fea mit lecture

8/20/2019 fea mit lecture

1/95

2.094—

Finite

Element

Analysis

of

Solids

and

Fluids

—Fall‘08—

MITOpenCourseWare


2/95

Contents

1

Large

displacement

analysis

of

solids/structures

3

1.1 Project Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

1.2 Large Displacement analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

1.2.1 Mathematical model/problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

1.2.2 Requirementstobefulfilledbysolutionattimet . . . . . . . . . . . . . . . . . . . 5

1.2.3 Finite Element Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

1.2.4 Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

2

Finite

element

formulation

of

solids

and

structures

7

2.1 Principle of Virtual Work . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

2.2 Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

3

Finite

element

formulation

for

solids

and

structures

10

4

Finite

element

formulation

for

solids

and

structures

14

5 F.E.displacement formulation,cont’d 19

6

Finite

element

formulation,

example,

convergence

23

6.1 Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23

6.1.1 F.E. model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23

6.1.2 Higher-order elements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25

7

Isoparametric

elements

28

8

Convergence

of

displacement-based

FEM

33

9

u/ p

formulation

37

10

F.E.

large

deformation/general

nonlinear

analysis

41

11

Deformation,

strain

and

stress

tensors

45

12TotalLagrangian formulation 49

1


3/95

MIT2.094 Contents

13

Total

Lagrangian

formulation,

cont’d

53

14

Total

Lagrangian

formulation,

cont’d

57

15

Field

problems

61

15.1 Heat transfer

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

61

15.1.1 Differential formulation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61

15.1.2 Principle of virtual temperatures . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62

15.2 Inviscid, incompressible, irrotational flow . . . . . . . . . . . . . . . . . . . . . . . . . . . . 64

16

F.E.

analysis

of

Navier-Stokes

fluids

65

17

Incompressible

fluid

flow

and

heat

transfer,

cont’d

71

17.1 Abstract body . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71

17.2 Actual 2D problem (channel flow) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71

17.3 Basic equations

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

72

17.4 Model problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72

17.5 FSI briefly . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 75

18

Solution

of

F.E.

equations

76

18.1 Slender structures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 79

19

Slender

structures

81

20

Beams,

plates,

and

shells

85

21

Plates

and

shells

90

2


4/95

2.094—FiniteElementAnalysisof SolidsandFluids Fall‘08

Lecture1- Largedisplacementanalysisof solids/structures

Prof. K.J.Bathe MITOpenCourseWare

1.1 ProjectExample

Physicalproblem

Reading:

Ch. 1 in

thetext

“Simple”

mathematical

model

analyticalsolution•

F.E.solution(s)•

More

complex

mathematical

model

holesincluded•

largedisp./largestrains•

F.E.solution(s)• ⇒

Howmanyfiniteelements?•

Weneedagooderrormeasure(especiallyforFSI)

“Even

more

complex”

mathematical

model

The“complexmathematicalmodel”includes FluidStructureInteraction(FSI).

YouwilluseADINAinyourprojects(andhomework)forstructuresandfluidflow.

3


5/95

MIT2.094 1.Largedisplacementanalysisof solids/structures

1.2 LargeDisplacementanalysis

Lagrangianformulations:

• TotalLagrangianformulation

•

Updated

Lagrangian

formulation

Reading:Ch. 6

1.2.1

Mathematical

model/problem

Given theoriginalconfigurationof thebody,

thesupportconditions,theappliedexternal loads,theassumedstress-strainlaw

Calculate thedeformations,strains,stressesof thebody.

Question Isthereauniquesolution? Yes,forinfinitesimalsmalldisplacement/strain. Notnecessarily

forlargedisplacement/strain.

Forexample:

Snap-through

problem

Thesame load. Twodifferentdeformedconfigurations.

4


6/95


Column problem,statics

NotphysicaltR is in“direction”of bendingmoment Not inequilibrium.

⇒

1.2.2

Requirements

to

be

fulfilled

by

solution

at

time

t

I. Equilibriumof stresses (Cauchystresses, forcesperunitarea in tV and on tS f )withtheappliedbodyforcestf B andsurfacetractionstf S f

II. Compatibility

III. Stress-strain law

1.2.3 FiniteElementMethod

I. Equilibriumconditionmeansnow

• equilibriumatthenodesof themesh

• equilibriumof eachfiniteelement

II. Compatibilitysatisfiedexactly

III. Stress-strain lawsatisfiedexactly

5


7/95


1.2.4 Notation

Cauchystresses(forceperunitareaattimet):

tτ ij

i,

j= 1,2,3 tτ ij

= tτ ji

(1.1)

6


8/95

�


Lecture2- Finiteelementformulationof solidsandstructures


Reading:

Ch. 1,Sec.AssumethatontS u

thedisplacementsarezero(andtS u

isconstant). Needtosatisfyattimet: 6.1-6.2

Equilibrium of Cauchystressestτ ij

withapplied loads•

tτ T = tτ 11

tτ 22

tτ 33

tτ 12

tτ 23

tτ 31

(2.1)

(Fori= 1,2,3)

tτ ij,j +tf i

B = 0 intV (sumover j) (2.2)

tτ ij

tnj

= tf iS f

on tS f

(sumover j) (2.3)

(e.g. tf iS f

= tτ i1

tn1

+ tτ i2

tn2

+ tτ i3

tn3

) (2.4)

And: tτ 11tn1 +

tτ 12tn2 =

tf 1S f

• Compatibility Thedisplacementstui needtobecontinuousandzeroontS u.

Stress-Strain

law •

tτ ij = functiontuj (2.5)

7


9/95

MIT2.094 2.Finiteelementformulationof solidsandstructures

2.1 Principleof VirtualWork∗

t V

tτ ij

teij

d

tV =

tV

tf Bi

ui

d

tV +

t S f

tf S f

i uS f

i dtS f

(2.6)

where

teij

withui

tS u

=

=

1

2

∂ui

∂

txj

0

+

∂uj

∂

txi

(2.7)

(2.8)

2.2 Example

Assume“planesectionsremainplane”

Principle

of

Virtual

Work

tV

tτ 11 te11 dtV

=

tV

tf B1 u1 dtV

+

tS f

tP r uS f 1

d

tS f (2.9)

Derivation

of (2.9)

tτ 11,1 +tf B1

=0 by(2.2)�

tτ 11,1

+ tf B1

u1

=0

∗orPrincipleof VirtualDisplacements

(2.10)

(2.11)

8


10/95

�

�

�

�

MIT2.094 2.Finiteelementformulationof solidsandstructures

Hence,

tτ 11,1 +tf 1

B u1 dtV =0 (2.12)

tV

tτ u1,1tτ 11

d V + u1 dtV =0 (2.13) 11u 1

t

t

S

S

u

f

− tV t

tV

tf 1

B

Sf tτ tS

te11 u1 11 f

where tτ 11|tS f

=tP r.

Thereforewehave

te11tτ 11

d

tV = u1

tf 1B

d

tV +u

S

1

f tP rtS f (2.14)

tV

tV

From(2.12)to(2.14)wesimplyusedmathematics. Hence,if (2.2)and(2.3)aresatisfied,then(2.14)musthold. If (2.14)holds,thenalso(2.2)and(2.3)hold!

Namely,from(2.14)

t

tS f

t B t S f u1,1

tτ 11d V = u1tτ 11

tS u− u1

tτ 11,1d V = u1tf 1

d V +u1tP r

tS f

(2.15)tV

tV

tV

or

B t

S f u1tτ 11,1 +

tf 1 d V +u1

tP r −tτ 11

tS f =0 (2.16)tV

Nowletu1

=x 1− txL

tτ 11,1

+tf 1

B ,where tL=lengthof bar.

Hencewemusthavefrom(2.16)

tτ 11,1 +tf 1

B =0 (2.17)

andthenalso

tP =tτ (2.18)r

11

9


11/95


Lecture3- Finiteelementformulationforsolidsandstructures


Reading:

Sec. 6.1-6.2Weneedtosatisfyattimet:

Equilibrium •

t∂ τ ij +tf B = 0 (i= 1,2,3)intV (3.1)t i∂ xj

tτ ij nt

j = f t S f

i

(i=1,2,3)ontS f (3.2)

• Compatibility

• Stress-strain law(s)

Principleof virtualdisplacements

tV

tτ ij

teij

dV t =

tV

uitf Bi

dV t +

tS f

ui|tS f

f t S f

i

dS t f

(3.3)

teij = 2

1

∂

∂ t

u

xi

j

+∂

∂utx

j

i

(3.4)

If (3.3)holdsforanycontinuousvirtualdisplacement(zeroon tS ),then(3.1)and(3.2)holdand• uviceversa.

RefertoEx. 4.2 inthetextbook. •

10


12/95

�

�

MIT2.094 3.Finiteelementformulationforsolidsandstructures

Major

steps

I. Take(3.1)andweighwithui:

tτ ij,j +tf i

B ui = 0. (3.5a)

II. Integrate(3.5a)overvolumetV :

tτ ij,j +tf i

B ui dtV =0 (3.5b)

tV

III. Usedivergencetheorem. Obtainaboundarytermof stressestimesvirtualdisplacementsontS =tS u ∪

tS f .

IV. But,ontS u

theui

=0andontS f

wehave(3.2)tosatisfy.

Result: (3.3).

Example

tτ te11

d V =

S tf

ui

f 1 dS f t

S f t (3.6)11

t

tV

One

element

solution:

11


13/95


u(r) =

tu(r) =

u(r) =

1 1(1+r)u1

+2 21

(1+r) tu1

+21 1

(1

+

r)

u1 +2 2

(1−r)u2

1

2(1−r) tu2

(1

−

r)

u2

(3.7)

(3.8)

(3.9)

Supposeweknowtτ 11,tV ,tS f ,

tu... use(3.6).

