skarpa grunty

Stateczność skarpy "Fundamentowanie 2"

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Ćwiczenie nr 3 z przedmiotu

FUNDAMENTOWANIE II Obliczanie stateczności skarpy metodą Felleniusa

Opracował: Konsultował: inż. Bartłomiej Durak dr inż. Krzysztof Wilk ________________________________________________________________________

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strona 1

Stateczność skarpy "Fundamentowanie 2" ________________________________________________________________________

METODA FELLENIUSA Metoda Felleniusa jest najstarszą z metod, które umożliwiają przeprowadzenie analizy stateczności dla różnych od prostoliniowej powierzchni poślizgu. Opracowana ona została na podstawie wyników badań Szwedzkiej Komisji Geotechnicznej, której prace prowadzone były w latach 1916-1925. Metoda ta wykorzystuje podział potencjalnej bryły osuwiskowej na bloki (paski) pionowe. Z powyższych względów metoda ta znana jest również pod nazwą metody Pettersona-Felleniusa lub metody szwedzkiej.W metodzie Felleniusa przyjęto następujące założenia: - powierzchnia poślizgu ma kształt walca cylindrycznego, - siły oddziaływania pomiędzy blokami są równoległe do podstawy bloku i nie wpływają na wartość reakcji normalnej do podstawy bloku oraz wartość sił oporu ścinania, - wskaźnik stateczności definiowany jest jako stosunek momentów sił biernych (utrzymujących równowagę) i sił czynnych (zsuwających).

Wypadkowa sił oddziaływania pomiędzy blokami wywołuje wprawdzie moment przy analizie pojedynczego bloku, ale ze względu na wewnętrzny charakter tych sił wywołany przez nie moment dla całej bryły względem dowolnego punktu powinien być równy zeru. metoda Felleniusa daje z reguły wyniki niższe niż inne metody analizy stateczności. W porównaniu z metodą Bishopa różnice te wynoszą od 5% do 20%, a niekiedy nawet do 60%. Zaniżone wartości wskaźników stateczności stawiają tą metodę w grupie metod bezpiecznych a nawet asekuracyjnych. Pomimo tego metoda ta jest często stosowana w praktyce, szczególnie wówczas, gdy sposób określania parametrów wytrzymałościowych ośrodka jest niezbyt dokładny. Dużą zaletą metody Felleniusa jest jej prostota. Jawna postać wzorów powoduje, że jej praktyczne wykorzystanie nie wymaga stosowania drogich programów obliczeniowych i komputerów.

1. Sprawdzenie statecznośći skarpy -grunt spoisty, promień R = 8.102m Założenia projektowe

H = 3.5m ; nachylenie 1:1,5 Rodzaj gruntu

Iły

IL = 0.0 ρB.n = 20.0

kN m

3

cu.n = 35kPa

ϕu.n = 6.5° ρD.n = 20.0

kN m

3

strona 2

Stateczność skarpy "Fundamentowanie 2"

strona 3

Stateczność skarpy "Fundamentowanie 2" ________________________________________________________________________

0

Hsk = 3.5m

01

γ1.sk = 26° 9

10

1

2

3

4

5

γ2.sk = 35°

H

8 7 6

Ti

4.5 ⋅Hsk = 15.75 m

Si Ni Gi

4,5H

________________________________________________________________________ 1.1 Pas nr 1 A - pole powierzchni danego pasa α - kąt nachylenia powierzchni poślizgu do terenu S - siła zsuwająca T - siła utrzymająca W - ciężar bryły pasa N - siła normalna

A1 = 0.048m

2

α1 = 11° 2

W1 = A1 ⋅ρB.n ⋅1m = 0.048 ⋅m ⋅20.0 ⋅

kN

⋅m = 0.96 ⋅kN 3 m kN 2 S1 = ρB.n ⋅A1 ⋅1m ⋅sin ( α1) = 20.0 ⋅ ⋅0.048 ⋅m ⋅m ⋅sin ( 11 ⋅°) = 0.18 ⋅kN 3 m kN 2 N1 = ρB.n ⋅A1 ⋅1m ⋅cos ( α1) = 20.0 ⋅ ⋅0.048 ⋅m ⋅m ⋅cos ( 11 ⋅°) = 0.94 ⋅kN 3 m T1 = N1 ⋅tan ( ϕu.n) + cu.n ⋅A1 = 0.94 ⋅kN ⋅tan ( 6.5 ⋅°) + 35 ⋅kPa ⋅0.048 ⋅m = 1.79 ⋅kN 2

1.2 Pas nr 2 A2 = 0.137m

2

α2 = 14°

strona 4

Stateczność skarpy "Fundamentowanie 2" 2

W2 = A2 ⋅ρB.n ⋅1m = 0.137 ⋅m ⋅20.0 ⋅

kN

⋅m = 2.74 ⋅kN 3 m kN 2 S2 = ρB.n ⋅A2 ⋅1m ⋅sin ( α2) = 20.0 ⋅ ⋅0.137 ⋅m ⋅m ⋅sin ( 14 ⋅°) = 0.66 ⋅kN 3 m kN 2 N2 = ρB.n ⋅A2 ⋅1m ⋅cos ( α2) = 20.0 ⋅ ⋅0.137 ⋅m ⋅m ⋅cos ( 14 ⋅°) = 2.66 ⋅kN 3 m T2 = N2 ⋅tan ( ϕu.n) + cu.n ⋅A2 = 2.66 ⋅kN ⋅tan ( 6.5 ⋅°) + 35 ⋅kPa ⋅0.137 ⋅m = 5.1 ⋅kN 2

1.3 Pas nr 3 A3 = 0.214m

2

α3 = 17° 2

W3 = A3 ⋅ρB.n ⋅1m = 0.214 ⋅m ⋅20.0 ⋅

kN

⋅m = 4.28 ⋅kN 3 m kN 2 S3 = ρB.n ⋅A3 ⋅1m ⋅sin ( α3) = 20.0 ⋅ ⋅0.214 ⋅m ⋅m ⋅sin ( 17 ⋅°) = 1.25 ⋅kN 3 m kN 2 N3 = ρB.n ⋅A3 ⋅1m ⋅cos ( α3) = 20.0 ⋅ ⋅0.214 ⋅m ⋅m ⋅cos ( 17 ⋅°) = 4.09 ⋅kN 3 m T3 = N3 ⋅tan ( ϕu.n) + cu.n ⋅A3 = 4.09 ⋅kN ⋅tan ( 6.5 ⋅°) + 35 ⋅kPa ⋅0.214 ⋅m = 7.96 ⋅kN 2

1.4 Pas nr 4 A4 = 0.278m

2

α4 = 21° 2

W4 = A4 ⋅ρB.n ⋅1m = 0.278 ⋅m ⋅20.0 ⋅

kN

⋅m = 5.56 ⋅kN 3 m kN 2 S4 = ρB.n ⋅A4 ⋅1m ⋅sin ( α4) = 20.0 ⋅ ⋅0.278 ⋅m ⋅m ⋅sin ( 21 ⋅°) = 1.99 ⋅kN 3 m kN 2 N4 = ρB.n ⋅A4 ⋅1m ⋅cos ( α4) = 20.0 ⋅ ⋅0.278 ⋅m ⋅m ⋅cos ( 21 ⋅°) = 5.19 ⋅kN 3 m T4 = N4 ⋅tan ( ϕu.n) + cu.n ⋅A4 = 5.19 ⋅kN ⋅tan ( 6.5 ⋅°) + 35 ⋅kPa ⋅0.278 ⋅m = 10.32 ⋅kN 2

