1. Analisi manuale del dominio di resistenza di un pilastro 30 x 30
Nella presente scheda si costruirà il dominio di resistenza di un pilastro in calcestruzzo armato Classe 25/30, con sezione pari a 30 x 30 [cm] ordito con 4 ferri Ø16 in acciaio Feb44k/s.
30 0 26 0
300
ε su = 0.01
ε yd
ε cu =0.0035 ε c1
f
ydf
cdCaratteristiche dei materiali
Calcestruzzo: R
ck= 30 [MPa]
25
f
ck= [MPa]
28 . 60 13 . 1
f 85 .
f
cd= 0 ⋅
ck= [MPa]
0035 .
cu
= 0
ε [-]
0020 .
1
0
c
=
ε [-]
Acciaio: f yk = 430 [MPa]
9 . 15 373 . 1
f
yd= 430 = [MPa]
200000
E
s= [MPa]
010 .
su
= 0
ε [-]
00187 . 200000 0
9 . 373 E
f
s
yd
=
yd= =
ε [-]
Armature: 4 φ 16
4 402 2 16 A
A
s=
s′ = ⋅ π ⋅
2= [mm
2]
PUNTO 1
30 0 26 0
300
ε su = 0.01
ε yd
ε cu =0.0035 ε c1
f
ydf
cdCondizioni
Deformazione dell’armatura inferiore: ε
s= ε
su= 0 . 01 Deformazione dell’armatura superiore: ε′
s= ε
su= 0 . 01 71 . 300 9 . 373 402 9 . 373 402 f A f A
N Rd = s ′ ⋅ yd + s ⋅ yd = ⋅ + ⋅ = [kN]
2 0 d h f
A 2 d
f h A
M
Rd=
s′ ⋅
yd⋅ − ′ +
s′ ⋅
yd⋅ − − = [kNm]
PUNTO 2
30 0 26 0
300
ε su = 0.01
ε yd
ε cu =0.0035 ε c1
f
ydf
cdCondizioni
Deformazione dell’armatura inferiore: ε
s= ε
su= 0 . 01 Deformazione dell’armatura superiore: ε
su: d = ε′
s: d ′
yd su
s
0 . 01 0 . 00154
260 40 d
d ′ ⋅ ε = ⋅ = < ε
= ε′
yd s s s s
Rd A E A f
N = ′ ⋅ ⋅ ε′ + ⋅
274 9 . 373 402 00154 . 0 200000 402
N
Rd= ⋅ ⋅ + ⋅ = [kN]
−
−
⋅
′ ⋅
′ +
−
⋅ ε′
⋅
′ ⋅
= 2
d h f
A 2 d
E h A
M
Rd s s s s yd915 . 2 2 40
9 300 . 373 402 2 40
00154 300 . 0 200000 402
M
Rd= ⋅ ⋅ ⋅ − + ⋅ ⋅ − − = [kN]
PUNTO 3
30 0 26 0
300
ε su = 0.01
ε yd
ε cu =0.0035 ε c1
f
ydf
cdCondizioni
Deformazione dell’armatura inferiore: ε
s= ε
su= 0 . 01 Deformazione nel calcestruzzo: ε
c= ε
c1= 0 . 002 Posizione dell’asse neutro: ε
su: ( d − x ) = ε
c1: x
( d x )
x
c1su
⋅ = ε ⋅ − ε
33 . 43 002 260
. 0 01 . 0
002 . d 0 x
1 c su
1
c
⋅ =
= + ε ⋅ + ε
= ε [mm]
Deformazione dell’armatura superiore: ε
su: ( d − x ) = ε′
s: ( x − d ′ )
yd su
s
0 . 01 0 . 000154
33 . 43 260
40 33 . 43 x
d d
x ⋅ = < ε
−
= − ε
− ⋅
− ′
= ε′
yd s cd s
s s
Rd A E x b f A f
N = − ′ ⋅ ⋅ ε′ − β ⋅ ⋅ ⋅ + ⋅
84 . 22 9 . 373 402 28 . 13 300 33 . 43 6667 . 0 000154 . 0 200000 402
N
Rd= − ⋅ ⋅ − ⋅ ⋅ ⋅ + ⋅ = [kN]
⋅ κ
−
⋅
⋅
⋅
⋅ β
−
−
−
⋅
′ ⋅
′ +
−
⋅ ε′
⋅
′ ⋅
−
= x
2 f h b 2 x
d h f
A 2 d
E h A
M
Rd s s s s yd cd( 150 40 ) 402 373 . 9 [ ( 260 150 ) ] 0 . 6667 43 . 33 300 13 . 28 ( 150 0 . 375 43 . 33 )
000154 . 0 200000 402
M
Rd= − ⋅ ⋅ ⋅ − + ⋅ ⋅ − − − ⋅ ⋅ ⋅ ⋅ − ⋅
