I podejscie projektowe 1 kombinacja
A1+M1+R1
Zadane w projekcie
V :=
k
570kN
H :=
k
42kN
M :=
k
48kNm
Q
:=
1k
530kN
Q
:=
2k
50kN
Q
:=
3k
43kNm
s :=
rozpiętość fundamentu
1
2.42m
h
:=
szer slupa
al
0.6m
s :=
Przerwa przy slupie
a
0.06m
kN
Gr := 20.5
ciezar gruntu
3
m
H
:=
wys. posadzki
pos
0.15m
H
:=
poziom posadzki
pos1
1m
h
:=
zaglebienie fundamentu
cal
2.15m
t := 0.3m
h :=
f
1m
B := 0.7m
L := 1.1m
e :=
b
0
Z tabelki
ϕ' := 21.6 ⋅deg
C' := 39.04kPa
kN
γ'
:=
miazszosc gruntu
mi
19.688
3
m
kN
γ' := 11.57
do obliczenia q' ze wzoru z tabeli 3
m
kN
γ := (21.0843)
ciezar gruntu
3
m
Z normy
γ
z normy A1 stale niekorzystne
Gnk
γ
:=
Gnk
1.35
γ
z normy A1 stale korzystne
Gk
:=
Gk
1
γ
z normy A1 zmienne niekorzystne Qnk
γ
:=
Qnk
1.5
γ
z normy A1 zmienne korzystne
Qk
γ
:=
Qk
0
γ
z tabelki z normy M1
ϕ'
γ
:=
ϕ'
1
γ
:=
C'
1
γ
z tabelki z normy R1
Rv
γ
:=
Rv
1
γ
z tabelki z normy R1
Rh
γ
:=
Rh
1
ϕ' :=
( ⋅
)
1
atan γϕ' tan(ϕ')
ϕ' =
1
0.377
Ciezar podloza dzialajacego na fundament kN
G :=
⋅ + (
+ ⋅ )⋅( − ) +
( − )⋅( − )
⋅
1
t L
hal 2 sa hf t
hf t s1 sa 25
B
3
m
G =
1
43.505 kN
−
)⋅ +
( − )⋅ ⋅( − ) ⋅ ⋅
2
hcal hf s1
hf t .5 s1 sa B Gr
G =
2
51.789 kN
G := (
−
−
−
)⋅ +
( − )⋅ ⋅( − ) ⋅ ⋅
3
hcal Hpos1 Hpos hf s1
s1 sa .5 hf t B Gr
G =
3
11.853 kN
G :=
⋅ ⋅
kN ⋅
4
Hpos s1 25
B
3
m
G =
4
6.353 kN
S := (
−
)⋅ ⋅ ⋅ + ⋅( − )⋅( − )⋅ ( −
) 2
⋅
2
hcal hf s1 .5 s1 .5 hf t s1 sa s1 sa
3
3
S =
2
4.667 m
S := (
−
−
−
)⋅ ⋅ ⋅ + ⋅( − )⋅( − )⋅ ( −
) 2⋅
3
hcal Hpos1 Hpos hf s1 .5 s1 .5 hf t s1 sa s1 sa
3
3
S =
3
1.3 m
F := (
−
)⋅ + ⋅( − )⋅( − )
2
hcal hf s1 .5 s1 sa hf t
2
F =
2
3.609 m
F := (
−
−
−
)⋅ + ( − )⋅( − )
3
hcal Hpos Hpos1 hf s1
s1 sa hf t
2
F =
3
1.652 m
Promienie na ktorych dziala ciezar podloza r :=
1
0
r :=
⋅
4
.5 s1
S2
r :=
⋅
+
2
.5 hal
F2
r =
4
1.21 m
r =
2
1.593 m
S2
r :=
+
3
.5hal
F2
r =
3
1.593 m
⋅
+
( +
+
+
)
d
γGnk Vk γGnk G1 G2 G3 G4
V =
d
922.725 kN
H :=
⋅
d
γGnk Hk
H =
d
56.7 kN
M :=
⋅
+
⋅
⋅
+
(− ⋅ + ⋅ + ⋅ )
d
γGnk Mk γGnk Hk hf γGnk G2 r2 G3 r3 G4 r4
M =
d
45.984 kN·m
1 Mimosrud
Md
e :=
l
Vd
L
e =
=
l
0.05 m
0.183 m
6
L
warunek
e ≤
=
l
1
6
spełniony
2 Naprezenia
V
⋅
d
Vd
6 e
l
σ
:=
⋅( + ⋅ ÷ )
:=
⋅ −
dmax
1
6 el L
σdmin
1
L ⋅B