Forelement1,

te11

=∂u

=B(1)u1

(3.10)∂

tx u2

T t for el. (1) B(1)T

tV te11t

τ 11d V −→ [u1 u2]

t V

tτ 11 dtV

(3.11)

=

tF (1)

for

el.

(1)

tF

(1)−→

[u1

u2] (3.12)tF ̂(1)

=U 1 U 2 U 3 (3.13)

0

u2 u1

where

tF ̂1(1)

= tF 2(1)

(3.14)

tF ̂2(1)

= tF 1(1)

(3.15)

Forelement2,similarly,

0

=U 1

U 2 U 3 F (2)

(3.16)

t ˆ

u2 u1

R.H.S.

(unknownreactionatleft)

U 1

U 2

U 3

0 (3.17)

t S f tS f

U T

f

· 1

Nowapply,

U

T =

1 0 0

(3.18)

then,

U T

= 0 1 0 (3.19)

then,

U

T

= 0 0 1 (3.20)

12


14/95


Thisgives,

tF̂

(1)

0

+

0tF̂

(2)

=

unknownreaction0

f

t

tS f 1

·tS f

(3.21)

We

write

that

as

tF = tR �

(3.22)tF

=fn tU 1,tU 2,

tU 3

(3.23)

13


15/95


Lecture4- Finiteelementformulationforsolidsandstructures


Weconsideredageneral3Dbody, Reading:Ch. 4

Theexactsolutionof themathematicalmodelmustsatisfytheconditions:

• Equilibrium withintV andontS f ,

Compatibility •

Stress-strain

law(s)

•

I. Differentialformulation

II. Variationalformulation(Principleof virtualdisplacements)(orweakformulation)

WedevelopedthegoverningF.E.equationsforasheetorbar

Weobtained

t

F

t

R =

(4.1)

wheretF isafunctionof displacements/stresses/materiallaw;andtR isafunctionof time.

Assume

for

now

linear

analysis: Equilibriumwithin0V andon0S f

,linearstress-strainlawandsmalldisplacementsyields

tF =K · tU (4.2)

Wewanttoestablish,

KU (t)=R (t) (4.3)

14


16/95


Consider

U ˆ T =

U 1 V 1 W 1 U 2 W N

(N nodes) (4.4)· · ·

whereU ˆ T isadistinctnodalpointdisplacementvector.

Note: forthemoment“removeS u

”

Wealsosay

U ˆ T = U 1 U 2 U 3 U n

(n= 3N ) (4.5)· · ·

Wenowassume (m)u

(m) (m)u =H (m)U ˆ , u = v (4.6a)w

whereH (m) is3xnandU ˆ isnx1.

�(m) =B(m)U ˆ (4.6b)

whereB(m) is6xn,and

�(m)T

=

�xx �yy �zz γ xy γ yz γ zx

∂v

∂u

e.g. γ xy = +∂x

∂y

Wealsoassume

u(m) = H (m)U ˆ (4.6c)

�(m) = B(m)U ˆ (4.6d)

15


17/95

�

�


Principleof VirtualWork:

T

�T τ dV = U ˆ f BdV V V

(4.7)canberewrittenas

(m)T

�(m)T

τ

(m)dV

(m) = U ˆ f B

(m)

dV

(m)

V (m) V (m)m m

Substitute(4.6a)to(4.6d).

T U ˆ B(m)

T

τ (m) dV (m) = V (m) m

U ˆT

H (m)T

f B(m)

dV

(m)

V

(m)m

τ

(m) =C (m)�(m) =C (m)B(m)U ˆ

Finally,

T U

̂� B(m)

T

C (m)B(m) dV (m) U ˆ =V (m)m

T

U ̂�

H (m)T

f B(m)

dV

(m)

V (m)m

with

�(m)T ˆ

T

B(m)T

=U

KU ˆ =R B

whereK isnxn,andR B isnx1.

Directstiffnessmethod:

K = K (m)

m

R B =

R B

(m)

m K (m) = B(m)

T C (m)B(m) dV

(m)

V (m)R B

(m)= H (m)

T f B

(m)

dV

(m)

V (m)

(4.7)

(4.8)

(4.9)

(4.10)

(4.11)

(4.12)

(4.13)

(4.14)

(4.15)

(4.16)

(4.17)

16


18/95


Example

4.5 textbook

E =Young’sModulus

Mathematicalmodel Planesectionsremainplane:

F.E.

model

U 1U

= U 2 (4.18)U 3

Element1

U 1

x xu(1)(x) = 1−100 100 0

U 2

(4.19)

U 3H (1)

17


19/95


�(1)xx(x)=

− 1100

1100

B(1)

0

U 1U 2U 3

(4.20)

Element2

u(2)(x)=

�(2)xx(x)=

0

0

1− x80x

80

H (2)

U

−180

1

80

B(2)

U

(4.21)

(4.22)

Then,

K =E

100

1−10

−110

000

+

13E

240

000

01

−1

0−11

(4.23)

where,

A

η=0

1


20/95


Lecture5- F.E.displacementformulation,cont’d


Forthecontinuum Reading:Ch. 4

Differentialformulation•

Variationalformulation(Principleof VirtualDisplacements) •

Next,weassumedinfinitesimalsmalldisplacement,Hooke’sLaw, linearanalysis

KU =R (5.1a)

(m) =

H (m)U u (5.1b)

K = K (m) (5.1c)

m

R = R (m)

(5.1d)B

m

�(m) =B(m)U (5.1e)

U

T = U 1

U 2

U n

, (n=alld.o.f. of elementassemblage) (5.1f)

· · ·

K (m) = B(m)T

C (m) B(m) dV (m) (5.1g)V (m)

R B(m

)

=

H (m)T

f B(m)

dV

(m)

(5.1h)V (m)

Surface

loads

Recallthatintheprincipleof virtualdisplacements,

“surface”loads= U S f

T

f S f dS f (5.2)S f

u

S

(m)

=H S (m)

U (5.3)

H S (m)

=H (m) (5.4)evaluated at the surface

19


21/95

MIT2.094 5.F.E.displacementformulation,cont’d

Substituteinto(5.2)

U

T H S

(m)T

f S (m)

dS (m) (5.5)S (m)

forelement(m)andonesurfaceof thatelement.

R (sm) = H S

(m)T

f S (m)

dS (m) (5.6)S (m)

Needtoaddcontributionsfromallsurfacesof allloadedexternalelements.

KU =R B

+R S

+R c

(5.7)

whereR c areconcentratednodalloads.

Assume

• (5.7)hasbeenestablishedwithoutanydisplacementboundaryconditions.

•

We,

however,

know

nodal

displacements

U b (rewriting

(5.7)).

KU =R ⇒

K aaK ba

K abK bb

U aU b

=

R aR b

(5.8)

SolveforU a:

K aaU a =R a −K abU b

whereU b

isknown!

Thenuse

(5.9)

K baU a +K bbU b =R b +R r (5.10)

whereR r areunknownreactions.

Example

4.6

textbook

τ xx E 1 ν 0 �xx

τ yy = ν 1 0 �yy (5.11)τ xy

1−ν 20 0 1−2

ν γ xy

20


22/95


u(x,y)v(x,y)

=H

u1u2

u3u4v1

v2

v3

v4

(5.12)

If wecansetthisrelationup,thenclearlywecangetH (1),H (2),H (3),H (4).

u(m) =H (m)U (5.13)

Alsowant�(m) =B(m)U . We want H . Wecouldproceedthisway

u(x,

y) =

a1 +

a2x

+

a3y

+

a4xy

(5.14)

v(x,y) =b1

+b2x+b3y+b4xy (5.15)

Expressa1. . .a4,b1. . .b4

intermsof thenodaldisplacementsu1. . .u4,v1. . .v4.

(e.g.) u(1,1)=a1

+a2

+a3

+a4

=u1.

h1(x,y) =14(1+x)(1+y) interpolation function

fornode1.

h2(x,y) =1

(1−x)(1+y)4

21


23/95


h3(x,y) =1(1−x)(1−y)4

h4(x,y) =1

(1+x)(1−y)4

u(x,y) =h1u1 +h2u2 +h3u3 +h4u4 (5.16)

v(x,y) =h1v1 +h2v2 +h3v3 +h4v4 (5.17)

u1

u2

u3u4v1v2v3v4

u(x,y)=

h1 h2 h3 h4 0 0 0 0 (5.18)v(x,y) 0 0 0 0 h1 h2 h3 h4

H (2x8)

Wealsowant,

u1u2u3u4v1v2

v3

v4

�xx h1,x h2,x h3,x h4,x 0 0 0 0 = 0 0 0 0 h1,y h2,y h3,y h4,y (5.19)�yy

γ xy

h1,y

h2,y

h3,y

h4,y

h1,x

h2,x

h3,x

h4,x

B (3x8)

�xx =∂u

∂x(5.20)

�yy =

∂v

∂y

(5.21)

γ xy =∂u

∂y+

∂v

∂x(5.22)

22


24/95

�


Lecture6- Finiteelementformulation,example,convergence


6.1 Example

Reading:

Ex. 4.6 in

thetext

t= 0.1, E, ν planestress

(6.1)KU =R ; R =R B

+R s

+R c

+R r

= K (m); K (m) = B(m)T

C (m) B(m) d V (m) (6.2)K V (m)m

R B = R (m) (m)

; R = H (m)T

f B(m) d V (m) (6.3)B BV (m)m

6.1.1

F.E.

model

u1

u2

u3

u4

v1

v2

v3

v4↓ ↓ ↓ ↓ ↓ ↓ ↓ ↓

u1× × × × × ←(6.4)K

el.

