I podejscie projektowe 2 kombinacja
A2+M2+R1
V :=
k
570
H :=
k
42
M :=
k
48
Q
:=
1k
530
Q
:=
2k
50
Q
:=
3k
43
s :=
rozpiętość fundamentu
1
2.42
h
:=
szer slupa
al
0.6
s :=
Przerwa przy slupie
a
0.06
Gr := 20.5
ciezar gruntu
H
:=
wys. posadzki
pos
0.15
H
:=
poziom posadzki
pos1
1
h
:=
zaglebienie fundamentu
cal
2.15
t := 0.3
h :=
f
1
B := 0.7
L := 1.1
e :=
b
0
Z tabelki
ϕ' := 21.6
C' := 39.04
γ'
:=
miazszosc gruntu
mi
19.688
γ' := 11.57
do obliczenia q' ze wzoru z tabeli γ := 21.0843
ciezar gruntu
Z normy
γ
z normy A2 stale niekorzystne Gnk
γ
:=
Gnk
1
γ
z normy A2 stale korzystne
Gk
γ
:=
Gk
1
z normy A2 zmienne niekorzystne Qnk
γ
:=
Qnk
1.3
γ
z normy A2 zmienne korzystne Qk
γ
:=
Qk
0
γ
z tabelki z normy M2
ϕ'
γ
:=
ϕ'
1.25
γ
:=
C'
1.25
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
0.476
Ciezar podloza dzialajacego na fundament G :=
⋅ + (
+ ⋅ )⋅( − ) +
( − )⋅( − ) ⋅
1
t L
hal 2 sa hf t
hf t s1 sa 25 B
G =
1
43.505
G := (
−
)⋅ +
( − )⋅ ⋅( − ) ⋅ ⋅
2
hcal hf s1
hf t .5 s1 sa B Gr
G =
2
51.789
G := (
−
−
−
)⋅ +
( − )⋅ ⋅( − ) ⋅ ⋅
3
hcal Hpos1 Hpos hf s1
s1 sa .5 hf t B Gr
G =
3
11.853
⋅ ⋅ ⋅
4
Hpos s1 25 B
G =
4
6.352
S := (
−
)⋅ ⋅ ⋅ + ⋅( − )⋅( − )⋅ ( − ) 2
⋅
2
hcal hf s1 .5 s1 .5 hf t s1 sa s1 sa
3
S =
2
4.667
S := (
−
−
−
)⋅ ⋅ ⋅ + ⋅( − )⋅( − )⋅ ( − ) 2⋅
3
hcal Hpos1 Hpos hf s1 .5 s1 .5 hf t s1 sa s1 sa
3
S =
3
1.3
F := (
−
)⋅ + ⋅( − )⋅( − )
2
hcal hf s1 .5 s1 sa hf t
F =
2
3.609
F := (
−
−
−
)⋅ + ( − )⋅( − )
3
hcal Hpos Hpos1 hf s1
s1 sa hf t
F =
3
1.652
Promienie na ktorych dziala ciezar podloza r :=
1
0
r :=
⋅
4
.5 s1
S2
r :=
⋅
+
2
.5 hal
F2
r =
4
1.21
r =
2
1.593
S2
r :=
+
3
.5hal
F2
r =
3
1.593
Obciazenia stale
V :=
⋅
+
( +
+
+
)
d
γGnk Vk γGnk G1 G2 G3 G4
V =
d
683.5
H :=
⋅
d
γGnk Hk
H =
d
42
M :=
⋅
+
⋅
⋅
+
(− ⋅ + ⋅ + ⋅ )
d
γGnk Mk γGnk Hk hf γGnk G2 r2 G3 r3 G4 r4
M =
d
34.062
1 Mimosrud
Md
e :=
l
Vd
L
e =
=
l
0.05
0.183
6
L
warunek
