1.0. Założenia i dane projektowe. |
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H = |
102 |
[m] |
H - wysokość komina. |
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Q = |
6700 |
[m3]/h |
Q - wydatek komina. |
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T = |
325 |
oC |
T - temperatura eksploatacji komina. |
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hf = |
2.5 |
[m] |
hf - głębokość posadownia. |
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hcz = |
2 |
[m] |
hcz - wysokość wyprowadzenia czopucha. |
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Beton: |
B37 |
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Stal: |
A-II |
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Strefa wiatrowa: I |
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* średnia prędkość gazów |
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V= |
10 |
[m/s] |
dla H>100m |
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* średnica przy wylocie |
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dw' = |
29.2073705590311 |
[m] |
wewnętrzna średnica drąży komina w szczycie komina |
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to dw = |
3.15 |
[m] |
średnica po zaokrągleniu do 5cm |
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dz = |
3.99 |
[m] |
zewnętrza średnica komina przy wierzchołku |
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tg= |
0.15 |
[m] |
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td= |
0.35 |
[m] |
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ti= |
0.15 |
[m] |
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to= |
0.12 |
[m] |
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* średnica przy podstawie |
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Dz = |
6.03 |
[m] |
średnica zewnetrzna komina przy podstawie |
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Dw = |
5.03 |
[m] |
średnica wewnętrzna komina przy podstawie |
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i = |
0.01 |
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zbieżność zewnętrzna komina na cełej wysokości |
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Wymiary czopucha: |
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Na wysokości hcz znajduje się czopuch. Jego przekrój nie powinien być mniejszy niż powierzchnia |
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wylotu komina Fcz>=Fw |
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Fcz -powierzchnia czopucha. |
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Fw -powierzchnia drąży u wylotu komina. |
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* przekrój wylotu |
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Fw =p*dw2 / 4 = |
7.79311327631118 |
[m2] |
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* przekrój czopucha |
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Fcz = a*1,5*a > Fw |
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Fcz = 1,25*Fw |
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a = |
2.54838401022936 |
[m] |
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Przyjęto a= |
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3.50 |
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Warstwy w przekroju komina : |
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1) pierścień żelbetowy, |
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2) pierścień wymurówki - cegła ceramiczna grubości 12 cm, |
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3) pierścień warstwy izolacyjnej - szkło piankowe czarne grubości 12 cm. |
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Dzielę komin na 14 segmentów, 13 z nich ma 10m a jeden ma 14m. |
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TABLICA NR 1 |
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Geometria komina |
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Numer |
Poziom |
Grubość |
Średn. zewn. |
Grubość |
Średn. zewn. |
Grubość |
Średn. zewn. |
Powierzchnia |
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segmentu |
przekroju |
płaszcza |
komina |
izolacji |
izolacji |
wymurówki |
wymurówki |
odniesienia A |
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[m] |
[cm] |
[m] |
[cm] |
[m] |
[cm] |
[m] |
[m2] |
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I |
102 |
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3.99 |
15 |
3.69 |
12 |
3.39 |
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100 |
15 |
4.03 |
15 |
3.73 |
12 |
3.43 |
56.14 |
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II |
100 |
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4.03 |
15 |
3.73 |
12 |
3.43 |
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90 |
15 |
4.23 |
15 |
3.93 |
12 |
3.63 |
41.3 |
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III |
90 |
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4.23 |
15 |
3.93 |
12 |
3.63 |
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80 |
20 |
4.43 |
15 |
4.03 |
12 |
3.73 |
43.3 |
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IV |
80 |
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4.43 |
15 |
4.03 |
12 |
3.73 |
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70 |
20 |
4.63 |
15 |
4.23 |
12 |
3.93 |
45.3 |
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V |
70 |
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4.63 |
15 |
4.23 |
12 |
3.93 |
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60 |
25 |
4.83 |
15 |
4.33 |
12 |
4.03 |
47.3 |
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VI |
60 |
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4.83 |
15 |
4.33 |
12 |
4.03 |
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50 |
25 |
5.03 |
15 |
4.53 |
12 |
4.23 |
49.3 |
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VII |
50 |
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5.03 |
15 |
4.53 |
12 |
4.23 |
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40 |
30 |
5.23 |
15 |
4.63 |
12 |
4.33 |
51.3 |
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VIII |
40 |
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5.23 |
15 |
4.63 |
12 |
4.33 |
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30 |
30 |
5.43 |
15 |
4.83 |
12 |
4.53 |
53.3 |
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IX |
30 |
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5.43 |
15 |
4.83 |
12 |
4.53 |
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20 |
35 |
5.63 |
15 |
4.93 |
12 |
4.63 |
55.3 |
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X |
20 |
35 |
5.63 |
15 |
4.93 |
12 |
4.63 |
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10 |
35 |
5.83 |
15 |
5.13 |
12 |
4.83 |
57.3 |
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XI |
10 |
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5.83 |
15 |
5.13 |
12 |
4.83 |
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0 |
35 |
6.03 |
15 |
5.33 |
12 |
5.03 |
59.3 |
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XII |
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XIII |
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XIV |
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SUMA A |
559.14 |
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Powierzchnię odniesienia Ai i-tego segmentu obliczam według wzoru : |
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Ai = hi*(dTrzonu z.ig + dTrzonu z.id)*0,5 |
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2.0. Zestawienie obciążeń działających na komin. |
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2.1. Ciężar własny komina. |
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2.1.1. Ciężar własny komina - wartości charakterystyczne. |
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Ciężar własny komina należy obliczyć w dwóch przypadkach dla fazy realizacji i eksploatacji. |
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W fazie realizacji uwzględnia się tylko ciężar trzonu żelbetowego, natomiast w fazie eksploatacji |
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uwzglednia się ciężar płaszcza, izolacji i wykładziny. Dla segmentu najwyższego należy |
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uwzględnić dodatkowo ciężar pochodzący od głowicy, a dla pozostałych segmentów ciężar |
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wsporników. |
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Ciężar własny płaszcza żelbetowego w i-tym segmencie obliczam wg wzoru : |
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Gi = p*ti*hi*(rdi + rgi - ti)*gbet |
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Gi - ciężar własny i-tego segmentu. |
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ti - grubość płaszcza żelbetowego i-tego segmentu. |
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hi - wysokość i-tego segmentu. |
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gbet - ciężar objętościowy betonu; przyjąłem |
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25 |
kN/m3. |
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rdi - promień dolny zewnętrzny i-tego segmentu. |
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rgi - promień górny zewnętrzny i-tego segmentu. |
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Ciężar własny izolacji termicznej w i-tym segmencie obliczam wg wzoru : |
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GIzol(i) = p*tIzol(i)*hi*(rdIzol(i) + rgIzol(i) - tIzol(i))*gIzol |
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GIzol(i) - ciężar izolacji termicznej w i-tym segmencie. |
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tIzol(i) - grubość izolacji termicznej w i-tego segmentu. |
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hi - wysokość i-tego segmentu. |
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gIzol - ciężar objętościowy izolacji termicznej; szkło piankowe - przyjąłem |
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4 |
kN/m3. |
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rdIzol(i) - promień zewnętrzny, dolny warstwy izolacji termicznej ułożonej w i-tym segmencie. |
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rgIzol(i) - promień zewnętrzny, górny warstwy izolacji termicznej ułożonej w i-tym segmencie. |
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Ciężar własny warstwy wykładziny w i-tym segmencie obliczam wg wzoru : |
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GWykł(i) = p*tWykł(i)*hi*(rdWykł(i) + rgWykł(i) - tWykł(i))*gWykł |
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GWykł(i) - ciężar wykładziny w i-tym segmencie. |
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tWykł(i) - grubość wykładziny w i-tego segmentu. |
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hi - wysokość i-tego segmentu. |
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gWykł - ciężar objętościowy warstwy wykładziny; cegła szamotowa - przyjąłem |
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19.5 |
kN/m3. |
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rdWykł(i) - promień zewnętrzny, dolny warstwy wykładziny ułożonej w i-tym segmencie. |
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rgWykł(i) - promień zewnętrzny, górny warstwy wykładziny ułożonej w i-tym segmencie. |
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Ciężar własny głowicy w szczytowym segmencie obliczam według wzoru : |
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Ggł = p*(dz + tg)*gbet*0,5*tg*hgł |
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Ggł - ciężar głowicy |
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tg - grubość płaszcza w najwyższym segmencie. |
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0.2 |
m |
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hgl - wysokość głowicy; hgl = |
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2.5 |
m |
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gbet - ciężar objętościowy betonu. |
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dz - średnica zewnętrzana komiana u wylotu. |
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Ciężar własny wsporników podwykładzinowych obliczamy ze wzoru : |
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Gwsp.i = p*(rgwsp.i - 0,5*awsp)*awspt*hwsp*gbet |
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Gwsi.i - ciężar wspornika i-tego segmentu |
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awsp - szerokość wspornika; przyjmuję a = |
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0.24 |
m |
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hwsp - wysokość wspornika; hwsp = |
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1.25 |
m |
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gbet - ciężar objętościowy betonu. |
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rgwsp,i - promień wewnętrzny, górny trzonu żelbetowego dla i-tego segmentu. |
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TABLICA NR 2 |
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Cieżar komina (wartości charakterystyczne) |
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Numer |
Poziom |
Ciężar |
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Ciężar segmentu |
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Faza montażu |
Faza eksploat. |
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segmentu |
przekroju |
trzonu |
Wspornik |
Izolacja |
Wymurówka |
Segment |
Cież. trzonu |
Cięż. całkow. |
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[m] |
[kN] |
[kN] |
[kN] |
[kN] |
[kN] |
[kN] |
[kN] |
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I |
144 |
82.27 |
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130 |
90.95 |
0.00 |
13.42 |
48.37 |
235.01 |
173.22 |
235.01 |
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II |
130 |
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120 |
468.88 |
41.12 |
69.37 |
250.68 |
830.04 |
683.22 |
1065.06 |
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III |
120 |
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110 |
648.74 |
43.47 |
72.19 |
261.71 |
1026.11 |
1375.43 |
2091.17 |
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IV |
110 |
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100 |
680.15 |
44.65 |
75.02 |
272.73 |
1072.56 |
2100.23 |
3163.73 |
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V |
100 |
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90 |
879.65 |
47.01 |
77.85 |
283.76 |
1288.26 |
3026.89 |
4451.99 |
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VI |
90 |
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80 |
918.92 |
48.18 |
80.68 |
294.79 |
1342.56 |
3993.99 |
5794.56 |
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VII |
80 |
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70 |
1138.04 |
50.54 |
83.50 |
305.82 |
1577.90 |
5182.57 |
7372.46 |
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VIII |
70 |
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60 |
1185.17 |
51.72 |
86.33 |
316.84 |
1640.06 |
6419.45 |
9012.51 |
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IX |
60 |
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50 |
1423.93 |
54.07 |
89.16 |
327.87 |
1895.03 |
7897.45 |
10907.54 |
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X |
50 |
