Zadanie 1
Wyznaczyć rozkład parcia czynnego (narysować wykres) wzdłuż ściany szczelinowej.
Parametry gruntu i wymiary ściany szczelinowej podano na rysunku poniżej.
Wyznaczenie wartości współczynników parcia czynnego Ka:
°
WARSTWA I: = 45° - = 45° - = 0,307


°
WARSTWA II: = 45° - = 45° - = 0,756

°
WARSTWA III: = 45° - = 45° - = 0,333

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Wyznaczenie wartości
parcia czynnego:
WARSTWA I:
= " 0 + " - 2 = 19 " 0 + 45 " 0,307 - 2 " 0 " 0,307 =
18,83
, = " ! + " - 2 = 19 " 1,5 + 45 " 0,307 - 2 " 0 " 0,307 =
22,58
WARSTWA II:
, = " ! + " - 2 = 19 " 1,5 + 45 " 0,756 - 2 " 40 " 0,756 =
- 14,0
, = " ! + " ! + " - 2 = 19 " 1,5 + 20 " 2,5 + 45 " 0,756 - 2 "
40 " 0,756 = 23,78
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WARSTWA III:
, = " ! + " ! + " - 2 = 19 " 1,5 + 20 " 2,5 + 45 "
0,333 - 2 " 0 " 0,333 = 47,17
, = " ! + " ! + " ! + " - 2 =
19 " 1,5 + 20 " 2,5 + 16 " 2 + 45 " 0,333 - 2 " 0 " 0,333 = 51,83
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Wyznaczenie wypadkowej parcia czynnego:
Wypadkowa parcia to pole powierzchni
wykresu parcia!
WARSTWA I:

= " + , " ! =


" 13,83 + 22,58 " 1,5 = 27,3

Położenie wypadkowej parcia:

! = " ! "

,
! = " ! " =
,
" , ,
" 1,5 " = 0,69
, ,
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Zadanie 2
Wyznaczyć rozkład parcia czynnego i biernego(narysować wykres) wzdłuż ściany
oporowej. Parametry gruntu i wymiary ściany oporowej podano na rysunku poniżej.
Wyznaczenie ciężaru objętościowego gruntu z uwzględnieniem wyporu wody:
= 1 - " -

= 1 - 0,388 " 25,98 - 9,81 = 9,90

= 1 - 0,36 " 25,98 - 9,81 = 10,45
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Wyznaczenie wartości współczynników parcia czynnego Ka:
, °
WARSTWA I: = 45° - = 45° - = 0,6


°
WARSTWA II: = 45° - = 45° - = 0,295

, °
WARSTWA III: = 45° - = 45° - = 0,28

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Wyznaczenie wartości parcia czynnego:
WARSTWA I:
= " 0 + " - 2 = 20,59 " 0 + 10 " 0,6 - 2 " 25 " 0,6 =
-32,67
, = " ! + " - 2 = 20,59 " 1,5 + 10 " 0,6 - 2 " 25 " 0,6
= -14,15
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WARSTWA II:
, = " ! + " - 2 = 20,59 " 1,5 + 10 " 0,295 - 2 " 25 " 0,295
= 12,06
, = " ! + " ! + " - 2
= 20,59 " 1,5 + 18,14 " 1,0 + 10 " 0,295 - 2 " 25 " 0,295 = 17,41

, = " ! + " ! + " ! + " - 2 =
= 20,59 " 1,5 + 18,14 " 1,0 + 9,9 " 0,5 + 10 " 0,295 - 2 " 25 " 0,295 = 18,90
Dodatkowo parcie wody na głębokości 3,0m:
, = " ! = 9,81 " 0,5 = 4,91
Sumaryczne parcie wody i gruntu na głębokości 3,0m:

, = , + , = 18,90 + 4,91 = 23,81
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WARSTWA III:

, = " ! + " ! + " ! + " - 2 =
= 20,59 " 1,5 + 18,14 " 1,0 + 9,9 " 0,5 + 10 " 0,28 - 2 " 0 " 0,28 = 17,94
Dodatkowo parcie wody na głębokości 3,0m:
, = " ! = 9,81 " 0,5 = 4,91
Sumaryczne parcie wody i gruntu na głębokości 3,0m:

