Temat: |
OBLICZENIE KĄTOWEJ ŚCIANY OPOROWEJ |
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Autorzy: |
Dariusz BOMBA, Robert DACKO |
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Opis programu: |
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Program sprawdza warunki stateczności kątowej ściany oporowej, której poziomy naziom |
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może być obciążony obciążeniem równomiernie rozłożonym. |
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Za ścianą należy przewidzieć warstwę zasypki z gruntu niespoistego np..piasku lub żwiru. |
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Warunki, które znajdują się na czerwonym tle muszą być wypełnione !!!. Jeśli którykolwiek z nich jest |
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nie spełniony, należy zmieniać parametry ściany aż do skutku. |
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Informacje dla użytkowników: |
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UWAGA !!! |
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-Wartości wejściowe do projektu należy wpisywać tylko w komórkach w żółtym kolorze !!! |
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[m] |
-Tekst oraz wartości liczbowe w kolorze zielonym są automatycznie przeliczane przez program i tu nie wolno nic zmieniać !!! |
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-Wartości,które program pokazuje na jasno niebieskim tle są charakterystyczne, przeliczane automatycznie i tu nie wolno nic zmieniać !!! |
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-Wartości,które program pokazuje na jasno brązowym tle są obliczeniowe, przeliczane automatycznie i tu nie wolno nic zmieniać !!! |
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DANE DO OBLICZEŃ: |
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WYMIARY ŚCIANY (wstawić w metrach) |
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a= |
0.2 |
[m] |
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b= |
0.15 |
[m] |
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c= |
0.9 |
[m] |
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d= |
0.2 |
[m] |
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e= |
0.2 |
[m] |
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f= |
2 |
[m] |
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g= |
0.2 |
[m] |
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h= |
0.2 |
[m] |
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k= |
4.35 |
[m] |
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m= |
0.25 |
[m] |
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p= |
1.2 |
[m] |
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qn= |
10 |
[kPa] |
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g(n)= |
17.5 |
[kN/m3] |
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gb= |
25 |
[kN/m3] |
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B= |
3.3 |
[m] |
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H= |
5 |
[m] |
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Składowe pionowe obciążenia wynoszą: |
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wartości charakterystyczne wynoszą: |
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wartości obliczeniowe wynoszą: |
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gf= |
0.9 |
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gf= |
1.1 |
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G1(n)= |
22.3 |
[kN] |
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G11(r)= |
20.0 |
[kN] |
G12(r)= |
24.5 |
[kN] |
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G2(n)= |
10.9 |
[kN] |
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G21(r)= |
9.8 |
[kN] |
G22(r)= |
12.0 |
[kN] |
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G3(n)= |
25.8 |
[kN] |
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G31(r)= |
23.2 |
[kN] |
G32(r)= |
28.3 |
[kN] |
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G4(n)= |
20.5 |
[kN] |
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G41(r)= |
18.4 |
[kN] |
G42(r)= |
22.5 |
[kN] |
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G5(n)= |
164.5 |
[kN] |
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G51(r)= |
148.1 |
[kN] |
G52r)= |
181.0 |
[kN] |
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G6(n)= |
20.0 |
[kN] |
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G61(r)= |
18.0 |
[kN] |
G62(r)= |
22.0 |
[kN] |
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Suma Gi(n)= |
263.9 |
[kN] |
Suma |
Gi1(r)= |
237.5 |
[kN] |
Gi2(r)= |
290.2 |
[kN] |
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Parametry geotechniczne zasypki znajdującej się za ścianą oporową: |
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ID(n)= |
0.4 |
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gf= |
0.9 |
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gf= |
1.1 |
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fu(n)= |
33 |
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fu1(r)= |
29.7 |
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fu2(r)= |
36.3 |
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gn(n)= |
17 |
[kN/m3] |
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gn1(r)= |
15.3 |
[kN/m3] |
gn2(r)= |
18.7 |
[kN/m3] |
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Parametry geotechniczne podłoża pod ścianą oporową: |
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wartości charakterystyczne wynoszą: |
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wartości obliczeniowe wynoszą: |
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1 warstwa |
