· x ﬁnite element method for 1d continua motivation - spatial truss structures real structures...

• X

finite element method for 1d continua

• motivation - spatial truss structures

real structures

mechanical modeling

weak form

• illustration of fem by means of a example

• computer oriented development of finite element method

• element tensors and element matrices

• structural equation of motion orstructural static equilibrium

• solution and post-processing

• examples

d.kuhl, wes.online, university of kasselÜ Ülinear computational structural mechanics • 2 1d finite element method

1d finite element methodÜ 1

• X

• model of a tensegrity exposition structureKUHL ET AL. (2011)

compression truss elements model of the structure detail of the structure

• mechanical model

1d continuum spatial truss structure

d.kuhl, wes.online, university of kasselÜ Ülinear computational structural mechanics • 2 1d finite element method

information about title picture ’1d fem’Ü 2

• X

man wka 60, helgoland man wka 60, helgoland sketch - wka 60

d.kuhl, wes.online, university of kasselÜ Ülinear computational structural mechanics • 2 1d finite element method • 2.1 motivation - 1d structures

motivation - towers of wind power plantsÜ 3

• X

loading tower heade1

e2

e3

M1M3

M2

F1F3

F2

normal loading tower heade1

e2

e3F1

loading of wind power plant towers

• loading of tower from gondola

dead load of gondola, rotor and hub (negative) F1

transverse force caused by cross wind F2

transverse force caused by rotor drag F3

yaw moment caused by non-symmetric wind M1

pitch moment caused by dead load of rotor and hub M2

rolling moment caused by rotor and generator moment M3

• direct loading of tower

dead load (negative) b1, p1

transverse force caused by wind q2, q3

• individual study of loading with normal forces

dead load (negative) b1, p1

single load from gondola F1



• X

e1e1

e3e3e3e3e3e3e3e3e3

dead

load

p1

dead

load

p1

dead

load

p1

dead

load

p1

dead

load

p1

dead

load

p1

dead

load

p1

dead

load

p1

dead

load

p1

F1F1F1F1F1F1F1F1F1

cros

sse

ctio

nA

cros

sse

ctio

nA

cros

sse

ctio

nA

cros

sse

ctio

nA

cros

sse

ctio

nA

cros

sse

ctio

nA

cros

sse

ctio

nA

cros

sse

ctio

nA

cros

sse

ctio

nA

normal loading

e1

e3e3e3e3e3e3e3e3e3

dead

load

p1(X

1)

F1F1F1F1F1F1F1F1F1

cros

sse

ctio

nA

(X1)

cros

sse

ctio

nA

(X1)

cros

sse

ctio

nA

(X1)

cros

sse

ctio

nA

(X1)

cros

sse

ctio

nA

(X1)

cros

sse

ctio

nA

(X1)

cros

sse

ctio

nA

(X1)

cros

sse

ctio

nA

(X1)

cros

sse

ctio

nA

(X1)

normal loading construction of tower



• X

e1e1

e3e3e3e3e3e3e3e3e3

dead

load

p1

dead

load

p1

dead

load

p1

dead

load

p1

dead

load

p1

dead

load

p1

dead

load

p1

dead

load

p1

dead

load

p1

F1F1F1F1F1F1F1F1F1cr

oss

sect

ionA

cros

sse

ctio

nA

cros

sse

ctio

nA

cros

sse

ctio

nA

cros

sse

ctio

nA

cros

sse

ctio

nA

cros

sse

ctio

nA

cros

sse

ctio

nA

cros

sse

ctio

nA

normal loading

e1

e3e3e3e3e3e3e3e3e3

dead

load

p1(X

1)

F1F1F1F1F1F1F1F1F1

cros

sse

ctio

nA

(X1)

cros

sse

ctio

nA

(X1)

cros

sse

ctio

nA

(X1)

cros

sse

ctio

nA

(X1)

cros

sse

ctio

nA

(X1)

cros

sse

ctio

nA

(X1)

cros

sse

ctio

nA

(X1)

cros

sse

ctio

nA

(X1)

cros

sse

ctio

nA

(X1)

normal loading construction of tower



• X

e1

e3e3e3e3e3e3e3e3e3

dead

load

p1

dead

load

p1

dead

load

p1

dead

load

p1

dead

load

p1

dead

load

p1

dead

load

p1

dead

load

p1

dead

load

p1

F1F1F1F1F1F1F1F1F1

cros

sse

ctio

nA

cros

sse

ctio

nA

cros

sse

ctio

nA

cros

sse

ctio

nA

cros

sse

ctio

nA

cros

sse

ctio

nA

cros

sse

ctio

nA

cros

sse

ctio

nA

cros

sse

ctio

nA

normal loading tower with a (almost) constant cross sectione.g. enercon e 36

• line load caused by dead load

constant cross section area

A = constant

line force

p1 = −ρ g A

volume force of one dimensional continua

b1 =p1

ρ A= −g

• single load at tower head (negative)

NEUMANN boundary condition - traction

t?1 =F1

A

• model problem with constant line force

volume force and load parameter

b1 = b with b = −g

model problem

e1

e3

E, ρ,A

L

ρb1, b1 = b = constant-t?1



• X

e1

e3e3e3e3e3e3e3e3e3

dead

load

p1

dead

load

p1

dead

load

p1

dead

load

p1

dead

load

p1

dead

load

p1

dead

load

p1

dead

load

p1

dead

load

p1

F1F1F1F1F1F1F1F1F1

cros

sse

ctio

nA

cros

sse

ctio

nA

cros

sse

ctio

nA

cros

sse

ctio

nA

cros

sse

ctio

nA

cros

sse

ctio

nA

cros

sse

ctio

nA

cros

sse

ctio

nA

cros

sse

ctio

nA

normal loading

tower with a (almost) constant cross sectione.g. enercon e 36

• line load caused by dead load


A = constant

line force

p1 = −ρ g A


b1 =p1

ρ A= −g

• single load at tower head (negative)

NEUMANN boundary condition - traction

t?1 =F1

A

• model problem with constant line force


b1 = b with b = −g

model problem

e1

e3

E, ρ,A

L

ρb1, b1 = b = constant-t?1



• X

?

??

?

6

?

6

dF1dF1dF1dF1dF1dF1dF1dF1dF1

dVdVdVdVdVdVdVdVdV

X1X1X1X1X1X1X1X1X1

LLLLLLLLL

e1e1e1e1e1e1e1e1e1

e2e2e2e2e2e2e2e2e2

e3e3e3e3e3e3e3e3e3

• rotor blade coordinate system

e1 - in direction of rotor blade e2 - in direction of generator shaft e3 = e1 × e2 - tangential direction

• centrifugal force of volume element dV

mass of volume element

dm = ρdV

number of revolutions of rotor n angular velocity of rotor ϕ2 = 2πn

distance generator shaft center and volume element X1

centrifugal force of volume element

dF1 = ϕ22 X1 ρdV

• resulting line force p1 (integration over A)

p1(X1) =

∫A

ϕ22 X1 ρ dX2dX3 = ϕ2

2 X1 ρ A(X1)

• comment: also other relevant forces (dead load, windloads) are acting on the rotor blade (not considered here)


motivation - rotor blades of wind power plantsÜ 7

• X

?

?

?

6

?

ϕϕϕϕϕϕϕϕϕ

XXXXXXXXX

uuuuuuuuu

uuuuuuuuu

dF = uρdVdF = uρdVdF = uρdVdF = uρdVdF = uρdVdF = uρdVdF = uρdVdF = uρdVdF = uρdV

6--

e3

e2

e1rb

• vector formulation of centrifugal force

distance vector generator shaft - volume element x

angular velocity vector of rotor

ϕ = [ϕ1 ϕ2 ϕ3]T

tangential velocity volume element

u = ϕ×X

acceleration volume element (constant rotation axisϕ/‖ϕ‖ and angular velocity ‖ϕ‖)

u =∂u

∂t= −ϕ× [ϕ×X]

centrifugal force volume element

dF = −ϕ× [ϕ×X] ρdV



• X

picture from HAU (2008) simplified rotor blade with a constant cross sectione.g. SMITH-PUTNAM wind power plant

• line load caused by centrifugal force


A = constant

line force

p1(X1) = ϕ22 ρ A X1


b1(X1) =p1(X1)

