1.0. ZEBRANIE OBCIĄŻEŃ. |
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1.1. Obciążenia stałe. |
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1.1.1. Obciążenie z pokrycia. |
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qk [kN/m2] |
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q [kN/m2] |
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- 2 x papa "500" na lepiku |
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2 * 0,05 |
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0,10 |
1,2 |
0,12 |
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- gładź cementowa gr.2cm |
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0,02 * 21,0 |
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0,42 |
1,3 |
0,55 |
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- styropian gr.4cm |
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0,04 * 0,45 |
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0,02 |
1,2 |
0,02 |
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- płyta łupinowa PŁ-12/P. |
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1,78 |
1,1 |
1,96 |
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2,32 |
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2,65 |
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1.1.2. Ciężar dźwigara. |
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qk [kN] |
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q [kN] |
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- dźwigar kablobetonowy KBOS-24/66 |
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92,00 |
1,1 |
101,20 |
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1.1.3. Ciężar belki podsuwnicowej. |
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qk [kN] |
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q [kN] |
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- belka podsuwnicowa KBP - 120/II |
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150,00 |
1,1 |
165,00 |
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- szyna typu SD65 |
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0,435 * 12,0 |
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5,22 |
1,1 |
5,74 |
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155,22 |
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170,74 |
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1.1.4. Ciężar ściany. |
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qk [kN] |
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q [kN] |
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- elementy ścienne |
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19,54 |
1,1 |
21,49 |
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- belka podwalinowa BPFF-1 |
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17,50 |
1,1 |
19,25 |
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37,04 |
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40,74 |
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1.1.5. Ciężar słupa. |
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qk [kN] |
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q [kN] |
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- część górna słupa |
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0,4*0,4*3,5*24,0 |
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13,44 |
1,1 |
14,78 |
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-wspornik cz.górnej słupa |
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0,15*0,4*24,0 |
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1,44 |
1,1 |
1,58 |
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- część dolna słupa |
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0,4*0,7*5,5*24,0 |
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36,96 |
1,1 |
40,66 |
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-wspornik cz.dolnej słupa |
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1,16*0,4*24,0 |
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11,14 |
1,1 |
12,25 |
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62,98 |
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69,27 |
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obciążenie na 1 mb słupa: |
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- część górna słupa |
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(14,78 + 1,58) / 3,8 = |
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7,11 |
kN/m |
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- część dolna słupa |
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(40,66 + 12,25) / 6,5 = |
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8,14 |
kN/m |
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1.2. Obciążenia zmienne. |
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1.2.1. Dopuszczalny moment od suwnicy. |
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M = f ( l, P, b ) |
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Pmax = |
131,0 |
kN |
- statyczny nacisk koła suwnicy |
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Q = |
125,0 |
kN |
stąd |
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1,2 |
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l = |
12,0 |
m |
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k = |
4,5 |
m |
- rozstaw kół |
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k = |
4,5 |
m < l /2 = |
6,0 |
m |
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Mmax = 2 Pmax * b * (0.5 l - 0.25 k)2 / l = |
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622,66 |
kNm |
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M = |
862,0 |
kNm |
-max moment przyjętej belki KBP - 120/II |
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1.2.2. Siły pionowe. |
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Rsmax = Pmax * b * gf + (Pmax * b * gf * (l - k)) / l Ł 2 Pmax * b * gf |
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1,1 |
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2 Pmax * b * gf = |
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345,84 |
kN |
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Rsmax =Pmax* b * gf+(Pmax * b * gf *(l - k))/l= |
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271,17 |
kN < |
345,84 |
kN |
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Rsmax = |
271,17 |
kN |
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Rsmin = Pmin * b * gf + (Pmin * b * gf * (l - k)) / l Ł 2 Pmin * b * gf |
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M = 232,0 * 1,1 = |
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255,2 |
kN |
- ciężar obliczeniowy suwnicy |
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Pmin = (M + Q - 2 Pmax) / 2 = |
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59,10 |
kN |
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2 Pmin * b * gf = |
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156,02 |
kN |
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Rsmin =Pmin * b * gf +(Pmin * b * gf * (l - k))/l= |
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126,77 |
kN < |
156,02 |
kN |
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Rsmin = |
126,77 |
kN |
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1.2.3. Siły poziome. |
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Hmax = Pmax / 10 = |
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13,10 |
kN |
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Hmin = Pmin / 10 = |
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5,91 |
kN |
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1.2.4. Obciążenie wiatrem. |
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Parcie i ssanie wiatru na powierzchnię boczną hali. |
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Ciśnienie charakterystyczne: |
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pk = qk * ce * c * b |
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Przyjęto II strefę: |
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qk = |
0,35 |
kPa |
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Wymiary obiektu: |
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H = |
10,8 |
m |
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B = |
24,0 |
m |
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L = |
60,0 |
m |
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Współczynnik ekspozycji dla H ( 10 do 20 ) m: |
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ce = |
0,8 + 0,02H = 0,8 + 0,02*10,8 = |
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1,02 |
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Współczynnik ze względuna niepodatność obiektu na działanie dynamiczne wiatru: |
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1,8 |
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wg PN 77/B-02011 |
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Określenie współczynnika aerodynamicznego c: |
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-0,7 |
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H / L = 10,8 / 60,0 = |
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0,18 |
< 2,0 |
