Modelli evolutivi per la verifica del rischio di edifici ...Quaderno 4 Primi concetti di analisi...

Odine degli Ingegneri della Provincia di PistoiaCorso sulla Vulnerabilità Sismica

Modelli evolutivi per la verifica del rischio di edifici esistenti

Quaderno 4Primi concetti di analisi nonlineare

Prof. Enrico SpaconeDipartimento di Ingegneria e Geologia

Università degli Studi “G. D’Annunzio” Chieti-Pescara

31 Maggio 2013

INTRODUCTION

� Three examples are presented hereafter to introduce nonlinear problems and nonlinear solution schemes

2

nonlinear solution schemes

ELASTIC BEAMLe = 3 m

P1, U1

P2,U2

=P KU

Example 1: Material Nonlinearity

3

ELASTO-PLASTICHINGE

3m

P3, U3

1

2

3

1

2

3

P

P

P

U

U

U

=

=

P

U

e 1=U

U1

U2

12

U

e 1

1

2

0

=

=

ID

Example 1

4

U3

3

4

1

2

e 2=U

U

ELEMENTS STRUCTURE

0

3

e 20

3

= =

ID

e 1=U

U1

U2

U3

12

U1

U2

U1 U2 U3

U3 2 3 2

b b b b

2 2

b b b be 1

b

3 2 3 2

12 6 12 6

L L L L

6 4 6 2

L L L LEI

12 6 12 6

L L L L

=

−

− = − − −

K

Example 1

5

3

4

1

2

U3

U3

U3

e 2=Ue 2

h

1 1k

1 1

=−

= − K

U3

U2

U1

U1 U2 U3

3 2 3 2

b b b b

2 2

b b b b

L L L L

6 2 6 4

L L L L

− − − −

3 2 2

b b b

b 2

b b b

h2

b b b

12 6 6

L L L

6 4 2EI

L L L

6 2 4k

L L L

=

+

K

ϕ

M

linear

Lb Hp:

Example 1

6

Lpl

θ = ϕ Lpl

θ

M

non-

linear

Lb Hp: curvature ϕ

constant over Lpl

Alternative 1

ELASTO-PLASTIC

2pl

dL d≤ ≤

Order of magnitude ofplastic hinge length

ϕ

M

linear

Lb

ϕ−ϕy

M

ϕy

My

Example 1

7

0-length

(θ −θy)= (ϕ−ϕ y)Lpl

θ−θy

M

non-

linear

LbConsider plastic

deformations only

Alternative 2

PLASTIC

ϕ

M

linear

Lbϕ

My

Example 1

8

0-length

ϕ

non-

linear

Lb

Procedure followed here

PLASTIC

ϕy

Strictly speaking this is not correctThe flexibility of the plastic hingelength is accounted for twice, both in the column element and in the hinge

This approach is followed to illustrate the nonlinear procedure

Example 1

9

EIel-b = 1013 N-mm2

EIel-h = 1013 N-mm2

EIpl-h = 5,2x1010 N-mm2

kel-h = EIel-cp/Lpl = 5 x1010 N-mm

kpl-h = EIpl-cp/Lpl = 2,6x108 N-mm

Lpl = 200 mm

30

40

50

60M

(k

N-m

)

(0.0009, 45)

(0.02, 50)

Example 1

10

EIel-b = 104 kN-m2

kel-h = 5 x104 kN-m

kpl-h = 2,6x102 kN-m

0

10

20

0 0,002 0,004 0,006 0,008 0,01 0,012 0,014 0,016 0,018 0,02

θθθθ (rad)

M (

kN

-m)

P=1 kN P1 (kN)

λ2P = 15

λ3P = 17

Example 1

11

Load History

P1 = λPλ= {7.5, 15, 17}

λ1P = 7.5

U1

U1

U2

u1

u2

Example 1

1

2

Unodal

Udispl.s

= U

12

U3

u3

u1

u2

u4

2

3

1

2

3

Udispl.s

U

Pnodal

Pforces

P

=

=

U

P

h 2 1 2

plastic

u u uhinge

rotation

θ = − =

U1

U2

tr h R

00 0

0 0= 0 0

0 00 0

0

⇒ = = ⇒ =

U P P P

u1

u2i=1LOAD STEP 1: λ1P = 7.5 kN

Example 1

13

U3unb R

0 00

7.5

0

0

7.5

0

0

=

=

∆ = − =

P

P P P P

u3

u1

u2

u4

7.5

R+ =P P 0

SIGN CONVENTION: When equilibrium is reached:

Resisting forces at node7.5

eq

ua

l a

nd

op

po

site

(fr

om

eq

uili

bri

um

)

P

Example 1

14

Resisting forcesat element end

7.5

eq

ua

l a

nd

op

po

site

(fr

om

eq

uili

bri

um

)

R- =P P 0

Convention used here:formally less correcteasier to represent

U1

U2

Initial stiffness

i=1

LOAD STEP 1: λ1P = 7.5 kN

0

10

20

30

40

50

60

0 0,002 0,004 0,006 0,008 0,01 0,012 0,014 0,016 0,018 0,02

M (

kN

-m)

Example 1

15

U3 b b b3 2 2

b b b

b b bel 2

b b b

b b bel h2

b b b

12EI 6EI 6EI

L L L

6EI 4EI 2EI

L L L

6EI 2EI 4EIk

L L L−

= =

+

K K

EIb = 104 kN-m2

kh = kel-h = EIel-h/Lpl = 5 x104 kN-m

Lb = 3 m

0 0,002 0,004 0,006 0,008 0,01 0,012 0,014 0,016 0,018 0,02

θθθθ (rad)

P

Kel

Graphic is purely indicativeAxes represent arrays!

Example 1

16

U

Punb

λ1P = 7.5

Axes represent arrays!

U1

U2

{ }-10.0081

0.0038

0.00045

∆ ∆ = − −

U = K P

u1


Example 1

17

U3h

b

h

0.00045

0.0081

0.0038

0.00045

00.0081

0.000450.0038

00.00045

0.00045

−

∆ = − −

=

− − = ⇓ θ = − −

U = U + U

U

U

u3

u1

u2

u4

P

Kel

Example 1

18

7.5

U

U

Determine resisting forces corresponding to

the current displacement vector U(STRUCTURE STATE DETERMINATION)

u1

u2

u3

Elements’ resisting forces(ELEMENT STATE DETERMINATION)

1) Column: linear elastic

2) Plastic hinge Mh = min(kel-h θh , M* + kpl-h θh) = -22.5 kN-m

b b b=P K U

i=1LOAD STEP 1: λ1P = 7.5 kN

u4

Example 1

19

u1

u2

2) Plastic hinge Mh = min(kel-h θh , M + kpl-h θh) = -22.5 kN-m

kel-h

22.5

kh = kel-h

0.00045

0

10

20

30

40

50

60

0 0,002 0,004 0,006 0,008 0,01 0,012 0,014 0,016 0,018 0,02

θθθθ (rad)

M (

kN

-m)

F

F*

F=kel u

kpl

kel

F=F*+kpl u

F = min(kel u , F* + kpl u)

Example 1

20

u

F = max(kel u , F* + kpl u)

7.5

b h

7.5

0 22.5

7.5 22.5

= =

− − P P

7.5

0

U1

U2


Example 1

21

b h

R

unb R

7.5 22.5

22.5

7.5

0

0

7.5 7.5 0

0 0 0

0 0 0

− −

=

∆ = − = − = =

P

P = P P P 0

7.5

22.5

22.5

22.5

U3

7.5

7.5

7.5

7.5

7.5

In equililbrium

External Forces on nodes

Resisting Forces on nodes

Element Forces

Example 1

22

7.522.5

7.5

22.5

22.57.5

7.5

22.5

22.57.5

22.57.5

7.5

22.5In equililbrium

In equililbrium

P

Kel

There is equilibrium between applied and

resisting forces

Example 1

23

7.5

U

∆U

Punb = 0

resisting forces

→ Load increment

Apply λ2P

P

P

Kel

λ2P = 15

Example 1

24

7.5

U

Punb

15

15

7.5

7.5

7.5

Not in equililbrium!!