ρ A= ϕ2

2 X1

• model problem with linearly changing line force


b1(X1) =2b

LX1 with b =

ϕ22 L

2

PUTMAN (1948): ’Power from the Wind’HAU (2008): ’Windkraftanlagen: Grundlagen, Technik, Ein-satz, Wirtschaftlichkeit’

model problem

e1

e3

E, ρ,A

L

b1 = 2bLX1

ρb1



• X

columns, railway station willhelmshöhe, kassel

columns, excentric high-rise, bochumcolumns, fridericianum, kasselsmokestack, university of kasselfloodlight mast, ruhrstadion bochum (rewirpowerstadion)columns, brandenburg gate, berlincolumns, marie-elisabeth-lüders house, berlintelevision tower, berlin


rod model for structural analysisÜ 10

• X

columns, railway station willhelmshöhe, kassel

columns, excentric high-rise, bochum

columns, fridericianum, kasselsmokestack, university of kasselfloodlight mast, ruhrstadion bochum (rewirpowerstadion)columns, brandenburg gate, berlincolumns, marie-elisabeth-lüders house, berlintelevision tower, berlin



• X

columns, railway station willhelmshöhe, kasselcolumns, excentric high-rise, bochum

columns, fridericianum, kassel

smokestack, university of kasselfloodlight mast, ruhrstadion bochum (rewirpowerstadion)columns, brandenburg gate, berlincolumns, marie-elisabeth-lüders house, berlintelevision tower, berlin



• X

columns, railway station willhelmshöhe, kasselcolumns, excentric high-rise, bochumcolumns, fridericianum, kassel

smokestack, university of kassel

floodlight mast, ruhrstadion bochum (rewirpowerstadion)columns, brandenburg gate, berlincolumns, marie-elisabeth-lüders house, berlintelevision tower, berlin



• X

columns, railway station willhelmshöhe, kasselcolumns, excentric high-rise, bochumcolumns, fridericianum, kasselsmokestack, university of kassel

floodlight mast, ruhrstadion bochum (rewirpowerstadion)

columns, brandenburg gate, berlincolumns, marie-elisabeth-lüders house, berlintelevision tower, berlin



• X

columns, railway station willhelmshöhe, kasselcolumns, excentric high-rise, bochumcolumns, fridericianum, kasselsmokestack, university of kasselfloodlight mast, ruhrstadion bochum (rewirpowerstadion)

columns, brandenburg gate, berlin

columns, marie-elisabeth-lüders house, berlintelevision tower, berlin



• X

columns, railway station willhelmshöhe, kasselcolumns, excentric high-rise, bochumcolumns, fridericianum, kasselsmokestack, university of kasselfloodlight mast, ruhrstadion bochum (rewirpowerstadion)columns, brandenburg gate, berlin

columns, marie-elisabeth-lüders house, berlin

television tower, berlin



• X

columns, railway station willhelmshöhe, kasselcolumns, excentric high-rise, bochumcolumns, fridericianum, kasselsmokestack, university of kasselfloodlight mast, ruhrstadion bochum (rewirpowerstadion)columns, brandenburg gate, berlincolumns, marie-elisabeth-lüders house, berlin

television tower, berlin



• X

columns, railway station willhelmshöhe, kasselcolumns, excentric high-rise, bochumcolumns, fridericianum, kasselsmokestack, university of kasselfloodlight mast, ruhrstadion bochum (rewirpowerstadion)columns, brandenburg gate, berlincolumns, marie-elisabeth-lüders house, berlintelevision tower, berlin



• X

e1

e3

dead

load

p1(X

1)

F1

=mg

cros

sse

ctio

nA

(X1)

m

normal loading

e1

e3

win

dlo

adq 3

(X1)

F3

cros

sse

ctio

nA

(X1)

transverse loading • transverse loading andtransverse deformation

• and normal force and normaldeformation

• can independently calculated(see beam models)

e1

e3

p1

=co

nsta

nt

F1

A=

cons

tant

column

e1

e3

p1≈

smal

l

F1

A=

cons

tant

truss element

e1

e3

p1≈

0

F1

A=

cons

tant

truss element



• X

• application of rod model for structural analysis of

columns, stanchions

masts, towers, smokestacks

pylons

towers and rotor blades of wind turbines

• transverse loading and transverse deformation / normal force /normal deformation can independently calculated(see beam models)

• truss element is included as special case

dead load of the bar is negligible compared with nodal forces

analysis of spatial and planar truss structures (forces anddeformation)



• X

d.kuhl, wes.online, university of kasselÜ Ülinear computational structural mechanics • 2 1d finite element method • 2.1 motivation - spatial structures

motivation - lattice tower of wind power plantsÜ 17

• X

balance mom.

constitutive law

kinematics

1d continuum & equilibrium

ρ u1 =σ11,1+ρ b1

σ11 = Eε11

ε11 = u1,1

−ρAu11 2

ρAb1dX1

N?1 N?

1

truss element

- -

differential line element

−ρAu1

σ11

Xσ

11+σ

11,1dX

1

A

ρAb1

dX1

-

• balance of linear momentum, CAUCHY equation of motion forX ∈ Ω (N1 = σ11A, p1 = ρb1A)1d continuum | truss element

ρu1 = σ11,1 + ρb1 ρAu1 =N1,1+p1

• DIRICHLET boundary conditions for X ∈ Γu

u1 = u?1

• NEUMANN boundary conditions for X ∈ Γσ(N1 = σ11A, n1 = 1)

σ11n1 = t?1 N1 = N?1

• initial conditions for X ∈ Ω

u1(t0) = u10, u1(t0) = u10

d.kuhl, wes.online, university of kasselÜ Ülinear computational structural mechanics • 2 1d finite element method • 2.2 modeling - strong form

mechanics of one-dimensional continuaÜ 18

• X

• differential equation within domain Ω

ρ u1 − ρ b1 =σ11,1 kinetics

?σ11 = Eε11 constitutive law

?ε11 = u1,1 kinematics

ρ u1 − ρ b1 =E u1,11 ∀ X1 ∈ Ω

• boundary conditions at DIRICHLET Γu and NEUMANN Γσ boundaries

u1 = u?1 ∀ X1 ∈ Γu σ11 n1 = σ11 = t?1 ∀ X1 ∈ Γσ

• initial conditions for displacement - or acceleration field

u1(t0) = u0 ∀ X1 ∈ Ω u1(t0) = u1 0 ∀ X1 ∈ Ω

• characterization

linear partial differential equation with respect to space (u1,11) and time (u1)


strong form - initial boundary value problemÜ 19

• X

boundary Γ=Γu∪Γσ ,Γu∩Γσ=0

body Ω

kinetics material kinematics

stresses

σ11(X1, t), σ011(X1)

balance of linear momentum

ρ u1 = σ11,1 + ρ b1

loadρ b1

initial stressesσ0

11(X1) = σ11(X1, 0)

NEUMANN bc

σ1 n1 = t?1

constitutive law

σ11 = E ε11 + σ011

strains

ε11(X1, t), εθ11(X1, t)

deformation

ε11 = u1,1 − εθ11

initial conditionsu1(X1, t0), (u1(X1, t0))

thermal strainsεθ11 = α [θ − θref]

DIRICHLET bc

u1 = u?1TONTI scheme(TONTI (1975)

STEIN&BARTHOLD (1996))


initial boundary-value problem of elastodynamicsÜ 20

• X

6

6

p1

N1

u1

line load

normal force

normal displacement

equilibrium, balance of linear momentum

constitutive law and kinematics

displacement field

p1(X1) = −N1,1(X1)

N1(X1) = Aσ11 = EAε11 = EAu1,1

N1(X1) = −∫p1(X1) dX1 + C

11

?

