Finite Element Methodsite.iugaza.edu.ps/marafa/files/FEM-Chapter-2-2017-18.pdfFinite Element Method Chapter 2 Introduction to the Stiffness Method k d f Elastic Spring Element: f =

Finite Element Method

Finite Element Method

Chapter 2

Introduction to the Stiffness

Method

k

d

f

Elastic Spring Element:

f = k d

Consider the same spring but as a part of a structure such that it is connected to

other springs at its ends.; so

points 1 & 2 nodes of element

where k = spring constant or stiffness of spring (N/m)

1 2

1 2

ˆ local axis, positive direction as indicated

ˆ ˆ ˆ, local nodal force, positive in x direction

ˆ ˆ, local nodal displacements (DOF at each node),

ˆ positive x in direction

x x

x x

x

f f

d d

Elastic Spring Element:

1 1

2 2

1 111 12

21 222 2

ˆ ˆˆ ˆ

ˆ ˆ

ˆ ˆ

ˆ ˆ

x x

x x

x x

x x

f df and d

f d

so

f dk k

k kf d

So , we can write the vectors

General Steps Applied to a Spring Structure:

Step 1. Discretize and Select Element Type:


Step 2. Select a displacement function:

In general, total # of the ai coefficients = total # of DOF associated with the element

xaau ˆˆ21

2

1ˆ1ˆa

axu

Evaluate

at each node;

11ˆ0ˆ adu x LaadLu x 212

ˆˆ

L

dda xx 12

2

ˆˆ

are called the shape functions (interpolation functions); noticing that

N1 = 1 at node 1 and N1 = 0 at node 2

N2 = 0 at node 1 and N2 = 1 at node 2, and

1 2

2 11

1 1

1 2

2 2

1 2

ˆ ˆ

ˆ ˆˆˆ ˆ

ˆ ˆˆ ˆˆ 1

ˆ ˆ

ˆ ˆ1 ,

x xx

x x

x x

u a a x

d du x d

L

d dx xu N N

L L d d

x xN and N

L L

1 2ˆN + N = 1 for any x along the element.

Shape Function


Step 3. Define Strain/Displacement & Stress/Strain Relationships:

For the linear spring, Hook’s law states (force/deformation

instead of stress/strain)

kT

)0(ˆ)(ˆ uLu

xx dd 12ˆˆ

1ˆ ˆNote that has a negative value because it in the opposite direction of

therefore; = positive value - (negative value).

xd x


Step 4. Derive the Element Stiffness Matrix and

Equations:Recall that we have three alternative methods that can be used to obtain

element stiffness matrix, namely;

1.Direct Equilibrium method.

2.Work or Energy methods (min. P.E. is the most common one).

3.Weighted Residual Methods (Galerkin’s is the most popular one).

For the spring elements, we will use the direct equilibrium approach:

1

2

1 2 1

2 2 1

ˆ

ˆ

ˆ ˆ ˆ( )

ˆ ˆ ˆ( )

x

x

x x x

x x x

f T

f T

T f k d d

T f k d d


Step 4. Derive the Element Stiffness Matrix and

Equations:

)ˆˆ(ˆ211 xxx ddkf

)ˆˆ(ˆ122 xxx ddkf

x

x

x

x

d

d

kk

kk

f

f

2

1

2

1

ˆ

ˆ

ˆ

ˆ

kk

kkk̂

the “local” stiffness matrix for a spring element is

So the element equations in matrix form are:


Step 5. Assemble Element Equations to Obtain Total Equations:

Next we assemble element equations to obtain the total (structure) equations.

dKF

N

e

eN

e

efFFkKK

1

)(

1

)(,

Where are now element stiffness and force matrices, respectively,

in a global frame of reference. If not in a global coordinate axes, element matrices

would have to be transformed into a global frame then may be assembled.

)()(,

eefandk


Step 6. Apply Boundary Conditions and Solve for Nodal

Displacements:

Next we impose boundary conditions to the total system Kd=F;

from which we obtain a reduced (smaller) system. This reduced system then can be

solved for the unknown d’s.

Step 7. Solve for Element Forces:

Back substituting the values of d’s into equation Kd=F will yield the element

forces.

Some properties of K

K is symmetric (as the case in each k)

K is singular (has no inverse) until sufficient boundary

conditions are imposes to remove singularity and prevent

rigid-body motion.