(2)

=

...

...

23


25/95

�

MIT2.094 6.Finiteelementformulation,example,convergence

Inpractice,

BT CBdV ; �=B

u1

...u4

v1

.

.

.

v4

K (6.5)=el

V

whereK is8x8andB is3x8.

AssumewehaveK (8x8)forel. (2)

U 1 U 2 U 3 U 4 U 5 U 11 U 18· · · · · ·↓ ↓ ↓ ↓ ↓ ↓ ↓ ↓ ↓× × × × × × × × ×

. . .. . .. . .

. . .. . .. . .

. . .. . .. . .

. . .. . .. . .

U 1←....

.

.

(6.6)

K =

assemblage

U 11←......

U 18←× × × × × × × × ×

Consider,

R S = H S T f S d S ; H S =H (6.7)

S

on surface

h1 h2 h3 h4 0 0 0 0 u(x,y)H =

0 0 0 0 h1 h2 h3 h4

←v(x,y)

(6.8)←

u1u2...

(6.9)u= u4v1...

v4

24


26/95


(6.10)H S =H y=+11

2

(1+x) 12

(1−x) 0 0 0 0 0 0=

(6.11)

0 0 0 0 1(1+x) 1(1−x) 0 02 2

From(6.7);

+1

=

=

1

(1+x) 021(1−x) 02

0

0(0.1) dx (6.12)R S

1(1+x) − p(x)2−1

10 (1−x) thickness2

0 0

− p0(0.1) − p0(0.1)

(6.13)R S

6.1.2 Higher-orderelements

Wanth1,h2,h3,h4,h5

u(x,y) =

i5=1hiui.

hi

=1atnodeiand0atallothernodes.

h5 =

1

(1

−

x2)(1

+

y)2

25


27/95


h1 =1

4(1+x)(1+y)−

1

2h5

h2 =1

4(1−x)(1+y)−

1

2h5

h3

=

1

4

(1

−

x)(1

−

y)

h1 =1

4(1+x)(1−y)

Note:

(6.14)

(6.15)

(6.16)

(6.17)

hi = 1

Wemusthave ihi =1tosatisfytherigidbodymodecondition.

u(x,y) = hiui

(6.18)i

Assumeallnodalpointdisplacements=u∗. Then,

u(x,y) = hiu∗

=u∗ hi =u∗ (6.19)i i

From(6.1),

K (m) U =R (6.20)m

B

(m)T

C

(m)

B

(m)

dV

(m)

U

=

R

(6.21)

V

(m)m

whereC (m)B(m)U =τ (m). (AssumewecalculatedU .)

B(m)T

τ

(m) d V

(m) =R (6.22)V

(m)m

F

(m) =R ; F (m) = B(m)

T τ

(m) d V

(m) (6.23)

V

(m)m

Twoproperties

I. Thesumof theF (m)’satanynode isequaltotheappliedexternalforces.

26


28/95


II. Everyelement isinequilibriumunderitsF (m)

Û

T

F

(m) =Û T

B(m)T

τ

(m) d

V

(m) (6.24)

V

(m)

=�(m)T

= �(m)T

τ

(m) d

V

(m) (6.25)

V

(m)

=0 (6.26)

whereU ˆ T

=virtualnodalpointdisplacement.

Applyrigidbodydisplacement.

If we move the element virtually in the rigid body modes, �(m) is zero. Therefore the virtual workobtainedduetovirtualmotionof theelementiszero. ThentheelementisinequilibriumunderitsF (m).

27


29/95


Lecture7- Isoparametricelements


Reading:

Sec. 5.1-5.3

WewantK =V

BT CB

dV ,R B

=V

H T f B dV . Uniquecorrespondence(x,y)⇔(r,s)

(r,s)arenaturalcoordinatesystemorisoparametriccoordinatesystem.

4

x= hixi

(7.1)i=1

4

y= hiyi (7.2)i=1

where

1h1

= (1 +r)(1+s) (7.3)4 1

h2 =

4

(1−r)(1+s) (7.4)

. . .

4

u(r,s) = hiui (7.5)i=1

4

v(r,s) = hivi (7.6)i=1

28


30/95

MIT2.094 7.Isoparametricelements

�=Bû ûT = u1

u2

v4

(7.7)· · ·

∂u ∂x

∂v =Bû (7.8)

∂y

∂u + ∂v∂y

∂x

∂ ∂x ∂y ∂ ∂r ∂r ∂r ∂x

= (7.9)∂ ∂x ∂y ∂

∂s ∂s ∂s ∂y

J

∂ ∂ ∂x ∂r

=

J −1

(7.10)∂ ∂

∂y ∂s

J mustbenon-singularwhichensuresthatthereisuniquecorrespondencebetween(x,y)and(r,s).

Hence, 1 1

K = BT CB tdet(J )drds (7.11)

−1 −1 d V

1

1

Also,R B

= H T f B tdet(J )drds (7.12)−1

−1

Numerical

integration

(Gauss

formulae)

(Ch.

5.5)

=t BT det(J ij)×(weighti, j) (7.13)i j

K

∼ijCBij

2x2Gaussintegration,

(i=1,2) (7.14)

( j=1,2) (weighti, j=1 inthiscase) (7.15)

29


31/95

�


9-node

element

9

x= hixi

(7.16)i=1

9

y= hiyi (7.17)

i=1

9

u= hiui

(7.18)i=1

9

v= hivi (7.19)i=1

Use3x3Gaussintegration

Forrectangularelements,J =const

Considerthefollowingelement,

Note,

here

we

could

use

hi(x,

y)

directly.

3 = 6 0J

= 2 3

(7.20)0 2

Then,wecandeterminethenumberof appropriateintegrationpointsbyinvestigatingthemaximumorderof BT CB .

Forarectangularelement,3x3Gauss integrationgivesexactK matrix. If theelement isdistorted,aK matrixwhich isstillaccurateenoughwillbeobtained,(if highenoughintegrationisused).

30


32/95


Convergence

Principle

of

virtual

work: Reading:

Sec. 5.5.5,

4.3�T C�

dV =R(u) (7.21)

V

Find u, solution, in V, vector space (any continuous function that satisfies boundary conditions),

satisfying

�T C�

dV = a(u,v) = (f ,v) forallv,anelementof V. (7.22)

V

bilinear

form R(v)

Example:

Finite

Element

problem Finduh

∈Vh,whereVh

isF.E.vectorspacesuchthat

a(uh,vh) = (f ,vh) ∀vh ∈Vh (7.23)

Sizeof Vh ⇒#of independentDOFs(hereit’s12).

Note:

a(w,w) >0forw∈V (w=0)

2x

(strain

energy

when

imposing

w)

Also,

a(wh,wh)>0forwh ∈Vh (Vh ⊂V,wh =0)

31


33/95


Property

I Define: eh =u−uh.

From(7.22),a(u,vh) = (f ,vh) (7.24)

From(7.23),a(uh,vh) = (f ,vh) (7.25)

Hence,

a(u−uh,vh)=0 (7.26)

a(eh,vh)=0 (7.27)

(errorisorthogonalinthatsensetoallvh inF.E.space).

PropertyII

(7.28)a(uh,uh)≤a(u,u)

Proof:

a(u,u) =a(uh

+eh,uh

+eh) (7.29)

=a(uh,uh) +

0byProp. I2a(uh,eh) +a(eh,eh) (7.30)

≥0

∴

a(u,u)≥a(uh,uh) (7.31)

32


34/95


Lecture8- Convergenceof displacement-basedFEM


(A) Find

u∈Vsuchthata(u,v) = (f ,v) ∀v∈V(Mathematicalmodel) (8.1)

a(v,v)>0 ∀v∈V, v= 0. (8.2)

where(8.2)impliesthatstructuresaresupportedproperly. E.g.

(B)

F.E.

Problem Find

uh ∈Vh suchthata(uh,vh)=(f ,vh) ∀vh ∈Vh (8.3)

a(vh,vh)>0 ∀vh ∈Vh,

Propertieseh =u−uh

vh =0 (8.4)

(I)

a(eh,

vh)

=

0

∀vh ∈

Vh (8.5)

(II)a(uh,uh)≤a(u,u) (8.6)

33


35/95

MIT2.094 8.Convergenceof displacement-based FEM

(C)

Assume Mesh “iscontainedin”Mesh

h1 h2

e.g. Mesh notcontainedinMesh

h1 h2

Weassume(C),butneedanotherproperty(independentof (C))

(III)

a(eh,eh)≤a(u−vh,u−vh) ∀vh

∈Vh

(8.7)

uh

minimizes! (Recalleh

=u−uh)

Proof: Pickwh ∈Vh.

a(eh +wh,eh +wh) =a(eh,eh) + 02a(eh,wh) ( )+a w wh, h

≥0

(8.8)

Equalityholdsfor(wh =0)

a(eh,eh)≤a(eh

+wh,eh

+wh) (8.9)

=a(u−uh

+wh,u−uh

+wh) (8.10)

Takewh =uh −vh.

a(eh,eh)≤a(u−vh,u−vh) (8.11)

Usingproperty(III)and(C),wecansaythatwewillconvergemonotonically,frombelow,toa(u,u):

34


36/95


Pascal

triangle

(2D)

(Ch. 4.3)

error indisplacement∼C ·

4-nodeelementiscompletetok=1.⇒

9-node

element

is

complete

to

k

=

2.