?u1(X1) =

1

EA

∫N1(X1) dX1 + C

21

+ NEUMANN and DIRICHLET boundary conditions for determination of C11 and C2

1


integration of deformation of bar (statics)Ü 21

• X

motivation - tower

e1

e3e3e3e3e3e3e3e3e3

dead

load

p1

dead

load

p1

dead

load

p1

dead

load

p1

dead

load

p1

dead

load

p1

dead

load

p1

dead

load

p1

dead

load

p1

F1F1F1F1F1F1F1F1F1

cros

sse

ctio

nA

cros

sse

ctio

nA

cros

sse

ctio

nA

cros

sse

ctio

nA

cros

sse

ctio

nA

cros

sse

ctio

nA

cros

sse

ctio

nA

cros

sse

ctio

nA

cros

sse

ctio

nA

model problem - tension bar loaded by ρb1 (without F1)

e1e2

e3

ρb1, b1 = b = constant

L

E, A, ρ, Ω = [0, L]

• balance of momentum(plus constitutive law and kinematics)

E u1,11 + ρ b1 = 0

• homogeneous DIRICHLET boundary conditions

u1(X1 = 0) = 0

• NEUMANN boundary condition (traction free surface)

t?1 = σ11(X1 = L) n1 = σ11(X1 =L) = 0

σ11(X1 =L) = E u1,1(X1 =L) → u1,1(X1 =L) = 0


example - tension barÜ 22

• X

tension bar loaded by ρb1

e1

e3

E, ρ,A

L


fe discretizationanalytical solution

ε11EρbL

u1EρbL2

X1/L0.0

0.2

0.4

0.6

0.8

1.0

1.2

0.0 1.0

solution of boundary value problem

• integration of balance of linear momentum

∂u1(X1)

∂X1= −

ρ b

E

∫dX1 + C1

1 = −ρ b

EX1 + C1

1

• NEUMANN boundary condition

∂u1(X1 = L)

∂X1= −

ρ b

EL+ C1

1 = 0 → C11 =

ρ b

EL

• normal strain - spatial derivative of displacement

∂u1(X1)

∂X1= ε11(X1) =

ρ b

E[L−X1]

• integration of normal strain (sparation and integration)

u1(X1) =ρ b

E

[ ∫L dX1 −

∫X1 dX1

]+ C2

1

=ρ b

E

[L X1 −

1

2X2

1

]+ C2

1

• DIRICHLET boundary condition

u1(X1 = 0) = C21 = 0 → C2

1 = 0


example - tension bar (load case i)Ü 23

• X


e1

e3

E, ρ,A

L



ε11EρbL

u1EρbL2

X1/L0.0

0.2

0.4

0.6

0.8

1.0

1.2

0.0 1.0

solution of boundary value problem

• displacement field

u1(X1) =ρ b

E

[L X1 −

1

2X2

1

]• displacements at selected positions

u1(L) =ρb

E

[L2 −

1

2L2

]=

1

2

ρbL2

E

u1(3L

4) =

ρb

E

[3

4L2 −

9

32L2]

=15

32

ρbL2

E

u1(2L

3) =

ρb

E

[2

3L2 −

4

18L2]

=4

9

ρbL2

E

u1(L

2) =

ρb

E

[1

2L2 −

1

8L2

]=

3

8

ρbL2

E

u1(L

3) =

ρb

E

[1

3L2 −

1

18L2]

=5

18

ρbL2

E

u1(L

4) =

ρb

E

[1

4L2 −

1

32L2]

=7

32

ρbL2

E



• X


e1

e3

E, ρ,A

L



ε11EρbL

u1EρbL2

X1/L0.0

0.2

0.4

0.6

0.8

1.0

1.2

0.0 1.0

postprocessing - calculation of strain and stress fields


u1(X1) =ρ b

E

[L X1 −

1

2X2

1

]• strain field - derivative of displacement field

ε11(X1) = u1,1(X1) =ρ b

E

[L−X1

]• stress field - constitutive law

σ11(X1) = E ε11 = ρ b[L−X1

]• derivative of stress field

σ11,1(X1) = E ε11,1 = −ρ b

• residuum (local error)solution of differential equation

σ11,1(X1) + ρ b1 = −ρ b+ ρ b = 0

local error is zero→ analytical solution is correct!



• X

displacement u1

X1/L

u1 EρbL2

0 0.25 0.75 10

0.1

0.2

0.3

0.4

0.5

0.6 12

1532

38

732

518

49

strain ε11

X1/L

ε11 EρbL

0 0.25 0.75 10

0.2

0.4

0.6

0.8

1.0

1.2

stress σ11

X1/L

σ11ρbL

0 0.25 0.75 10

0.2

0.4

0.6

0.8

1.0

1.2

residuum σ11,1 + ρb1

X1/L

σ11,1 + ρb1ρb

0 0.25 0.75 1−1.2

−0.8

−0.4

0.0

0.4

0.8

1.2


example - tension barÜ 26

• X

displacement u1

strain ε11

X1

u1

u1

u1 + δu1

δu1 δu1

X1

ε11

ε11

ε11 + δε11

δε11

δε11

• balance of momentum and NEUMANN boundary condition

ρ u1 − ρ b1 − σ11,1 = 0 ∀X1 ∈ Ω

σ11 n1 − t?1 = 0 ∀X1 ∈ Γσ

• multiplication with virtual displacement δu1

δu1 [ρ u1 − ρ b1 − σ11,1] = 0

δu1 [σ11 n1 − t?1] = 0

• properties of virtual displacements δu1

δu1 satisfies DIRICHLET boundary conditionsδu1 = 0 ∀X1 ∈ Γu

δu1 satisfies field conditionδε11 = δu1,1 ∀X1 ∈ Ω

δu1 is arbitrary (not equal zero)

δu1 is infinite small

d.kuhl, wes.online, university of kasselÜ Ülinear computational structural mechanics • 2 1d finite element method • 2.3 pvw - weak form

weak form of continuum mechanics - 1dÜ 27

• X

displacement u1

strain ε11

X1

u1

u1

u1 + δu1

δu1 δu1

X1

ε11

ε11

ε11 + δε11

δε11

δε11

• integration over domain Ω (length)

L∫0

δu1 [ρ u1 − ρ b1 − σ11,1] dX1 = 0

δu1 [σ11 n1 − t?1] = 0

• partial integration of stress term

L∫0

δu1 σ11,1 dX1 =[δu1 σ11 n1︸︷︷︸

1

]L0−L∫

0

δu1,1︸︷︷︸δε11

σ11 dX1

• inclusion into integrated balance of momentumNEUMANN boundary conditions at left and right boundaryaddition of both termsL∫

0

δu1 [ρ u1 − ρ b1] dX1 +

L∫0

δε11 σ11 dX1

−[δu1 σ11 n1

]L0

+[δu1 σ11 n1

]L0−[δu1 t

?1

]L0

= 0

• rearrangement


weak form of continuum mechanics - 1dÜ 28

• X

displacement u1

strain ε11

X1

u1

u1

u1 + δu1

δu1 δu1

X1

ε11

ε11

ε11 + δε11

δε11

δε11

• principle of virtual workL∫

0

δu1 ρ u1 dX1 +

L∫0

δε11 σ11 dX1

=[δu1 t

?1

]L0

+

L∫0

δu1 ρ b1 dX1

• principle of virtual work

δWdyn + δWint = δWext

L∫0

δ

• virtual work of inertia forces

δWdyn =

L∫0

δu1 ρ u1 dX1

• internal virtual work

δWint =

L∫0

δε11 σ11 dX1

• external virtual work

δWext =[δu1 t

?1

]L0

+

L∫0

δu1 ρ b1 dX1


principle of virtual work - 1dÜ 29

• X

displacement u1

X1/L

u1 EρbL2

0 10

0.1

0.2

0.3

0.4

0.5

0.6

strain ε11

X1/L

ε11 EρbL

0 10

0.2

0.4

0.6

0.8

1.0

1.2

model problem - tension bar loaded by ρb1

e1e2

e3


L

E, A, ρ, Ω = [0, L]

• principle of virtual work, statics u1 = 0

DIRICHLET and NEUMANN

boundary conditions δu1(0) = 0 and t?1(L) = 0

L∫0

δε11 σ11 dX1 =

L∫0

δu1 ρ b1 dX1

• linear approximation of solution(ansatz of displacement field)

u1(X1) =u2

1

LX1

displacement ansatz u1

X1

u1

node 1 node 2

u21u1(X1)

strain approximation ε11

X1

ε11

node 1 node 2

ε11(X1)


tension bar - approximated solutionÜ 30

• X


X1

u1

node 1 node 2

u21u1(X1)


X1

ε11

node 1 node 2

ε11(X1)

• linear approximation of solution(ansatz of displacement field)

u1(X1) =u2

1

LX1

• constant approximation of normal strain

ε11(X1) =∂u1(X1)

∂X1=u2

1

L

• constant approximation of normal stress

σ11(X1) = E ε11(X1) =E u2

1

L

• linear approximation of virtual displacement(analogous to real displacement)

δu1(X1) =δu2

1

LX1

• constant approximation of virtual normal strain

δε11(X1) =∂δu1(X1)

∂X1=δu2

1

L



• X

• principle of virtual work including approximations

L∫0

δε11 σ11 dX1 =

L∫0

δu1 ρ b1 dX1

• inclusion of approximations

L∫0

δu21

L

E u21

LdX1 =

L∫0

δu21

LX1 ρ b1 dX1

• integration - real and virtual nodal displacements u21 and δu2

1 are independent of X1

δu21

E

Lu2

1 = δu21

ρ b L

2δu2

1

[E

Lu2

1 −ρ b L

2

]= 0

• δu21 is arbitrary→ term in brackets is zero

E

Lu2

1 −ρ b L

2= 0 u2

1 =ρ b

2 EL2

• approximated solution is at node 2 identical to analytical solution



• X

postprocessingcalculation of approximated displacement, strain and stress fields


u1(X1) =u2

1

LX1 =

ρ b L

2 EX1

• strain field - derivative of displacement field

ε11(X1) = u1,1(X1) =ρ b L

2 E

• stress field - constitutive law

σ11(X1) = E ε11(X1) =ρ b L

2

• derivative of stress field

σ11,1(X1) = E ε11,1(X1)

• residuum (local error) - solution of differential equation

σ11,1(X1) + ρ b = 0 + ρ b = ρ b

local error is not zero→ approximated solution is not correct for every X1!



