Energy Method Easy

8/13/2019 Energy Method Easy

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CIVL 7117 Finite Elements Methods in Structural Mechanics Page 43

Introduction to the Stiffness(Displacement) Method

Introduction

This section introduces some of the basic concepts on which the direct stiff-

ness methodis based. The linear spring is simple and an instructive tool to illus-

trate the basic concepts. The steps to develop a finite element model for a linear

spring follow our general 8 step procedure.

1. Discretize and Select Element Types- Linear spring elements2. Select a Displacement Function- Assume a variation of the dis-

placements over each element.

3. Define the Strain/Displacement and Stress/Strain Relationships-

We next illustrate how to assemble the total stiffness matrix for a struc-

ture comprising an assemblage of spring elements by using elementary

concepts of equilibrium and compatibility.

4. Derive the Element Stiffness Matrix and Equations- Define the stiff-

ness matrix for an element and then consider the derivation of the stiff-ness matrix for a linear-elastic spring element.

5. Assemble the Element Equations to Obtain the Global or Total

Equations and Introduce Boundary Conditions- We then show how

the total stiffness matrix for the problem can be obtained by superimpos-

ing the stiffness matrices of the individual elements in a direct manner.

The term direct stiffness methodevolved in reference to this method.

After establishing the total structure stiffness matrix, impose boundary

conditions (both homogeneous and nonhomogeneous).6. Solve for the Unknown Degrees of Freedom (or Generalized Dis-

placements)- Solve for the nodal displacements.

7. Solve for the Element Strains and Stresses- The reactions and inter-

nal forces association with the bar element.

8. Interpret the Results


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Step 1 - Select Element Type

Consider the linear spring shown below. The spring is of length Land is sub-

jected to a nodal tensile force, Tdirected along the x axis.

Step 2 - Select a Displacement Function

A displacement function u is assumed.

xaau 21+=

In general, the number of coefficients in the displacement function is equal to the

total number of degrees of freedom associated with the element. We can write

the displacement function in matrix forms as:

[ ]

=

2

11a

axu

We can express u as a function of the nodal displacements d by evaluating u at each node and solving for a1and a2.

11)0( adxux===

122)( aLadLxu

x +===


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Solving for a2:

L

dda xx 12

2

=

Substituting a1and a2into u gives:

x

xx dxL

ddu

1

12

+

=

In matrix form:

=

x

x

d

d

L

x

L

xu

2

1

1

Or in another form:

[ ]

=x

x

d

dNNu

2

1

21

where N1and N2are defined as:

The functions Niare called interpolation functionsbecause they describe how

the assumed displacement function varies over the domain of the element. In this

case the interpolation functions are linear.

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

Tensile forces produce a total elongation (deformation) of the spring. For lin-

ear springs, the force T and the displacementare related by Hookes law:

kT=

L

xN

11 = L

xN

2 =


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where deformation of the spring is given as:

)0()( uLu =

xx dd

12 =

Step 4 - Derive the Element Stiffness Matrix and Equations

We can now derive the spring element stiffness matrix as follows:

Rewrite the forces in terms of the nodal displacements:

( )xxx ddkfT 121 ==

( )xxx

ddkfT122

==

We can write the last two force-displacement relationships in matrix form as:

=

x

x

x

x

d

d

k

k

k

k

f

f

2

1

2

1

This formulation is valid as long as the spring deforms along the x axis. The co-efficient matrix of the above equation is called the local stiffness matrixk:

=

k

k

k

kk

Tfx =

1 Tf

x=

2


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Step 5 - Assemble the Element Equations

and Introduce Boundary Conditions

The global stiffness matrixand the global force vectorare assembled using

the nodal force equilibrium equations, and force/deformation and compatibilityequations.

[ ] =

==N

e

ekK1

)(K

{ } =

==N

e

efF1

)(F

where kand fare the element stiffness and force matrices expressed in globalcoordinates.

