 0 0 14 0 14 7

0 0 0 0 0 0 0 00 0 0 0 0 0 0 0375,000

[ ]0.91 0 14 0 0 28 0 28 14

12 0 0 0 0 80 12 80

20 14 0 0 28 12 48 26

12 7 0 0 14 80 26 87

− −

− −

=− −

− −

− − −

− − −

u v u v u v u v

k




Using the superposition, the total global stiffness matrix is:

1 1 2 2 3 3 4 4

48 0 28 14 0 26 20 12

0 87 12 80 26 0 14 7

28 12 48 26 20 14 0 0

14 80 26 87 12 7 0 0375,000[ ]

0.91 0 26 20 12 48 0 28 14

26 0 14 7 0 87 12 80

20 14 0 0 28 12 48 26

12 7 0 0 14 80 26 87

− − −

− − −

− − −

− −=

− − −

− − −

− − −

− − −

u v u v u v u v

k

The governing global matrix equations are:

[ ] =F K d

1

1

2

2

48 0 28 14 0 26 20 12

0 87 12 80 26 0 14 7

28 12 48 26 20 14 0 0

14 80 26 87 12 7 0 0

0 26 20 12 48 0 28 1426 0 14 7 0 87 12 80

20 14 0 0 28 12 48 26

12 7 0 0 14 80 26 87

375,000

0.915,0000

500

0

− − −

− − −

− − −

− −

− − −− − −

− − −

− − −

=

x

y

x

y

R

R

R

R

lb

lb

1

1

2

2

3

3

4

4

x

y

x

y

x

y

x

y

d

d

d

d

d d

d

d

Applying the boundary conditions:

1 1 2 20= = = = x y x y

d d d d

The governing equations are:

3

3

4

4

5,000 48 0 28 14

0 0 87 12 80375,000

0.915,000 28 12 48 26

0 14 80 26 87

x

y

x

y

d lb

d

d lb

d

− − = − −

− −




Solving the equations gives:

( )

3

3 6

4

4

609.6

4.210

663.7

104.1

−

=

x

y

x

y

d

d in

d

d

The exact solution for the displacement at the free end of the one-dimensional

bar subjected to a tensile force is:

6

6

(10,000)20670 10

10(30 10 )δ

−= = = ××

PLin

AE

The in-plane stress can be related to displacements by:

[ ][ ] =s D B d

( )2

1 0 0 0 0

1 0 0 0 02 (1 )

0 0 0.5 1

ν β β β

σ ν γ γ γ ν

ν γ β γ β γ β

= − −

ix

iy

i j m

jx

i j m

jy

i i j j m m

mx

my

d

d

d E

d A

d d

Element 1:

( )

1

1

1 3 2

3

1 3 223

1 1 3 3 2 2

2

2

1 0 0 0 0

1 0 0 0 02 (1 )

0 0 0.5 1

ν β β β

σ ν γ γ γ ν


= −

−

x

y

x

y

x

y

d

d

d E

d A

d

d




6 6

0.0

0.01 0.3 0 0 0 10 0 10 0

609.630(10 )(10 )0.3 1 0 0 20 0 0 0 20

0.96(200) 4.20 0 0.35 20 0 0 10 20 10

0.00.0

σ

σ

τ

−

− = −

− −

x

y

xy

The stresses are:

1,005

301

2.4

σ

σ

τ

=

x

y

xy

psi

psi

psi

Element 2:

( )

1

1

1 4 3

4

1 4 324

1 1 4 4 3 3

3

3

1 0 0 0 0

1 0 0 0 02 (1 )

0 0 0.5 1

ν β β β

σ ν γ γ γ ν


= − −

x

y

x

y

x

y

d

d

d E

d A

d

d

6 6

0.0

0.01 0.3 0 10 0 10 0 0 0

663.730(10 )(10 )0.3 1 0 0 0 0 20 0 20

0.96(200) 104.10 0 0.35 0 10 20 10 20 0

609.6

4.2

σ

σ

τ

−

= −

−

x

y

xy

The stresses are:

995

1.2

2.4

σ

σ

τ

= − −

x

y

xy

psi

psi

psi




The principal stresses and principal angle are:

2

2

1

995 1.2 995 1.2( 2.4) 995

2 2

− + = + + − =

s psi

22

2

995 1.2 995 1.2( 2.4) 1.1

2 2σ

− + = − + − = −

psi

11 2( 2.4)0

2 995 1.2θ − − = ≈ +

o

p tan



Problems

16. Do problems 6.5, 6.6, 6.9, 6.10, and 6.13 on pages 301 - 306 in yourtextbook “A First Course in the Finite Element Method” by D. Logan.

17. Rework the plane stress problem given on page 291 in your textbook “A

First Course in the Finite Element Method” by D. Logan using SAP2000 to

do analysis. Start with the simple two element model. Continuously refine

your discretization by a factor of two each time until your FEM solution is in

agreement with the exact solution for both displacements and stress. Howmany elements did you need?

caso placa 4

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