7. Sprawdzenie warunku nośności na przesuw

γGn 1.35

:=

γϕ 1

:=

γqu 1

:=

γRv 1.4

:=

D 1.8m

:=

γpos 17.658

kN
m

3



:=

γGk 1.0

:=

γc 1

:=

γγ 1.0

:=

γRh 1.1

:=

z

0.4m

:=

γQn 1.5

:=

γcu 1

:=

c

0.05m

:=

γżelb 25

kN
m

3



:=

γQk 1.0

:=

hp 0.2m

:=

γposadz 24

kN
m

3

:=

Vk 990kN

:=

Qv.k 680kN

:=

Hk 263kN

:=

QH.k 166kN

:=

Mk

35

- kN m



:=

Qm.k

68

- kN m



:=

7.1. Obciążenia fundamentu

hf1 0.5 m



:=

aSL 0.6 m



:=

aSB 0.4 m



:=

Lmin1 aSL 2 hf1 c

-

(

)



+

2c

+

1.6 m



=

:=

Bmin1 aSB 2 hf1 c

-

(

)



+

2c

+

1.4 m



=

:=

LL

1.6 m



:=

BB 1.45 m



:=

LL Lmin1



1

=

BB Bmin1



1

=

LL
BB

1.103

=

av 0 m



:=

s1

LL

2

aSL

2

-









av

+

0.5m

=

:=

s2

LL

2

aSL

2

-









av

-

0.5m

=

:=

V11.k

LL BB



hf1



aSL aSB



D hf1

-

(

)



+





1.472 m

3



=

:=

V22.k BB LL



D



LL BB



hf1



aSL aSB



D hf1

-

(

)



+





-

s1

( )

BB



z



+

2.994 m

3



=

:=

V33.k s1 BB



hp



0.145 m

3



=

:=

G11.k γżelb V11.k



36.8 kN



=

:=

G22.k γpos V22.k



52.868 kN



=

:=

G33.k γposadz V33.k



3.48 kN



=

:=

Σ

Gi G11.k G22.k

+

G33.k

+

93.148 kN



=

:=

av1 0m

:=

av2

LL

2

aSL

2

-









2

0.5 aSL



(

)

+

av

+

















-

0.55

-

m

=

:=

γGk 1

=

γQn 1.5

=

av3

0.5 LL



0.5aSL

-

(

)

av

+

2

0.5aSL

(

)

av

-

+









0.55 m

=

:=

γQkk 0

:=

γGn 1.35

=

M

G11.k av1



G22.k av2



+

G33.k av3



+

27.163

-

kN m





=

:=

Vf.H.d γGk Vk



γQkk 0.7



Qv.k



+

γGk ΣGi



+

1.083 10

3



kN



=

:=

Hf.H.d γGn Hk



γQn 1QH.k

+

604.05 kN



=

:=

Mf.H.d γGn Mk



γQn 0.7

 Qm.k

+

γGk M



+

av γGk Vk



0.7 Qv.k



γQkk



+

(

)



+

hf1 γGn Hk



γQn 1 QH.k



+

(

)



+

:=

Mf.H.d 156.212 kN m





=

7.2. Sprawdzenie mimośrodu

eL.H.d

Mf.H.d

Vf.H.d

0.144m

=

:=

eL.H.d

LL

6



1

=

warunek spełniony

7.3 Sprawdzenie naprężeń

σmax.H.d

Vf.H.d
LL BB











1

6 eL.H.d



LL

+











719.371 kPa



=

:=

σmin.H.d

Vf.H.d
LL BB











1

6 eL.H.d



LL

-











214.377 kPa



=

:=

7.4 Wymiary efektywne fundamentu

B'H BB 1.45m

=

:=

L'H LL 2 eL.H.d



-

1.312m

=

:=

A'H B'H L'H



1.902m

2

=

:=

7.5 Sprawdzenie warunku na przesuw
7.5.2 bez odpływu ( dla gruntu spoistego pod fundamentem)

Cu

18 kPa



:=

A'

9.76 m

2



:=

Vf.H.d 1.083 10

3



kN



=

Ac A' 9.76m

2

=

:=

Rd Ac Cu



175.68 kN



=

:=

Rd 0.4 Vf.H.d





1

=

warunek spełniony

parcie czynne

σa

i

γi hi



qi

+

(

)















Ka



2 C'



Ka



-

:=

γi

Ka tg 45

φ

'

2

-









2



:=

tg

parcie bierne

σp

j

γj hj



qj

+

(

)















Kp



2 C'



Kp



-

:=

γj