Main diagonal elements of K are always positive.

k̂k̂ T

0)k̂(det 22 kk

Note The consequence is that the matrix is NOT invertible. It is not

possible to invert it to obtain the displacements. Why?

The spring is not constrained in space and hence it can attain

multiple positions in space for the same nodal forces

e.g.,

2

2-

4

3

22-

2-2

f̂

f̂

2

2-

2

1

22-

2-2

f̂

f̂

2x

1x

2x

1x

Problem

Analyze the behavior of the system composed of the two springs

loaded by external forces as shown above

k1 k2

F1x F2x F3xx

Given

F1x , F2x ,F3x are external loads. Positive directions of the forces

are along the positive x-axis

k1 and k2 are the stiffnesses of the two springs

Problem of Springs System

SolutionStep 1: In order to analyze the system we break it up into smaller

parts, i.e., “elements” connected to each other through “nodes”

k1k2F1x F2x F3x

x

1 2 3

Element 1 Element 2

Node 1d1x d2x

d3x

Unknowns: nodal displacements d1x, d2x, d3x,

Solution

Step 2: Analyze the behavior of a single element (spring)

k1k2F1x F2x F3x

x

1 2 3

Element 1 Element 2

Node 1d1x d2x

d3x

Two nodes: 1, 2

Nodal displacements:

Nodal forces:

Spring constant: k

1xd̂ 2xd̂

1xf̂2xf̂

d̂

2x

1x

k̂f̂

2x

1x

d̂

d̂

kk-

k-k

f̂

f̂

Solution

Step 3: Now that we have been able to describe the behavior of

each spring element, lets try to obtain the behavior of the original

structure by assembly

Split the original structure into component elements

)1()1()1(

d̂

(1)

2x

(1)

1x

k̂

11

11

f̂

(1)

2x

(1)

1x

d̂

d̂

kk-

k-k

f̂

f̂

)2()2()2(

d̂

(2)

2x

(2)

1x

k̂

22

22

f̂

(2)

2x

(2)

1x

d̂

d̂

kk-

k-k

f̂

f̂

Eq (3) Eq (4)

Element 1k11 2

(1)

1xd̂(1)

1xf̂ (1)

2xf̂(1)

2xd̂

Element 2

k22 3

(2)

1xd̂(2)

1xf̂(2)

2xf̂(2)

2xd̂

To assemble these two results into a single description of the

response of the entire structure we need to link between the local

and global variables.

Question 1: How do we relate the local (element) displacements

back to the global (structure) displacements?

k1k2F1x F2x F3x

x

1 2 3

Element 1 Element 2

Node 1d1x d2x

d3x

3x

(2)

2x

2x

(2)

1x

(1)

2x

1x

(1)

1x

dd̂

dd̂d̂

dd̂

Eq (5)

Hence, equations (3) and (4) may be rewritten as

2x

1x

11

11

(1)

2x

(1)

1x

d

d

kk-

k-k

f̂

f̂

3x

2x

22

22

(2)

2x

(2)

1x

d

d

kk-

k-k

f̂

f̂

Eq (6) Eq (7)

Or, we may expand the matrices and vectors to obtain

d

3x

2x

1x

k̂

11

11

f̂

(1)

2x

(1)

1x

d

d

d

000

0kk-

0kk

0

f̂

f̂

)1()1(

ee

d

x3

2x

1x

k̂

22

22

f̂

(2)

2x

(2)

1x

d

d

d

kk-0

kk0

000

f̂

f̂

0

)2()2(

ee

Expanded element stiffness matrix of element 1 (local)

Expanded nodal force vector for element 1 (local)

Nodal load vector for the entire structure (global)

e)1(

k̂e)1(

f̂

d

Question 2: How do we relate the local (element) nodal forces back

to the global (structure) forces? Draw 5 FBDs

0f̂-F:3nodeAt

0f̂f̂-F:2nodeAt

0f̂-F:1nodeAt

(2)

2x3x

(2)

1x

(1)

2x2x

(1)

1x1x

© 2002 Brooks/Cole Publishing / Thomson Learning™

k1k2F1x F2x F3x

x

1 2 3

d1x d2xd3x

A B C D

(1)

1xf̂ (1)

2xf̂(2)

1xf̂ (2)

2xf̂2xF1xF 3xF

2 3

In vector form, the nodal force vector (global)

(2)

2x

(2)

1x

(1)

2x

(1)

1x

3x

2x

1x

f̂

f̂f̂

f̂

F

F

F

F

Recall that the expanded element force vectors were

(2)

