II podejscie projektowe
A1+M1+R2
Zadane w projekcie
Vk
570kN
:=
Hk
42kN
:=
Mk
48kNm
:=
Q1k
530kN
:=
Q2k
50kN
:=
Q3k
43kNm
:=
s1
2.42m
:=
rozpiętość fundamentu
hal
0.6m
:=
szer slupa
sa
0.06m
:=
Przerwa przy slupie
Gr
20.5
kN
m
3
:=
ciezar gruntu
Hpos
0.15m
:=
wys. posadzki
Hpos1
1m
:=
poziom posadzki
hcal
2.15m
:=
zaglebienie fundamentu
t
0.3m
:=
hf
1m
:=
B
0.7m
:=
L
1.1m
:=
eb
0
:=
Z tabelki
ϕ
ϕ
ϕ
ϕ'
21.6 deg
⋅
:=
C'
39.04 kPa
:=
γ
γ
γ
γ'mi
19.688
kN
m
3
:=
miazszosc gruntu
γ
γ
γ
γ'
11.57
kN
m
3
:=
do obliczenia q' ze wzoru z tabeli
γ
γ
γ
γ
21.0843
(
)
kN
m
3
:=
ciezar gruntu
Z normy
γ
γ
γ
γGnk
z normy A1 stale niekorzystne
γ
γ
γ
γGnk
1.35
:=
γ
γ
γ
γGk
z normy A1 stale korzystne
γ
γ
γ
γGk
1
:=
γ
γ
γ
γQnk
z normy A1 zmienne niekorzystne
γ
γ
γ
γQnk
1.5
:=
γ
γ
γ
γQk
z normy A1 zmienne korzystne
γ
γ
γ
γQk
0
:=
γ
γ
γ
γϕ
ϕ
ϕ
ϕ'
z tabelki z normy M1
γ
γ
γ
γϕ
ϕ
ϕ
ϕ'
1
:=
γ
γ
γ
γC'
1
:=
γ
γ
γ
γRv z tabelki z normy R2
γ
γ
γ
γRv
1.4
:=
γ
γ
γ
γRh z tabelki z normy R2
γ
γ
γ
γRh
1.1
:=
ϕ
ϕ
ϕ
ϕ'1
atan
γ
γ
γ
γϕ
ϕ
ϕ
ϕ' tan ϕ
ϕ
ϕ
ϕ'
(
)
⋅
(
)
:=
ϕ
ϕ
ϕ
ϕ'1 0.377
=
Ciezar podloza dzialajacego na fundament
G1
t L
⋅
hal 2 sa
⋅
+
(
)
hf t
−
(
)
⋅
+
hf t
−
(
)
s1 sa
−
(
)
⋅
+
25
kN
m
3
B
⋅
:=
G1 43.505 kN
=
G2
hcal hf
−
(
)
s1
⋅
hf t
−
(
)
.5
⋅
s1 sa
−
(
)
⋅
+
B
⋅ Gr
⋅
:=
G2 51.789 kN
=
G3
hcal Hpos1
−
Hpos
−
hf
−
(
)
s1
⋅
s1 sa
−
(
)
.5
⋅
hf t
−
(
)
⋅
+
B
⋅ Gr
⋅
:=
G3 11.853 kN
=
G4
Hpos s1
⋅
25
⋅
kN
m
3
B
⋅
:=
G4 6.353 kN
=
S2
hcal hf
−
(
)
s1
⋅
.5
⋅
s1
⋅
.5 hf t
−
(
)
⋅
s1 sa
−
(
)
⋅
s1 sa
−
(
)
2
3
⋅
⋅
+
:=
S2 4.667m
3
=
S3
hcal Hpos1
−
Hpos
−
hf
−
(
)
s1
⋅
.5
⋅
s1
⋅
.5 hf t
−
(
)
⋅
