DYNAMIKA BUDOWLI
Projekt komina stalowego
DANE OGÓLNE:
Wysoko
ść
komina:
H
140m
:=
Promie
ń
wewn
ę
trzny komina:
rk
2.5m
:=
Ś
rednica zewn
ę
trzna:
dz
2 rk
⋅
:=
dz 5 m
=
Lokalizacja:
Gda
ń
sk
Rodzaj terenu:
C
DANE MATERIAŁOWE:
STAL St3SX:
Wytrzymało
ść
obliczeniowa stali:
fd
205MPa
:=
Moduł spr
ęż
ysto
ś
ci podłu
ż
nej:
E
205GPa
:=
G
ę
sto
ść
stali:
ρ
s
7850
kg
m
3
:=
Ci
ęż
ar jednostkowy:
γ
s
ρ
s g
⋅
:=
γ
s
76.982
kN
m
3
⋅
=
WYKŁADZINA CEGLANA:
Grubo
ść
okładziny:
gc
0.2m
:=
Wysoko
ść
okładziny:
H1
0.6 H
⋅
:=
H1 84m
=
Ci
ęż
ar jednostkowy wykładziny:
ρ
w
1800
kg
m
3
:=
γ
w
ρ
w g
⋅
:=
γ
w
17.652
kN
m
3
⋅
=
ZESTAWIENIE OBCI
Ąś
ENIA WIATREM:
Strefa obci
ąż
enia wiatrem: II (Gda
ń
sk)
Warto
ść
charakterystyczna ci
ś
nienia pr
ę
dko
ś
ci wiatru:
qk
350Pa
:=
Współczynnik oporu aerodynamicznego (przekrój kołowy)
Cz
1.3
:=
Współczynnik ekspozycji (teren C):
dla wysoko
ś
ci 0 - 30m:
•
Ce1
0.7
:=
dla wysoko
ś
ci 30 - 100m:
•
Ce2 z
( )
0.5
0.007 z
⋅
+
:=
Ce2 100
(
)
1.2
=
dla wysoko
ś
ci 100 - 140m:
•
Ce3 z
( )
0.75
0.0045 z
⋅
+
:=
Ce3 140
(
)
1.38
=
Współczynnik działania porywów wiatru:
β
3
:=
Zakładamy:
d
2 rk
⋅
0.4m
+
:=
d
5.4 m
=
Współczynnik bezpiecze
ń
stwa:
γ
fw
1.3
:=
Obci
ąż
enie wiatrem na jednostk
ę
długo
ś
ci komina:
pk1
qk Ce1
⋅
Cz
⋅
β
⋅
d
⋅
:=
pk1 5.16
kN
m
⋅
=
p1
pk1
γ
fw
⋅
:=
p1 6.708
kN
m
⋅
=
pk2
qk Ce2 100
(
)
⋅
Cz
⋅
β
⋅
d
⋅
:=
pk2 8.845
kN
m
⋅
=
p2
pk2
γ
fw
⋅
:=
p2 11.499
kN
m
⋅
=
pk3
qk Ce3 140
(
)
⋅
Cz
⋅
β
⋅
d
⋅
:=
pk3 10.172
kN
m
⋅
=
p3
pk3
γ
fw
⋅
:=
p3 13.224
kN
m
⋅
=
Strona -1-
DYNAMIKA BUDOWLI
Projekt komina stalowego
WST
Ę
PNE PRZYJ
Ę
CIE WYMIARÓW PRZEKROJU KOMINA:
Ś
rednie obci
ąż
enie od obci
ąż
enia trapezowego:
Obci
ąż
enie charakterystyczne:
pksr
0.5 pk2 pk3
+
(
)
⋅
:=
pksr 9.509
kN
m
⋅
=
Obci
ąż
enie obliczeniowe:
psr
γ
fw pksr
⋅
:=
psr 12.361
kN
m
⋅
=
Przyj
ę
cie grubo
ś
ci blachy z warunku no
ś
no
ś
ci:
•
Mmax
psr H
2
⋅
2
:=
Mmax 121139.437 kNm
⋅
=
W
Mmax
fd
:=
W
590924.081 cm
3
⋅
=
tk1
1m
:=
Given
π
64
dz
4
dz 2 tk1
