3. WYMIAROWANIE STOPY POD SLUPEM ZEWNETRZNYM.
Obciazenia stopy:
N1.max
654kN
:=
M1.odp
26kN m
⋅
:=
Q31
20kN
:=
N1.odp
560kN
:=
M1.max
29kN m
⋅
:=
Q32
22kN
:=
system znakowania sil:
Wstepny dobór wymiarow stopy:
bs
35cm
:=
hs
35cm
:=
przyjeto:
Bst
1.5m
:=
Lst
2.0m
:=
hst
60cm
:=
wysokosc stopy
Pole powierzchni stopy:
Ast
Bst Lst
⋅
:=
Bst Lst
⋅
1.5 m
⋅
2.0 m
⋅
⋅
=
3 m
2
=
wskaznik wytrzymalosci
Wz
Bst Lst
2
⋅
6
:=
Bst Lst
2
⋅
6
1.5 m
⋅
2.0 m
⋅
(
)
2
⋅
6
=
1 m
3
=
sredni mimosrod:
esr
M1.odp Q31 hst
⋅
+
N1.max
M1.max Q32 hst
⋅
+
N1.odp
+
1
2
⋅
:=
esr 6.7cm
=
Przyjeto
esr
7cm
:=
Ciezar gruntu:
γ
gr
18
kN
m
3
:=
Dmin
2m
:=
minimalna glebokosc posadowienia
Ciezar stopy + gruntu powyzej:
G
Bst Lst
⋅
hst
⋅
25
⋅
kN
m
3
Bst Lst
⋅
Dmin hst
−
(
)
⋅
γ
gr
⋅
+
:=
G
120.6 kN
=
Gd
1.2 G
⋅
:=
Gd 144.7kN
=
wart obliczeniowa
kombinacja obciazen 1:
••••
kombinacja obciazen 2:
••••
N1
654kN
:=
M1
26kN m
⋅
:=
Q1
20kN
:=
N2
560kN
:=
M2
29kN m
⋅
:=
Q2
22kN
:=
eL1
N1 esr
⋅
M1
−
Q1 hst
⋅
−
N1 Gd
+
:=
eL2
N2 esr
⋅
M2
−
Q2 hst
⋅
−
N2 Gd
+
:=
eL1
654 kN
⋅
7 cm
⋅
⋅
26 kN
⋅
m
⋅
−
20 kN
⋅
60 cm
⋅
⋅
−
654 kN
⋅
145 kN
⋅
+
=
eL2
560 kN
⋅
7 cm
⋅
⋅
29 kN
⋅
m
⋅
−
22 kN
⋅
60 cm
⋅
⋅
−
560 kN
⋅
145 kN
⋅
+
=
eL1 0.97 cm
=
eL2
0.43
−
cm
=
qmaxL2
N2 Gd
+
Ast
N2 Gd
+
Wz
eL2
⋅
+
:=
qmaxL1
N1 Gd
+
Ast
N1 Gd
+
Wz
eL1
⋅
−
:=
qmaxL2
560 kN
⋅
208 kN
⋅
+
3 m
2
⋅
560 kN
⋅
208 kN
⋅
+
1 m
3
⋅
0.39
−
(
) cm
⋅
⋅
+
=
qmaxL1
654 kN
⋅
208 kN
⋅
+
3 m
2
⋅
654 kN
⋅
208 kN
⋅
+
1 m
3
⋅
0.9 cm
⋅
⋅
−
=
qmaxL1 0.259MPa
=
qmaxL2 0.232MPa
=
28
qminL2
N2 Gd
+
Ast
N2 Gd
+
Wz
eL2
⋅
−
:=
qminL1
N1 Gd
+
Ast
N1 Gd
+
Wz
eL1
⋅
+
:=
qminL2
560 kN
⋅
208 kN
⋅
+
3 m
2
⋅
560 kN
⋅
208 kN
⋅
+
1 m
3
⋅
0.39
−
(
) cm
⋅
⋅
−
=
qminL1
654 kN
⋅
208 kN
⋅
+
3 m
2
⋅
654 kN
⋅
208 kN
⋅
+
1 m
3
⋅
0.9 cm
⋅
⋅
+
=
qminL2 0.238MPa
=
qminL1 0.274MPa
=
Ekstremalne naprezenia pod stopa:
qmaxL
max qmaxL1 qmaxL2
,
(
)
:=
qmaxL 0.26 MPa
=
qminL
max qminL1 qminL2
,
(
)
:=
qminL 0.274MPa
=
Obliczanie odporu granicznego podloza gruntowego: piasek.
