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1.0. PŁYTA |
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W projekcie przyjęto płytę o grubości i rozpętości pomiędzy podporami: |
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hf = |
9 |
cm |
leff = L = |
2,3 |
m |
bż = |
0,2 |
m |
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1.1 Zebranie obciążeń (na 1mb płyty żelbetowej) |
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Nazwa obciążenia |
Grubość |
Ciężar objętościowy |
Obc. Charakteryst. |
γf |
Obc. Obliczeniowe |
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[m] |
[kN/m3] |
[kN/mb] |
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[kN/mb] |
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lastrico |
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0,03 |
22 |
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0,66 |
1,2 |
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0,792 |
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styropian |
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0,02 |
0,45 |
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0,009 |
1,2 |
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0,0108 |
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płyta żelbet. |
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0,09 |
25 |
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2,25 |
1,1 |
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2,475 |
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tynk cem.-wap. |
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0,015 |
19 |
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0,285 |
1,3 |
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0,3705 |
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SUMA |
qk = |
3,204 |
- |
q = |
3,6483 |
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Obc.użytkowe |
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pk = |
4,4 |
1,3 |
p = |
5,72 |
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Obciążenie zmienne długotrwałe: |
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pklt = |
pk *ψd *1m = |
3,52 |
kPa |
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dla budownictwa przemysłowego ψd = |
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0,8 |
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1.2 Schemat statyczny |
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Płyta sześcioprzęsłowa, obliczenia wykonano dla belki pięcioprzęsłowej za pomocą tablic Winklera. |
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Obciążenie zmienne jest dowolnie ustawione na belce. |
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1.3 Momenty zginające w wybranych miejscach od obciążenia zewnętrznego. |
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1.3.1. Wzór ogólny na oblicznianie momentów przy pomocy tablic Winklera: |
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gdzie: |
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αij - |
współczynniki z tablic Winklera |
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leff - |
długość obliczeniowe równa L |
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1.3.2. Momenty przęsłowe: |
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MABmax = |
(0,078*3,6483+0,1*5,72)*2,32 = |
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4,531 |
kNm |
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MABmin = |
(0,078*43,6483-0,026*5,72)*2,32 = |
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0,719 |
kNm |
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MBCmax = |
(0,033*3,6483+0,079*5,72)*2,32 = |
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3,027 |
kNm |
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MBCmin = |
(0,033*3,6483-0,046*5,72)*2,32 = |
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-0,755 |
kNm |
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MCCmax = |
(0,046*3,6483+0,086*5,72)*2,32 = |
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3,490 |
kNm |
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MCCmin = |
(0,046*3,6483-0,040*5,72)*2,32 = |
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-0,323 |
kNm |
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1.3.3. Momenty odpowiadające: |
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odp MB = |
(-0,105*3,6483-0,053*5,72)*2,32 = |
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-3,63 |
kNm |
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odp MC = |
(-0,079*3,6483-0,040*5,72)*2,32 = |
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-2,74 |
kNm |
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1.3.4. Momenty podporowe: |
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min MB = |
(-0,105*3,6483-0,119*5,72)*2,32 = |
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-5,63 |
kNm |
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min MC = |
(-0,079*3,6483-0,111*5,72)*2,32 = |
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-4,88 |
kNm |
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1.3.5. Momenty krawędziowe: |
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QBL = |
(-0,605*3,6483-0,620*5,72)*2,3 = |
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-13,23332945 |
kN |
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QBP = |
(0,526*3,6483+0,598*5,72)*2,3 = |
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12,281 |
kN |
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QCL = |
(-0,474*3,6483-0,577*5,72)*2,3 = |
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-11,568 |
kN |
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QCP = |
(0,500*3,6483+0,591*5,72)*2,3 = |
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11,971 |
kN |
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kraw MB = |
- (|min MB| - |0,5 * bż * minQB|) = |
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-4,399 |
kNm |
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kraw MC = |
- (|min MC| - |0,5 * bż * minQC|) = |
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-3,727 |
kNm |
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1.3.6. Wartości uśrednione momentów: |
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uśr MAB = (minMAB +0,5*(odp MB))/2 = |
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-0,548 |
kNm |
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uśr MBC = (minMBC +0,5*(odp MB +odp MC))/2 = |
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-1,969 |
kNm |
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uśr MCC = (minMCC + odp MC)/2 = |
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-1,529 |
kNm |
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1.4. Wymiarowanie płyty: |
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1.4.1 DANE wyjściowe |
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Obliczenie otuliny zbrojenia: |
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c = |
20 |
mm |
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Do obliczeń przyjęto pręty Φ |
6 |
mm |
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d =90-20-8/2 = |
67 |
mm |
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d = |
0,067 |
m |
, wysokość użyteczna przekroju z projektu wstępnego |
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b = |
1 |
m |
, szerokość pasma obliczeniowego |
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beton |