29 . 33
M
Rd= [kNm]
Si noti che per ε
c= ε
c1= 0 . 02 i valori di β e κ si determinano nel seguente modo:
Coefficiente di riempimento: 0 . 002 0 . 6667
3 250000 002
. 0 3 500
250000
500 ⋅ ε
c1− ⋅ ε
2c1= ⋅ − ⋅
2=
=
β [-]
Coefficiente di baricentro: 0 . 375
3 002 . 0 500
1 002 . 0 125 3 500
1 125
1 c
1
c
=
−
⋅
−
= ⋅
− ε
⋅
− ε
= ⋅
κ [-]
PUNTO 4 (rottura bilanciata)
30 0 26 0
300
ε su = 0.01
ε yd
ε cu =0.0035 ε c1
f
ydf
cdCondizioni
Deformazione dell’armatura inferiore: ε
s= ε
su= 0 . 01 Deformazione nel calcestruzzo: ε
c= ε
cu= 0 . 0035 Posizione dell’asse neutro: ε
su: ( d − x ) = ε
cu: x
( d x )
x
cusu
⋅ = ε ⋅ − ε
41 . 67 0035 260
. 0 01 . 0
0035 . d 0 x
cu su
cu
⋅ =
= + ε ⋅ + ε
= ε [mm]
Deformazione dell’armatura superiore: ε
su: ( d − x ) = ε′
s: ( x − d ′ )
yd su
s
0 . 01 0 . 001423
41 . 67 260
40 41 . 67 x
d d
x ⋅ = < ε
−
= − ε
− ⋅
− ′
= ε′
In questo caso il coefficiente di riempimento vale β
0= 0 . 80 e il coefficiente di baricentro vale κ
0= 0 . 40 .
yd s cd 0
s s s
Rd A E x b f A f
N = − ′ ⋅ ⋅ ε′ − β ⋅ ⋅ ⋅ + ⋅
95 . 178 9 . 373 402 28 . 13 300 41 . 67 80 . 0 001423 . 0 200000 402
N
Rd= − ⋅ ⋅ − ⋅ ⋅ ⋅ + ⋅ = − [kN]
⋅ κ
−
⋅
⋅
⋅
⋅ β
−
−
−
⋅
′ ⋅
′ +
−
⋅ ε′
⋅
′ ⋅
−
= x
2 f h b 2 x
d h f
A 2 d
E h A
M
Rd s s s s yd 0 cd 0( 150 40 ) 402 373 . 9 [ ( 260 150 ) ] 0 . 80 67 . 41 300 13 . 28 ( 150 0 . 40 67 . 41 )
001423 . 0 200000 402
M
Rd= − ⋅ ⋅ ⋅ − + ⋅ ⋅ − − − ⋅ ⋅ ⋅ ⋅ − ⋅
55 . 55
M
Rd= [kNm]
PUNTI 5
Da questo punto in poi si potrebbero assumere arbitrariamente delle deformazioni dell’aratura inferiore, comprese tra 01
.
su
= 0
ε e ε yd in modo da ottenere una maggiore accuratezza del dominio di resistenza . Questo è possibile perché il meccanismo di rottura è governato dalla deformazione limite del calcestruzzo e non più dalla deformazione limite dell’acciaio.
30 0 26 0
300
ε su = 0.01
ε yd
ε cu =0.0035 ε c1
f
ydf
cdCondizioni
Deformazione dell’armatura inferiore: ε su < ε s = 0 . 008 < ε yd “assunta arbitrariamente”
Deformazione nel calcestruzzo: ε
c= ε
cu= 0 . 0035 Posizione dell’asse neutro: ε
s: ( d − x ) = ε
cu: x
( d x )
x
cus
⋅ = ε ⋅ −
ε
13 . 79 0035 260
. 0 008 . 0
0035 . d 0
x
cu s
cu
⋅ =
= + ε ⋅ + ε
= ε [mm]
Deformazione dell’armatura superiore: ε
s: ( d − x ) = ε′
s: ( x − d ′ )
yd s
s
0 . 008 0 . 00173
13 . 79 260
40 13 . 79 x
d d
x ⋅ = < ε
− −
= ε
− ⋅
− ′
= ε′
In questo caso il coefficiente di riempimento vale β
0= 0 . 80 e il coefficiente di baricentro vale κ
0= 0 . 40 .