e ≤
=
l
1
6
spełniony
V
⋅
d
Vd
6 e
l
σ
:=
⋅( + ⋅ ÷ )
:=
⋅ −
dmax
1
6 el L
σdmin
1
L ⋅B
L ⋅B
L
3
σ
=
×
=
dmax
1.129
10
σdmin 646.372
σ
σ
warunek
σdmax
σ
=
dmax
1.747
wynik
≤ 3 = 1 spełniony
σdmin
σdmin
3. Efektywne pole podstawy fundamentu B' := B − 2 ⋅eb
B' = 0.7
L' := L − 2 ⋅el
L' = 1
A' := B'⋅L'
A' = 0.7
N - wspolczynniki podloza π γ
⋅
ϕ'
ϕ'
ϕ
ϕ' tan( ϕ')
1
1
N :=
⋅
+
⋅
⋅
+
q
e
tan 45
γ
tan 45
ϕ
2
'
2
N =
q
2.33
N :=
( − )⋅
⋅
γ
2 Nq 1 tan(ϕ') γϕ'
N
=
γ
1.371
−
N := (
− )
1
⋅
c
Nq 1 tan(ϕ') γϕ'
N =
c
2.579
−
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 podsrawy fundamentu B'
S :=
+
⋅
( )
q
1
sin ϕ'
1
L'
S =
q
1.321
B'
S :=
−
γ
1
.3
L'
S
=
γ
0.79
(S ⋅ − )
q Nq
1
S :=
c
(N − )
q
1
S =
c
1.562
i -- wspolczynnik wplywu nachylenia wypadkowej obciazen L'
2 + B'
m :=
l
L'
1 + B'
m =
l
1.412
ml
H
d
i :=
−
q
1
1
V +
⋅
⋅ ⋅
d
A' γC' C'
tan(ϕ') ⋅γ
ϕ'
i =
q
0.905
l 1
Hd
i :=
−
γ
1
1
V +
⋅
⋅ ⋅
d
A' γC' C'
tan(ϕ') ⋅γ
ϕ'
i
=
γ
0.844
1 + iq
i :=
−
c
iq
N ⋅
⋅
c tan(ϕ') γϕ'
i =
c
0.528
−
q' - naprezenia w podlozu D ---- zagłębienie fundamentu D := 2.15
q' := D⋅γ'
q' = 24.875
R :=
⋅( ⋅
⋅
⋅ ⋅ ⋅ +
⋅
⋅
⋅
⋅
+
⋅
⋅ ⋅
⋅
⋅
⋅ )
k
A' C' γC' Nc bc Sc ic q' Nq bq Sq iq .5 γ'mi L' Nγ bγ Sγ iγ
R =
k
114.859
Rk
R :=
d
γRv
R =
=
d
780
Vd 683.5
R ≥
=
d
Vd 1
warunek
spełniony
Vd
N :=
⋅100
Rd
N = 87.628
Warunek na przesuniec ie fundamentu Obciazenia stale
V :=
⋅
+
( +
+
+
)
d
γGk Vk γGk G1 G2 G3 G4
V =
d
683.5
H :=
⋅
d
γGnk Hk
H =
d
42
M :=
⋅
+
⋅
⋅
+
(− ⋅ + ⋅ + ⋅ )
d
γGnk Mk γGnk Hk hf γGk G2 r2 G3 r3 G4 r4
M =
d
34.062
I przesuniecie fundamentu w poziomie grunt- beton V =
d
683.5
z normy PN-B_03010 tabela 3
μ := 0.35
z kolumny chropowata na podstawie kata tarcia wewnetrznego
R
:=
⋅
hk
Vd μ
R
=
hk
239.225
Rhk
R
:=
hd
γRh
R
=
=
hd
239.225
Hd 42
warunek
spełniony
R
>
=
hd
Hd 1
II Przesuniecie w gruncie pod fundamentem R
:=
⋅
⋅
+
⋅ ⋅ ⋅
hk
Vd tan(ϕ') γϕ' γC' C' L B
R
=
hk