2x

(2)

1x

)2((1)

2x

(1)

1x)1(

f̂

f̂

0

f̂ and

0

f̂

f̂

f̂ee

Hence, the global force vector is simply the sum of the expanded

element nodal force vectors

ee )2()1(

3x

2x

1x

f̂f̂

F

F

F

F

dk̂f̂ anddk̂f̂(2)e)2((1)e)1(

ee

But we know the expressions for the expanded local force vectors

from Eqs (6) and (7)

dk̂k̂dk̂dk̂f̂f̂F(2)e(1)e(2)e(1)e)2()1(

ee

Hence

dKF

matricesstiffnesselementexpandedofsum

matrixstiffnessGlobalK

vectorntdisplacemenodalGlobald

vectorforcenodalGlobalF

For our original structure with two springs, the global stiffness

matrix is

22

2211

11

k̂

22

22

k̂

11

11

kk-0

kkkk-

0kk

kk-0

kk0

000

000

0kk-

0kk

K

)2()1(

ee

NOTE

1. The global stiffness matrix is symmetric

2. The global stiffness matrix is singular

The system equations imply

3x22x23x

3x22x211x12x

2x11x11x

3x

2x

1x

22

2211

11

3x

2x

1x

dkd-kF

dkd)kk(d-kF

dkdkF

d

d

d

kk-0

kkkk-

0kk

F

F

F

These are the 3 equilibrium equations at the 3 nodes.

dKF

0f̂-F:3nodeAt

0f̂f̂-F:2nodeAt

0f̂-F:1nodeAt

(2)

2x3x

(2)

1x

(1)

2x2x

(1)

1x1x

k1k2F1x F2x F3x

x

1 2 3

d1x d2xd3x

A B C D


(1)

1xf̂ (1)

2xf̂(2)

1xf̂ (2)

2xf̂2xF1xF 3xF

2 3

(1)

1x2x1x11x f̂ddkF

(2)

1x

(1)

2x

3x2x22x1x1

3x22x211x12x

f̂f̂

ddkddk

dkd)kk(d-kF

(2)

2x3x2x23x f̂dd-kF

Notice that the sum of the forces equal zero, i.e., the structure is in

static equilibrium.

F1x + F2x+ F3x =0

Given the nodal forces, can we solve for the displacements?

To obtain unique values of the displacements, at least one of the

nodal displacements must be specified.

Direct assembly of the global stiffness matrix

k1k2F1x F2x F3x

x

1 2 3

Element 1 Element 2d1x d2x

d3x

Global

Element 1k11 2

(1)

1xd̂(1)

1xf̂ (1)

2xf̂(1)

2xd̂

Element 2

k22 3

(2)

1xd̂(2)

1xf̂(2)

2xf̂(2)

2xd̂

Local

Node element connectivity chart : Specifies the global node

number corresponding to the local (element) node numbers

ELEMENT Node 1 Node 2

1 1 2

2 2 3

Global node number

Local node number

11

11)1(

kk-

k-kk̂

Stiffness matrix of element 1

d1x

d2x

d2xd1x

Stiffness matrix of element 2

22

22)2(

kk-

k-kk̂

d2x

d3x

d3xd2x

Global stiffness matrix

22

2211

11

kk-0

k-kkk-

0k-k

K d2x

d3x

d3xd2x

d1x

d1x

Examples: Problems 2.1 and 2.3 of Logan

Imposition of boundary conditionsConsider 2 cases

Case 1: Homogeneous boundary conditions (e.g., d1x=0)

Case 2: Nonhomogeneous boundary conditions (e.g., one of the

nodal displacements is known to be different from zero)

Homogeneous boundary condition at node 1

k1=500N/m k2=100N/m F3x=5Nx1

2 3

Element 1 Element 2d1x=0 d2x

d3x

System equations

1 1

2

3

500 -500 0

-500 600 -100 0

0 -100 100 5

x x

x

x

d F

d

d

Note that F1x is the wall reaction which is to be computed as part

of the solution and hence is an unknown in the above equation

Writing out the equations explicitly

2x 1

2 3

2 3

-500d

600 100 0

100 100 5

x

x x

x x

F

d d

d d

0

Eq(1)

Eq(2)

Eq(3)

Eq(2) and (3) are used to find d2x and d3x by solving

Note use Eq(1) to compute 1 2x=-500d 5xF N

2

3

2

3

600 100 0

100 100 5

0.01

0.06

x

x

x

x

d

d

d m

d m

NOTICE: The matrix in the above equation may be obtained from

the global stiffness matrix by deleting the first row and column

500 -500 0

-500 600 -100

0 -100 100

600 100

100 100

NOTICE:

1. Take care of homogeneous boundary conditions

by deleting the appropriate rows and columns from the

global stiffness matrix and solving the reduced set of

equations for the unknown nodal displacements.