B30 |
→ |
fcd = |
16,7 |
MPa |
= |
16700 |
kPa |
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stal |
A-III |
→ |
fyd = |
350 |
MPa |
= |
350000 |
kPa |
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αcc = |
1,0 |
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fyk = |
410 |
MPa |
= |
410000 |
kPa |
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1.4.2. Stany graniczne nośności z uwagi na zginanie: |
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Obliczenie względnej wysokośc strefy ściskanej: |
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Sprawdzenie czy przekrój wymaga dozbrojenia w strefie ściskanej: |
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, z Tablicy 9 normy PN-B-03264:2002 |
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ξeff.lim = |
0,53 |
dla stali AIII |
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TO JEST NIEPOTRZEBNE !!!! |
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Pole powierzchni przekroju zbrojenia: |
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Warunki konstrukcyjne dla płyt jednokierunkowych: |
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CC |
uśr MCC = |
1,529 |
1,0 |
0,067 |
0,020 |
0,0206 |
0,659 |
0,10 |
3 |
0,848 |
0,127 |
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Maksymalny rozstaw zbrojenia dla płyty |
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góra |
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Φ 8 |
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o grubości 9 cm wynosi: |
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AB |
uśr MAB = |
0,548 |
1,0 |
0,067 |
0,007 |
0,0073 |
0,235 |
0,04 |
2 |
0,565 |
0,084 |
0,9 * 12cm = |
8,1 |
cm |
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góra |
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Φ 8 |
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12cm gdy hf<10cm |
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Φ - maksymalna średnica pręta |
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BC |
uśr MBC = |
1,969 |
1,0 |
0,067 |
0,026 |
0,0266 |
0,851 |
0,13 |
4 |
1,131 |
0,169 |
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góra |
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Φ 8 |
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dg - maksymalny wymiar ziarn kruszywa (16 mm) |
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8 |
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sI = max |
20 |
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Minimalny rozstaw prętów zbrojenia głównego wynosi 21 mm |
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21 |
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= |
1,105 |
cm2 |
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→ minimalne zbrojenie elementu |
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zginanego wynosi: |
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= |
0,871 |
cm2 |
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As1,min = |
1,105 |
cm2 |
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Sprawdzenie czy wymagane jest dozbrojenie górą: |
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Wc = b*h2/6 = |
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0,001350 |
m3 |
dla betonu B30 |
fctm = |
2,6 |
MPa |
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= |
2600 |
kPa |
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Mcr = Wc * fctm = |
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3,510 |
kNm |
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|uśr MAB| = |
0,548 |
< |
3,510 |
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Brak konieczności zbrojenia płyty górą |
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Zestawienie wyników w poniższej tabeli: |
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|uśr MBC| = |
1,969 |
< |
3,510 |
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Brak konieczności zbrojenia płyty górą |
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M [kNm] |
b [m] |
d [m] |
μcs |
ξeff |
AS1 [cm2] |
ρL [%] |
ilość zbr. |
AS1 [cm2] |
ρL [%] |
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[sztuki] |
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|uśr MCC| = |
1,529 |
< |
3,510 |
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Brak konieczności zbrojenia płyty górą |
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AB |
MABmax = |
4,531 |
1,0 |
0,067 |
0,060 |
0,0624 |
1,995 |
0,30 |
10 |
2,827 |
0,422 |
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dół |
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Φ 6 |
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1.4.3. Stan graniczny użytkowania: |
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BC |
MBCmax = |
3,027 |
1,0 |
0,067 |
0,040 |
0,0412 |
1,318 |
0,20 |
6 |
1,696 |
0,253 |
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dół |
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Φ 6 |
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1.4.3.1. Szerokość rozwarcia rys prostopadłych: |
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CC |
MCCmax = |
3,490 |
1,0 |
0,067 |
0,047 |
0,0477 |
1,525 |
0,23 |
6 |
1,696 |
0,253 |
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dół |
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Φ 6 |
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d'=d+bż/6 |
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B |
min MB = |
5,627 |
1,0 |
0,1003 |
0,033 |
0,0341 |
1,630 |
0,16 |
― |
― |
― |
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gdzie z załącznika D normy PN-B-03264:2002: |
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kraw MB = |
4,399 |
1,0 |
0,067 |
0,059 |
0,0605 |
1,934 |
0,29 |
8 |
2,262 |
0,338 |
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d'=d+bż/6 |
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Φ 6 |
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ζ = |
0,9 |
jeżeli |
ρL ≤ 0,5 % |
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C |
min MC = |
4,883 |
1,0 |
0,1003 |
0,029 |
0,0295 |
1,411 |
0,14 |
― |
― |
― |
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ζ = |
0,85 |
jeżeli |
0,5 < ρL ≤ 1 % |
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kraw MC = |
3,727 |
1,0 |
0,067 |
0,050 |
0,0510 |
1,631 |
0,24 |
8 |
2,262 |
0,338 |
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Φ 6 |
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ζ = |
0,8 |
jeżeli |
ρL > 1 % |
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POD TABELKĄ POWINNA BYĆ INFORMACJA O Asmin I KSIeff |
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1.4.3.2. Momenty podporowe - wartości charakterystyczne: |
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min MB = |
(-0,105*3,204-0,12*3,52)*2,32 = |
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-4,01 |
kNm |
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min MC = |
(-0,079*3,204-0,111*3,52)*2,32 = |
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-3,41 |
kNm |
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1.4.3.3. Momenty krawędziowe - wartości charakterystyczne: |
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QBL = |
(-0,605*3,204-0,620*3,52)*2,3 = |
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-9,478 |
kN |
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QBP = |
(0,526*3,204+0,598*3,52)*2,3 = |