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40 |
1478.90 |
55.25 |
91.99 |
338.90 |
1965.04 |
9431.61 |
12872.58 |
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XI |
40 |
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30 |
1533.88 |
57.61 |
95.76 |
353.60 |
2040.85 |
11023.10 |
14913.43 |
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XII |
30 |
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20 |
0.00 |
-2.83 |
0.00 |
0.00 |
-2.83 |
11020.28 |
14910.60 |
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XIII |
20 |
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10 |
0.00 |
-2.83 |
0.00 |
0.00 |
-2.83 |
11017.45 |
14907.77 |
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XIV |
10 |
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0 |
0.00 |
-2.83 |
0.00 |
0.00 |
-2.83 |
11014.62 |
14904.95 |
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Suma kontrolna |
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14904.95 |
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2.1.2. Ciężar własny komina - wartości obliczeniowe. |
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Przyjęto następujące współczynniki bezpieczeństwa : |
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gf1 = |
1.1 |
- dla trzonu, korony komina i wsporników. |
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gf2 = |
1.2 |
- dla ciężaru własnego warstwy wykładziny. |
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gf3 = |
1.3 |
- dla ciężaru wlasnego izolacji termicznej. |
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TABLICA NR 3 |
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Cieżar komina (wartości obliczeniowe) |
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Numer |
Poziom |
Ciężar |
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Ciężar segmentu |
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Faza montażu |
Faza eksploat. |
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segmentu |
przekroju |
trzonu |
Wspornik |
Izolacja |
Wymurówka |
Segment |
Cież. trzonu |
Cięż. całkow. |
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[m] |
[kN] |
[kN] |
[kN] |
[kN] |
[kN] |
[kN] |
[kN] |
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I |
144 |
90.50 |
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|
130 |
100.04 |
0.00 |
17.45 |
58.05 |
266.03 |
190.54 |
266.03 |
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II |
130 |
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120 |
515.77 |
45.23 |
90.18 |
300.82 |
951.99 |
751.54 |
1218.03 |
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III |
120 |
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110 |
713.61 |
47.82 |
93.85 |
314.05 |
1169.33 |
1512.97 |
2387.36 |
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IV |
110 |
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|
|
|
100 |
748.17 |
49.11 |
97.53 |
327.28 |
1222.09 |
2310.26 |
3609.45 |
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V |
100 |
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90 |
967.61 |
51.71 |
101.20 |
340.51 |
1461.03 |
3329.57 |
5070.49 |
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VI |
90 |
|
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80 |
1010.81 |
53.00 |
104.88 |
353.75 |
1522.43 |
4393.38 |
6592.92 |
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VII |
80 |
|
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70 |
1251.85 |
55.59 |
108.55 |
366.98 |
1782.97 |
5700.82 |
8375.89 |
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VIII |
70 |
|
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60 |
1303.68 |
56.89 |
112.23 |
380.21 |
1853.01 |
7061.40 |
10228.91 |
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IX |
60 |
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50 |
1566.32 |
59.48 |
115.91 |
393.44 |
2135.15 |
8687.20 |
12364.06 |
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X |
50 |
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40 |
1626.80 |
60.78 |
119.58 |
406.68 |
2213.83 |
10374.77 |
14577.89 |
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XI |
40 |
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30 |
1687.27 |
63.37 |
124.48 |
424.32 |
2299.44 |
12125.41 |
16877.33 |
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XII |
30 |
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20 |
0.00 |
-3.11 |
0.00 |
0.00 |
-3.11 |
12122.30 |
16874.22 |
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XIII |
20 |
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10 |
0.00 |
-3.11 |
0.00 |
0.00 |
-3.11 |
12119.19 |
16871.11 |
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XIV |
10 |
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0 |
0.00 |
-3.11 |
0.00 |
0.00 |
-3.11 |
12116.08 |
16868.00 |
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Suma kontrolna |
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16868.00 |
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2.2. Obciążenie wiatrem. |
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Zgodnie z punktem 3.2.2 PN-88/B - 03004 obciążenie charakterystyczne pk wywołane działaniem |
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wiatru na komin jest określone wzorem : |
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pk = qk*Ce*Cx*b*gd |
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qk - charakterystyczne ciśnienie wiatru wg rozdziału 3 normy PN-77/B - 02011, zwiększone o 20%. |
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Ce - współczynnik ekspozycji wg rozdziału 4 PN-77/B - 02011. |
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Cx - współczynnik aerodynamiczny wg punktu 2.4 i załącznika 1 PN-77/B - 02011. |
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b - współczynnik działania porywów wiatru wg punktu 3.2.3 PN-88/B - 03004 oraz punktu 2.5 |
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i załącznika 2 PN-77/B - 02011. |
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gd - współczynnik ujmujący konsekwencje założeń modelowych; przyjmuje następujące wartości : |
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gd = 1,35 dla kominów nizszych niż 100m, |
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gd = 1,30 dla kominów o wysokości od 100 do 250 m, |
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gd = 1,25 dla kominów o wysokości powyzej 250m |
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Dla komina o H = 144 m przyjmuję gd = 1,3 |
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2.2.1. Sprawdzenie podatności komina. |
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T = K*H02*((ΣA0i*gi)/(E*I0*g))0,5 |
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Oznaczenia : |
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T - podstawowy okres drgań własnych komina. |
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A0i - pole powierzchni przekroju poprzecznego danej warstwy komina na poziomie 0, czyli w miejscu |
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połączenia płaszcz żelbetowego z fundamentem. |
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H0 - wysokość trzonu komina ponad fundamentem (H0 = H). |
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gi - ciężar objętościowy danego materiału. |
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E - moduł sprężystości materiału, z którego wykonany jest płaszcz; Eb = |
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32 |
GPa |
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I0 - moment bezwładności tylko płaszcza w poziomie 0. |
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K - współczynnik uwzględniający wpływ zbieżności grubości ścianek komina i średnicy zewnętrznej. |
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A0t = (p*(Dz2 - (Dz - 2*gt)2))/4 = |
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0.00 |
m2 |
gbet = |
25 |
kN/m3 |
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A0i = (p*(Dz2 - (Dz - 2*gi)2))/4 = |
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0.00 |
m2 |
gIzol = |
4 |
kN/m3 |
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A0w = (p*(Dz2 - (Dz - 2*gw)2))/4 = |
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0.00 |
m2 |
gWykł = |
19.5 |
kN/m3 |
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ΣA0i*gi = A0t*gt + A0i*gi + A0w*gw = |
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0.00 |
kN/m |
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I0 = (p*(Dz4 - (Dz - 2*gt)4)) / 64 = |
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0.00 |
m4 |
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H = |
144 |
m |
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K = A(λ) - B(λ)*e(-C(λ)*μ) = |
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#DIV/0! |
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A(λ) = (2,37*λ + 0,91)0,5 = |
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#DIV/0! |
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B(λ) = (0,36*λ + 0,02)0,5 = |
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#DIV/0! |
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C(λ) = 4,58*λ2 - 5,7*λ + 3,92 = |
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#DIV/0! |
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λ = g2 / g1 = |
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#DIV/0! |
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μ = d2 / d1 = |
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#DIV/0! |
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T = K*H02*((ΣA0i*gi)/(E*I0*g))0,5 = |
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#DIV/0! |
s |
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Δ - dekrement tłumienia dla konst.żelbetowych; Δ = |
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0.12 |
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Na podstawie obliczonego okresu drgań własnych T, Δ i rysunku nr 1 PN - 77/B - 02011 określam |
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podatność budowli - komin jest budowlą podatną. |
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2.2.2. Określenie współaczynnika ekspozycji Ce. |
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Obiekt znajduje się w terenie typu A. Współczynnik ekspozycji zależny jest od wysokości nad |
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poziomem terenu : |
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1) dla z ≤ 10m => Ce = 1,0 |
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2) dla z = 10 ÷ 20 => Ce = 0,8 + 0,02*z |
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3) dla z = 20 ÷ 40 => Ce = 0,9 + 0,015*z |
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4) dla z = 40 ÷ 100 => Ce = 1,23 + 0,0067*z |
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5) dla z = 100 ÷ 280 => Ce = 1,5 + 0,004*z |
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TABLICA NR 4 |
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Współczynnik ekspozycji i zmiana jego wartości |
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L.p. |
zi [m] |
Cei |
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1 |
144 |
2.076 |
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2 |
130 |
2.02 |
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3 |
120 |
1.98 |
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4 |
110 |
1.94 |
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5 |
100 |
1.9 |
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6 |
90 |
1.833 |
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7 |
80 |
1.766 |
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8 |
70 |
1.699 |
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9 |
60 |
1.632 |
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10 |
50 |
1.565 |
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11 |
40 |
1.5 |
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12 |
30 |
1.35 |
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13 |
20 |
1.2 |
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14 |
10 |
1 |
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15 |
0 |
1 |
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2.2.3. Określenie charakterystycznego ciśnienia wiatru qk. |
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Komin zlokalizowany jest w strefie wiatrowej I. |
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Według PN - 77/B - 02011 wynosi ono: qkPN = |
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250 |
Pa = |
0.25 |
kPa |
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Ostatecznie ciśnienie prędkości wiatru wynosi : |
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qk = 1,2*qkPN = |
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300 |
Pa = |
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kPa |
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2.2.4. Współczynnik aerodynamiczny Cx. |
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Dśr. - średnia średnicy. |
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Dśr. = ΣAi / H = |
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3.88291666666667 |
m |
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H / Dśr. = |
37.09 |
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Dla H / Dśr. ≤ 25 Cx obliczamy ze wzoru : |
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Cx = 0,7*(1-0,25*lg(25(Dśr./H))) = |
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0.73 |
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2.2.5. Współczynnik działania porywów wiatru b. |
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b = 1+ ψ*((r/Ce)*(kb+kr))^0,5 |
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Oznaczenia : |
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ψ - współczynnik szczytowej wartości obciążenia. |
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r - współczynnik chropowatości terenu; r = |
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0.08 |
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kb - współczynnik oddziaływania turbulentnego o częstościach pozarezonansowych |
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(o okresie różnym od okresu drgań własnych). |
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kr - współczynnik oddziaływania turbulentnego o częstościach rezonansowych z częstościami |
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drgań własnych konstrukcji. |
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ψ = (2*ln(600*n))0,5 + 0,577/(2(ln(600*n))0,5 ≤ 4,0 |
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n = 1/T = |
#DIV/0! |
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ψ = |
#DIV/0! |
≤ 4,0 |
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kb = A*(lnH)2 + B*lnH + C = |
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0.97 |
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A = - 0,042/(28,8*ξ + 1) = |
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-0.02 |
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B = - ξ/(2,65*ξ + 0,24) = |
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-0.09 |
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C = 2,29 - 0,12*ξ + (ξ - 1,29)/(24,5*ξ + 3,48) = |
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1.98 |
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ξ = Dśr./H= |