⇒

hk+1 (8.12)

(C isaconstantdeterminedbytheexactsolution,materialproperty...)

errorinstresses∼C hk (8.13)·

errorinstrainenergy∼C h2k ( theseC aredifferent) (8.14)· ←

Hence,

E −E h

=C h2k (roughlyequalto) (8.15)·

Bytheory,

log(E −E h)=logC + 2klogh (8.16)

35


37/95


Byexperiment,wecanevaluate log(E −E h)fordifferentmeshesandplotlog(E −E h)vs. logh

Weneedtousegradedmeshesif wehavehighstressgradients.

Example Consideranalmost incompressiblematerial:

�V

= vol. strain (8.17)

or

∇·v→ verysmallorzero (8.18)

Wecan“see”difficulties:

p=−κ�V

κ= bulkmodulus (8.19)

Asthematerialbecomesincompressible(ν = 0.3 0.4999)→

κ

→ ∞

p finitenumber (8.20)�V

0→

→

(Smallerrorin�V resultsinhugeerroronpressureasκ→ ∞,theconstantC in(8.15)canbeverylargelocking)⇒

36


38/95


Lecture9- u/ pformulation


Wewanttosolve Reading:Sec. 4.4.3

I. Equilibrium

τ ij,j +f iB

=0 inVolume(9.1)S f τ ijnj =f i

onS f

II. Compatibility

III. Stress-strain law

Usetheprincipleof virtualdisplacements

�T C�

dV =R (9.2)

V

Werecognizethatif ν 0.5→

�V

0 (�V

=�xx

+�yy

+�zz ) (9.3)→

E

κ=3(1−2ν )

→ ∞ (9.4)

p=−κ�V mustbeaccuratelycomputed (9.5)

Solution

τ ij

=κ�V

δ ij

+ 2G�ij

(9.6)

where

1 i= jδ ij = Kroneckerdelta = (9.7)

0 i= j

Deviatoricstrains:

�V �ij =�ij − 3

δ ij (9.8)

τ kkτ ij =− pδ ij + 2G�ij p=− 3

(9.9)

(9.2)becomes

�T C � dV + �V

κ�V

dV =R (9.10)

V V

�T C � dV − �T V pdV =R (9.11)

V V

37


39/95

� =

MIT2.094 9.u/ pformulation

Weneedanotherequationbecausewenowhaveanotherunknown p.

p+κ�V

=0 (9.12)

p( p+κ�V

) dV =0 (9.13)

V −

V

p �V

+κ p

dV =0 (9.14)

Foranelement,

u=Hû (9.15)

� =BDû (9.16)

�V =BV û (9.17)

p=H p p̂ (9.18)

Planestrain(�zz4/1element

=0)Example

Reading:

Ex. 4.32 in

thetext

�V =�xx +�yy (9.19)

131

3

(�xx +�yy )�xx −

�yy − (�xx +�yy )

(9.20) γ xy(�xx +

�yy )− 13

Note:

�zz

=0but� =0!zz

p=H p p̂ =[1]{ p0} (9.21)

p(x,y) = p0

(9.22)

Weobtainfrom(9.11)and(9.14)

K uu K up û R

K pu K pp p̂ =

0(9.23)

BT

V

K uu

= DC BD

dV (9.24a)

K up

=− BV T H p

dV (9.24b) V

K pu

=− H pT

BV

dV (9.24c) V

1K pp =− H p

T

κH p dV (9.24d)

V

38


40/95

�


In practice, we use elements that use pressure interpolations per element, not continuous betweenelements. Forexample:

Then ,unlessν = 0.5(whereK pp =0),wecanusestaticcondensationonthepressuredof’s.

Use p̂ equationstoeliminate p̂ fromtheû equations.

K uu

−K upK −1

pp

K pu

û =R (9.25)

(Inpractice,ν canbe0.499999...)

The“bestelement”isthe9/3element. (9nodesfordisplacementand3pressuredof’s).

p(x,y)= p0

+ p1x+ p2y (9.26)

The

inf-sup

condition Reading:

Sec. 4.5

=�V

Volq h∇·vh dVol

inf sup ≥β >0 (9.27) q vh h qh∈Qh vh∈Vh

for

normalization

Qh: pressurespace.

If “this”holds,theelementisoptimalforthedisplacementassumptionused(ellipticitymustalsobesatisfied).

Note:

infimum=largest lowerbound

supremum=leastupperbound

Forexample,

inf {1,2,4}= 1

sup{1,2,4}= 4

inf {x∈R; 0< x


41/95


For (entry [3,1] inmatrix)assumethecircledentryistheminimum(inf)of .

Also,allentries inthematrixnotshownarezero.

Case

1

β h = 0

0 uh|i

=0 (fromthebottomequation)·

α uh

+ 0 ph

= Rh

(fromthetopequation)⇒

· |i

· |j

|i

=0

⇒noequationfor ph|j

spuriouspressure! (anypressuresatisfiesequation)⇒

Case2 β h =small=�

�·uh|i

= 0 ⇒ uh|i

= 0

∴�· ph|j

+uh|i

·α= Rh|i

i ph|j =

Rh| displ. = 0as�issmall⇒

�

⇒pressure large→

The behavior of given mesh when bulk modulus increases: locking, large pressures. See Example4.39textbook.

40


42/95


Lecture10- F.E.largedeformation/generalnonlinearanalysis


Wedeveloped

t V

tτ ij

teij

d

V t = Rt

Reading:

Ch. 6

(10.1)

et ij =1

2

∂ui∂ xt j

+∂uj∂ xt i

(10.2)

t

V

tτ ij

δ teij

d

V t = Rt (10.3)

δ teij =1

2

∂ (δui)

∂ txj+

∂ (δuj)

∂ txi

(≡ teij) (10.4)

In

FEA:

tF = tR (10.5)

Inlinearanalysis

tF =K tU KU =R (10.6)⇒

Ingeneralnonlinearanalysis,weneedto iterate. Assumethesolutionisknown“attimet”

t

0

tx

= x+ u (10.7)

HencetF isknown. Thenweconsider

t+∆t t+∆tF

=

R

(10.8)

Considerthe loads(appliedexternalloads)tobedeformation-independent,e.g.

41


43/95

•

MIT2.094 10.F.E.largedeformation/generalnonlinearanalysis

Thenwecanwrite

t+∆tF =tF +F (10.9)t+∆tU

=tU +U (10.10)

whereonlytF andtU areknown.

=tK ∆U , tK =tangentstiffnessmatrixattimet (10.11)F ∼

From(10.8),

tK ∆U =t+∆tR −tF (10.12)

WeusethistoobtainanapproximationtoU . WeobtainamoreaccuratesolutionforU (i.e. t+∆tF )using

t+∆tK (i−1)∆U (i) =t+∆tR −t+∆tF (i−1) (10.13)t+∆t

U

(i)

=

t+∆t

U

(i−

1)

+ ∆U (i)

(10.14)

Also,

t+∆tF

(0) =tF (10.15)

t+∆tK (0) =tK (10.16)

t+∆tU

(0) =tU (10.17)

Iteratefori= 1,2,3. . .untilconvergence. Convergenceisreachedwhen

∆U (i) < �D (10.18)2

t+∆tR

t+∆tF

(i−1) (10.19)

−

< �F 2

Note:

a2 = (ai)2

i

∆U (i) =U i=1,2,3...

∆U (1) in(10.13)is∆U in(10.12).

(10.13) isthe full Newton-Raphsoniteration.

Howwecould(inprinciple)calculatetK

Process

IncreasethedisplacementtU i

by�,withnoincrementforalltU j

, j=i

calculatet+�F •

• thei-thcolumnin tK = (t+�F −tF )/�=∂ tF

.t∂ U i

42


44/95


So, perform this process for i = 1,2,3, . . . , n, where n is the total number of degrees of freedom.Pictorially,

tK =

. .. .. .

. .. .. . · · ·

. .. .

. .

Ageneraldifficulty: wecannot“simply”incrementCauchystresses.

t+∆tτ

referred

to

area

at

time

t

+ ∆t• ij

• tτ ij referredtoareaattimet.

Wedefineanewstressmeasure,2nd Piola - Kirchhoff stress ,t+∆0

tS ij

,where0intheleadingsubscriptreferstooriginalconfiguration. Then,

t+∆t

0S ij

=0tS ij +0S ij (10.20)

Thestrainmeasureenergy-conjugatetothe2ndP-Kstress0tS ij istheGreen-Lagrangestrain 0

t�ij

Then,

0

t

0tS ij δ 0

t�ij d V = R (10.21)0V

Also,

t+∆t t+∆t 0 t+∆tS δ

� d V = R (10.22)0 ij 0 ij0V

Example

43


45/95


tF =tR (10.23)

t+∆t t+∆tF = R (10.24)

(everytimeit isinequilibrium)

(10.13)and(10.14)give:

i

=

1,

t+∆tK (0) ∆U (1) =t+∆tR t+∆tF (0) ≡fn(tU ) (10.25)−

t+∆tU

(1) =t+∆tU (0) + ∆U (1) (10.26)

i=2,

t+∆tK (1) ∆U (2) =t+∆tR −t+∆tF (1) (10.27)

t+∆tU

(2) =t+∆tU (1) + ∆U (2) (10.28)

44


46/95


Lecture11- Deformation,strainandstresstensors


Westatedthatweuse Reading:Ch. 6

tV

tτ ij δ eij d V t =

0V

tS

t� 0V =tR (11.1)t 0 ij δ 0 ij d

The

deformation

gradient Weuse txi

=0xi

+tui

∂

tx1 ∂ tx1 ∂

tx1∂ 0x1 ∂

0x2 ∂ 0x3

∂ tx2 ∂ t

x2 ∂ t

x2

tX 0

(11.2)= 0 0 0∂ ∂ ∂ x x x1 2 3

∂

x

t

3 ∂ xt

3 ∂ xt

3

∂

x0 1 ∂ x0

2 ∂ x0

3

td x1 dtx (11.3)td x= 2

td x30d x1

d0x 0 (11.4)