1.5 Pas nr 5 A5 = 0.328m

2

α5 = 24°

kN 2 W5 = A5 ⋅ρB.n ⋅1m = 0.328 ⋅m ⋅20.0 ⋅ ⋅m = 6.56 ⋅kN 3 m

strona 5

Stateczność skarpy "Fundamentowanie 2" S5 = ρB.n ⋅A5 ⋅1m ⋅sin ( α5) = 20.0 ⋅

kN 3

⋅0.328 ⋅m ⋅m ⋅sin ( 24 ⋅°) = 2.67 ⋅kN 2

m kN 2 N5 = ρB.n ⋅A5 ⋅1m ⋅cos ( α5) = 20.0 ⋅ ⋅0.328 ⋅m ⋅m ⋅cos ( 24 ⋅°) = 5.99 ⋅kN 3 m T5 = N5 ⋅tan ( ϕu.n) + cu.n ⋅A5 = 5.99 ⋅kN ⋅tan ( 6.5 ⋅°) + 35 ⋅kPa ⋅0.328 ⋅m = 12.16 ⋅kN 2

1.6 Pas nr 6 A6 = 0.363m

2

α6 = 27° 2

W6 = A6 ⋅ρB.n ⋅1m = 0.363 ⋅m ⋅20.0 ⋅

kN

⋅m = 7.26 ⋅kN 3 m kN 2 S6 = ρB.n ⋅A6 ⋅1m ⋅sin ( α6) = 20.0 ⋅ ⋅0.363 ⋅m ⋅m ⋅sin ( 27 ⋅°) = 3.3 ⋅kN 3 m kN 2 N6 = ρB.n ⋅A6 ⋅1m ⋅cos ( α6) = 20.0 ⋅ ⋅0.363 ⋅m ⋅m ⋅cos ( 27 ⋅°) = 6.47 ⋅kN 3 m T6 = N6 ⋅tan ( ϕu.n) + cu.n ⋅A6 = 6.47 ⋅kN ⋅tan ( 6.5 ⋅°) + 35 ⋅kPa ⋅0.363 ⋅m = 13.44 ⋅kN 2

1.7 Pas nr 7 A7 = 0.381m

2

α7 = 31° 2

W7 = A7 ⋅ρB.n ⋅1m = 0.381 ⋅m ⋅20.0 ⋅

kN

⋅m = 7.62 ⋅kN 3 m kN 2 S7 = ρB.n ⋅A7 ⋅1m ⋅sin ( α7) = 20.0 ⋅ ⋅0.381 ⋅m ⋅m ⋅sin ( 31 ⋅°) = 3.92 ⋅kN 3 m kN 2 N7 = ρB.n ⋅A7 ⋅1m ⋅cos ( α7) = 20.0 ⋅ ⋅0.381 ⋅m ⋅m ⋅cos ( 31 ⋅°) = 6.53 ⋅kN 3 m T7 = N7 ⋅tan ( ϕu.n) + cu.n ⋅A7 = 6.53 ⋅kN ⋅tan ( 6.5 ⋅°) + 35 ⋅kPa ⋅0.381 ⋅m = 14.08 ⋅kN 2

1.8 Pas nr 8 A8 = 0.381m

2

α8 = 35° 2

W8 = A8 ⋅ρB.n ⋅1m = 0.381 ⋅m ⋅20.0 ⋅

kN m

S8 = ρB.n ⋅A8 ⋅1m ⋅sin ( α8) = 20.0 ⋅

kN m

3

3

⋅m = 7.62 ⋅kN

⋅0.381 ⋅m ⋅m ⋅sin ( 35 ⋅°) = 4.37 ⋅kN 2

strona 6

Stateczność skarpy "Fundamentowanie 2" N8 = ρB.n ⋅A8 ⋅1m ⋅cos ( α8) = 20.0 ⋅

kN m

3

⋅0.381 ⋅m ⋅m ⋅cos ( 35 ⋅°) = 6.24 ⋅kN 2

T8 = N8 ⋅tan ( ϕu.n) + cu.n ⋅A8 = 6.24 ⋅kN ⋅tan ( 6.5 ⋅°) + 35 ⋅kPa ⋅0.381 ⋅m = 14.05 ⋅kN 2

1.9 Pas nr 9 A9 = 0.360m

2

α9 = 39° 2

W9 = A9 ⋅ρB.n ⋅1m = 0.360 ⋅m ⋅20.0 ⋅

kN m

S9 = ρB.n ⋅A9 ⋅1m ⋅sin ( α9) = 20.0 ⋅

kN m

N9 = ρB.n ⋅A9 ⋅1m ⋅cos ( α9) = 20.0 ⋅

3

⋅m = 7.2 ⋅kN

⋅0.360 ⋅m ⋅m ⋅sin ( 39 ⋅°) = 4.53 ⋅kN 2

kN m

3

3

⋅0.360 ⋅m ⋅m ⋅cos ( 39 ⋅°) = 5.6 ⋅kN 2

T9 = N9 ⋅tan ( ϕu.n) + cu.n ⋅A9 = 5.6 ⋅kN ⋅tan ( 6.5 ⋅°) + 35 ⋅kPa ⋅0.360 ⋅m = 13.24 ⋅kN 2

1.10 Pas nr 10 A10 = 0.313m

2

α10 = 43° 2

W10 = A10 ⋅ρB.n ⋅1m = 0.313 ⋅m ⋅20.0 ⋅

kN m

S10 = ρB.n ⋅A10 ⋅1m ⋅sin ( α10) = 20.0 ⋅

kN m

N10 = ρB.n ⋅A10 ⋅1m ⋅cos ( α10) = 20.0 ⋅

3

3

⋅0.313 ⋅m ⋅m ⋅sin ( 43 ⋅°) = 4.27 ⋅kN 2

kN m

⋅m = 6.26 ⋅kN

3

⋅0.313 ⋅m ⋅m ⋅cos ( 43 ⋅°) = 4.58 ⋅kN 2

T10 = N10 ⋅tan ( ϕu.n) + cu.n ⋅A10 = 4.58 ⋅kN ⋅tan ( 6.5 ⋅°) + 35 ⋅kPa ⋅0.313 ⋅m = 11.48 ⋅kN 2

1.11 Pas nr 11 A11 = 0.236m

2

α11 = 47°

kN 2 W11 = A11 ⋅ρB.n ⋅1m = 0.236 ⋅m ⋅20.0 ⋅ ⋅m = 4.72 ⋅kN 3 m S11 = ρB.n ⋅A11 ⋅1m ⋅sin ( α11) = 20.0 ⋅

kN m

N11 = ρB.n ⋅A11 ⋅1m ⋅cos ( α11) = 20.0 ⋅

3

⋅0.236 ⋅m ⋅m ⋅sin ( 47 ⋅°) = 3.45 ⋅kN

kN m

3

2

⋅0.236 ⋅m ⋅m ⋅cos ( 47 ⋅°) = 3.22 ⋅kN 2

strona 7

Stateczność skarpy "Fundamentowanie 2" T11 = N11 ⋅tan ( ϕu.n) + cu.n ⋅A11 = 3.22 ⋅kN ⋅tan ( 6.5 ⋅°) + 35 ⋅kPa ⋅0.236 ⋅m = 8.63 ⋅kN 2