• X

displacements u1 and u1

X1/L

u1 EρbL2

0 0.25 0.75 10

0.1

0.2

0.3

0.4

0.5

0.6 12

strain ε11 and ε11

X1/L

ε11 EρbL

0 0.25 0.75 10

0.2

0.4

0.6

0.8

1.0

1.2

stress σ11 and σ11

X1/L

σ11ρbL

0 0.25 0.75 10

0.2

0.4

0.6

0.8

1.0

1.2

residuum σ11,1 + ρb1 and σ11,1 + ρb1

X1/L

σ11,1 + ρb1ρb

0 0.25 0.75 1−1.2

−0.8

−0.4

0.0

0.4

0.8

1.2



• X

• check principle of virtual work

introducing approximations δu1(X1), δu1,1(X1) and u1(X1)

using nodal displacement u21 =

ρb2E

L2

.L∫

0

δu21

L

E u21

LdX1 −

L∫0

δu21

LX1 ρ b dX1 = δu2

1

[ρ b2L−

ρ b

2L]

= 0

• approximated solution fulfills principle of virtual work even if differential equation is not fulfilled!

• principle of virtual work is basis for development of numerical methods for mechanics

• development of finite element method is based on principle of virtual work

• possibilities to improve quality of approximated solution

subdivision of domain in smaller sub domains using linear ansatz functions (h method)

using higher order polynomials for approximations (p method)



• X

displacement u1

X1/L

u1 EρbL2

0 10

0.1

0.2

0.3

0.4

0.5

0.6

strain ε11

X1/L

ε11 EρbL

0 10

0.2

0.4

0.6

0.8

1.0

1.2

model problem - tension bar loaded by ρb1

e1e2

e3


L

E, A, ρ, Ω = [0, L]

• principle of virtual work

L∫0

δε11 σ11 dX1 =

L∫0

δu1 ρ b1 dX1

• piecewise linear approximation of displacement field

domain 1

u1(X1) =2 u2

1

LX1

domain 2

u1(X1) = u21 +

2[u31 − u2

1]

L[X1 −

L

2]

u1(X1) = 2u21 − u3

1 +2[u3

1 − u21]

LX1


u1

node 1 node 2 node 3

domain 1 domain 2

u31u2

1

u1(X1)


ε11


domain 1 domain 2

ε11(X1)

d.kuhl, wes.online, university of kasselÜ Ülinear computational structural mechanics • 2 1d finite element method • 2.4 finite element concept

tension bar - finite element solutionÜ 36

• X


u1


domain 1 domain 2

u31u2

1

u1(X1)


ε11


domain 1 domain 2

ε11(X1)

• piecewise constant strain approximation domains 1 and 2

ε11(X1) =2 u2

1

Lε11(X1) =

2[u31 − u2

1]

L

• piecewise constant stress approximation σ11 = Eε11

domains 1 and 2

σ11(X1) = E2 u2

1

Lσ11(X1) = E

2[u31 − u2

1]

L

• piecewise linear approximation of virtual displacement field domain 1

δu1(X1) =2 δu2

1

LX1

domain 2

δu1(X1) = 2δu21 − δu3

1 +2[δu3

1 − δu21]

LX1

• approximation of virtual strain domains 1 and 2

δε11(X1) =2 δu2

1

Lδε11(X1) =

2[δu31 − δu2

1]

L



• X

• principle of virtual work including approximations

L∫0

δε11 σ11 dX1 =

L∫0

δu1 ρ b1 dX1

• internal virtual work - considering two domains, inducing approximations, integration

L∫0

δε11 σ11 dX1 =

L/2∫0

δε11 σ11 dX1 +

L∫L/2

δε11 σ11 dX1

=

L/2∫0

2δu21

L

2Eu21

LdX1 +

L∫L/2

2[δu31 − δu2

1]

L

2E[u31 − u2

1]

LdX1

=4Eδu2

1u21

L2

L/2∫0

dX1 +4E[δu3

1 − δu21][u3

1 − u21]

L2

L∫L/2

dX1

=2Eδu2

1u21

L+

2E[δu31 − δu2

1][u31 − u2

1]

L

= δu21

2E

Lu2

1 +2E[δu3

1u31 − δu3

1u21 − δu2

1u31 + δu2

1u21]

L



• X

• rearrangement and transformation in matrix format

L∫0

δε11σ11 dX1 = δu21

2E

Lu2

1 + δu31

2E

Lu3

1 − δu31

2E

Lu2

1 − δu21

2E

Lu3

1 + δu21

2E

Lu2

1

=

δu21

δu31

︸︷︷︸δu

·

4EL

−2EL

−2EL

2EL

︸︷︷︸

K

u21

u31

︸︷︷︸u

= δu · K u

• definition of

system displacement vector u,

virtual system displacement vector δu

and system stiffness matrix K.

u =

u21

u31

, δu =

δu21

δu31

K =

4EL

−2EL

−2EL

2EL



• X

• external virtual work - considering two domains, inducing approximations, integration

L∫0

δu1 ρb1 dX1 =

L/2∫0

δu1 ρb1 dX1 +

L∫L/2

δu1 ρb1 dX1

=

L/2∫0

2 δu21

LX1 ρb dX1 +

L∫L/2

[2δu2

1−δu31+

2[δu31−δu2

1]

LX1

]ρb dX1

= δu21

2ρb

L

L/2∫0

X1 dX1 + [2δu21 − δu3

1] ρb

L∫L/2

dX1

+ [δu31−δu2

1]2ρb

L

L∫L/2

X1 dX1

= δu21

ρb

L

[[X1]2

]L/20

+ [2δu21 − δu3

1] ρbL

2

+ [δu31 − δu2

1]ρb

L

[[X1]2

]LL/2

= δu21

ρbL

4+ [2δu2

1 − δu31]ρbL

2

+ [δu31 − δu2

1]3ρbL

4



• X

• rearrangement and transformation in matrix format

L∫0

δu1 ρ b1 dX1 = δu21

[ρbL

4+ ρbL−

3ρbL

4

]+ δu3

1

[−ρbL

2+

3ρbL

4

]

= δu21

ρbL

2+ δu3

1

ρbL

4=

δu21

δu31

︸︷︷︸δu

·

ρbL2ρbL

4

︸︷︷︸r

= δu · r

• definition of

virtual system displacement vector δu and system load vector r

x

δu =

δu21

δu31

, r =

ρbL2ρbL

4

• principle of virtual displacements including approximationsδu2

1

δu31

︸︷︷︸δu

·

4EL

−2EL

−2EL

2EL

︸︷︷︸

K

u21

u31

︸︷︷︸u

=

δu21

δu31

︸︷︷︸δu

·

ρbL2ρbL

4

︸︷︷︸r



• X

• rearrangementδu21

δu31

︸︷︷︸δu

·

4EL

−2EL

−2EL

2EL

︸︷︷︸

K

u21

u31

︸︷︷︸u

−

ρbL2ρbL

4

︸︷︷︸r

=

0

0

︸︷︷︸0

• for arbitrary virtual displacements δu21 and δu3

1

(or for arbitrary virtual system displacement vector δu) 4EL

−2EL

−2EL

2EL

︸︷︷︸

K

u21

u31

︸︷︷︸u

=

ρbL2ρbL

4

︸︷︷︸r

• system stiffness relation

K u = r

• solution of linear system of equations u21 und u3

1 4EL

−2EL

0 2EL

u21

u31

=

ρbL2ρbL

, u31 =

ρb

2EL2

u21 =

L

4E

[ρbL

2+

2E

L

ρb

2EL2

]=

3ρb

8EL2



• X

postprocessing - calculation of approximated displacement, strain and stress fields• Verschiebungsfeld

domain 1

u1(X1) =2 u2

1

LX1 =

3ρbL

4EX1

domain 2

u1(X1) = 2u21 − u3

1 −2[u2

1 − u31]

LX1 =

ρb

4EL2 +

ρb

4EL X1 =

ρb

4EL [L+X1]

• strain field - derivative of displacement fields in domains 1 and 2

ε11(X1) =3ρbL

4Eε11(X1) =

ρb

4EL

• stress field an derivative of stress field in domains 1 and 2

σ11(X1) = E ε11(X1) =ρ b L

2σ11,1(X1) = E ε11,1(X1)

• residuum (local error) - solution of differential equation

σ11,1(X1) + ρ b1 = 0 + ρ b = ρ b

local error is not zero→ approximated solution is not correct for every X1!



