03 One Dimensional Problems

8/22/2019 03 One Dimensional Problems

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One Dimensional Problems

Lecture Notes

Dr Mohd Afendi

Universiti Malaysia Perlis

ENT467 Finite Element Analysis

Chapter 3

2

Introduction

Total potential energy, stress-strain and strain-displacement relationships are used in developing

FEM for one dimensional problems. Same basic procedure for two- and three-

dimensional problems.

For one-dimensional problems, the vector is afunction ofx.

u=u(x), =(x), = (x), T=T(x), f=f(x)

Stress-strain and strain-displacement relations are:

=du

Edx

3

Introduction

Three types of loading:

Body force, f (self weight due to gravity)

Traction, T (frictional resistance, viscous drag)

Point load, Pi.

Figure 1

4

Finite Element Modelling

In one dimensional problems, every node ispermitted to displace only in the xdirection.

Each node has only one degree of freedom (dof).

Element division, numbering scheme.

Figure 2

5


Displacement along each dof is denoted by Q=[Q1,

Q2,,Qn]T

and global node vector, F=[F1,,Fn]T

.

Figure 3

6


Global numbering.

Element connectivity.

Figure 4

Figure 5


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7

Coordinates andShape Functions

Natural or intrinsic coordinate system, :

Figure 6

12 1

2( ) 1x x

x x

8


The unknown displacement field within an elementwill be interpolated by a linear distribution.

Approximation becomes more accurate if moreelements are considered.

Figure 7

9


Linear shape function:

Linear displacement:

where q1 and q2 are nodal displacement

In matrix notation: u= Nq N = [N1, N2]T and q = [q1, q2]T

1 2

1 1( ) , ( )

2 2N N

1 1 2 2u N q N q

10


Stress derivation.

Strain-displacement relation.

Also, since:

We have:

chain rule

du du d

dx d dx

differentiate1

2 1 2 1

2 2( ) 1

dx x

x x dx x x

1 1 2 2 1 2

1 1

2 2u N q N q q q

1 2

2

du q q

d

11


Then, substitute into previous equation:

This equation can be written as:

= Bq

matrix B is the element strain-displacement matrix.

The stress, from Hookes law, is:

=EBq

1 22 1

1( )q qx x

2 1

1= [ 1 1]

x x

B

12

Example 3.1

Referring to the following figure:

(a) Evaluate , N1and N2at point P.

(b) Ifq1= 0.003 in and q2= -0.005 in,determine the value of the displacement qatpoint P

x2 = 36 inx = 24 inx1 = 20 in

21 P


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Example 3.1

Answer:

Using Eq. 3.4 the coordinate of P is given by:

Using Eq. 3.7a we get:

1 2

x1 = 20 in

P

x2 = 36 in

x = 24 in

5.01)2024(16

2p 25.075.0 21 NandN

inup 001.0)005.0(25.0)003.0(75.0 14

Problem 3.1

Consider the bar in Fig. 3.1. Cross-section Ae= 1.2 in2, and

Young modulus E = 30x106 psi. If q1= 0.02 and q2 = 0.025

in determine the following: The displacement at point P

The strain and stress

The element stiffness matrix

The strain energy in the element

1 2

x1 = 15 in

P

x2 = 23 in

x = 20 in

Figure 3.1

15

Problem 3.1

Answer

(a) q=0.023125 in

(b) = 0.000625 and = 18750 psi

16

Potential-Energy Approach

General expression for potential-energy approach:

Since the continuum has been discretized into finiteelements, the expression for becomes:

T T T1 u f u T u P2

Ti iV V S

i

dV dV dS

T T1

2

Ti i

e e ee e e i

Adx u fAdx u Tdx Q P

17


Then, the equation can be written as:

where Ueis the element strain energy.

T Te i i

e ee e e i

U u fAdx u Tdx Q P

1

2

Te e

U Adx

18


Element Stiffness Matrix

Consider the strain energy term.

Substitute for = EBqand =Bqinto above equation.

From previous equation:

1

2

Te e

U Adx

T T T T1 1or [ ]2 2

e ee eU E Adx U E Adx q B Bq q B B q

2 1 or2 2

elx xdx d dx d


4/8

19


The element strain energy Ueis now written as:

where Eeis Youngs modulus. Note that , then:

which result in:

1T T

1

1

2 2

ee e e

lU A E d

q B B q

1

12d

T 21 1 1 [ 1 1]12e e e ee

U A l E l

q q

T1 1 11 12

e ee

e

A EU

l

q q

20



where the element stiffness matrix ke is given by:

T1

2

eeU q k q

1 11 1

e e e

e

E A

l

21


Force terms

Element body force term:


Integrals of the shape functions:

T1 1 2 2( )ee e

u fAdx A f N q N q dx

1T T

2

e e

ee e

A f N dxu fAdx

A f N dx

q

11 1

1

21

12 2 2

1

2 2 2

e ee

e e

e

l lN dx d

l lN dx d

22


The body force term can be written as:

or

where the element body force vector fe is:

T T 112e

ee

Au fAdx l f q

T T e

eu fAdx q f

112e e

e

Al ff

23


The element traction force term:T

1 1 2 2( )e eu Tdx N q N q Tdx

1T T

2

e

ee

T N dxu Tdx

T N dx

q

T T e

eu Tdx q T

112e eTlT

24


At this stage, element matrices ke, feand Tehave

been obtained. Then, the total potential energy canbe written as:

where Kis the global stiffness matrix, F is the global loadvectorand Q is the global displacement vector.