1.12 Pas nr 12 A12 = 0.236m

2

α12 = 47°

kN 2 W12 = A12 ⋅ρB.n ⋅1m = 0.236 ⋅m ⋅20.0 ⋅ ⋅m = 4.72 ⋅kN 3 m S12 = ρB.n ⋅A12 ⋅1m ⋅sin ( α12) = 20.0 ⋅

kN m

N12 = ρB.n ⋅A12 ⋅1m ⋅cos ( α12) = 20.0 ⋅

3

⋅0.236 ⋅m ⋅m ⋅sin ( 47 ⋅°) = 3.45 ⋅kN

kN m

3

2

⋅0.236 ⋅m ⋅m ⋅cos ( 47 ⋅°) = 3.22 ⋅kN 2

T12 = N12 ⋅tan ( ϕu.n) + cu.n ⋅A12 = 3.22 ⋅kN ⋅tan ( 6.5 ⋅°) + 35 ⋅kPa ⋅0.236 ⋅m = 8.63 ⋅kN 2

 ( T1 + T2 + T3 + T4 + T5 + T6 + T7 + T8 + T9 + T10 + T11 + T12)    ( S1 + S2 + S3 + S4 + S5 + S6 + S7 + S8 + S9 + S10 + S11 + S12) 

n = 

1.79 ⋅kN + 5.1 ⋅kN + 7.96 ⋅kN + 10.32 ⋅kN + 12.16 ⋅kN + 13.44 ⋅kN + 14.08 ⋅kN + 14.05 ⋅kN + 13.24 ⋅kN + 11.48 ⋅kN + 8.63 ⋅kN + 8.63 ⋅kN n = 0.18 ⋅kN + 0.66 ⋅kN + 1.25 ⋅kN + 1.99 ⋅kN + 2.67 ⋅kN + 3.3 ⋅kN + 3.92 ⋅kN + 4.37 ⋅kN + 4.53 ⋅kN + 4.27 ⋅kN + 3.45 ⋅kN + 3.45 ⋅kN

= 3.551 > 1.

M0 = S1 + S2 + S3 + S4 + S5 + S6 + S7 + S8 + S9 + S10 + S11 + S12 M0 = 0.18 ⋅kN + 0.66 ⋅kN + 1.25 ⋅kN + 1.99 ⋅kN + 2.67 ⋅kN + 3.3 ⋅kN = 34.04 ⋅kN + 3.92 ⋅kN + 4.37 ⋅kN + 4.53 ⋅kN + 4.27 ⋅kN + 3.45 ⋅kN + 3.45 ⋅kN Mu = T1 + T2 + T3 + T4 + T5 + T6 + T7 + T8 + T9 + T10 + T11 + T12 Mu = 1.79 ⋅kN + 5.1 ⋅kN + 7.96 ⋅kN + 10.32 ⋅kN + 12.16 ⋅kN + 13.44 ⋅kN = 103.62 ⋅kN + 14.08 ⋅kN + 14.05 ⋅kN + 13.24 ⋅kN + 11.48 ⋅kN Mu M0

=

103.62 ⋅kN = 3.044 > 1.

34.04 ⋅kN

________________________________________________________________________ 2. Sprawdzenie statecznośći skarpy -grunt spoisty, promień R = 6.100m Założenia projektowe

H = 3.5m ; nachylenie 1:1,5 Rodzaj gruntu IL = 0.0

Iły cu.n = 35kPa

strona 8

Stateczność skarpy "Fundamentowanie 2" ρB.n = 20.0

kN m

3

ϕu.n = 6.5° ρD.n = 20.0

kN m

3

________________________________________________________________________

0

Hsk = 3.5m

01

1

2

3

4

5

m +1

γ1.sk = 26° H

 1− HK  iλ =  9 10  8 V + A' c' ctgΦ '  d  7 6

Ti Si Ni Gi

γ2.sk = 35°

4.5 ⋅Hsk = 15.75 m

4,5H

________________________________________________________________________ 2.1 Pas nr 1

A - pole powierzchni danego pasa α - kąt nachylenia powierzchni poślizgu do terenu S - siła zsuwająca T - siła utrzymająca W - ciężar bryły pasa N - siła normalna

A1 = 0.192m

2

α1 = −10°

kN 2 W1 = A1 ⋅ρB.n ⋅1m = 0.192 ⋅m ⋅20.0 ⋅ ⋅m = 3.84 ⋅kN 3 m

strona 9

Stateczność skarpy "Fundamentowanie 2" S1 = ρB.n ⋅A1 ⋅1m ⋅sin ( α1) = 20.0 ⋅

kN 3

⋅0.192 ⋅m ⋅m ⋅sin ( −10 ⋅°) = −0.67 ⋅kN 2

m kN 2 N1 = ρB.n ⋅A1 ⋅1m ⋅cos ( α1) = 20.0 ⋅ ⋅0.192 ⋅m ⋅m ⋅cos ( −10 ⋅°) = 3.78 ⋅kN 3 m T1 = N1 ⋅tan ( ϕu.n) + cu.n ⋅A1 = 3.78 ⋅kN ⋅tan ( 6.5 ⋅°) + 35 ⋅kPa ⋅0.192 ⋅m = 7.15 ⋅kN 2

2.2 Pas nr 2 A2 = 0.552m

2

α2 = −10°

kN 2 W2 = A2 ⋅ρB.n ⋅1m = 0.552 ⋅m ⋅20.0 ⋅ ⋅m = 11.04 ⋅kN 3 m kN 2 S2 = ρB.n ⋅A2 ⋅1m ⋅sin ( α2) = 20.0 ⋅ ⋅0.552 ⋅m ⋅m ⋅sin ( −10 ⋅°) = −1.92 ⋅kN 3 m kN 2 N2 = ρB.n ⋅A2 ⋅1m ⋅cos ( α2) = 20.0 ⋅ ⋅0.552 ⋅m ⋅m ⋅cos ( −10 ⋅°) = 10.87 ⋅kN 3 m T2 = N2 ⋅tan ( ϕu.n) + cu.n ⋅A2 = 10.87 ⋅kN ⋅tan ( 6.5 ⋅°) + 35 ⋅kPa ⋅0.552 ⋅m = 20.56 ⋅kN 2

2.3 Pas nr 3 A3 = 0.868m

2

α 3 = − 4° 2

W3 = A3 ⋅ρB.n ⋅1m = 0.868 ⋅m ⋅20.0 ⋅

kN

⋅m = 17.36 ⋅kN 3 m kN 2 S3 = ρB.n ⋅A3 ⋅1m ⋅sin ( α3) = 20.0 ⋅ ⋅0.868 ⋅m ⋅m ⋅sin ( −4 ⋅°) = −1.21 ⋅kN 3 m kN 2 N3 = ρB.n ⋅A3 ⋅1m ⋅cos ( α3) = 20.0 ⋅ ⋅0.868 ⋅m ⋅m ⋅cos ( −4 ⋅°) = 17.3 ⋅kN 3 m T3 = N3 ⋅tan ( ϕu.n) + cu.n ⋅A3 = 17.3 ⋅kN ⋅tan ( 6.5 ⋅°) + 35 ⋅kPa ⋅0.868 ⋅m = 32.35 ⋅kN 2