L ⋅B
L
3
σ
=
×
=
dmax
1.524
10 kPa
σdmin 872.602 kPa
σ
σ
warunek
dmax
dmax
= 1.747
wynik
≤ 3 = 1
spełniony
σdmin
σdmin
3. Efektywne pole podstawy fundamentu B' := B − 2 ⋅eb
B' = 0.7 m
L' := L − 2 ⋅el
L' = 1 m
A' := B'⋅L'
2
A' = 0.7 m
⋅
ϕ'
ϕ'
ϕ
ϕ' tan( ϕ')
1
1
N :=
⋅
+
⋅
⋅
+
q
e
tan 45
γ
tan 45
ϕ
2
'
2
N =
q
23.814
N :=
( − )⋅
⋅
γ
2 Nq 1 tan(ϕ') γϕ'
N
=
γ
18.066
N := (
− )
1
⋅
c
Nq 1 tan(ϕ') γϕ'
N =
c
57.622
b - wspolczynnik nachylenia podstawy fundamentu α := 0
b := ( − ⋅
⋅
)
q
1
α tan(ϕ') γϕ'
b =
q
1
b :=
γ
bq
b
=
γ
1
1 − bq
b :=
−
c
bq
N ⋅
⋅
c tan(ϕ') γϕ'
b =
c
1
s - wspolczynnik ksztaltu podstawy fundamentu B'
S :=
+
⋅
( )
q
1
sin ϕ'
1
L'
S =
q
1.258
B'
S :=
−
γ
1
.3
L'
S
=
γ
0.79
(S ⋅ − )
q Nq
1
S :=
c
(N − )
q
1
S =
c
1.269
i -- wspolczynnik wplywu nachylenia wypadkowej obciazen L'
2 + B'
m :=
l
L'
1 + B'
m =
l
1.412
m
l
Hd
i :=
−
q
1
1
V +
⋅
⋅ ⋅
d
A' γC' C'
tan(ϕ') ⋅γ
ϕ'
i =
q
0.92
m +
l 1
Hd
i :=
−
γ
1
1
V +
⋅
⋅ ⋅
d
A' γC' C'
tan(ϕ') ⋅γ
ϕ'
i
=
γ
0.868
1 + iq
i :=
−
c
iq
N ⋅
⋅
c tan(ϕ') γϕ'
i =
c
0.836
q' - naprezenia w podlozu
D ---- zagłębienie fundamentu D := 2.15 ⋅m
q' := D⋅γ'
q' = 24.875 kPa
⋅( ⋅
⋅
⋅ ⋅ ⋅ +
⋅
⋅
⋅
⋅
+
⋅
⋅ ⋅
⋅
⋅
⋅ )
k
A' C' γC' Nc bc Sc ic q' Nq bq Sq iq .5 γ'mi L' Nγ bγ Sγ iγ
3
R =
×
k
2.237
10 kN
Rk
R :=
d
γRv
3
R =
×
=
d
2.237
10 kN
Vd 922.725 kN
warunek
R ≥
=
d
Vd 1
spełniony
Vd
N :=
⋅100
Rd
N = 41.255
Wykorzystanie fundamentu wynosi N = 41.255 %
Warunek na przesuniec ie fundamentu Obciazenia stale
V :=
⋅
+
( +
+
+
)
d
γGk Vk γGk G1 G2 G3 G4
V =
d
683.5 kN
H :=
⋅
d
γGnk Hk
H =
d
56.7 kN
M :=
⋅
+
⋅
⋅
+
(− ⋅ + ⋅ + ⋅ )
d
γGnk Mk γGnk Hk hf γGk G2 r2 G3 r3 G4 r4
M =
d
65.562 kN·m
I przesuniecie fundamentu w poziomie grunt- beton V =
d
683.5 kN
μ := 0.35
z normy PN-B_03010 tabela 3
z kolumny chropowata na podstawie kata tarcia wewnetrznego
R
:=
⋅
hk
Vd μ
R
=
hk
239.225 kN
R
:=
hd
γRh
R
=
=
hd
239.225 kN
Hd 56.7 kN
warunek
R
>
=
hd
Hd 1
spełniony
II Przesuniecie w gruncie pod fundamentem R
:=
⋅
⋅
+
⋅ ⋅ ⋅
hk
Vd tan(ϕ') γϕ' γC' C' L B
R
=
hk
300.677 kN
Rhk
R
:=
hd
γRh
R
=
=
hd
300.677 kN
Hd 56.7 kN
warunek
R
>
=
hd
Hd 1
spełniony
h := 2.7m
gr. spoisty
h
h ≤ B = 0
b :=
=
1
b
4
1
0.675 m
h
h > B = 1