⋅
−
(
)
4
−
⋅
0.5 dz
⋅
W
−
0
=
t1
Find tk1
( )
:=
t1 30.655 mm
⋅
=
Przyj
ę
cie grubo
ś
ci blachy z warunku dopuszczalnego ugi
ę
cia:
•
tk2
1m
:=
Given
pksr H
4
⋅
8 E
⋅
π
64
dz
4
dz 2 tk2
⋅
−
(
)
4
−
⋅
⋅
H
150
−
0
=
t2
Find tk2
( )
:=
t2 50.102 mm
⋅
=
t
max t1 t2
,
( )
:=
t
50.102 mm
⋅
=
Przyj
ę
te wymiary przekroju komina:
•
d
z
t
d
w
t
t
55mm
:=
dz 5 m
=
dw
dz 2 t
⋅
−
:=
dw 4.89 m
⋅
=
Moment bezwładno
ś
ci przekroju komina:
I
π
64
dz
4
dw
4
−
⋅
:=
I
2.612 m
4
⋅
=
Ugi
ę
cie komina:
u
pksr H
4
⋅
8E I
⋅
:=
u
85.273 cm
⋅
=
Ugi
ę
cie dopuszczalne:
H
150
0.933 m
=
Strona -2-
DYNAMIKA BUDOWLI
Projekt komina stalowego
MODEL OBLICZENIOWY KOMINA:
m
0
m
1
m
2
m
3
m
4
3
5
3
5
3
5
3
5
1
7
.5
3
5
3
5
3
5
1
7
.5
1
4
0
m
0
m
3
m
2
m
1
3
1
.5
3
5
m
4
hm
H
4
:=
hm 35m
=
h
hm
hm
hm
hm
2
:=
h
35
35
35
17.5
m
=
hmokl
31.5m
:=
OBLICZENIE MACIERZY MAS:
m1
π
dz
2
4
dw
2
4
−
⋅
hm
⋅
ρ
s
⋅
π
dw
2
4
dw 2 gc
⋅
−
(
)
2
4
−
⋅
hm
⋅
ρ
w
⋅
+
:=
m1 4.204 10
5
×
kg
=
m2
π
dz
2
4
dw
2
4
−
⋅
hm
⋅
ρ
s
⋅
π
dw
2
4
dw 2 gc
⋅
−
(
)
2
4
−
⋅
hmokl
⋅
ρ
w
⋅
+
:=
m2 4.018 10
5
×
kg
=
m3
π
dz
2
4
dw
2
4
−
⋅
hm
⋅
ρ
s
⋅
:=
m3 2.348 10
5
×
kg
=
m4
π
dz
2
4
dw
2
4
−
⋅
hm
2
⋅
ρ
s
⋅
:=
m4 1.174 10
5
×
kg
=
M
m1
0
0
0
0
m2
0
0
0
0
m3
0
0
0
0
m4
:=
M
4.204
10
5
×
0
0
0
0
4.018
10
5
×
0
0
0
0
2.348
10
5
×
0
0
0
0
1.174
10
5
×
kg
=
Strona -3-
DYNAMIKA BUDOWLI
Projekt komina stalowego
OBLICZENIE MACIERZY PODATNO
Ś
CI:
M1
M2
M3
M4
P
P
P
P
P
1
:=
Mu i()
P i
⋅
hm
⋅
:=
M1
Mu 1
( )
:=
M1
35 m
=
M2
Mu 2
( )
:=
M2
70 m
=
M3
Mu 3
( )
:=
M3
105 m
=
M4
Mu 4
( )
:=
M4
140 m
=
Obliczenie współczynników macierzy podatno
ś
ci:
•
δ
1 1
,
1
E I
⋅
1
3
M1
⋅
M1
⋅
hm
⋅
⋅
:=
δ
1 1
,
2.669
10
5
−
×
m
kN
⋅
=
δ
2 2
,
1
E I
⋅
1
3
M2
⋅
M2
⋅
2
⋅
hm
⋅
:=
δ
2 2
,
2.135
10
4
−
×
m
kN
⋅
=
δ
3 3
,
1
E I
⋅
1
3
M3
⋅
M3
⋅
3
⋅
hm
⋅
:=
δ
3 3
,
7.206
10
4
−
×
m
kN
⋅
=
δ
4 4
,
1
E I
⋅
1
3
M4
⋅
M4
⋅
4
⋅
hm
⋅
:=
δ
4 4
,
1.708
10
3
−
×
m
kN
⋅
=
δ
1 2
,
1
E I
⋅
hm
6
2 M1
⋅
M2
⋅
2 0
⋅
M2
2
⋅
+
M1
M2
2
⋅
+
0 M2
⋅
+
⋅
⋅
:=
δ
1 2
,
6.673
10
5