2. Both displacements and forces CANNOT be known at

the same node. If the displacement at a node is known, the

reaction force at that node is unknown (and vice versa)

Imposition of boundary conditions…contd.

Nonhomogeneous boundary condition: Node 1 has nonzero

displacement of 0.06 m)

k1=500N/m k2=100N/mx1

2 3Element 1 Element 2

d1x=0.06m d2xd3x

F3x=5N

System equations

1 1

2

3

500 -500 0

-500 600 -100 0

0 -100 100 5

x x

x

x

d F

d

d

Note that now d2x , d3x and F1x are not known.

0.06

1 11 12 1

2 21 22 2

=Q K K d

Q K K d

1

2

3

500 -500 0 0.06

-500 600 -100 0

0 -100 100 5

x

x

x

F

d

d

2

3

0.07

0.12

x

x

d

d m

Now solve for d2x and d3x

2x

3

2x

3

d600 100 0 5000.06

100 100 5 0

d600 100 30

100 100 5

x

x

d

d

solve for F1x

2 1

1

500 0.06 500 0

500 0.06 500 0.07 5

x x

x

d F

F N

Recap of what we did

Step 1: Divide the problem domain into non overlapping regions (“elements”)

connected to each other through special points (“nodes”)

Step 2: Describe the behavior of each element ( )

Step 3: Describe the behavior of the entire body (by “assembly”).

This consists of the following steps

1. Write the force-displacement relations of each spring in expanded form

d̂k̂f̂

d̂k̂f̂ ee

Recap of what we did…contd.

2. Relate the local forces of each element to the global forces at the nodes

(use FBDs and force equilibrium).

Finally obtain

Where the global stiffness matrix

e

f̂F

dKF

e

kK

Recap of what we did…contd.

Apply boundary conditions by partitioning the matrix and vectors

2

1

2

1

2221

1211

F

F

d

d

KK

KK

Solve for unknown nodal displacements

1212222 dKFdK

Compute unknown nodal forces

2121111 dKdKF

Physical significance of the stiffness matrix

In general, we will have a stiffness

matrix of the form

(assume for now that we do not know

k11, k12, etc)

333231

232221

131211

kkk

kkk

kkk

K

3

2

1

3

2

1

333231

232221

131211

F

F

F

d

d

d

kkk

kkk

kkkThe finite element

force-displacement

relations:

k1k2F1x F2x F3x

x

1 2 3

Element 1 Element 2d1x d2x

d3x


The first equation is

313

212

111

kF

kF

kF

1313212111 Fdkdkdk Force equilibrium

equation at node 1

What if d1=1, d2=0, d3=0 ?

Force along node 1 due to unit displacement at node 1



While nodes 2 and 3 are held fixed

Similarly we obtain the physical significance of the other

entries of the global stiffness matrix

Columns of the global stiffness matrix


ijk = Force at node ‘i’ due to unit displacement at node ‘j’

keeping all the other nodes fixed

In general

This is an alternate route to generating the global stiffness matrix

e.g., to determine the first column of the stiffness matrix

Set d1=1, d2=0, d3=0

k1k2F1 F2 F3

x

1 2 3

Element 1 Element 2d1 d2

d3

Find F1=?, F2=?, F3=?


For this special case, Element #2 does not have any contribution.

Look at the free body diagram of Element #1

x

k1

(1)

1xd̂

(1)

1xf̂ (1)

2xf̂

(1)

2xd̂

(1) (1) (1)

2x 1 2x 1x 1 1ˆ ˆ ˆf (d d ) (0 1)k k k

(1) (1)

1x 2x 1ˆ ˆf f k


F1

F1 = k1d1 = k1=k11

F2 = -F1 = -k1=k21

F3 = 0 =k31

(1)

1xf̂

Force equilibrium at node 1

(1)

1 1x 1ˆF =f k

Force equilibrium at node 2

(1)

2xf̂

F2

(1)

