DANE DO PROJEKTU |
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Różnica osiadań pomiędzy stopami (ławami) |
x= |
5 |
cm |
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Siła |
Qn1= |
980 |
kN |
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Siła |
Qn2= |
1070 |
kN |
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Rzędna terenu |
RzT= |
0 |
m |
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Tab. 1. Zestawienie wartości cech wiodących. |
Tab. 2. Zestawienie parametrów geotechnicznych. Metoda B wg. PN-81/B-03020 |
Rzędna posadowienia fundamentu (poniżej RzT) |
Rz pf= |
1,3 |
m p.p.t. |
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Lp. |
1 |
2 |
3 |
4 |
5 |
6 |
7 |
8 |
9 |
10 |
11 |
12 |
13 |
14 |
15 |
16 |
17 |
18 |
19 |
20 |
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ID(n) |
IL(n) |
gm |
gm |
gm,prz |
fu(n) |
rs(n) |
cu(n) |
fu(r) |
rs(r) |
cu(r) |
Spąg warstwy I |
Rz wI= |
2,5 |
m p.p.t. |
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Warstwa nr I |
ID |
0,52 |
0,60 |
0,60 |
0,58 |
0,58 |
0,53 |
0,53 |
0,52 |
0,54 |
0,55 |
0,53 |
0,58 |
0,58 |
0,56 |
0,61 |
0,57 |
0,59 |
0,59 |
0,52 |
0,61 |
Warstwa nr I |
ID |
0,56 |
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1,06 |
0,94 |
0,90 |
33,50 |
1,70 |
0,00 |
30,15 |
1,53 |
0,00 |
Spąg warstwy II |
Rz wII= |
4,7 |
m p.p.t. |
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Warstwa nr II |
IL |
0,52 |
0,59 |
0,59 |
0,58 |
0,58 |
0,52 |
0,53 |
0,55 |
0,55 |
0,55 |
0,58 |
0,58 |
0,57 |
0,52 |
0,53 |
0,55 |
0,57 |
0,57 |
0,57 |
0,59 |
Warstwa nr II |
IL |
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0,56 |
1,04 |
0,98 |
0,90 |
9,00 |
2,00 |
8,00 |
8,10 |
1,80 |
7,20 |
Warstwa I |
Ps |
fu(n)= |
33,5 |
rs= |
1,7 |
cu(n)= |
0 |
Warstwa nr III |
ID |
0,61 |
0,70 |
0,70 |
0,61 |
0,62 |
0,63 |
0,63 |
0,63 |
0,65 |
0,66 |
0,67 |
0,69 |
0,69 |
0,69 |
0,69 |
0,70 |
0,61 |
0,62 |
0,62 |
0,70 |
Warstwa nr III |
ID |
0,66 |
|
1,05 |
0,95 |
0,90 |
34,00 |
1,85 |
0,00 |
30,60 |
1,67 |
0,00 |
Warstwa II |
Gp |
fu(n)= |
9 |
rs= |
2 |
cu(n)= |
8 |
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Warstwa III |
Ps |
fu(n)= |
34 |
rs= |
1,85 |
cu(n)= |
0 |
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Współcznynnik korekcyjny |
m= |
0,81 |
[-] |
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Obliczenia dla pierwszej stopy fundamentowej. |
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Podłoże jednorodne |
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Wartość obliczeniowa obciążenia |
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Qr= |
1176 |
[kN] |
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Przyjęto fundament o wymiarach |
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B= |
2,5 |
[m] |
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L= |
3 |
[m] |
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Przyjęte wartości |
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ND= |
5,26 |
[-] |
Marcin:
Jeżeli podłoże ma różne kąty tarcia wew. należy zrobić średnią ważoną z tych warstw
dla kąta= |
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iD= |
1 |
cu(r)= |
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ru(r)= |
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NC= |
13,1 |
[-] |
18,63 |
iC= |
1 |
3,74 |
1,67 |
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NB= |
1,04 |
[-] |
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iB= |
1 |
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Wartość obliczeniowa oporu granicznego |
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Qf= |
2590,33740264 |
[kN] |
B<L |
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mQf= |
2098,1732961384 |
[kN] |
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Warunek pierwszego stanu granicznego |
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1176 |
< |
2098,1732961384 |
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TAK |
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Obliczenia dla drugiej stopy fundamentowej. |
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Podłoże jednorodne |
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Wartość obliczeniowa obciążenia |
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Qr= |
1284 |
[kN] |
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Przyjęto fundament o wymiarach |
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B= |
2,5 |
[m] |
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L= |
3 |
[m] |
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Przyjęte wartości |
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ND= |
5,26 |
[-] |
Marcin:
Jeżeli podłoże ma różne kąty tarcia wew. należy zrobić średnią ważoną z tych warstw
dla kąta= |
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iD= |
1 |
cu(r)= |
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ru(r)= |
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NC= |
13,1 |
[-] |
18,63 |
iC= |
1 |
3,74 |
1,67 |
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NB= |
1,04 |
[-] |
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iB= |
1 |
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Wartość obliczeniowa oporu granicznego |
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Qf= |
2603,65153464 |
[kN] |
B<L |
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mQf= |
2108,9577430584 |
[kN] |
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Warunek pierwszego stanu granicznego |
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1284 |
< |
2108,9577430584 |
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TAK |
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