(11.5)

d x2

0d x3

Impliesthat

dtx=0

tX d0x

=

(0tX

is frequently denoted by0tF

or simply F , but we use F forforcevector)

WewillalsousetherightCauchy-Greendeformationtensor

tC tX T tX = (11.6)0 0 0

Some

applications

45


47/95

�

�

�

�

MIT2.094 11.Deformation,strainandstresstensors

Thestretchof afiber(tλ): 2t t

t� 2

dxT dx d stλ = = (11.7)

d0xT d0x d0s

Thelengthof afiberis

0

0 0d s = d xT d x12 (11.8)

�

�

� 2 d0xT tX T tX d0xtλ

=0

0

0

0 ,

from(11.5) (11.9)d s d s·

Express

0 0 0d

x= d s n (11.10)

0

0n

= unitvector intodirectionof d x (11.11)

tλ

2 =

0n

T 0tC

0n

(11.12)

⇒

∴

tλ= 0nT 0

tC

0n

12 (11.13)

Also,�

d x̂tT

·�

dxt

=�

d ŝt

�

dst

costθ, (a·b=abcosθ) (11.14)

From(11.5),

costθ =

d

x̂0 T ˆtX T 0

�

tX 0 d x0

d ˆtsdst

=d

ŝ0 n̂0 T tC 0 n0 d

s0

d ˆts·dst

ˆtX 0 ≡tX 0

(11.15)

(11.16)

∴

0n̂T 0

tC 0ncostθ=

tλ̂tλ(11.17)

Also,

0ρtρ=det0

tX

(seeEx. 6.5) (11.18)

Example Reading:Ex. 6.6 in

thetext

46


48/95

�


1 0 0h1

=4

(1+ x1)(1+ x2) (11.19)

...

t

0

txi = xi + ui (11.20)

4

= hktxi

k, (i= 1,2) (11.21)

k=1

where txik

arethenodalpointcoordinatesattimet(tx11

=2, tx21

= 1.5)

Thenweobtain

tX =

1 5 +0x2 1 +0x1 (11.22)0 4 12

(1+0x2)12

(9+0x1)

At 0x1

= 0, 0x2

=0,

1 5 10tX

0x

=0x2=0=

41 9

(11.23)i

2 2

The

Green-Lagrange

Strain

1� 1� 0t� =

2 0tX T 0

tX −I =

2 0tC

−I (11.24)

t

0 t t∂ xi

∂ xi

+ ui

∂ ui

= =

δ ij

+

(11.25)

∂ 0x ∂ 0x ∂ 0xj j j

Wefindthat

1� t� = tu + tu + tu tu , sumoverk= 1,2,3 (11.26)0 ij 0 i,j 0 j,i 0 k,i0 k,j2where

t∂ u0tui,j = 0

i (11.27)

∂ xj

47


49/95

�


Polar

decomposition

of

tX

tR tU = 000 (11.28)

tX

isarotationmatrix,suchthat

0

where tR

tR T

0

tU

0

0

tR

isasymmetricmatrix(stretch)

0

=I (11.29)

and

Ex.

6.9

textbook

√ 3

41 0−√

32tX

=

Then, � 2tC

tX T tX

tU = = 0

t� tU =

0

0

2

0

0

0

0

(11.30)3 301 222

(11.31)

(11.32)1 2

I −

2

Thisshows,byanexample,thatthecomponentsof theGreen-Lagrangestrainareindependentof arigid-bodyrotation.

48


50/95

�


Lecture12- TotalLagrangianformulation


Wediscussed:

t

tX =

∂

∂ x

xi

j

⇒ dtx= tX d0x, d0x=�

tX

−1

dtx (12.1)0 0 0 0

tC = tX T tX (12.2)0

0 0

d0x=0tX d

tx

where 0tX =�

0tX

−1

=∂

t

0xi (12.3)∂ xj

TheGreen-Lagrangestrain:

1

� 1

�

0

t� = 0

tX T 0

tX −I = 0

tC −I (12.4)

2 2

Polardecomposition:

1� 2

0tX

=0tR 0

tU ⇒0

t� =2 0

tU −I (12.5)

Wesee,physicallythat:

wheredt+∆tx anddtx arethesame lengthsthe components of the G-L strain do not⇒

change.

Note

in

FEA

0 0 kxi

= hk xi

k

foranelement (12.6)t t kui

= hk ui

k

t

0

t

xi =

xi +

ui →

for

any

particle

(12.7)

Hencefortheelement

t

0

k

t kxi = hk xi + hk ui (12.8)

k k

= hk0

xik

+tuik

(12.9)k

= hktxk

k (12.10)k

49


51/95

MIT2.094 12.TotalLagrangianformulation

E.g.,k= 4

2ndPiola-Kirchhoff stress

0ρ

0

tS =

tρ

0t X

tτ

0tX

T → componentsalso independentof arigidbodyrotation (12.11)

Then

t 0 t ttS ijδ �ijd V =tτ ijδ eijd V = R (12.12)0 0 t

0V tV

Wecanusean incrementaldecompositionof stress/strain.

t+∆0

tS =0

tS +0S (12.13)

t+∆tS = tS +0S (12.14)0

ij

0 ij ij

t+∆t �= t� +0� (12.15)0 0

t+∆

0

t�ij =0t�ij +0�ij (12.16)

Assumethesolutioniskownattimet,calculatethesolutionattimet+ ∆t. Hence,weapply(12.12)attimet+ ∆t:

t+∆t t+∆t 0

t+∆tS δ

� d V = R (12.17)0

ij

0

ij0V

Lookatδ �0

t

ij

:

1� t t t t tδ � =δ u + u + u u (12.18a)0 ij 0 i,j 0 j,i 0 k,i0 k,j2t t1 ∂δui

∂δuj

∂δuk

∂ u ∂ u ∂δukt k k

δ � =

+

+

+

(12.18b)

0

ij 2

∂ 0x ∂

0x ∂ 0x ·

∂ 0x ∂ 0x ·

∂ 0xj i i j i j

1� δ �0

tij

=2

δ 0ui,j

+δ 0uj,i

+δ 0uk,i

0

tuk,j

+0

tuk,iδ 0uk,j

(12.18c)

Wehave

t+∆t�ij −t�ij =0�ij (12.19)0 0

0�ij

=0eij

+0ηij

(12.20)

50


52/95

�

�

�


where 0eij isthe linear incrementalstrain,0ηij isthenonlinearincrementalstrain,and

1 t t

0eij

= 0ui,j

+0uj,i

+0uk,i

0uk,j

+0uk,i

0uk,j

(12.21)2

initial displ. effect

1

0ηij =

0uk,i0uk,j (12.22)2

where

∂uk t+∆t t

0uk,j

= 0

,

uk

= uk

−

uk

(12.23)∂ xj

Note

t+∆t tδ � =δ � ∵δ � =0whenchangingtheconfigurationatt+ ∆t (12.24)ij 0 ij 0 ij

From(12.17):

� �

tS δ ij +δ η d V 0 ij +0S ij 0e 0 ij0

0V

= tS + + tS ijδ η + ijδ η d V (12.25)0 ijδ 0eij 0S ijδ 0eij 0 0 ij 0S 0 ij0

0V

=t+∆tR (12.26)

Linearization

tt K U K

0V

0S 0e +tS ijδ η

d V = R −0V

tS 0eijd V (12.27)

ijδ ij 0 0 ij0

t+∆t

0 ijδ 0

L

U 0 NL0 tF 0

Weuse,

0S ij �0C ijrs0ers (12.28)

Wearriveat,withthefiniteelementinterpolations,

0

tK L

+0

tK NL

U =t+∆tR − 0

tF (12.29)

whereU isthenodaldisplacementincrement.

51


53/95


Lefthandsideasbeforebutusing(k−1)andrighthandsideis

t+∆t t+∆t t+∆t (k−1) 0= R −0V

S δ �

d V (12.30)0 ij 0 ij

gives

t+∆tR −

t+∆0tF

(k−1)(12.31)

InthefullN-Riteration,weuse

t+∆t (k−1) t+∆t (k−1)∆U (k) t+∆t

t+∆t (k−1)0

K L

+0

K NL

= R −0

F (12.32)

52


54/95

�


Lecture13- TotalLagrangianformulation,cont’d


Exampletrusselement. Recall:

Principleof virtualdisplacementsappliedatsometimet+ ∆t:

t+∆t t+∆t t+∆tτ ijδ t+∆teijd V = R (13.1)t+∆tV

t+∆t t+∆t 0 t+∆tS δ

� δ V = R (13.2)0 ij 0 ij0V

t+∆0

tS ij

=0

tS ij

+0S ij

(13.3)

t+∆0t�ij =0

t�ij

+0�ij

(13.4)

0�ij =0eij +0ηij (13.5)

where0

tS ij

and 0

t�ij

areknown,but0S ij

and 0�ij

arenot.