2.4 Pas nr 4 A4 = 1.142m

2

α4 = 2°

kN 2 W4 = A4 ⋅ρB.n ⋅1m = 1.142 ⋅m ⋅20.0 ⋅ ⋅m = 22.84 ⋅kN 3 m kN 2 S4 = ρB.n ⋅A4 ⋅1m ⋅sin ( α4) = 20.0 ⋅ ⋅1.142 ⋅m ⋅m ⋅sin ( 2 ⋅°) = 0.8 ⋅kN 3 m kN 2 N4 = ρB.n ⋅A4 ⋅1m ⋅cos ( α4) = 20.0 ⋅ ⋅1.142 ⋅m ⋅m ⋅cos ( 2 ⋅°) = 22.83 ⋅kN 3 m T4 = N4 ⋅tan ( ϕu.n) + cu.n ⋅A4 = 22.83 ⋅kN ⋅tan ( 6.5 ⋅°) + 35 ⋅kPa ⋅1.142 ⋅m = 42.57 ⋅kN 2

strona 10

Stateczność skarpy "Fundamentowanie 2"

2.5 Pas nr 5 A5 = 1.37m

2

α5 = 8°

kN 2 W5 = A5 ⋅ρB.n ⋅1m = 1.37 ⋅m ⋅20.0 ⋅ ⋅m = 27.4 ⋅kN 3 m kN 2 S5 = ρB.n ⋅A5 ⋅1m ⋅sin ( α5) = 20.0 ⋅ ⋅1.37 ⋅m ⋅m ⋅sin ( 8 ⋅°) = 3.81 ⋅kN 3 m kN 2 N5 = ρB.n ⋅A5 ⋅1m ⋅cos ( α5) = 20.0 ⋅ ⋅1.37 ⋅m ⋅m ⋅cos ( 8 ⋅°) = 27.13 ⋅kN 3 m T5 = N5 ⋅tan ( ϕu.n) + cu.n ⋅A5 = 27.13 ⋅kN ⋅tan ( 6.5 ⋅°) + 35 ⋅kPa ⋅1.37 ⋅m = 51.04 ⋅kN 2

2.6 Pas nr 6 A6 = 1.560m

2

α6 = 14°

kN 2 W6 = A6 ⋅ρB.n ⋅1m = 1.560 ⋅m ⋅20.0 ⋅ ⋅m = 31.2 ⋅kN 3 m kN 2 S6 = ρB.n ⋅A6 ⋅1m ⋅sin ( α6) = 20.0 ⋅ ⋅1.560 ⋅m ⋅m ⋅sin ( 14 ⋅°) = 7.55 ⋅kN 3 m kN 2 N6 = ρB.n ⋅A6 ⋅1m ⋅cos ( α6) = 20.0 ⋅ ⋅1.560 ⋅m ⋅m ⋅cos ( 14 ⋅°) = 30.27 ⋅kN 3 m T6 = N6 ⋅tan ( ϕu.n) + cu.n ⋅A6 = 30.27 ⋅kN ⋅tan ( 6.5 ⋅°) + 35 ⋅kPa ⋅1.560 ⋅m = 58.05 ⋅kN 2

2.7 Pas nr 7 A7 = 1.70m

2

α7 = 20°

kN 2 W7 = A7 ⋅ρB.n ⋅1m = 1.70 ⋅m ⋅20.0 ⋅ ⋅m = 34 ⋅kN 3 m kN 2 S7 = ρB.n ⋅A7 ⋅1m ⋅sin ( α7) = 20.0 ⋅ ⋅1.70 ⋅m ⋅m ⋅sin ( 20 ⋅°) = 11.63 ⋅kN 3 m kN 2 N7 = ρB.n ⋅A7 ⋅1m ⋅cos ( α7) = 20.0 ⋅ ⋅1.70 ⋅m ⋅m ⋅cos ( 20 ⋅°) = 31.95 ⋅kN 3 m T7 = N7 ⋅tan ( ϕu.n) + cu.n ⋅A7 = 31.95 ⋅kN ⋅tan ( 6.5 ⋅°) + 35 ⋅kPa ⋅1.70 ⋅m = 63.14 ⋅kN 2

2.8 Pas nr 8 A8 = 1.79m

2

α8 = 27°

strona 11

Stateczność skarpy "Fundamentowanie 2" 2

W8 = A8 ⋅ρB.n ⋅1m = 1.79 ⋅m ⋅20.0 ⋅

kN m

S8 = ρB.n ⋅A8 ⋅1m ⋅sin ( α8) = 20.0 ⋅

kN m

N8 = ρB.n ⋅A8 ⋅1m ⋅cos ( α8) = 20.0 ⋅

3

3

⋅1.79 ⋅m ⋅m ⋅sin ( 27 ⋅°) = 16.25 ⋅kN 2

kN m

⋅m = 35.8 ⋅kN

3

⋅1.79 ⋅m ⋅m ⋅cos ( 27 ⋅°) = 31.9 ⋅kN 2

T8 = N8 ⋅tan ( ϕu.n) + cu.n ⋅A8 = 31.9 ⋅kN ⋅tan ( 6.5 ⋅°) + 35 ⋅kPa ⋅1.79 ⋅m = 66.28 ⋅kN 2

2.9 Pas nr 9 A9 = 1.74m

2

α9 = 33° 2

W9 = A9 ⋅ρB.n ⋅1m = 1.74 ⋅m ⋅20.0 ⋅

kN m

S9 = ρB.n ⋅A9 ⋅1m ⋅sin ( α9) = 20.0 ⋅

kN m

N9 = ρB.n ⋅A9 ⋅1m ⋅cos ( α9) = 20.0 ⋅

3

3

⋅1.74 ⋅m ⋅m ⋅sin ( 33 ⋅°) = 18.95 ⋅kN 2

kN m

⋅m = 34.8 ⋅kN

3

⋅1.74 ⋅m ⋅m ⋅cos ( 33 ⋅°) = 29.19 ⋅kN 2

T9 = N9 ⋅tan ( ϕu.n) + cu.n ⋅A9 = 29.19 ⋅kN ⋅tan ( 6.5 ⋅°) + 35 ⋅kPa ⋅1.74 ⋅m = 64.23 ⋅kN 2

2.10 Pas nr 10 A10 = 1.43m

2

α10 = 40° 2

W10 = A10 ⋅ρB.n ⋅1m = 1.43 ⋅m ⋅20.0 ⋅

kN m

S10 = ρB.n ⋅A10 ⋅1m ⋅sin ( α10) = 20.0 ⋅

3

kN m

N10 = ρB.n ⋅A10 ⋅1m ⋅cos ( α10) = 20.0 ⋅

3

⋅1.43 ⋅m ⋅m ⋅sin ( 40 ⋅°) = 18.38 ⋅kN

kN m

⋅m = 28.6 ⋅kN

3

2

⋅1.43 ⋅m ⋅m ⋅cos ( 40 ⋅°) = 21.91 ⋅kN 2

T10 = N10 ⋅tan ( ϕu.n) + cu.n ⋅A10 = 21.91 ⋅kN ⋅tan ( 6.5 ⋅°) + 35 ⋅kPa ⋅1.43 ⋅m = 52.55 ⋅kN 2