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B / L = 24,0 / 60,0 = |
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0,40 |
< 1,0 |
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-0,4 |
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-0,7 |
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Parcie: |
pk = 0,35 * 1,02 * 0,7 * 1,8 = |
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0,45 |
kPa |
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p0 = gf * pk = 1,3 * 0,45 = |
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0,58 |
kPa |
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Ssanie |
sk = 0.35 * 1.02 * (-0.4) * 1.8 = |
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-0,26 |
kPa |
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s0 = gf * sk = 1.3 * (-0.26) = |
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-0,33 |
kPa |
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1.2.5. Obciążenie śniegiem. |
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Przyjęto I strefę: |
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Qk = |
0,7 |
kPa |
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Nachylenie połaci dachu < 15 => |
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c = |
0,8 |
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sk = Qk * c = |
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0,56 |
kPa |
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s0 = gf * sk = 1,4 * 0,56 = |
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0,78 |
kPa |
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1.3. Schemat statyczny |
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1
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EJ = Ą |
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EJ1 |
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EJ1 |
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EJ2 |
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EJ2 |
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4 |
4 |
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24.0 |
m |
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Beton B 20 |
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b = |
0,4 |
m |
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E = |
27000 |
MPa |
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h1 = |
0,4 |
m |
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h2 = |
0,7 |
m |
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J = b * h3 / 12 |
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J1 = |
0,002133333333333 |
m4 |
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EJ1 = |
57,60 |
kNm2 |
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J2 = |
0,011433333333333 |
m4 |
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EJ2 = |
308,70 |
kNm2 |
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1.4. Obciążanie schematów statycznych. |
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1.4.1. Obciążenia stałe - schemat 1. |
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Reakcja z dachu: |
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a) obciążenie z dachu: |
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2,65 |
kN/m2 |
(poz 1.1.1.) |
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b) ciężar dźwigara: |
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101,20 |
kN |
(poz 1.1.2.) |
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R1 = (101,20 + 24*12*2,65)*0.5 = |
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431,5664 |
kN |
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Reakcja krótkiego wspornika: |
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a) ciężar belki podsuwnicowej: |
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170,74 |
kN |
(poz 1.1.3) |
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R2 = |
170,74 |
kN |
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Mimośrody działania sił skupionych: |
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R1 |
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e1 = (0,4 +0,1)/2 - 0,2 = |
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0,05 |
m. |
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M1 = R1 * e1 = |
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21,58 |
kNm |
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R2 |
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ciężar cz. górnej słupa:14,78+1,58 = |
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16,37 |
kN |
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e2 = (1,4-0,35)-(1,4-0,4)/2 = |
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0,55 |
m. |
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R1* |
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e3 = 0,35-0,2 = |
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0,15 |
m. |
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e3 |
e2 |
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R1* = R1 + 16,37 = |
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447,93 |
kN |
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M2 = - R1* * e3 + R2 * e2 = |
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26,72 |
kNm |
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ciężar części górnej słupa: |
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q1 = |
7,11 |
kN/m. |
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(poz. 1.1.) |
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ciężar części dolnej słupa: |
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q2 = |
8,14 |
kN/m. |
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(poz. 1.1.) |
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Schemat 1A |
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R1 |
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R1 |
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M1 |
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q1
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q1
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R2 |
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R2 |
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q2
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q2
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1.4.2. Obciążenie śniegiem - schemat 1B |
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s = 12 * s0 = |
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9,41 |
kNm |
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M = s0 * 0.5 * B * e1 = |
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0,47 |
kNm |
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Schemat 1B |
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s |
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M |
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M |
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1.4.3. Obciążenie siłą pionową wywołaną suwnicą - schemat 2A i 2B. |
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Reakcje pionowe obciążenia zmiennego: |
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Rsmax = |
271,17 |
kN |
(poz 1.2.2.) |
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Rsmin = |
126,77 |
kN |
(poz 1.2.2.) |
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Mmax = Rsmax * e2 = |
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149,14 |
kNm |
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Mmin = Rsmin * e2 = |
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69,72 |
kNm |
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Schemat 2A |
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Schemat 2B |
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Rmax |
Rmin |
Rmin |
Rmax |
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Mmax |
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1.4.4. Obciążanie siłą poziomą wywołaną hamowaniem wózka suwnicy |
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- schematy 3A, 3B, 3C, 3D. |
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Reakcje poziome obciążenie zmiennego: |
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Hmax = |
13,10 |
kN |
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Hmin = |
5,91 |
kN |
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Schemat 3A |
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Schemat 3B |
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Hmax |
Hmin |
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Hmax |
Hmin |
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Schemat 3C |
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Schemat 3D |
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Hmin |
Hmax |
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Hmin |
Hmax |
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1.4.5. Obciążenie wiatrem - schemat 4A, 4B. |
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Parcie wiatru: |
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p0 = |
0,58 |
kPa |
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Ssanie wiatru: |
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s0 = |
-0,33 |
kPa |
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p =10,8 * 0,58 = |