••••
γ
nr
17
kN
m
3
:=
φ
ur
33deg
:=
Cur
0Pa
:=
K2
K1
tg
δ
2
Q2
N2 Gd
+
:=
Q2
N2 Gd
+
22 kN
⋅
560 kN
⋅
145 kN
⋅
+
=
0.0312
=
tg
δ
1
Q1
N1 Gd
+
:=
Q1
N1 Gd
+
20 kN
⋅
654 kN
⋅
145 kN
⋅
+
=
0.025
=
tan
φ
ur
( )
0.649
=
tan
φ
ur
( )
0.649
=
tg
δ
2
tan
φ
ur
( )
0.048
=
tg
δ
1
tan
φ
ur
( )
0.039
=
Wspolczynniki nosnosci:
ND
26.1
:=
iD
0.6
:=
NC
38.6
:=
iC
0.58
:=
NB
12.2
:=
iB
0.38
:=
Zredukowane wymiary stopy:
L'st
Lst 2 eL1
⋅
−
:=
L'st 1.981 m
=
Bst 1.5 m
=
Nosnosc podloza gruntowego:
QfNL
Bst L'st
⋅
1
0.3
Bst
L'st
⋅
+
NC
⋅
Cur
⋅
iC
⋅
1
1.5
Bst
L'st
⋅
+
ND
⋅
γ
nr
⋅
Dmin
⋅
iD
⋅
+
1
0.25
Bst
L'st
⋅
−
NB
⋅
γ
nr
⋅
L'st
⋅
iB
⋅
+
⋅
:=
QfNL 3754.8 kN
=
N1 Gd
+
0.81 QfNL
⋅
<
1
=
Warunek nosnosci spelniony!
N1 Gd
+
799 kN
=
0.81 QfNL
⋅
3041.3 kN
=
29
3.1. Sprawdzenie stopy na przebicie.
Sprawdzenie stopy na przebicie w kierunku "L".
••••
e
sr
qmaxL 0.26 MPa
=
qminL 0.27 MPa
=
qsrL
qmaxL qminL
+
2
:=
qmaxL qminL
+
2
0.26 MPa
⋅
0.27 MPa
⋅
+
2
=
0.265 MPa
=
ast
6cm
:=
grubosc otulenia od strony gruntu
dst
hst ast
−
:=
hst ast
−
60 cm
⋅
6 cm
⋅
−
=
0.54 m
=
uzyteczna wysokosc przekroju stopy
bmL
bs dst
+
:=
bs dst
+
35 cm
⋅
0.54 m
⋅
+
=
0.89 m
=
fctd
1.0MPa
:=
obliczeniowa wytrzymalosc na rozciaganie betonu B25
AL
Bst
Lst
2
hs
2
−
dst
−
esr
+
⋅
:=
AL 0.532 m
2
=
(rysunek powyzej)
NRdL
fctd bmL
⋅
dst
⋅
:=
fctd bmL
⋅
dst
⋅
1.0 MPa
⋅
0.89 m
⋅
⋅
0.54 m
⋅
⋅
=
0.481 MN
=
-nosnosc stopy
NSdL
qmaxL AL
⋅
:=
qmaxL AL
⋅
0.26 MPa
⋅
0.53 m
2
⋅
⋅
=
0.138 MN
=
-sila parcia gruntu
NSdL NRdL
<
1
=
warunek nosnosci spelniony!!
30
Sprawdzenie stopy na przebicie w kierunku "B".
••••
qsrB
NSd
Ast
:=
NSd
N1 Gd
+
Ast
654 kN
⋅
145 kN
⋅
+
3 m
2
⋅
=
0.266 MPa
=
ast
6cm
:=
dst 0.54 m
=
fctd 1 MPa
=
bmB
hs dst
+
:=
hs dst
+
35 cm
⋅
0.54 m
⋅
+
=
0.89 m
=
AB
Lst
Bst
2
bs
2
−
dst
−
⋅
:=
AB 0.07 m
2
=
(rysunek powyzej)
NRdB
fctd bmB
⋅
dst
⋅
:=
fctd bmB
⋅
dst
⋅
1.0 MPa
⋅
0.89 m
⋅
⋅
0.54 m
⋅
⋅
=
0.481 MN
=
-nosnosc stopy
NSdB
qsrB AB
⋅
:=
qsrB AB
⋅
0.266 MPa
⋅
0.07 m
2
⋅
⋅
=
0.019 MN
=
-sila parcia gruntu
NSdB NRdB
<
1
=
warunek nosnosci spelniony!!