2.11 Pas nr 11 A11 = 1.00m

2

α11 = 48°

kN 2 W11 = A11 ⋅ρB.n ⋅1m = 1.00 ⋅m ⋅20.0 ⋅ ⋅m = 20 ⋅kN 3 m

strona 12

Stateczność skarpy "Fundamentowanie 2" S11 = ρB.n ⋅A11 ⋅1m ⋅sin ( α11) = 20.0 ⋅

kN m

N11 = ρB.n ⋅A11 ⋅1m ⋅cos ( α11) = 20.0 ⋅

3

⋅1.00 ⋅m ⋅m ⋅sin ( 48 ⋅°) = 14.86 ⋅kN 2

kN m

3

⋅1.00 ⋅m ⋅m ⋅cos ( 48 ⋅°) = 13.38 ⋅kN 2

T11 = N11 ⋅tan ( ϕu.n) + cu.n ⋅A11 = 13.38 ⋅kN ⋅tan ( 6.5 ⋅°) + 35 ⋅kPa ⋅1.00 ⋅m = 36.52 ⋅kN 2

2.12 Pas nr 12 A12 = 0.38m

2

α12 = 57°

kN 2 W12 = A12 ⋅ρB.n ⋅1m = 0.38 ⋅m ⋅20.0 ⋅ ⋅m = 7.6 ⋅kN 3 m S12 = ρB.n ⋅A12 ⋅1m ⋅sin ( α12) = 20.0 ⋅

kN m

N12 = ρB.n ⋅A12 ⋅1m ⋅cos ( α12) = 20.0 ⋅

3

⋅0.38 ⋅m ⋅m ⋅sin ( 57 ⋅°) = 6.37 ⋅kN

kN m

3

2

⋅0.38 ⋅m ⋅m ⋅cos ( 57 ⋅°) = 4.14 ⋅kN 2

T12 = N12 ⋅tan ( ϕu.n) + cu.n ⋅A12 = 4.14 ⋅kN ⋅tan ( 6.5 ⋅°) + 35 ⋅kPa ⋅0.38 ⋅m = 13.77 ⋅kN 2

 ( T1 + T2 + T3 + T4 + T5 + T6 + T7 + T8 + T9 + T10 + T11 + T12)    ( S1 + S2 + S3 + S4 + S5 + S6 + S7 + S8 + S9 + S10 + S11 + S12) 

n = 

7.15 ⋅kN + 20.56 ⋅kN + 32.35 ⋅kN + 42.57 ⋅kN + 51.04 ⋅kN + 58.05 ⋅kN + 63.14 ⋅kN + 66.28 ⋅kN + 64.23 ⋅kN + 52.55 ⋅kN + 36.52 ⋅kN + 13.77 ⋅kN n = −0.67 ⋅kN + −1.92 ⋅kN + −1.21 ⋅kN + 0.8 ⋅kN + 3.81 ⋅kN + 7.55 ⋅kN + 11.63 ⋅kN + 16.25 ⋅kN + 18.95 ⋅kN + 18.38 ⋅kN + 14.86 ⋅kN + 6.37 ⋅kN

= 5.361 > 1

M0 = S1 + S2 + S3 + S4 + S5 + S6 + S7 + S8 + S9 + S10 + S11 + S12 M0 = −0.67 ⋅kN − 1.92 ⋅kN − 1.21 ⋅kN + 0.8 ⋅kN + 3.81 ⋅kN + 7.55 ⋅kN + 11.63 ⋅kN + 16.25 ⋅kN + 18.95 ⋅kN + 18.38 ⋅kN + 14.86 ⋅kN + 6.37 ⋅kN

= 94.8 ⋅kN

Mu = T1 + T2 + T3 + T4 + T5 + T6 + T7 + T8 + T9 + T10 + T11 + T12 Mu = 7.15 ⋅kN + 20.56 ⋅kN + 32.35 ⋅kN + 42.57 ⋅kN + 51.04 ⋅kN + 58.05 ⋅kN + 63.14 ⋅kN + 66.28 ⋅kN + 64.23 ⋅kN + 52.55 ⋅kN + 36.52 ⋅kN + 13.77 ⋅kN Mu M0

=

= 508.21

508.21 ⋅kN = 5.361 > 1.

94.8 ⋅kN

________________________________________________________________________ 3. Sprawdzenie statecznośći skarpy -grunt lużny,płaszczyzna poślizgu równoległa do

strona 13

Stateczność skarpy "Fundamentowanie 2" płaszczyzny skarpy. Kąt nachylenia skarpy 34 deg Założenia projektowe

H = 3.5m ; nachylenie 1:1,5 Rodzaj gruntu

Piasek drobny

ID = 0.5 ρB.n = 16.5

kN m

3

cu.n = 0kPa

ϕu.n = 30.5° ρD.n = 16.5

kN m

3

________________________________________________________________________

________________________________________________________________________

strona 14

Stateczność skarpy "Fundamentowanie 2"

3.1 Pas nr 1

A - pole powierzchni danego pasa α - kąt nachylenia powierzchni poślizgu do terenu S - siła zsuwająca T - siła utrzymająca W - ciężar bryły pasa N - siła normalna

A1 = 0.144m

2

α1 = 0°

kN 2 W1 = A1 ⋅ρB.n ⋅1m = 0.144 ⋅m ⋅16.5 ⋅ ⋅m = 2.376 ⋅kN 3 m kN 2 S1 = ρB.n ⋅A1 ⋅1m ⋅sin ( α1) = 16.5 ⋅ ⋅0.144 ⋅m ⋅m ⋅sin ( 0 ⋅°) = 0 ⋅kN 3 m kN 2 N1 = ρB.n ⋅A1 ⋅1m ⋅cos ( α1) = 16.5 ⋅ ⋅0.144 ⋅m ⋅m ⋅cos ( 0 ⋅°) = 2.38 ⋅kN 3 m T1 = N1 ⋅tan ( ϕu.n) + cu.n ⋅A1 = 2.38 ⋅kN ⋅tan ( 30.5 ⋅°) + 0 ⋅kPa ⋅0.144 ⋅m = 1.4 ⋅kN 2

3.2 Pas nr 2 A2 = 0.431m

2

α2 = 0° 2

W2 = A2 ⋅ρB.n ⋅1m = 0.431 ⋅m ⋅16.5 ⋅

kN m

S2 = ρB.n ⋅A2 ⋅1m ⋅sin ( α2) = 16.5 ⋅

kN 3

3

⋅m = 7.112 ⋅kN

⋅0.431 ⋅m ⋅m ⋅sin ( 0 ⋅°) = 0 ⋅kN 2

m kN 2 N2 = ρB.n ⋅A2 ⋅1m ⋅cos ( α2) = 16.5 ⋅ ⋅0.431 ⋅m ⋅m ⋅cos ( 0 ⋅°) = 7.11 ⋅kN 3 m T2 = N2 ⋅tan ( ϕu.n) + cu.n ⋅A2 = 7.11 ⋅kN ⋅tan ( 30.5 ⋅°) + 0 ⋅kPa ⋅0.431 ⋅m = 4.19 ⋅kN 2

