Argomento Tensioni Totali e Tensioni efficaci Redattore Dott. Ing. Simone Caffè Riferimento ENV 1997 – 1 – 1
SCHEDE APPLICATIVE DI GEOTECNICA
Scheda 1 Pagina 1di 3
STRATIGRAFIA DEL TERRENO
La potenza del primo strato è pari a 5 [m], mentre quella del secondo strato è pari a 10 [m]. La falda è alta 5 [m].
CALCOLO DELLE TENSIONI TOTALI VERTICALI [ ]1 1
yy=γdry⋅y σ
(
y1 0)
0yy = =
σ [kPa]
(
y 5)
[ ]1 h1 17 5 85 1 dryyy = =γ ⋅ = ⋅ =
σ [kPa]
[ ] [ ]
2 2 dry 1 1
dry
yy =γ ⋅h +γ ⋅y σ
(
y 0)
[ ]1 h1 17 5 85 2 dryyy = =γ ⋅ = ⋅ =
σ [kPa]
(
y 5)
[ ] h [ ]2 h2 17 5 16 5 165dry 1 1
dry 2
yy = =γ ⋅ +γ ⋅ = ⋅ + ⋅ =
σ [kPa]
[ ] [ ]2 2
(
[ ]wet2 water)
3 water 31 dry dry1
yy=γ ⋅h +γ ⋅h + γ −γ ⋅y +γ ⋅y
σ
(
y 0)
[ ] h [ ]2 h2 17 5 16 5 165dry 1 1
dry 3
yy = =γ ⋅ +γ ⋅ = ⋅ + ⋅ =
σ [kPa]
( )
[ ] [ ](
[ ])
(
y 5)
17 5 16 5(
21 10)
5 10 5 270y y
h h
5 y
3 yy
3 water 3 water wet2 2 2
1 dry dry1 3
yy
=
⋅ +
⋅
− +
⋅ +
⋅
=
= σ
⋅ γ +
⋅ γ
− γ +
⋅ γ +
⋅ γ
=
=
σ [kPa]
CALCOLO DELLE TENSIONI EFFICACI VERTICALI [ ]1 1
yy =γdry⋅y σ′
(
y1 0)
0yy = =
σ′ [kPa]
(
y 5)
[ ]1 h1 17 5 85 1 dryyy = =γ ⋅ = ⋅ =
σ′ [kPa]
[ ] [ ]
2 2 1 dry dry1
yy =γ ⋅h +γ ⋅y σ′
(
y 0)
[ ]1 h1 17 5 85 dry2
yy = =γ ⋅ = ⋅ =
σ′ [kPa]
(
y 5)
[ ] h [ ]2 h2 17 5 16 5 1651 dry dry1 2
yy = =γ ⋅ +γ ⋅ = ⋅ + ⋅ =
σ′ [kPa]
[ ] [ ]
(
[ ]2 water)
3wet 2 2
dry 1 1
dry
yy=γ ⋅h +γ ⋅h + γ −γ ⋅y
σ′
(
y 0)
[ ] h [ ]2 h2 17 5 16 5 1651 dry dry1 3
yy = =γ ⋅ +γ ⋅ = ⋅ + ⋅ =
σ′ [kPa]
( )
[ ] [ ](
[ ])
(
yy3 55)
17 5h16 5(
h21 10)
5 220 yyy
3 water wet2 2 2
dry 1 1
dry 3
yy
=
⋅
− +
⋅ +
⋅
=
= σ′
⋅ γ
− γ +
⋅ γ +
⋅ γ
=
=
σ′ [kPa]
[ ]
[ ]
[ ]
[ ]
[ ]= ° φ
= γ
= γ
30 m kN 20
m kN 17
1
3 wet1
3 dry1
Falda
y1
y2
y3
y
x
[
3]
water =10kNm γ
[ ]
[ ]
[ ]
[ ]
[ ]= ° φ
= γ
= γ
26 m kN 21
m kN 16
2
3 wet2
3 dry2
Argomento Tensioni Totali e Tensioni efficaci Redattore Dott. Ing. Simone Caffè Riferimento ENV 1997 – 1 – 1
SCHEDE APPLICATIVE DI GEOTECNICA
Scheda 1 Pagina 2di 3
CALCOLO DELLE TENSIONI TOTALI ORIZZONTALI
