Esempio di Calcolo della Portata di un micropalo φφφφ220 [mm] - Dott. Ing. Simone Caffè
Stratigrafia del terreno
Potenza Profondità γγγγ [kN/m3] φ [°]φ [°] φ [°]φ [°] c' [kPa] γγγγ' [kN/m3]
h1 3,00 3,00 16,00 20,00 30 6,00 Strato 1
h2 6,50 9,50 17,30 31,90 5 7,30 Strato 2
h3 6,00 15,50 18,00 30,00 5 8,00 Strato 3
h4 4,50 20,00 17,00 28,00 5 7,00 Strato 4
Tutto il palo si trova completamente sottofalda Pressione geostatica:
(z 0.00) 0.00
0
v = =
σ′ [kPa]
(z h1 3.00) 1 z 6 3.00 18 1
v = = =γ′⋅ = ⋅ =
σ′ [kPa]
(z h2 6.50) 1 h1 2 z 6 3 7.3 6.5 65.45 2
v = = =γ′⋅ +γ′⋅ = ⋅ + ⋅ =
σ′ [kPa]
(z h3 6.00) 1 h1 2 h2 3 z 6 3 7.3 6.5 8 6 113.45 3
v = = =γ′⋅ +γ′ ⋅ +γ′ ⋅ = ⋅ + ⋅ + ⋅ =
σ′ [kPa]
(z h4 4.50) 1 h1 2 h2 3 h3 4 z 6 3 7.3 6.5 8 6 7 4.50 144.95 4
v = = =γ′⋅ +γ′ ⋅ +γ′ ⋅ +γ′ ⋅ = ⋅ + ⋅ + ⋅ + ⋅ =
σ′ [kPa]
Andamento della pressione geostatica σσσσ'(z)
0 20 40 60 80 100 120 140 160
0 2 4 6 8 10 12 14 16 18 20
[m]
[kPa]
Portata alla punta secondo la teoria di Berezantzev:
Angolo di attrito efficace Kishida (1967): φ′=φ−3°=28−3=25 [-]
Dal grafico che mette in relazione angolo di attrito, e valore del coefficiente di capacità portante Nq in funzione del rapporto L/D (Viggiani – pag. 377) si ottiene:
50 9 . 90 22 . 0 20 D
L = = > → Nq=6
( q v.max)
2
P N
4
P = π⋅D ⋅ ⋅σ′
(6 144.95) 33
4 22 .
PP= π⋅0 2 ⋅ ⋅ = [kN]
Portata laterale:
( )z D ( )z dz D [( z)K tan c ] dz
P
1 i 1
i h
0
1 i 1 i s 1 i vi h
0
L =π⋅ ⋅ τ ⋅ =π⋅ ⋅ σ′ +γ′ ⋅ ⋅ ⋅ φ +α⋅ ′ ⋅
+ +
+ + +
00 . 1
Ks = per tenere in conto della modalità di iniezione dei micropali
Andamento della tensione laterale ττττ(z)
0 10 20 30 40 50 60 70 80 90
0 2 4 6 8 10 12 14 16 18 20
[m]
[kPa]
(z 3.00) D [ z tan 0.5 c] dz
P
h1
0
1 1
1
L = =π⋅ ⋅ γ′⋅ ⋅ φ + ⋅ ′ ⋅
(z 3.00) 0.22 [6 z tan( )20 0.5 30] dz 37.89
P
3 0
L = =π⋅ ⋅ ⋅ ⋅ + ⋅ ⋅ = [kN]
(z 6.50) D [( z) tan 0.5 c ] dz
P
h2
0
2 2
2 1 v
L = =π⋅ ⋅ σ′ +γ′ ⋅ ⋅ φ + ⋅ ′ ⋅
(z 6.50) 0.22 [(18 7.3 z) tan( )31.9 0.5 5] dz 127.91
P
5 . 6 0
L = =π⋅ ⋅ + ⋅ ⋅ + ⋅ ⋅ = [kN]
(z 6.00) D [( z) tan 0.5 c ] dz
P
h3
0
3 3
3 2 v
L = =π⋅ ⋅ σ′ +γ′ ⋅ ⋅ φ + ⋅ ′ ⋅
(z 6.00) 0.22 [(65.45 8 z) tan( )30 0.5 5] dz 224.53
P
6 0
L = =π⋅ ⋅ + ⋅ ⋅ + ⋅ ⋅ = [kN]
(z 4.50) D [( z) tan 0.5 c ] dz
P
h4
0
4 4
4 3 v
L = =π⋅ ⋅ σ′ +γ′ ⋅ ⋅ φ + ⋅ ′ ⋅
(z 4.50) 0.22 [(113.45 7 z) tan( )28 0.5 5] dz 221.44
P
5 . 4 0
L = =π⋅ ⋅ + ⋅ ⋅ + ⋅ ⋅ = [kN]
( + + + )=
+
= +
=
i Li P
TOT P P 33 37.89 127.91 224.53 221.44 644.77
P [kN]
τ αc τ(z)
0 [m] 0,00 + 15,00 = 15,00
3 [m] 6,55 + 15,00 = 21,55
3 [m] 11,20 + 2,50 = 13,70
9,5 [m] 40,74 + 2,50 = 43,24
9,5 [m] 37,78 + 2,50 = 40,28
15,5 [m] 65,50 + 2,50 = 68,00
15,5 [m] 60,32 + 2,50 = 62,82
20 [m] 77,07 + 2,50 = 79,57
00 . 0 z=
=
⋅
′=
⋅ α
= τ
15 30 5 . 0 c
0 0 [kPa]
00 . 3 z=
=
⋅
′=
⋅ α
= φ
⋅
⋅ γ′
= τ
= φ
⋅
⋅ γ′
= τ
15 30 5 . 0 c
20 . 11 tan h
55 . 6 tan h
2 1 1 1
1 1 1
1 [kPa]
50 . 9
z= ( )
( )
=
⋅
′=
⋅ α
= φ
⋅
⋅ γ′
+ σ′
= τ
= φ
⋅
⋅ γ′
+ σ′
= τ
5 . 2 5 5 . 0 c
78 . 37 tan h
74 . 40 tan h
3 2 2 1 v 2
2 2 2 1 v
2 [kPa]
50 . 15 z=
( )
( )
=
⋅
′=
⋅ α
= φ
⋅
⋅ γ′
+ σ′
= τ
= φ
⋅
⋅ γ′
+ σ′
= τ
5 . 2 5 5 . 0 c
32 . 60 tan h
50 . 65 tan h
4 3 3 2 v 3
3 3 3 2 v 3
[kPa]
00 . 20
z= ( )
=
⋅
′=
⋅ α
= φ
⋅
⋅ γ′
+ σ′
= τ
5 . 2 5 5 . 0 c
07 . 77 tan
h4 4
4 3 v
4 [kPa]