3.3 Pas nr 3 A3 = 0.571m

2

α3 = 34° 2

W3 = A3 ⋅ρB.n ⋅1m = 0.571 ⋅m ⋅16.5 ⋅

kN m

3

⋅m = 9.421 ⋅kN

strona 15

Stateczność skarpy "Fundamentowanie 2" S3 = ρB.n ⋅A3 ⋅1m ⋅sin ( α3) = 16.5 ⋅

kN 3

⋅0.571 ⋅m ⋅m ⋅sin ( 34 ⋅°) = 5.27 ⋅kN 2

m kN 2 N3 = ρB.n ⋅A3 ⋅1m ⋅cos ( α3) = 16.5 ⋅ ⋅0.571 ⋅m ⋅m ⋅cos ( 34 ⋅°) = 7.8 ⋅kN 3 m T3 = N3 ⋅tan ( ϕu.n) + cu.n ⋅A3 = 7.8 ⋅kN ⋅tan ( 30.5 ⋅°) + 0 ⋅kPa ⋅0.571 ⋅m = 4.59 ⋅kN 2

3.4 Pas nr 4 A4 = 0.5711m

2

α4 = 34°

kN 2 W4 = A4 ⋅ρB.n ⋅1m = 0.5711 ⋅m ⋅16.5 ⋅ ⋅m = 9.423 ⋅kN 3 m kN 2 S4 = ρB.n ⋅A4 ⋅1m ⋅sin ( α4) = 16.5 ⋅ ⋅0.5711 ⋅m ⋅m ⋅sin ( 34 ⋅°) = 5.27 ⋅kN 3 m kN 2 N4 = ρB.n ⋅A4 ⋅1m ⋅cos ( α4) = 16.5 ⋅ ⋅0.5711 ⋅m ⋅m ⋅cos ( 34 ⋅°) = 7.81 ⋅kN 3 m T4 = N4 ⋅tan ( ϕu.n) + cu.n ⋅A4 = 7.81 ⋅kN ⋅tan ( 30.5 ⋅°) + 0 ⋅kPa ⋅0.5711 ⋅m = 4.6 ⋅kN 2

3.5 Pas nr 5 A5 = 0.571m

2

α5 = 34° 2

W5 = A5 ⋅ρB.n ⋅1m = 0.571 ⋅m ⋅16.5 ⋅

kN

⋅m = 9.421 ⋅kN 3 m kN 2 S5 = ρB.n ⋅A5 ⋅1m ⋅sin ( α5) = 16.5 ⋅ ⋅0.571 ⋅m ⋅m ⋅sin ( 34 ⋅°) = 5.27 ⋅kN 3 m kN 2 N5 = ρB.n ⋅A5 ⋅1m ⋅cos ( α5) = 16.5 ⋅ ⋅0.571 ⋅m ⋅m ⋅cos ( 34 ⋅°) = 7.81 ⋅kN 3 m T5 = N5 ⋅tan ( ϕu.n) + cu.n ⋅A5 = 7.81 ⋅kN ⋅tan ( 30.5 ⋅°) + 0 ⋅kPa ⋅0.571 ⋅m = 4.6 ⋅kN 2

3.6 Pas nr 6 A6 = 0.571m

2

α6 = 34° 2

W6 = A6 ⋅ρB.n ⋅1m = 0.571 ⋅m ⋅16.5 ⋅

kN m

3

⋅m = 9.421 ⋅kN

strona 16

Stateczność skarpy "Fundamentowanie 2" S6 = ρB.n ⋅A6 ⋅1m ⋅sin ( α6) = 16.5 ⋅

kN 3

⋅0.571 ⋅m ⋅m ⋅sin ( 34 ⋅°) = 5.27 ⋅kN 2

m kN 2 N6 = ρB.n ⋅A6 ⋅1m ⋅cos ( α6) = 16.5 ⋅ ⋅0.571 ⋅m ⋅m ⋅cos ( 34 ⋅°) = 7.81 ⋅kN 3 m T6 = N6 ⋅tan ( ϕu.n) + cu.n ⋅A6 = 7.81 ⋅kN ⋅tan ( 30.5 ⋅°) + 0 ⋅kPa ⋅0.571 ⋅m = 4.6 ⋅kN 2

3.7 Pas nr 7 A7 = 0.571m

2

α7 = 34° 2

W7 = A7 ⋅ρB.n ⋅1m = 0.571 ⋅m ⋅16.5 ⋅

kN

⋅m = 9.421 ⋅kN 3 m kN 2 S7 = ρB.n ⋅A7 ⋅1m ⋅sin ( α7) = 16.5 ⋅ ⋅0.571 ⋅m ⋅m ⋅sin ( 34 ⋅°) = 5.27 ⋅kN 3 m kN 2 N7 = ρB.n ⋅A7 ⋅1m ⋅cos ( α7) = 16.5 ⋅ ⋅0.571 ⋅m ⋅m ⋅cos ( 34 ⋅°) = 7.81 ⋅kN 3 m T7 = N7 ⋅tan ( ϕu.n) + cu.n ⋅A7 = 7.81 ⋅kN ⋅tan ( 30.5 ⋅°) + 0 ⋅kPa ⋅0.571 ⋅m = 4.6 ⋅kN 2

3.8 Pas nr 8 A8 = 0.571m

2

α8 = 34° 2

W8 = A8 ⋅ρB.n ⋅1m = 0.571 ⋅m ⋅16.5 ⋅

kN m

S8 = ρB.n ⋅A8 ⋅1m ⋅sin ( α8) = 16.5 ⋅

kN m

N8 = ρB.n ⋅A8 ⋅1m ⋅cos ( α8) = 16.5 ⋅

3

⋅m = 9.421 ⋅kN

⋅0.571 ⋅m ⋅m ⋅sin ( 34 ⋅°) = 5.27 ⋅kN 2

kN m

3

3

⋅0.571 ⋅m ⋅m ⋅cos ( 34 ⋅°) = 7.81 ⋅kN 2

T8 = N8 ⋅tan ( ϕu.n) + cu.n ⋅A8 = 31.9 ⋅kN ⋅tan ( 30.5 ⋅°) + 0 ⋅kPa ⋅0.571 ⋅m = 18.79 ⋅kN 2

3.9 Pas nr 9 A9 = 0.431m

2

α9 = 34° 2

W9 = A9 ⋅ρB.n ⋅1m = 0.431 ⋅m ⋅16.5 ⋅

kN m

S9 = ρB.n ⋅A9 ⋅1m ⋅sin ( α9) = 16.5 ⋅

kN m

3

3

⋅m = 7.112 ⋅kN

⋅0.431 ⋅m ⋅m ⋅sin ( 34 ⋅°) = 3.98 ⋅kN 2

strona 17

Stateczność skarpy "Fundamentowanie 2" N9 = ρB.n ⋅A9 ⋅1m ⋅cos ( α9) = 16.5 ⋅

kN m

3

⋅0.431 ⋅m ⋅m ⋅cos ( 34 ⋅°) = 5.9 ⋅kN 2

T9 = N9 ⋅tan ( ϕu.n) + cu.n ⋅A9 = 5.9 ⋅kN ⋅tan ( 30.5 ⋅°) + 0 ⋅kPa ⋅0.431 ⋅m = 3.48 ⋅kN 2