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8,718 |
kN |
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QCL = |
(-0,474*3,204-0,577*3,52)*2,3 = |
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-8,164 |
kN |
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QCP = |
(0,500*3,204+0,591*3,52)*2,3 = |
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8,469 |
kN |
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kraw MB = |
- (|min MB| - |0,5 * bż * minQB|) = |
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-3,142 |
kNm |
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kraw MC = |
- (|min MC| - |0,5 * bż * minQC|) = |
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-2,589 |
kNm |
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Przęsło AB |
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Msd = MABmax = |
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(0,078*3,204+0,1*3,52)*2,32 = |
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3,184 |
kNm |
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ρL = |
0,4 |
% |
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σs = |
197743,657735823 |
kPa |
<= |
410000 |
kPa |
warunek nie przekroczony |
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Przęsło BC |
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Msd = MBCmax = |
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(0,033*3,204+0,079*3,52)*2,32 = |
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2,0304 |
kNm |
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ρL = |
0,3 |
% |
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σs = |
210153,612606997 |
kPa |
<= |
410000 |
kPa |
warunek nie przekroczony |
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Przęsło CC |
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Msd = MCCmax = |
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(0,046*3,204+0,086*3,52)*2,32 = |
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2,381 |
kNm |
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ρL = |
0,3 |
% |
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σs = |
246451,3 |
kPa |
<= |
410000 |
kPa |
warunek nie przekroczony |
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Podpora B |
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Msd =kraw MB = |
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- (|min MB| - |0,5 * bż * minQB|) = |
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3,142 |
kNm |
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ρL = |
0,3 |
% |
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σs = |
243941,092790722 |
kPa |
<= |
410000 |
kPa |
warunek nie przekroczony |
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Podpora C |
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Msd = kraw MC = |
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- (|min MC| - |0,5 * bż * minQC|) = |
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2,589 |
kNm |
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ρL = |
0,3 |
% |
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σs = |
201016,618046497 |
kPa |
<= |
410000 |
kPa |
warunek nie przekroczony |
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1.4.3.4. Warunek ugięcia płyty: |
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Sprawdzamy przesło AB - skrajne przesło belki ciagłej |
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leff = L = |
2,3 |
m |
d = |
0,067 |
m |
σs = |
197743,657735823 |
kPa |
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As/bd = |
0,422 |
% |
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leff / d = |
28 |
, na podstawie Tablicy 13 normy PN-B-03264:2002 |
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34,328 |
<= |
35,399 |
kPa |
Warunek ugięcia nie przekroczony. |
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Na podstawie analizy załącznika D1 normy PN-B-03264:2002 wynika, że maksymalna średnica zbrojenia |
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to Φ 12 mm, natomiast największa zaprojektowana to Φ 6 mm → maksymalna szerokośc rys |
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wynosząca 0,3 mm nie zostanie przekroczona. |
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next page |
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end |
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Część nr 3 |
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85% |
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2.0. ŻEBRO |
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L = |
2,3 |
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qk = |
3,204 |
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q = |
3,6483 |
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pk = |
4,4 |
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p = |
5,72 |
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Żebro: |
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h = |
0,41 |
m |
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hf = |
0,09 |
m |
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bż = bw = |
0,2 |
m |
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h - hf = |
0,32 |
m |
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OBLICZENIA PRZEPROWADZAMY DLA PRZEDOSTATNIEGO ŻEBRA, NAJBARDZIEJ OBCIĄŻONEGO: |
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2.1 Zebranie obciążeń: |
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a) obciążenia stałe |
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→ reakcje od obciążeń stałych: |
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Ciężar płyty z pozycji 1.1 (qk ,q) |
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charakterystyczne: |
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RBqk = |
1,132*3,204*2,3 |
= |
8,342 |
kN/m |
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obliczeniowe: |
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RBq = |
1,132*3,6482*2,3 |
= |
9,499 |
kN/m |
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→ ciężar własny żebra z tynkiem: |
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charakterystyczne: |
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gk = |
0,32*0,2*24+0,015*(0,2+2*032)*19 |
= |
1,775 |
kN/m |
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obliczeniowe: |
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g = |
0,32*0,2*24+0,015*(0,2+2*0,32)*19*1,3 |
= |
2,001 |
kN/m |
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SUMA OBCIĄŻEŃ STAŁCH: |
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Obciążenia charakterystyczne: |
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qk = |
gk + RBk = |
10,117 |
kN/m |
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Obciążenia obliczeniowo: |
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q = |
g + RB = |
11,500 |
kN/m |
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b) obciążenia zmienne: |
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Obciążenia użytkowe płyty z pozycji 1.1 (pk, p) |
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charakterystyczne: |
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RBpk = |
1,218*4,4*2,3 |
= |
12,326 |
kN/m |
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obliczeniowe: |
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RBp = |
1,218*5,72*2,3 |
= |
16,024 |
kN/m |
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2.2. Schemat statyczny (obwiednie) |