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6,29 |
kN/m. |
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s =10,88*(-0,33) = |
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-3,59 |
kN/m. |
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Schemat 4A |
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Schemat 4B |
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p |
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s
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2.0. OBLICZENIE ZBROJENIA W SŁUPIE. |
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2.1. Dane dla betonu B20 i stali AIII . |
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Ea = |
210000 |
MPa |
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Eb = |
27000 |
MPa |
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Rb = |
11500 |
kN/m2 |
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Ra = Rac = |
350000 |
kN/m2 |
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0,60 |
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Sbgr = xgr * (1 - xgr / 2) = |
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0,42 |
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ld = |
6,50 |
m |
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l0d = 1.6 * ld = |
10,40 |
m |
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lg = |
3,80 |
m |
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l0g = 1.6 * lg = |
6,08 |
m |
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2.2. Zbrojenie w przekroju 4- 4. |
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2.2.1. Obliczenia dla Mmax. |
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Dane: |
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M = |
591,21861 |
kNm |
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b = |
0,4 |
m |
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N = |
2127,78 |
kN |
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h = |
0,7 |
m |
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a = a' = |
0,03 |
m |
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h0 = |
0,67 |
m |
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es = M / N = |
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0,28 |
m |
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en = |
0,02 |
m |
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e0 = es + en = |
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0,30 |
m |
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l0d / h = |
14,86 |
> 1.0 => słup wiotki h > 1.0 |
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Przyjęto: |
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1,1 |
i duży mimośród |
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e = h * e0 = |
0,33 |
m |
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ea = e + h / 2 - a = |
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0,65 |
m |
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Ma0 = N * ea = |
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1378,0 |
kNm |
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Fac1 = (Ma0 - Sbgr * b * h02 * Rb) / (Rac * (h0 - a') = |
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22,80 |
cm2 |
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Famin = 0.002 * b * h0 = |
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5,36 |
cm2 |
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Fac1 > Famin |
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przyjęto: |
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Fac1 = |
22,80 |
cm2 |
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Sb = (Ma0 - Fac1 * Rac * (h0 - a')) / (Rb * b * h02 ) = |
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0,420 |
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_____
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0,60 |
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Ponieważ: |
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0,60 |
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ł 2a / h0 = |
0,09 |
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Fa1 = (x *b *h0 *Rb + Fac1 *Rac - N) / Ra = |
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14,84 |
cm2 |
> Famin |
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przyjęto: |
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Fa1 = |
14,84 |
cm2 |
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Rzeczywista wartość współczynnika h: |
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Nkr = 6.4 * E / l0d2 * [Jb / kd * (0.11 / (0.1 - e0/h0) + 0.1) + n * Ja] |
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Jb = (b * h03 ) / 12 = |
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0,010 |
m4 |
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2,0 |
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kd = 1 + 0.5 * Nd / N * jp = |
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2,0 |
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n = Ea / Eb = |
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7,78 |
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Ja = (h / 2 - a)2 * Fa1 + (h /2 - a')2 * Fac1 = |
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2,34 |
m4 |
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Nkr = 6.4*E / l0d2*[Jb / kd*(0.11 / (0.1 + e0/h0)+0.1) + n*Ja] = |
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29018 |
kN |
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1,08 |
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Uzykane h jest zbliżone do h założonego, obliczenia można uznać |
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za poprawne. |
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2.2.2. Obliczenia dla Mmin. |
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Dane: |
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M = |
-814,24739 |
kNm |
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b = |
0,4 |
m |
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N = |
1838,98 |
kN |
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h = |
0,7 |
m |
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a = a' = |
0,03 |
m |
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h0 = |
0,67 |
m |
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es = M / N = |
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0,44 |
m |
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en = |
0,02 |
m |
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e0 = es + en = |
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0,46 |
m |
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l0d / h = |
14,86 |
> 1.0 => słup wiotki h > 1.0 |
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Przyjęto: |
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1,1 |
i duży mimośród |
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e = h * e0 = |
0,51 |
m |
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ea = e + h / 2 - a = |
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0,83 |
m |
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Ma0 = N * ea = |
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1524,6 |
kNm |
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Fac2 = (Ma0 - Sbgr * b * h02 * Rb) / (Rac * (h0 - a') = |
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29,35 |
cm2 |
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Fac2 > Famin |
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przyjęto: |
Fac2 = |
29,35 |
cm2 |
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Sb = (Ma0 - Fac2 * Rac * (h0 - a')) / (Rb * b * h02 ) = |
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0,42 |
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_____ |
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0,60 |
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Ponieważ: |
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0,60 |
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0,09 |
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Fa2 = (x *b *h0 *Rb + Fac2 *Rac - N) / Ra = |
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29,64 |
cm2 |
> Famin |
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Rzeczywista wartość współczynnika h: |
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Nkr = 6.4 * E / l02 * [Jb / kd * (0.11 / (0.1 - e0/h0) + 0.1) + n * Ja] |
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Jb = (b * h03 ) / 12 = |
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0,01 |
m4 |
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2,0 |
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kd = 1 + 0.5 * Nd / N * jp = |
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2,0 |
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n = Ea / Eb = |
|
7,78 |
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Ja = (h / 2 - a)2 * Fa2 + (h /2 - a')2 * Fac2 = |
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3,01 |
m4 |
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Nkr = 6.4*E / l0d2*[Jb / kd*(0.11 / (0.1 + e0/h0)+0.1) + n*Ja] = |
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|
37345 |
kN |