s1 sa
−
(
)
⋅
s1 sa
−
(
)
2
3
⋅
⋅
+
:=
S3 1.3 m
3
=
F2
hcal hf
−
(
)
s1
⋅
.5 s1 sa
−
(
)
⋅
hf t
−
(
)
⋅
+
:=
F2 3.609 m
2
=
F3
hcal Hpos
−
Hpos1
−
hf
−
(
)
s1
⋅
s1 sa
−
(
)
hf t
−
(
)
⋅
+
:=
F3 1.652 m
2
=
Promienie na ktorych dziala ciezar podloza
r1
0
:=
r4
.5 s1
⋅
:=
r2
.5 hal
⋅
S2
F2
+
:=
r4 1.21 m
=
r2 1.593 m
=
r3
.5hal
S2
F2
+
:=
r3 1.593 m
=
Obciazenia stale
Vd
γ
γ
γ
γGnk Vk
⋅
γ
γ
γ
γGnk G1 G2
+
G3
+
G4
+
(
)
+
:=
Vd 922.725 kN
=
Hd
γ
γ
γ
γGnk Hk
⋅
:=
Hd 56.7 kN
=
Md
γ
γ
γ
γGnk Mk
⋅
γ
γ
γ
γGnk Hk
⋅
hf
⋅
+
γ
γ
γ
γGnk G2
−
r2
⋅
G3 r3
⋅
+
G4 r4
⋅
+
(
)
+
:=
Md 45.984 kN·m
=
1 Mimosrud
el
Md
Vd
:=
el 0.05 m
=
L
6
0.183 m
=
warunek
spełniony
el
L
6
≤
1
=
2 Naprezenia
σ
σ
σ
σdmax
Vd
L B
⋅
1
6 el
⋅
L
÷
+
(
)
⋅
:=
σ
σ
σ
σdmin
Vd
L B
⋅
1
6 el
⋅
L
−
⋅
:=
σ
σ
σ
σdmax 1.524 10
3
×
kPa
=
σ
σ
σ
σdmin 872.602 kPa
=
warunek
spełniony
σ
σ
σ
σdmax
σ
σ
σ
σdmin
1.747
=
wynik
σ
σ
σ
σdmax
σ
σ
σ
σdmin
3
≤
1
=
3. Efektywne pole podstawy fundamentu
B'
B
2 eb
⋅
−
:=
B'
0.7 m
=
L'
L
2 el
⋅
−
:=
L'
1 m
=
A'
B' L'
⋅
:=
A'
0.7 m
2
=
N - wspolczynniki podloza
Nq
e
π
π
π
π γ
γ
γ
γ ϕ
ϕ
ϕ
ϕ' tan ϕ
ϕ
ϕ
ϕ'
(
)
⋅
tan 45
ϕ
ϕ
ϕ
ϕ'1
2
+
⋅
γ
γ
γ
γϕ
ϕ
ϕ
ϕ'
⋅
tan 45
ϕ
ϕ
ϕ
ϕ'1
2
+
⋅
:=
Nq 23.814
=
Nγγγγ
2 Nq 1
−
(
)
tan
ϕ
ϕ
ϕ
ϕ'
(
)
⋅
γ
γ
γ
γϕ
ϕ
ϕ
ϕ'
⋅
:=
Nγγγγ 18.066
=
Nc
Nq 1
−
(
)
1
tan
ϕ
ϕ
ϕ
ϕ'
(
)
γ
γ
γ
γϕ
ϕ
ϕ
ϕ'
⋅
:=
Nc 57.622
=
b - wspolczynnik nachylenia podstawy fundamentu
α
α
α
α
0
:=
bq
1
α
α
α
α tan ϕ
ϕ
ϕ
ϕ'
(
)
⋅
γ
γ
γ
γϕ
ϕ
ϕ
ϕ'
⋅
−
(
)
:=
bq 1
=
bγγγγ
bq
:=
bγγγγ 1
=
bc
bq
1
bq
−
Nc tan ϕ
ϕ
ϕ
ϕ'
(
)
⋅
γ
γ
γ
γϕ
ϕ
ϕ
ϕ'
⋅
−
:=
bc 1
=
s - wspolczynnik ksztaltu podsrawy fundamentu
Sq
1
B'
L'