1� 0eij = 2 0

ui,j +0uj,i +0tuk,i0uk,j +0

tuk,j0uk,i (13.6)

1� 0ηij = 0uk,i0uk,j (13.7)2

Substituteinto(13.2)andlinearizetoobtain

0 0 t+∆t 0δ 0e 0C 0ers + ijδ ηijd V = δ 0eij ijd V 0V

ij ijrs d V 0V

0tS 0 R −

0V 0tS (13.8)

F.E.

discretization

gives

0

tK L

+0

tK NL

∆U =t+∆tR −0

tF (13.9)

53


55/95

MIT2.094 13.TotalLagrangianformulation,cont’d

tK

tBT tB

0

0 L = 0 L 0C 0 Ld V (13.10)0V

T

0tK = tB tS tBNL d V (13.11)0 NL 0 NL 0 0

0V

matrix

T 0tF = tB tS ˆ d V (13.12)0 0

L

00V

vector

Theiteration(fullNewton-Raphson)is

t+∆tK

(i−1)

+t+∆t

K (i−1)

∆U (i) =t+∆tR t+∆t

F

(i−1)

(13.13)0 L 0 NL − 0

t+∆tU (i) =t+∆tU (i−1) + ∆U (i) (13.14)

Truss

element

example (p. 545)

Herewehavetoonlydealwith0

tS 11, 0e11, 0η11

t

0e11

=∂u1

+∂ uk

∂uk(13.15)

∂ 0x1

∂ 0x1

·∂ 0x1

1 ∂uk

∂uk

0η11

=2 ∂ 0x1

·∂ 0x1

(13.16)

Weareafter

1u1

u1

tB 2 tB0e11 =0 L

2

=0

L

û (13.17)u12u2

2

ui = hkuki (13.18)

k=1

2

t t kui = hk ui (13.19)k=1

54


56/95

�

�


t t

=∂u1

+∂ u1 ∂u1 +

∂ u2 ∂u2 (13.20a)0e11 ∂ 0x1

∂ 0x1

∂ 0x1

∂ 0x1

∂ 0x1

2 0L cosθ−0L (13.20b)t + ∆L=u1

2

0Lt + ∆L sinθ (13.20c)=u2

1−1 0 1 0 û0e11

=0L

0L+ ∆L 1ûcosθ−1 −1 0 1 0+ ·

0L

0L

∂

tu1

0∂ x1

0L+ ∆L 1 û

=0tBLû (13.20e)

Hence,

0

−1 0 1sin

θ

(13.20d)

+

·0L

0L

∂

tu2

∂ x01

0e11 =0L+ ∆L

(0L)2−cosθ −sinθ cosθ sinθ û (13.20f)

where the boxed quantity above equals 0

tBL. In small strain but large rotation analysis we assume∆L0L,

1 θcos− −sinθ cosθ sinθ û (13.20g)0e11

=0L

1 ∂u1

∂u1

∂u2

∂u2+ (13.21a)0 0 0 0x x

=0η11 2 ∂ ∂ ∂ ∂ x x1 1 1 1

1δ η =0

11 2 1

∂δu1

∂u1

∂u1

∂δu1

∂δu2

∂u2

∂u2

∂δu2+ + + (13.21b)0 0

0 0

0 0

0 0x x∂

∂

∂

∂

∂

∂

∂

∂ x x x x x x1

1

1

1

1

1

1

∂δu1

∂u1

∂δu2

∂u2

= 0 0 + 0 0 (13.21c)∂ x1 ∂ x1 ∂ x1 ∂ x1

∂u1tS 00

0∂δu1 ∂δu 2 ∂ 11tS x1 (13.21d)11δ η =0 11 0 0 0 0

tS

∂u2∂ ∂ x x1 1011

∂ x1

tS 0

∂u10 1 −1 0 1 0

0∂

x1 = û (13.21e)∂u2 0L −1 0 1∂

0x1

BNL

55

0


57/95


0C =E (13.22)

tS ˆ = tS (13.23)0 0 11

Assumesmallstrains

EAtK =0 0L

cos2θ cosθsinθ −cos2θ −cosθsinθsin2θ −sinθcosθ −sin2θ

cos2θ sinθcosθ

=

sym sin2θ

tK (13.24)0 L

1 0 −1 00tP

0

L

0 1 −1

0 + −1 0 10 −1 0 1

tK 0 NL

Whenθ=0,0tK L doesn’tgivestiffnesscorrespondingtou2

2,but0tK NL does.

56


58/95

�


Lecture14- TotalLagrangianformulation,cont’d


Trusselement. 2Dand3Dsolids.

t+∆t t+∆t t+∆tτ ij

δ t+∆teijd V = R (14.1)t+∆tV

t+∆t t+∆t 0

t+∆t

0S ijδ 0�ijδ V = R (14.2)0V

linearization

δ ij

δ V + tS ij

δ ηij

δ V = tS ij

δ V (14.3)0V

0C ijrs

0ers 0e 0

0V 0

00 t+∆tR −

0V 0

ij

δ 0e 0

Note:

tδ =δ �0eij

0

ij

varyingwithrespecttotheconfigurationattimet.

F.E.

discretization

0 0

k

t t k

t+∆t t+∆t kxi = hk xi xi = hk xi xi

= hk xi

(14.4a)k k k

t t k t+∆t t+∆t k kui = hk ui ui = hk ui

ui = hkui (14.4b)k k

k

(14.4)

into

(14.3)

gives

0tK L +0

tK NL U =t+∆tR − 0

tF (14.5)

57


59/95


Truss

∆L 1smallstrainassumption:0L

0E AtK

=

0 0L

cos2θ cosθsinθ −cos2θ −cosθsinθcosθsinθ sin2θ −sinθcosθ −sin2θ

cos2θ sinθcosθ

=−cos2θ

−cosθsinθ

−cosθsinθ−sin2θ

(14.6)sinθcosθ sin2θ

1 0 −1 00tP

0L

0 1 −10

+−1 0 1

0 −1 0 1

(noticethatthebothmatricesaresymmetric)

cosθ sinθu1 u1 (14.7)=v1

−sinθ cosθ v1

Correspondingtotheuandv displacementswehave:

0E AtK = (14.8)0L

1 0 −1 00

1 0 −1 00

+ tP 0L

0 0 0 0 1 −10 (14.9)= −1 0 1 00 0

−1 0 100 0 −1 0 1

58

0


60/95

�


tP 0 tP Q L = ∆ Q= ∆ (14.10)0L

· ⇒ ·

tP where the boxed term is the stiffness. In axial direction, 0L is not very important because usually

E 0A tP tP 0L

0L

. But, inverticaldirection, 0L

is important.

−cosθ

tF tP

−sinθ

0

=

cosθ

(14.11)

sinθ

2D/3D

(e.g.

Table

6.5)

2D:

1� 2 � 20�11

= 0u1,1 +0tu1,1 0u 1,1 +0tu2,1 0u2,1

+2 0

u1,1

+ 0u2,1

(14.12)

0e11 0η11

0�22

= (14.13)· · ·

0�12

= (14.14)· · ·

(Axisymmetric)

0�33 = ? (14.15)

1� 2 t� =2

tU −I (14.16)0 0

× × 0× × 0

0

tU 2 = 0 0 × (14.17)

(tλ

↑

)2

t

0 td s 2π x1

+ u1tλ= =

0 0d s 2π x1

(14.18)

t

= 1 +u1

0x1

2 0

t�33

=2

11 + 0

tu

x1 −1 1

(14.19) 2tu1 1 tu1= +0x1

2 0x1

59


61/95


2tu +u1 1 tu +u1t+∆t 1 1� = + (14.20)0 33 0

0x1

2 x1

2u1 tu u1 1

u10�33 =

t+∆0

t�33

−0t�33 = 0x1

+ 0x1

1

· 0x1+

2 0x1 (14.21)

How

do

we

assess

the

accuracy

of

an

analysis? Reading:

Sec. 4.3.6

• Mathematicalmodel∼u

• F.E.solution∼uh

Findu−uhandτ −τ h.

References

[1] T.SussmanandK.J.Bathe.“Studiesof FiniteElementProcedures- onMeshSelection.”Computers

&

Structures ,

21:257–264,

1985.

[2] T. Sussman and K. J. Bathe. “Studies of Finite Element Procedures - Stress Band Plots and theEvaluationof FiniteElementMeshes.”Journal of Engineering Computations ,3:178–191,1986.

60


62/95


Lecture15- Fieldproblems


Heattransfer,incompressible/inviscid/irrotationalflow,seepageflow,etc. Reading:Sec. 7.2-7.3

Differentialformulation•

Variationalformulation•

Incrementalformulation•

F.E.discretization•

15.1 Heattransfer

AssumeV constantfornow:

S

=

S θ ∪

S q s

θ(x,y,z,t)isunkownexceptθ|S θ

=θ pr. Inaddition, q |S q isalsoprescribed.

15.1.1

Differential

formulation

I. HeatflowequilibriuminV andonS q.

II. Constitutive lawsq x =−k∂θ .∂x

∂θ

q y

=−k (15.1)∂y

∂θ

q z =−k (15.2)∂z

III. Compatibility: temperaturesneedtobecontinuousandsatisfytheboundaryconditions.

61


63/95

| |

MIT2.094 15.Fieldproblems

Heatflowequilibriumgives

∂ ∂θ

∂ ∂θ

∂ ∂θ Bk

+ k + k =−q (15.3)∂x

∂x

∂y

∂y

∂z

∂z

whereq B istheheatgeneratedperunitvolume. Recall1Dcase:

unitcross-section

dV =dx (1) (15.4)·

q x

−q x+dx

+q Bdx=0 (15.5)

∂q xq |x − q |x +

∂xdx

+q Bdx=0 (15.6)

∂

∂θ B

−∂x

−k∂x

dx+q dx =0 (15.7)

∂

∂θ Bk =−q (15.8)

∂x

∂x

We

also

need

to

satisfy

k

∂θ

=q S (15.9)∂n

onS q.