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1,05 |
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Uzykane h jest zbliżone do h założonego, obliczenia można uznać |
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za poprawne. |
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2.2.3. Obliczenia dla Nmax. |
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Dane: |
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M = |
-808,33139 |
kNm |
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b = |
0,4 |
m |
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N = |
2127,78 |
kN |
|
h = |
0,7 |
m |
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a = a' = |
0,03 |
m |
|
h0 = |
0,67 |
m |
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es = M / N = |
|
0,38 |
m |
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en = |
0,02 |
m |
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e0 = es + en = |
|
0,40 |
m |
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l0d / h = |
14,86 |
> 1.0 => słup wiotki h > 1.0 |
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Przyjęto: |
|
1,1 |
i duży mimośród |
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e = h * e0 = |
0,44 |
m |
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ea = e + h / 2 - a = |
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0,76 |
m |
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Ma0 = N * ea = |
|
1616,9 |
kNm |
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Fac3 = (Ma0 - Sbgr * b * h02 * Rb) / (Rac * (h0 - a') = |
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33,46 |
cm2 |
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Fac3 > Famin |
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przyjęto: |
Fac3 = |
33,46 |
cm2 |
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Sb = (Ma0 - Facmin * Rac * (h0 - a')) / (Rb * b * h02 ) = |
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0,42 |
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_____ |
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0,60 |
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Ponieważ: |
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0,60 |
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0,09 |
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Fa3 = (x *b *h0 *Rb + Fac3 *Rac - N) / Ra = |
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25,50 |
cm2 |
> Famin |
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Rzeczywista wartość współczynnika h: |
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Nkr = 6.4 * E / l0d2 * [Jb / kd * (0.11 / (0.1 - e0/h0) + 0.1) + n * Ja] |
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Jb = (b * h03 ) / 12 = |
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0,010 |
m4 |
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2,0 |
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kd = 1 + 0.5 * Nd / N * jp = |
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2,0 |
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n = Ea / Eb = |
|
7,78 |
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Ja = (h / 2 - a)2 * Fa3 + (h /2 - a')2 * Fac3 = |
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3,43 |
m4 |
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Nkr = 6.4*E / l02*[Jb / kd*(0.11 / (0.1 + e0/h0)+0.1) + n*Ja] = |
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|
42586 |
kN |
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1,05 |
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Uzykane h jest zbliżone do h założonego, obliczenia można uznać |
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za poprawne. |
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2.2.4. Dobranie zbrojenia. |
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Fa = max (Fa1, Fa2, Fa3) = |
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29,64 |
cm2 |
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Fac = max (Fac1, Fac2, Fac3) = |
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|
33,4638611160714 |
cm2 |
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Przyjęte zbrojenie: |
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- po stronie rozciąganej: |
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|
5 f |
28 |
|
Fa = |
30,79 |
cm2 |
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|
- po stronie ściskanej |
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5 f |
32 |
|
Fac = |
40,21 |
cm2 |
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pręty pośrednie: |
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2 f 12 |
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strzemiona: |
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45,0 |
cm |
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Na odcinku d1 = b = 70 cm licząc od podstawy słupa rozstaw strzemion |
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zagęszczono do 8 cm. |
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W miejscu łączenia prętów rozstaw strzemion zagęszczono . |
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Minimalna długość zakotwienia: la = 45 d = |
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135,0 |
cm |
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2.3. Zbrojenie w przekroju 3 - 3. |
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2.3.1. Obliczenia dla Mmax. |
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Dane: |
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M = |
180,935436 |
kNm |
|
b = |
0,4 |
m |
|
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N = |
2048,38 |
kN |
|
h = |
0,7 |
m |
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a = a' = |
0,03 |
m |
|
h0 = |
0,67 |
m |
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es = M / N = |
|
0,09 |
m |
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en = |
0,02 |
m |
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e0 = es + en = |
|
0,11 |
m |
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l0d / h = |
14,86 |
> 1.0 => słup wiotki h > 1.0 |
|
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|
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|
|
Przyjęto: |
|
1,4 |
i duży mimośród |
|
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|
e = h * e0 = |
0,15 |
m |
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ea = e + h / 2 - a = |
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|
0,47 |
m |
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Ma0 = N * ea = |
|
966,15 |
kNm |
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Fac1 = (Ma0 - Sbgr * b * h02 * Rb) / (Rac * (h0 - a') = |
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|
4,41 |
cm2 |
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Famin = 0.002 * b * h0 = |
|
|
5,36 |
cm2 |
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Fac1 < Famin |
|
przyjęto: |
Fac1 = Famin = |
|
5,36 |
cm2 |
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Sb = (Ma0 - Fac1 * Rac * (h0 - a')) / (Rb * b * h02 ) = |
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0,41 |
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|
_____
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0,58 |
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Ponieważ: |
|
|
0,58 |
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|
0,09 |
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Fa1 = (x *b *h0 *Rb + Fac1 *Rac - N) / Ra = |
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|
-2,52 |
cm2 |
< Famin |
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|
przyjęto: |
|
Fa1 = Famin = |
|
5,36 |
cm2 |
|
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|
Rzeczywista wartość współczynnika h: |
|
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Nkr = 6.4 * E / l0d2 * [Jb / kd * (0.11 / (0.1 - e0/h0) + 0.1) + n * Ja] |
|
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Jb = (b * h03 ) / 12 = |
|
|
0,010 |
m4 |
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|
2,0 |
|
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|
kd = 1 + 0.5 * Nd / N * jp = |
|
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|
2,0 |
|
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|
|
|
n = Ea / Eb = |
|
7,78 |
|
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|
Ja = (h / 2 - a)2 * Fa1 + (h /2 - a')2 * Fac1 = |
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|
0,55 |
m4 |
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Nkr = 6.4*E / l0d2*[Jb / kd*(0.11 / (0.1 + e0/h0)+0.1) + n*Ja] = |
|
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|
6825,1 |
kN |
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|
1,429 |
|
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|
|
Uzykane h jest zbliżone do h założonego, obliczenia można uznać |
|
|
|
|
|
|
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|
|
za poprawne. |
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|
2.3.2. Obliczenia dla Mmin. |
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Dane: |
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|
M = |
-360,432564 |
kNm |
|
b = |
0,4 |
m |
|
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|
|
N = |
1759,58 |