−
×
m
kN
⋅
=
δ
1 3
,
1
E I
⋅
hm
6
2 M1
⋅
M3
⋅
2 0
⋅
2M3
3
⋅
+
M1
2M3
3
⋅
+
0 M3
⋅
+
⋅
⋅
:=
δ
1 3
,
1.068
10
4
−
×
m
kN
⋅
=
δ
1 4
,
1
E I
⋅
hm
6
2 M1
⋅
M4
⋅
2 0
⋅
3M4
4
⋅
+
M1
3M4
4
⋅
+
0 M4
⋅
+
⋅
⋅
:=
δ
1 4
,
1.468
10
4
−
×
m
kN
⋅
=
δ
2 3
,
1
E I
⋅
2hm
6
2 M2
⋅
M3
⋅
2 0
⋅
M3
3
⋅
+
M2
M3
3
⋅
+
0 M3
⋅
+
⋅
⋅
:=
δ
2 3
,
3.737
10
4
−
×
m
kN
⋅
=
δ
2 4
,
1
E I
⋅
2hm
6
2 M2
⋅
M4
⋅
2 0
⋅
M4
2
⋅
+
M2
M4
2
⋅
+
0 M4
⋅
+
⋅
⋅
:=
δ
2 4
,
5.338
10
4
−
×
m
kN
⋅
=
δ
3 4
,
1
E I
⋅
3hm
6
2 M3
⋅
M4
⋅
2 0
⋅
M4
4
⋅
+
M3
M4
4
⋅
+
0 M4
⋅
+
⋅
⋅
:=
δ
3 4
,
1.081
10
3
−
×
m
kN
⋅
=
δ
2 1
,
δ
1 2
,
:=
δ
3 1
,
δ
1 3
,
:=
δ
4 1
,
δ
1 4
,
:=
δ
3 2
,
δ
2 3
,
:=
δ
4 2
,
δ
2 4
,
:=
δ
4 3
,
δ
3 4
,
:=
Strona -4-
DYNAMIKA BUDOWLI
Projekt komina stalowego
δ
2.669
10
5
−
×
6.673
10
5
−
×
1.068
10
4
−
×
1.468
10
4
−
×
6.673
10
5
−
×
2.135
10
4
−
×
3.737
10
4
−
×
5.338
10
4
−
×
1.068
10
4
−
×
3.737
10
4
−
×
7.206
10
4
−
×
1.081
10
3
−
×
1.468
10
4
−
×
5.338
10
4
−
×
1.081
10
3
−
×
1.708
10
3
−
×
m
kN
⋅
=
OBLICZENIE MACIERZY SZTYWNO
Ś
CI:
K
δ
1
−
:=
K
2.348
10
8
×
1.475
−
10
8
×
5.562
10
7
×
9.27
−
10
6
×
1.475
−
10
8
×
1.792
10
8
×
1.197
−
10
8
×
3.245
10
7
×
5.562
10
7
×
1.197
−
10
8
×
1.236
10
8
×
4.558
−
10
7
×
9.27
−
10
6
×
3.245
10
7
×
4.558
−
10
7
×
2.009
10
7
×
kg
s
2
=
OBLICZENIE CZ
Ę
STO
Ś
CI DRGA
Ń
WŁASNYCH:
M
δ
⋅
0.011
0.027
0.025
0.017
0.028
0.086
0.088
0.063
0.045
0.15
0.169
0.127
0.062
0.215
0.254
0.201
s
2
=
ω
eigenvals M
δ
⋅
(
)
:=
ω
0.44667
0.01692
0.00228
0.00083
s
2
=
ω
k
1
1
ω
1
:=
ω
k
1
1.496 Hz
⋅
=
ω
k
2
1
ω
2
:=
ω
k
2
7.688 Hz
⋅
=
ω
k
3
1
ω
3
:=
ω
k
3
20.952 Hz
⋅
=
ω
k
4
1
ω
4
:=
ω
k
4
34.668 Hz
⋅
=
CZ
Ę
STOTLIWO
ŚĆ
I OKRES DRGA
Ń
WŁASNYCH:
fk
ω
k
2
π
⋅
:=
fk
0.238
1.224
3.335
5.518
Hz
⋅
=
Tk
1
fk
:=
Tk
4.199
0.817
0.3
0.181
s
=
Strona -5-
DYNAMIKA BUDOWLI
Projekt komina stalowego
OBLICZENIE WSPÓŁCZYNNIKA DZIAŁANIA PORYWÓW WIATRU:
n
fk
1 1
,
:=
n
0.238 Hz
⋅
=
Współczynnik warto
ś
ci szczytowej:
•
ψ
2 ln 600 n
⋅
(
)
⋅
0.577
2 ln 600 n
⋅
(
)
⋅
+
:=
ψ
3.333
=
