SIMULAZIONE DI UN IMPATTO Dott. Ing. Simone Caffè
Analisi Time History di una trave 60x40 cm in Calcestruzzo C30/37, con carico concentrato nel nodo 2 pari a 100 kN e carico da impatto pari a 10 kN. Gli effetti sono valutati in termini di spostamento verticale del DOF 2.
Definizione del materiale:
E
c:= 33000 MPa ⋅
Coordinate Nodali:
x
1:= 0 m ⋅ z
1:= 0 m ⋅ x
2:= 3 m ⋅ z
2:= 0 m ⋅ x
3:= 6 m ⋅ z
3:= 0 m ⋅ Sezione della trave:
h := 60 cm ⋅ b := 40 cm ⋅ A
t:= h b ⋅ = 0.24 m 2
I
tb h ⋅ 3
12 = 0.01 m 4 :=
L
t:= x
22 − x
12 = 3 m Rigidezze delle trave:
K
xE
c⋅ A
tL
tm
⋅ N = 2640000000 :=
K
z112 E ⋅
c⋅ I
tL
t3
m
⋅ N = 105600000
:= K
z26 E ⋅
c⋅ I
tL
t2
1
⋅ N = 158400000 :=
K
r14 E ⋅
c⋅ I
tL
t1 N m ⋅
⋅ = 316800000
:= K
r22 E ⋅
c⋅ I
tL
t1 N m ⋅
⋅ = 158400000
:=
Matrici di rigidezza locali:
K
1_locK
x0 0 K
x− 0 0
0 K
z1K
z20 K
z1− K
z20 K
z2K
r10 K
z2− K
r2K
x− 0 0 K
x0 0
0 K
z1− K
z2− 0 K
z1K
z2− 0 K
z2K
r20 K
z2− K
r1
:=
K
2_loc:= K
1_locMatrici di rigidezza globali:
K
1_glob:= K
1_locK
2_glob:= K
2_locC
11 0 0 0 0 0
0 1 0 0 0 0
0 0 1 0 0 0
0 0 0 1 0 0
0 0 0 0 1 0
0 0 0 0 0 1
0 0 0 0 0 0
0 0 0 0 0 0
0 0 0 0 0 0
:= C
20 0 0 0 0 0
0 0 0 0 0 0
0 0 0 0 0 0
1 0 0 0 0 0
0 1 0 0 0 0
0 0 1 0 0 0
0 0 0 1 0 0
0 0 0 0 1 0
0 0 0 0 0 1
:=
K
1:= C
1T ⋅ K
1_glob⋅ C
1K
2:= C
2T ⋅ K
2_glob⋅ C
2Matrice di rigidezza della struttura:
K
tot:= K
1+ K
2OC 0 0 0 0
0 0 0 0
0 0 0 0
1 0 0 0
0 1 0 0
0 0 1 0
0 0 0 0
0 0 0 0
0 0 0 1
:=
K
ridOC K ⋅
tot⋅ OC T
5280000000 0 0 0
0 211200000
0 158400000
0 0 633600000 158400000
0 158400000 158400000 316800000
=
:=
Matrice delle masse:
F
j:= 100 kN ⋅ m
jF
jg = 10197.16 kg :=
m
1_glob0 0 0
0
0
0 0 0 0
0
0
0 0 0 0
0
0
0 0 0 0
0.5 m
j⋅ kg 0
0 0 0 0
0
0.5 m
j⋅ kg 0
0 0 0
0
0
0.001
0 0 0 0 0 0
0 0 0 0 0 0
0 0 0 0 0 0
0 0 0 5098.58
0 0
0 0 0 0 5098.58
0 0 0 0 0 0 0
= :=
m
2_glob0.5
m
j⋅ kg 0
0 0 0 0
0
0.5 m
j⋅ kg 0 0 0 0
0
0
0.001 0 0 0
0
0
0 0 0 0
0
0
0 0 0 0
0
0
0 0 0 0.001
5098.58 0 0 0 0 0
0 5098.58
0 0 0 0
0 0 0 0 0 0
0 0 0 0 0 0
0 0 0 0 0 0
0 0 0 0 0 0
= :=
m
1:= C
1T ⋅ m
1_glob⋅ C
1m
2:= C
2T ⋅ m
2_glob⋅ C
2Matrice completa delle masse:
M
tot:= m
1+ m
2Matrice ridotta delle masse:
M