• X

displacements u1 and u1

X1/L

u1 EρbL2

0 0.25 0.75 10

0.1

0.2

0.3

0.4

0.5

0.6 12

38

strains ε11 and ε11

X1/L

ε11 EρbL

0 0.25 0.75 10

0.2

0.4

0.6

0.8

1.0

1.2

stresses σ11 and σ11

X1/L

σ11ρbL

0 0.25 0.75 10

0.2

0.4

0.6

0.8

1.0

1.2

residuals σ11,1 + ρb1 and σ11,1 + ρb1

X1/L

σ11,1 + ρb1ρb

0 0.25 0.75 1−1.2

−0.8

−0.4

0.0

0.4

0.8

1.2



• X

displacement u1

strain ε11

u1

X1

ε11 σ11 = Eε11

X1

analytical solutionfe solution

displacement u1

strain ε11

u1

X1

ε11 σ11 = Eε11

X1

displacement u1

strain ε11

u1

X1

ε11 σ11 = Eε11

X1


tension bar - convergence of fe solutionÜ 45

• X


u1

s-node 1 system node 2 s-node 3

element 1 element 2

u31u2

1

u1(X1)


ε11


element 1 element 2

ε11(X1)

• subdivision of domain Ω in subdomains or elements Ωe withe ∈ [1, 2]

• generation of elements with linear approximations ofdisplacement field

• derivation of approximations for stress, strain, virtualdisplacment and virtual strain fields

• application of principle of virtual displacements within domain Ω

using approximations (δWdyn + δWint =δWext)

• rearrangement: moving virtual and real nodal displacements(δu2

1, δu31, u2

1 and u31) outside integrals (δui1 to left hand side,

ui1 to right hand side)

• integration of virtual works of hole domain using summationover elements

• transformation to matrix format, application for arbitrary virtualdisplacements

• solution of system stiffness relation

• postprocessing calculating approximated displacement, strainand stress fields


main aspect of present finite element solutionÜ 46

• X


u1


element 1 element 2

u31u2

1

u1(X1)


ε11


element 1 element 2

ε11(X1)

postive aspects of present finite element solution

• approximative solution of elasticity can be found by usingsimple mathematics (statement is also valid for differentialequations which can not be solved analytically)

• approximated numerical solution converges to the analyticalsolution, using an increased number of elements (h method) oran increased polynomial degree (p method)

• values interesting for engineers (strain and stress) can becalculated by post processing procedure

negative aspects of present finite element solution

• only approximation of solution which should be evaluated byengineer

• ansatz functions have been hardly developed, no systematicformulation included

• hand made rearrangement of approximated weak forms areindividual for every problem

• procedure is not well suited for computational implementation


plus and minus of present fe solutionÜ 47

• X


u1


element 1 element 2

u31u2

1

u1(X1)


ε11


element 1 element 2

ε11(X1)

opitimzation of finite element conceptcomputer oriented formulation

• strict separation of structural and element levels

• systematic design of displacement approximation on elementlevel

• generation of weak form on the element levelδW e

dyn + δW eint =δW e

ext

• systematic integration of virtual work terms and derivation ofelement matrices and vectors

• generation of structural equation of motion (structural stiffnessrelation) by assembling of elements

• postprocessing on element level


computer oriented formulation of femÜ 48

• X

phenomenological levels of the finite element method

’tetraeder’ in bottrop

selected truss element

partial domain

fe system

finite truss element

GAUSS point

structural level

component level

material point level

system of equations

finite elements

constitutive law

K u = r

ke ue = re

N1 = EA ε11

reality discretized model


finite elements - levels of observationsÜ 49

• X

phenomenological levels of the finite element method

composite bridge

piece of road

concrete

fe-system

volume element

GAUSS point

structural level

component level

material point level

system of equations

finite elements

constitutive law

K u = r

ke ue = re

σ = C : ε

reality discretized model


finite elements - levels of observationsÜ 50

• X

system level (structure) element level (finite element)

subdivision of domain Ω in elements Ωediplacement approximation and resulting ap-proximations

separation of real and virtual nodal displace-ments in virtual work terms

formulation of approximated weak form usingelement matrices and vectors

formulation of approximated weak form usingsystem matrices and vectors by assembly con-sidering DIRICHLET boundary conditions

generation of system equation of motion orstructural stiffness relation using fundamentallemma of variational calculus

solution of structural displacement vector separation of element displacements

postprocessing: approximation of field quanti-ties

graphical interpretation of approximated fieldquantities


finite element method conceptÜ 51

• X

analytical solution

system, structure or domain Ω

u1

concept

• subdivision of structure in finite elements e• definition of structural nodal displacements uj1 as sampling points of piecewise linear

displacement approximations u1

• definition of elements, element nodes and element nodal displacements uei1• design of displacement approximation within elements e

subdivision in elements, structural nodal displacements and displacement approximation

u1

u1

u1

u11 = 0

u21

u31

u41

1 2 3 4a e c

associated elemental nodal displacements and displacement approximation

u21

u31

u41

1 2 3 4

u1

ua11

ua21

1 2a

u1

ue11

ue21

1 2e

u1uc11

uc21

1 2c

d.kuhl, wes.online, university of kasselÜ Ülinear computational structural mechanics • 2 1d finite element method • 2.5 displacement approximation

design of displacement approximationÜ 52

• X

disp. approx. element e

u1

ue11

ue21

1 2e

disp. approx. element e

two parts of ue1

u1

ue11

ue21

1 2e

u1

ue11

ue21

1 2e

approximation to node 1

approximation to node 2

u1 = u11 + u2

1

ue11

u11

1 2e

ue21

u21

1 2e

shape function N1

shape function N2

u1 = N1 ue11 +N2 ue21

ue11

N1

1

1 2e

ue21

N2

1

1 2e

d.kuhl, wes.online, university of kasselÜ Ülinear computational structural mechanics • 2 1d finite element method • 2.5 displacement approximation

design of displacement approximationÜ 53

• X

constant b1linear element

element load vector

re=

[re11

re21

]=ρb1L

2

[1

1

]1 2e

ρ, L

b1 b1

1 2e

re11 re21

linear b1linear element

element load vector

re=ρL6

[2be11 +be21

be11 +2be21

]1 2e

ρ, L

be11

be21

1 2e

re11 re21

concept

• substitution of distributedloads by nodal forces yieldingsame virtual work as originaldistributed load

realization

• evaluating integrals forelement load vectors or’tensors’

constant b1linear element

element load vector

re=

[re11

re21

]=ρb1L

2

[1

1

]1 2e

ρ, L

b1 b1

1 2e

re11 re21

linear b1linear element

element load vector

re=ρL6

[2be11 +be21

be11 +2be21

]1 2e

ρ, L

be11

be21

1 2e

re11 re21

constant b1quadratic element

element load vector

re=

re11

re21

re31

=ρb1L

6

1

4

1

1 32

ρ, L

b1 b1

1 32

re11 re31re21

d.kuhl, wes.online, university of kasselÜ Ülinear computational structural mechanics • 2 1d finite element method • 2.7 approximation weak form

consistent element load vectorsÜ 70

• X

• element stiffness ’tensor’

keij =

1∫−1

N i,1(ξ1) E Nj

,1(ξ1)L

2dξ1

• element stiffness matrix for p = 1

ke =

[ke11 ke12

ke21 ke22

]=E

L

[1 −1

−1 1

]

• element load ’tensor’

rei1 =

1∫−1

N i(ξ1) ρ b1(ξ1)L

2dξ1

• element load vector

re =

[re11

re21

]

• element mass ’tensor’

meij =

1∫−1

N i(ξ1) ρ Nj(ξ1)L

2dξ1

• element mass matrix for p = 1

me =

[me11 me12

me21 me22

]=ρL

6

[2 1

1 2

]

• element load vector for p = 1

and constant b1

re =

[re11

re21

]=ρ b1 L

2

[1

1

]

• element load vector for für p = 1

and linear b1

re =

[re11

re21

]=ρ L

6

[2be11 + be21

be11 + 2be21

]