External Forces on nodes

Resisting Forces on nodes

Element Forces

Example 1

25

7.522.5

7.5

22.5

22.57.5

7.5

22.5

22.57.5

22.57.5

7.5

22.5In equililbrium

In equililbrium

U1

U2

R

7.5 15

0 0

0 0

= =

P P

u1

u2i=1LOAD STEP 2: λ2P = 15 kN

Example 1

26

U3{ }

unb R

el

-1

7.5

0

0

0.0081

0.0038

0.00045

∆ = − =

=

∆ ∆ = − −

P = P P P

K K

U = K P

u3

u1

u2

u4

P

15

Kel

Example 1

27

7.5

U

Punb

∆U


the current displacement vector U(STRUCTURE STATE DETERMINATION)

U1

U2

0.0162

0.00765

0.0009

∆ = − −

U = U + U

u1


Example 1

28

U3

h

b

h

0.0009

00.0162

0.00090.00765

00.0009

0.0009

−

=

− − = ⇓ θ = − −

U

U

u3

u1

u2

u4

u1

u2

u3

Elements’ resisting forces

(ELEMENT STATE DETERMINATION)


2) Plastic hinge

i=1

Mh = min(kel-h θh , M* + kpl-h θh) = -45 kN-m

b b b=P K U

LOAD STEP 2: λ2P = 15 kN

u4

Example 1

29

u1

u2

2) Plastic hinge

kel-h

45

kh = kel-h


0.0009

0

10

20

30

40

50

60

0 0,002 0,004 0,006 0,008 0,01 0,012 0,014 0,016 0,018 0,02

θθθθ (rad)

M (

kN

-m)

15

b h

15

0 45

15 45

45

15

= =

− −

P P

15

0

U1

U2

i=1LOAD STEP 2: λ2P = 15 kN

Example 1

30

R

unb R

15

0

0

15 15 0

0 0 0

0 0 0

=

∆ = = − = − =

P

P P P P

15

45

45

45

U3

P

Kel

15 Punb=0

Example 1

There is equilibrium between applied and

resisting forces

31

7.5

U

resisting forces

→ Load increment

Apply λ3P

U1

U2

R

15 17

0 0

0 0

= =

P P

u1


Example 1

32

U3

{ }

unb R

el

-1

2

0

0

0.00216

0.001

0.00012

∆ = − =

=

∆ ∆ = − −

P = P P P

K K

U = K P

u3

u1

u2

u4

P

Ke

15Punb

λ3P = 17

Example 1

33

7.5

U

∆U


the current displacement vector U(STATE DETERMINATION)

U1

U2

0.01836

0.00867

∆ = −

U = U + U

u1


Example 1

34

U3

h

b

h

0.00102

00.01836

0.001020.00867

00.00102

0.00102

−

=

− − = ⇓ θ = − −

U

U

u3

u1

u2

u4

u1

u2

u3




2) Plastic hinge

i=1

b b b=P K U


Mh = min(kel-h θh , M* + kpl-h θh) = -45,03 kN-m

u4

Example 1

350

10

20

30

40

50

60

0 0.005 0.01 0.015 0.02 0.025

rotazione θcp

Mo

me

nto

(kN

-m)

u1

u2

2) Plastic hinge

45,03

kpl-hkh = kpl-h

Mh = min(kel-h θh , M* + kpl-h θh) = -45,03 kN-m

0.00102

17

b h

17

0 45.0312

17 45.0312

51

= =

− −

P P

17

0

U1

U2


Example 1

36

R

unb R

17

0

5.9688

17 17 0

0 0 0

0 5.9688 5.9688

=

∆ = − = − = −

P

P = P P P

17

45.03

45.03

51

U3

There is no equilibrium between applied and

resisting forces

U1

U2

pl

0.06887

=

K K

u1


Example 1

37

U3

{ }-10.06887

0.0229

0.0229

0.08723

0.0316

0.0240

∆ ∆ = − −

= + ∆ = − −

U = K P

U U U

u3

u1

u2

u4

P

Kep15 Punb

17

Example 1

38

7.5

U

∆U

u1

u2

u3




2) Plastic hinge

i=2

b b b=P K U



u4

Example 1

0

10

20

30

40

50

60

0 0.005 0.01 0.015 0.02 0.025

rotazione θcp

Mo

me

nto

(kN

-m)

39

u1

u2

2) Plastic hinge

51 Kh = kpl-h


0.024

17

b h

17

0 51

17 51

= =

− − P P

17

0

U1

U2


Example 1

40

R

unb R

17 51

51

17

0

0

17 17 0

0 0 0

0 0 0

− −

=

∆ = − = − =

P

P = P P P

-17

51

-51

51

U3

P

15

Punb=017

Example 1

41

7.5

U

P

(16.2,15)