, = , + , = 17,94 + 4,91 = 22,85
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, = " ! + " ! + " ! + " ! + " - 2 =
= 20,59 " 1,5 + 18,14 " 1,0 + 9,9 " 0,5 + 10,45 " 2,0 + 10 " 0,28 - 2 " 0 " 0,28
= 23,79
Dodatkowo parcie wody na głębokości 5,0m:
, = " ! = 9,81 " 2,5 = 24,53
Sumaryczne parcie wody i gruntu na głębokości 5,0m:

, = , + , = 23,79 + 24,53 = 48,32
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, = " ! + " ! + " ! + " ! + " ! + " - 2 =
= 20,59 " 1,5 + 18,14 " 1,0 + 9,9 " 0,5 + 10,45 " 2,0 + 10,45 " 4,0 + 10 " 0,28 - 2
" 0 " 0,28 = 35,49
Dodatkowo parcie wody na głębokości 9,0m:
, = , = " ! = 9,81 " 2,5
= 24,53
- ż ę ! ś
Sumaryczne parcie wody i gruntu na głębokości 9,0m:

, = , + , = 35,49 + 24,53 = 60,02
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Wyznaczenie odporu gruntu (parcia biernego):
Wyznaczenie współczynnika parcia biernego Kp:
, °
= 45° + = 45° + = 3,567

Wyznaczenie wartości parcia biernego:
, = Å‚ " h " K + 2c K = 10,45 " 4,0 " 3,567 + 2 " 0 " 3,567 =

149,10 kPa
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Badanie wytrzymałości na ścinanie gruntu spoistego w aparacie AB.
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250.0
Ä = 0.6094Ć + 20.313
200.0
150.0
100.0
50.0
0.0
0 50 100 150 200 250 300 350
à [kPa]
= 20,3
= 31,4°
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Ä [kPa]
WYTRZYMAA
OŚĆ GRUNTU NA ŚCINANIE
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WYTRZYMAA
OŚĆ GRUNTU NA ŚCINANIE 
KONSTRUKCJA KOA
A MOHRA
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Zadanie 8
Obliczyć kąt tarcia wewnętrznego gruntu niespoistego jeżeli wyniki badań w
aparacie trójosiowego Å›ciskania wynoszÄ… Ã1=800 kPa, Ã2=200 kPa.
+ 800 + 200
= =
2 2
= 500
- 800 - 200
= =
2 2
= 300
300
= = = 0,6
500
= 0,6 = 36,7°
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Zadanie 9

W aparacie trójosiowego Å›ciskania dla Ã1= 300 kPa
= = " 15°
okreÅ›lono nastÄ™pujÄ…ce parametry: Ć=15°, c=o kPa.

Obliczyć wartość Ã3.
= 0,259

=



=

2 = +
2 = -
2 = +
2 " 0,259 = -
2 = +
0,518 = -
2,518 = 2
2,518 = 2 " 300
= 238,28
= 2 - = 2 " 238,28 - 300 = 176,6
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Zadanie 13
W aparacie trójosiowego ściskania wykonano dwa badania próbek tego samego gruntu.
Otrzymano następujące wyniki:
dla badania 1: Ã3=50 kPa, Ã1=250 kPa
dla badania 2: Ã3=150 kPa, Ã1=450 kPa.
Obliczyć parametry Ć i c badanego gruntu.
- 250 - 50
= = = 100
2 2
- 450 - 150
= = = 150
2 2
+ 250 + 50
= = = 150
2 2
+ 450 + 150
= = = 300
2 2
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"
" " - " "
"
=
" "
" - ( )
2 " + - + + 7500
= arcsin = = 19,3°

22500
2 " + - +
" " - " "
1 ( " )
= "

" "
" - ( )

1 + + - + +
=
"


2 " + - +
1 1125000
= " = 53,0
19,3° 22500
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