sin
ϕ
ϕ
ϕ
ϕ'1
( )
⋅
+
:=
Sq 1.258
=
Sγγγγ
1
.3
B'
L'
−
:=
Sγγγγ 0.79
=
Sc
Sq Nq
⋅
1
−
(
)
Nq 1
−
(
)
:=
Sc 1.269
=
i -- wspolczynnik wplywu nachylenia wypadkowej obciazen
ml
2
L'
B'
+
1
L'
B'
+
:=
ml 1.412
=
iq
1
Hd
Vd A' γγγγC'
⋅
C'
⋅
1
tan
ϕ
ϕ
ϕ
ϕ'
(
)
γ
γ
γ
γϕ
ϕ
ϕ
ϕ'
⋅
⋅
+
−
ml
:=
iq 0.92
=
iγγγγ
1
Hd
Vd A' γγγγC'
⋅
C'
⋅
1
tan
ϕ
ϕ
ϕ
ϕ'
(
)
γ
γ
γ
γϕ
ϕ
ϕ
ϕ'
⋅
⋅
+
−
ml 1
+
:=
iγγγγ 0.868
=
ic
iq
1
iq
+
Nc tan ϕ
ϕ
ϕ
ϕ'
(
)
⋅
γ
γ
γ
γϕ
ϕ
ϕ
ϕ'
⋅
−
:=
ic 0.836
=
q' - naprezenia w podlozu
D ---- zagłębienie fundamentu
D
2.15m
:=
q'
D
γ
γ
γ
γ'
⋅
:=
q'
24.875 kPa
=
Rk
A'
γ
γ
γ
γC' C'
⋅
Nc
⋅
bc
⋅
Sc
⋅
ic
⋅
q' Nq
⋅
bq
⋅
Sq
⋅
iq
⋅
+
.5
γ
γ
γ
γ'mi
⋅
L'
⋅
Nγγγγ
⋅
bγγγγ
⋅
Sγγγγ
⋅
iγγγγ
⋅
+
(
)
⋅
:=
Rk 2.237 10
3
×
kN
=
Rd
Rk
γ
γ
γ
γRv
:=
Rd 1.598 10
3
×
kN
=
Vd 922.725 kN
=
warunek
spełniony
Rd Vd
≥
1
=
N
Vd
Rd
100
⋅
:=
N
82.795
=
Wykorzystanie fundamentu wynosi
N
82.795
=
%
Warunek na przesuniec ie fundamentu
Obciazenia stale
Vd
γ
γ
γ
γGk Vk
⋅
γ
γ
γ
γGk G1 G2
+
G3
+
G4
+
(
)
+
:=
Vd 683.5 kN
=
Hd
γ
γ
γ
γGnk Hk
⋅
:=
Hd 56.7 kN
=
Md
γ
γ
γ
γGnk Mk
⋅
γ
γ
γ
γGnk Hk
⋅
hf
⋅
+
γ
γ
γ
γGk G2
−
r2
⋅
G3 r3
⋅
+
G4 r4
⋅
+
(
)
+
:=
Md 65.562 kN·m
=
I przesuniecie fundamentu w poziomie grunt- beton
Vd 683.5 kN
=
z normy PN-B_03010 tabela 3
z kolumny chropowata na podstawie
kata tarcia wewnetrznego
μ
μ
μ
μ
0.35
:=
Rhk
Vd μ
μ
μ
μ
⋅
:=
Rhk 239.225 kN
=
Rhd
Rhk
γ
γ
γ
γRh
:=
Rhd 217.477 kN
=
Hd 56.7 kN
=
warunek
spełniony
Rhd Hd
>
1
=
II Przesuniecie w gruncie pod fundamentem
Rhk
Vd tan ϕ
ϕ
ϕ
ϕ'
(
)
⋅
γ
γ
γ
γϕ
ϕ
ϕ
ϕ'
⋅
γ
γ
γ
γC' C'
⋅
L
⋅ B
⋅
+
:=
Rhk 300.677 kN
=
Rhd
Rhk
γ
γ
γ
γRh
:=
Rhd 273.343 kN
=