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IL(n)= |
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ID(n)= |
1 |
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gf= |
0.9 |
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gf= |
1.1 |
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fu(n)= |
1 |
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fu1(r)= |
0.9 |
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fu2(r)= |
1.1 |
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cu(n)= |
1 |
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cu1(r)= |
0.9 |
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cu2(r)= |
1.1 |
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gb(n)= |
17 |
[kN/m3] |
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gb1(r)= |
15.3 |
[kN/m3] |
gb2(r)= |
18.7 |
[kN/m3] |
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ND= |
18.4 |
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NB= |
7.5 |
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OBCIĄŻENIA DZIAŁAJĄCE NA ŚCIANĘ OPOROWĄ: |
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Obciążenie naziomu: |
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qn= |
10 |
[kPa] |
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qr1= |
8 |
[kPa] |
qr2= |
12 |
[kPa] |
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Współczynnik parcia granicznego, czynnego, gruntu przy założeniu idealnie gładkiej powierzchni ściany: |
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wg wzoru nr 3 PN-83/B-03010 |
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Ka= |
0.29 |
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Wysokość zastępcza uwzględniająca wpływ obciążenia naziomu: |
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hz= |
0.6 |
[m] |
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Jednostkowe parcie gruntu wynosi: |
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1) dla z= |
0 |
[m] |
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ea= |
2.95 |
[kN/m] |
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2) dla z= |
5 |
[m] |
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ea= |
28.01 |
[kN/m] |
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Wypadkowa parcia granicznego gruntu wynosi: |
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1) Wypadkowa parcia gruntu pochodząca od obciążenia naziomu: |
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Ea1= |
14.74 |
[kN] |
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Ea1(r)= |
17.69 |
[kN] |
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2) Wypadkowa parcia gruntu : |
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Ea2= |
62.65 |
[kN] |
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Ea2(r)= |
75.17 |
[kN] |
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SPRAWDZENIE STANÓW GRANICZNYCH PODŁOŻA: |
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1) Sprawdzenie stateczności na obrót względem przedniej krawędzi ściany: |
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mo= |
0.8 |
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Mo(r)= |
184.25 |
[kNm] |
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Mu(r)= |
462.59 |
[kNm] |
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mo x Mu(r)= |
370.0719 |
>Mo(r)= |
184.25 |
[kNm] |
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WARUNEK SPEŁNIONY |
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2) Przesunięcie w poziomie posadowienia fundamentu: |
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Qt(r)= |
92.86 |
[kN] |
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m= |
0.8 |
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mt= |
0.95 |
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Qtf = |
189.972 |
[kN] |
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Qt(r)= |
92.86 |
< mt x Qtf = |
180.4734 |
[kN] |
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WARUNEK SPEŁNIONY |
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3) Przesunięcie w podłożu: |
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Qtf = |
135.44 |
[kN] |
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Qt(r) = |
92.86 |
< mt x Qtf= |
128.670691593815 |
[kN] |
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WARUNEK SPEŁNIONY |
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4) Wypieranie gruntu spod płyty fundamentowej: |
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Nr= |
290.24 |
[kN] |
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m= |
0.9 |
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L=L= |
1 |
[m] |
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B/L= |
0 |
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Dmin= |
1.2 |
[m] |
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eB= |
0.63 |
[m] |
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Bśr= |
2.03 |
[m] |
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tgdB= |
0.32 |
[o] |
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tgfu (n)= |
0.65 |
[o] |
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tgdu(n)/tgfu (n)= |
0.49 |
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iB= |
0.24 |
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iD= |
0.45 |
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QfNB= |
422.18 |
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m= |
0.8 |
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mxQfNB= |
337.74 |
>Nr= |
290.24 |
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WARUNEK SPEŁNIONY |
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