1D ELEMENTS

• Simplest type of FE problems• All object are 1D• All forces are 1D• All stresses / strains are 1D

BASICS

• u=u(x) : Deformations• ε=ε(x) : Strain• σ=σ(x) : Stress• f=f(x) : Body forces• T=T(x) : Tractive forces• P=P(x) : Point loads• σ=Eε, ε=du/dx

Problem

P1 f

P2

Discritization

76 6 65 5544433

322

211

Node 2Node 1ElemNo.

Element connectivity Matrix

INDIVIDUAL ELEMENTS

x1 x2 :Coordinatesξ=-1 ξ=1 : Local coordinatesq1 q2 : Deformations

(At the nodes)

q2:Deformation

q1 within the element

Linear Interpolation N1, N2: Shape functions

node1 node2

LINEAR SHAPE FUNCTIONS

N1 N2

u = N1 q1 + N2 q2 = (1 - ξ) / 2 * q1+ (1 + ξ) / 2 * q2

Relationship between local and global coordinates-

ξ = -1+ 2*(x-x1)/(x2-x1)

N1 =(1 - ξ) / 2 N2 =(1 + ξ) / 2

STRAIN:

ε = du/dx

= du/dξ * dξ/dx

u=N1q1+N2q2

dξ/dx = 2/(x2-x1)

= 2/Le

du/dξ = (-q1+q2)/2

therefore-

ε = 1/le [-1 1][q1 q2]T

= Bq

N1 =(1 - ξ) / 2 N2 =(1 + ξ) / 2

Where B is the element strain matrix and

B== 1/le [-1 1]

As B is constant, this element is CONSTANT STRAIN ELEMENT

That means strain inside the element does not vary.

STRESS:

σ = EBq

P.E. approach

↓↓↓↓

−−−∈

−−−∈=Π

∑∑ ∑ ∫∫∫∑

∑∫∫∫

iie e L

T

L

T

L

T

e

iiL

T

L

T

L

T

PQTdxuAdxFuAdx

PuTdxuAdxFuAdx

σ

σ

21

21

STRAIN ENERGY FORCED TRACTIVE POINT FIELD FORCE FORCE LOADS

Total energy of the body is the sum of P.E, of all Elements.

ELEMENT STRAIN ENERGY

qEA qle

T2

1

1111

⎥⎦

⎤⎢⎣

⎡−

−=

∫ ∈=L

Te AdxU σ2

1

∫=L

TT EBqAdxBq21

qBdxBEAqL

TT⎥⎦⎤

⎢⎣⎡= ∫2

1

[ ] [ ]

∫

∫

⎥⎦

⎤⎢⎣

⎡−

−=

⎥⎦⎤

⎢⎣⎡ −−=

qdxl

EAq

qdxEAq

e

T

L

Tl

Te

11111

1111.

221

12

12

⎥⎦

⎤⎢⎣

⎡−

−=

=

1111

21

ee

e

eT

LAEk

matrixstiffnesselementisk

qkqUe

CONTRIBUTION OF THE FORCES TO PE

BODY FORCES

dxNfAqfAdxNq TTTT∫ ∫=

⎥⎦

⎤⎢⎣

⎡=

∫∫

dxNdxN

fAq T

2

1

node1 node2

1N1

2/2/)(.

12

1

eLxxtriangletheunderareadxN

=−==∫

x1x2

2tf 2tf

⎥⎦

⎤⎢⎣

⎡=

11

2eT lfAq

vectorforcebodyelementfq eT ==

⎥⎦

⎤⎢⎣

⎡=11

2eT fAlq

Thus, body force on element gets split equally attwo nodes.

TRACTIVE FORCES

2tT 2tT

∫ ∫= TdxNqTdxu TTT

⎥⎦

⎤⎢⎣

⎡=

∫∫

dxNdxN

TqT

2

1

eT

eT

Tq

Tlq

=

⎥⎦

⎤⎢⎣

⎡=

11

2[TOTAL TRACTIVE FORCE ON THE ELEMENT CAN BE ASSUMED TO BE SPLIT EQUALLY AT THE NODES]