3.10 Pas nr 10 A10 = 0.144m

2

α10 = 34° 2

W10 = A10 ⋅ρB.n ⋅1m = 0.144 ⋅m ⋅16.5 ⋅

kN m

S10 = ρB.n ⋅A10 ⋅1m ⋅sin ( α10) = 16.5 ⋅

kN m

N10 = ρB.n ⋅A10 ⋅1m ⋅cos ( α10) = 16.5 ⋅

3

3

⋅0.144 ⋅m ⋅m ⋅sin ( 34 ⋅°) = 1.33 ⋅kN

kN m

⋅m = 2.376 ⋅kN

3

2

⋅0.144 ⋅m ⋅m ⋅cos ( 34 ⋅°) = 1.97 ⋅kN 2

T10 = N10 ⋅tan ( ϕu.n) + cu.n ⋅A10 = 1.97 ⋅kN ⋅tan ( 30.5 ⋅°) + 0 ⋅kPa ⋅0.144 ⋅m = 1.16 ⋅kN 2

 ( T1 + T2 + T3 + T4 + T5 + T6 + T7 + T8 + T9 + T10)    ( S1 + S2 + S3 + S4 + S5 + S6 + S7 + S8 + S9 + S10) 

n = 

1.4 ⋅kN + 4.19 ⋅kN + 4.59 ⋅kN + 4.6 ⋅kN + 4.6 ⋅kN + 4.6 ⋅kN + 4.6 ⋅kN + 66.28 ⋅kN + 3.48 ⋅kN + 1.16 ⋅kN + 36.52 ⋅kN + 13.77 ⋅kN n = 0 ⋅kN + 0 ⋅kN + 5.27 ⋅kN + 5.27 ⋅kN + 5.27 ⋅kN + 5.27 ⋅kN + 5.27 ⋅kN + 16.25 ⋅kN + 3.98 ⋅kN + 1.33 ⋅kN + 14.86 ⋅kN + 6.37 ⋅kN

= 2.166 > 1

M0 = S1 + S2 + S3 + S4 + S5 + S6 + S7 + S8 + S9 + S10 M0 = 0 ⋅kN + 0 ⋅kN + 5.27 ⋅kN + 5.27 ⋅kN + 5.27 ⋅kN + 5.27 ⋅kN + 5.27 ⋅kN + 16.25 ⋅kN + 3.98 ⋅kN + 1.33 ⋅kN + 14.86 ⋅kN + 6.37 ⋅kN

= 69.14 ⋅kN

Mu = T1 + T2 + T3 + T4 + T5 + T6 + T7 + T8 + T9 + T10 + T11 + T12 Mu = 1.4 ⋅kN + 4.19 ⋅kN + 4.59 ⋅kN + 4.6 ⋅kN + 4.6 ⋅kN + 4.6 ⋅kN + 4.6 ⋅kN ... = 149.79 ⋅kN + 66.28 ⋅kN + 3.48 ⋅kN + 1.16 ⋅kN + 36.52 ⋅kN + 13.77 ⋅kN Mu M0

=

149.79 ⋅kN = 2.166 69.14 ⋅kN

. > 1.

________________________________________________________________________ 4.Sprawdzenie statecznośći skarpy -grunt lużny,płaszczyzna poślizgu równoległa do płaszczyzny skarpy. Kąt nachylenia skarpy 26 deg Założenia projektowe

strona 18

Stateczność skarpy "Fundamentowanie 2"

H = 3.5m ; nachylenie 1:1,5 Rodzaj gruntu

Piasek drobny

ID = 0.5 ρB.n = 16.5

kN m

3

cu.n = 0kPa

ϕu.n = 30.5° ρD.n = 16.5

kN m

3

________________________________________________________________________

________________________________________________________________________

4.1 Pas nr 1 A - pole powierzchni danego pasa α - kąt nachylenia powierzchni poślizgu do terenu S - siła zsuwająca T - siła utrzymająca W - ciężar bryły pasa N - siła normalna

strona 19

Stateczność skarpy "Fundamentowanie 2" A1 = 0.049m

2

α1 = 26°

kN 2 W1 = A1 ⋅ρB.n ⋅1m = 0.049 ⋅m ⋅16.5 ⋅ ⋅m = 0.809 ⋅kN 3 m kN 2 S1 = ρB.n ⋅A1 ⋅1m ⋅sin ( α1) = 16.5 ⋅ ⋅0.049 ⋅m ⋅m ⋅sin ( 26 ⋅°) = 0.35 ⋅kN 3 m kN 2 N1 = ρB.n ⋅A1 ⋅1m ⋅cos ( α1) = 16.5 ⋅ ⋅0.049 ⋅m ⋅m ⋅cos ( 26 ⋅°) = 0.73 ⋅kN 3 m T1 = N1 ⋅tan ( ϕu.n) + cu.n ⋅A1 = 0.73 ⋅kN ⋅tan ( 30.5 ⋅°) + 0 ⋅kPa ⋅0.049 ⋅m = 0.43 ⋅kN 2

4.2 Pas nr 2 A2 = 0.148m

2

α2 = 26° 2

W2 = A2 ⋅ρB.n ⋅1m = 0.148 ⋅m ⋅16.5 ⋅

kN

⋅m = 2.442 ⋅kN 3 m kN 2 S2 = ρB.n ⋅A2 ⋅1m ⋅sin ( α2) = 16.5 ⋅ ⋅0.148 ⋅m ⋅m ⋅sin ( 26 ⋅°) = 1.07 ⋅kN 3 m kN 2 N2 = ρB.n ⋅A2 ⋅1m ⋅cos ( α2) = 16.5 ⋅ ⋅0.148 ⋅m ⋅m ⋅cos ( 26 ⋅°) = 2.19 ⋅kN 3 m T2 = N2 ⋅tan ( ϕu.n) + cu.n ⋅A2 = 2.19 ⋅kN ⋅tan ( 30.5 ⋅°) + 0 ⋅kPa ⋅0.148 ⋅m = 1.29 ⋅kN 2

4.3 Pas nr 3 A3 = 0.246m

2

α3 = 26°

kN 2 W3 = A3 ⋅ρB.n ⋅1m = 0.246 ⋅m ⋅16.5 ⋅ ⋅m = 4.059 ⋅kN 3 m kN 2 S3 = ρB.n ⋅A3 ⋅1m ⋅sin ( α3) = 16.5 ⋅ ⋅0.246 ⋅m ⋅m ⋅sin ( 26 ⋅°) = 1.78 ⋅kN 3 m kN 2 N3 = ρB.n ⋅A3 ⋅1m ⋅cos ( α3) = 16.5 ⋅ ⋅0.246 ⋅m ⋅m ⋅cos ( 26 ⋅°) = 3.6 ⋅kN 3 m T3 = N3 ⋅tan ( ϕu.n) + cu.n ⋅A3 = 3.6 ⋅kN ⋅tan ( 30.5 ⋅°) + 0 ⋅kPa ⋅0.246 ⋅m = 2.12 ⋅kN 2

4.4 Pas nr 4 A4 = 0.345m

2

α4 = 26° 2

W4 = A4 ⋅ρB.n ⋅1m = 0.345 ⋅m ⋅16.5 ⋅

kN m

3

⋅m = 5.692 ⋅kN

strona 20

Stateczność skarpy "Fundamentowanie 2" S4 = ρB.n ⋅A4 ⋅1m ⋅sin ( α4) = 16.5 ⋅

kN 3

⋅0.345 ⋅m ⋅m ⋅sin ( 26 ⋅°) = 2.5 ⋅kN 2

m kN 2 N4 = ρB.n ⋅A4 ⋅1m ⋅cos ( α4) = 16.5 ⋅ ⋅0.345 ⋅m ⋅m ⋅cos ( 26 ⋅°) = 5.12 ⋅kN 3 m T4 = N4 ⋅tan ( ϕu.n) + cu.n ⋅A4 = 5.12 ⋅kN ⋅tan ( 30.5 ⋅°) + 0 ⋅kPa ⋅0.345 ⋅m = 3.02 ⋅kN 2