ridOC M ⋅
tot⋅ OC T
10197.16 0 0 0
0 10197.16
0 0
0 0 0 0
0 0 0 0
=
:=
Determinazione degli autovalori:
ORIGIN := 1
Λ sort eigenvals M
rid−1 ⋅ K
rid
11835.23 517791.12 204794291867.4 428805717009.01
= :=
λ
1Λ
1 = 11835.23 :=
λ
2Λ
2 = 517791.12 :=
λ
3Λ
3 = 204794291867.4 :=
λ
4Λ
4 = 428805717009.01 :=
Determinazione delle frequenze proprie di vibrazione:
ω
1:= λ
1= 108.79 f
1ω
12 π ⋅ = 17.31
:= T
1:= f
1−1 = 0.058
ω
2:= λ
2= 719.58 f
2ω
22 π ⋅ = 114.52
:= T
2:= f
2−1 = 0.009
ω
3:= λ
3= 452542.03 f
3ω
32 π ⋅ = 72024.3
:= T
3:= f
3−1 = 0
ω
4:= λ
4= 654832.59 f
4ω
42 π ⋅ = 104219.84
:= T
4:= f
4−1 = 0
Determinazione degli autovettori:
Ψ
1eigenvec M
rid−1 ⋅ K
rid
, λ
1
0
− 0.86
− 0.12 0.49
=
:= Ψ
2eigenvec M
rid−1 ⋅ K
rid
, λ
2
1 0 0 0
= :=
Ψ
3eigenvec M
rid−1 ⋅ K
rid
, λ
3
0
− 0 0.58
− 0.82
=
:= Ψ
4eigenvec M
rid−1 ⋅ K
rid
, λ
4
0 0 0.58 0.82
=
:=
Ψ Ψ
11
Ψ
12
Ψ
13
Ψ
14
Ψ
21
Ψ
22
Ψ
23
Ψ
24
Ψ
31
Ψ
32
Ψ
33
Ψ
34
Ψ
41
Ψ
42
Ψ
43
Ψ
44
0
− 0.86
− 0.12 0.49
1 0 0 0
0
− 0 0.58
− 0.82 0 0 0.58 0.82
= :=
Determinazione della matrice di smorzamento:
Rapporto di smorzamento al critico:
ξ := 0.05
Determinazione dei coefficienti α e β:
ω
1= 108.79
α
2 ω ⋅
1⋅ ( 10 ω ⋅
1) ⋅ ( 10 ω ⋅
1⋅ ξ − ω
1⋅ ξ )
10 ω ⋅
1( ) 2 − ω
12 = 9.89
:=
β 2
10 ω ⋅
1⋅ ξ − ω
1⋅ ξ
( )
10 ω ⋅
1( ) 2 − ω
12
⋅ = 8.3564 × 10
−5
:=
C
ridα M ⋅
rid+ β K ⋅
rid542067.52 0 0 0
0 118498.48
0 13236.53
0 0 52946.15 13236.53
0 13236.53 13236.53 26473.07
= :=
Azione impulsiva:
Modulo dell'azione impulsiva:
F
0:= 10 kN ⋅
Andamento della forza impulsiva nel tempo:
f
TH( ) t t
0.1 if 0 < t ≤ 0.1 0.5 if 0.1 < t ≤ 0.105 0.5 if 0.105 < t ≤ 0.15
− 100 ⋅ t + 15.5
( ) if 0.15 < t ≤ 0.155 0 otherwise
:=
0 0.2 0.4 0.6 0.8
−
1
−
0.5 0.5
1
Funzione Time History - Impulso
f
TH( ) t
t
EQUAZIONE DEL MOTO:
Disaccoppiamento delle equazioni del moto:
Passo 1: Normalizzazione degli autovettori:
ψ
1Ψ
1max Ψ
11
Ψ
1,
2Ψ
1,
3Ψ
1,
4( )
0
− 1
− 0.14 0.57
= :=
ψ
2Ψ
2max Ψ
21
Ψ
2,
2Ψ
2,
3Ψ
2,
4( )
1 0 0 0
= :=
ψ
3Ψ
3max Ψ
31
Ψ
3,
2Ψ
3,
3Ψ
3,
4( )
0
− 0 0.71
− 1
= :=
ψ
4Ψ
4max Ψ
41
Ψ
4,
2Ψ
4,
3Ψ
4,
4( )
0 0 0.71
1
=
:=
μ
1:= ψ
1T ⋅ M
rid⋅ ψ
1= 10197.16 μ
2:= ψ
2T ⋅ M
rid⋅ ψ
