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