Współczynnik oddziaływania turbulentnego o cz
ę
sto
ś
ciach pozarezonansowych:
•
L
dz
:=
ξ
L
H
:=
ξ
0.036
=
Ce
Ce3 140
(
)
:=
Ce 1.38
=
r
0.1
:=
A
0.042
−
28.8
ξ
⋅
1
+
:=
A
0.021
−
=
B
ξ
−
2.65
ξ
⋅
0.24
+
:=
B
0.107
−
=
C
2.29
0.12
ξ
⋅
−
ξ
1.29
−
24.5
ξ
⋅
3.48
+
+
:=
C
1.998
=
kb
A
ln
H
m
2
⋅
B ln
H
m
⋅
+
C
+
:=
kb 0.965
=
Współczynnik oddziaływania turbulentnego o cz
ę
sto
ś
ciach rezonansowych:
•
Charakterystyczna pr
ę
dko
ść
wiatru:
Strefa II
Vk
24
:=
Pr
ę
dko
ść
wiatru na poziomie całkowitej wysoko
ś
ci budowli:
VH
Vk
Ce
⋅
:=
VH 28.194
=
Współczynnik zmniejszaj
ą
cy rezonansowe oddziaływanie porywów:
KL
π
3
1
1
8 n
⋅
H
⋅
3 VH
⋅
m
⋅
+
⋅
1
1
10 n
⋅
L
⋅
VH m
⋅
+
⋅
:=
KL 0.177
=
Współczynnik energii porywów:
x
1200 n
⋅
VH
:=
x
10.136
=
Ko
x
2
1
x
2
+
(
)
4
3
:=
Ko 0.211
=
Logarytmiczny dekrement tłumienia drga
ń
:
∆
0.02
:=
kr
2
π
⋅
KL
⋅
Ko
⋅
∆
:=
kr 11.738
=
β
1
ψ
r
Ce
kb kr
+
(
)
⋅
⋅
+
:=
β
4.198
=
Strona -6-
DYNAMIKA BUDOWLI
Projekt komina stalowego
ZESTAWIENIE OBCI
Ąś
E
Ń
DLA NOWEGO WSPÓŁCZYNNIKA
β
:
Obci
ąż
enie wiatrem na jednostk
ę
długo
ś
ci komina:
pk1
qk Ce1
⋅
Cz
⋅
β
⋅
d
⋅
:=
pk1 7.22
kN
m
⋅
=
p1
pk1
γ
fw
⋅
:=
p1 9.387
kN
m
⋅
=
pk2
qk Ce2 100
(
)
⋅
Cz
⋅
β
⋅
d
⋅
:=
pk2 12.378
kN
m
⋅
=
p2
pk2
γ
fw
⋅
:=
p2 16.091
kN
m
⋅
=
pk3
qk Ce3 140
(
)
⋅
Cz
⋅
β
⋅
d
⋅
:=
pk3 14.235
kN
m
⋅
=
p3
pk3
γ
fw
⋅
:=
p3 18.505
kN
m
⋅
=
Ś
rednie obci
ąż
enie od obci
ąż
enia trapezowego:
Obci
ąż
enie charakterystyczne:
pksr
0.5 pk2 pk3
+
(
)
⋅
:=
pksr 13.306
kN
m
⋅
=
Obci
ąż
enie obliczeniowe:
psr
γ
fw pksr
⋅
:=
psr 17.298
kN
m
⋅
=
SPRAWDZENIE WARUNKÓW:
Sprawdzenie warunku ze wzgl
ę
du na no
ś
no
ść
:
•
Mmax
psr H
2
⋅
2
:=
Mmax 169521.011 kNm
⋅
=
W
Mmax
fd
:=
W
826931.762 cm
3
⋅
=
tk1
1m
:=
Given
π
64
dz
4
dz 2 tk1
⋅
−
(
)
4
−
⋅
0.5 dz
⋅
W
−
0
=
t1
Find tk1
( )
:=
t1 43.223 mm
⋅
=
Sprawdzenie warunku ze wzgl
ę
du na ugi
ę
cie:
•
tk2
1m
:=
Given
pksr H
4
⋅
8 E
⋅
π
64
dz
4
dz 2 tk2
⋅
−
(
)
4
−
⋅
⋅
H
150
−
0
=
t2
Find tk2
( )
:=
t2 71 mm
⋅
=
t
max t1 t2
,
( )
:=