15.1.2

Principle

of

virtual

temperatures

θ

∂ k

∂θ+ +q B =0 (15.10)

∂x ∂x· · ·

(θ =0andθ tobecontinuous.)S θ

θ

∂ k

∂θ+ +q B dV =0 (15.11)

V

∂x ∂x· · ·

62


64/95

�

�


Transformusingdivergencetheorem(seeEx4.2,7.1)

S qθ

T kθ dV = θq BdV + θ q S dS q

(15.12)V V S q

heat flow

∂θ ∂x

θ = ∂θ (15.13)

∂y

∂θ

∂z

k 0 0

k= 0 k 0 (15.14)

0 0 k

Convection

boundary

condition

q

S =h θe −θS (15.15)

where

θe is

the

given

environmental

temperature.

Radiation

q

S =κ∗

(θr)4

−�

θS 4

(15.16)

=κ∗

(θr)2 +�

θS 2

�

θr +θS �

θr −θS

(15.17)

=κ θr −θS (15.18)

whereκ=κ(θS )andθr isgiventemperatureof source. Attimet+ ∆t

θT t+∆tkt+∆tθdV = θ t+∆tq BdV + θ

S t+∆tq S dS q (15.19)V V S q

Let t+∆tθ =tθ+θ (15.20)

or t+∆tθ(i) =t+∆tθ(i−1) + ∆θ(i) (15.21)

with t+∆tθ(0) =tθ (15.22)

From(15.19)

θT t+∆t

(i−1)∆θ(i)k dV V

= θt+∆tq BdV − θT t+∆tk(i−1)t+∆tθ(i−1)dV (15.23)

V V

S (i−

1)

∆θS (i)

θ

S t+∆th(i−1) t+∆t e t+∆t+

θ θ

+

dS q−S q

wherethe∆θS (i)

termwouldbemovedtothe left-handside.

Weconsideredtheconvectionconditions

θS t+∆th

�

t+∆tθe −t+∆tθS

dS q (15.24)S q

Theradiationconditionswouldbeincludedsimilarly.

63


65/95


F.E.

discretization

t+∆t t+∆tθ =H 1x4

· θ̂4x1 for4-node2Dplanarelement

t+∆tθ t+∆tθ̂2x1 =B2x4 · 4x1t+∆tθS =H S t+∆tθ̂·

For(15.23)

t+∆t (i−1)∆θ(i) gives t+∆t (i−1)θ

T k

dV = BT k B dV ∆θ̂(i)⇒

V V

4x2 2x2 2x4 4x1

θt+∆tq

BdV

H T t+∆tq

BdV ⇒

V V

θT t+∆tk(i−1)t+∆tθ(i−1)dV

BT t +∆tk(i−1)BdV

t+∆tθ̂(i−1)⇒ V V known

S q

θS T t+∆th(i−1) t+∆tθe − t+∆tθS

(i−1)

+ ∆θS (i)

dS q =⇒

H S T t+∆th(i−1) H S t+∆tθ̂e t+∆tθ̂(i−1)+∆θ̂(i)dS q

−

S q 4x1

1x4 4x1

4x1

4x1

15.2

Inviscid,

incompressible,

irrotational

flow

2Dcase: vx,vy

arevelocitiesinxandy directions.

∇·v= 0

or∂vx

+∂vy

=0 (incompressible)∂x

∂y

∂vx

∂vy

∂y

−∂x

=0 (irrotational)

Usethepotentialφ(x,y),

∂φ

∂φ

vx =

vy =

∂x

∂y

∂ 2φ ∂ 2φ+ = 0 inV ⇒

∂x2 ∂y 2

(Sameastheheattransferequationwithk=1,q B =0)

64

(15.25)

(15.26)

(15.27)

(15.28)

(15.29)

(15.30)

(15.31)

Reading:

Sec. 7.3.2

(15.32)

(15.33)

(15.34)

(15.35)

(15.36)


66/95


Lecture16- F.E.analysisof Navier-Stokesfluids


Incompressibleflowwithheattransfer

Reading:

Sec.Werecallheattransferforasolid: 7.1-7.4,

Table7.3

Governing

differential

equations

(kθ,i),i +q B = 0 inV (16.1)

isprescribed,k =q S ∂θ

(16.2)θ∂nS θ

S q S q

Sθ ∪

Sq = S

Sθ ∩

Sq =

∅

(16.3)

Principle

of

virtual

temperatures

θ,ikθ,idV = θq B

dV + θ

S q

S dS q

(16.4)V

V S q

forarbitrarycontinuousθ(x1, x2, x3)zeroonSθ

Forafluid,weusetheEulerianformulation.

65


67/95

MIT2.094 16.F.E.analysisof Navier-Stokesfluids

ρc pvθ|x

−

ρc pvθ|x

+∂

∂x

(ρc pvθ)dx

+ conduction + etc (16.5)

Ingeneral3D,wehaveanadditionaltermforthe lefthandsideof (16.1):

−∇

·

(ρc pvθ)=−ρc p∇·(vθ)=−ρc p( ∇·v)θ−ρc p(v·∇)θ

term (A)

(16.6)

where∇·v=0inthe incompressiblecase.

∇·v=vi,i

=div(v)=0 (16.7)

So(16.1)becomes

(kθ,i),i +q B =ρc pθ,ivi

⇒(kθ,i),i +

�

q

B −ρc pθ,ivi

=0 (16.8)

Principle

of

virtual

temperatures

is

now

(use

(16.4))

V

θ,ikθ,idV +

V

θ(ρc pθ,ivi)dV =

V

θq BdV +

S q

θS

q S dS q

(16.9)

Navier-Stokes

equations

• Differential form

τ ij,j +f Bi =ρvi,jvj (16.10)

withρvi,jvj liketerm(A) in(16.6)=ρ(v·∇)v inV .

τ ij

=− pδ ij

+2µeij

eij

=1

2

∂vi∂xj

+∂vj∂xi

(16.11)

• Boundary conditions (needbemodifiedforvariousflowconditions)

τ ijnj =f S f i

onS f (16.12)

Mostly used as f n = τ nninflowconditions).

= prescribed, f t = unknown with possibly∂vn

∂n= ∂vt

∂n= 0 (outflow or

Andvi prescribedonS v,andSv ∪Sf = SandSv ∩Sf =∅.

66


68/95


Variational

form •

viρvi,j

vj

dV + eij

τ ij

dV = vif i

BdV + vi

S f f i

S f dS f

(16.13)V

V V S f

p∇·vdV =0 (16.14)V

F.E.

solution •

Weinterpolate(x1, x2, x3),vi,vi,θ,θ, p, p. Goodelementsare

×: linearpressure: biquadraticvelocities◦

(Q2, P 1),9/3element

9/4celement

Bothsatisfytheinf-supcondition.

Soingeneral,

Example:

ForS f e.g.

τ nn = 0,∂vt

=0; (16.15)∂n

67


69/95

�


and ∂vn issolvedfor. Actually,wefrequently justset p=0.∂t

Frequentlyused isthe4-nodeelementwithconstantpressure

Itdoesnotstrictlysatisfythe inf-supcondition. Oruse Reading:Sec. 7.4

3-nodeelementwithabubblenode.Satisfies inf-supcondition

1D

case

of

heat

transfer

with

fluid

flow, v= constant Reading:

Sec. 7.4.3

Re

=

vL

Pe

=

vL

α

=

k

(16.16)

ν

α

ρc p

Differential

equations •

kθ =ρc pθv (16.17)

θ|x=0 =θL θ|x=L =θR (16.18)

Innon-dimensionalform Reading:p. 683

1(nowθ andθ arenon-dimensional) (16.19)θ =θ

Pe

exp Pexθ−θL L

−1⇒

θR −

θL

=

exp

(Pe)

−

1

(16.20)

68


70/95


• F.E.discretization

θ =Peθ (16.21)

1

0

θθdx+Pe

1

0

θθdx=0+{effectof boundaryconditions =0here}

Using2-nodeelementsgives

(16.22)

1 Pe

(h∗)2(θi+1

−2θi

+θi−1) =2h∗

(θi+1

−θi−1) (16.23)

vL

Pe= (16.24)α

Define

h

vh

Pee =Pe = (16.25)·L α

Pee Pee−1−

2θi−1 + 2θi +

2−1 θi+1

=0 (16.26)

whatishappeningwhenPee islarge? Assumetwo2-nodeelementsonly.

θi−1 =0 (16.27)

θi+1 =1 (16.28)

1 Peeθi =

21−

2(16.29)

69


71/95


1 Peeθi

=2

1−2

(16.30)

ForPee >2,wehavenegative θi

(unreasonable).