0.026964699074074 |
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kr = (2*p*Kl*Ko)/Δ = |
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#DIV/0! |
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Kl - współczynnik zmniejszający oddziaływanie rezonansowe porywów ze względu na rozmiary |
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konstrukcji. |
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K0 - współczynnik energii porywów o częstościach rezonansowych. |
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Kl = (p/3)*(1/(1+ (8*n*H)/(3*VH))*(1/(1+ (10*n*Dśr.)/VH)) = |
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#DIV/0! |
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VH = Vk*Ce0,5 = |
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28.82 |
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Vk = |
20 |
[m/s] |
dla strefy wiatrowej I |
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Ce = Ce1 = |
2.076 |
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K0 = X2/(1+X2)4/3 = |
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#DIV/0! |
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x = (1200*n)/VH = |
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#DIV/0! |
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b = 1+ ψ*((r/Ce)/(kb+kr))^0,5 = |
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#DIV/0! |
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Jeśli b < 2, to przyjmuję b = |
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2 |
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2.2.6. Obliczenie charakterystycznego obciążenia wiatrem. |
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pk = qk*Ce*Cx*b*gd |
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Dla komina o H = 144 m przyjmuję gd = |
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1.3 |
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TABLICA NR 5 |
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Wartości obciążenia komina wiatrem |
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Nr |
Poziom |
pk |
pkśr. |
F. montażu |
F. eksploatacji |
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segmentu |
przekroju |
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[m] |
[kPa] |
[kPa] |
p [kPa] |
p [kPa] |
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I |
144 |
1.182 |
1.166 |
0.933 |
1.399 |
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130 |
1.150 |
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II |
130 |
1.150 |
1.139 |
0.911 |
1.367 |
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120 |
1.127 |
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III |
120 |
1.127 |
1.116 |
0.893 |
1.339 |
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110 |
1.105 |
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IV |
110 |
1.105 |
1.093 |
0.875 |
1.312 |
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100 |
1.082 |
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V |
100 |
1.082 |
1.063 |
0.850 |
1.275 |
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90 |
1.044 |
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VI |
90 |
1.044 |
1.025 |
0.820 |
1.230 |
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80 |
1.006 |
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VII |
80 |
1.006 |
0.986 |
0.789 |
1.184 |
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70 |
0.967 |
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VIII |
70 |
0.967 |
0.948 |
0.759 |
1.138 |
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60 |
0.929 |
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IX |
60 |
0.929 |
0.910 |
0.728 |
1.092 |
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50 |
0.891 |
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X |
50 |
0.891 |
0.873 |
0.698 |
1.047 |
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40 |
0.854 |
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XI |
40 |
0.854 |
0.811 |
0.649 |
0.974 |
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30 |
0.769 |
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XII |
30 |
0.769 |
0.726 |
0.581 |
0.871 |
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20 |
0.683 |
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XIII |
20 |
0.683 |
0.626 |
0.501 |
0.752 |
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10 |
0.569 |
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XIV |
10 |
0.569 |
0.569 |
0.456 |
0.683 |
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0 |
0.569 |
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TABLICA NR 6 |
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Obliczenie sił wewn.wywołanych powiewem wiatru |
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Nr |
Powierz. |
Obc.w.seg. |
p [kPa] |
p [kPa] |
F. Montażu |
F. Ekspl. |
F. Montażu |
F. Ekspl. |
M.od wiatru |
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segmentu |
odniesien. |
W = A*pkśr |
Faza |
Faza |
W = A*p |
W = A*p |
M.od wiatru |
M.od wiatru |
charakt. |
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A [m2] |
[kN] |
montażu |
eksploatacji |
[kN] |
[kN] |
[kNm] |
[kNm] |
[kNm] |
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I |
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56.14 |
65.464 |
0.933 |
1.399 |
52.371 |
78.557 |
366.599 |
549.898 |
458.248 |
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II |
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41.3 |
47.031 |
0.911 |
1.367 |
37.624 |
56.437 |
1078.433 |
1617.650 |
1348.041 |
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III |
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43.3 |
48.322 |
0.893 |
1.339 |
38.658 |
57.986 |
2171.678 |
3257.517 |
2714.597 |
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IV |
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45.3 |
49.522 |
0.875 |
1.312 |
39.618 |
59.427 |
3656.299 |
5484.449 |
4570.374 |
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V |
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47.3 |
50.268 |
0.850 |
1.275 |
40.214 |
60.321 |
5540.080 |
8310.120 |
6925.100 |
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VI |
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49.3 |
50.513 |
0.820 |
1.230 |
40.410 |
60.615 |
7826.982 |
11740.473 |
9783.728 |
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VII |
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51.3 |
50.605 |
0.789 |
1.184 |
40.484 |
60.726 |
10518.353 |
15777.530 |
8761.852 |
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VIII |
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53.3 |
50.544 |
0.759 |
1.138 |
40.435 |
60.653 |
13614.320 |
20421.481 |
17017.900 |
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IX |
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|
|
55.3 |
50.331 |
0.728 |
1.092 |
40.265 |
60.398 |
17113.790 |
25670.685 |
21392.237 |
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X |
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57.3 |
49.998 |
0.698 |
1.047 |
39.999 |
59.998 |
21014.578 |
31521.867 |
26268.223 |
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XI |
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59.3 |
48.114 |
0.649 |
0.974 |
38.491 |
57.737 |
25307.815 |
37961.722 |
31634.769 |
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XII |
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0 |
0.000 |
0.581 |
0.871 |
0.000 |
0.000 |
29793.507 |
44690.260 |
37241.884 |
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XIII |
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0 |
0.000 |
0.501 |
0.752 |
0.000 |
0.000 |
34279.199 |
51418.798 |
42848.999 |
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XIV |
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0 |
0.000 |
0.456 |
0.683 |
0.000 |
0.000 |
38764.891 |
58147.336 |
48456.114 |
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Σ |
560.712 |
|
Σ |
448.569 |
672.854 |
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2.2.7. Sprawdzenie wpływu ugięcia II rzędu. |
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Jeśli współczynnik a ≥ 0,35, to muszę uwzględnić wpływ ugięcia II rzędu. |
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a = H0*(N0 / (E*I0))0,5 |
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H0 = H |
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N0 - całkowite pionowe obciążenie ciężarem własnym komina (do poziomu połączenia z fundam.). |
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I0 - moment bezwładności przekroju płaszcza w poziomie połączenia z płyta fundamentową. |
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E - moduł sprężystości materiału, z którego wykonany jest płaszcz; Eb = |
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32 |
GPa |
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I0 = (p*(Dz4 - (Dz - 2*gt)4)) / 64 = |
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0.000 |
m4 |
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N0 = |
11014.62 |
kN |
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a = H0*(N0/(E*I0))0,5 = |
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#DIV/0! |
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Ponieważ współczynnik a =0,434 > 0,35 to w dalszych obliczeniach trzeba uwzględniać wpływ ugięcia II rzedu. |
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2.2.8. Uwzględnienie wpływu ugięcia II rzędu na wartość momentów wywołanych powiewem wiatru. |
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TABLICA NR 7 |
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Nr |
Poziom |
f |
M0 |
MII |
MII |
MII |
MI + MII |
MI + MII |
MI + MII |
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segmentu |
przekroju |
|
|
|
(montaż) |
(ekspl.) |
|
(montaż) |
(ekspl.) |
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|
[m] |
[-] |
[kNm] |
[kNm] |
[kNm] |
[kNm] |
[kNm] |
[kNm] |
[kNm] |
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I |
144 |
|
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130 |
0.239 |
458.248 |
#DIV/0! |
#DIV/0! |
#DIV/0! |
#DIV/0! |
#DIV/0! |
#DIV/0! |
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II |
130 |
|
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|
120 |
0.097 |
1348.041 |
#DIV/0! |
#DIV/0! |
#DIV/0! |
#DIV/0! |
#DIV/0! |
#DIV/0! |
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III |
120 |
|
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|
110 |
0.029 |
2714.597 |
#DIV/0! |
#DIV/0! |
#DIV/0! |
#DIV/0! |
#DIV/0! |
#DIV/0! |
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IV |
110 |
|
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|
100 |
0.001 |
4570.374 |
#DIV/0! |
#DIV/0! |
#DIV/0! |
#DIV/0! |
#DIV/0! |
#DIV/0! |
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V |
100 |
|
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|
90 |
0.007 |
6925.100 |
#DIV/0! |
#DIV/0! |
#DIV/0! |
#DIV/0! |
#DIV/0! |
#DIV/0! |
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VI |
90 |
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|
80 |
0.041 |
9783.728 |
#DIV/0! |
#DIV/0! |
#DIV/0! |
#DIV/0! |
#DIV/0! |
#DIV/0! |
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VII |
80 |
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|
70 |
0.095 |
8761.852 |
#DIV/0! |
#DIV/0! |
#DIV/0! |
#DIV/0! |
#DIV/0! |
#DIV/0! |
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VIII |
70 |
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|
60 |
0.165 |
17017.900 |
#DIV/0! |
#DIV/0! |
#DIV/0! |
#DIV/0! |
#DIV/0! |
#DIV/0! |
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IX |
60 |
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50 |
0.243 |
21392.237 |
#DIV/0! |
#DIV/0! |
#DIV/0! |
#DIV/0! |
#DIV/0! |
#DIV/0! |
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X |
50 |
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|
40 |
0.323 |
26268.223 |
#DIV/0! |
#DIV/0! |
#DIV/0! |
#DIV/0! |
#DIV/0! |
#DIV/0! |
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XI |
40 |
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30 |
0.400 |
31634.769 |
#DIV/0! |
#DIV/0! |
#DIV/0! |
#DIV/0! |
#DIV/0! |
#DIV/0! |
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XII |
30 |
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|
20 |
0.467 |
37241.884 |
#DIV/0! |
#DIV/0! |
#DIV/0! |
#DIV/0! |
#DIV/0! |
#DIV/0! |
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XIII |
20 |
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|
10 |
0.518 |
42848.999 |
#DIV/0! |
#DIV/0! |
#DIV/0! |
#DIV/0! |
#DIV/0! |
#DIV/0! |
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XIV |
10 |
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0 |
0.546 |
48456.114 |
#DIV/0! |
#DIV/0! |
#DIV/0! |
#DIV/0! |
#DIV/0! |
#DIV/0! |
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2.3. Obciążenia termiczne komina. |
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Temperaturę gazów tw technologicznych należy przyjmować zgodnie z danymi technologicznymi, |
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uwzględniając możliwość jej awaryjnego podwyższenia. |
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Podwyższam awaryjnie temperaturę na wlocie o 20%. |
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Tmax |
= |
325 |
oC |
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tw =1,2*Tmax = |
|
390 |
oC |
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Wyliczam współczynnik przenikalności ciepła dla przegrody cylindrycznej o n warstwach ze wzoru: |
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1/k = 1/an + Σ((gi/λi)*ki*(R/ri)) + 1/a0 |
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, czyli dla rozpatrywanego przypadku będzie on wyglądał następująco : |
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1/k = 1/an + (g1/λ1)*k1*(R/r1) + (g2/λ2)*k2*(R/r2)+ (g3/λ3)*k3*(R/r3) + 1/a0 |
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an - współczynnik napływu ciepła; an = 8 + Vs = |
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18 |
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Vs = |
10 |
m/s |
- prędkość przepływu gazów w kominie. |
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a0 - współczynnik odpływu ciepła; a0 = |
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24 |
zimą lub |
8 |
latem. |
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r1 - promień zewnętrzny wykładziny [m]. |
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r2 - promień zewnętrzny izolacja [m]. |
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r3 = R - promień zewnętrzny trzonu komina [m]. |
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g1, g2, g3 - poszczególne warstwy komina. |
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λ1, λ2, λ3 - współczynniki przewodności poszczególnych warstw. |
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k1, k2, k3 - współczynniki poprwakowe uwzględniajace zakrzywienie ściany; oblicza się je na |
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podstawie zależności : |
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ki = (R/ri)0,47 |
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W cele uzyskania maksymalnych temperatur przyjmujemy : |
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1) w zimie tz = |
