modelli evolutivi per la verifica del rischio di edifici ...quaderno 4 primi concetti di analisi...
TRANSCRIPT
-
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
u2i=1LOAD STEP 1: λ1P = 7.5 kN
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
i=1LOAD STEP 1: λ1P = 7.5 kN
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
Determine resisting forces corresponding to
the current displacement vector U(STRUCTURE STATE DETERMINATION)
-
U1
U2
0.0162
0.00765
0.0009
∆ = − −
U = U + U
u1
u2i=1LOAD STEP 2: λ2P = 15 kN
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)
1) Column: linear elastic
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
Mh = min(kel-h θh , M* + kpl-h θh) = -45 kN-m
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
u2i=1LOAD STEP 3: λ3P = 17 kN
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
Determine resisting forces corresponding to
the current displacement vector U(STATE DETERMINATION)
-
U1
U2
0.01836
0.00867
∆ = −
U = U + U
u1
u2i=1LOAD STEP 3: λ3P = 17 kN
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
Elements’ resisting forces
(ELEMENT STATE DETERMINATION)
1) Column: linear elastic
2) Plastic hinge
i=1
b b b=P K U
LOAD STEP 3: λ3P = 17 kN
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
i=1LOAD STEP 3: λ3P = 17 kN
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
u2i=2LOAD STEP 3: λ3P = 17 kN
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
Elements’ resisting forces
(ELEMENT STATE DETERMINATION)
1) Column: linear elastic
2) Plastic hinge
i=2
b b b=P K U
LOAD STEP 3: λ3P = 17 kN
Mh = min(kel-h θh , M* + kpl-h θh) = -51 kN-m
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
Mh = min(kel-h θh , M* + kpl-h θh) = -51 kN-m
0.024
-
17
b h
17
0 51
17 51
= =
− − P P
17
0
U1
U2
i=2LOAD STEP 1: λ3P = 17 kN
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
u2i=1LOAD STEP 1: λ1P = 7.5 kN
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
LOAD STEP 1: λ1P = 7.5 kN
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
LOAD STEP 1: λ1P = 7.5 kN
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
i=1LOAD STEP 1: λ1P = 7.5 kN
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
LOAD STEP 1: λ1P = 7.5 kN
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
i=2LOAD STEP 1: λ1P = 7.5 kN
Elements’ resisting forces
Column: linear elastic
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
i=2LOAD STEP 1: λ1P = 7.5 kN
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
u2i=3LOAD STEP 1: λ1P = 7.5 kN
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
i=3LOAD STEP 1: λ1P = 7.5 kN
Example 2
Elements’ resisting forces
Column: linear elastic
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
i=3LOAD STEP 1: λ1P = 7.5 kN
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
LOAD STEP 1: λ1P = 7.5 kN
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
i=15LOAD STEP 1: λ1P = 7.5 kN
Example 2
Elements’ resisting forces
Column: linear elastic
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
i=15LOAD STEP 1: λ1P = 7.5 kN
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
u2i=1LOAD STEP 1: λ1P = 7.5 kN
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
LOAD STEP 1: λ1P = 7.5 kN
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
i=1LOAD STEP 1: λ1P = 7.5 kN
K = 3.14 104 kN-m
Example 2
Elements’ resisting forces
Column: linear elastic
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
i=1LOAD STEP 1: λ1P = 7.5 kN
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
There is no equilibrium between applied and resisting forces
Apply Punb
-
U1
U2
{ }-1tan
0.00071
0.00024
0.00024
∆ ∆ = − −
U = K P
u1
u2
i=2
LOAD STEP 1: λ1P = 7.5 kN
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
i=2LOAD STEP 1: λ1P = 7.5 kN
K = 2.5 104 kN-m
Example 2
Elements’ resisting forces
Column: linear elastic
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
i=2LOAD STEP 1: λ1P = 7.5 kN
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�
There is no equilibrium between applied and resisting forces
Apply Punb
-
U1
U2
{ }-1tan
0.00009
0.00003
0.00003
∆ ∆ = − −
U = K P
u1
u2
i=3
LOAD STEP 1: λ1P = 7.5 kN
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
i=3LOAD STEP 1: λ1P = 7.5 kN
K = 2.43 1010 N-mm
Example 2
Elements’ resisting forces
Column: linear elastic
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
i=3LOAD STEP 1: λ1P = 7.5 kN
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�
There is no equilibrium between applied and resisting forces
Apply Punb
-
U1
U2
{ }-1tan
0.0000014
0.000000474
0.000000474
∆ ∆ = − −
U = K P
u1
u2
i=4
LOAD STEP 1: λ1P = 7.5 kN
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
i=4LOAD STEP 1: λ1P = 7.5 kN
K = 2.4289 104 kN-m
Example 2
Elements’ resisting forces
Column: linear elastic
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
i=4LOAD STEP 1: λ1P = 7.5 kN
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