314.767
−
Rhk
R
:=
hd
γRh
warunek
R
=
=
hd
314.767
Hd 42
spełniony
R
>
=
hd
Hd 1
gr. spoisty
h
h ≤ B = 0
b :=
=
1
b
4
1
0.675
h
h > B = 1
b :=
=
2
b
3
2
0.9
gr. niespoisty
h
h ≤ B = 0
b :=
=
3
b
3
3
0.9
2h
h > B = 1
b :=
=
4
b
3
4
0.667
Wybieramy b odpowiednia dla nas dla gruntu na wysokosci 2B
L :=
+
1
L
b2
L =
1
2
B :=
+
1
B
b2
B =
1
1.6
γ' :=
1
10.39
γ' :=
2
16.68
γ' :=
3
20.15
γ'5
średnia ważona gruntu pod fundamentem (γ' +
+
)
1
γ'2 γ'3
γ' :=
5
3
G :=
⋅
⋅ ⋅
5
L1 B1 h γ'5
G =
5
135.994
V :=
⋅
+
( +
+
+
+
)
d
γGk Vk γGk G1 G2 G3 G4 G5
V =
d
819.493
H :=
⋅
d
γGnk Hk
H =
d
42
M :=
⋅
+
⋅
⋅
+
(− ⋅ + ⋅ + ⋅ ) + ⋅ ⋅
d
γGnk Mk γGnk Hk hf γGk G2 r2 G3 r3 G4 r4
Hd h γGk
M =
d
147.462
e :=
b
0
e :=
l
Vd
L1 = 0.333
e =
l
0.18
6
L
warunek
1
e ≤
=
l
1
spełniony
6
Naprezenia
Naprezenia
B' := B −
⋅
1
2 eb
B' = 1.6
L' := L −
⋅
1
2 el
L' = 1.64
V
⋅
d
Vd
6 e
l
σ
:=
⋅( + ⋅ ÷ )
:=
⋅ −
dmax
1
6 el L'
σdmin
1
L'⋅B'
L'⋅B'
L'
σ
=
=
dmax
517.857
σdmin 106.713
σdmax = 4.853
σdmin
:=
⋅
+
⋅(
+
+
+
) +
⋅
1d
γGnk Vk γGnk G1 G2 G3 G4
γQnk Q1k
3
V
=
×
1d
1.372
10
H
:=
⋅
+
⋅
1d
γGnk Hk γQnk Q2k
H
=
1d
107
M
:=
⋅
+
⋅
⋅
+
(− ⋅ + ⋅ + ⋅ ) +
⋅
+
⋅
⋅
1d
γGnk Mk γGnk Hk hf γGnk G2 r2 G3 r3 G4 r4
γQnk Q3k γQnk Q2k h
M
=
1d
154.962
1 Mimosrud
M1d
e
:=
1l
V1d
L
e
=
=
1l
0.113
0.183
6
warunek
L
e
≤
=
1l
1
spełniony
6
⋅
1d
V1d
6 e
1l
σ
:=
⋅( + ⋅
÷ )
:=
⋅ −
d1max
1
6 e1l L
σd1min
1
L ⋅B
L ⋅B
L
3
σ
=
×
=
d1max
2.88
10
σd1min 684.743
σdmax
σdmax
warunek
= 1.747
wynik
≤ 3 = 1
σdmin
σdmin
spełniony
3. Efektywne pole podstawy fundamentu B' :=
− ⋅
1
B
2 eb
B' =
1
0.7
L' :=
− ⋅
1
L
2 e1l
L' =
1
0.874
A' :=
⋅
1
B'1 L'1
A' =
1
0.612
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
2.33
N
:= (
− )⋅
⋅
1γ
2 N1q 1 tan(ϕ') γϕ'
N
=
1γ
1.371
−
N
:= (
− )
1
⋅
1c
N1q 1 tan(ϕ')⋅γϕ'
N
=
1c
2.579
−
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 podsrawy fundamentu B'
1
S
:=
+
⋅
( )
1q
1
sin ϕ'1