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-25 |
oC |
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2) w lecie tz = |
|
35 |
oC |
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Spadek temperatur na poszczególnych warstwach oblicza się nastepująco : |
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Δti = k*(gi/λi)*ki*(R/ri)*Δti |
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Δti - róznica między temperaturą wewnętrzną a temperaturą zewnetrzą. |
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Spadek temperatury związany ze współczynnikiem napływu powietrza an obliczamy : |
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Δtan = k*(1/an)*Δti |
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Spadek temperatury związany ze współczynnikiem odpływu powietrza a0 obliczamy : |
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Δta0 = k*(1/a0)*Δti |
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λp = |
1.51 |
W/(m*K) |
współczynnik przewodności płaszcza |
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λi = |
0.07 |
W/(m*K) |
współczynnik przewodności izolacji |
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λw = |
0.96 |
W/(m*K) |
współczynnik przewodności wymurówki |
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W obliczeniach termicznych kominów o średnicy D ≥ 5,0 m można pominąć wpływ zakrzywienia |
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ściany, tzn. ki = 1,0. |
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Przyjmujemy, iż temperatura gazów w trzonie obniża się wraz ze wzrostem wysokości |
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komina 0,5oC / 1m. |
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TABLICA NR 8 |
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Obliczenie różnicy temperatur - okres letni. |
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Numer |
Poziom |
Grubość |
Grubość |
Grubość |
Promień zewn. |
Promień zewn. |
Promień zewn. |
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k |
tw |
Δt |
Spadek temperatur |
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Temp.z. |
Tem.z. |
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segmentu |
przekroju |
płaszcza |
izolacji |
wymurówki |
płaszcza |
izolacji |
wymurówki |
rn/ri |
rn/rf |
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[oC] |
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płasz. |
powiet. |
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rn = R |
ri |
rf |
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W/ |
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Δtn |
Δti |
Δtf |
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[m] |
[m] |
[m] |
[m] |
[m] |
[m] |
[m] |
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m2*K |
[oC] |
[oC] |
[oC] |
[oC] |
[oC] |
[oC] |
[oC] |
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I |
144 |
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130 |
0.2 |
0.12 |
0.12 |
2.015 |
1.865 |
1.715 |
1.080 |
1.175 |
0.430 |
323 |
288 |
16.42 |
229.57 |
19.636225183041 |
57.38 |
35 |
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II |
130 |
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120 |
0.2 |
0.12 |
0.12 |
2.115 |
1.965 |
1.815 |
1.076 |
1.165 |
0.434 |
330 |
295 |
16.96 |
236.27 |
18.6516951536879 |
58.12 |
35 |
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III |
120 |
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110 |
0.25 |
0.12 |
0.12 |
2.215 |
2.015 |
1.865 |
1.099 |
1.188 |
0.420 |
335 |
300 |
20.88 |
237.63 |
18.7209969020206 |
57.77 |
35 |
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IV |
110 |
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100 |
0.25 |
0.12 |
0.12 |
2.315 |
2.115 |
1.965 |
1.095 |
1.178 |
0.422 |
340 |
305 |
21.31 |
241.50 |
18.9535671177759 |
58.24 |
35 |
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V |
100 |
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90 |
0.3 |
0.12 |
0.12 |
2.415 |
2.165 |
2.015 |
1.115 |
1.199 |
0.410 |
345 |
310 |
25.23 |
242.82 |
19.0237129897876 |
57.93 |
35 |
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VI |
90 |
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80 |
0.3 |
0.12 |
0.12 |
2.515 |
2.265 |
2.115 |
1.110 |
1.189 |
0.411 |
350 |
315 |
25.74 |
246.61 |
19.257359591737 |
58.39 |
35 |
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VII |
80 |
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70 |
0.35 |
0.12 |
0.12 |
2.615 |
2.315 |
2.165 |
1.130 |
1.208 |
0.400 |
355 |
320 |
29.67 |
247.89 |
19.3274555195449 |
58.11 |
35 |
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VIII |
70 |
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60 |
0.35 |
0.12 |
0.12 |
2.715 |
2.415 |
2.265 |
1.124 |
1.199 |
0.402 |
360 |
325 |
30.26 |
251.61 |
19.561271597125 |
58.57 |
35 |
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IX |
60 |
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50 |
0.4 |
0.12 |
0.12 |
2.815 |
2.465 |
2.315 |
1.142 |
1.216 |
0.391 |
365 |
330 |
34.21 |
252.84 |
19.6306818475916 |
58.32 |
35 |
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X |
50 |
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40 |
0.4 |
0.12 |
0.12 |
2.915 |
2.565 |
2.415 |
1.136 |
1.207 |
0.393 |
370 |
335 |
34.88 |
256.49 |
19.8640118430383 |
58.77 |
35 |
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XI |
40 |
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30 |
0.4 |
0.12 |
0.12 |
3.015 |
2.665 |
2.515 |
1.131 |
1.199 |
0.395 |
375 |
340 |
35.53 |
260.15 |
20.1004576526687 |
59.22 |
35 |
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XII |
30 |
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20 |
0.45 |
0.12 |
0.12 |
0 |
0 |
0 |
#DIV/0! |
#DIV/0! |
#DIV/0! |
380 |
345 |
#DIV/0! |
#DIV/0! |
#DIV/0! |
#DIV/0! |
35 |
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XIII |
20 |
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10 |
0.45 |
0.12 |
0.12 |
0 |
0 |
0 |
#DIV/0! |
#DIV/0! |
#DIV/0! |
385 |
350 |
#DIV/0! |
#DIV/0! |
#DIV/0! |
#DIV/0! |
35 |
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XIV |
10 |
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0 |
0.45 |
0.12 |
0.12 |
0 |
0 |
0 |
#DIV/0! |
#DIV/0! |
#DIV/0! |
390 |
355 |
#DIV/0! |
#DIV/0! |
#DIV/0! |
#DIV/0! |
35 |
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TABLICA NR9 |
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Obliczenie różnicy temperatur - okres zimowy. |
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Numer |
Poziom |
Grubość |
Grubość |
Grubość |
Promień zewn. |
Promień zewn. |
Promień zewn. |
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k |
tw |
Δt |
Spadek temperatur |
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Temp.z. |
Tem.z. |
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segmentu |
przekroju |
płaszcza |
izolacji |
wymurówki |
płaszcza |
izolacji |
wymurówki |
rn/ri |
rn/rf |
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[oC] |
|
płasz. |
powiet. |
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rn = R |
ri |
rf |
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W/ |
|
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Δtn |
Δti |
Δtf |
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[m] |
[m] |
[m] |
[m] |
[m] |
[m] |
[m] |
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m2*K |
[oC] |
[oC] |
[oC] |
[oC] |
[oC] |
[oC] |
[oC] |
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I |
144 |
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130 |
0.2 |
0.12 |
0.12 |
2.015 |
1.865 |
1.715 |
1.080 |
1.175 |
0.446 |
323 |
348 |
20.57 |
287.71 |
24.6097068795141 |
-9.90 |
-25 |
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II |
130 |
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120 |
0.2 |
0.12 |
0.12 |
2.115 |
1.965 |
1.815 |
1.076 |
1.165 |
0.450 |
330 |
355 |
21.18 |
294.99 |
23.2876175855794 |
-9.46 |
-25 |
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III |
120 |
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110 |
0.25 |
0.12 |
0.12 |
2.215 |
2.015 |
1.865 |
1.099 |
1.188 |
0.436 |
335 |
360 |
25.96 |
295.51 |
23.2806837565365 |
-9.75 |
-25 |
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IV |
110 |
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100 |
0.25 |
0.12 |
0.12 |
2.315 |
2.115 |
1.965 |
1.095 |
1.178 |
0.437 |
340 |
365 |
26.43 |
299.54 |
23.5088271205847 |
-9.48 |
-25 |
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V |
100 |
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90 |
0.3 |
0.12 |
0.12 |
2.415 |
2.165 |
2.015 |
1.115 |
1.199 |
0.424 |
345 |
370 |
31.18 |
300.06 |
23.5081741077714 |
-9.74 |
-25 |
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VI |
90 |
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80 |
0.3 |
0.12 |
0.12 |
2.515 |
2.265 |
2.115 |
1.110 |
1.189 |
0.426 |
350 |
375 |
31.73 |
304.00 |
23.7390655332176 |
-9.47 |
-25 |
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VII |
80 |
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70 |
0.35 |
0.12 |
0.12 |
2.615 |
2.315 |
2.165 |
1.130 |
1.208 |
0.414 |
355 |
380 |
36.45 |
304.52 |
23.7428565306253 |
-9.71 |
-25 |
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VIII |
70 |
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60 |
0.35 |
0.12 |
0.12 |
2.715 |
2.415 |
2.265 |
1.124 |
1.199 |
0.416 |
360 |
385 |
37.09 |
308.38 |
23.9751519496562 |
-9.44 |
-25 |
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IX |
60 |
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50 |
0.4 |
0.12 |
0.12 |
2.815 |
2.465 |
2.315 |
1.142 |
1.216 |
0.405 |
365 |
390 |
41.80 |
308.88 |
23.9820448244844 |
-9.66 |
-25 |
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X |
50 |
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40 |
0.4 |
0.12 |
0.12 |
2.915 |
2.565 |
2.415 |
1.136 |
1.207 |
0.406 |
370 |
395 |
42.51 |
312.67 |
24.214775877748 |
-9.40 |
-25 |
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XI |
40 |
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30 |
0.4 |
0.12 |
0.12 |
3.015 |
2.665 |
2.515 |
1.131 |
1.199 |
0.408 |
375 |
400 |
43.22 |
316.46 |
24.4514778525041 |
-9.14 |
-25 |
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XII |
30 |
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20 |
0.45 |
0.12 |
0.12 |
0 |
0 |
0 |
#DIV/0! |
#DIV/0! |
#DIV/0! |
380 |
405 |
#DIV/0! |
#DIV/0! |
#DIV/0! |
#DIV/0! |
-25 |
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XIII |
20 |
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10 |
0.45 |
0.12 |
0.12 |
0 |
0 |
0 |
#DIV/0! |
#DIV/0! |
#DIV/0! |
385 |
410 |
#DIV/0! |
#DIV/0! |
#DIV/0! |
#DIV/0! |
-25 |
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XIV |
10 |
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0 |
0.45 |
0.12 |
0.12 |
0 |
0 |
0 |
#DIV/0! |
#DIV/0! |
#DIV/0! |
390 |
415 |
#DIV/0! |
#DIV/0! |
#DIV/0! |
#DIV/0! |
-25 |
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Warunek Δtmax<30°C nie jest spełniony, przy obliczaniu zbrojenia należy uwzględnić wpływ obc. termicznego. |
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Projekt wykonywany jest do celów dydaktycznych, dlatego też przy obliczaniu zbrojenia pomijamy wpływ obc. termincznego. |
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3.0. Sprawdzenie nośności komina. |
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Obliczenia trzonu projektowanego komina przeprowadzono zgodnie z metodą stanów granicznych. |
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Minimalny stopień zbrojenia (komin projektuje się na minimalny stopień zbrojenia) : |
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1) kierunek pionowy - rmin = μ =(4,2*fck) / (100*fyk)>=0,3% |
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2) kierunek poziomy - rhmin = μh = (2,1*fck) / (100*fyk) >=0,35% |
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fck - charakterystyczna wytrzymałośc betonu na ściskanie. |
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fyk - charakterystyczna granica plastyczności stali. |
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3.1. Sprawdzenie nośności komina w stanie montażu. |
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W stadium montażu naprężenia w betonie muszą spełnić następujący warunek - |
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wg ptk. 5.2.2 PN - 88/B - 03004 : |
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σb = (N/Ab)*B ≤ 0,4*fck |
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natomiast naprężenia w stali muszą spełniać warunek : |
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σa = σb*C ≤ 0,6*fyk. |
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N - siła ściskająca prostopadła do przekroju. |
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Ab - pole powierzchni przekroju betonu brutto. |
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B, C - współczynniki uwzględniajace mimośrodowe przyłożenie siły (wg załącznika 6) |
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PN - 88/B - 03004) : |
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Ab = 2*π*rs*ti |
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e0/rs = 0,5*((0,5*sin2a - a + π*(1+ n*μ)) / (sina + (-a + π*(1 + n*μ)*cosa)) = M/N |
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n = Es/E = |
6.25 |
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E = |
32 |
GPa |
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fck = |
30000 |
kPa |
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Es = |
200 |
GPa |
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fyk = |
355000 |
kPa |
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ti - grubość ścianki na danym poziomie. |
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μ - stopień zbrojenia pionowego komina. |
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M - moment zginjący w danym przekroju. |
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TABLICA NR 10 |
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Sprawdzenie nośności komina w stadium montażu. |
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Numer |
Poziom |
Pr.zewn. |
Pr. wewn. |
Grubość |
Obc. |
Powierz. |
Siła |
Ciężar |
Mom.od |
rs |
e0 |
e0/rs |
B |
C |
Ab |
σb |
σbdop |
σs |
σsdop |
segmentu |
przekroju |
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|
płaszcza |
wiatrem |
odniesien. A |
pozioma |
segm. |
wiatru |
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μ=0,3% |
μ=0,3% |
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[ - ] |
t.Z6-1 |
t.Z6-2 |
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[m] |
[m] |
[m] |
[m] |
[kPa] |
[m2] |
[kN] |
[kN] |
[kNm] |
[m] |
[m] |
M/N |
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|
[m2] |
[kPa] |
[kPa] |
[kPa] |
[kPa] |
I |
144 |
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130 |
2.015 |
1.865 |
0.15 |
0.933 |
56.14 |
52.371 |
190.54 |
#DIV/0! |
1.94 |
#DIV/0! |
#DIV/0! |
1.957 |
0.019 |
1.828 |
203.942 |
12000 |
3.875 |
213000 |
II |
130 |
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120 |
2.115 |
1.965 |
0.15 |
0.911 |
41.3 |
37.624 |
751.54 |
#DIV/0! |
2.04 |
#DIV/0! |
#DIV/0! |
1.957 |
0.019 |
1.923 |
764.965 |
12000 |
14.534 |
213000 |
III |
120 |
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|
110 |
2.215 |
2.015 |
0.2 |
0.893 |
43.3 |
38.658 |
1512.97 |
#DIV/0! |