1d continuum elementÜ 71

• X

• approximated weak form of element e

δW edyn + δW e

int = δW eext

• approximated weak form of element e using element matrices and vectors

δue ·me ue + δue · ke ue = δue · re

• example - graphical interpretation of δue · ke ue using FALK matrix scheme

EL

EL

−EL

−EL

δue11 δue21

ue11

ue21

• rows of ke correspond to virtual displacements δuei1• columns of ke correspond to real displacements uej1

-EL

EL

−EL

−EL

δue11

δue21

ue11 ue21

virt

ual

noda

lele

men

tdi

spla

cem

ents

realnodal elementdisplacements

elementstiffness matrix


approximation of weak formÜ 72

• X


δW edyn + δW e

int = δW eext



• example - graphical interpretation of δue · ke ue using FALK matrix scheme

EL

EL

−EL

−EL

δue11 δue21

ue11

ue21

• rows of ke correspond to virtual displacements δuei1• columns of ke correspond to real displacements uej1

-EL

EL

−EL

−EL

δue11

δue21

ue11 ue21

vir t

ual

noda

l ele

men

tdi

spla

cem

ents

realnodal elementdisplacements

elementstiffness matrix


approximation of weak formÜ 72

• X


δW edyn + δW e

int = δW eext



• approximated weak form of system / structure

δWdyn + δWint = δWext

• approximated weak form of system with

virtual structural displacement vector δu and structural displacement vector u structural acceleration vector u and structral mass matrix M structural stiffness matrix K and structural load vector r

x

δu ·M u+ δu · K u = δu · r

• relation between element and structural quantities? basis is given by summation of elementvirtual works

δWint,dyn,ext =NE∑e=1

δW eint,dyn,ext

d.kuhl, wes.online, university of kasselÜ Ülinear computational structural mechanics • 2 1d finite element method • 2.8 assembly and solution