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Obliczenia wykonano dla schematu belki czteroprzesłowej: |
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L = leff = |
6,3 |
m |
br = |
0,5 |
m |
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2.2.1. Momenty przęsłowe: |
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Maksymalne: |
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MABmax = |
(0,077*15,077+0,1*21,924)*6,62 = |
98,743 |
kNm |
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MBCmax = |
(0,036*15,077+0,074*21,924)*6,62 = |
63,494 |
kNm |
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Minimalne: |
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MABmin = |
(0,077*15,077-0,021*21,924)*6,62 = |
21,788 |
kNm |
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MBCmin = |
(0,036*15,077-0,044*21,924)*6,62 = |
-11,553 |
kNm |
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2.2.2. Momenty podporowe: |
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min MB = |
(-0,107*15,077-0,121*21,924)*6,62 = |
-125,79 |
kNm |
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min MC = |
(-0,071*15,077-0,107*21,924)*6,62 = |
-100,46 |
kNm |
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2.2.3. Momenty odpowiadające: |
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odp MB = |
(-0,107*15,077+0,061*21,924)*6,62 = |
-10,041 |
kNm |
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odp MC = |
(-0,071*15,077-0,056*21,924)*6,62 = |
-68,021 |
kNm |
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2.2.4. Momenty krawędziowe: |
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QA = |
(0,393*15,077+0,446*21,924)*6,6 = |
73,496 |
kN |
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QBL = |
(-0,607*15,077-0,621*21,924)*6,6 = |
-106,666 |
kN |
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QBP = |
(0,536*15,077+0,603*21,924)*6,6 = |
101,522 |
kN |
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QCL = |
(-0,464*15,077-0,571*21,924)*6,6 = |
-91,259 |
kN |
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QCP = |
(0,464*15,077+0,571*21,924)*6,6 = |
91,259 |
kN |
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kraw MB = |
- (|min MB| - |0,5 * bż * minQB|) = |
-100,411 |
kNm |
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kraw MC = |
- (|min MC| - |0,5 * bż * minQC|) = |
-77,642 |
kNm |
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2.2.6. Obwiednia momentów zginających: |
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2.2.7. Obwiednia sił tnących: |
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2.3.1. Obliczenie szerokości płyty współpracującej z żebrem beff |
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bs = |
2,3 |
m |
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hf = |
0,09 |
m |
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bw = br = |
0,2 |
m |
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h = |
0,41 |
m |
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l0 - odcinek, na którym występuje moment jednakowego znaku |
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, ponieważ mamy do czynienia z żebrem przedostatnim to: |
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l0 = |
0,85*leff = |
5,355 |
m |
dla przęsła skrajnego |
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l0 = |
0,7*leff = |
4,41 |
m |
dla przęsła wewnętrznego |
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beff1 (beff2) ≤ 0,1 l0 |
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beff1 (beff2) ≤ 0,5 bs |
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poprawić |
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beff1 (beff2) ≤ 6 hf |
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2.3.1.1. beff dla przęsła skrajnego: |
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beff1 (beff2) |
≤ |
0,5355 |
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beff1 (beff2) |
≤ |
1,15 |
→ |
beff1 = |
0,5355 |
m |
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beff1 (beff2) |
≤ |
0,54 |
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beff = |
2 * beff1 + bw = |
|
1,271 |
m |
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2.3.1.2. beff dla przęsła wewnętrznego: |
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beff1 (beff2) |
≤ |
0,441 |
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beff1 (beff2) |
≤ |
1,15 |
→ |
beff1 = |
0,441 |
m |
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beff1 (beff2) |
≤ |
0,54 |
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beff = |
2 * beff1 + bw = |
|
0,882 |
m |
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2.3.2. Przyjęcie wymiarów d i d': |
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Obliczenie otuliny zbrojenia: |
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cmin = |
25 |
mm |
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Do obliczeń przyjęto pręty: |
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dla prętów głównych Φ = |
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22 |
mm |
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As = |
3,80132711084365 |
cm2 |
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dla strzemion Φst = |
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8 |
mm |
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a1 = |
cmin + Φst + Φ/2 = |
|
44 |
mm |
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d = h - a1 = |
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0,366 |
m |
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d' = |
d + br/6 = |
|
0,399 |
m |
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2.3.3. Wyznaczenie rodzaju przekroju: |
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Dla przęsła skrajnego: |
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Mf = |
425,532 |
kNm |
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Dla przęsła wewnętrznego: |
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Mf = |
613,210 |
kNm |
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→ |
przekrój pozornie teowy, gdy: |
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Msd ≤ Mf |
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→ |
przekrój pozornie teowy, gdy: |
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Msd > Mf |
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Przęsło AB |
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Msd |
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Mf |
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98,743 |
<= |
425,532 |
|
Przekrój pozornie teowy. |
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Przęsło BC |
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Msd |
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Mf |
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63,494 |
<= |
613,210 |
|
Przekrój pozornie teowy. |
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2.3.4. Stan graniczny nośności (SGN) z uwagi na zginanie: |
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STAL: |
A-III |
25G2S |
→ |
fyk = |
395000 |
kPa |
= |
395 |
MPa |
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fyd = |
350000 |
kPa |
= |
350 |
MPa |
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BETON: |
B25 |
|
→ |
fctm = |
2600 |
kPa |
= |
2,6 |
MPa |
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fcd = |
16700 |
kPa |
= |