2.115 |
#DIV/0! |
#DIV/0! |
1.957 |
0.019 |
2.658 |
1114.041 |
12000 |
21.167 |
213000 |
IV |
110 |
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100 |
2.315 |
2.115 |
0.2 |
0.875 |
45.3 |
39.618 |
2310.26 |
#DIV/0! |
2.215 |
#DIV/0! |
#DIV/0! |
1.957 |
0.019 |
2.783 |
1624.304 |
12000 |
30.862 |
213000 |
V |
100 |
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90 |
2.415 |
2.165 |
0.25 |
0.850 |
47.3 |
40.214 |
3329.57 |
#DIV/0! |
2.29 |
#DIV/0! |
#DIV/0! |
1.957 |
0.019 |
3.597 |
1811.441 |
12000 |
34.417 |
213000 |
VI |
90 |
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80 |
2.515 |
2.265 |
0.25 |
0.820 |
49.3 |
40.410 |
4393.38 |
#DIV/0! |
2.39 |
#DIV/0! |
#DIV/0! |
1.957 |
0.019 |
3.754 |
2290.194 |
12000 |
43.514 |
213000 |
VII |
80 |
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70 |
2.615 |
2.315 |
0.3 |
0.789 |
51.3 |
40.484 |
5700.82 |
#DIV/0! |
2.465 |
#DIV/0! |
#DIV/0! |
1.957 |
0.019 |
4.646 |
2401.101 |
12000 |
45.621 |
213000 |
VIII |
70 |
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60 |
2.715 |
2.415 |
0.3 |
0.759 |
53.3 |
40.435 |
7061.40 |
#DIV/0! |
2.565 |
#DIV/0! |
#DIV/0! |
1.957 |
0.019 |
4.835 |
2858.202 |
12000 |
54.306 |
213000 |
IX |
60 |
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50 |
2.815 |
2.465 |
0.35 |
0.728 |
55.3 |
40.265 |
8687.20 |
#DIV/0! |
2.64 |
#DIV/0! |
#DIV/0! |
1.957 |
0.019 |
5.806 |
2928.321 |
12000 |
55.638 |
213000 |
X |
50 |
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40 |
2.915 |
2.565 |
0.35 |
0.698 |
57.3 |
39.999 |
10374.77 |
#DIV/0! |
2.74 |
#DIV/0! |
#DIV/0! |
1.957 |
0.019 |
6.026 |
3369.542 |
12000 |
64.021 |
213000 |
XI |
40 |
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30 |
3.015 |
2.665 |
0.35 |
0.649 |
59.3 |
38.491 |
12125.41 |
#DIV/0! |
2.84 |
#DIV/0! |
#DIV/0! |
1.957 |
0.019 |
6.245 |
3799.453 |
12000 |
72.190 |
213000 |
XII |
30 |
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20 |
0 |
0 |
0 |
0.581 |
0 |
0.000 |
12122.30 |
#DIV/0! |
0 |
#DIV/0! |
#DIV/0! |
1.957 |
0.019 |
0.000 |
#DIV/0! |
12000 |
#DIV/0! |
213000 |
XIII |
20 |
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10 |
0 |
0 |
0 |
0.501 |
0 |
0.000 |
12119.19 |
#DIV/0! |
0 |
#DIV/0! |
#DIV/0! |
1.957 |
0.019 |
0.000 |
#DIV/0! |
12000 |
#DIV/0! |
213000 |
XIV |
10 |
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0 |
0 |
0 |
0 |
0.456 |
0 |
0.000 |
12116.08 |
#DIV/0! |
0 |
#DIV/0! |
#DIV/0! |
1.957 |
0.019 |
0.000 |
#DIV/0! |
12000 |
#DIV/0! |
213000 |
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3.2. Sprawdzenie nośności komina w stanie eksploatacji. |
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W stadium eksploatacji naprężenia w betonie muszą spełnić następujący warunek : |
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σb = (N/Ab)*B ≤ 0,65*fck |
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natomiast naprężenia w stali muszą spełniać warunek : |
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σa = σb*C ≤ 0,7*fyk |
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Wszystkie oznaczenia analogiczne jak w fazie montażu. |
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TABLICA NR 11 |
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Sprawdzenie nośności komina w stadium eksploatacji. |
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Numer |
Poziom |
Pr.zewn. |
Pr. wewn. |
Grubość |
Obc. |
Powierz. |
Siła |
Ciężar |
Mom.od |
rs |
e0 |
e0/rs |
B |
C |
Ab |
σb |
σbdop |
σs |
σsdop |
segmentu |
przekroju |
|
|
płaszcza |
wiatrem |
odniesien. A |
pozioma |
segm. |
wiatru |
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|
|
μ=0,3% |
μ=0,3% |
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[ - ] |
t.Z6-1 |
t.Z6-2 |
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[m] |
[m] |
[m] |
[m] |
[kPa] |
[m2] |
[kN] |
[kN] |
[kNm] |
[m] |
[m] |
M/N |
|
|
[m2] |
[kPa] |
[kPa] |
[kPa] |
[kPa] |
I |
144 |
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|
130 |
2.015 |
1.865 |
0.15 |
1.399 |
56.14 |
78.557 |
266.03 |
#DIV/0! |
1.94 |
#DIV/0! |
#DIV/0! |
1.957 |
0.019 |
1.828 |
284.745 |
19500 |
5.410 |
248500 |
II |
130 |
|
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|
|
120 |
2.115 |
1.965 |
0.15 |
1.367 |
41.3 |
56.437 |
1218.03 |
#DIV/0! |
2.04 |
#DIV/0! |
#DIV/0! |
1.957 |
0.019 |
1.923 |
1239.784 |
19500 |
23.556 |
248500 |
III |
120 |
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|
|
110 |
2.215 |
2.015 |
0.2 |
1.339 |
43.3 |
57.986 |
2387.36 |
#DIV/0! |
2.115 |
#DIV/0! |
#DIV/0! |
1.957 |
0.019 |
2.658 |
1757.875 |
19500 |
33.400 |
248500 |
IV |
110 |
|
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|
|
100 |
2.315 |
2.115 |
0.2 |
1.312 |
45.3 |
59.427 |
3609.45 |
#DIV/0! |
2.215 |
#DIV/0! |
#DIV/0! |
1.957 |
0.019 |
2.783 |
2537.748 |
19500 |
48.217 |
248500 |
V |
100 |
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|
|
90 |
2.415 |
2.165 |
0.25 |
1.275 |
47.3 |
60.321 |
5070.49 |
#DIV/0! |
2.29 |
#DIV/0! |
#DIV/0! |
1.957 |
0.019 |
3.597 |
2758.576 |
19500 |
52.413 |
248500 |
VI |
90 |
|
|
|
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|
|
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|
|
|
|
|
|
80 |
2.515 |
2.265 |
0.25 |
1.230 |
49.3 |
60.615 |
6592.92 |
#DIV/0! |
2.39 |
#DIV/0! |
#DIV/0! |
1.957 |
0.019 |
3.754 |
3436.773 |
19500 |
65.299 |
248500 |
VII |
80 |
|
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|
|
70 |
2.615 |
2.315 |
0.3 |
1.184 |
51.3 |
60.726 |
8375.89 |
#DIV/0! |
2.465 |
#DIV/0! |
#DIV/0! |
1.957 |
0.019 |
4.646 |
3527.800 |
19500 |
67.028 |
248500 |
VIII |
70 |
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|
|
60 |
2.715 |
2.415 |
0.3 |
1.138 |
53.3 |
60.653 |
10228.91 |
#DIV/0! |
2.565 |
#DIV/0! |
#DIV/0! |
1.957 |
0.019 |
4.835 |
4140.298 |
19500 |
78.666 |
248500 |
IX |
60 |
|
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|
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|
|
|
|
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|
|
|
50 |
2.815 |
2.465 |
0.35 |
1.092 |
55.3 |
60.398 |
12364.06 |
#DIV/0! |
2.64 |
#DIV/0! |
#DIV/0! |
1.957 |
0.019 |
5.806 |
4167.734 |
19500 |
79.187 |
248500 |
X |
50 |
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40 |
2.915 |
2.565 |
0.35 |
1.047 |
57.3 |
59.998 |
14577.89 |
#DIV/0! |
2.74 |
#DIV/0! |
#DIV/0! |
#DIV/0! |
#DIV/0! |
6.026 |
#DIV/0! |
19500 |
#DIV/0! |
248500 |
XI |
40 |
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30 |
3.015 |
2.665 |
0.35 |
0.974 |
59.3 |
57.737 |
16877.33 |
#DIV/0! |
2.84 |
#DIV/0! |
#DIV/0! |
#DIV/0! |
#DIV/0! |
6.245 |
#DIV/0! |
19500 |
#DIV/0! |
248500 |
XII |
30 |
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20 |
0 |
0 |
0 |
0.871 |
0 |
0.000 |
16874.22 |
#DIV/0! |
0 |
#DIV/0! |
#DIV/0! |
#DIV/0! |
#DIV/0! |
0.000 |
#DIV/0! |
19500 |
#DIV/0! |
248500 |
XIII |
20 |
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10 |
0 |
0 |
0 |
0.752 |
0 |
0.000 |
16871.11 |
#DIV/0! |
0 |
#DIV/0! |
#DIV/0! |
#DIV/0! |
#DIV/0! |
0.000 |
#DIV/0! |
19500 |
#DIV/0! |
248500 |
XIV |
10 |
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0 |
0 |
0 |
0 |
0.683 |
0 |
0.000 |
16868.00 |
#DIV/0! |
0 |
#DIV/0! |
#DIV/0! |
#DIV/0! |
#DIV/0! |
0.000 |
#DIV/0! |
19500 |
#DIV/0! |
248500 |
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4.0. Sprawdzenie stanu granicznego użytkowania. |
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4.1. Sprawdzenie ugięcia wierzchołka. |
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Zakładam sprężyste wychylenie komina oraz osiadanie równomierne terenu. |
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yw = (ΣWi*Ho3) / (4*Eb*Io) ≤ yw,dop = Ho/200 |
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Wi - suma sił poziomych od obciążeń charakterystycznych. |
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H = H0 = |
102 |
m |
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I0 - moment bezwładności przekroju płaszcza w poziomie połączenia z płyta fundamentową. |
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Eb - moduł sprężystości materiału, z którego wykonany jest płaszcz; Eb = |
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= |
32 |
GPa |
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I0 = (p*(Dz4 - (Dz - 2*gt)4)) / 64 = |
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0.00 |
m4 |
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ΣWi = |
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560.712 |
kN |
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yw = |
#DIV/0! |
m |
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yw,dop = |
0.51 |
m |
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yw ≤ yw,dop => |
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Warunek jest spełniony. |
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4.2. Sprawdzenie możliwości powstania rys w trzonie komina żelbetowego.. |
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Jeśli : |
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1) Mt+Mv< M1 - nie ma zarysowań, |
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2) Mt+Mv > M1 - komin zarysowany. |
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Mt - moment zginający w rozpatrywanym przekroju od obciążeń termicznych. |
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Mt = (at*Δt*Eb*I1) / g |
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Mv= moment zginający wywołany obciążeniami dodatkowymi (Mv=0) |
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at - współczynnik rozszerzalności termicznej betonu; at = |
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1E-05 |
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I1 - moment bezwładności przekroju niezarysowanego. |
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W1 - wskażnik wytrzymałości ptrzekroju niezarysowanego. |
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Δt - różnica temperatur między zewnętrzną i wewnętrzną powierzchnią trzonu. |
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I1 = (b*g3)/12 |
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W1 = (b*g2)/6 |
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g - grubość trzonu w danym miejscu. |
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M1 = W1*(f'ctk' - N/A1) |
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f'ctk' = (0,3*(fGc,cube + 8)2/3 * (2,6+24*g)) / (1,0+40*g) |
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fGc,cube = |
37 |
MPa |
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TABLICA NR 12 |
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Sprawdzenie możliwości powstania rys. |
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Nr |
Poziom |
rs |
Grubość |
Pole pow. |
W1 |
I1 |
Δtn |
N |
Mt |
f'ctk' |
M1 |
Mt+Mv<M1? |
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segm. |
przekroju |
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płaszcza |
przekroju |
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niezarys.A1 |
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[m] |
[m] |
[m] |
[m2] |
[m3] |
[m4] |
[oC] |
[MN] |
[kNm] |
[MPa] |
[kNm] |
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I |
144 |
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spełnione |
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130 |
1.940 |
0.15 |
0.15 |
0.004 |
0.000 |
20.575 |
0.235 |
12.345 |
3.362 |
12.600 |
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II |
130 |
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nie spełnione |
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120 |
2.040 |
0.15 |
0.15 |
0.004 |
0.000 |
21.176 |
1.065 |
12.705 |
3.362 |
12.580 |
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III |
120 |
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nie spełnione |
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110 |
2.115 |
0.20 |
0.20 |
0.007 |
0.001 |
25.963 |
2.091 |
27.694 |
3.121 |
20.735 |
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IV |
110 |
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nie spełnione |
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100 |
2.215 |
0.20 |
0.20 |
0.007 |
0.001 |
26.430 |
3.164 |
28.192 |
3.121 |
20.699 |
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V |
100 |
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nie spełnione |
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90 |
2.290 |
0.25 |
0.25 |
0.010 |
0.001 |
31.175 |
4.452 |
51.959 |
2.967 |
30.724 |
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VI |
90 |
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nie spełnione |
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80 |
2.390 |
0.25 |
0.25 |
0.010 |
0.001 |
31.730 |
5.795 |
52.883 |
2.967 |
30.668 |
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VII |
80 |
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nie spełnione |
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70 |
2.465 |
0.30 |
0.30 |
0.015 |
0.002 |
36.450 |
7.372 |
87.481 |
2.861 |
42.549 |
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VIII |
70 |
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nie spełnione |
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60 |
2.565 |
0.30 |
0.30 |
0.015 |
0.002 |
37.089 |
9.013 |
89.013 |
2.861 |
42.467 |
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IX |
60 |
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nie spełnione |
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50 |
2.640 |
0.35 |
0.35 |
0.020 |
0.004 |
41.796 |
10.908 |
136.533 |
2.783 |
56.190 |
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X |
50 |
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nie spełnione |
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40 |
2.740 |
0.35 |
0.35 |
0.020 |
0.004 |
42.514 |
12.873 |
138.879 |
2.783 |
56.075 |
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XI |
40 |
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nie spełnione |
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30 |
2.840 |
0.35 |
0.35 |
0.020 |
0.004 |
43.224 |
14.913 |
141.200 |
2.783 |
55.956 |
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XII |
30 |
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#DIV/0! |
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20 |
0.000 |
0.00 |
0.00 |
0.000 |
0.000 |
#DIV/0! |
14.911 |
#DIV/0! |
9.868 |
#DIV/0! |
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XIII |
20 |
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#DIV/0! |
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10 |
0.000 |
0.00 |
0.00 |
0.000 |
0.000 |
#DIV/0! |
14.908 |
#DIV/0! |
9.868 |
#DIV/0! |
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XIV |
10 |
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#DIV/0! |
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0 |
0.000 |
0.00 |
0.00 |
0.000 |
0.000 |
#DIV/0! |
14.905 |
#DIV/0! |
9.868 |
#DIV/0! |
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5.0. Wymiarowanie zbrojenia pionowego i poziomego trzonu komina żelbetowego. |
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Kominy przemysłowe wymiarujemy na minimalny stopień zbrojenia. |
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Minimalny stopień zbrojenia: |
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Kierunek pionowy: |
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rmin = µ = 4,2*fck/(100*fyk) = 4,2*30/(100*355) = |