(87.23,17)

Example 1

42

(8.1,7.5)

U

P

(16.2,15)

(87.23,17)

Example 1

43

U

(8.1,7.5)

P=1 kNP1 (kN)

λ P = 16

The load path is not known: A load history is the applied

Example 1

44

Load history

P1 = λPλ= {2, 4, 6, 8, 10 ,12, 14, 16, …}

λ1P = 2

U1

λ8P = 16

λ2P = 4

λ3P = 6

λ4P = 8

λ5P = 10

λ6P = 12

λ7P = 14

P=1 kN

12

14

16

18

20

22

(k

N)

Example 1

45

Load history

P1 = λPλ= {0, 20}

0

2

4

6

8

10

0 50 100 150 200

U1 (mm)

P1 (

kN

)

P=1 kN

12

14

16

18

20

22

(k

N)

Example 1

46

Load history

P1 = λPλ= {0, 10, 20}

0

2

4

6

8

10

0 50 100 150 200

U1 (mm)

P1 (

kN

)

P=1 kN

12

14

16

18

20

22

(k

N)

Example 1

47

Load history

P1 = λPλ= {0, 5, 10, 15, 20}

0

2

4

6

8

10

0 50 100 150 200

U1 (mm)

P1 (

kN

)

P=1 kN

12

14

16

18

20

22

(k

N)

Example 1

48

Load history

P1 = λPλ= {0, 2, 4, 6, 8, 10, 12, 14, 16, 18, 20}

0

2

4

6

8

10

0 50 100 150 200

U1 (mm)

P1 (

kN

)

PLASTIC HINGE

Phenomenological

nonlinear M-θ modelP, U

Example 2

49

rotation = u2-u1

u1

u2

P, U

SYSTEM RESPONSE(closed form solution)

12

14

16

18

Example 2

50

0

2

4

6

8

10

12

0 10 20 30 40 50 60 70 80 90

U1 (mm)

P1 (

kN

)

U1

U2

b h R

00 0

0 0= 0 0

0 00 0

0

⇒ = = ⇒ =

U P P P

u1


Example 2

51

U3unb R

0 00

7.5

0

0

7.5

0

0

=

∆ = = − =

P

P P P P

u3

u1

u2

u4

10

20

30

40

50

60

M (

kN

-m)

U1

U2

Initial stiffness

i=1


Example 2

0

0 0,005 0,01 0,015 0,02 0,025

θcp

52

U3 b b b3 2 2

b b b

el b b0 el 2

b b b

b b bel h2

b b b

12EI 6EI 6EI

L L L

6EI 4EI 2EI

L L L

6EI 2EI 4EIk

L L L−

= =

+

K K

EIb = 104 kN-m2

kh = kel-h = EIel-h/Lpl = 5 x104 kN-m

Lb = 3 m

U1

U2

{ }-10

0.0081

0.0038

0.00045

∆ ∆ = − −

U = K P

u1

u2

i=1


Example 2

53

U3h

b

h

0.00045

0.0081

0.0038

0.00045

00.0081

0.000450.0038

00.00045

0.00045

−

∆ = − −

=

− − = ⇓ θ = − −

U = U + U

U

U

u3

u1

u2

u4

u1

u2Elements’ resisting forces

Column: linear elastic

Plastic hinge Mh = -15.07 kN-m

b b b=P K U


Example 2

54

u3

u1

u2

0

10

20

30

40

50

60

0 0.005 0.01 0.015 0.02 0.025

rotazione θcp

Mo

men

to (

kN

-m)