3.2. Wymiarowanie stopy na zginanie:
3.2.1. Zginanie w kierunku XY (kierunek "L"):
cL
Lst
2
hs
2
−
:=
Lst
2
hs
2
−
esr
+
2.0 m
⋅
2
35 cm
⋅
2
−
7 cm
⋅
+
=
0.9 m
=
- wysieg wspornika
31
q1L
qmaxL
qmaxL qminL
−
Lst
cL
⋅
−
:=
qmaxL
qmaxL qminL
−
Lst
cL
⋅
−
0.26 MPa
⋅
0.26 MPa
⋅
0.27 MPa
⋅
−
2.0 m
⋅
0.9 m
⋅
⋅
−
=
0.265 MPa
=
- gestosc obc na krawedzi 1
qsr1L
qmaxL q1L
+
2
:=
qmaxL q1L
+
2
0.26 MPa
⋅
0.265 MPa
⋅
+
2
=
0.263 MPa
=
- srednia gestosc obc na lewo od krawedzi 1
MSdL
qsr1L
cL
2
6
⋅
2 Bst
⋅
bs
+
(
)
⋅
:=
qsr1L
cL
2
6
⋅
2 Bst
⋅
bs
+
(
)
⋅
0.263 MPa
⋅
0.9 m
⋅
(
)
2
6
⋅
2 1.5 m
⋅
⋅
35 cm
⋅
+
(
)
⋅
=
118.9 kN m
⋅
=
- moment zginajacy na krawedzi 1
fyd
210MPa
:=
stal A-I
As1L
MSdL
0.9 dst
⋅
fyd
⋅
:=
MSdL
0.9 dst
⋅
fyd
⋅
119 kN m
⋅
⋅
0.9 0.54 m
⋅
⋅
210 MPa
⋅
⋅
=
11.66 cm
2
=
powierzchnia zbrojenia:
As1L.prov
7
π
16mm
(
)
2
4
⋅
:=
As1L.prov 14.07cm
2
=
przyjmuje 7
φφφφ
16 co 20cm
3.2.2 Zginanie w kierunku XZ (kierunek "B"):
W kierunku tym gestosc obc jest rozlozona rownomiernie:
qsrB
N1 Gd
+
Ast
:=
N1 Gd
+
Ast
654 kN
⋅
145 kN
⋅
+
3 m
2
⋅
=
0.266 MPa
=
cB
Bst
2
bs
2
−
:=
Bst
2
bs
2
−
1.5 m
⋅
2
35 cm
⋅
2
−
=
0.575 m
=
AB
cB hs
+
(
)
cB
⋅
:=
cB hs
+
(
)
cB
⋅
0.575 m
⋅
35 cm
⋅
+
(
) 0.575 m
⋅
⋅
=
0.532 m m
=
pole powierzchni zakreskowanego
pola (pole obc.)
32
MSdB
qsrB
cB
2
6
⋅
2 Lst
⋅
hs
+
(
)
⋅
:=
qsrB
cB
2
6
⋅
2 Lst
⋅
hs
+
(
)
⋅
0.266 MPa
⋅
0.575 m
⋅
(
)
2
6
⋅
2 2.0 m
⋅
⋅
35 cm
⋅
+
(
)
⋅
=
63.8 kN m
⋅
=
- moment zginajacy na krawedzi 1
As1B
MSdB
0.9 dst
⋅
fyd
⋅
:=
MSdB
0.9 dst
⋅
fyd
⋅
64 kN m
⋅
⋅
0.9 0.54 m
⋅
⋅
210 MPa
⋅
⋅
=
6.27 cm
2
=
powierzchnia zbrojenia:
As1B.prov
10
π
12mm
(
)
2
4
⋅
:=
As1B.prov 11.31cm
2
=
przyjmuje 15
φ12
φ12
φ12
φ12
co 20cm
33