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0.0035 |
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Kierunek poziomy: |
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rminh = µh = 2,1*fck/(100*fyk) = 2,1*20/(100*220) = |
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0.0018 |
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Jednocześnie PN-88/B-03004 w pkt. 7.2.2 mowi, iż minimalny stopień zbrojenia poziomego |
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obliczony z powyższych wzorów nie powinien być mniejszy niż 0,35% przy temperaturze |
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odprowadzanych gazów rzędu 100-300. |
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Maksymalny rozstaw zbrojenia pionowego nie powinien przekraczać 30cm, natomiast |
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zbrojenia poziomego 20cm. Minimalne otulenie zbrojenia pionowego wynosi 3cm. |
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TABLICA NR 13 |
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Zbrojenie pionowe płaszcza komina. |
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Nr |
Poziom |
rs |
Obwód |
Ab |
Grubość |
St. Zbrojenia |
As |
średnica |
Liczba prętów |
Pole pow |
Ostatecznie przyjęte zbrojenie pionowe |
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Pole pow |
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|
segm. |
przekroju |
|
dla rs |
pole betonu |
płaszcza |
ρ |
pole zbrojenia |
Ø |
n |
n prętów |
L. Pretów |
Rozstaw |
L. Pretów |
Rozstaw |
zbrojenia |
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|
[m] |
[m] |
[m2] |
[m] |
[-] |
[cm2] |
[mm] |
[-] |
[cm2] |
zewnętrz |
[cm] |
wewnętrz |
[cm] |
[cm2] |
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|
I |
144 |
|
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|
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|
130 |
1.940 |
12.19 |
1.83 |
0.15 |
0.0035 |
64.9 |
12 |
105 |
118.8 |
70 |
17 |
55 |
22 |
141.4 |
|
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|
II |
130 |
|
|
|
|
|
|
|
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|
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|
120 |
2.040 |
12.82 |
1.92 |
0.15 |
0.0035 |
68.2 |
12 |
110 |
124.4 |
73 |
17 |
57 |
22 |
147.4 |
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III |
120 |
|
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|
|
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|
110 |
2.115 |
13.29 |
2.66 |
0.2 |
0.0035 |
94.3 |
12 |
141 |
159.5 |
94 |
14 |
59 |
23 |
173.0 |
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IV |
110 |
|
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|
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|
100 |
2.215 |
13.92 |
2.78 |
0.2 |
0.0035 |
98.8 |
12 |
147 |
166.3 |
98 |
14 |
61 |
23 |
179.8 |
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V |
100 |
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90 |
2.290 |
14.39 |
3.60 |
0.25 |
0.0035 |
127.7 |
12 |
181 |
204.7 |
121 |
12 |
63 |
23 |
207.7 |
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VI |
90 |
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80 |
2.390 |
15.02 |
3.75 |
0.25 |
0.0035 |
133.2 |
12 |
188 |
212.6 |
125 |
12 |
65 |
23 |
215.3 |
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VII |
80 |
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|
70 |
2.465 |
15.49 |
4.65 |
0.3 |
0.0035 |
164.9 |
12 |
225 |
254.5 |
150 |
10 |
75 |
21 |
254.5 |
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VIII |
70 |
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60 |
2.565 |
16.12 |
4.83 |
0.3 |
0.0035 |
171.6 |
12 |
233 |
263.5 |
155 |
10 |
78 |
21 |
263.5 |
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IX |
60 |
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50 |
2.640 |
16.59 |
5.81 |
0.35 |
0.0035 |
206.1 |
12 |
273 |
308.8 |
182 |
9 |
91 |
18 |
308.8 |
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X |
50 |
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40 |
2.740 |
17.22 |
6.03 |
0.35 |
0.0035 |
213.9 |
12 |
282 |
318.9 |
188 |
9 |
94 |
18 |
318.9 |
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XI |
40 |
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30 |
2.840 |
17.84 |
6.25 |
0.35 |
0.0035 |
221.7 |
12 |
291 |
329.1 |
194 |
9 |
97 |
18 |
329.1 |
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XII |
30 |
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20 |
0.000 |
0.00 |
0.00 |
0 |
0.0035 |
0.0 |
12 |
336 |
380.0 |
224 |
0 |
112 |
0 |
380.0 |
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XIII |
20 |
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10 |
0.000 |
0.00 |
0.00 |
0 |
0.0035 |
0.0 |
12 |
346 |
391.3 |
231 |
0 |
115 |
0 |
391.3 |
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XIV |
10 |
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|
|
0 |
0.000 |
0.00 |
0.00 |
0 |
0.0035 |
0.0 |
12 |
355 |
401.5 |
237 |
0 |
118 |
0 |
401.5 |
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Zbrojenie pionowe trzonu komina, przy założeniu rozkładu zbrojenia w stosunku 2/3 do 1/3 oraz maksymalnego dopuszczalnego rozstawu prętów wewnętrznych co 30cm. |
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TABLICA NR 14 |
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Zbrojenie poziome płaszcza komina: |
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Nr |
Poziom |
rs |
Grubość |
Ab |
St. Zbrojenia |
As |
średnica |
Liczba prętów |
Pole. pow. |
Rozstaw |
Pole pow |
|
|
|
|
|
|
|
|
segm. |
przekroju |
|
płaszcza |
pole betonu |
ρ |
pole zbrojenia |
Ø |
n |
zbroj "n" prętów |
|
zbrojenia |
|
|
|
|
|
|
|
|
|
|
[m] |
[m] |
[m2] |
[-] |
[cm2] |
[mm] |
[-] |
[cm2] |
[cm] |
[cm2] |
|
|
|
|
|
|
|
|
I |
144 |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
130 |
1.940 |
0.15 |
0.15 |
0.0035 |
5.3 |
12 |
7 |
7.9 |
14 |
7.9 |
|
|
|
|
|
|
|
|
II |
130 |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
120 |
2.040 |
0.15 |
0.15 |
0.0035 |
5.3 |
12 |
7 |
7.9 |
14 |
7.9 |
|
|
|
|
|
|
|
|
III |
120 |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
110 |
2.115 |
0.2 |
0.20 |
0.0035 |
7.1 |
12 |
8 |
9.0 |
13 |
9.0 |
|
|
|
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|
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|
|
IV |
110 |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
100 |
2.215 |
0.2 |
0.20 |
0.0035 |
7.1 |
12 |
8 |
9.0 |
13 |
9.0 |
|
|
|
|
|
|
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|
V |
100 |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
90 |
2.290 |
0.25 |
0.25 |
0.0035 |
8.9 |
12 |
10 |
11.3 |
10 |
11.3 |
|
|
|
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|
|
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|
VI |
90 |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
80 |
2.390 |
0.25 |
0.25 |
0.0035 |
8.9 |
12 |
10 |
11.3 |
10 |
11.3 |
|
|
|
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VII |
80 |
|
|
|
|
|
|
|
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|
|
70 |
2.465 |
0.3 |
0.30 |
0.0035 |
10.6 |
12 |
11 |
12.4 |
9 |
12.4 |
|
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VIII |
70 |
|
|
|
|
|
|
|
|
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|
60 |
2.565 |
0.3 |
0.30 |
0.0035 |
10.6 |
12 |
11 |
12.4 |
9 |
12.4 |
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IX |
60 |
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50 |
2.640 |
0.35 |
0.35 |
0.0035 |
12.4 |
12 |
13 |
14.7 |
8 |
14.7 |
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X |
50 |
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|
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|
40 |
2.740 |
0.35 |
0.35 |
0.0035 |
12.4 |
12 |
13 |
14.7 |
8 |
14.7 |
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XI |
40 |
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30 |
2.840 |
0.35 |
0.35 |
0.0035 |
12.4 |
12 |
13 |
14.7 |
8 |
14.7 |
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XII |
30 |
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20 |
0.000 |
0 |
0.00 |
0.0035 |
0.0 |
12 |
15 |
17.0 |
7 |
17.0 |
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XIII |
20 |
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10 |
0.000 |
0 |
0.00 |
0.0035 |
0.0 |
12 |
15 |
17.0 |
7 |
17.0 |
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XIV |
10 |
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0 |
0.000 |
0 |
0.00 |
0.0035 |
0.0 |
12 |
15 |
17.0 |
7 |
17.0 |
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6.0. Sprawdzenie stateczności komina żelbetowego. |
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Sprawdzenie stateczności komina żelbetowego przeprowadzam dla wartości obliczeniowych |
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sił wewnętrznych zarówno dla stadium montażu jak i dla stadium eksploatacji. |
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Zgodnie z punktem 5.3. PN-88/B-03004 należy wyznaczyć tzw. wspołczynnik wyboczeniowy: |
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φw = Pkr/No ≥ 2,5 |
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gdzie: |
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Pkr - siła krytyczna wyznaczona wg Z4 PN-88/B-03004 |
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No - całkowite pionowe obciążenie ciężarem własnym komina w poziomie |
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górnej powierzchni fundamentu |
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Wyznaczenie siły krytycznej Pkr |
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Siłę krytyczną wyznaczam na podstaie wzoru Z4-3 PN-88/B-03004: |
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Pkr = (p^2*E*In)/(4*Ho^2)*[1+(I2-I1)/I1*a1^2/Ho^2]^-1*...*[1+(In-In-1)/In-1*(an-1^2/Ho^2)]^-1 |
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gdzie: |
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ai, i=0, 1, ..., n-1 - rzędna dolnego [rzekroju i-tego segmentu liczona względem |
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wierzchołka komina [m] |
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Ii, i =0, 1, ..., n-1 - moment bezwładności trzonu żelbetowego dolnego przekroju i-tego |
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segmentu , |
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Ho = ao - wysokośc komina żelbetowego ponad fundamentem [m] |
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E - moduł Young'a betonu płaszcza żelbetowego komina |
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Ii =p*(Dzi4 - dzi4)/64 |
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gdzie: |
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Rzi - promień zewnętrzny trzonu żelbetowego komina w i-tym przekroju |
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rzi - promień wewnętrzny trzonu żelbetowego komina w i-tym przekroju |
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Dzi - średnica zewnętrzna trzonu żelbetowego komina w i-tym przekroju |
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dzi - średnica wewnętrzna trzonu żelbetowego komina w i-tym przekroju |
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TABLICA NR 15 |
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Obliczenia stateczności komina |
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Nr |
Poziom |
Grubość |
ai |
Promień |
Promień |
Moment |
Składnik |
No |
No |
Siła krytyczna |
Wsp. Wyboczeniowy φw |
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segm. |
przekroju |
płaszcza |
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zewnętrzny |
wewnętrzny |
bezwładności |
[1+(In-In-1)/In-1* |
Faza realizacji |
Faza eksploat. |
Pkr |
Faza |
Faza |
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[m] |
[m] |
[m] |
[m] |
[m4] |
(an-1^2/Ho^2)]^-1 |
[kN] |
[kN] |
[kN] |
realizacji |
eksploatacji |
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I |
144 |
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130 |
0.15 |
14 |
2.015 |
1.865 |
3.446 |
0.997 |
190.542 |
266.035 |
26070.909 |
136.825 |
97.998 |
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II |
130 |
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120 |
0.15 |
24 |
2.115 |
1.965 |
4.006 |
0.974 |
751.540 |
1218.025 |
29513.520 |
39.271 |
24.231 |
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III |
120 |
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110 |
0.2 |
34 |
2.215 |
2.015 |
5.958 |
0.984 |
1512.971 |
2387.358 |
43179.685 |
28.540 |
18.087 |
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IV |
110 |
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100 |
0.2 |
44 |
2.315 |
2.115 |
6.842 |
0.934 |
2310.257 |
3609.452 |
46292.850 |
20.038 |
12.825 |
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V |
100 |
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90 |
0.25 |
54 |
2.415 |
2.165 |
9.460 |
0.963 |
3329.574 |
5070.486 |
69146.290 |
20.767 |
13.637 |
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VI |
90 |
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80 |
0.25 |
64 |
2.515 |
2.265 |
10.752 |
0.889 |
4393.384 |
6592.920 |
62271.409 |
14.174 |
9.445 |
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VII |
80 |
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70 |
0.3 |
74 |
2.615 |
2.315 |
14.169 |
0.938 |
5700.824 |
8375.894 |
76944.106 |
13.497 |
9.186 |
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VIII |
70 |
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60 |
0.3 |
84 |
2.715 |
2.415 |
15.959 |
0.844 |
7061.397 |
10228.907 |
73118.302 |
10.355 |
7.148 |
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IX |
60 |
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50 |
0.35 |
94 |
2.815 |
2.465 |
20.320 |
0.909 |
8687.199 |
12364.058 |
84641.857 |
9.743 |
6.846 |
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X |
50 |
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40 |
0.35 |
104 |
2.915 |
2.565 |
22.711 |
0.895 |
10374.772 |
14577.888 |
84637.346 |
8.158 |
5.806 |
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XI |
40 |
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30 |
0.35 |
114 |
3.015 |
2.665 |
25.282 |
-4.014 |
12125.413 |
16877.330 |
-378188.773 |
-31.190 |
-22.408 |
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XII |
30 |
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20 |
0 |
124 |
0 |
0 |
0.000 |
#DIV/0! |
12122.303 |
16874.220 |
#DIV/0! |
#DIV/0! |
#DIV/0! |
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XIII |
20 |
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10 |
0 |
134 |
0 |
0 |
0.000 |
#DIV/0! |
12119.192 |
16871.110 |
#DIV/0! |
#DIV/0! |
#DIV/0! |
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XIV |
10 |
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0 |
0 |
144 |
0 |
0 |
0.000 |
- |
12116.082 |
16867.999 |
#DIV/0! |
#DIV/0! |
#DIV/0! |
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Stateczność komina jest zachowana. |
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7.0. Posadowienie komina. |
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Sprawdzenie nośności podłoża gruntowego pod kominem wykonuje metodą uproszczoną. |
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Istotą tej metody jest zastąpienie fundamentu kołowego rownoważnym fundamentem kwadratowymi |