Assembly of the global stiffness matrix and loadvector:

T T1

2 Q KQ Q F

( )

e

eee

e

K k

F f + T P


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Galerkin Approach

Galerkins variational form for one dimensionalproblem:

On the discretized region:

T T( ) 0T i iL L Li

Adx fAdx Tdx P

T T( ) 0T i ie e ee e e i

E Adx fAdx Tdx P

26

Galerkin Approach

Summary.

The variational form can be written as:

Or

where

T T T 0e e e i ie e e i

P k q f T

T ( ) 0 KQ F

1 11 1

112

112

e e e

e

e e e

e e

E A

lA l f

Tl

k

f

T

Element bodyforce vector

Element stiffnessmatrix

Element tractionforce

27

Penalty Approach

Penalty approach is used to handle the boundaryconditions.

In this approach, global stiffness K is modified byadding a large number Cto the first diagonalelement which has specified boundary condition.

Similarly, global load vector is also modified byadding Cand boundary condition number.

Consider a displacement boundary condition:

Q1= a1

28

Penalty Approach

The modified stiffness matrix and modified loadvector are given by:

It is that the value ofCis:

Reaction force: R= -CQ

11 12 1 1 1 1

21 22 2 2 2

1 2

( ) NN

N NN N NN

K C K K Q F CaK K K Q F

Q FK K K

4max 10ijC K 11

i Nj N

29

Example 3.4

Consider the bar shown in fig. E3.4. An axial

load P = 200x103

N is applied as shown. Usingthe penalty approach for handling boundaryconditions do the following:

(a) Determine the nodal displacements

(b) Determine the stress in each material

(c) Determine the reaction forces

Ex 3.4

30

300 mm 400 mm

P

1 2

12

3

Aluminum

A1 = 2400 mm2

E1 = 70 x 109 N/m2

Steel

A2 = 600 mm2

E2 = 200 x 109 N/m2


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3131

Ex. 3.4

(a) Determine the nodal displacements

11

11

300

24001070 31k

11

11

400

60010200 32k

1 21

2

Global dof

2 3

2

3

11

11

e

eee

l

AEk

3232

Ex. 3.4

Structural stiffness matrix K is assembled from k1and k2

Global load vector

11

11

300

24001070 31k

11

11

400

60010200 32k

30.330.30

30.386.056.5

056.556.5

106K

TF 010200,0 3

3333

Ex. 3.4

Global load vector is

dofs 1 and 3 are fixed

Using penalty approach, a large number of C isadded to the first and third diagonal element of K

30.330.30

30.386.056.5

056.556.5

106K

TF 010200,0 3

410max ijKC

46 101086.0 C

max

30.860030.30

30.386.056.5

056.556.8600

106K

3434

Ex. 3.4

Finite element equation are given by

46 101086.0 C

0

10200

0

30.860030.30

30.386.056.5

056.556.8600

3

3

2

1

Q

Q

Q

NNNNNN

N

N

F

F

CaF

Q

Q

Q

KKK

KKK

KKCK

2

11

3

2

21

22221

11211

3535

Ex. 3.4

The solution of displacement

0

10200

0

30.860030.30

30.386.056.5

056.556.86003

3

2

1

Q

Q

Q

mmQ T66 101127.823257.0101432.15

3636

EBQ

Ex. 3.4

(b) the stresses in each element

Using Eq. 3.15 and Eq. 3.16

MPa27.5423257.0

101432.1511

300

11070

66

1

111

12

xx

B

MPa29.116101127.8

23257.011

400

110200

6

3

2


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Ex. 3.4

(c) Determine the reaction force at the support

Using Eq. 3.78

3610

11 1023.130101432.151086.0 CQR

riaQCR iPP ,...,2,1)( 11

3610

33 1077.69101127.81086.0 CQR

38

Quadratic Shape Functions

In some problems, use of quadratic interpolationleads to more accurate results.

Consider a typical three-node quadratic element.

Transformation of the coordinate system:

Figure 8

3

2 1

2( )x x

x x

39


In -coordinate, quadratic shape function N1, N2and N3 will be introduced as:

Figure 9

11 2

12 2

3

( ) (1 )

( ) (1 )

( ) (1 )(1 )

N

N

N

40


The displacement field within the element:

Figure 10

1 1 2 2 3 3

oru N q N q N q

u

Nq

41


The strain is given by:

Then,

The stress is given by:

du du d dx d dx

2 1

2 1 2 1 2, , 2 or =

2 2x x

q Bq

E Bq

42


The element stiffness matrix:

The element body force vector:

The element traction-force vector:

7 1 81 7 83 8 8 16

e e e

e

E Al

k

1/ 61/ 62 / 3

ee eA l f

f

1/ 61/ 62 / 3

eel T

T


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Temperature Effects

For thermal stress problems.

Initial strain: 0 = T

The element load vector:

This equation can be simplified as:

1 T0 1

2

e ee e

lE A d

B

2 1

1

1e e e eE A l T

x x

2 1

[ 1 1]

x x

B

44

Temperature Effects

The global load vector:

The desired element load vector:

The stress in each element:

Or

( )e e e

e

F f T P

( )E T Bq

2 1

[ 1 1]E

E Tx x

q

2 1

11

e e e eE A l T

x x

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