t
71 mm
⋅
=
Przyj
ę
te wymiary przekroju komina:
•
t
72mm
:=
dz 5 m
=
dw
dz 2 t
⋅
−
:=
dw 4.856 m
⋅
=
Strona -7-
DYNAMIKA BUDOWLI
Projekt komina stalowego
Moment bezwładno
ś
ci przekroju komina:
I
π
64
dz
4
dw
4
−
⋅
:=
I
3.385 m
4
⋅
=
Ugi
ę
cie komina:
u
pksr H
4
⋅
8E I
⋅
:=
u
92.093 cm
⋅
=
Ugi
ę
cie dopuszczalne:
H
150
93.333 cm
⋅
=
OBLICZENIE MACIERZY MAS:
m1
π
dz
2
4
dw
2
4
−
⋅
hm
⋅
ρ
s
⋅
π
dw
2
4
dw 2 gc
⋅
−
(
)
2
4
−
⋅
hm
⋅
ρ
w
⋅
+
:=
m1 4.906 10
5
×
kg
=
m2
π
dz
2
4
dw
2
4
−
⋅
hm
⋅
ρ
s
⋅
π
dw
2
4
dw 2 gc
⋅
−
(
)
2
4
−
⋅
hmokl
⋅
ρ
w
⋅
+
:=
m2 4.721 10
5
×
kg
=
m3
π
dz
2
4
dw
2
4
−
⋅
hm
⋅
ρ
s
⋅
:=
m3 3.063 10
5
×
kg
=
m4
π
dz
2
4
dw
2
4
−
⋅
hm
2
⋅
ρ
s
⋅
:=
m4 1.531 10
5
×
kg
=
M
m1
0
0
0
0
m2
0
0
0
0
m3
0
0
0
0
m4
:=
M
4.906
10
5
×
0
0
0
0
4.721
10
5
×
0
0
0
0
3.063
10
5
×
0
0
0
0
1.531
10
5
×
kg
=
OBLICZENIE CZ
Ę
STO
Ś
CI DRGA
Ń
WŁASNYCH:
M
δ
⋅
0.013
0.032
0.033
0.022
0.033
0.101
0.114
0.082
0.052
0.176
0.221
0.166
0.072
0.252
0.331
0.262
s
2
=
ω
eigenvals M
δ
⋅
(
)
:=
ω
0.57178
0.02059
0.0028
0.00102
s
2
=
ω
k
1
1
ω
1
:=
ω
k
1
1.322 Hz
⋅
=
ω
k
2
1
ω
2
:=
ω
k
2
6.97 Hz
⋅
=
Strona -8-
DYNAMIKA BUDOWLI
Projekt komina stalowego
ω
k
3
1
ω
3
:=
ω
k
3
18.908 Hz
⋅
=
ω
k
4
1
ω
4
:=
ω
k
4
31.388 Hz
⋅
=
CZ
Ę
STOTLIWO
ŚĆ
I OKRES DRGA
Ń
WŁASNYCH:
fk
ω
k
2
π
⋅
:=
fk
0.21
1.109
3.009
4.996
Hz
⋅
=
Tk
1
fk
:=
Tk
4.751
0.902
0.332
0.2
s
=
A1
1m
:=
-amplituda masy pierwszej
Id
1
0
0
0
0
1
0
0
0
0
1
0
0
0
0
1
:=
AMPLITUDY W PIERWSZEJ POSTACI DRGA
Ń
WŁASNYCH:
Am1
δ
M
⋅
ω
k
1
2
⋅
Id
−
:=
Am1
0.977
−
0.057
0.092
0.126
0.055
0.824
−
0.309
0.441
0.057
0.2
0.614
−
0.579
0.039
0.143
0.289
0.543
−
=
Z1
Am1
1 2
,
Am1
2 2
,
Am1
3 2
,
Am1
1 3
,
Am1
2 3
,
Am1
3 3
,
Am1
1 4
,
Am1
2 4
,
Am1
3 4
,
1
−
Am1
−
(
)
1 1
,
Am1
−
(
)
2 1
,
Am1
−
(
)
3 1
,
⋅
:=
A1
1
Z1
1
Z1
2
Z1
3
:=
A1
1
3.474
6.713
10.22
=
Strona -9-
DYNAMIKA BUDOWLI
Projekt komina stalowego
AMPLITUDY W DRUGIEJ POSTACI DRGA
Ń
WŁASNYCH:
Am2
δ
M
⋅
ω
k
2
2
⋅
Id
−
:=
Am2
0.364
−
1.59
2.544
3.498
1.53
3.897
8.569