b :=
=
2
b
3
2
0.9 m
gr. niespoisty
h
h ≤ B = 0
b :=
=
3
b
3
3
0.9 m
2h
h > B = 1
b :=
=
4
b
3
4
0.667
Wybieramy b odpowiednia dla nas dla gruntu na wysokosci 2B
L :=
+
1
L
b1
L =
1
1.775 m
B :=
+
1
B
b1
B =
1
1.375 m
γ' :=
1
6.87
γ' :=
2
10.39
γ' :=
3
10.3
średnia ważona gruntu pod fundamentem (γ' +
+
)
1
γ'2 γ'3 kN
γ' :=
5
3
3
m
G :=
⋅
⋅ ⋅
5
L1 B1 h γ'5
G =
5
60.537 kN
V :=
⋅
+
( +
+
+
+
)
d
γGk Vk γGk G1 G2 G3 G4 G5
V =
d
744.037 kN
H :=
⋅
d
γGnk Hk
H =
d
56.7 kN
M :=
⋅
+
⋅
⋅
+
(− ⋅ + ⋅ + ⋅ ) + ⋅ ⋅
d
γGnk Mk γGnk Hk hf γGk G2 r2 G3 r3 G4 r4
Hd h γGk
M =
d
218.652 kN·m
e :=
b
0
Md
e :=
l
Vd
L1 = 0.296m
e =
l
0.294 m
6
L
warunek
1
e ≤
=
l
1
spełniony
6
Naprezenia
Naprezenia
B' := B −
⋅
1
2 eb
B' = 1.375 m
L' := L −
⋅
1
2 el
L' = 1.187 m
V
⋅
d
Vd
6 e
l
σ
:=
⋅( + ⋅ ÷ )
:=
⋅ −
dmax
1
6 el L'
σdmin
1
L'⋅B'
L'⋅B'
L'
3
σ
=
×
=
dmax
1.133
10 kPa
σdmin
221.112
−
kPa
σ
σ
warunek
σdmax
σ
=
dmax
5.123
−
wynik
≤ 3 = 1 spełniony
σdmin
σdmin
3
N = 1 × 10
kN kN := 10 N kN = 1 kN
kNm := 1kN⋅1m
1
1
= 1
2
2
m
m
3
N
10 kN
Pa = 1 ⋅
MPa :=
° := deg
kPa = 1 kPa
2
2
m
1m
:=
⋅
+
⋅(
+
+
+
) +
⋅
1d
γGnk Vk γGnk G1 G2 G3 G4
γQnk Q1k
3
V
=
×
1d
1.718
10 kN
H
:=
⋅
+
⋅
1d
γGnk Hk γQnk Q2k
H
=
1d
131.7 kN
M
:=
⋅
+
⋅
⋅
+
(− ⋅ + ⋅ + ⋅ ) ...
1d
γGnk Mk γGnk Hk hf γGnk G2 r2 G3 r3 G4 r4
+ γ
⋅
+
⋅
⋅
Qnk Q3k
γQnk Q2k hf
M
=
1d
185.484 kN·m
1 Mimosrud
M1d
e
:=
1l
V1d
L
e
=
=
1l
0.108 m
0.183 m
6
L
warunek
e
≤
=
1l
1
6
spełniony
2 Naprezenia
V
⋅
1d
V1d
6 e
1l
σ
:=
⋅( + ⋅
÷ )
:=
⋅ −
d1max
1
6 e1l L
σd1min
1
L ⋅B
L ⋅B
L
3
σ
=
×
=
d1max
3.545
10 kPa
σd1min 916.876 kPa
σ
warunek
d1max = 3.866
spelniony
σd1min
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
2
A' =
1
0.619 m
N - wspolczynniki podloza π⋅ γ
⋅
ϕ'
ϕ'
ϕ
ϕ' tan( ϕ')
1
1
N
:=
⋅
+
⋅
⋅
+
1q
e
tan 45
γ
tan 45
ϕ
2
'
2
N
=
1q
23.814
N
:= (
− )⋅
⋅
1γ
2 N1q 1 tan(ϕ') γϕ'
N
=
1γ
18.066
N
:= (
− )
1
⋅
1c
N1q 1 tan(ϕ')⋅γϕ'
N
=
1c
57.622
b - wspolczynnik nachylenia podstawy fundamentu α := 0
b
:= ( − ⋅
⋅
)
1q
1
α tan(ϕ') γϕ'
b
=
1q
1
b
:=
1γ
b1q
b
=
1γ
1
1 − b1q
b
:=
−
1c
b1q
N
⋅
⋅
1c tan(ϕ') γϕ'
b
=
1c
1
s - wspolczynnik ksztaltu podstawy fundamentu B'
1
S
:=
+
⋅
( )
1q
1
sin ϕ'1
L'
1
S
=
1q
1.291