∑−−−=Π PqTqfqqkq Te

Te

Te

Te 2

1

∑ ∑ ∑−=Π=Π eT

eT

e Fqqkq21

⎥⎦⎤

⎢⎣⎡ +⋅+=

2222eeee

eTlfAlTlfAlF

POTENTIAL ENERGY

Potential energy for an element is-

Total Potential energy-

CONSIDER A SET OF ELEMENTS

Element No. Node 1 Node 2

1 1 2

2 2 3

3 3 4

4 4 5

5 5 6

1 e1 2 e2 3 e3 4 e4 5 e5 6

Q1 Q2 Q3 Q4 Q5 Q6

ELEMENT STIFFNESS MATRIX

[ ] [ ] [ ] [ ]TTTT qqqqqqqq 54433221

[ ]TnqqqqQ −−−= 321

L⎥⎦

⎤⎢⎣

⎡−

−⎥⎦

⎤⎢⎣

⎡−

−⎥⎦

⎤⎢⎣

⎡−

−1111

1111

1111

3

33

2

22

1

11

lAE

lAE

lAE

GLOBAL DISPLACEMENT MATRIX

ELEMENT DISPLACEMENT MATRIX

⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

−+−

−+−

−−−−+−

−−−

=

n

nn

lAE

lAE

lAE

lAE

lAE

lAE

lAE

lAE

lAE

lAE

lAE

lAE

nlAE

lAE

K

.

.

.

00

0

0

00

4

44

3

33

3

33

3

33

3

33

2

22

2

22

2

22

2

22

1

11

1

11

1

11

1

11

2 L2 3 i Li I+11 L1

GLOBAL STIFFNESS MATRIX1 2 3 4 5 6 .......... n

K = 1⎡k111 k1

12 0 0 0 0 ..... 0⎤2⎜k1

21 k122+k2

11 k212 0 0 0 ..... 0 ⎜

3⎜0 k221 k2

22+k311 k3

12 0 0 ..... 0 ⎜4⎜0 0 k3

21 k322+k4

11 k412 0 ..... 0 ⎜

5..

n ⎣0 0 ....... km21 km

22⎦where

ki11, ki

12, ki21, ki

22: stiffness matrix elements of element number i.

PROPERTIES OF K

• Symmetric• Banded• Can be sparse (if numbering is not

appropriate)• Is N X N (where N is the number of nodes

in a 1 D problem)

SPARSE ‘K’ MATRIX

1 6 2 3 4 5

543

26

Node 2

4 3 2

61

Node 1

543

21

Elem No.

THE RESULTING K MATRIX: SPARSE AND NON-BANDED

1 2 3 4 5 6

K = 1⎡k111 0 0 0 0 k1

12 ⎤2⎜0 k2

22+k311 k3

12 0 0 k212 ⎜

3⎜0 k321 k4

11+k322 k4

12 0 0 ⎜4⎜0 0 k4

21 k422+k5

11 k512 0 ⎜

5⎜0 0 0 k521 k5

22 0 ⎜6⎣k1

21 k212 0 0 0 k1

22 + k211⎦

ASSEMBLING GLOBAL FORCE MATRICES FROM ELEMENT MATRICES

∑∑ +=++= PFP)T(f ie

ie

ie

Where : force on node 1 of element i: force on node 2 of element I: point load on node number i

T441

e32

e

331e

22e

212e

12e

111e

)...]PF(F

)PF)(FPF)(FP[(FF

++

+++++=

i1eF

i2eFiP

[ ][ ]

MATRIXFORCEGLOBALfffffffF

fifiOR

MATRIXFORCEELEMENT

TllfATllfA

T

iiiiii

...

2222

41323122211211

21

+++=

⎥⎦⎤

⎢⎣⎡ ++

TOTAL PE USING GLOBAL MATRICES

∑ ∑ ∑ ∑−−−=∏ iieTeT

eT uPTqfqqkq ******2/1

iieT

eT uPFqqkq∑ ∑ ∑−−= *****2/1

FQQKQ TT ****2/1 −=

where

Q = [q1 q2 q3 q4 .... qn]T

K : Global stiffness matrix

F : Global Force matrix

( ) ∫ ∑∫∫ =−−−∈s

T

v

T

v

T PTdSfdVdV

APPROACHSGALERKIN

0

'

φφσφσ

( )

( )

( )[ ]T

BN

dxd

x

21 φφ

φφ

φφ

φφ

=Ψ

Ψ=⇒∈Ψ=

=⇒∈

=

EBqBqNqu

=∈=

=

σ

DONEWORKINTERNAL

( )

[ ]

Ψ=

Ψ⎥⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡−

−=

Ψ=

Ψ=

Ψ=∈

∫

∫∫

eT

e

eT

Tee

T

Te

T

eTTT

kq

lEAq

BBlEAq

dxBBEAq

dxAEBBqdV

1111

φσν

eT

eeT

eeT

TeeT

Te

T

eTTT

F

lAf

dN

dNlAf

dNlAf

dxNfA

dxfANfdV

Ψ=

⎥⎦

⎤⎢⎣

⎡Ψ=

⎥⎥⎦

⎤

⎢⎢⎣

⎡Ψ=

Ψ=

Ψ=

Ψ=

∫∫

∫

∫

∫ ∫

11

2

2

2

2

1

ξ

ξ

ξ

φν

Virtual work dueTo body forces-

[ ]Teee

TT TlTwhereTTdx

DONEWORKTRACTIVESIMILARLY

112

=Ψ=∫ν

φ

Thus Galerkin's Equation Becomes

( )

∑ ∑∑∑ ∑

∫ ∫ ∫

=Ψ−Ψ⇒

=Ψ−Ψ−Ψ−Ψ

=Σ−−−∈

e ee

Te

T

Te

Te

Te

T

T

S S

TTT

FkqPTfkq

Ptdsfdvdv

00

0φφφφσν

[ ]nii ΨΨΨΨ + LLLL 11

⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

⎥⎦

⎤⎢⎣

⎡ Ψ+

Ψ−Ψ−Ψ

+

++

.