12.242
1.588
5.559
9.721
16.081
1.092
3.971
8.04
11.706
=
Z2
Am2
1 2
,
Am2
2 2
,
Am2
3 2
,
Am2
1 3
,
Am2
2 3
,
Am2
3 3
,
Am2
1 4
,
Am2
2 4
,
Am2
3 4
,
1
−
Am2
−
(
)
1 1
,
Am2
−
(
)
2 1
,
Am2
−
(
)
3 1
,
⋅
:=
A2
1
Z2
1
Z2
2
Z2
3
:=
A2
1
1.706
0.314
2.514
−
=
AMPLITUDY W TRZECIEJ POSTACI DRGA
Ń
WŁASNYCH:
Am3
δ
M
⋅
ω
k
3
2
⋅
Id
−
:=
Am3
3.681
11.703
18.725
25.747
11.263
35.043
63.075
90.107
11.69
40.915
77.908
118.362
8.037
29.225
59.181
92.52
=
Z3
Am3
1 2
,
Am3
2 2
,
Am3
3 2
,
Am3
1 3
,
Am3
2 3
,
Am3
3 3
,
Am3
1 4
,
Am3
2 4
,
Am3
3 4
,
1
−
Am3
−
(
)
1 1
,
Am3
−
(
)
2 1
,
Am3
−
(
)
3 1
,
⋅
:=
A3
1
Z3
1
Z3
2
Z3
3
:=
A3
1
0.024
−
0.968
−
0.983
=
AMPLITUDY W TRZECIEJ POSTACI DRGA
Ń
WŁASNYCH:
Am4
δ
M
⋅
ω
k
4
2
⋅
Id
−
:=
Am4
11.9
32.249
51.599
70.949
31.038
98.321
173.812
248.302
32.213
112.747
216.441
326.161
22.147
80.534
163.08
256.707
=
Z4
Am4
1 2
,
Am4
2 2
,
Am4
3 2
,
Am4
1 3
,
Am4
2 3
,
Am4
3 3
,
Am4
1 4
,
Am4
2 4
,
Am4
3 4
,
1
−
Am4
−
(
)
1 1
,
Am4
−
(
)
2 1
,
Am4
−
(
)
3 1
,
⋅
:=
Strona -10-
DYNAMIKA BUDOWLI
Projekt komina stalowego
A4
1
Z4
1
Z4
2
Z4
3
:=
A4
1
1.147
−
1.289
0.805
−
=
MACIERZ AMPLITUD:
A
A1
1
A2
1
A3
1
A4
1
A1
2
A2
2
A3
2
A4
2
A1
3
A2
3
A3
3
A4
3
A1
4
A2
4
A3
4
A4
4
:=
A
1
1
1
1
3.474
1.706
0.024
−
1.147
−
6.713
0.314
0.968
−
1.289
10.22
2.514
−
0.983
0.805
−
=
WYMUSZENIE HARMONICZNE:
Siła wymuszaj
ą
ca:
Ya=Pp*sin(wt)
Pp
15kN
:=
w
0.8
ω
k
1
⋅
:=
w
1.058 Hz
⋅
=
tz
1s
:=
P0
Pp sin w tz
⋅
( )
⋅
:=
P0 13.07 kN
⋅
=
hwym
0.5 H
⋅
:=
hwym 70m
=
1
7
.5
3
5
3
5
3
5
1
7
.5
m
0
m
1
m
2
m
3
m
4
3
1
.5
3
5
P
P
P
P
M4
M3
M2
M1
Mp
P
0
P
0
Moment od siły wymuszaj
ą
cej:
Mp
P0 hwym
⋅
:=
Mp
914.931 kNm
⋅
=
Momenty od sił jednostkowych:
M1
35 m
=
M2
70 m
=
M3
105 m
=
M4
140 m
=
∆
1p
1
E I
⋅
hm
6
2 M1
⋅
Mp
⋅
2 0
⋅
Mp
2
⋅
+
M1
Mp
2
⋅
+
0 Mp
⋅
+
⋅
⋅
:=
∆
1p
6.731
10
4
−
×
m
=
∆
2p
1
E I
⋅
2hm
6
2 M2
⋅
Mp
⋅
2 0
⋅
0
⋅
+
M2 0
⋅
+
0 Mp
⋅
+
(
)
⋅
⋅
:=
∆
2p
2.154
10
3
−
×
m
=
Strona -11-
DYNAMIKA BUDOWLI
Projekt komina stalowego
∆
3p
1
E I
⋅
2hm
6
2 M3