yd s cd 0
s s s
Rd A E x b f A f
N = − ′ ⋅ ⋅ ε′ − β ⋅ ⋅ ⋅ + ⋅
241 9 . 373 402 28 . 13 300 13 . 79 80 . 0 00173 . 0 200000 402
N
Rd= − ⋅ ⋅ − ⋅ ⋅ ⋅ + ⋅ = − [kN]
⋅ κ
−
⋅
⋅
⋅
⋅ β
−
−
−
⋅
′ ⋅
′ +
−
⋅ ε′
⋅
′ ⋅
−
= x
2 f h b 2 x
d h f
A 2 d
E h A
M
Rd s s s s yd 0 cd 0( 150 40 ) 402 373 . 9 [ ( 260 150 ) ] 0 . 80 79 . 13 300 13 . 28 ( 150 0 . 40 79 . 13 )
00173 . 0 200000 402
M
Rd= − ⋅ ⋅ ⋅ − + ⋅ ⋅ − − − ⋅ ⋅ ⋅ ⋅ − ⋅
68 . 61
M
Rd= [kNm]
Condizioni
Deformazione dell’armatura inferiore: ε su < ε s = 0 . 005 < ε yd “assunta arbitrariamente”
Deformazione nel calcestruzzo: ε
c= ε
cu= 0 . 0035 Posizione dell’asse neutro: ε
s: ( d − x ) = ε
cu: x
( d x )
x
cus
⋅ = ε ⋅ −
ε
06 . 107 0035 260
. 0 005 . 0
0035 . d 0 x
cu s
cu
⋅ =
= + ε ⋅ + ε
= ε [mm]
Deformazione dell’armatura superiore: ε
s: ( d − x ) = ε′
s: ( x − d ′ )
yd s
s
0 . 005 0 . 00219
06 . 107 260
40 06 . 107 x
d d
x ⋅ = > ε
− −
= ε
− ⋅
− ′
=
ε′ “snervata”
In questo caso il coefficiente di riempimento vale β
0= 0 . 80 e il coefficiente di baricentro vale κ
0= 0 . 40 .
yd s cd 0
yd s
Rd A f x b f A f
N = − ′ ⋅ − β ⋅ ⋅ ⋅ + ⋅
22 . 341 9 . 373 402 28 . 13 300 06 . 107 80 . 0 9 . 373 402
N
Rd= − ⋅ − ⋅ ⋅ ⋅ + ⋅ = − [kN]
⋅ κ
−
⋅
⋅
⋅
⋅ β
−
−
−
⋅
′ ⋅
′ +
−
⋅
′ ⋅
−
= x
2 f h b 2 x
d h f
A 2 d
f h A
M
Rd s yd s yd 0 cd 0( 150 40 ) 402 373 . 9 [ ( 260 150 ) ] 0 . 80 107 . 06 300 13 . 28 ( 150 0 . 40 107 . 06 )
9 . 373 402
M
Rd= − ⋅ ⋅ − + ⋅ ⋅ − − − ⋅ ⋅ ⋅ ⋅ − ⋅
64 . 69
M
Rd= [kNm]
Condizioni
Deformazione dell’armatura inferiore: ε su < ε s = 0 . 0035 < ε yd “assunta arbitrariamente”
Deformazione nel calcestruzzo: ε
c= ε
cu= 0 . 0035 Posizione dell’asse neutro: ε
s: ( d − x ) = ε
cu: x
( d x )
x
cus
⋅ = ε ⋅ −
ε
130 0035 260
. 0 0035 . 0
0035 . d 0
x
cu s
cu
⋅ =
= + ε ⋅ + ε
= ε [mm]
Deformazione dell’armatura superiore: ε
s: ( d − x ) = ε′
s: ( x − d ′ )
yd s
s
0 . 0035 0 . 00242
130 260
40 130 x
d d
x ⋅ = > ε
−
= − ε
− ⋅
′
= −
ε′ “snervata”
In questo caso il coefficiente di riempimento vale β
0= 0 . 80 e il coefficiente di baricentro vale κ
0= 0 . 40 .