assembly of elements to structureÜ 73

• X

u21

u31

u41

1 2 3 4

u1

ua11

ua21

1 2a

u1

ue11

ue21

1 2e

u1uc11

uc21

1 2c

finite element system with element nodal displacements

ke11 ke12

ke21 ke22

δue11

δue21

ue11 ue21

δW eint

ka11 ka12

ka21 ka22

δua11

δua21

ua11 ua2

1

δWaint

kc11 kc12

kc21 kc22

δuc11

δuc21

uc11 uc21

δW cintδWint = + +

u21

u31

u41

1 2 3 4

u1

ua11

ua21

1 2a

u1

ue11

ue21

1 2e

u1uc11

uc21

1 2c

relation between system and element nodal displacements

δWint = + +

ke11 ke12

ke21 ke22

δue11

δue21

ue11 ue21

δW eint

ka11 ka12

ka21 ka22

δua11

δua21

ua11 ua2

1

δWaint

kc11 kc12

kc21 kc22

δuc11

δuc21

uc11 uc21

δW cint

δu21

δu31

u21 u3

1

0

δu21

0 u21

δu31

δu41

u31 u4

1


addition of element virtual worksÜ 74

• X

u21

u31

u41

1 2 3 4

u1

ua11

ua21

1 2a

u1

ue11

ue21

1 2e

u1uc11

uc21

1 2c


ke11 ke12

ke21 ke22

δue11

δue21

ue11 ue21

δW eint

ka11 ka12

ka21 ka22

δua11

δua21

ua11 ua2

1

δWaint

kc11 kc12

kc21 kc22

δuc11

δuc21

uc11 uc21

δW cint

δWint = + +

u21

u31

u41

1 2 3 4

u1

ua11

ua21

1 2a

u1

ue11

ue21

1 2e

u1uc11

uc21

1 2c


δWint = + +

ke11 ke12

ke21 ke22

δue11

δue21

ue11 ue21

δW eint

ka11 ka12

ka21 ka22

δua11

δua21

ua11 ua2

1

δWaint

kc11 kc12

kc21 kc22

δuc11

δuc21

uc11 uc21

δW cint

δu21

δu31

u21 u3

1

0

δu21

0 u21

δu31

δu41

u31 u4

1



• X

u21

u31

u41

1 2 3 4

u1

ua11

ua21

1 2a

u1

ue11

ue21

1 2e

u1uc11

uc21

1 2c


ke11 ke12

ke21 ke22

δue11

δue21

ue11 ue21

δW eint

ka11 ka12

ka21 ka22

δua11

δua21

ua11 ua2

1

δWaint

kc11 kc12

kc21 kc22

δuc11

δuc21

uc11 uc21

δW cintδWint = + +

u21

u31

u41

1 2 3 4

u1

ua11

ua21

1 2a

u1

ue11

ue21

1 2e

u1uc11

uc21

1 2c


δWint = + +

ke11 ke12

ke21 ke22

δue11

δue21

ue11 ue21

δW eint

ka11 ka12

ka21 ka22

δua11

δua21

ua11 ua2

1

δWaint

kc11 kc12

kc21 kc22

δuc11

δuc21

uc11 uc21

δW cint

δu21

δu31

u21 u3

1

0

δu21

0 u21

δu31

δu41

u31 u4

1



• X

u21

u31

u41

1 2 3 4

u1

ua11

ua21

1 2a

u1

ue11

ue21

1 2e

u1uc11

uc21

1 2c

generation of structural stiffness matrix by assembly

δWint = + +

ke11 ke12

ke21 ke22

δue11

δue21

ue11 ue21

δW eint

ka11 ka12

ka21 ka22

δua11

δua21

ua11 ua2

1

δWaint

kc11 kc12

kc21 kc22

δuc11

δuc21

uc11 uc21

δW cint

δu21

δu31

u21 u3

1

0

δu21

0 u21

δu31

δu41

u31 u4

1

0

δu21

δu31

δu41

0 u21 u3

1 u41

K11 K12 K13 K14

K21 K22 K23 K24

K31 K32 K33 K34

K41 K42 K43 K44

δWint =

K22 K23 K24

K32 K33 K34

K42 K43 K44

ka22+ke11ka22+ke11 ke12 0

ke21 ke22+kc11 kc12

0 kc21 kc22

δWint = δu · K u



• X

u21

u31

u41

1 2 3 4

u1

ua11

ua21

1 2a

u1

ue11

ue21

1 2e

u1uc11

uc21

1 2c


δWint = + +

ke11 ke12

ke21 ke22

δue11

δue21

ue11 ue21

δW eint

ka11 ka12

ka21 ka22

δua11

δua21

ua11 ua2

1

δWaint

kc11 kc12

kc21 kc22

δuc11

δuc21

uc11 uc21

δW cint

δu21

δu31

u21 u3

1

0

δu21

0 u21

δu31

δu41

u31 u4

1

0

δu21

δu31

δu41

0 u21 u3

1 u41

K11 K12 K13 K14

K21 K22 K23 K24

K31 K32 K33 K34

K41 K42 K43 K44

δWint =

K22 K23 K24

K32 K33 K34

K42 K43 K44

ka22+ke11

ka22+ke11 ke12 0

ke21 ke22+kc11 kc12

0 kc21 kc22

δWint = δu · K u



• X

u21

u31

u41

1 2 3 4

u1

ua11

ua21

1 2a

u1

ue11

ue21

1 2e

u1uc11

uc21

1 2c


δWint = + +

ke11 ke12

ke21 ke22

δue11

δue21

ue11 ue21

δW eint

ka11 ka12

ka21 ka22

δua11

δua21

ua11 ua2

1

δWaint

kc11 kc12

kc21 kc22

δuc11

δuc21

uc11 uc21

δW cint

δu21

δu31

u21 u3

1

0

δu21

0 u21

δu31

δu41

u31 u4

1

0

δu21

δu31

δu41

0 u21 u3

1 u41

K11 K12 K13 K14

K21 K22 K23 K24

K31 K32 K33 K34

K41 K42 K43 K44

δWint =

K22 K23 K24

K32 K33 K34

K42 K43 K44

ka22+ke11ka22+ke11

ke12 0

ke21 ke22+kc11 kc12

0 kc21 kc22

δWint = δu · K u



• X

u21

u31

u41

1 2 3 4

u1

ua11

ua21

1 2a

u1

ue11

ue21

1 2e

u1uc11

uc21

1 2c


δWint = + +

ke11 ke12

ke21 ke22

δue11

δue21

ue11 ue21

δW eint

ka11 ka12

ka21 ka22

δua11

δua21

ua11 ua2

1

δWaint

kc11 kc12

kc21 kc22

δuc11

δuc21

uc11 uc21

δW cint

δu21

δu31

u21 u3

1

0

δu21

0 u21

δu31

δu41

u31 u4

1

0

δu21

δu31

δu41

0 u21 u3

1 u41

K11 K12 K13 K14

K21 K22 K23 K24

K31 K32 K33 K34

K41 K42 K43 K44

δWint =

K22 K23 K24

K32 K33 K34

K42 K43 K44

ka22+ke11ka22+ke11 ke12

0

ke21 ke22+kc11 kc12

0 kc21 kc22

δWint = δu · K u



• X

u21

u31

u41

1 2 3 4

u1

ua11

ua21

1 2a

u1

ue11

ue21

1 2e

u1uc11

uc21

1 2c


δWint = + +

ke11 ke12

ke21 ke22

δue11

δue21

ue11 ue21

δW eint

ka11 ka12

ka21 ka22

δua11

δua21

ua11 ua2

1

δWaint

kc11 kc12

kc21 kc22

δuc11

δuc21

uc11 uc21

δW cint

δu21

δu31

u21 u3

1

0

δu21

0 u21

δu31

δu41

u31 u4

1

0

δu21

δu31

δu41

0 u21 u3

1 u41

K11 K12 K13 K14

K21 K22 K23 K24

K31 K32 K33 K34

K41 K42 K43 K44

δWint =

K22 K23 K24

K32 K33 K34

K42 K43 K44


ke21 ke22+kc11 kc12

0 kc21 kc22

δWint = δu · K u



• X

u21

u31

u41

1 2 3 4

u1

ua11

ua21

1 2a

u1

ue11

ue21

1 2e

u1uc11

uc21

1 2c


δWint = + +

ke11 ke12

ke21 ke22

δue11

δue21

ue11 ue21

δW eint

ka11 ka12

ka21 ka22

δua11

δua21

ua11 ua2

1

δWaint

kc11 kc12

kc21 kc22

δuc11

δuc21

uc11 uc21

δW cint

δu21

δu31

u21 u3

1

0

δu21

0 u21

δu31

δu41

u31 u4

1

0

δu21

δu31

δu41

0 u21 u3

1 u41

K11 K12 K13 K14

K21 K22 K23 K24

K31 K32 K33 K34

K41 K42 K43 K44

δWint =

K22 K23 K24

K32 K33 K34

K42 K43 K44


ke21

ke22+kc11 kc12

0 kc21 kc22

δWint = δu · K u



• X

u21

u31

u41

1 2 3 4

u1

ua11

ua21

1 2a

u1

ue11

ue21

1 2e

u1uc11

uc21

1 2c


δWint = + +

ke11 ke12

ke21 ke22

δue11

δue21

ue11 ue21

δW eint

ka11 ka12

ka21 ka22

δua11

δua21

ua11 ua2

1

δWaint

kc11 kc12

kc21 kc22

δuc11

δuc21

uc11 uc21

δW cint

δu21

δu31

u21 u3

1

0

δu21

0 u21

δu31

δu41

u31 u4

1

0

δu21

δu31

δu41

0 u21 u3

1 u41

K11 K12 K13 K14

K21 K22 K23 K24

K31 K32 K33 K34

K41 K42 K43 K44

δWint =

K22 K23 K24

K32 K33 K34

K42 K43 K44


ke21 ke22+kc11

kc12

0 kc21 kc22

δWint = δu · K u



• X

u21

u31

u41

1 2 3 4

u1

ua11

ua21

1 2a

u1

ue11

ue21

1 2e

u1uc11

uc21

1 2c


δWint = + +

ke11 ke12

ke21 ke22

δue11

δue21

ue11 ue21

δW eint

ka11 ka12

ka21 ka22

δua11

δua21

ua11 ua2

1

δWaint

kc11 kc12

kc21 kc22

δuc11

δuc21

uc11 uc21

δW cint

δu21

δu31

u21 u3

1

0

δu21

0 u21

δu31

δu41

u31 u4

1

0

δu21

δu31

δu41

0 u21 u3

1 u41

K11 K12 K13 K14

K21 K22 K23 K24

K31 K32 K33 K34

K41 K42 K43 K44

δWint =

K22 K23 K24

K32 K33 K34

K42 K43 K44


ke21 ke22+kc11 kc12

0 kc21 kc22

δWint = δu · K u



• X

1

23

1 2 3 41 21 1 22 1 23

counting of structural degrees of freedom→ equation number

element numbers and element node numbers

?

1 2 3

1

2

3

4

1

2

3

0

--

?

spatial direction

stru

ctur

alno

de

id

equation number?

1 2 3

1

2

1

2

2

3

3

4

-

?

element number

elem

entn

ode

it

structural node number

storage of assembly information

1

1

2

3

4

1

2

3

0

id

equation number (position in K, M, r)

input: element no. e

and node no. i

out: sturc. node

in: sturc. nodei

e

1 2 3

1

2

1

2

2

3

3

4it

read-out assembly information

• structural nodes and displacements (id)

vertical: structural node numbers

horizontal: spatial direction (here just 1)

array: equation numbers dof’s

• element and structural nodes (it)

vertical: element node numbers

horizontal: element numbers

array: structural node number


computer-oriented assembly procedureÜ 76

• X

NE = 4, p = 1, NEQ = 4

rb rb rb rb rb1 2 3 4

1 2 3 4 51 2 3 4

equation number

structural node numberelement number

1

1

2

3

4

5

1

2

3

4

0

-

?

spatial direction

stru

ctur

alno

de

id

1 2 3 4

1

2

1

2

2

3

3

4

4

5

-

?

element number

elem

entn

ode

it

NE = 8, p = 1, NEQ = 8

rb rb rb rb rb rb rb rb rb1 2 3 4 5 6 7 8

1 2 3 4 5 6 7 8 91 2 3 4 5 6 7 8

1

1

2

3

4

5

6

7

8

9

1

2

3

4

5

6

7

8

0

-

?

spatial direction

stru

ctur

alno

deid

1 2 3 4 5 6 7 8

1

2

1

2

2

3

3

4

4

5

5

6

6

7

7

8

8

9

-

?

element number

elem

entn

ode

it

NE = 4, p = 2, NEQ = 8

rb rb rb rb rb rb rb rb rb1 2 3 4 5 6 7 8

1 2 3 4 5 6 7 8 91 2 3 4

1

1

2

3

4

5

6

7

8

9

1

2

3

4

5

6

7

8

0

-

?

spatial direction

stru

ctur

alno

de

id

1 2 3 4

1

2

3

1

2

3

3

4

5

5

6

7

7

8

9

-

?

element number

elem

entn

ode

it


computer-oriented assembly procedureÜ 77

• X

• assemly is element wise addition of stiffness ’tensors’ into the structural stiffness matrix(at the correct position)

• assembly of structural mass matrix and structural load vector is analogous to the elementstiffness matrix

• approximated weak form of structure / system

δu ·M u+ δu · K u = δu · r

• reformualtion

δu · [M u+ K u− r] = 0

• fundamental lemma of variational calculus (arbitrary virtual displacements)

M u+ K u = r

ordinary second order (time) differential equation eigen value analysis and time integration

• time independent case - structural stiffness relation

K u = r

linear system of equations for solution of structural displacement vector u


structural equation of motionÜ 78

• X

e1

e3e3e3e3e3e3e3e3e3

dead

load

p1

dead

load

p1

dead

load

p1

dead

load

p1

dead

load

p1

dead

load

p1

dead

load

p1

dead

load

p1

dead

load

p1

F1F1F1F1F1F1F1F1F1

cros

sse

ctio

nA

cros

sse

ctio

nA

cros

sse

ctio

nA

cros

sse

ctio

nA

cros

sse

ctio

nA

cros

sse

ctio

nA

cros

sse

ctio

nA

cros

sse

ctio

nA

cros

sse

ctio

nA

motivation - normal loading tower


e1

e3

E, ρ,A

L


fe discretization

rbrbu11 =0 u2

1

ua11 ua2

1a

system / structural level

element level

one element

• element stiffness matrix with element length La = L

ka =E

La

[1 −1

−1 1

]• element load vector for constant b1 = b

ra =ρ b La

2

[1

1

]• structural stiffness relation Ku = r

=

EL

−EL

−EL

EL

u11

u21

ρbL2

ρbL2

• homogeneous DIRICHLET boundary condition(deleting first line and first column)

=EL

u21

ρbL2

• linear system of equations for u21

solution identical to first finite element application


tension bar - finite element solution iiÜ 79

• X

e1

e3e3e3e3e3e3e3e3e3

dead

load

p1

dead

load

p1

dead

load

p1

dead

load

p1

dead

load

p1

dead

load

p1

dead

load

p1

dead

load

p1

dead

load

p1

F1F1F1F1F1F1F1F1F1

cros

sse

ctio

nA

cros

sse

ctio

nA

cros

sse

ctio

nA

cros

sse

ctio

nA

cros

sse

ctio

nA

cros

sse

ctio

nA

cros

sse

ctio

nA

cros

sse

ctio

nA

cros

sse

ctio

nA



e1

e3

E, ρ,A

L


fe discretization

rbrbu11 =0 u2

1

ua11 ua2

1a


element level

one element


ka =E

La

[1 −1

−1 1


ra =ρ b La

2

[1

1

]