16,7 |
MPa |
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fctd = |
1000 |
kPa |
= |
1 |
MPa |
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fck = |
20000 |
kPa |
= |
20 |
MPa |
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Obliczenie względnej wysokośc strefy ściskanej: |
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Sprawdzenie czy przekrój wymaga dozbrojenia w strefie ściskanej: |
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, z Tablicy 9 normy PN-B-03264:2002 |
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ξeff.lim = |
0,53 |
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Pole powierzchni przekroju zbrojenia: |
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Wyznaczenie zbrojenia minimalnego: |
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Zestawienie wyników w poniższej tabeli: |
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PRZYJĘTO |
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przekrój |
|
Msd |
d/d' |
beff/bw |
μeff |
ξeff |
AS1 |
ρL [%] |
Φ |
ilość zbr. |
AS1 |
ρL |
= |
1,2527 |
cm2 |
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|
[kNm] |
[cm] |
[cm] |
[-] |
[-] |
[cm2] |
|
[mm] |
[sztuki] |
[cm2] |
[%] |
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As1,min = |
1,2527 |
cm2 |
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AB |
MABmax = |
98,743 |
36,6 |
127,1 |
0,035 |
0,0354 |
7,847 |
0,17 |
22 |
3 |
11,404 |
0,245 |
= |
0,952 |
cm2 |
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BC |
MBCmax = |
63,494 |
36,6 |
88,2 |
0,032 |
0,0327 |
5,039 |
0,16 |
22 |
2 |
7,603 |
0,236 |
Wyniki z powyższych obliczeń zostały zestawione w kolejnej tabeli: |
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→ |
→ |
→ |
→ |
→ |
→ |
→ |
→ |
→ |
→ |
→ |
B |
min MB = |
125,792 |
39,9 |
20,0 |
0,236 |
0,2736 |
10,426 |
1,31 |
― |
― |
― |
― |
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2.3.5. Długość zakotwienia pręta: |
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kraw MB = |
100,411 |
36,6 |
20,0 |
0,224 |
0,2576 |
8,997 |
1,23 |
22 |
5 |
19,007 |
2,597 |
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C |
min MC = |
100,457 |
39,9 |
20,0 |
0,189 |
0,2108 |
8,034 |
1,01 |
― |
― |
― |
― |
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kraw MC = |
77,642 |
36,6 |
20,0 |
0,174 |
0,1920 |
6,705 |
0,92 |
22 |
3 |
11,404 |
1,558 |
gdzie: |
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lb - podstawowa długość zakotwienia |
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fbd - przyczepność obliczeniowa, wartość z tablicy 24 |
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As1,min = |
1,2527 |
cm2 |
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fcd = |
2,3 |
MPa |
, dla prętów żebrowanych Φ ≤ 32mm i betonu B25 |
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ξeff.lim = |
0,53 |
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lb = |
83,70 |
cm |
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2.4. Wymiarowanie żebra z uwagi na ścinanie: |
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2.4.1. Zestawienie sił tnących: |
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odległość od podpory skąd odczytujemu siłę tnącą: |
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br /2 + d = |
0,616 |
m |
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Wartości sił tnacych zostały wyznaczone na podstawie wykresu obwiedni |
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sił tnących wykonanego w programie RM - WIN |
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max QA = |
81,08 |
kN |
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min QBL = |
127,60 |
kN |
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max QBP = |
117,91 |
kN |
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min QCL = |
106,28 |
kN |
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2.4.2. Wymiarowanie strzemion: |
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VRd1= |
nośnośc elementu bez zbrojenia poprzecznego |
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VRd2= |
nośnośc elementu z uwagi na beton |
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VRd3= |
nośność elementu z uwagi na zbrojenie poprzeczne |
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Vsd= |
siła obliczeniowa |
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Przyjęto do obliczeń strzemiona dwucięte 8 mm ze stali A-I. |
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fyd = |
210 |
MPa |
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Asw1= |
1,00528 |
m2 |
= |
0,000100528 |
m2 |
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fyk = |
240 |
MPa |
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Wyznaczenie maksymalnego rozstawu strzemion: |
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0,275 |
m |
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0,4 |
m |
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Przyjęto maksymalny rozstaw strzemion 27 cm |
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0,001491 |
= |
0,1491 |
% |
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2.4.2.1. Podpora A |
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k = 1,6 - d = |
1,234 |
m |
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0,010386 |
> |
0,01 |
Przyjmujemy ρL = |
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0,01 |
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VRd1= |
50,58 |
kN |
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Vsd |
> |
VRd1 |
odcinek drugiego rodzaju |
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Vsd = max QA = |
81,08 |
kN |
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Wyznaczenie długości odcinka drugiego rodzaju. |
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Odczytane z obwiedni sił tnących wygenerowanej przez program RM-WIN |
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Odcinek ten ma długość: |
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c = |
1,45 |
m |
, mierzone od osi podpory A |
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Wyznaczenie zbrojenia z uwagi na ścinanie: |
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Przyjmujemy: |
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ctg Θ = |
1 |
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, współczynnik zmiejszajacy wytrzymałość betonu na ściskanie |
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v = |
0,552 |
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z=0,9*d = |
0,329 |
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VRd2= |
303,65 |
kN |
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Vsd |
< |
VRd2 |
Warunek jest spełniony |
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Vsd = |
81,08 |
kN |
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Wyznaczenie potrzebnego rozstawu strzemion: |
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|
|
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|
|
Vsd=VRd3= |
81,08 |
|
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|
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|
s1= |
0,086 |
m |
,przyjęto rozstaw co |
|
0,085 |
m |
= |
8,5 |
cm |
|
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|
0,005861 |
= |
0,5861 |
% |
Warunek spełniony |
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|
2.4.2.2. Podpora BL |
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|
VRd1= |
50,58 |