0.00045

15.07

u4

7.5b h

7.5

0 15.066586

7.5 15.066586

22.5

7.5

= =

− −

P P

7.5

0

U1

U2

i=1PLASTIC HINGE 1: λ1P = 7.5 kN

Example 2

55

R

unb R

7.5

0

7.433414

7.5 7.5 0

0 0 0

0 7.433414 7.433414

=

∆ = = − = − = −

P

P P P P

7.5

15.07

15.07

22.5

U3

There is no equilibrium between applied and resisting forces

Apply Punb

U1

U2

{ }-10

0.00044

0.00015

0.00015

∆ ∆ = − −

U = K P

u1

u2

i=2


Example 2

56

U3h

b

h

0.00015

0.00854

0.00397

0.000599

00.00854

0.0005990.00397

00.000599

0.000599

−

∆ = − −

=

− − = ⇓ θ = − −

U = U + U

U

U

u3

u1

u2

u4

u1

u2




Plastic hinge Mh = -19.43 kN-m

b b b=P K U

Example 2

57

u3

u1

u2

0

10

20

30

40

50

60

0 0.005 0.01 0.015 0.02 0.025

rotazione θcp

Mo

men

to (

kN

-m)

0.000599

19.43

u4

7.5b h

7.5

0 19.430188

7.5 19.430188

22.5

7.5

= =

− −

P P

7.5

0

U1

U2


Example 2

58

R

unb R

7.5

0

3.069812

7.5 7.5 0

0 0 0

0 3.069812 3.069812

=

∆ = = − = − = −

P

P P P P

-7.5

19.43

-19.43

22.5

U3

=2 =1

unb unbNote thati i

U1

U2

{ }-10

0.00018

0.00006

0.00006

∆ ∆ = − −

U = K P

u1


Example 2

59

U3h

b

h

0.00006

0.00873

0.00403

0.00066

00.00873

0.000660.00403

00.00066

0.00066

−

∆ = − −

=

− − = ⇓ θ = − −

U = U + U

U

U

u3

u1

u2

u4

u1

u2

Mh = -21.06 kN-m

b b b=P K U


Example 2



Plastic hinge

60

u3

u1

u2

0

10

20

30

40

50

60

0 0.005 0.01 0.015 0.02 0.025

rotazione θcp

Mo

men

to (

kN

-m)

0.00066

21.06

u4

7.5b h

7.5

0 21.060194

7.5 21.060194

22.5

7.5

= =

− −

P P

7.5

0

U1

U2


Example 2

61

R

unb R

7.5

0

1.439806

7.5 7.5 0

0 0 0

0 1.439806 1.439806

=

∆ = = − = − = −

P

P P P P

7.5

21.06

21.06

22.5

U3

=3 =2

unb unbNote thati i

U1

U2

{ }-10

0.00000000007

0.00000000002

0.00000000002

∆ ∆ = − −

U = K P

u1

u2

i=15


Example 2

62

U3h

b

h

0.00000000002

0.0089

0.00409

0.00072

00.0089

0.000720.00409

00.00072

0.00072

−

∆ = − −

=

− − = ⇓ θ = − −

U = U + U

U

U

u3

u1

u2

u4

u1

u2

Mh = -22.5 kN-m

h h h=P K U


Example 2



Plastic hinge

63

u3

u1

u2

0

10

20

30

40

50

60

0 0.005 0.01 0.015 0.02 0.025

rotazione θcp

Mo

men

to (

kN

-m)

0.00072

22.5

u4

7.5

7.5

0

U1

U2


b h

7.5

0 22.499999

7.5 22.499999

22.5

7.5

= =

− −

P P

Example 2

64

7.5

22.5

22.5

22.5

U3

Small enough!!

R

unb R

7.5

0

.00000113

7.5 7.5 0

0 0 0

0 .00000113 .00000113

=

= − = − = −

P

P P P

=15

unbNote thati ≈P 0

There is equilibrium between

applied and resisting forces Apply λ2P

Convergence was very slow because initial stiffness was used

Example 2

5,00

6,00

7,00

8,00

-m)

K-initial

65

0,00

1,00

2,00

3,00

4,00

5,00

1 3 5 7 9 11 13 15

Pu

nb

(3)

(kN

-

iteration (i)

The tangent stiffness does not change:

Example 2

66

The tangent stiffness does not change:Advantage: K is inverted only onceDisadvantage: convergence is slow

What if the stiffness is updated at every step (tangent stiffness)?

Tangent stiffness (Newton-Raphson)

Example 2

67

It should be much faster!