Calcolo del coefficiente di spinta a riposo:
[ ] [ ]
( )
[ ]==1−−senφφ[ ]==1−−sen
( )
26==0.5616k
50 . 0 30 sen 1 sen 1 k
2 02
1 01
[-]
[ ] [ ]
1 1 1 dry xx =k0 ⋅γ ⋅y σ
(
y1 0)
0xx = =
σ [kPa]
(
y 5)
k[ ] [ ]1 h1 0.50 17 5 42.50 1 dry1 0
xx = = ⋅γ ⋅ = ⋅ ⋅ =
σ [kPa]
[ ]
(
[ ]1 1 [ ]dry2 2)
2 dry
xx=k0 ⋅γ ⋅h +γ ⋅y σ
(
y 0)
k[ ] [ ]1 h1 0.5616 17 5 47.74 2 dry0 2
xx = = ⋅γ ⋅ = ⋅ ⋅ =
σ [kPa]
(
y 5)
k[ ](
[ ] h [ ]2 h2)
0.5616(
17 5 16 5)
92.6641 dry dry1 02 2
xx = = ⋅ γ ⋅ +γ ⋅ = ⋅ ⋅ + ⋅ =
σ [kPa]
[ ]
[
[ ] [ ]2 2(
[ ]wet2 water)
3]
water 31 dry 1 dry 2
xx=k0 ⋅γ ⋅h +γ ⋅h +γ −γ ⋅y +γ ⋅y
σ
(
y 0)
k[ ][
[ ] h [ ]2 h2]
0.5616(
17 5 16 5)
92.6641 dry dry1 02 3
xx = = ⋅γ ⋅ +γ ⋅ = ⋅ ⋅ + ⋅ =
σ [kPa]
( )
[ ][
[ ] [ ](
[ ]) ]
(
yy3 55)
k0.5616[
17h5 16 5h(
21 10)
5]
10 y5 173.55 yxx
3 water 3 water wet2 2 2
1 dry dry1 02 3
xx
=
⋅ +
⋅
− +
⋅ +
⋅
⋅
=
= σ
⋅ γ +
⋅ γ
− γ +
⋅ γ +
⋅ γ
⋅
=
=
σ [kPa]
Argomento Tensioni Totali e Tensioni efficaci Redattore Dott. Ing. Simone Caffè Riferimento ENV 1997 – 1 – 1
SCHEDE APPLICATIVE DI GEOTECNICA
Scheda 1 Pagina 3di 3
CALCOLO DELLE TENSIONI EFFICACI ORIZZONTALI
Calcolo del coefficiente di spinta a riposo:
[ ] [ ]
( )
[ ]==1−−senφφ[ ]==1−−sen
( )
26==0.5616k
50 . 0 30 sen 1 sen 1 k
2 02
1 01
[-]
[ ] [ ]
1 1 1 dry xx =k0 ⋅γ ⋅y σ′
(
y1 0)
0xx = =
σ′ [kPa]
(
y 5)
k[ ] [ ]1 h1 0.50 17 5 42.50 1 dry1 0
xx = = ⋅γ ⋅ = ⋅ ⋅ =
σ′ [kPa]
[ ]
(
[ ]1 1 [ ]dry2 2)
2 dry
xx=k0 ⋅γ ⋅h +γ ⋅y σ′
(
y 0)
k[ ] [ ]1 h1 0.561617 5 47.74 2 dry0 2
xx = = ⋅γ ⋅ = ⋅ ⋅ =
σ′ [kPa]
(
y 5)
k[ ](
[ ] h [ ]2 h2)
0.5616(
17 5 16 5)
92.6641 dry dry1 02 2
xx = = ⋅ γ ⋅ +γ ⋅ = ⋅ ⋅ + ⋅ =
σ′ [kPa]
[ ]
[
[ ] [ ]2 2(
[ ]wet2 water)
3]
1 dry 1 dry 2
xx=k0 ⋅γ ⋅h +γ ⋅h +γ −γ ⋅y σ′
(
y3 0)
k[ ]02[
[ ]dry1 h1 [ ]dry2 h2]
0.5616(
17 5 16 5)
92.664xx = = ⋅γ ⋅ +γ ⋅ = ⋅ ⋅ + ⋅ =
σ′ [kPa]
( )
[ ][
[ ] [ ](
[ ]) ]
(
yy3 55)
0k.5616[
17h5 16 5h(
21 10)
5]
123y.55xx
3 water wet2 2 2
1 dry dry1 02 3
xx
=
⋅
− +
⋅ +
⋅
⋅
=
= σ′
⋅ γ
− γ +
⋅ γ +
⋅ γ
⋅
=
=
σ′ [kPa]
Discontinuità