4.5 Pas nr 5 A5 = 0.444m

2

α5 = 26°

kN 2 W5 = A5 ⋅ρB.n ⋅1m = 0.444 ⋅m ⋅16.5 ⋅ ⋅m = 7.326 ⋅kN 3 m kN 2 S5 = ρB.n ⋅A5 ⋅1m ⋅sin ( α5) = 16.5 ⋅ ⋅0.444 ⋅m ⋅m ⋅sin ( 26 ⋅°) = 3.21 ⋅kN 3 m kN 2 N5 = ρB.n ⋅A5 ⋅1m ⋅cos ( α5) = 16.5 ⋅ ⋅0.444 ⋅m ⋅m ⋅cos ( 26 ⋅°) = 6.58 ⋅kN 3 m T5 = N5 ⋅tan ( ϕu.n) + cu.n ⋅A5 = 6.58 ⋅kN ⋅tan ( 30.5 ⋅°) + 0 ⋅kPa ⋅0.444 ⋅m = 3.88 ⋅kN 2

4.6 Pas nr 6 A6 = 0.542m

2

α6 = 26°

kN 2 W6 = A6 ⋅ρB.n ⋅1m = 0.542 ⋅m ⋅16.5 ⋅ ⋅m = 8.943 ⋅kN 3 m kN 2 S6 = ρB.n ⋅A6 ⋅1m ⋅sin ( α6) = 16.5 ⋅ ⋅0.542 ⋅m ⋅m ⋅sin ( 26 ⋅°) = 3.92 ⋅kN 3 m kN 2 N6 = ρB.n ⋅A6 ⋅1m ⋅cos ( α6) = 16.5 ⋅ ⋅0.542 ⋅m ⋅m ⋅cos ( 26 ⋅°) = 8.04 ⋅kN 3 m T6 = N6 ⋅tan ( ϕu.n) + cu.n ⋅A6 = 8.04 ⋅kN ⋅tan ( 30.5 ⋅°) + 0 ⋅kPa ⋅0.542 ⋅m = 4.74 ⋅kN 2

4.7 Pas nr 7 A7 = 0.641m

2

α7 = 26°

kN 2 W7 = A7 ⋅ρB.n ⋅1m = 0.641 ⋅m ⋅16.5 ⋅ ⋅m = 10.576 ⋅kN 3 m kN 2 S7 = ρB.n ⋅A7 ⋅1m ⋅sin ( α7) = 16.5 ⋅ ⋅0.641 ⋅m ⋅m ⋅sin ( 26 ⋅°) = 4.64 ⋅kN 3 m kN 2 N7 = ρB.n ⋅A7 ⋅1m ⋅cos ( α7) = 16.5 ⋅ ⋅0.641 ⋅m ⋅m ⋅cos ( 26 ⋅°) = 9.51 ⋅kN 3 m

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Stateczność skarpy "Fundamentowanie 2" T7 = N7 ⋅tan ( ϕu.n) + cu.n ⋅A7 = 9.51 ⋅kN ⋅tan ( 30.5 ⋅°) + 0 ⋅kPa ⋅0.641 ⋅m = 5.6 ⋅kN 2

4.8 Pas nr 8 A8 = 0.622m

2

α8 = 26°

kN 2 W8 = A8 ⋅ρB.n ⋅1m = 0.622 ⋅m ⋅16.5 ⋅ ⋅m = 10.263 ⋅kN 3 m S8 = ρB.n ⋅A8 ⋅1m ⋅sin ( α8) = 16.5 ⋅

kN m

N8 = ρB.n ⋅A8 ⋅1m ⋅cos ( α8) = 16.5 ⋅

3

⋅0.622 ⋅m ⋅m ⋅sin ( 26 ⋅°) = 4.5 ⋅kN 2

kN m

3

⋅0.622 ⋅m ⋅m ⋅cos ( 26 ⋅°) = 9.22 ⋅kN 2

T8 = N8 ⋅tan ( ϕu.n) + cu.n ⋅A8 = 9.22 ⋅kN ⋅tan ( 30.5 ⋅°) + 0 ⋅kPa ⋅0.622 ⋅m = 5.43 ⋅kN 2

4.9 Pas nr 9 A9 = 0.382m

2

α9 = 26°

kN 2 W9 = A9 ⋅ρB.n ⋅1m = 0.382 ⋅m ⋅16.5 ⋅ ⋅m = 6.303 ⋅kN 3 m S9 = ρB.n ⋅A9 ⋅1m ⋅sin ( α9) = 16.5 ⋅

kN m

N9 = ρB.n ⋅A9 ⋅1m ⋅cos ( α9) = 16.5 ⋅

3

⋅0.382 ⋅m ⋅m ⋅sin ( 26 ⋅°) = 2.76 ⋅kN 2

kN m

3

⋅0.382 ⋅m ⋅m ⋅cos ( 26 ⋅°) = 5.67 ⋅kN 2

T9 = N9 ⋅tan ( ϕu.n) + cu.n ⋅A9 = 5.67 ⋅kN ⋅tan ( 30.5 ⋅°) + 0 ⋅kPa ⋅0.382 ⋅m = 3.34 ⋅kN 2

4.10 Pas nr 10 A10 = 0.127m

2

α10 = 26°

kN 2 W10 = A10 ⋅ρB.n ⋅1m = 0.127 ⋅m ⋅16.5 ⋅ ⋅m = 2.095 ⋅kN 3 m S10 = ρB.n ⋅A10 ⋅1m ⋅sin ( α10) = 16.5 ⋅

kN m

N10 = ρB.n ⋅A10 ⋅1m ⋅cos ( α10) = 16.5 ⋅

3

⋅0.127 ⋅m ⋅m ⋅sin ( 26 ⋅°) = 0.92 ⋅kN

kN m

3

2

⋅0.127 ⋅m ⋅m ⋅cos ( 26 ⋅°) = 1.88 ⋅kN 2

T10 = N10 ⋅tan ( ϕu.n) + cu.n ⋅A10 = 1.88 ⋅kN ⋅tan ( 30.5 ⋅°) + 0 ⋅kPa ⋅0.127 ⋅m = 1.11 ⋅kN 2

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Stateczność skarpy "Fundamentowanie 2" n =

( T1 + T2 + T3 + T4 + T5 + T6 + T7 + T8 + T9 + T10) ( S1 + S2 + S3 + S4 + S5 + S6 + S7 + S8 + S9 + S10)

0.43 ⋅kN + 1.29 ⋅kN + 2.12 ⋅kN + 3.02 ⋅kN + 3.88 ⋅kN + 4.74 ⋅kN ... + 5.6 ⋅kN + 5.43 ⋅kN + 3.24 ⋅kN + 1.11 ⋅kN n = = 1.203 > 1 0.35 ⋅kN + 1.07 ⋅kN + 1.78 ⋅kN + 2.5 ⋅kN + 3.21 ⋅kN + 3.92 ⋅kN ... + 4.64 ⋅kN + 4.5 ⋅kN + 2.76 ⋅kN + 0.92 ⋅kN

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