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i przyjęcie rzeczywistego rozkładu naprężeń w poziomie posadowienia przyjmując, że |
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fundament jest bryła sztywną. Następnie wyznaczamy dla takiego fundamentu pasmo najbardziej |
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obciążone o szerokości B' i dla niego sprawdzamy warunek normowy (wg Z1-1 PN-81/B-03020): |
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Nr ≤ m*QfNB |
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gdzie: |
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Nr - obliczeniowa wartość pionowej składowej obciążenia [kN] |
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m - współczynnik korekcyjny, wg pkt. 3.3.7 PN-81/B-03020 |
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QfNB - pionowa składowa obliczeniowego oporu granicznego podłoża gruntowego [kN] |
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7.1. Wymiary fundamentu - płyta kołowa. |
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Df = |
20 |
m |
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d1 = |
10.24 |
m |
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Dz = |
8.84 |
m |
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Dw = |
7.54 |
m |
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h1 = |
1.5 |
m |
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h2 = |
3 |
m |
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h3 = |
1 |
m |
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hf = |
4 |
m |
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Rf = |
10 |
m |
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Rz = |
4.42 |
m |
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r1 = |
5.12 |
m |
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Rw = |
3.77 |
m |
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7.2. Zebranie obciążeń. |
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* ciężar komina |
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GNkE = |
14904.95 |
kN |
GNoE = |
16868.00 |
kN |
stadium eksploatacji |
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GNkM = |
11014.62 |
kN |
GNoM = |
12116.08 |
kN |
stadium montażu |
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* siła pozioma od wiatru |
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Wk = |
560.712 |
kN |
WE = |
672.854 |
kN |
stadium eksploatacji |
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WM = |
448.569 |
kN |
stadium montażu |
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* moment od obciążenia poziomego |
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Mk = |
#DIV/0! |
kNm |
ME = |
#DIV/0! |
kNm |
stadium eksploatacji |
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MM = |
#DIV/0! |
kNm |
stadium montażu |
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* moment dodatkowy od obciążenia wiatrem |
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MDk = Wk*(hf+0,1) = |
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2298.917 |
kNm |
MDE =WE*(hf+0,1) = |
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2758.701 |
kNm |
stadium eksploatacji |
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MDM =WM*(hf+0,1) = |
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1839.134 |
kNm |
stadium montażu |
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* siła pozioma od parcia gruntu (charakterystyka gruntu w pkt 7.4) |
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Do wyznaczenia współczynników wpływu nachylenia potrzebujemy następujących parametrów: |
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Współczynnik parcia Ka = tg2 (45-Φ/2) = |
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0.283 |
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Ciężar piasku średniego w warunkach wilgotnych γ [kN/m3] = |
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18.150 |
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Rozkład parcia od gruntu jest trójkątny i wynosi w poziomie 0.00 = 0kPa i na poziomie głębokości |
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posadowienia danej w temacie i wynoszącej hf = 4m, wynosi e1 = γ*hf*Ka = 18,15*4*0,283= |
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20.546 |
kN |
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Wypadkowa parcia od gruntu E = |
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25.682 |
kN |
W wartości obliczeniowej E`= |
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30.819 |
kN |
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* moment od wypadkowej parcia gruntu |
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Mgk= |
34.243 |
kNm |
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Mgo= |
41.092 |
kNm |
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* ciężar fundamentu |
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g = |
25 |
kN/m3 |
gf = |
1.1 |
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V1 = p*Rf2*h1 = |
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471.239 |
kN |
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V2 = p*(h2-h1)*(Rf2+Rf*R1+R12)/3 = |
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278.682 |
kN |
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V3 = p*Rw2*h3 = |
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14.884 |
kN |
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V4 = p*(hf-h2)*(R12+R1*Rz+Rz2)/3 = |
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71.609 |
kN |
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Vf = V1+V2+V4-V3 = |
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806.646 |
kN |
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Gfk = Vf*g = |
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20166.142 |
kN |
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|
|
Gfo = Gfk*gf = |
|
22182.757 |
kN |
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* ciężar gruntu na fundamencie |
|
|
g = |
18.15 |
kN/m3 |
gf = |
1.2 |
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|
|
V5 = p*Rf2*hf - Vf - V3 = |
|
435.107650369043 |
kN |
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|
Ggk = V5*g = |
|
7897.204 |
kN |
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|
|
Ggo = Ggk*gf = |
|
9476.645 |
kN |
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|
* ciężar izolacji fundamentu - piasek drobny |
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g = |
17 |
kN/m3 |
gf = |
1.3 |
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V3 = |
14.884 |
kN |
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Gizfk = V3*g = |
|
253.023 |
kN |
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Gizfo = Gizfk*gf = |
|
328.930 |
kN |
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* ciężar posadzki betonowej gpos = |
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|
0.1 |
m g = |
24 |
kN/m3 |
gf = |
1.2 |
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Vpos =p*Rf2*gpos= |
|
31.416 |
kN |
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Gposk = Vpos*g = |
|
753.982 |
kN |
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Gposo = Gposk*gf = |
|
904.779 |
kN |
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* ciężar podlewki betonowej gpod = |
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|
0.1 |
m g = |
24 |
kN/m3 |
gf = |
1.2 |
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Vpod =p*Rf2*gpod= |
|
31.416 |
kN |
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Gpodk = Vpod*g = |
|
753.982 |
kN |
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Gpodo = Gpodk*gf = |
|
904.779 |
kN |
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SIŁY W POZIOMIE POSADOWIENIA |
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* siła pionowa |
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NkE =GNkE+Gfk+Ggk+Gizfk+Gposk+Gpodk = |
|
|
|
44729.28 |
kN |
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NkM =GNkM+Gfk+Ggk+Gizk+Gposk+Gpodk = |
|
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|
40838.95 |
kN |
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NoE =GNoE+Gfo+Ggo+Gizo+Gposo+Gpodo = |
|
|
|
50665.89 |
kN |
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NoM =GNkoM+Gfo+Ggo+Gizo+Gposo+Gpodo = |
|
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|
45913.97 |
kN |
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* siła pozioma |
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Tk = Wk+E = |
|
586.394 |
kN |
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ToE = WoE+E' = |
|
703.673 |
kN |
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ToM = WoM+E' = |
|
479.388 |
kN |
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* moment zginający |
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Mk = Mk + MDk + Mgk= |
|
#DIV/0! |
kNm |
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ME = ME + MDE + Mgo= |
|
#DIV/0! |
kNm |
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MM = MM + MDM + Mgo= |
|
#DIV/0! |
kNm |
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7.3. Kształt zastępczy fundamentu kołowego. |
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Płytę kołową zastępują równoważnym jej kwadratem. |
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L=B |
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Df |
|
B |
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Fo = p*Df2/4 |
|
|
Fkw = B2 |
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B = L = (p*Df2/4)^0,5 = |
17.725 |
m |
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Szerokość pasma najbardziej obciążonego B': |
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B' = 0,225*B = |
|
3.988 |
m |
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7.4. Obliczenie oporu granicznego gruntu QfNB. |
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Odpór graniczny gruntu obliczam wg wzoru: |
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w którym: |
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NC, NB, ND - współczynniki nośności, wyznaczone w zależności od wartości |
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F = Fu z nomogramu na rys. Z1-1, lub z tabl. Z1-1, lub wg wzorów normowych: |
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ND = e^(p*tgF)*(tg(p/4+F/2))^2 |
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NC = (ND - 1)*ctgF |
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NB = 0,75*(ND - 1)*tgF |
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cu( r ) - obliczeniowa spójność gruntu [kPa] |
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iC, iD, iB - współczynniki wpływu nachylenia wypadkowej obciażenia |
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γD, γB - obliczeniowe ciężary objętościowe gruntu |
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Bˉ - szerokość zredukowana prostokątnej podstawy fundamentu |
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(wymiar krótszego boku) [m], wg wzoru: |
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Bˉ = B - 2*eB |
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Lˉ - długość zredukowana prostokątnej podstawy fundamentu |
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(wymiar dłuższego boku) [m], wg wzoru: |
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Lˉ = L - 2*eL |
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eB, eL - mimośród działania obciążenia, odpowiednio w kierunku równoległym |
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do szerokości B i długości L podstawy, (B ≤ L) [m] |
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Dmin - głębokość posadowienia mierzona od najniższego poziomu przyległego |
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terenu [m] |
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F - efektywny kąt tarcia wewnętrznego gruntu |
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PARAMETRY GRUNTU: |
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Ps: |
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ID = |
0.6 |
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f = |
34 |
o |
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g = |
18.15 |
kN/m3 |
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eL = eB = |
MoE / NoE = |
#DIV/0! |
m |
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#DIV/0! |
m |
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Dmin = hf = |
4 |
m |
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ND(32) = |
29.44 |
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NB(32) = |
14.39 |
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NC(32) = |
42.16 |
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W rozpatrywanym przypadku nie występuje spójność gruntu, zatem wyrażenie ze spónością wynosi 0, |
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a wzór na QfNB |
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tgdB = ToE / NoE = |
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0.0139 |
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tgdB /tgf1 = |
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0.0206 |
iD = |
0.95 |
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iB = |
0.9 |
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QfNB |
= |
#DIV/0! |
kN |
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7.5. Obliczenie naprężeń pod stopą. |
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* pole płyty F |
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F = B*L = |
314.159 |
m2 |
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* wskaźnik wytrzymałości W* |
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W* = B*L2/ 6 = |
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928.055 |
m3 |
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NAPRĘŻENIA: |
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s1 = NoE/F + ME/W = |
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#DIV/0! |
kN/m2 |
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s2 = NoE/F - ME/W = |
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#DIV/0! |
kN/m2 |
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s3 =s2 + (s1 - s2)*(B-B')/B = |
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#DIV/0! |
kN/m2 |
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ssr = (s1 + s3)/2 = |
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#DIV/0! |
kN/m2 |
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Sprawdzenie warunku naprężeń dopuszczalnych pod płytą |
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ssr*B`*B < m*QfNB |
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gdzie m = 0,9 * 0,7 = |
0.63 |
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ssr*B`*B = |
#DIV/0! |
< m*QfNB |
= |
#DIV/0! |
kN |
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Warunek spełniony |
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7.6. Sprawdzenie warunku nacisków stopy na grunt. |
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Przy sprawdzaniu tych warunków powracamy do fundamentu okrągłego. |
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Zmieni się wskaźnik wytrzymałości W* →W |
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W czasie eksploatacji i remontu komina każdorazowo powinny być spełnione |
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następujące warunki: |