kN |
|
h = |
0,7 |
m |
|
|
|
|
a = a' = |
0,03 |
m |
|
h0 = |
0,67 |
m |
|
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|
es = M / N = |
|
0,20 |
m |
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|
en = |
0,02 |
m |
|
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e0 = es + en = |
|
0,22 |
m |
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|
|
|
|
|
|
l0d / h = |
14,86 |
> 1.0 => słup wiotki h > 1.0 |
|
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|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
Przyjęto: |
|
1,2 |
i duży mimośród |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
e = h * e0 = |
0,27 |
m |
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|
|
|
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|
|
ea = e + h / 2 - a = |
|
|
0,59 |
m |
|
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|
|
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|
|
|
|
|
|
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|
|
Ma0 = N * ea = |
|
1037,8 |
kNm |
|
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|
|
|
|
|
|
|
|
|
|
|
|
|
|
Fac2 = (Ma0 - Sbgr * b * h02 * Rb) / (Rac * (h0 - a') = |
|
|
|
|
|
7,61 |
cm2 |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
Fac2 > Famin |
|
|
|
przyjęto: |
Fac2 = |
7,61 |
cm2 |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
Sb = (Ma0 - Fac2 * Rac * (h0 - a')) / (Rb * b * h02 ) = |
|
|
|
|
|
0,42 |
|
|
|
|
|
_____
|
|
|
|
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|
0,60 |
|
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|
|
|
|
|
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|
|
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|
|
|
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|
|
|
Ponieważ: |
|
|
0,60 |
|
|
0,09 |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
Fa2 = (x *b *h0 *Rb + Fac2 *Rac - N) / Ra = |
|
|
|
|
10,17 |
cm2 |
> Famin |
|
|
|
|
|
|
|
|
|
|
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|
|
|
przyjęto: |
|
Fa2 = |
10,17 |
cm2 |
|
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|
|
|
|
|
|
Rzeczywista wartość współczynnika h: |
|
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|
|
Nkr = 6.4 * E / l0d2 * [Jb / kd * (0.11 / (0.1 - e0/h0) + 0.1) + n * Ja] |
|
|
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|
|
|
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|
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|
|
Jb = (b * h03 ) / 12 = |
|
|
0,010 |
m4 |
|
|
|
|
|
|
|
2,0 |
|
|
|
|
|
|
|
|
|
kd = 1 + 0.5 * Nd / N * jp = |
|
|
|
2,0 |
|
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|
|
|
|
n = Ea / Eb = |
|
7,78 |
|
|
|
|
|
|
|
|
Ja = (h / 2 - a)2 * Fa2 + (h /2 - a')2 * Fac2 = |
|
|
|
|
0,78 |
m4 |
|
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|
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|
|
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|
|
|
|
|
|
|
|
Nkr = 6.4*E / l0d2*[Jb / kd*(0.11 / (0.1 + e0/h0)+0.1) + n*Ja] = |
|
|
|
|
|
|
9691,6 |
kN |
|
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|
|
|
|
|
|
|
|
|
|
|
|
|
|
1,22 |
|
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|
|
|
|
Uzykane h jest zbliżone do h założonego, obliczenia można uznać |
|
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|
za poprawne. |
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|
2.3.3. Obliczenia dla Nmax. |
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Dane: |
|
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|
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|
|
|
|
M = |
-134,615564 |
kNm |
|
b = |
0,4 |
m |
|
|
|
|
N = |
2048,38 |
kN |
|
h = |
0,7 |
m |
|
|
|
|
a = a' = |
0,03 |
m |
|
h0 = |
0,67 |
m |
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es = M / N = |
|
0,07 |
m |
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en = |
0,02 |
m |
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e0 = es + en = |
|
0,09 |
m |
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l0d / h = |
14,86 |
> 1.0 => słup wiotki h > 1.0 |
|
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|
Przyjęto: |
|
1,4 |
i duży mimośród |
|
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e = h * e0 = |
0,12 |
m |
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ea = e + h / 2 - a = |
|
|
0,44 |
m |
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|
|
|
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|
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|
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Ma0 = N * ea = |
|
901,30 |
kNm |
|
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Fac3 = (Ma0 - Sbgr * b * h02 * Rb) / (Rac * (h0 - a') = |
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|
1,52 |
cm2 |
|
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|
|
Fac3 < Famin |
|
przyjęto: |
|
Fac3 = Famin = |
|
5,36 |
cm2 |
|
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|
Sb = (Ma0 - Facmin * Rac * (h0 - a')) / (Rb * b * h02 ) = |
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0,38 |
|
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|
_____
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0,51 |
|
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Ponieważ: |
|
|
0,51 |
|
|
0,09 |
|
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|
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|
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Fa3 = (x *b *h0 *Rb + Fac3 *Rac - N) / Ra = |
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|
|
-8,55 |
cm2 |
< Famin |
|
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|
przyjęto: |
|
Fa3= Famin = |
|
5,36 |
cm2 |
|
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|
|
Rzeczywista wartość współczynnika h: |
|
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|
Nkr = 6.4 * E / l0d2 * [Jb / kd * (0.11 / (0.1 - e0/h0) + 0.1) + n * Ja] |
|
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|
|
Jb = (b * h03 ) / 12 = |
|
|
0,010 |
m4 |
|
|
|
|
|
|
|
2,0 |
|
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|
|
kd = 1 + 0.5 * Nd / N * jp = |
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|
2,0 |
|
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|
|
n = Ea / Eb = |
|
7,78 |
|
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|
|
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|
Ja = (h / 2 - a)2 * Fa3 + (h /2 - a')2 * Fac3 = |
|
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|
|
0,55 |
m4 |
|
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|
|
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|
|
|
|
Nkr = 6.4*E / l0d2*[Jb / kd*(0.11 / (0.1 + e0/h0)+0.1) + n*Ja] = |
|
|
|
|
|
|
6823,8 |
kN |
|
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|
1,43 |
|
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|
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|
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|
|
|
|
|
|
|
|
|
|
Uzykane h jest zbliżone do h założonego, obliczenia można uznać |
|
|
|
|
|
|
|
|
|
|
za poprawne. |
|
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|
2.3.4. Dobranie zbrojenia. |
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Fa = max (Fa1, Fa2, Fa3) = |
|
|
|
10,17 |
cm2 |
|
|
|
|
|
Fac = max (Fac1, Fac2, Fac3) = |
|
|
|
7,61 |
cm2 |
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Przyjęte zbrojenie: |
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- po stronie rozciąganej: |
|
|
5 f |
20 |
|
Fa = |
15,71 |
cm2 |
|
|
- po stronie ściskanej |
|
|
5 f |
20 |
|
Fac = |
15,71 |
cm2 |
|
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|
pręty pośrednie: |
|
|
2 f 12 |
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strzemiona: |
|
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|
|
30,0 |
cm |
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|
|
W miejscu łączenia prętów rozstaw strzemion zagęszczono . |
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|
2.4. Zbrojenie w przekroju 2 - 2. |
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|
|
2.4.1. Obliczenia dla Mmax. |
|
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|
Dane: |
|
|
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|
|
|
|
|
|
M = |
215,5978 |
kNm |
|
b = |
0,4 |
m |
|
|
|
|
N = |
1150,13 |
kN |
|
h = |
0,4 |
m |
|
|
|
|
a = a' = |
0,03 |
m |
|
h0 = |
0,37 |
m |
|
|
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|
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|
|
|
|
|
es = M / N = |
|
0,19 |
m |
|
|
|
|
|
|
|
en = |
0,02 |
m |
|
|
|
|
|
|
|
|
e0 = es + en = |
|
0,21 |
m |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
l0g / h = |
15,20 |
> 1.0 => słup wiotki h > 1.0 |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
Przyjęto: |
|
1,10 |
i duży mimośród |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
e = h * e0 = |
0,23 |
m |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
ea = e + h / 2 - a = |
|
|
0,40 |
m |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
Ma0 = N * ea = |
|
457,98 |
kNm |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
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|
|
Famin = 0.002 * b * h0 = |
|
|
2,96 |
cm2 |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