kN |
|
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|
|
Vsd |
> |
VRd1 |
odcinek drugiego rodzaju |
|
|
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|
|
|
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|
|
Vsd = min QBL = |
127,60 |
kN |
|
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|
k = 1,6 - d = |
1,234 |
m |
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|
Wyznaczenie długości odcinka drugiego rodzaju. |
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Odczytane z obwiedni sił tnących wygenerowanej przez program RM-WIN |
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Odcinek ten ma długość: |
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|
> |
0,01 |
Przyjmujemy ρL = |
|
0,01 |
|
|
|
c = |
2,7 |
m |
, mierzone od osi podpory A |
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Strefę przypodporową dzielimy na dwa odcinki: |
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c1 = |
1 |
m |
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c2 = |
1,7 |
m |
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|
Odcinek |
c1 = |
1 |
m |
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|
Vsd=VRd3= |
127,60 |
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|
s1= |
0,054 |
m |
,przyjęto rozstaw co |
|
0,055 |
m |
= |
5,5 |
cm |
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|
0,009223 |
= |
0,9223 |
% |
Warunek spełniony |
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|
Odcinek |
c2 = |
1,7 |
m |
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|
Z obwiedni sił tnących w RM - WIN |
|
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|
Vsd=VRd3= |
90,672 |
kN |
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|
s2 = |
0,077 |
m |
,przyjęto rozstaw co |
|
0,075 |
m |
= |
7,5 |
cm |
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|
0,006554 |
= |
0,6554 |
% |
Warunek spełniony |
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|
2.4.2.3. Podpora BP |
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|
VRd1= |
50,58 |
kN |
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|
Vsd |
> |
VRd1 |
odcinek drugiego rodzaju |
|
|
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|
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|
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|
|
|
|
|
|
Vsd = max QBP = |
117,91 |
kN |
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|
Wyznaczenie długości odcinka drugiego rodzaju. |
|
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|
|
|
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|
|
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|
|
|
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|
|
|
Odczytane z obwiedni sił tnących wygenerowanej przez program RM-WIN |
|
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|
Odcinek ten ma długość: |
|
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|
c = |
2,45 |
m |
, mierzone od osi podpory A |
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|
Strefę przypodporową dzielimy na dwa odcinki: |
|
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|
c1 = |
1 |
m |
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|
c2 = |
1,45 |
m |
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|
|
Odcinek |
c1 = |
1 |
m |
|
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|
Vsd=VRd3= |
117,91 |
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|
s1= |
0,059 |
m |
,przyjęto rozstaw co |
|
0,060 |
m |
= |
6,0 |
cm |
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|
0,008522 |
= |
0,8522 |
% |
Warunek spełniony |
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|
|
Odcinek |
c2 = |
1,45 |
m |
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|
Z obwiedni sił tnących w RM - WIN |
|
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|
|
|
Vsd=VRd3= |
80,980 |
kN |
|
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|
s2 = |
0,086 |
m |
,przyjęto rozstaw co |
|
0,086 |
m |
= |
8,6 |
cm |
|
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|
0,005853 |
= |
0,5853 |
% |
Warunek spełniony |
|
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|
2.4.2.4. Podpora CL |
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|
|
VRd1= |
50,58 |
kN |
|
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|
|
Vsd |
> |
VRd1 |
odcinek drugiego rodzaju |
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Vsd = min QCL = |
106,28 |
kN |
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Wyznaczenie długości odcinka drugiego rodzaju. |
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Odczytane z obwiedni sił tnących wygenerowanej przez program RM-WIN |
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Odcinek ten ma długość: |
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c = |
2,10 |
m |
, mierzone od osi podpory A |
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Strefę przypodporową dzielimy na dwa odcinki: |
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c1 = |
1 |
m |
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c2 = |
1,10 |
m |
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Odcinek |
c1 = |
1 |
m |
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Vsd=VRd3= |
106,28 |
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s1= |
0,065 |
m |
,przyjęto rozstaw co |
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0,065 |
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6,5 |
cm |
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0,007682 |
= |
0,7682 |
% |
Warunek spełniony |
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Odcinek |
c2 = |
1,1 |
m |
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Z obwiedni sił tnących w RM - WIN |
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Vsd=VRd3= |
69,350 |
kN |
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s2 = |
0,100 |
m |
,przyjęto rozstaw co |
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0,100 |
m |
= |
10,0 |
cm |
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0,005013 |
= |
0,5013 |
% |
Warunek spełniony |
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2.5. Zbrojenie podłużne. |
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Ftd- |
obliczeniowa siła rozciągająca w zbrojeniu podłużnym |
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w przekroju sprawdzanym na ścinanie |
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Msd - moment w odległości d +br/2 od podpory |
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wpływ siły poprzecznej |
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Warunek: |
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ctgΘ |
= |
1 |
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Podpora |
Msd |
Vsd |
z |
ΔFtd |
As1 |
fyd |
Ftd |
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As1 * fyd |
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A |
55,662 |
81,08 |
0,329 |
40,541 |
11,404 |
350000 |
209,521 |
<= |
399,139 |
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BL |
101,396 |
127,60 |
0,329 |
63,7995 |
19,007 |
350000 |
371,620 |
<= |
665,232 |
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BP |
109,625 |
117,91 |
0,329 |
58,9535 |
19,007 |
350000 |
391,756 |
<= |
665,232 |
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CL |
78,947 |
106,28 |
0,329 |
53,1385 |
11,404 |
350000 |
292,808 |
<= |
399,139 |
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Wszystkie warunki zostały spełnione |