B'
1
S
:=
−
1γ
1
.3
L'
1
S
=
1γ
0.762
(S ⋅
− )
1q N1q
1
S
:=
1c
(N − )
1q
1
S
=
1c
1.304
i -- wspolczynnik wplywu nachylenia wypadkowej obciazen L'1
2 + B'1
m
:=
l1
L'1
1 + B'1
m
=
l1
1.442
m
l1
H1d
i
:= −
1q
1
1
V
+
⋅
⋅ ⋅
1d
A'1 γC' C'
tan(ϕ') ⋅γ
ϕ'
i
=
1q
0.895
m +
l1 1
H1d
i
:= −
1γ
1
1
V
+
⋅
⋅ ⋅
1d
A'1 γC' C'
tan(ϕ') ⋅γ
ϕ'
i
=
1γ
0.829
1 + i1q
i
:=
−
1c
i1q
N ⋅
c tan(ϕ') γϕ'
i
=
1c
0.812
:=
⋅( ⋅
⋅
⋅
⋅
⋅
+
⋅
⋅
⋅
⋅
+
⋅
⋅ ⋅
⋅
⋅
⋅
)
1k
A'1 C' γC' N1c b1c S1c i1c q' N1q b1q S1q i1q .5 γ'mi L' N1γ b1γ S1γ i1γ
3
R
=
×
1k
1.968
10 kN
R1k
R
:=
1d
γRv
3
3
R
=
×
=
×
1d
1.968
10 kN
V1d 1.718 10 kN
R
≥
=
1d
V1d 1
warunek
spełniony
V1d
N :=
⋅
1
100
R1d
N =
1
87.304
Wykorzystanie fundamentu wynosi N =
1
87.304%
Warunek na przesuniec ie fundamentu Obciazenia zmienne
V
:=
⋅
+
⋅(
+
+
+
) +
⋅
1d
γGk Vk γGk G1 G2 G3 G4
γQk Q1k
V
=
1d
683.5 kN
H
:=
⋅
+
⋅
1d
γGnk Hk γQnk Q2k
H
=
1d
131.7 kN
M
:=
⋅
+
⋅
⋅
+
(− ⋅ + ⋅ + ⋅ ) +
⋅
...
1d
γGnk Mk γGnk Hk hf γGk G2 r2 G3 r3 G4 r4
γQnk Q3k
+ γ
⋅
⋅
Qnk Q2k hf
M
=
1d
205.062 kN·m
I przesuniecie fundamentu w poziomie grunt- beton V
=
1d
683.5 kN
μ := 0.35
z normy PN-B_03010 tabela 3
z kolumny chropowata na podstawie kata tarcia wewnetrznego
R
:=
⋅
1hk
V1d μ
R
=
1hk
239.225 kN
R
:=
1hd
γRh
R
=
=
1hd
239.225 kN
H1d 131.7 kN
warunek
R
>
=
1hd
H1d 1
spełniony
II Przesuniecie w gruncie pod fundamentem R
:=
⋅
⋅
+
⋅ ⋅ ⋅
1hk
Vd tan(ϕ') γϕ' γC' C' L B
R
=
1hk
300.677 kN
R1hk
R
:=
1hd
γRh
R
=
=
1hd
300.677 kN
H1d 131.7 kN
warunek
R
>
=
1hd
H1d 1
spełniony
:=
⋅
+
⋅(
+
+
+
+
) +
⋅
1d
γGk Vk γGk G1 G2 G3 G4 G5
γQk Q1k
V
=
1d
744.037 kN
H
:=
⋅
+
⋅
1d
γGnk Hk γQnk Q2k
H
=
1d
131.7 kN
M
:=
⋅
+
⋅
⋅
+
(− ⋅ + ⋅ + ⋅ ) +
⋅
...
1d
γGnk Mk γGnk Hk hf γGk G2 r2 G3 r3 G4 r4
γQnk Q3k
+ γ
⋅
⋅
+
⋅ ⋅
Qnk Q2k hf
Hd h γGk
M
=
1d
358.152 kN·m
e
:=
1b
0
M1d
e
:=
1l
V1d
L1
e
=
=
1l
0.481 m
0.296 m
6
L1
warunek
e
≤
=
1l
0
6
spełniony
Naprezenia
B' :=
− ⋅
1
B1 2 e1b
B' =
1
1.375 m
L' :=
− ⋅
1
L1 2 e1l
L' =
1
0.812 m
V
⋅
1d
V1d
6 e
1l
σ
:=
⋅( + ⋅
÷
)
:=
⋅
−
d1max
1
6 e1l L'1 σd1min
1
L' ⋅
⋅
1 B'1
L'1 B'1
L'
1
3
3
σ
=
×
=
×
d1max
3.035
10 kPa
σd1min
1.703
−
10 kPa
σd1max
σd1max
warunek
= 1.783
−
wynik
≤ 3 = 1
σd1min
σd1min
spełniony