yd s cd 0
yd s
Rd A f x b f A f
N = − ′ ⋅ − β ⋅ ⋅ ⋅ + ⋅
34 . 414 9 . 373 402 28 . 13 300 130 80 . 0 9 . 373 402
N
Rd= − ⋅ − ⋅ ⋅ ⋅ + ⋅ = − [kN]
⋅ κ
−
⋅
⋅
⋅
⋅ β
−
−
−
⋅
′ ⋅
′ +
−
⋅
′ ⋅
−
= x
2 f h b 2 x
d h f
A 2 d
f h A
M
Rd s yd s yd 0 cd 0( 150 40 ) 402 373 . 9 [ ( 260 150 ) ] 0 . 80 130 300 13 . 28 ( 150 0 . 40 130 )
9 . 373 402
M
Rd= − ⋅ ⋅ − + ⋅ ⋅ − − − ⋅ ⋅ ⋅ ⋅ − ⋅
67 . 73
M
Rd= [kNm]
PUNTO 6
30 0 26 0
300
ε su = 0.01
ε yd
ε cu =0.0035 ε c1
f
ydf
cdCondizioni
Deformazione dell’armatura inferiore: ε s = ε yd = 0 . 00187 Deformazione nel calcestruzzo: ε
c= ε
cu= 0 . 0035 Posizione dell’asse neutro: ε
s: ( d − x ) = ε
cu: x
( d x )
x
cus
⋅ = ε ⋅ −
ε
46 . 169 0035 260
. 0 00187 . 0
0035 . d 0
x
cu s
cu
⋅ =
= + ε ⋅ + ε
= ε [mm]
Deformazione dell’armatura superiore: ε
c: x = ε′
s: ( x − d ′ )
yd s
s
0 . 00187 0 . 00267
46 . 169 260
40 46 . 169 x
d d
x ⋅ = > ε
−
= − ε
− ⋅
′
= −
ε′ “snervata”
In questo caso il coefficiente di riempimento vale β
0= 0 . 80 e il coefficiente di baricentro vale κ
0= 0 . 40 .
yd s cd 0
yd s
Rd A f x b f A f
N = − ′ ⋅ − β ⋅ ⋅ ⋅ + ⋅
1 . 540 9 . 373 402 28 . 13 300 46 . 169 80 . 0 9 . 373 402
N
Rd= − ⋅ − ⋅ ⋅ ⋅ + ⋅ = − [kN]
⋅ κ
−
⋅
⋅
⋅
⋅ β
−
−
−
⋅
′ ⋅
′ +
−
⋅
′ ⋅
−
= x
2 f h b 2 x
d h f
A 2 d
f h A
M
Rd s yd s yd 0 cd 0( 150 40 ) 402 373 . 9 [ ( 260 150 ) ] 0 . 80 169 . 46 300 13 . 28 ( 150 0 . 40 169 . 46 )
9 . 373 402
M
Rd= − ⋅ ⋅ − + ⋅ ⋅ − − − ⋅ ⋅ ⋅ ⋅ − ⋅
47 . 77
M
Rd= [kNm]
PUNTO 7
30 0 26 0
300
ε su = 0.01
ε yd
ε cu =0.0035 ε c1
f
ydf
cdCondizioni
Deformazione dell’armatura inferiore: ε
s= 0 Deformazione nel calcestruzzo: ε
c= ε
cu= 0 . 0035
Posizione dell’asse neutro: ε
s: ( d − x ) = ε
cu: x
( d x )
x
cus
⋅ = ε ⋅ −
ε
260 0035 260
. 0
0035 . d 0 x
cu s
cu
⋅ = ⋅ =
ε + ε
= ε [mm]
Deformazione dell’armatura superiore: ε
c: x = ε′
s: ( x − d ′ )
yd c
s
0 . 0035 0 . 00296
260 40 260 x
d
x − ′ ⋅ ε = − ⋅ = > ε
=
ε′ “snervata”
In questo caso il coefficiente di riempimento vale β
0= 0 . 80 e il coefficiente di baricentro vale κ
0= 0 . 40 .
cd 0
yd s
Rd A f x b f
N = − ′ ⋅ − β ⋅ ⋅ ⋅
979 28 . 13 300 260 80 . 0 9 . 373 402
N
Rd= − ⋅ − ⋅ ⋅ ⋅ = − [kN]
⋅ κ
−
⋅
⋅
⋅
⋅ β
−
′
−
⋅
⋅
′
−
= x
2 f h b x 2 d
f h A
M
Rd s yd 0 cd 0( 150 40 ) 0 . 80 260 300 13 . 28 ( 150 0 . 40 260 )
9 . 373 402
M
Rd= − ⋅ ⋅ − − ⋅ ⋅ ⋅ ⋅ − ⋅
65 . 54
M
Rd= [kNm]
PUNTO 8
30 0 26 0
300
ε su = 0.01
ε yd
ε cu =0.0035 ε c1
f
ydf
cdCondizioni
Deformazione nel calcestruzzo: ε
c= ε
cu= 0 . 0035
Posizione dell’asse neutro: x = h = 300 [mm]
Deformazione dell’armatura superiore: ε
c: x = ε′
s: ( x − d ′ )
yd c
s
0 . 0035 0 . 00303
300 40 300 x
d
x − ′ ⋅ ε = − ⋅ = > ε
=
ε′ “snervata”
Deformazione dell’armatura inferiore: ε
c: x = ε
s: ( x − d )
yd c
s