Step 6 - Solve for the Nodal Displacements

Solve the displacements by imposing the boundary conditions and solving the

following set of equations:

KdF=

Step 7 - Solve for the Element Forces

Once the displacements are found, the forces in each element may be calcu-

lated from:

kT=


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Example 1 - Spring Problem

Consider the following two-spring system shown below:

where the element axis x coincides with the global axisx.

For element 1:

=

x

x

x

x

d

d

k

k

k

k

f

f

3

1

1

1

1

1

3

1

For element 2:

=

x

x

x

x

d

d

k

k

k

k

f

f

2

3

2

2

2

2

2

3

Both continuity and compatibility require that both elements remain connected atnode 3.

)2(

3

)1(

3 xx dd =

We can write the nodal equilibrium equation at each node as:

)1(

11 xx fF =

)2(

22 xx

fF =

)2(

3

)1(

33 xxx ffF +=


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Therefore the force-displacement equations for this spring system are:

xxx dkdkF

31111 =

xxx dkdkF

22322 +=

xxxxx dkdkdkdkF

223231113 ++=

In matrix form the above equations are:

+

=

x

x

x

x

x

x

d

d

d

kkkk

kk

kk

F

F

F

3

2

1

2121

22

11

3

2

1

0

0

or

KdF=

where Fis the global nodal force vector, dis called the global nodal dis-

placement vector, and Kis called the global stiffness matrix.

Assembling the Total Stiffness Matrix by Superposition

Consider the spring system defined in the last example:

The elemental stiffness matrices may be written for each element. For element 1:


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=

1

1

1

1)1(

31

k

k

k

kk

ddxx

For element 2:

=

2

2

2

2)2(

23

k

k

k

kk

ddxx

Write the stiffness matrix in global format for element 1 as follows:

=

)1(

3

)1(

2

)1(

1

)1(

3

)1(

2

)1(

1

1

101

000101

x

x

x

x

x

x

f

ff

d

dd

k

For element 2:

=

)2(

3

)2(

2

)2(

1

)2(

3

)2(

2

)2(

1

2

110110

000

x

x

x

x

x

x

ff

f

dd

d

k

Apply the force equilibrium equations at each node.

=

+

x

x

x

x

x

x

x

F

F

F

f

f

f

f

3

2

1

)2(

3

)2(

2

)1(

3

)1(

10

0

The above equations gives:

=

+

x

x

x

x

x

x

F

F

F

d

d

d

kkkk

kk

kk

3

2

1

3

2

1

2121

22

11

0

0


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To avoid the expansion of the each elemental stiffness matrix, we can use a

more direct, shortcut form of the stiffness matrix. For element 1:

=1

1

1

1)1(

31

k

k

k

kk

ddxx

For element 2:

=

2

2

2

2)2(

23

k

k

k

kk

ddxx

The global stiffness matrix may be constructed by directly adding terms associ-

ated with the degrees of freedom in k(1)and k(2)into their corresponding locations

in the Kas follows:

+

=

21

2

1

2

2

1

1

321

0

0

kk

k

k

k

k

k

k

dddxxx

K

Boundary Conditions

In order to solve the equations defined by the global stiffness matrix, we must

apply some form of constraints or supports or the structure will be free to move

as a rigid body.

Boundary conditions are of two general types: homogeneous boundary

conditions(the most common) occur at locations that are completely prevented

from movement; nonhomogeneous boundaryconditions occur where finite


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non-zero values of displacement are specified, such as the settlement of a sup-

port.

Consider the equations we developed for the two-spring system. We will con-

sider node 1 to be fixed d1x= 0. The equations describing the elongation of the

spring system become:

=

+

x

x

x

x

x

F

F

F

d

d

kkkk

kk

kk

3

2

1

3

2

2121

22

110

0

0

Expanding the matrix equations gives:

xx dkF 311 =

xxx dkdkF

22322 +=

( ) xxx

dkkdkF321223

++=

The second and third equation may be written in matrix form as:

=

+

x

x

x

x

F

F

d

d

kk

k

k

k

3

2

3

2

21

2

2

2

Once we have solved the above equations for the unknown nodal displacements,

we can use the first equation in the original matrix to find the support reaction.

xx dkF

311 =

For homogeneous boundary conditions, we can delete the row and column cor-

responding to the zero-displacement degrees-of-freedom.