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7.6.1. Naciski w trakcie montażu, wartości charakterystyczne. |
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q = N/F - M/W > 0 |
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W = |
785.398 |
m3 |
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N= NkM-Ggk-Gizk-Gposk = |
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31934.74 |
kN |
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M= MM = |
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#DIV/0! |
kNm |
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q = |
#DIV/0! |
kN/m2 > 0 |
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Warunek spełniony. |
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7.6.2. Naciski w trakcie eksploatacji. |
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qmax = N/F + M/W = |
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#DIV/0! |
kN/m2 |
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qmin = N/F - M/W = |
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#DIV/0! |
kN/m2 |
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N = NkE = |
44729.28 |
kN |
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M = ME = |
#DIV/0! |
kNm |
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qmax / qmin = |
#DIV/0! |
≤ |
5 |
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Warunek spełniony. |
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7.7. Sprawdzenie osiadań stopy fundamentowej. |
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Sprawdzenie osiadań fundamentu. Osiadanie fundamentu komina będziemy |
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sprawdzali dla działania całkowitego charakterystycznego obciążenia stałego sa : |
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sa = qsr*w*Df*(1-n2) / Eo ≤ sdop = 80 mm |
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gdzie: |
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qsr - obciążenie sprowadzone do obciążenia równomiernie rozłożonego |
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qsr = NkE / F = |
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142.378 |
kN/m2 |
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ω - współczynnik kształtu płyty fundamentowej |
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ω = |
0.79 |
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Df - średnica płyty fundamentowej |
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Df = |
20 |
m |
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υ - współczynnik Poissona |
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υ = |
0.3 |
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Eo - moduł pierwotnego odkształcenia gruntu w kPa |
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Eo = |
93000 |
kPa |
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sa = |
22.012 |
mm < |
sdop = 80 mm |
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7.8. Wyznaczenie sił wewnętzrnych w płycie fundamentowej. |
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Schemat statyczny płyty fundamentowej stanowi kołowa płyta obciążona oddziaływaniem gruntu, obwodowo podparta |
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przez trzon komina. Przy obliczaniu sił wewnętrznych płytę traktujemy jako sprężystą, izotropową, cienką i o stałej grubości. |
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Obciążenie stanowi oddziaływanie podłoża gruntowego, które przyjmujemy jako liniowe. |
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Rozkład tego obciążenia jest następujący: |
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Obciążenie symetryczne p0 od obciążenia ciężarem własnym komina |
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Obciążenie antysymetryczne pa od obciążenia wiatrem |
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Momenty promieniowe i pierścieniowe od obciążeń symetrycznych |
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Mrs = (po*a2/16) * X |
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Mts = (po*a2/16) * X` |
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gdzie: |
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p0 - średni nacisk komina na grunt |
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a - odległość od osi do połowy trzonu żelbetowego |
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X, X' - współczynniki wg specjalnych nomogramów |
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(są one funkcjami współczynników β i ρ ) |
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po = N/F |
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N – jest to ciężar komina w stadium eksp. bez uwzględnienia ciężaru płyty i |
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gruntu spoczywającego na tej płycie |
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F - pole fundamentu |
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N = NoE - Gfo - Ggo = |
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19006.49 |
kN |
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po = |
60.4995268666667 |
kN/m2 |
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β = R/a |
2.44200244200244 |
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a = (Rz + Rw)/ 2 = |
4.095 |
m |
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ρ = r/a |
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Momenty promieniowe i pierścieniowe od obciążeń antysymetrycznych |
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Mra = (pa*a2/16) * Y |
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Mta = (pa*a2/16) * Y` |
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gdzie: |
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pa - maksymalny nacisk na grunt spowodowany oddziaływaniem wiatru |
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a - odległość od osi do połowy trzonu żelbetowego |
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Y, Y' - współczynniki wg specjalnych nomogramów |
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(są one funkcjami współczynników β i ρ ) |
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pa = M/W |
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M - jest to moment zginający w poziomie posadowienia fundamentu |
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W - wskaźnik wytrzymałości wyznaczony jak dla fundamentu okrągłego |
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M = |
#DIV/0! |
kNm |
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pa = |
#DIV/0! |
kN/m2 |
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β = R/a |
2.44200244200244 |
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a = (Rz + Rw)/ 2 = |
4.095 |
m |
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ρ = r/a |
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TABLICA NR 16 |
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Momenty promieniowe w płycie fundamentowej o śr. 20m |
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Obciążenie symatryczne |
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Obciążenie antysymetryczne |
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Lp. |
r |
ρ = r/a |
X |
po*a2/16 |
Mrs |
Y |
pa*a2/16 |
Mra |
Mr |
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[m] |
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[m] |
[kN] |
[kNm] |
[m] |
[kN] |
[kNm] |
[kNm] |
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1 |
0.5 |
0.122 |
13 |
63.407 |
824.296 |
5 |
#DIV/0! |
#DIV/0! |
#DIV/0! |
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2 |
1.5 |
0.366 |
14 |
63.407 |
887.703 |
7.5 |
#DIV/0! |
#DIV/0! |
#DIV/0! |
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3 |
2.5 |
0.611 |
15 |
63.407 |
951.111 |
12 |
#DIV/0! |
#DIV/0! |
#DIV/0! |
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4 |
3.5 |
0.855 |
16 |
63.407 |
1014.518 |
17.5 |
#DIV/0! |
#DIV/0! |
#DIV/0! |
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5 |
4.5 |
1.099 |
13 |
63.407 |
824.296 |
15 |
#DIV/0! |
#DIV/0! |
#DIV/0! |
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6 |
5.5 |
1.343 |
6 |
63.407 |
380.444 |
12.5 |
#DIV/0! |
#DIV/0! |
#DIV/0! |
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7 |
6.5 |
1.587 |
3 |
63.407 |
190.222 |
9 |
#DIV/0! |
#DIV/0! |
#DIV/0! |
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8 |
7.5 |
1.832 |
1.5 |
63.407 |
95.111 |
5 |
#DIV/0! |
#DIV/0! |
#DIV/0! |
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9 |
8.5 |
2.076 |
0.1 |
63.407 |
6.341 |
1.5 |
#DIV/0! |
#DIV/0! |
#DIV/0! |
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10 |
9.5 |
2.320 |
0 |
63.407 |
0.000 |
0 |
#DIV/0! |
#DIV/0! |
#DIV/0! |
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TABLICA NR 17 |
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Momenty pierścieniowe w płycie fundamentowej o śr. 20m |
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Obciążenie symatryczne |
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Obciążenie antysymetryczne |
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Lp. |
r |
ρ = r/a |
X' |
po*a2/16 |
Mts |
Y' |
pa*a2/16 |
Mta |
Mt |
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[m] |
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[m] |
[kN] |
[kNm] |
[m] |
[kN] |
[kNm] |
[kNm] |
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1 |
0.5 |
0.122 |
13.5 |
63.407 |
856.000 |
0.5 |
#DIV/0! |
#DIV/0! |
#DIV/0! |
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2 |
1.5 |
0.366 |
13.7 |
63.407 |
868.681 |
1.5 |
#DIV/0! |
#DIV/0! |
#DIV/0! |
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3 |
2.5 |
0.611 |
14 |
63.407 |
887.703 |
2 |
#DIV/0! |
#DIV/0! |
#DIV/0! |
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4 |
3.5 |
0.855 |
14.5 |
63.407 |
919.407 |
3 |
#DIV/0! |
#DIV/0! |
#DIV/0! |
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5 |
4.5 |
1.099 |
14 |
63.407 |
887.703 |
3.5 |
#DIV/0! |
#DIV/0! |
#DIV/0! |
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6 |
5.5 |
1.343 |
12 |
63.407 |
760.889 |
3.5 |
#DIV/0! |
#DIV/0! |
#DIV/0! |
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7 |
6.5 |
1.587 |
11 |
63.407 |
697.481 |
3.2 |
#DIV/0! |
#DIV/0! |
#DIV/0! |
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8 |
7.5 |
1.832 |
9 |
63.407 |
570.666 |
2.8 |
#DIV/0! |
#DIV/0! |
#DIV/0! |
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9 |
8.5 |
2.076 |
7.5 |
63.407 |
475.555 |
2.1 |
#DIV/0! |
#DIV/0! |
#DIV/0! |
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10 |
9.5 |
2.320 |
6 |
63.407 |
380.444 |
1.8 |
#DIV/0! |
#DIV/0! |
#DIV/0! |
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TABLICA NR 18 |
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Beton B-37 |
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Stal A-II |
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Zbrojenie płyty fund. D=20m w kierunku promieniowym. |
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fcd= |
20000 |
kPa |
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fyd= |
310000 |
kPa |
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Średn. prętów |
Ilość |
Rozstaw |
Pole pow |
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Lp. |
r |
h |
d |
Asmin=0,0015*b*d |
Mr |
μeff |
ζ |
As |
zbrojeniowych |
prętów |
prętów na |
przyjętego zbr. |
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[m] |
[m] |
[m] |
[cm2] |
[kNm] |
[-] |
[-] |
[cm2/mb] |
[mm] |
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1mb [cm] |
[cm2] |
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1 |
0.5 |
3 |
2.9 |
43.5 |
#DIV/0! |
#DIV/0! |
#DIV/0! |
#DIV/0! |
32 |
#DIV/0! |
17 |
#DIV/0! |
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2 |
1.5 |
3 |
2.9 |
43.5 |
#DIV/0! |
#DIV/0! |
#DIV/0! |
#DIV/0! |
32 |
#DIV/0! |
17 |
#DIV/0! |
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3 |
2.5 |
3 |
2.9 |
43.5 |
#DIV/0! |
#DIV/0! |
#DIV/0! |
#DIV/0! |
32 |
#DIV/0! |
17 |
#DIV/0! |
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4 |
3.5 |
3 |
2.9 |
43.5 |
#DIV/0! |
#DIV/0! |
#DIV/0! |
#DIV/0! |
32 |
#DIV/0! |
17 |
#DIV/0! |
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5 |
4.5 |
3.9 |
2.9 |
43.5 |
#DIV/0! |
#DIV/0! |
#DIV/0! |
#DIV/0! |
32 |
#DIV/0! |
17 |
#DIV/0! |
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6 |
5.5 |
2.89 |
2.79 |
41.85 |
#DIV/0! |
#DIV/0! |
#DIV/0! |
#DIV/0! |
32 |
#DIV/0! |
17 |
#DIV/0! |
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7 |
6.5 |
2.58 |
2.48 |
37.2 |
#DIV/0! |
#DIV/0! |
#DIV/0! |
#DIV/0! |
32 |
#DIV/0! |
20 |
#DIV/0! |
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8 |
7.5 |
2.27 |
2.17 |
32.55 |
#DIV/0! |
#DIV/0! |
#DIV/0! |
#DIV/0! |
32 |
#DIV/0! |
20 |
#DIV/0! |
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9 |
8.5 |
1.96 |
1.86 |
27.9 |
#DIV/0! |
#DIV/0! |
#DIV/0! |
#DIV/0! |
32 |
#DIV/0! |
25 |
#DIV/0! |
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10 |
9.5 |
1.66 |
1.56 |
23.4 |
#DIV/0! |
#DIV/0! |
#DIV/0! |
#DIV/0! |
32 |
4 |
25 |
32.170 |
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TABLICA NR 19 |
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Pole 1 pręta |
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Zbrojenie płyty fund. D=20m w kierunku pierścieniowym. |
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A= |
8.04247719318987 |
cm^2 |
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Średn. prętów |
Ilość |
Rozstaw |
Pole pow |
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Lp. |
r |
h |
d |
Asmin=0,0015*b*d |
Mt |
μeff |
ζ |
As |
zbrojeniowych |
prętów |
prętów na |
przyjętego zbr. |
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[m] |
[m] |
[m] |
[cm2] |
[kNm] |
[-] |
[-] |
[cm2/mb] |
[mm] |
|
1mb [cm] |
[cm2] |
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1 |
0.5 |
3 |
2.9 |
43.5 |
#DIV/0! |
#DIV/0! |
#DIV/0! |
#DIV/0! |
32 |
#DIV/0! |
17 |
#DIV/0! |
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2 |
1.5 |
3 |
2.9 |
43.5 |
#DIV/0! |
#DIV/0! |
#DIV/0! |
#DIV/0! |
32 |
#DIV/0! |
17 |
#DIV/0! |
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3 |
2.5 |
3 |
2.9 |
43.5 |
#DIV/0! |
#DIV/0! |
#DIV/0! |
#DIV/0! |
32 |
#DIV/0! |
17 |
#DIV/0! |
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4 |
3.5 |
3 |
2.9 |
43.5 |
#DIV/0! |
#DIV/0! |
#DIV/0! |
#DIV/0! |
32 |
6 |
17 |
48.255 |
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5 |
4.5 |
3.9 |
2.9 |
43.5 |
#DIV/0! |
#DIV/0! |
#DIV/0! |
#DIV/0! |
32 |
6 |
17 |
48.255 |
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6 |
5.5 |
2.89 |
2.79 |
41.85 |
#DIV/0! |
#DIV/0! |
#DIV/0! |
#DIV/0! |
32 |
#DIV/0! |
17 |
#DIV/0! |
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7 |
6.5 |
2.58 |
2.48 |
37.2 |
#DIV/0! |
#DIV/0! |
#DIV/0! |
#DIV/0! |
32 |
6 |
17 |
48.255 |
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8 |
7.5 |
2.27 |
2.17 |
32.55 |
#DIV/0! |
#DIV/0! |
#DIV/0! |
#DIV/0! |
32 |
#DIV/0! |
20 |
#DIV/0! |
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9 |
8.5 |
1.96 |
1.86 |
27.9 |
#DIV/0! |
#DIV/0! |
#DIV/0! |
#DIV/0! |
32 |
5 |
20 |
40.212 |
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10 |
9.5 |
1.66 |
1.56 |
23.4 |
#DIV/0! |
#DIV/0! |
#DIV/0! |
#DIV/0! |
32 |
5 |
20 |
40.212 |
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