.

.

1

1

11

i

ii

iii

i

iii

i

iii

i

iii

QQ

Q

lAE

lAE

lAE

lAE

LL

Q

lAE

lAE

lAE

lAE

lAE

lAE

lAE

lAE

lAE

lAE

lAE

i

ii

i

ii

i

ii

i

ii

i

ii

i

ii

⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

+−

−+

−−

+−

−

+

++

−

−−

1

11

1

11

2

22

1

11

1

11

1

11

1

11

In global form-

approachEPFQKQQ

approachsGalerkinFKQ

TT

TT

........21

'........0

−=Π

=Ψ−Ψ

TO SOLVE THE PROBLEM• Find Deformations Qi’s

1. Define Boundary conditions2. Apply minimization of PE

• Find strains ε = B * qFind stresses σ = E ε = E B q

SOLVING FOR Q

[ ][ ]

⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢

⎣

⎡

=

=

=

nnnn

n

n

Tn

Tn

KKK

KKKKKK

K

FFFF

QQQQ

21

22221

11211

.....21

.....21

.

.....

FQKQQ

FQFQFQFQ

QKQQKQQKQQKQ

QKQQKQQKQQKQQKQQKQQKQQKQ

TT

nn

nnnnnnnnnnn

nn

nn

−=

+++−

⎟⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜⎜

⎝

⎛

+++

++++++++

=Π

21

).....(

.......

..........

21

332211

32211

22323222221212

11313121211111

BOUNDARY CONDITIONS

CONSTRAINTPOINTSINGLE−

...],0,0..[

....

52

2,1 21

==

==

QQge

aQaQ pp

LET US CONSIDER AS AN EXAMPLE

Q1 = a1

(

)...()........

.

.

...

...21

332211

2211

23231313

22323222221212

11313121211111

nn

nnnnnnnn

nn

nn

FQFQFQFaQKQaKQaKQ

QKQaKaQKQQKQQKQaKQ

QKaQKaQKaaKa

BECOMESA

++++−+++

+++++++++

+++++=Π

:.. PRINICIPLEEPMINIMUM

.0).....

.....(21

0

0).....

.....(21

0

0

33333232113

3333322131

3

22332222112

2323222121

2

⎪⎩

⎪⎨⎧

=−+++

++++=

∂Π∂

⎪⎩

⎪⎨⎧

=−+++

++++=

∂Π∂

=∂

Π∂

FQKQKKQaK

QKQKKQaKQ

FQKQKKQaK

QKQKKQaKQ

Q

nn

nn

nn

nn

i

⎪⎩

⎪⎨⎧

=−+++

+++++=

∂Π∂

0).....

.....(21

0

.

332211

332211

inniiii

niniii

i FQKQKKQaK

QKQKKQaKQ

111

11

1313

1212

3

2

432

3343332

2242322

113322

13133333322

12122233222

.

...

.....

...

............

...................

FQK

aKF

aKFaKF

Q

QQ

KKKK

KKKKKKKK

aKFKQKQKQ

aKFKQKQKQaKFKQKQKQ

nnnnnnnn

n

n

iiinnii

nn

nn

=⇒

⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢

⎣

⎡

−

−−

=

⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢

⎣

⎡

⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢

⎣

⎡

−=+++

−=+++−=+++

K’= OBTAINED FROM K BY DELETING 1ST ROW AND COLUMN

Q’= OBTAINED FROM Q BY DELETING Q1

F’ = OBTAINED FROM F BY DELETING F1 AND SUBTRACTING Ki1a1FROM FI

IF INSTEAD OF Q1 = a1, WE HAD Qi=ai THE SAME STEPS SHALL BE CARRIED PUT BY DOING THESE OPERATIONS ON ith ROW AND COLUMN.

REACTION AT THE SUPPORT (NODE 1 IF Q1 = a1)

113132121111

111313212111

........