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2.6. SGU - stan graniczny użytkowania. |
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2.6.1. Sprawdzenie ugięcia. |
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2.6.1.1. Wyznaczenie kombinacji obciążęń charakterystycznych. |
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gk= |
10,117 |
kN/mb |
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pk= |
12,326 |
kN/mb |
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0,8*pk= |
9,861 |
kN/mb |
na podstawie tab.2 PN-82/B-02003 współczynnik |
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obciążenie długotrwałe dla budownictwa przemysłowego |
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L = leff = |
6,6 |
m |
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Wykres momentów zginających wykonany w programie RM-WIN |
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MBC = |
71,499 |
kNm |
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2.6.1.2. Przęsło AB - sprawdzenie ugięcia |
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As1 = |
11,404 |
cm2 |
= |
0,0011404 |
m2 |
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ρL = |
0,245 |
% |
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d= |
0,366 |
m |
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ζ= |
0,85 |
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f yk= |
395000 |
kPa |
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|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
Sprawdzenie naprężeń: |
|
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|
|
|
|
|
|
|
|
|
|
|
|
|
|
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|
|
|
|
|
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|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
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|
|
|
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|
|
|
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|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
σS= |
301019,35 |
kPa |
= |
301,019 |
MPa |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
301,019 |
<= |
395 |
MPa |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
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|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
Warunek na naprężenia jest spełniony |
|
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|
|
|
|
|
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|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
Sprawdzenie smukłości przekroju: |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
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|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
18,033 |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
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|
|
|
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|
|
|
|
|
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|
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|
|
|
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|
|
|
|
|
|
|
|
|
|
|
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|
|
|
|
|
|
|
|
Z interpolacji wynika: |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
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|
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|
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|
|
|
|
|
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|
|
|
|
|
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|
|
|
|
|
|
|
23,254 |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
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|
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|
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|
|
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|
|
|
|
|
|
|
23,254 |
> |
18,033 |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
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|
|
|
|
|
|
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|
|
|
|
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|
|
|
Warunek na smukłość spełniony |
|
|
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|
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|
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|
|
|
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|
2.6.1.3. Przęsło BC - sprawdzenie ugięcia |
|
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|
|
|
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|
|
|
|
|
|
As1 = |
7,603 |
cm2 |
= |
0,0007603 |
m2 |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
ρL = |
0,236 |
% |
|
|
|
|
|
|
|
|
|
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|
|
|
|
|
|
|
|
|
|
|
d= |
0,366 |
m |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
ζ= |
0,9 |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
f yk= |
395000 |
kPa |
|
|
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|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
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|
|
|
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|
|
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|
|
|
|
|
|
|
Sprawdzenie naprężeń: |
|
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|
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|
|
|
|
|
|
|
|
|
|
|
σS= |
285503,30 |
kPa |
= |
285,503 |
MPa |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
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|
|
|
|
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|
|
|
|
|
|
|
|
|
|
|
285,503 |
<= |
395 |
MPa |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
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|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
Warunek na naprężenia jest spełniony |
|
|
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|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
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|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
Sprawdzenie smukłości przekroju: |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
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|
|
|
|
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|
|
|
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|
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|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
18,033 |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
Z interpolacji wynika: |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
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|
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|
|
|
|
|
|
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|
|
|
|
|
|
|
|
|
|
|
|
24,518 |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
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|
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|
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|
|
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|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
24,518 |
> |
18,033 |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
Warunek na smukłość spełniony |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
2.6.2. Sprawdzenie rys prostopadłych. |
|
|
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|
|
|
|
|
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|
|
|
|
|
|
|
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|