⋅
Mp
⋅
2
M3
3
⋅
0
⋅
+
M3 0
⋅
+
M3
3
Mp
⋅
+
⋅
⋅
:=
∆
3p
3.769
10
3
−
×
m
=
∆
4p
1
E I
⋅
2hm
6
2 M4
⋅
Mp
⋅
2
M4
2
⋅
0
⋅
+
M4 0
⋅
+
M4
2
Mp
⋅
+
⋅
⋅
:=
∆
4p
5.385
10
3
−
×
m
=
δ
'
1 1
,
δ
1 1
,
1
M
1 1
,
w
2
⋅
−
:=
δ
'
1 1
,
1.795
−
10
6
−
×
s
2
kg
=
δ
'
2 2
,
δ
2 2
,
1
M
2 2
,
w
2
⋅
−
:=
δ
'
2 2
,
1.679
−
10
6
−
×
s
2
kg
=
δ
'
3 3
,
δ
3 3
,
1
M
3 3
,
w
2
⋅
−
:=
δ
'
3 3
,
2.197
−
10
6
−
×
s
2
kg
=
δ
'
4 4
,
δ
4 4
,
1
M
4 4
,
w
2
⋅
−
:=
δ
'
4 4
,
4.126
−
10
6
−
×
s
2
kg
=
Given
X1
1kN
:=
X2
1kN
:=
X3
1kN
:=
X4
1kN
:=
δ
'
1 1
,
X1
⋅
δ
1 2
,
X2
⋅
+
δ
1 3
,
X3
⋅
+
δ
1 4
,
X4
⋅
+
∆
1p
−
=
δ
2 1
,
X1
⋅
δ
'
2 2
,
X2
⋅
+
δ
2 3
,
X3
⋅
+
δ
2 4
,
X4
⋅
+
∆
2p
−
=
δ
3 1
,
X1
⋅
δ
3 2
,
X2
⋅
+
δ
'
3 3
,
X3
⋅
+
δ
3 4
,
X4
⋅
+
∆
3p
−
=
δ
4 1
,
X1
⋅
δ
4 2
,
X2
⋅
+
δ
4 3
,
X3
⋅
+
δ
'
4 4
,
X4
⋅
+
∆
4p
−
=
SIŁY BEZWŁADNO
Ś
CI:
X1
X2
X3
X4
Find X1 X2
,
X3
,
X4
,
(
)
:=
B
X1
X2
X3
X4
:=
B
0.915
2.959
3.571
2.656
kN
⋅
=
SPECTRUM ODPOWIEDZI:
Wektor sił pionowych:
•
Q
M
1 1
,
M
2 2
,
M
2 2
,
M
2 2
,
g
⋅
:=
Q
4.811
10
3
×
4.63
10
3
×
4.63
10
3
×
4.63
10
3
×
kN
⋅
=
Strona -12-
DYNAMIKA BUDOWLI
Projekt komina stalowego
Macierze współczynników drga
ń
:
•
K1
K1
i 1
,
Q
i
A
i j
,
⋅
←
j
1 4
..
∈
for
i
1 4
..
∈
for
:=
K2
K2
i 1
,
Q
i
A
i j
,
⋅
A
T
( )
i j
,
⋅
←
j
1 4
..
∈
for
i
1 4
..
∈
for
:=
η
η
i j
,
A
i j
,
K1
i 1
,
K2
i 1
,
⋅
←
j
1 4
..
∈
for
i
1 4
..
∈
for
:=
K1
4.916
10
4
×
1.164
−
10
4
×
4.552
10
3
×
3.727
−
10
3
×
kN
⋅
=
K2
4.916
10
4
×
1.336
10
4
×
5.869
10
3
×
3
10
3
×
kN
⋅
=
η
1
0.872
−
0.776
1.242
−
3.474
1.487
−
0.018
−
1.425
6.713
0.273
−
0.751
−
1.602
−
10.22
2.191
0.762
1
=
Przyspieszeniowe spectrum odpowiedzi:
•
Sa
1 1
,
0.3
ω
k
1
⋅
mm
s
:=
Sa
1 1
,
3.967
10
4
−
×
m
s
2
=
Sa
2 1
,
0.2
ω
k
1
⋅
0.1
ω
k
2
⋅
+
mm
s
:=
Sa
2 1
,
9.614
10
4
−
×
m
s
2
=
Sa
3 1
,
0.2
ω
k
1
⋅
0.1
ω
k
2
⋅
+
0.03
ω
k
3
⋅
+
mm
s
:=
Sa
3 1
,
1.529
10
3
−
×
m
s
2
=
Sa
4 1
,
0.1
ω
k
1
⋅
mm
s
:=
Sa
4 1
,
1.322
10
4
−
×
m
s
2
=
Wektor obci
ąż
e
ń
:
•
P
P
i j
,
Q
i
η
i j
,
⋅
Sa
i
g
⋅
←
j
1 4
..