2 2x 1ˆF =f k

Of course, F3=0


Hence the first column of the stiffness matrix is

1 1

2 1

3 0

F k

F k

F

To obtain the second column of the stiffness matrix, calculate the

nodal reactions at nodes 1, 2 and 3 when d1=0, d2=1, d3=0

1 1

2 1 2

3 2

F k

F k k

F k

Check that


To obtain the third column of the stiffness matrix, calculate the

nodal reactions at nodes 1, 2 and 3 when d1=0, d2=0, d3=1

1

2 2

3 2

0F

F k

F k

Check that

Steps in solving a problem

Step 1: Write down the node-element connectivity table

linking local and global displacements

Step 2: Write down the stiffness matrix of each element

Step 3: Assemble the element stiffness matrices to form the

global stiffness matrix for the entire structure using the

node element connectivity table

Step 4: Incorporate appropriate boundary conditions

Step 5: Solve resulting set of reduced equations for the

unknown displacements

Step 6: Compute the unknown nodal forces

Example 1

Given: For the spring system shown,

K1 = 100 N/mm, K2 = 200 N/mm,

K3 = 300 N/mm

P = 500 N,

Find:

(a) the global stiffness matrix

(b) displacements of nodes 3 and 4

(c) the reaction forces at nodes 1 and 2

(d) the force in the spring 2

Solution

(a) the global stiffness matrix

1

100 0 100 0 1

100 100 1 0 0 0 0 2

100 100 3 100 0 100 0 3

0 0 0 0 4

k

2

0 0 0 0 1

200 200 3 0 0 0 0 2

200 200 4 0 0 200 200 3

0 0 200 200 4

k

3

0 0 0 0 1

300 300 4 0 300 0 300 2

300 300 2 0 0 0 0 3

0 300 0 300 4

k

K = k1 + k2 + k3

100 0 100 0 1

0 300 0 300 2

100 0 300 200 3

0 300 200 500 4

K

K=F

(b) displacements of nodes 3 and 4

d1 = d2 = 0., F3 = 0 and F4 = P = 500

3 3

4 4

10

300 200 0. 11

200 500 500 15

11

d d

d d

1 1

2 2

3 3

4 4

100 0 100 0

0 300 0 300

100 0 300 200

0 300 200 500

d F

d F

d F

d F

1

12

2

0

100 0 100 0 0 1000

0 300 0 300 1110

100 0 300 200 0 450011

110 300 200 500 15 500

11

F

FF

F

(c) the reaction forces at nodes 1 and 2

(d) the force in the spring 2

22 2 3 3

22 2 4 4

2 2

3 3

2 2

4 4

10 1000

200 200 3 11 11

200 200 4 15 1000

11 11

k k d f

k k d f

f f

f f

Example 2

Compute the global stiffness matrix of the assemblage of springs

shown above

1000 1000 0 0

1000 1000 2000 2000 0K

0 2000 2000 3000 3000

0 0 3000 3000

d3xd2xd1xd4x

d2x

d3x

d1x

d4x


22 3 4

Example 3

1 1

1 1 2 3 2 3

2 3 2 3

k -k 0

K -k k k k - k k

0 - k k k k

Compute the global stiffness matrix of the assemblage of springs

shown above


3

Given: For the spring system shown,

Find:

(a) Nodal displacements

(b) The reactions

(c) Forces in each element

Solution

1

100 100 0 0 1

100 100 1 100 100 0 0 2

100 100 2 0 0 0 0 3

0 0 0 0 4

k

Example 3

2

0 0 0 0 1

50 50 2 0 50 50 0 2

50 50 3 0 50 50 0 3

0 0 0 0 4

k

3

0 0 0 0 1

50 50 2 0 50 0 50 2

50 50 4 0 0 0 0 3

0 50 0 50 4

k

1 1

2 2

3 3

4 4

100 100 0 0 1 100 100 0 0 1

100 200 50 50 2 100 200 50 50 2

0 50 50 0 3 0 50 50 0 3

0 50 0 50 4 0 50 0 50 4

d F

d FK

d F

d F

1

2

2

3

4

0 100 0 0 0

4000 200 0 02

0 50 0 0 0

0 50 0 0 0

F

dd mm

F

F

(a) Nodal displacements

d1 = d3 = d4 = 0., and F2 = P = 400

(b) The reactions

1

1

3

3

4

4

0 100 0 0 0200

4000 200 0 0 2100

0 50 0 0 0100

0 50 0 0 0

FF N

F NF

F NF

(c) Forces in each element

1 1

1 1

2 2

2 2

2 2

3 3

3 3

2 2

4 4

1

200100 100 0

200100 100 2

2

50 50 2 100

50 50 0 100

3

50 50 2 100

50 50 0 100

Element

f f

f f

Element

f f

f f

Element

f f

f f

Example

For the spring system with arbitrarily numbered nodes and

elements, as shown above, find the global stiffness matrix.