L'
1
S
=
1q
1.367
B'
1
S
:=
−
1γ
1
.3
L'
1
S
=
1γ
0.76
(S ⋅
− )
1q N1q
1
S
:=
1c
(N − )
1q
1
S
=
1c
1.643
i -- wspolczynnik wplywu nachylenia wypadkowej obciazen L'1
2 + B'1
m
:=
l1
L'1
1 + B'1
m
=
l1
1.445
ml1
H
1d
i
:= −
1q
1
1
V
+
⋅
⋅ ⋅
1d
A'1 γC' C'
tan(ϕ') ⋅γ
ϕ'
i
=
1q
0.885
l1 1
H1d
i
:= −
1γ
1
1
V
+
⋅
⋅ ⋅
1d
A'1 γC' C'
tan(ϕ') ⋅γ
ϕ'
i
=
1γ
0.813
1 + i1q
i
:=
−
1c
i1q
N ⋅
c tan(ϕ') γϕ'
i
=
1c
0.533
−
R
:=
⋅( ⋅
⋅
⋅
⋅
⋅
+
⋅
⋅
⋅
⋅
+
⋅
⋅ ⋅
⋅
⋅
⋅
)
1k
A'1 C' γC' N1c b1c S1c i1c q' N1q b1q S1q i1q .5 γ'mi L' N1γ b1γ S1γ i1γ
R
=
1k
105.207
R1k
R
:=
1d
γRv
R
=
=
1d
815
V1d 691
R
≥
=
1d
V1d 1
warunek
spełniony
V1d
N :=
⋅
1
100
R1d
N =
1
84.785
:=
⋅
+
⋅(
+
+
+
) +
⋅
1d
γGk Vk γGk G1 G2 G3 G4
γQk Q1k
V
=
1d
683.5
H
:=
⋅
+
⋅
1d
γGnk Hk γQnk Q2k
H
=
1d
107
M
:=
⋅
+
⋅
⋅
+
(− ⋅ + ⋅ + ⋅ ) +
⋅
+
⋅
⋅
1d
γGnk Mk γGnk Hk hf γGk G2 r2 G3 r3 G4 r4
γQnk Q3k γQnk Q2k hf
M
=
1d
154.962
I przesuniecie fundamentu w poziomie grunt- beton V
=
1d
683.5
z normy PN-B_03010 tabela 3
μ := 0.35
z kolumny chropowata na podstawie kata tarcia wewnetrznego
R
:=
⋅
1hk
V1d μ
R
=
1hk
239.225
Rhk
R
:=
1hd
γRh
R
=
=
1hd
239.225
H1d 107
warunek
spełniony
R
>
=
1hd
H1d 1
II Przesuniecie w gruncie pod fundamentem R
:=
⋅
⋅
+
⋅ ⋅ ⋅
1hk
Vd tan(ϕ') γϕ' γC' C' L B
R
=
1hk
314.767
−
R1hk
R
:=
1hd
γRh
R
=
=
1hd
314.767
H1d 107
warunek
spełniony
R
>
=
1hd
H1d 1
:=
⋅
+
⋅(
+
+
+
+
) +
⋅
1d
γGk Vk γGk G1 G2 G3 G4 G5
γQk Q1k
V
=
1d
819.493
H
:=
⋅
+
⋅
1d
γGnk Hk γQnk Q2k
H
=
1d
107
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
154.962
e
:=
1b
0
e
:=
1l
V1d
L1
e
=
=
1l
0.189
0.333
6
warunek
spełniony
L1
e
≤
=
1l
1
6
Naprezenia
B' :=
− ⋅
1
B1 2 e1b
B' =
1
1.6
L' :=
− ⋅
1
L1 2 e1l
L' =
1
1.622
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
σ
=
=
d1max
536.741
σd1min 94.879
σd1max = 5.657
σd1min
1
0.476
d
780
R
:=
:=
1d
815 V1d
691
R
R
:=
⋅
hk
hd
( 1
− )
γ
R
:=
⋅( 1
− )
Rh
1hd
γRh