Hd 56.7 kN
=
warunek
spełniony
Rhd Hd
>
1
=
h
2.7m
:=
gr. spoisty
h
B
≤
0
=
b1
h
4
:=
b1 0.675m
=
h
B
>
1
=
b2
h
3
:=
b2 0.9 m
=
gr. niespoisty
h
B
≤
0
=
b3
h
3
:=
b3 0.9 m
=
h
B
>
1
=
b4
2h
3
:=
b4 0.667
=
Wybieramy b odpowiednia dla nas dla gruntu na wysokosci 2B
L1
L
b1
+
:=
L1 1.775 m
=
B1
B
b1
+
:=
B1 1.375 m
=
γ
γ
γ
γ'1
6.87
:=
γ
γ
γ
γ'2
10.39
:=
γ
γ
γ
γ'3
10.3
:=
γ'5
średnia ważona gruntu pod fundamentem
γ
γ
γ
γ'5
γ
γ
γ
γ'1 γγγγ'2
+
γ
γ
γ
γ'3
+
(
)
3
kN
m
3
:=
G5
L1 B1
⋅
h
⋅ γ
γ
γ
γ'5
⋅
:=
G5 60.537 kN
=
Vd
γ
γ
γ
γGk Vk
⋅
γ
γ
γ
γGk G1 G2
+
G3
+
G4
+
G5
+
(
)
+
:=
Vd 744.037 kN
=
Hd
γ
γ
γ
γGnk Hk
⋅
:=
Hd 56.7 kN
=
Md
γ
γ
γ
γGnk Mk
⋅
γ
γ
γ
γGnk Hk
⋅
hf
⋅
+
γ
γ
γ
γGk G2
−
r2
⋅
G3 r3
⋅
+
G4 r4
⋅
+
(
)
+
Hd h
⋅ γ
γ
γ
γGk
⋅
+
:=
Md 218.652 kN·m
=
eb
0
:=
el
Md
Vd
:=
L1
6
0.296 m
=
el 0.294m
=
warunek
spełniony
el
L1
6
≤
1
=
Naprezenia
Naprezenia
B'
B1 2 eb
⋅
−
:=
B'
1.375 m
=
L'
L1 2 el
⋅
−
:=
L'
1.187 m
=
σ
σ
σ
σdmax
Vd
L' B'
⋅
1
6 el
⋅
L'
÷
+
(
)
⋅
:=
σ
σ
σ
σdmin
Vd
L' B'
⋅
1
6 el
⋅
L'
−
⋅
:=
σ
σ
σ
σdmax 1.133 10
3
×
kPa
=
σ
σ
σ
σdmin 421.112 kPa
=
warunek
spełniony
σ
σ
σ
σdmax
σ
σ
σ
σdmin
2.69
=
σ
σ
σ
σdmax
σ
σ
σ
σdmin
3
≤
1
=
N
1
10
3
−
×
kN
=
kN
10
3
N
:=
kN
1 kN
=
kNm
1kN 1
⋅ m
:=
1
m
2
1
1
m
2
=
Pa
1
N
m
2
⋅
=
MPa
10
3
kN
1m
2
:=
°
deg
:=
kPa
1 kPa
=
Obciazenia zmienne
V1d
γ
γ
γ
γGnk Vk
⋅
γ
γ
γ
γGnk G1 G2
+
G3
+
G4
+
(
)
⋅
+
γ
γ
γ
γQnk Q1k
⋅
+
:=
V1d 1.718 10
3
×
kN
=
H1d
γ
γ
γ
γGnk Hk
⋅
γ
γ
γ
γQnk Q2k
⋅
+
:=
H1d 131.7 kN
=
M1d
γ
γ
γ
γGnk Mk
⋅
γ
γ
γ
γGnk Hk
⋅
hf
⋅
+
γ
γ
γ
γGnk G2
−
r2
⋅
G3 r3
⋅
+
G4 r4
⋅
+
(
)
+
γ
γ
γ
γQnk Q3k
⋅
γ
γ
γ
γQnk Q2k
⋅
hf
⋅
+
+
...