Lets again look at the equations we developed for the two-spring system.

However, this time we will consider a nonhomogeneous boundary condition (at

node 1 d1x= . The equations describing the elongation of the spring system be-

come:


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=

+

x

x

x

x

x

F

F

F

d

d

kkkk

kk

kk

3

2

1

3

2

2121

22

11

0

0

Expanding the matrix equations gives:

xx dkkF

3111 =

xxx dkdkF

32222 =

( ) xxx

dkkdkkF3212213

++=

By considering the second and third equations because they have known nodal

forces we get:

xxx dkdkF

32222 =

( ) xxx

dkkdkkF3212213

++=+

Writing the above equations in matrix form gives:

+=

+

13

2

3

2

21

2

2

2

kFF

dd

kkk

kk

x

x

x

x

For nonhomogeneous boundary conditions, we must transfer the terms from the

stiffness matrix to the right-hand-side force vector before solving for the unknown

displacements.

Once we have solved the above equations for the unknown nodal displace-

ments, we can use the first equation in the original matrix to find the support re-

action.

xx dkkF

3111 =


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Example 2 -Spring Problem

Consider the following three-spring system:

The elemental stiffness matrices for each element are :

=

1

1

1

11000)1(

31

k

=

1

1

1

12000)2(

43

k

=

1

1

1

13000)3(

24

k

Using the concept of superposition (the direct stiffness method), the global stiff-

ness matrix is:

=

5000

2000

3000

0

2000

3000

0

1000

3000

0

3000

0

0

1000

0

1000

K

The global force-displacement equations are:

=

x

x

x

x

x

x

x

x

F

F

F

F

d

d

d

d

4

3

2

1

4

3

2

1

5000

2000

3000

0

2000

3000

0

1000

3000

0

3000

0

0

1000

0

1000


13/27


We have homogeneous boundary conditions at nodes 1 and 2 (d1x= 0 and d2x=

0). Deleting the first two rows and the first two columns and substituting for the

known force at node 4 (F4x= 5000) gives:

=

5000

0

5000

2000

2000

3000

4

3

x

x

d

d

Solving for d3xand d4xgives:

.

11

15.

11

1043

indindxx

==

To obtain the global forces, substitute the displacement in the force-displacement

equations.

=

1115

1110

4

3

2

1

0

0

5000

2000

3000

0

2000

3000

0

1000

3000

0

3000

0

0

1000

0

1000

x

x

x

x

F

F

F

F

Solving for the forces gives:

lbFlbFxx

11

000,45

11

000,1021

==

lbFFxx

11

000,550

43 ==

Next, use the local element equations to obtain the force in each spring.For element 1:

=

1110

3

10

1000

1000

1000

1000

x

x

f

f

The local forces are:


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lbflbfxx

11

000,1011

000,1031

==

A free-body diagram of the spring element 1 is shown below.

For element 2:

=

1115

1110

4

3

2000

2000

2000

2000

x

x

f

f


lbflbfxx

11

000,1011

000,1043

==


For element 3:

=

03000

3000

3000

3000

11

15

2

4

x

x

f

f


lbflbfxx

11

000,4511

000,4524

==


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Consider the following four-spring system:

The spring constant k= 200 kN/m and the displacement = 20 mm. Therefore,

the elemental stiffness matrices are:

mkNkkkk /1

1

1

1200)4()3()2()1(

====

Using superposition (the direct stiffness method), the global stiffness matrix is:

=

200

200

0

0

0

200

400

200

0

0

0

200

400

200

0

0

0

200

400

200

0

0

0

200

200

K


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=

x

x

x

x

x

x

x

x

x

x

F

FF

F

F

d

dd

d

d

5

4

3

2

1

5

4

3

2

1

200

2000

0

0

200

400200

0

0

0

200400

200

0

0

0200

400

200

0

00

200

200

Applying the boundary conditions (d1x= 0 and d5x= 20 mm) and the known

forces (F2x, F3x, and F4xequal to zero) gives:

=

x

x

x

x

F

d

d

d

5

4

3

2

0

0

0

02.020020000

2004002000

0200400200

00200400

Rearranging the first three equations gives:

=

4

0

0

400

200

0

200

400

200

0

200

400

4

3

2

x

x

x

d

d

d

Solving for d2x, d3x, and d4xgives:

mdmdmdxxx

015.001.0005.0432

===

Solving for the forces F1xand F5xgives:

kNFx

0.1)005.0(2001

==

kNFx

0.1)02.0(200)015.0(2005

=+=


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Next, use the local element equations to obtain the force in each spring.

For element 1:

=

005.0

0

200

200

200

200

2

1

x

x

f

f


kNfkNfxx

0.10.121

==

For element 2:

=

01.0

005.0

200

200

200

200

3

2

x

x

f

f


kNfkNfxx

0.10.132

==

For element 3:

=

015.0

01.0

200

200

200

200

4

3

x

x

f

f


kNfkNfxx

0.10.143

==

For element 4:

=

02.0015.0

200200

200200

5

4

x

x

ff


kNfkNfxx

0.10.154

==


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Consider the following spring system:

The boundary conditions are:

0431

=== xxx

ddd

The compatibility condition at node 2 is:

xxxx dddd

2

)3(

2

)2(

2

)1(

2 ===

Using the direct stiffness method: the elemental stiffness matrices for each ele-

ment are:

=1

1

1

1)1(

21

k

k

k

kk

=2

2

2

2)2(

32

k

k

k

kk

=

3

3

3

3)3(

42

k

k

k

kk


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Using the concept of superposition (the direct stiffness method), the global stiff-

ness matrix is:

++

=

3

3

2

2

3

2

321

1

1

1

0

0

0

0

0

0

4321

k

k

k

k

k

k

kkkk

kk

K

Applying the boundary conditions (d1x= d3x = d4x = 0) and the known forces (F2x

= P) gives:

( ) xdkkkP

2321 ++=

Solving the equation gives:

321

2kkk

Pd

x++

=

Solving for the forces gives:

xxxxxx dkFdkFdkF

234223211 ===

Potential Energy Approach to Derive Spring Element Equations

One of the alternative methods often used to derive the element equations

and the stiffness matrix for an element is based on theprinciple of minimumpotential energy. This method has the advantage of being more general than

the methods involving nodal and element equilibrium equations, along with the

stress/strain law for the element.

Thus, the principle of minimum potential energy is more adaptable for the de-

termination of element equations for complicated elements (those with large

numbers of degrees of freedom) such as the plane stress/strain element, the axi-


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symmetric stress element, the plate bending element, and the three-dimensional

solid stress element.

Total Potential Energy

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

and the potential energy of the external forces :

+=Up

Strain energy is the capacity of the internal forces (or stresses) to do work

through deformations (strains) in the structure; is the capacity of forces such

as body forces, surface traction forces, and applied nodal forces to do work

through the deformation of the structure.

Recall the force-displacement relationship for a linear spring:

kxF=

The differential internal work (or strain energy) dUin the spring is the internal

force multiplied by the change in displacement which the force moves through:

( )dxkxFdxdU ==

The total strain energy is:

( ) 20 2

1kxdxkxdUU

x

L

===

The strain energy is the area under the force-displacement curve. The potential

energy of the external forces is the work done by the external forces:


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Fx=

Therefore, the total potential energy is:

Fxkxp = 2

2

1

The concept of a stationary valueof a function Gis shown below:

The function Gis expressed in terms ofx. To find a value ofxyielding a station-

ary value of G(x), we use differential calculus to differentiate Gwith respect tox

and set the expression equal to zero.