Solution:

First we construct the following

which specifies the global node numbers corresponding

to the local node numbers for each element.

Then we can write the element stiffness matrices as

follows :

Potential Energy Approach to Derive Spring

Element EquationsThe alternative methods used to derive the element equations are:

1. Equilibrium method.

2. Energy methods (minimum potential, virtual work)

3. Weighted residual methods (Galerkin’s)

Energy Method (minimum potential energy):

npp dddd ,.....,,, 321

Potential Energy Approach to Derive Spring

Element Equations

Energy Method (minimum potential energy):

npp dddd ,.....,,, 321

Total potential energy is defined as the sum of the internal strain

energy U and the potential energy of the external forces

Up

Internal strain energy, U, is the work done by internal forces (or

stresses) through deformations (strains) in the structure.

Potential energy of the external forces, , is the potential energy

which is lost when work is done by external forces (body forces, surface

traction forces, and applied nodal forces)

W

As an example, consider the linear spring shown:The differential internal strain energy,

dU = internal force x change in displacement through which the

force moves

Or dU = F dx and since F = k x

then

dU = k x dx

therefore, the total strain energy is

xFW

U is area under the force-deformation curve.

2120

1 12 2( )

x

U k x dx k x

U kx x F x

Also, the potential energy of the external force

can be written as:

For a function G(x)The stationary value of a function G(x) can be a maximum,

a minimum, or a neutral point on G(x).

From differential calculus, any stationary value x of G(x)

must satisfy .

0dx

dG

xFxkp 2

21

npp dddd ,.....,,, 321

Then the first variation of has the general form:

n

n

ppp

p dd

dd

dd

2

2

1

1

To satisfy , all coefficients associated with the must be zero independently.

Thus,

0),,2,1(0

dorni

d

p

i

p

NE

e

e

pp

1

)(

For the linear spring element shown

212 1 1 1 2 22

2 212 1 2 1 1 1 2 22

ˆ ˆ ˆ ˆ ˆ ˆ( )

ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ( 2 )

p x x x x x x

p x x x x x x x x

k d d f d f d

k d d d d f d f d

12 1 12

1

12 1 22

2

0 1,2

ˆ ˆ ˆ( 2 2 ) 0ˆ

ˆ ˆ ˆ( 2 2 ) 0ˆ

p

i

p

x x x

x

p

x x x

x

id

k d d fd

k d d fd

The total potential energy is

Simplifying, we get

In matrix form, we write the element equations as:

x

x

x

x

f

f

d

d

kk

kk

2

1

2

1

ˆ

ˆ

ˆ

ˆ

xxx fddk 121ˆ)ˆˆ(

xxx fddk 221ˆ)ˆˆ(

It can be shown that the total potential energy of an entire

structure can be minimized with respect to each nodal degree of

freedom and this minimization will result in the same total

equations of the structure as those obtained by the direct stiffness

method.

For the linear-elastic spring subjected to a force shown, evaluate the

potential energy for various displacement values and show that the

minimum potential energy also corresponds to the equilibrium position

of the spring.

Example 4

We now illustrate the minimization of p through

standard mathematics

Example 4

We now search for the minimum value of p for various values of spring

deformation x.

We observe that p has a minimum value at x = 2:00 in. This deformed

position also corresponds to the equilibrium position.

Example 4

Obtain the total potential energy of the spring assemblage and find its

minimum value. The procedure of assembling element equations can

then be seen to be obtained from the minimization of the total

potential energy.

Example 4

Example 4

Example 4

When we apply the boundary conditions and substitute F3x = 0

and F4x = 5000 lb into the above equation, the solution is

identical to that of Example 2.1.

HW

Fifth Edition:

P 2.9, 2.10, 2.17,2.18 and 2.19

Documents

Finite Element Methodsite.iugaza.edu.ps/marafa/files/FEM-Chapter-2-2017-18.pdfFinite Element Method Chapter 2 Introduction to the Stiffness Method k d f Elastic Spring Element: f =