:=
M1d 185.484 kN·m
=
1 Mimosrud
e1l
M1d
V1d
:=
e1l 0.108 m
=
L
6
0.183 m
=
warunek
spełniony
e1l
L
6
≤
1
=
2 Naprezenia
σ
σ
σ
σd1max
V1d
L B
⋅
1
6 e1l
⋅
L
÷
+
(
)
⋅
:=
σ
σ
σ
σd1min
V1d
L B
⋅
1
6 e1l
⋅
L
−
⋅
:=
σ
σ
σ
σd1max 3.545 10
3
×
kPa
=
σ
σ
σ
σd1min 916.876 kPa
=
warunek
spełniony
σ
σ
σ
σdmax
σ
σ
σ
σdmin
1.747
=
wynik
σ
σ
σ
σdmax
σ
σ
σ
σdmin
3
≤
1
=
3. Efektywne pole podstawy fundamentu
B'1
B
2 eb
⋅
−
:=
B'1 0.7 m
=
L'1
L
2 e1l
⋅
−
:=
L'1 0.884 m
=
A'1
B'1 L'1
⋅
:=
A'1 0.619 m
2
=
N - wspolczynniki podloza
N1q
e
π
π
π
π γ
γ
γ
γ ϕ
ϕ
ϕ
ϕ'
⋅
tan
ϕ
ϕ
ϕ
ϕ'
(
)
⋅
tan 45
ϕ
ϕ
ϕ
ϕ'1
2
+
⋅
γ
γ
γ
γϕ
ϕ
ϕ
ϕ'
⋅
tan 45
ϕ
ϕ
ϕ
ϕ'1
2
+
⋅
:=
N1q 23.814
=
N1γγγγ
2 N1q 1
−
(
)
tan
ϕ
ϕ
ϕ
ϕ'
(
)
⋅
γ
γ
γ
γϕ
ϕ
ϕ
ϕ'
⋅
:=
N1γγγγ 18.066
=
N1c
N1q 1
−
(
)
1
tan
ϕ
ϕ
ϕ
ϕ'
(
)
γ
γ
γ
γϕ
ϕ
ϕ
ϕ'
⋅
⋅
:=
N1c 57.622
=
b - wspolczynnik nachylenia podstawy fundamentu
α
α
α
α
0
:=
b1q
1
α
α
α
α tan ϕ
ϕ
ϕ
ϕ'
(
)
⋅
γ
γ
γ
γϕ
ϕ
ϕ
ϕ'
⋅
−
(
)
:=
b1q 1
=
b1γγγγ
b1q
:=
b1γγγγ 1
=
b1c
b1q
1
b1q
−
N1c tan ϕ
ϕ
ϕ
ϕ'
(
)
⋅
γ
γ
γ
γϕ
ϕ
ϕ
ϕ'
⋅
−
:=
b1c 1
=
s - wspolczynnik ksztaltu podsrawy fundamentu
S1q
1
B'1
L'1
sin
ϕ
ϕ
ϕ
ϕ'1
( )
⋅
+
:=
S1q 1.291
=
S1γγγγ
1
.3
B'1
L'1
−
:=
S1γγγγ 0.762
=
S1c
S1q N1q
⋅
1
−
(
)
N1q 1
−
(
)
:=
S1c 1.304
=
i -- wspolczynnik wplywu nachylenia wypadkowej obciazen
ml1
2
L'1
B'1
+
1
L'1
B'1
+
:=
ml1 1.442
=
i1q
1
H1d
V1d A'1 γγγγC'
⋅
C'
⋅
1
tan
ϕ
ϕ
ϕ
ϕ'
(
)
γ
γ
γ
γϕ
ϕ
ϕ
ϕ'
⋅
⋅
+
−
ml1
:=
i1q 0.895
=
i1γγγγ
1
H1d
V1d A'1 γγγγC'
⋅
C'
⋅
1
tan
ϕ
ϕ
ϕ
ϕ'
(
)
γ
γ
γ
γϕ
ϕ
ϕ
ϕ'
⋅
⋅
+
−
ml1 1
+
:=
i1γγγγ 0.829
=
i1c
i1q
1
i1q
+
Nc tan ϕ
ϕ
ϕ
ϕ'
(
)
γ
γ
γ
γϕ
ϕ
ϕ
ϕ'
⋅
−
:=
i1c 0.812
=
R1k
A'1 γγγγC' C'
⋅
N1c
⋅
b1c
⋅
S1c
⋅
i1c
⋅
q' N1q
⋅
b1q
⋅
S1q
⋅
i1q
⋅
+
.5
γ
γ
γ
γ'mi
⋅
L'
⋅
N1γγγγ
⋅
b1γγγγ
⋅
S1γγγγ
⋅
i1γγγγ
⋅
+
(
)
⋅
:=
R1k 1.968 10
3
×
kN
=
R1d
R1k
γ
γ
γ
γRv
:=
R1d 1.718 10
3
×
kN
=
V1d 1.405 10
3
×
kN
=
R1d V1d
≥
1
=
warunek