2= 10197.16 μ
3:= ψ
3T ⋅ M
rid⋅ ψ
3= 0 μ
4:= ψ
4T ⋅ M
rid⋅ ψ
4= 0
Ψ
1_norm1
μ
1ψ
1⋅
0
− 0.01
− 0 0.01
=
:= Ψ
2_norm1
μ
2ψ
2⋅
0.01 0 0 0
= :=
Ψ
3_norm1
μ
3ψ
3⋅
0
− 0 15.81
22.36
−
=
:= Ψ
4_norm1
μ
4ψ
4⋅
0 0 15.81 22.36
= :=
Ψ
normΨ
1_norm1
Ψ
1_norm2
Ψ
1_norm3
Ψ
1_norm4
Ψ
2_norm1
Ψ
2_norm2
Ψ
2_norm3
Ψ
2_norm4
Ψ
3_norm1
Ψ
3_norm2
Ψ
3_norm3
Ψ
3_norm4
Ψ
4_norm1
Ψ
4_norm2
Ψ
4_norm3
Ψ
4_norm4
0
− 0.01
− 0 0.01
0.01 0 0 0
0
− 0 15.81
22.36
−
0 0 15.81 22.36
= :=
Passo 2: Determinazione delle matrici principali:
Matrice Principale di Massa:
M
normΨ
normT ⋅ M
rid⋅ Ψ
norm1 0 0 0
0 1 0 0
0 0 1 0
0 0 0 1
= :=
Matrice Principale di Rigidezza:
K
normΨ
normT ⋅ K
rid⋅ Ψ
norm11835.23 0 0 0
0 517791.12
0 0
0 0 204794291867.4
0
0 0 0 428805717009.01
= :=
Matrice Principale di Smorzamento:
C
normΨ
normT ⋅ C
rid⋅ Ψ
norm10.88 0 0 0
0 53.16
0 0
0 0 17113433.47
0
0 0
− 0 35832716.96
=
:=
Passo 3: Equazioni del moto nelle coordinate principali:
Vettore della forza impulsiva:
F
0_adm0 0 0 0 F
0N 0 0 0 0
:=
F
impulsiva( ) t := OC F ⋅
0_adm⋅ f
TH( ) t
Prima equazione del moto:
Given
M
norm〈 〉 1
1 ⋅ q
2x''( ) t C
norm〈 〉 1
1 ⋅ q
2x'( ) t
+ K
norm〈 〉 1
1 ⋅ q
2x( ) t
+ Ψ
norm〈 〉 1 T
F
impulsiva( ) t
⋅
=
q
2x( ) 0 = 0 q
2x'( ) 0 = 0
q
2x:= Odesolve t 3 ( , )
Seconda equazione del moto:
Given
M
norm〈 〉 2
2 ⋅ q
2z''( ) t C
norm〈 〉 2
2 ⋅ q
2z'( ) t
+ K
norm〈 〉 2
2 ⋅ q
2z( ) t
+ Ψ
norm〈 〉 2 T
F
impulsiva( ) t
⋅
=
q
2z( ) 0 = 0 q
2z'( ) 0 = 0 q
2z:= Odesolve t 3 ( , )
Terza equazione del moto:
Given
M
norm〈 〉 3
3 ⋅ q
2r''( ) t C
norm〈 〉 3
3 ⋅ q
2r'( ) t
+ K
norm〈 〉 3
3 ⋅ q
2r( ) t
+ Ψ
norm〈 〉 3 T
F
impulsiva( ) t
⋅
=
q
2r( ) 0 = 0 q
2r'( ) 0 = 0
Quarta equazione del moto:
Given
M
norm4
〈 〉
4 ⋅ q
3r''( ) t C
norm〈 〉 4
4 ⋅ q
3r'( ) t
+ K
norm4
〈 〉
4 ⋅ q
3r( ) t
+ Ψ
norm4
〈 〉 T
F
impulsiva( ) t
⋅
=
q
3r( ) 0 = 0 q
3r'( ) 0 = 0 q
3r:= Odesolve t 3 ( , )
Passo 4: Soluzione delle equazioni del moto nelle coordinate fisiche:
Χ t ( ) Ψ
normq
2x( ) t q
2z( ) t q
2r( ) t q
3r( ) t
⋅ :=
Spostamenti del nodo 2:
u
x2( ) t Χ t ( ) := 1 u
z2( ) t Χ t ( )
:= 2 u
r2( ) t Χ t ( )
:= 3
Rotazione del nodo 3:
u
r3( ) t Χ t ( ) := 4
0 0.2 0.4 0.6 0.8
−