Fac1 = (Ma0 - Sbgr * b * h02 * Rb) / (Rac * (h0 - a')) = |
|
|
|
|
|
16,11 |
cm2 |
> Famin |
|
|
|
|
|
|
|
|
|
|
|
|
|
Ponieważ: |
|
|
0,600 |
|
|
0,16 |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
Fa1 = (x *b *h0 *Rb + Fac1 *Rac - N) / Ra = |
|
|
|
|
12,43 |
cm2 |
> Famin |
|
|
|
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|
|
Rzeczywista wartość współczynnika h: |
|
|
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|
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|
|
|
|
|
|
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|
|
|
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|
|
|
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|
|
|
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|
|
|
Nkr = 6.4 * E / l0g2 * [Jb / kd * (0.11 / (0.1 - e0/h0) + 0.1) + n * Ja] |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
Jb = (b * h03 ) / 12 = |
|
|
0,0017 |
m4 |
|
|
|
|
|
|
|
2,0 |
|
|
|
|
|
|
|
|
|
kd = 1 + 0.5 * Nd / N * jp = |
|
|
|
2,0 |
|
|
|
|
|
|
n = Ea / Eb = |
|
7,78 |
|
|
|
|
|
|
|
|
Ja = (h / 2 - a)2 * Fa1 + (h /2 - a')2 * Fac1 = |
|
|
|
|
0,47 |
m4 |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
Nkr = 6.4*E / l0g2*[Jb / kd*(0.11 / (0.1 + e0/h0)+0.1) + n*Ja] = |
|
|
|
|
|
|
16929 |
kN |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
1,07 |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
Uzykane h jest zbliżone do h założonego, obliczenia można uznać |
|
|
|
|
|
|
|
|
|
|
za poprawne. |
|
|
|
|
|
|
|
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|
|
|
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|
|
|
2.4.2. Obliczenia dla Mmin. |
|
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|
|
|
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|
|
|
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|
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|
|
|
|
|
|
|
|
|
|
Dane: |
|
|
|
|
|
|
|
|
|
|
|
M = |
-159,9992 |
kNm |
|
b = |
0,4 |
m |
|
|
|
|
N = |
1150,13 |
kN |
|
h = |
0,4 |
m |
|
|
|
|
a = a' = |
0,03 |
m |
|
h0 = |
0,37 |
m |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
es = M / N = |
|
0,14 |
m |
|
|
|
|
|
|
|
en = |
0,02 |
m |
|
|
|
|
|
|
|
|
e0 = es + en = |
|
0,16 |
m |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
l0g / h = |
15,20 |
> 1.0 => słup wiotki h > 1.0 |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
Przyjęto: |
|
1,1 |
i duży mimośród |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
e = h * e0 = |
0,18 |
m |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
ea = e + h / 2 - a = |
|
|
0,35 |
m |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
Ma0 = N * ea = |
|
396,82 |
kNm |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
Fac2 = (Ma0 - Sbgr * b * h02 * Rb) / (Rac * (h0 - a') = |
|
|
|
|
|
11,12 |
cm2 |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
Fac2 > Famin |
|
|
|
przyjęto: |
Fac2 = |
11,12 |
cm2 |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
Sb = (Ma0 - Fac2 * Rac * (h0 - a')) / (Rb * b * h02 ) = |
|
|
|
|
|
0,420 |
|
|
|
|
|
_____ |
|
|
|
|
|
|
|
|
|
|
|
|
0,60 |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
Ponieważ: |
|
|
0,60 |
|
|
0,16 |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
Fa2 = (x *b *h0 *Rb + Fac2 *Rac - N) / Ra = |
|
|
|
|
7,44 |
cm2 |
> Famin |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
przyjęto: |
|
Fa2 = |
7,44 |
cm2 |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
Rzeczywista wartość współczynnika h: |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
Nkr = 6.4 * E / l0g2 * [Jb / kd * (0.11 / (0.1 - e0/h0) + 0.1) + n * Ja] |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
Jb = (b * h03 ) / 12 = |
|
|
0,0017 |
m4 |
|
|
|
|
|
|
|
2,0 |
|
|
|
|
|
|
|
|
|
kd = 1 + 0.5 * Nd / N * jp = |
|
|
|
2,0 |
|
|
|
|
|
|
n = Ea / Eb = |
|
7,78 |
|
|
|
|
|
|
|
|
Ja = (h / 2 - a)2 * Fa2 + (h /2 - a')2 * Fac2 = |
|
|
|
|
0,32 |
m4 |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
Nkr = 6.4*E / l0g2*[Jb / kd*(0.11 / (0.1 + e0/h0)+0.1) + n*Ja] = |
|
|
|
|
|
|
11687 |
kN |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
1,109 |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
Uzykane h jest zbliżone do h założonego, obliczenia można uznać |
|
|
|
|
|
|
|
|
|
|
za poprawne. |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
2.4.3. Obliczenia dla Nmax. |
|
|
|
|
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|
Dane: |
|
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|
|
|
|
|
|
M = |
-139,0102 |
kNm |
|
b = |
0,4 |
m |
|
|
|
|
N = |
1150,13 |
kN |
|
h = |
0,4 |
m |
|
|
|
|
a = a' = |
0,03 |
m |
|
h0 = |
0,37 |
m |
|
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|
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|
es = M / N = |
|
0,12 |
m |
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|
|
en = |
0,02 |
m |
|
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|
e0 = es + en = |
|
0,14 |
m |
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|
l0g / h = |
15,20 |
> 1.0 => słup wiotki h > 1.0 |
|
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|
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|
|
Przyjęto: |
|
1,2 |
i duży mimośród |
|
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|
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|
|
e = h * e0 = |
0,16 |
m |
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|
|
ea = e + h / 2 - a = |
|
|
0,33 |
m |
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|
Ma0 = N * ea = |
|
381,84 |
kNm |
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|
|
Fac3 = (Ma0 - Sbgr * b * h02 * Rb) / (Rac * (h0 - a') = |
|
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|
|
9,86 |
cm2 |
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|
|
Fac3 > Famin |
|
|
|
przyjęto: |
Fac3 = |
9,86 |
cm2 |
|
|
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|
|
|
Sb = (Ma0 - Facmin * Rac * (h0 - a')) / (Rb * b * h02 ) = |
|
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|
0,42 |
|
|
|
|
|
_____
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|
0,60 |
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|
Ponieważ: |
|
|
0,60 |
|
|
0,16 |
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|
|
Fa3 = (x *b *h0 *Rb + Fac3 *Rac - N) / Ra = |
|
|
|
|
6,18 |
cm2 |
> Famin |
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|
przyjęto: |
|
Fa3 = |
6,18 |
cm2 |
|
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|
Rzeczywista wartość współczynnika h: |
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|
Nkr = 6.4 * E / l0g2 * [Jb / kd * (0.11 / (0.1 - e0/h0) + 0.1) + n * Ja] |
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|
Jb = (b * h03 ) / 12 = |
|
|
0,0017 |
m4 |
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|
2,0 |
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|
kd = 1 + 0.5 * Nd / N * jp = |
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|
2,0 |
|
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|
|
n = Ea / Eb = |
|
7,78 |
|
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|
Ja = (h / 2 - a)2 * Fa3 + (h /2 - a')2 * Fac3 = |
|
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|
0,29 |
m4 |
|
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|
Nkr = 6.4*E / l0g2*[Jb / kd*(0.11 / (0.1 + e0/h0)+0.1) + n*Ja] = |
|
|
|
|
|
|
10363 |
kN |
|
|
|
|
|
1,12 |
|
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|
Uzykane h jest zbliżone do h założonego, obliczenia można uznać |
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|
za poprawne. |
|
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|
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|
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|
|
|
2.4.4. Dobranie zbrojenia. |
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|
Fa = max (Fa1, Fa2, Fa3) = |
|
|
|
12,43 |
cm2 |
|
|
|
|
|
Fac = max (Fac1, Fac2, Fac3) = |
|
|
|
16,11 |
cm2 |
|
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|
Przyjęte zbrojenie: |
|
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|
|
|
|
|
|
- po stronie rozciąganej: |
|
|
5 f |
20 |
|
Fa = |
15,71 |
cm2 |
|
|
- po stronie ściskanej |
|
|
5 f |
22 |
|
Fac = |
19,01 |
cm2 |
|
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|
pręty pośrednie: |
|
|
2 f 12 |
|
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|
|
|
strzemiona: |
|
|
|
|
33,0 |
cm |
|
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|
|
W miejscu łączenia prętów rozstaw strzemion zagęszczono . |
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|
2.5. Zbrojenie w przekroju 1 - 1. |
|
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|
|
2.5.1. Obliczenia dla Mmin (dla Mmax i Nmax są takie same). |
|
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|
Dane: |
|
|
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|
|
|
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|
|
|
|
M = |
-44,1 |
kNm |
|
b = |
0,4 |
m |
|
|
|
|
N = |
1088,98 |
kN |
|
h = |
0,4 |
m |
|
|
|
|
a = a' = |
0,03 |
m |
|
h0 = |
0,37 |
m |
|
|
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|
|
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|
|
es = M / N = |
|
0,04 |
m |
|
|
|
|
|
|
|
en = |
0,02 |
m |
|
|
|
|