spełniony
N1
V1d
R1d
100
⋅
:=
N1 81.781
=
Wykorzystanie fundamentu wynosi
N1 81.781
=
%
Obciazenia zmienne
V1d
γ
γ
γ
γGk Vk
⋅
γ
γ
γ
γGk G1 G2
+
G3
+
G4
+
(
)
⋅
+
γ
γ
γ
γQk Q1k
⋅
+
:=
V1d 683.5 kN
=
H1d
γ
γ
γ
γGnk Hk
⋅
γ
γ
γ
γQnk Q2k
⋅
+
:=
H1d 131.7 kN
=
M1d
γ
γ
γ
γGnk Mk
⋅
γ
γ
γ
γGnk Hk
⋅
hf
⋅
+
γ
γ
γ
γGk G2
−
r2
⋅
G3 r3
⋅
+
G4 r4
⋅
+
(
)
+
γ
γ
γ
γQnk Q3k
⋅
+
γ
γ
γ
γQnk Q2k
⋅
hf
⋅
+
:=
M1d 205.062 kN·m
=
I przesuniecie fundamentu w poziomie grunt- beton
V1d 683.5 kN
=
z normy PN-B_03010 tabela 3
z kolumny chropowata na podstawie
kata tarcia wewnetrznego
μ
μ
μ
μ
0.35
:=
R1hk
V1d μ
μ
μ
μ
⋅
:=
R1hk 239.225 kN
=
R1hd
Rhk
γ
γ
γ
γRh
:=
R1hd 217.477 kN
=
H1d 131.7 kN
=
warunek
spełniony
R1hd H1d
>
1
=
II Przesuniecie w gruncie pod fundamentem
R1hk
Vd tan ϕ
ϕ
ϕ
ϕ'
(
)
⋅
γ
γ
γ
γϕ
ϕ
ϕ
ϕ'
⋅
γ
γ
γ
γC' C'
⋅
L
⋅ B
⋅
+
:=
R1hk 300.677 kN
=
R1hd
R1hk
γ
γ
γ
γRh
:=
R1hd 273.343 kN
=
H1d 131.7 kN
=
warunek
spełniony
R1hd H1d
>
1
=
V1d
γ
γ
γ
γGk Vk
⋅
γ
γ
γ
γGk G1 G2
+
G3
+
G4
+
G5
+
(
)
⋅
+
γ
γ
γ
γQk Q1k
⋅
+
:=
V1d 744.037 kN
=
H1d
γ
γ
γ
γGnk Hk
⋅
γ
γ
γ
γQnk Q2k
⋅
+
:=
H1d 131.7 kN
=
M1d
γ
γ
γ
γGnk Mk
⋅
γ
γ
γ
γGnk Hk
⋅
hf
⋅
+
γ
γ
γ
γGk G2
−
r2
⋅
G3 r3
⋅
+
G4 r4
⋅
+
(
)
+
γ
γ
γ
γQnk Q3k
⋅
+
γ
γ
γ
γQnk Q2k
⋅
hf
⋅
+
:=
M1d 358.152 kN·m
=
e1b
0
:=
e1l
M1d
V1d
:=
e1l 0.294 m
=
L1
6
0.296 m
=
warunek
spełniony
e1l
L1
6
≤
1
=
Naprezenia
B'1
B1 2 e1b
⋅
−
:=
B'1 1.375 m
=
L'1
L1 2 e1l
⋅
−
:=
L'1 1.187 m
=
σ
σ
σ
σd1max
V1d
L'1 B'1
⋅
1
6 e1l
⋅
L'1
÷
+
(
)
⋅
:=
σ
σ
σ
σd1min
V1d
L'1 B'1
⋅
1
6 e1l
⋅
L'1
−
⋅
:=
σ
σ
σ
σd1max 1.133 10
3
×
kPa
=
σ
σ
σ
σd1min 421.112 kPa
=
warunek
spełniony
σ
σ
σ
σd1max
σ
σ
σ
σd1min
2.69
=
σ
σ
σ
σd1max
σ
σ
σ
σd1min
3
≤
1
=
R1d
1718kN
:=
V1d
1405kN
:=
Vd
1322.725kN
:=
Hd h
⋅ γ
γ
γ
γGk
⋅
+
e1l
el
:=
σ
σ
σ
σdmin
421.112kPa
:=
σ
σ
σ
σd1min
421.112kPa
:=
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