0=dx

dG

We can replace Gwith the total potential energy pand the coordinatexwith a

discrete value di. To minimize pwe first take the variationof p(we will not

cover the details of variational calculus):

n

n

ppp

p dddddd

++

+

= ...22

1

1

The principle states that equilibrium exist when the didefine a structure state

such that p= 0 for arbitrary admissible variations difrom the equilibrium state.

An admissible variationis one in which the displacement field still satisfies the

boundary conditions and interelement continuity.


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To satisfy p= 0, all coefficients associated with dimust be zero independently,

therefore:

{ }0,,2,10 =

==

dorni

d

p

i

p L


Consider the following linear-elastic spring system subjected to a force of

1,000 lb. Evaluate the potential energy for various displacement values and showthat the minimum potential energy also corresponds to the equilibrium position of

the spring.


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

and the potential energy of the external forces :

The variation of pwith respect toxis:

0=

= x

x

p

p

Since xis arbitrary and might not be zero, then:

0=

x

p

Using our express for p, we get:

( )xlbxinlbFxkxp 1000)/(50021

21 22 ==

.0.210005000 inxxx

p ===

If we had plotted the total potential energy function pfor various values of de-

formation we would get:

Fx=22

1kxU=

+=Up


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Lets derive the spring element equations and stiffness matrix using the principal

of minimum potential energy. Consider the linear spring subjected to nodal forces

shown below:

The total potential energy p is:

( ) xxxxxxp

dfdfddk2211

2

12

2

1=

Expanding the above express gives:

( ) xxxxxxxxp dfdfddddk 22112

121

2

22

21 =

Minimizing the total potential energy p:

210 toid

i

p ==


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( ) xxx

pfddk

d112

1

222

10 +==

( ) xxx

pfddk

d

212

2

22

2

10 ==

In matrix form the above equations are:

=

x

x

x

x

d

d

k

k

k

k

f

f

2

1

2

1

This result is identical to stiffness matrix we obtained using equilibrium.

Example 6 - Spring Problem

Obtain the total potential energy of the spring system shown below and find its

minimum value.

The potential energy p for element 1 is:

( ) xxxxxxp

dfdfddk3311

2

131

)1( 2

1=


( ) xxxxxxp

dfdfddk4433

2

342

)2(

2

1=


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( ) xxxxxxp

dfdfddk4422

2

423

)3(

2

1=

The total potential energy p for the spring system is:

=

=3

1

)(

e

e

pp

Minimizing the total potential energy p:

)1(

111311

0xxx

x

pfdkdk

d+==

)3(

24323

2

0xxx

x

pfdkdk

d==

)2(

3

)1(

332421131

3

0xxxxxx

x

pffdkdkdkdk

d+==

)3(

4

)2(

443233242

4

0xxxxxx

x

p ffdkdkdkdkd

+==

In matrix form:

+

+=

+

+

)3(

4

)2(

4

)2(

3

)1(

3

2

1

4

3

2

1

32

2

3

2

21

1

3

3

1

10

0

0

0

0

0

xx

xx

x

x

x

x

x

x

ff

ff

f

f

d

d

d

d

kk

k

k

k

kk

k

k

k

k

k

Using the following force equilibrium equations:

xx Ff

1

)1(

1 =

xx Ff

2

)3(

2 =


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=

x

x

x

x

x

x

x

x

F

F

F

F

d

d

d

d

4

3

2

1

4

3

2

1

5000

2000

3000

0

2000

3000

0

1000

3000

0

3000

0

0

1000

0

1000

The above equations are identical to those we obtained through the direct stiff-

ness method.

Problems:

5. Do problems 2.2, 2.9, and 2.17on pages 59 - 62 in your textbook A First

Course in the Finite Element Method by D. Logan.

xxx Fff

4

)3(

4

)2(

4 =+xxx Fff 3

)2(

3

)1(

3 =+

Energy Method Easy

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