∈
for
i
1 4
..
∈
for
:=
P
0.195
0.396
−
0.56
0.078
−
0.676
0.675
−
0.013
−
0.089
1.307
0.124
−
0.542
−
0.1
−
1.989
0.995
0.55
0.062
kN
⋅
=
Suma warto
ś
ci sił bezwładno
ś
ci od postaci drga
ń
na poszczególnych kierunkach mas:
•
B
B
i 1
,
1
4
j
P
i j
,
( )
2
∑
=
←
j
1 4
..
∈
for
i
1 4
..
∈
for
:=
B
2.482
1.272
0.954
0.167
kN
⋅
=
Strona -13-
DYNAMIKA BUDOWLI
Projekt komina stalowego
WYMIARY FUNDAMENTU:
Wysoko
ść
fundamentu:
hf
1.2m
:=
Ci
ęż
ar fundamentu:
γ
z
25
kN
m
3
:=
ρ
z
γ
z
g
:=
ρ
z
2.549
10
3
×
kg
m
3
=
Maksymalne warto
ś
ci sił:
Mmax 1.695 10
5
×
kNm
⋅
=
Tmax
Mmax
H
:=
Tmax 1.211 10
3
×
kN
⋅
=
Moment w podstawie fundamentu:
Mf
Mmax Tmax hf
⋅
+
:=
Mf 1.71 10
5
×
kNm
⋅
=
WARUNEK NA NAPR
Ęś
ENIA EKSTREMALNE:
mp
1500kg
:=
masa siły wymuszaj
ą
cej
a
1m
:=
Given
1
4
i
M
i i
,
∑
=
M
1 1
,
2
+
mp
+
a a
⋅
hf
⋅
ρ
z
⋅
+
g
⋅
a
2
6
Mf
a
3
⋅
−
0
=
Find a
( )
26.92 m
=
Przyj
ę
to wymiar fundamentu:
•
a
27m
:=
Masa fundamentu:
•
mf
a a
⋅
hf
⋅
ρ
z
⋅
:=
mf 2.23 10
6
×
kg
⋅
=
Siła osiowa:
•
N
1
4
i
M
i i
,
∑
=
mp
+
mf
+
g
⋅
:=
N
3.583
10
4
×
kN
⋅
=
Moment obracaj
ą
cy w podstawie fundamentu:
•
Mo
Mf
:=
Mo 1.71 10
5
×
kNm
⋅
=
Moment utrzymuj
ą
cy w podstawie fundamentu:
•
Mu
g a
2
hf
⋅
ρ
z
⋅
⋅
hf
2
⋅
N
a
2
⋅
+
:=
Mu 4.968 10
5
×
kNm
⋅
=
Mo Mu
<
Strona -14-
DYNAMIKA BUDOWLI
Projekt komina stalowego
JEDNOSTKI:
ORIGIN
1
:=
mm
1
10
3
−
×
m
=
kN
1000N
:=
Nm
N m
⋅
:=
MPa
MN
m
2
:=
t
1000kg
:=
Pa
N
m
2
:=
MN
1000kN
:=
kNm
kN m
⋅
:=
GPa
GN
m
2
:=
GN
1000MN
:=
MNn
MN m
⋅
:=
kPa
kN
m
2
:=
GNm
GN m
⋅
:=
Strona -15-
DYNAMIKA BUDOWLI
Projekt komina stalowego
Mmaxk
pksr H
2
⋅
2
:=
Mmaxk 93184.182 kNm
⋅
=
u
1
E I
⋅
1
4
Mmaxk
⋅
140
⋅
m 140
⋅
m
⋅
:=
u
85.273 cm
⋅
=
Strona -16-
DYNAMIKA BUDOWLI
Projekt komina stalowego
ms i
( )
π
dz
2
4
dw
2
4
−
⋅
h
i
( )
⋅
ρ
s
⋅
π
dw
2
4
dw 2 gc
⋅
−
(
)
2
4
−
⋅
h
i
( )
⋅
ρ
w
⋅
+
:=
Strona -17-
DYNAMIKA BUDOWLI
Projekt komina stalowego
n
n
Hz
:=
Strona -18-