• structural stiffness relation Ku = r

=

EL

−EL

−EL

EL

u11

u21

ρbL2

ρbL2


=EL

u21

ρbL2





• X

e1

e3e3e3e3e3e3e3e3e3

dead

load

p1

dead

load

p1

dead

load

p1

dead

load

p1

dead

load

p1

dead

load

p1

dead

load

p1

dead

load

p1

dead

load

p1

F1F1F1F1F1F1F1F1F1

cros

sse

ctio

nA

cros

sse

ctio

nA

cros

sse

ctio

nA

cros

sse

ctio

nA

cros

sse

ctio

nA

cros

sse

ctio

nA

cros

sse

ctio

nA

cros

sse

ctio

nA

cros

sse

ctio

nA



e1

e3

E, ρ,A

L


fe discretization

rbrbu11 =0 u2

1

ua11 ua2

1a


element level

one element


ka =E

La

[1 −1

−1 1


ra =ρ b La

2

[1

1


=

EL

−EL

−EL

EL

u11

u21

ρbL2

ρbL2


=EL

u21

ρbL2





• X


e1

e3

E, ρ,A

L


fe discretization

rbrb rbrb0 u21 u3

1

ua11 ua2

1 ub11 ub21a b

two elements

• element stiffness matrices, element lengths Le = L/2

ka =2E

L

[1 −1

−1 1

]kb =

2E

L

[1 −1

−1 1


ra =ρ b L

4

[1

1

]rb =

ρ b L

4

[1

1

]

• structural stiffness relation Ku = r

=

2EL

−2EL

−2EL

2EL

+ 2EL

−2EL

−2EL

2EL

u11

u21

u31

ρbL4

ρbL4 +

ρbL4

ρbL4

• homogeneous DIRICHLET boundary condition

=

4EL−2EL

−2EL

2EL

u21

u31

ρbL2

ρbL4

u31 =

1

2

ρbL2

E

u21 =

3

8

ρbL2

E



• X


e1

e3

E, ρ,A

L


fe discretization

rbrb rbrb0 u21 u3

1

ua11 ua2

1 ub11 ub21a b

two elements

• element stiffness matrices, element lengths Le = L/2

ka =2E

L

[1 −1

−1 1

]kb =

2E

L

[1 −1

−1 1


ra =ρ b L

4

[1

1

]rb =

ρ b L

4

[1

1


=

2EL

−2EL

−2EL

2EL

+ 2EL

−2EL

−2EL

2EL

u11

u21

u31

ρbL4

ρbL4 +

ρbL4

ρbL4

• homogeneous DIRICHLET boundary condition

=

4EL−2EL

−2EL

2EL

u21

u31

ρbL2

ρbL4

u31 =

1

2

ρbL2

E

u21 =

3

8

ρbL2

E



• X

u11 = 0 u5

1 = u?1

finite element model of coaxial tensile bar

DIRICHLET displacementfixed support

structural node1 2 3 4 5

equation number0 1 2 3 −1

− counting prescribeddisplacements

element 12 3

4

5

1

1

2

3

4

5

1

2

3

−1

0

-

?

direction

str.

node id

1 2 3 4 5

1

2

1

2

2

3

3

4

2

4

4

5

-

?

element no.

ele.

node

it

• structural nodes and displacements (id)

vertical: structural node numbers horizontal: spatial direction (here just 1) array: equation numbers dof’s

− positive: unknown displacements− negative: prescribed displacements

• element and structural nodes (it)

vertical: element node numbers horizontal: element numbers array: structural node number

d.kuhl, wes.online, university of kasselÜ Ülinear computational structural mechanics • 2 1d finite element method • 2.11 examples

computer-oriented assemblyÜ 129

• X


e1

e3

E, ρ,A

L


fe discretization

constant line force tension bar loaded by ρb1

e1

e3

E, ρ,A

L

b1 = 2bLX1

ρb1

fe discretization

linear line force tension bar loaded by ρb1

e1

e3

E, ρ,A

L

b1 =[1− cos

[2πL

X1

]]b

ρb1

fe discretization

cosinus line forcedata

L = 1.0 m

A = 1.0× 10−4 m2

ρ = 1.0× 104 kg/m3

b = 1.0× 101 m/s2

E = 2.0× 1011 N/m2

load case i load case ii

homework: In the present homework a finite element program for the static analysis of one dimen-sional continua should be developed using you favorite programming language. Therefore, linear(p = 1), one dimensional continuum elements should be applied. The correct implementation ofthe finite element and finite element procedure on the structural level should be verified by meansof above sketched model problems. These examples are described by a truss loaded by a con-stant (load case i) and a linear (load case ii) line load. They should be analyzed using usingNE = 1, 2, 4, 8, 16, 32 finite elements for the discretization of the truss. For these reasons the follow-ing working stages are proposed:

• develop a finite element routine for calculation of the element stiffness ’tensors’ keij and theconsistent load ’tensors’ rei for i, j ∈ [1, 2] for load case i

• develop a finite element procedure for the discretization with only one finite element NE = 1,implement an assembly procedure and the solution procedure for the linear system of equationswith NEQ = 1. Check the correctness of the solution by means of load case i

• extend your finite element program for analyses with NE = 2, 4, 8, 16, 32 finite elements, yieldingNEQ = 1, 2, 4, 8, 16, 32 unknowns, and check your solutions for load case i

• extend your finite element program for analyses of load case ii with NE = 1, 2, 4, 8, 16, 32 finiteelements

• extend your finite element program by a post-processing procedure, calculating theapproximations of the displacement u1, stress σ11 and residuum σ11,1 + ρb1

• calculate the local (at position X1) and global (of the hole system) displacement errors withrespect to the analytical solution• plot diagrams of the displacements, stresses, the residuum and the local displacement error• plot a double logarithmic diagram of the global error as function of the element length and,

additionally, as function of the numbers of degrees of freedom of the system NEQ

The homework should be uploaded until December 19, 2013 on the moodle course. Your submissionshould include

• a brief report documenting your results in form of diagrams• and your program code

d.kuhl, wes.online, university of kasselÜ Ülinear computational structural mechanics • 2 1d finite element method • 2.12 homework

truss element loaded by line forcesÜ 130

• X

In the present homework a finite element program for the static analysis of one dimensional continuashould be developed using you favorite programming language. Therefore, linear (p = 1), onedimensional continuum elements should be applied. The correct implementation of the finite elementand finite element procedure on the structural level should be verified by means of above sketchedmodel problems. These examples are described by a truss loaded by a constant (load case i) anda linear (load case ii) line load. They should be analyzed using using NE = 1, 2, 4, 8, 16, 32 finiteelements for the discretization of the truss. For these reasons the following working stages areproposed:

• develop a finite element routine for calculation of the element stiffness ’tensors’ keij and theconsistent load ’tensors’ rei for i, j ∈ [1, 2] for load case i

• develop a finite element procedure for the discretization with only one finite element NE = 1,implement an assembly procedure and the solution procedure for the linear system of equationswith NEQ = 1. Check the correctness of the solution by means of load case i

• extend your finite element program for analyses with NE = 2, 4, 8, 16, 32 finite elements, yieldingNEQ = 1, 2, 4, 8, 16, 32 unknowns, and check your solutions for load case i

• extend your finite element program for analyses of load case ii with NE = 1, 2, 4, 8, 16, 32 finiteelements

• extend your finite element program by a post-processing procedure, calculating theapproximations of the displacement u1, stress σ11 and residuum σ11,1 + ρb1

• calculate the local (at position X1) and global (of the hole system) displacement errors withrespect to the analytical solution

• plot diagrams of the displacements, stresses, the residuum and the local displacement error• plot a double logarithmic diagram of the global error as function of the element length and,

additionally, as function of the numbers of degrees of freedom of the system NEQ

The homework should be uploaded until December 19, 2013 on the moodle course. Your submissionshould include

• a brief report documenting your results in form of diagrams• and your program code

d.kuhl, wes.online, university of kasselÜ Ülinear computational structural mechanics • 2 1d finite element method • 2.12 homework

truss element loaded by line forcesÜ 131

· x ﬁnite element method for 1d continua motivation - spatial truss structures real structures...

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