70


72/95


Lecture17- Incompressiblefluidflowandheattransfer,cont’d


17.1 Abstractbody

Reading:

Sec. 7.4

Fluid

Flow

Heat

transfer

Sv, Sf

Sθ, Sq

Sv

∪Sf

Sθ

∪Sq

=S =S Sv ∩Sf = 0 Sθ ∩Sq = 0

17.2 Actual2Dproblem(channelflow)

71


73/95

MIT2.094 17.Incompressiblefluidflowandheattransfer,cont’d

17.3 Basicequations

P.V.

velocities

viρvi,jvjdV + τ ijeijdV = vif iB

dV + viS f f i

S f dS f (17.1)V V V S f

Continuity

pvi,idV =0 (17.2)V

P.V.

temperature

θρc pθ,ividV + θ,ikθ,idV = θq B

dV + θ

S q

S dS

(17.3)V

V V S q

F.E.

solution

xi

= hkxki

(17.4)

vi

= hkvik (17.5)

θ= hkθk (17.6)

p= h̃k pk (17.7)

v

F (u) =R u=

p

nodalvariables (17.8)⇒

θ

17.4 Modelproblem

1Dequation,

dθ

d2θ

ρc pv =k (17.9)dx

dx2

(v isgiven,unitcrosssection)

Non-dimensionalform(Section7.4)

dθ

d2θPe = (17.10)

dx

dx2

72


74/95

�


vL

kPe= , α= (17.11)

α

ρc p

θ∗ isnon-dimensional

θ

−

θL

exp

PeLx

−

1

θR

−θL

=exp(Pe)−1

(17.12)

(17.10) inF.E.analysisbecomes

dθ dθ dθθPe dV + dV =0 (17.13)

dx dx dxV V

Discretizedby linearelements:

hh∗ =

L

ξ ξ

θ(ξ ) = 1− θi−1

+ θi

(17.14)h∗ h∗

Fornodei:

Pee Pee −θi−1 −

2θi−1 + 2θi −θi+1 +

2θi+1 =0 (17.15)

where

vh

hPee = =Pe (17.16)

α L

Thisresultisthesameasobtainedbyfinitedifferences

1(θi+1 −2θi +θi−1) (17.17)θ =

i (h∗)2

θ =θi+1 −θi−1

(17.18)i 2h∗

73


75/95

�


Consideredθi+1 =1,θi−1 =0. Then

θi

= 1−(Pee/2)

(17.19)2

PhysicallyunrealisticsolutionwhenPee >2. Forthisnottohappen,weshouldrefinethemesh—averyfinemeshwouldberequired. Weuse“upwinding”

dθ=

θi −θi−1(17.20)

h∗dx i

Theresultis

(−1−Pee)θi−1 +(2+Pee)θi −θi+1 =0 (17.21)

Verystable,e.g.

θi−1

=0=

1(17.22)

θi+1

= 1⇒θi

2+Pee

Unfortunately

it

is

not

that

accurate.

To

obtain

better

accuracy

in

the

interpolation

for

θ,

use

the

function

exp PexL

−1

exp(Pe)−1(17.23)

TheresultisPee dependent:

This impliesflow-conditionbasedinterpolation. Weusesuch interpolationfunctions—seereferences.

References

[1] K.J. Bathe and H. Zhang. “A Flow-Condition-Based Interpolation Finite Element Procedure for

Incompressible

Fluid

Flows.”

Computers

&

Structures ,

80:1267–1277,

2002.

[2] H. Kohno and K.J. Bathe. “A Flow-Condition-Based Interpolation Finite Element Procedure forTriangularGrids.”International Journal for Numerical Methods in Fluids ,51:673–699,2006.

74


76/95


17.5 FSIbriefly

Lagrangianformulationforthestructure/solid

Arbitrary

Lagrangian-Eulerian

(ALE)

formulation Letf beavariableof aparticle(e.g. f =θ).

Consider1D

f ̇∂f

∂f

= + vparticle

∂t

∂x

(17.24)

where

v

is

the

particle

velocity.

For

a

mesh

point,

f ∗ =∂f

∂f

+ vm

(17.25)mesh point ∂t ∂x

wherevm

f ̇

isthemeshpointvelocity. Hence,

∂f =f ∗ +

∂x(v−vm) (17.26)

particle mesh point

Use(17.26) inthemomentumandenergyequationsanduse forceequilibriumandcompatibilityattheFSIboundarytosetupthegoverningF.E.equations.

References

[1] K.J.Bathe,H.ZhangandM.H.Wang.“FiniteElementAnalysisof IncompressibleandCompressibleFluidFlowswithFreeSurfacesandStructuralInteractions.” Computers & Structures ,56:193–213,1995.

[2] K.J. Bathe, H. Zhang and S. Ji. “Finite Element Analysis of Fluid Flows Fully Coupled withStructuralInteractions.”Computers & Structures ,72:1–16,1999.

[3] K.J.BatheandH.Zhang.“FiniteElementDevelopmentsforGeneralFluidFlowswithStructuralInteractions.”International Journal for Numerical Methods in Engineering ,60:213–232,2004.

75


77/95


Lecture18- Solutionof F.E.equations


Instructures, Reading:Sec. 8.4

F (u, p)=R . (18.1)

Inheattransfer,

F (θ) =Q (18.2)

Influidflow,

F (v, p,θ) =R (18.3)

Instructures/solids

F

(m)

t (m)T tˆ(m) 0 (m)F = = 0BL

0S d V (18.4)0V (m)

m m

Elasticmaterials

Example p. 590textbook

76


78/95


79/95

MIT2.094 18.Solutionof F.E.equations

Plasticity

• yieldcriterion

flowrule•

• hardeningrule

ttτ

=t−∆tτ + dτ (18.16)t−∆t

Solution

of

(18.1)

(similarly

(18.2)

and

(18.3))

Newton-Raphson FindU ∗ asthezeroof f (U ∗)

t+∆t t+∆tR f (U ∗) = (18.17)F −

·t+∆t (i−1)U

U ∗ −t+∆tU (i−1) +H.O.T. (18.18)∂ f t+∆t

(i−1) +=f U

∂ U

where t+∆tU (i−1) isthevaluewe justcalculatedandanapproximationtoU ∗.

Assume t+∆tR isindependentof thedisplacements.

·∆U (i) (18.19)∂ t+∆tF t+∆tR

t+∆tF

(i−1)0= − −

∂ U t+∆t (i−1)U

Weobtain

t+∆tK (i−1)∆U (i) =t+∆tR −t+∆tF (i−1) (18.20)

(18.21)∂

t+∆tF

∂ F t+∆tK (i−1) = =∂ U

t+∆t (i−1) ∂ U t+∆t (i−1)U U

Physically

∆t+∆t

F 1(i−1)

t+∆t (i−1)K 11 = (18.22)∆u

78


80/95


Pictorially

for

a

single

degree

of

freedom

system

i=1; tK ∆u(1) =t+∆tR−tF (18.23)

i=2; t+∆tK (1)∆u(2) =t+∆tR t+∆tF (1) (18.24)−

Convergence Use

∆U (i)2

< � (18.25)

a2 = (ai)

2

(18.26)

i

But,if incrementaldisplacementsaresmall ineveryiteration,needtoalsouse

t+∆tR −t+∆tF (i−1)2 < �R (18.27)

18.1 Slenderstructures

(beams,plates,shells)

t

1 (18.28)Li

79


81/95


Beam

t 1e.g.L

= 100

(4-nodeel.) The element does not have curvature we have aspuriousshearstrain

→

(9-node

el.)

Wedonothaveashear(better)→But, still for thin structures, it has problems→

likeill-conditioning.

We need to use beam elements. For curved structures also spurious membrane strain can be⇒present.

80


82/95


Lecture19- Slenderstructures


Beamanalysis, tL

1(e.g. tL

= 11100

,1000

,· · ·) Reading:Sec. 5.4,

6.5

(plane

stress)

2L 0

(19.1)J =0

1

t

2

(1−r) (1+s) (19.2)h2 =41

(1−r)(1−s) (19.3)h3

=4

∗ ∗=v v32

3

2

Beamtheoryassumptions(Timoshenkobeamtheory):

u∗

u∗

(19.4)

(19.5)

(19.6)

=v2t

=u2

+ θ22t

=

u2

−

2

θ2

∗ ∗ ∗ ∗u v u v3322

1 2L

0 − 14 (1−s)2L

0

2(1 )−rt

(1+s)− 4

0

etc (19.7)B∗ =1

42(1 )−rt

(1+s)

1

42(1 )−rt

2(1−s)

0 −

1 2L

− 142(1 )−rt

81

− 14− 4 L

14


83/95

MIT2.094 19.Slenderstructures

u2

v2

θ2 ∂u

∂x1 t0− s

∼

L 2L

Bbeam =

· · ·0

(19.8)0 0 0 ∂v

∂y

0 −1 −1(1−r)∂y

∂x

L

2

∂u + ∂v

1v(r)=

2(1−r)v2

1 st u(r)=

2(1−r)u2

−4

(1−r)θ2

atr=−1,

v(−1)=v2

st

u(−1)=−2

θ2

+u2

Kinematics is

1u(r)=

2(1−r)u2

resultsinto�xx

∂u 2 1

→

�xx

=∂r

·

L

=−L

st

u(r,s) =−4

(1−r)θ2

resultsinto�xx,γ xy

st

�xx =→2L

∂u 2 1

γ xy =∂s

·t

=−2

(1−r)

1v(r)=

2(1−r)v2

resultsintoγ xy

1→γ xy

=−L

(19.9)

(19.10)

(19.11)

(19.12)

(19.13)

(19.14)

(19.15)

(19.16)

(19.17)

(19.18)

(19.19)

82


84/95

MIT2.094 19.Slenderstructures

Forapurebendingmoment,wewant

1 1 −

Lv2 −

2(1−r)θ2 =0 (19.20)

forallr! Impossible (exceptforv2 =θ2 =0) So,theelementhasaspuriousshearstrain!⇒ ⇒

Beam

kinematics (Timoshenko,Reissner-Mindlin)

dwγ

=dx

−β (19.21)

1I = bt3 (19.22)

12

Principle

of

virtual

work

EI

L

0

dβ

dx

dβ

dx

dx+AS

G

L

0

dw

dx

−β

dw

dx

−β

dx=

L

0

pwdx (19.23)

As

=kA=kbt (19.24)

Tocalculatek Reading:

A

1

2G(τ a)

2dA=

AS

1

2G

V

As

2

dAs (19.25)

p. 400

where

τ a is

the

actual

shear

stress:

τ a

=3

2·

V

A

�

Documents

fea mit lecture