FQKQKQKQKRRFQKQKQKQK

nn

nn

−++++=⇒+=++++

ELIMINATION METHOD

ENALTY APPROACH

•

ATTACH A SPRING OF STIFFNESS C

DEFLECT THE FIXED END BY a1

DEFLECTION OF NODE 1-Q1

e1 e5e2 e3 e4wallC

q1

a1

11, aQC →∞→

11 aQ −=δ

( ) [ ]SPRINGofEPaQCU s ..21 2

11 −=

.

....

...(21

)(21

21)(2

1

22323222221212

11313121211111

211

211

+++++

+++++

+−=

−+−=Π

nn

nn

TT

QKQQKQQKQQKQ

QKQQKQQKQQKQ

aQC

FQKQQaQC

-DEFLECTION OF SPRING

AS

)...()...

332211

22323222221212

nn

nn

FQFQFQFQQKQQKQQKQQKQ

++++−++++

0.....0

0.....0

0.....0

0

333333221313

223232221212

11313221111111

=−+++→=∂

Π∂

=−+++→=∂

Π∂

=−++++−→=∂Π∂

=∂Π∂

FQKQKKQQKQ

FQKQKKQQKQ

FQKQKQKQKCaCQQ

Q

nn

nn

nn

i

0.....0 332211 =−+++→=∂

Π∂nnnnnnn

n

FQKQKKQQKQ

⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢

⎣

⎡ +

=

⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢

⎣

⎡ +

nnnnnnn

n

n

F

FF

CaF

Q

QQQ

KKKK

KKKKKKKKKKKCK

.

...

....

..

..

..

3

2

11

3

2

1

321

3333231

2232221

13131211

Multi point constraints-

FQQQCKQQ

approachpenaultybyitsolvecanweaaexamplefor

QQ

Tpp

Tm

pp

−−++=Π

=−

=+

)(21

21

2

02211

21

02211

βββ

βββ

Since C has very large value, P.E. takes minimum valueWhen (β1Qp1+ β1Qp1-β0) is minimum.

⎥⎥⎦

⎤

⎢⎢⎣

⎡

++

++→

⎥⎥⎦

⎤

⎢⎢⎣

⎡22222112

21212

111

2212

2111

βββ

βββ

CkCk

CkCk

kk

kk

pppp

pppp

pppp

pppp

The modified stiffness and force matrices are-

⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

+

+→

⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

202

101

2

1

ββ

ββ

CF

CF

F

F

p

p

p

p

Reactions at support are given by-

)(

)(

0221122

0221111

ββββ

ββββ

−+−=

−+−=

ppp

ppp

QQCR

QQCR

STEPS INVOLVED IN SOLVING A 1D FE PROBLEM

• Make the Geometric Model

• Make an FE Mesh

• Define the Loading

• Develop Element and Global Matrices

• Define Boundary Conditions

• Develop Modified Global Matrices

• Solving Using Numerical Techniques

A SIMPLE PROBLEM

6” 5.25”l1=12”

12” Elem 1:1-2

P P=100lbE=3x107 psi l2=12”ρ=0.2836lb/in3

12” t=1” Elem_2:2-3

3” 3.75”

k1 = AE /L ⎡1 -1 ⎤ = 15.75 * 107 / 12 ⎡1 -1 ⎤⎣-1 1 ⎦ ⎣-1 1 ⎦

k2 = AE /L ⎡1 -1 ⎤ = 11.25 * 107 / 12 ⎡1 -1 ⎤⎣-1 1 ⎦ ⎣-1 1 ⎦

f1 = ρ A L / 2 [1 1]T = 5.25 * 12 *0.2836 / 2 [1 1]T

f2 = ρ A L / 2 [1 1]T = 3.75 * 12 *0.2836 / 2 [1 1]T

GLOBAL MATRICES

K = 3x107 / 12 ⎡ 5.25 -5.25 0 ⎤⎜-5.25 9.00 -3.75 ⎜⎣0 -3.75 3.75⎦

F = [8.93 15.31+100 6.38]T

BOUNDARY CONDITIONS

Q1 = 0

MODIFIED MATRICESK’= 3x107 / 12 ⎡ 5.25 -5.25 0 ⎤

⎜-5.25 9.00 -3.75 ⎜⎣0 -3.75 3.75⎦

= 3x107/12⎡9 -3.75 ⎤⎣-3.75 3.75⎦

F’= [8.93 (15.31+100-K21*a1) (6.38-K31*a1)]T

= [115.31 6.38]T

Q’ = [Q2 Q3]T

FINAL EQUATIONSK’Q’ = F’

Solving we get: Q’ = [0.9272 0.9953] x 10-5”

Q = [ 0 0.9272 0.9953] x 10-5”

STRESSES:

σ1 = EB1q = 3x107 * 1/12 [1 -1][0 0.9272x10-5]T

= 23.18 psiσ2 = EB2q = 3x107*1/12 [1 -1][0.9273 0.9953]T*10-5

= 1.70 psi