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|
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|
|
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|
|
|
|
|
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|
|
|
|
|
|
|
|
|
|
|
|
|
Obciążenie charakterystyczne (bez długotrwałych): |
|
|
|
|
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|
|
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|
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|
|
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|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
gk= |
10,117 |
kN/mb |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
pk= |
12,326 |
kN/mb |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
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|
|
Z wykresu obwiedni (Program RM-WIN) |
|
|
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|
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|
|
|
|
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|
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|
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|
|
|
|
|
|
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|
|
|
|
|
|
|
|
MB = |
137,755 |
kNm |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
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|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
MC = |
108,891 |
kNm |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
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|
|
|
2.6.2.1. Podpora B |
|
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|
|
|
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|
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|
|
|
|
|
|
|
|
|
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|
|
|
|
|
|
|
As1 = |
19,007 |
cm2 |
= |
0,001900663555422 |
m2 |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
ρL = |
2,597 |
% |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
d= |
0,366 |
m |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
ζ= |
0,8 |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
f yk= |
395000 |
kPa |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
Sprawdzenie naprężeń: |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
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|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
σS= |
247531,82930976 |
kPa |
= |
247,532 |
MPa |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
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|
|
|
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|
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|
|
|
|
|
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|
|
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|
|
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|
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|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
247,532 |
< |
395 |
MPa |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
Warunek na naprężenia spełniony |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
2.6.2.2. Podpora C |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
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|
|
|
|
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|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
As1 = |
11,404 |
cm2 |
= |
0,001140398133253 |
m2 |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
ρL = |
1,558 |
% |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
d= |
0,366 |
m |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
ζ= |
0,8 |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
f yk= |
395000 |
kPa |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
Sprawdzenie naprężeń: |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
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|
|
|
|
|
|
|
σS= |
326110,224981175 |
kPa |
= |
326,110 |
MPa |
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326,110 |
< |
395 |
MPa |
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Warunek na naprężenia spełniony |
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2.6.3. Sprawdzenie rys ukośnych. |
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szerokość rys ukośnych: |
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gdzie: |
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Naprężenia: |
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Współczynnik: |
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Stopień zbrojenia: |
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ES= |
200 |
GPa |
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fck= |
20000 |
kPa |
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1 |
, współczynnik przyczepności, dla strzemion ze stali gładkiej |
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Musi być spełniony warunek: |
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Zestawienie wielkości rys: |
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przekrój |
Vsd |
τ |
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λ |
wk |
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wk,lim |
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[kN] |
[kN/m2] |
[mm] |
[mm] |
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[mm] |
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A |
60,043 |
820,260 |
0,005913411764706 |
450,952 |
0,051 |
< |
0,3 |
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BL |
95,328 |
1302,295 |
0,009138909090909 |
291,793 |
0,054 |
< |
0,3 |
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BP |
87,442 |
1194,563 |
0,008377333333333 |
318,319 |
0,054 |
< |
0,3 |
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CL |
78,265 |
1069,194 |
0,007681860236505 |
347,138 |
0,052 |
< |
0,3 |
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Wartości Vsd z programu RM-WIN |
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Warunki rys ukośnych = 0,3 mm zostały spełnione |
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Analiza załącznika D1 |
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Tablicy D.1. PN-B-03264:2002 Maksymalna średnica prętów |
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zbrojenia [w milimetrach] |
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ρI [%] |
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σS[Mpa] |
0,75 |
1 |
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225 |
32 |
32 |
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250 |
28 |
32 |
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275 |
22 |
32 |
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300 |
20 |
28 |
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Na podstawie analizy załącznika D1 normy PN-B-03264:2002 wynika, że maksymalna średnica zbrojenia |
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to Φ 28 mm, natomiasy największa zaprojektowana to Φ 22 mm → maksymalna szerokośc rys |
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wynosząca 0,3 mm nie zostanie przekroczona. |
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2.6.4. Połączenie prętów na zakład |
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(wg 8.1.4.3. PN-B-03264:2002) |
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2 |
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ls= |
167,4 |
cm = |
1,7 |
m |
, długość wymaganego zakłądu |
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