U1

U2

b h R

00 0

0 0= 0 0

0 00 0

0

⇒ = = ⇒ =

U P P P

u1


Example 2

68

U3unb R

0 00

7.5

0

0

7.5

0

0

=

∆ = = − =

P

P P P P

u3

u1

u2

u4

U1

U2

{ }-1tan tan 0

0.0081

0.0038 for 1,

0.00045

i

∆ ∆ = − = = −

U = K P K K

u1

u2

i=1


Example 2

69

U3h

b

h

0.00045

0.0081

0.0038

0.00045

00.0081

0.000450.0038

00.00045

0.00045

−

∆ = − −

=

− − = ⇓ θ = − −

U = U + U

U

U

u3

u1

u2

u4

u1

u2

Mh = -15.07 kN-m

b b b=P K U


K = 3.14 104 kN-m

Example 2



Plastic hinge

70

u3

u1

u2

0

10

20

30

40

50

60

0 0.005 0.01 0.015 0.02 0.025

rotazione θcp

Mo

men

to (

kN

-m)

0.00045

15.07

Kh,tan= 3.14 104 kN-m

u4

7.5b h

7.5

0 15.066586

7.5 15.066586

22.5

7.5

= =

− −

P P

7.5

0

U1

U2


Example 2

71

R

unb R

7.5

0

7.433414

7.5 7.5 0

0 0 0

0 7.433414 7.433414

=

∆ = = − = − = −

P

P P P P

7.5

15.07

15.07

22.5

U3


Apply Punb

U1

U2

{ }-1tan

0.00071

0.00024

0.00024

∆ ∆ = − −

U = K P

u1

u2

i=2


Example 2

72

U3h

b

h

0.00024

0.00881

0.00406

0.000687

00.00881

0.0006870.00406

00.000687

0.000687

−

∆ = − −

=

− − = ⇓ θ = − −

U = U + U

U

U

u3

u1

u2

u4

u1

u2

Mh = -21.74 kN-m

b b b=P K U


K = 2.5 104 kN-m

Example 2



Plastic hinge

73

u3

u1

u2

0

10

20

30

40

50

60

0 0.005 0.01 0.015 0.02 0.025

rotazione θcp

Mo

men

to (

kN

-m)

0.000687

21.74

Kh,tan= 2.5 104 kN-m

u4

7.5b h

7.5

0 21.743466

7.5 21.743466

22.5

7.5

= =

− −

P P

7.5

0

U1

U2


Example 2

74

R

unb R

7.5

0

0.756533

7.5 7.5 0

0 0 0

0 0.756533 0.756533

=

∆ = = − = − = −

P

P P P P

7.5

21.74

21.74

22.5

U3

=2 =1

unb unbNote thati iP P�


Apply Punb

U1

U2

{ }-1tan

0.00009

0.00003

0.00003

∆ ∆ = − −

U = K P

u1

u2

i=3


Example 2

75

U3h

b

h

0.00003

0.0089

0.00409

0.000717

00.0089

0.0007170.00400

00.000717

0.000717

−

∆ = − −

=

− − = ⇓ θ = − −

U = U + U

U

U

u3

u1

u2

u4

u1

u2

Mh = -22.49 kN-m

b b b=P K U


K = 2.43 1010 N-mm

Example 2



Plastic hinge

76

u3

u1

u2

0

10

20

30

40

50

60

0 0.005 0.01 0.015 0.02 0.025

rotazione θcp

Mo

men

to (

kN

-m)

0.000717

22.49

Kh,tan= 2.43 1010 N-mm

u4

7.5b h

7.5

0 22.488482

7.5 22.488482

22.5

7.5

= =

− −

−

P P

7.5

0

U1

U2


Example 2

77

R

unb R

7.5

0

0.011518

7.5 7.5 0

0 0 0

0 0.011518 0.011518

−

= −

∆ = = − = − = −

P

P P P P

7.5

22.49

22.49

22.5

U3

=3 =2

unb unbNote thati iP P�


Apply Punb

U1

U2

{ }-1tan

0.0000014

0.000000474

0.000000474

∆ ∆ = − −

U = K P

u1

u2

i=4


Example 2

78

U3h

b

h

0.000000474

0.0089

0.00409

0.00072

00.0089

0.000720.00409

00.00072

0.00072

−

∆ = − −

=

− − = ⇓ θ = − −

U = U + U

U

U

u3

u1

u2

u4

u1

u2

Mh = -22.5 kN-m

b b b=P K U


K = 2.4289 104 kN-m

Example 2



Plastic hinge

79

u3

u1

u2

0

10

20

30

40

50

60

0 0.005 0.01 0.015 0.02 0.025

rotazione θcp

Mo

men

to (

kN

-m)

0.00072

22.5

Kh,tan= 2.4289 104 kN-m

u4

7.5b h

7.5

0 22.499999

7.5 22.499999

22.5

7.5

= =

− −

P P

7.5

0

U1

U2


Example 2

80

R

unb R

7.5

0

0.0000028

7.5 7.5 0

0 0 0

0 0.0000028 0.0000028

=

∆ = = − = − = −

P

P P P P

7.5

22.5

22.5

22.5

U3

Small enough!!