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|
|
|
e0 = es + en = |
|
0,06 |
m |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
l0g / h = |
15,20 |
> 1.0 => słup wiotki h > 1.0 |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
Przyjęto: |
|
1,55 |
i duży mimośród |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
e = h * e0 = |
0,09 |
m |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
ea = e + h / 2 - a = |
|
|
0,26 |
m |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
Ma0 = N * ea = |
|
287,24 |
kNm |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
Fac = (Ma0 - Sbgr * b * h02 * Rb) / (Rac * (h0 - a') = |
|
|
|
|
|
1,91 |
cm2 |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
Fac < Famin |
|
przyjęto: |
|
Fac = Famin = |
|
2,96 |
cm2 |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
Sb = (Ma0 - Fac * Rac * (h0 - a')) / (Rb * b * h02 ) = |
|
|
|
|
|
0,40 |
|
|
|
|
|
_____ |
|
|
|
|
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|
|
|
|
|
|
|
0,55 |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
Ponieważ: |
|
|
0,55 |
|
|
0,16 |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
Fa = (x *b *h0 *Rb + Fac *Rac - N) / Ra = |
|
|
|
|
-1,25 |
cm2 |
< Famin |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
przyjęto: |
|
Fa = Famin = |
|
2,96 |
cm2 |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
Rzeczywista wartość współczynnika h: |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
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|
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|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
Nkr = 6.4 * E / l0g2 * [Jb / kd * (0.11 / (0.1 - e0/h0) + 0.1) + n * Ja] |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
Jb = (b * h03 ) / 12 = |
|
|
0,0017 |
m4 |
|
|
|
|
|
|
|
2,0 |
|
|
|
|
|
|
|
|
|
kd = 1 + 0.5 * Nd / N * jp = |
|
|
|
2,0 |
|
|
|
|
|
|
n = Ea / Eb = |
|
7,78 |
|
|
|
|
|
|
|
|
Ja = (h / 2 - a)2 * Fa2 + (h /2 - a')2 * Fac2 = |
|
|
|
|
0,09 |
m4 |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
Nkr = 6.4*E / l0g2*[Jb / kd*(0.11 / (0.1 + e0/h0)+0.1) + n*Ja] = |
|
|
|
|
|
|
3112,5 |
kN |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
1,54 |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
Uzykane h jest zbliżone do h założonego, obliczenia można uznać |
|
|
|
|
|
|
|
|
|
|
za poprawne. |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
2.5.2. Dobranie zbrojenia. |
|
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|
|
|
|
|
|
|
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|
|
|
|
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|
|
Fa = |
2,96 |
cm2 |
|
|
|
|
|
|
|
|
Fac = |
2,96 |
cm2 |
|
|
|
|
|
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|
|
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|
|
|
|
|
|
|
|
|
|
Przyjęte zbrojenie: |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
- po stronie rozciąganej: |
|
|
3 f |
12 |
|
Fa = |
3,39 |
cm2 |
|
|
- po stronie ściskanej |
|
|
3 f |
12 |
|
Fac = |
3,39 |
cm2 |
|
|
|
|
|
|
|
|
|
|
|
|
|
pręty pośrednie: |
|
|
2 f 12 |
|
|
|
|
|
|
|
strzemiona: |
|
|
|
|
18,0 |
cm |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
W miejscu łączenia prętów rozstaw strzemion zagęszczono . |
|
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|
3.0. OBLICZENIE ZBROJENIA WE WSPORNIKU POD |
|
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|
|
BELKĘ PODSUWNICOWĄ. |
|
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|
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|
|
|
Dane: |
h = |
1,0 |
m |
|
|
|
|
|
|
|
h1 = |
0,40 |
m |
|
|
|
R |
|
|
h2 = |
0,70 |
m |
|
|
|
|
|
b = |
0,40 |
m |
|
|
|
|
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|
|
|
|
|
|
|
|
|
|
|
a = |
0,05 |
m |
|
|
|
|
|
|
h0 = h - a = |
0,95 |
m |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
x = |
1,4 |
m |
|
|
|
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|
|
y = |
0,4 |
m |
|
|
|
|
|
|
ap = |
0,68 |
m |
|
|
|
|
|
|
|
|
|
|
|
Rbz = |
900,0 |
kN/m2 |
- dla betonu B20 |
|
|
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|
|
|
|
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|
|
|
|
|
|
|
|
|
R = |
271,17 |
kN |
( poz. 1.4.3.) |
|
|
|
|
|
|
|
H = |
13,10 |
kN |
( poz. 1.4.4.) |
|
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|
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|
|
|
|
|
|
|
|
Warunki: |
|
ap Ł h0 = |
0,95 |
m |
|
|
|
|
|
|
|
|
R Ł 2 * Rbz * b * h01 |
|
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|
|
h02 = [(h - y)(ap - h2 + h1)] / (x - h2) + y = |
|
|
|
|
0,73 |
m |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
przyjęto: |
|
h01 = h02 = |
0,73 |
m |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
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|
|
|
2 * Rbz * b * h01 = |
522,514285714286 |
kN > |
R = |
271,17 |
kN |
|
|
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|
|
|
|
Warunki są spełnione. |
|
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|
|
3.1. Obliczenie zbrojenia poziomego i ukośnego. |
|
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|
|
z = 0.85 * h0 = |
|
0,81 |
m |
|
|
|
|
|
|
|
z2 = 0.85 * h02 = |
|
0,62 |
m |
|
|
|
|
|
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|
|
Przyjęto stal AII o średnicy f 16, adop = 0.3 mm => Raw = |
|
|
|
|
|
|
310000 |
kN/m2 |
|
|
|
|
|
|
|
|
|
|
|
|
|
Fa = 1 / Raw * (0.8 * R * ap / z + H) = |
|
|
|
|
7,32 |
cm2 |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
F0 = 0.3 * R / Raw * ap / z * 1 + (z2 / ap)2 = |
|
|
|
3,48 |
cm2 |
|
|
|
|
|
|
|
|
|
|
|
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|
|
|
Fmin = 0.002 * b * h0 = |
|
|
7,60 |
cm2 |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
Fa + F0 = |
10,80 |
cm2 > |
Fmin = |
7,60 |
cm2 |
|
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|
|
3.2. Obliczenie strzemion. |
|
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|
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|
|
|
|
|
|
|
|
Fs ł (0.7 * R + H) / Raw - Fa = |
|
|
|
0,77 |
cm2 |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
Fs ł 0.25 * F0 = |
|
0,87 |
cm2 |
|
|
|
|
|
|
|
|
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|
|
|
|
|
|
|
|
|
przyjęto: |
Fs = |
0,87 |
cm2 |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
3.3. Dobranie zbrojenia. |
|
|
|
|
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przekrój: |
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dobrane zbrojenie: |
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Fa = |
7,32 |
cm2 |
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4 f |
16 |
Fa = |
8,04 |
cm2 |
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F0 = |
3,48 |
cm2 |
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2 f |
16 |
F0 = |
4,02 |
cm2 |
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Fs = |
0,87 |
cm2 |
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4 f |
6 |
Fs = |
1,13 |
cm2 |
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rozstaw strzemion: |
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s Ł 12 * dm = |
19,20 |
cm |
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s Ł |
15,00 |
cm |
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s Ł 0.25 * h = |
25,00 |
cm |
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Przyjęto rozstaw strzemion co 15.0 cm. |
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4.0. STOPA FUNDAMENTOWA (KIELICHOWA). |
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Dane: |
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L = |
3,80 |
m |
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W = |
1,0 |
m |
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B = |
3,40 |
m |
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h0 = W - 0.05 = |
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0,95 |
m |
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a1 = |
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0,70 |
m |
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b1 = |
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0,40 |
m |
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a2 = a1+2*0.08 = |
|
0,86 |
m |
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b2 = b1+2*0.08 = |
|
0,56 |
m |
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a3 = a2+2*0.10 = |
|
1,06 |
m |
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b3 = b2+2*0,40 = |
|
1,36 |
m |
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4.1. Obciążenia. |
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Ciężar słupa: |
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Gs = |
69,27 |
kN |
( poz. 1.1.5.) |
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Ciężar ściany: |
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Nśc = |
21,49 |
kN |
(poz. 1.1.4.) |
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Obciążenia z konstrukcji: |
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M = |
591,21861 |
kNm |
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N = |
2127,78 |
kN |
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T = |
94,34067 |
kN |
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Zamiana ciężaru ściany na siłę działającą osiowo: |