=4 =3

unb unb

=4

unb

Note that i i

i ≈

P P

P 0

�

There is equilibrium between

applied and resisting forces Apply λ2P

COMPARISON BETWEEN CONVERGENCE SPEEDS

Example 2

5,00

6,00

7,00

8,00

Pu

nb

(3)

(kN

-m)

K-initial

K-tangent

81

0,00

1,00

2,00

3,00

4,00

1 3 5 7 9 11 13 15

Pu

nb

(3)

(kN

iteration

NONLINEAR GEOMETRY: P-∆ ∆ ∆ ∆ EFFECT

H

P P

H ∆

P∆

Example 3

30

40

50

60

hin

ge

(kN

-m)

NL law: Example 2

NL law: Example 2

83

Nonlinear hinge law: from Example 2

0

10

20

0 0,005 0,01 0,015 0,02

Mh

ing

e

rotation θθθθcp

NL law: Example 2

P

H∆y

θh

( )-

M x = -Pv - Hx

EIv" = -Pv Hx

P Hxv" + v = -

Example 3

84

L

x

v

xθh

( )

( )

( ) hh

sin cos

0 0

cos

1 2

2

1

P Hxv" + v = -

EI EI

P P Hv x = C x C x x

EI EI P

v 0 = C

H P +v' L = C

P PL

EI EI

θθ

+ −

⇒ =

⇒ =

( ) h sincos

tan

H P + P Hv x = x x

EI PP PL

EI EI

PL

H P H

θ

θ

−P

H∆y

θcp

Example 3

EQUILIBRIUM IN THE DEFORMED CONFIGURATION

85

( ) h

h

h

tan

tan

LH P HEI= v L = + L LP EI PP P

EI EI

from equilibrium M = HL+ P

M HL=

P

θ∆ −

∆

−∆

L

x

v

xθcp

Get closed form solution ∆

hh

tan

tan

PL

PEIM = H + P LEIP P

EI EI

P

θ

Example 3

EQUILIBRIUM IN THE DEFORMED CONFIGURATION

86

h h

h

h h

tan

Assign

P

EIH = M PP

LEI

M - HL

P

M H

θ

θ

−

∆ =

⇒ ⇒ ⇒ ∆

H

P

Example 3

872740kN

4

π=

2

cr-el 2

EIP =

L

� 7.3.1 ... Second order effests

P (forces from all stories above)

V (total floor load)

dr θ < 0,1

0,1< θ < 0,2

Secondo order effects

are neglected

Horizontal seismic action

Effects are incremented by

Nonlinear geometry in NTC 2008

88

V (total floor load)

θ = P dr/V hh

V

0,1< θ < 0,2 Effects are incremented by1/(1-θ)

Not allowedθ > 0,3

0,2 < θ < 0,3 No comment

αP1

αP2 ∆

α

1st order linear elastic analysis

2nd order linear elastic analysis

Bifurcation

Conclusions

89∆

1st order inelastic analysis

2nd order inelastic analysis

Elastic Analysis – Materials are all linear elasticInelastic Analysis – Materials are inelastic

First order analysis – Equilibrium in the underformed configurationSecond order analysis – Equilibrium in the deformed

Material

Geometry

Conclusions

90

Second order analysis – Equilibrium in the deformed configuration (large displacements, small, moderate or finite deformations)

Structural collapse is typically associated with loads that lead materials into the inelastic range, and with displacements that lead to structural instability at collapse

Modelli evolutivi per la verifica del rischio di edifici ...Quaderno 4 Primi concetti di analisi...

Documents

Transcript of Modelli evolutivi per la verifica del rischio di edifici ...Quaderno 4 Primi concetti di analisi...