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M' = Nśc * (a2 + 0.08) / 2 = |
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10,10 |
kNm |
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N' = Nśc = |
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21,49 |
kN |
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Obliczenie głębokości kielicha: |
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e = M / N = |
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0,31 |
m < |
a1 / 2 = |
0,35 |
m |
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la = |
1,35 |
m |
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( poz. 2.2.4.) |
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hk = 1.5 * a1 = |
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1,05 |
m |
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hk = la = |
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1,35 |
m |
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przyjęto: hk = |
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1,35 |
m |
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wysokość stopy: h = hk + W + 0.05 = |
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2,4 |
m |
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Ciężar stopy: |
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Gst = [L * B * W + a3 * b3 * (h - W)] * 24.0 * 1.1 = |
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394,37 |
kN |
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Ciężar gruntu na stopie: |
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Gg = (L * B - a3 * b3) * (h - W) * 18.0 * 1.3= |
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376,03 |
kN |
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Nacisk całkowity na podłoże gruntowe: |
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G = N + N' + Gs + Gst + Gg = |
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2988,9 |
kN |
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M = M - M' + T * h = |
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807,53 |
kNm |
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G1 = Gst + Gg = |
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770,40 |
kN |
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4.2. Sprawdzenie naprężeń pod stopą. |
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Naprężenia dopuszczalne gruntu: |
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qgr = |
240 |
kN/m2 |
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qgr * 1.2 = |
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288,0 |
kN/m2 |
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e = M / G = |
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0,27 |
m < |
L / 6 = |
0,63 |
m |
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qmin = G / (B * L) * (1 - 6 * e / L) = |
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132,654471810331 |
kN/m2 |
< qgr * 1.2 |
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qmax = G / (B * L) * (1 + 6 * e / L) = |
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330,031212400196 |
kN/m2 |
< qgr * 1.2 |
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-warunki są spełnione |
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4.3. Obliczenie zbrojenia poziomego stopy na zginanie. |
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Fmin = 0.0009 * 1.0 * h0 = |
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8,55 |
cm2 |
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4.3.1. Kierunek równoległy do L (ABCD). |
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q'r min = qmin - G1 / (L * B) = |
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73,03 |
kN/m2 |
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q'r max = qmax - G1 / (L * B) = |
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270,402580821248 |
kN/m2 |
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C = (L - a2) / 2 = |
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1,47 |
m |
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q'r śr = q' r max - [(q'r max - q'r min) * C] / (2 * L) = |
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232,225763891366 |
kN/m2 |
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M = [q'r śr * C2 * (2 * B + b1)] / 6 = |
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602,179983831424 |
kNm |
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Sb = M / (1.0 * h02 * Rb) = |
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0,06 |
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0,97 |
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FaL = M / (z * h * Ra) = |
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7,39 |
cm2 |
< Fmin = |
8,55 |
cm2 |
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4.3.2. Kierunek równoległy do B (DCEF). |
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q'r śr = (q'r max + q'r min) / 2 = |
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171,714210526316 |
kN/m2 |
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M = q'r śr * [(B - b1)2 * (2 * L + a1) / 24] = |
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534,460480263158 |
kNm |
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Sb = M / (1.0 * h02 * Rb) = |
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0,05 |
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0,97 |
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FaB = M / (z * h * Ra) = |
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6,54 |
cm2 |
< Fmin = |
8,55 |
cm2 |
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4.3.3. Przyjęcie zbrojenia (na 1 mb). |
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FaL = |
8,55 |
cm2 |
12 f |
10 |
FarzL = |
9,42 |
cm2 |
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FaB = |
8,55 |
cm2 |
12 f |
10 |
FarzB = |
9,42 |
cm2 |
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4.3.4. Sprawdzenie stopy na przebicie. |
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q'r max * F Ł Rbz * bśr * h0 |
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bśr = (b1 + b3 + 0.2) / 2 = |
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0,65 |
m |
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x = (L - a1 - 2 * W) / 2 = |
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0,55 |
m |
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F = B * x = |
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1,87 |
m2 |
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q'r max * F = |
505,652826135734 |
kN < |
Rbz * bśr * h0 = |
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558,6 |
kN |
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- warunek jest spełniony |
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4.4. Obliczenie zbrojenia kielicha stopy. |
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4.4.1. Zbrojenie pionowe. |
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M = |
591,21861 |
kNm |
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N = |
2127,78 |
kN |
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e = M / N = |
0,31 |
m > 0.3 * a3 = |
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0,303 |
m |
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d = a3 - e = |
0,75 |
m |
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a = |
0,04 |
m |
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ba = a3 - 2*a = |
0,98 |
m |
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Fa1 = [N * (e - a3 + d + 0.5 * ba)] / (Ra * ba) = |
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30,40 |
cm2 |
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4.4.2. Zbrojenie poziome. |
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Q = |
49,5 |
kN |
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e = M / N = |
0,31 |
m > b1 / 2 = |
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0,20 |
m |
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Mn = M + Q * hk - 0.7 * N * e = |
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184,51 |
kNm |
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Mk = 0.8 * (M + Q * hk - 0.5 * N * b1) = |
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185,990088 |
kNm > |
Mn |
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h0 = a3 - a = |
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1,02 |
m |
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Sb = Mn / (1.0 * h02 * Rb) = |
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0,015 |
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__ |
___ |
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0,99 |
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Fa2 = Mn / (z * a3 * Ra) = |
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5,01 |
cm2 |
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4.4.3. Przyjęcie zbrojenia. |
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Fa1 = |
30,40 |
cm2 |
12 f |
18 |
FarzL = |
30,54 |
cm2 |
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Fa2 = |
5,01 |
cm2 |
4 f |
14 |
FarzB = |
6,16 |
cm2 |
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