PŁYTA STROPOWA |
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1.1 |
Zestawienie obciążeń [kN/m2] |
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Obciążenia stałe |
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Obciążenie charakterys. |
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Obciążenie obl. |
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Lastriko |
(0,02*22) |
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0,44 |
1,3 |
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0,572 |
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Gładź cementowa |
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(0,03*21) |
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0,63 |
1,3 |
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0,819 |
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Izolacja styropianem (0,03*0,45) |
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0,0135 |
1,3 |
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0,0175 |
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Ciężar własny płyty (0,08*25) |
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2 |
1,1 |
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2,2 |
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Tynk cem-wap. (0,015*19) |
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0,285 |
1,3 |
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0,3705 |
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Razem= |
gk= |
3,369 |
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go= |
3,979 |
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Obciążenie zmienne |
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pk |
8 |
1,2 |
p0= |
9,6 |
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Razem= |
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11,369 |
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13,579 |
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Obciążenie zastępcze [kN/m2] |
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gp=g0+0,5p0=3,979+0,5*9,6 |
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gp= |
8,779 |
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pp=0,5p0=0,5*9,6 |
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pp= |
4,8 |
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pkd=0,8*pk=0,8*8 |
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pkd= |
6,4 |
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pkk=0,2*pk=0,2*8 |
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pkk= |
1,6 |
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Schemat belki pięcioprzesłowej |
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1
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1,6 |
1,82 |
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1,82 |
1,82 |
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1,6 |
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Obwiednie momentów [kNm] |
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Obwiednie sił tnących [kN] |
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Mmax= a*gp*l2+b*pp*l2 |
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Qmax,min=a*gp*l+b*pp*l |
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Mmin= a*gp*l2+b*pp*l2 |
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M1max= |
0,0781*8,779*1,62+0,1*4,8*1,62= |
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2,984038144 |
kNm |
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M1min= |
0,0781*8,779*1,62-0,0263*4,8*1,62= |
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1,432063744 |
kNm |
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M2max= |
0,0331*8,779*1,822+0,0787*4,8*1,822= |
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2,21382564676 |
kNm |
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M2min= |
0,0331*8,779*1,822-0,0461*4,8*1,822= |
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0,22956555076 |
kNm |
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M3max= |
0,0462*8,779*1,822+0,0855*4,8*1,822= |
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2,70288461352 |
kNm |
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M3min= |
0,0462*8,779*1,822-0,0395*4,8*1,822= |
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0,71544461352 |
kNm |
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MBmax= |
(-0,105)*8,779*1,712-0,013*4,8*1,712= |
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-4,3656666795 |
kNm |
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MBmin= |
(-0,105)*8,779*1,712+0,119*4,8*1,712= |
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-2,5129569195 |
kNm |
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MCmax= |
(-0,079)*8,779*1,822+0,018*4,8*1,822= |
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-3,66722860107604 |
kNm |
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MCmin= |
(-0,079)*8,779*1,822-0,111*4,8*1,822= |
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-2,0110938484 |
kNm |
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QAP= |
0,395*8,779*1,6+0,447*4,8*1,6= |
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8,981288 |
kN |
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QBL= |
(-0,606)*8,779*1,6-0,62*4,8*1,6= |
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-13,2737184 |
kN |
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QBP= |
0,526*8,779*1,82+0,598*4,8*1,82= |
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13,62844028 |
kN |
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QCL= |
(-0,474)*8,887*1,82-0,576*4,8*1,82= |
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-12,60540372 |
kN |
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QCP= |
0,5*8,887*1,82+0,591*4,8*1,82= |
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13,151866 |
kN |
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Dobór materiałów (kN/cm2) |
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Klasa betonu - B25 |
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fcd= |
1,33 |
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trd= |
0,026 |
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Stal klasy A-III |
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fyd= |
35 |
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xeff,lim=0,53 |
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Wymiarowanie płyty na zginanie (cm) |
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h= |
8 |
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a1= |
2 |
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b= |
100 |
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bwż= |
20 |
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h1= |
11,3333333333333 |
d |
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h |
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d= |
5,5 |
a
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d1= |
8,5 |
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b |
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H= |
40 |
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Mmax |
Mmax' |
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h |
d |
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d1 h1 |
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H |
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bwż |
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ZGINANIE-obliczenia |
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Podpora B |
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1) |
MBmax= |
432,9 |
(kNcm) |
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SM=0 |
a*fcd*b*x*(d1-0,5x)-Msd=0 |
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SM=0 |
0,85*1,33*100*x*(8,55-0,5x)-Msd=0 |
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x1=16,64>d1 |
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x2=0,46 |
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0,46 |
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eeff=xeff/d=0,054<eeff.lim=0,53 |
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Przekrój pojedyńczo zbrojony |
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As=Msd/((d-0,5xeff)*fyd) |
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As=432,9/((8,55-0,5*0,368)*35) |
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As1= |
1,49559509414407 |
(cm2) |
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Przyjmuję 5 f 6 o As1= |
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1,4 |
(cm2) |
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2) |
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MBkr'= |
1,971 |
(kNm) = |
197,1 |
(kNcm) |
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SM=0 |
a*fcd*b*x*(d-0,5x)-Msd=0 |
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SM=0 |
0,85*1,33*100*x*(5,5-0,5x)-Msd=0 |
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x1=10,67>d |
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x2=0,327 |
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eeff=xeff/d=0,059<eeff.lim=0,53 |
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Przekrój pojedyńczo zbrojony |
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As=Msd/((d-0,5xeff)*fyd) |
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As=197,1/((5,5-0,5*0,327)*35) |
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As2= |
1,0238961038961 |
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Przyjmuję 4f 6 o As1= |
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1,13 |
(cm2) |
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Podpora C |
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1) |
MCmax= |
392,1 |
(kNcm) |
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SM=0 |
a*fcd*b*x*(d1-0,5x)-Msd=0 |
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SM=0 |
0,85*1,33*100*x*(8,55-0,5x)-Msd=0 |
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x1=16,68>d1 |
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x2=0,416 |
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eeff=xeff/d=0,048<eeff.lim=0,53 |
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Przekrój pojedyńczo zbrojony |
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As=Msd/((d-0,5xeff)*fyd) |
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As=392,1/((8,55-0,5*0,416)*35) |
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As1= |
1,31798319327731 |
(cm2) |
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Przyjmuję 5 f 6 o As1= |
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1,4 |
(cm2) |
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2) |
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MCkr'= |
1,698 |
(kNm) = |
169,8 |
(kNcm) |
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SM=0 |
a*fcd*b*x*(d-0,5x)-Msd=0 |
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SM=0 |
0,85*1,33*100*x*(5,5-0,5x)-Msd=0 |
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x1=10,72>d |
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x2=0,28 |
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0,28 |
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eeff=xeff/d=0,05<eeff.lim=0,53 |
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Przekrój pojedynczo zbrojony |
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As=Msd/((d-0,5xeff)*fyd) |
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As=169,8/((5,5-0,5*0,28)*35) |
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As2= |
0,905117270788912 |
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Przyjmuję 5 f 6 o As1= |
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1,4 |
(cm2) |
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Przęsło 1 |
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M1max= |
298,4 |
(kNcm) |
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SM=0 |
a*fcd*b*x*(d1-0,5x)-Msd=0 |
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SM=0 |
1*1,33*100*x*(6-0,5x)-Msd=0 |
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x1=11,61>d1 |
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x2=0,38 |
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xeff=0,8*x=0,8*0,38= |
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0,304 |
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eeff=xeff/d=0,051<eeff.lim=0,53 |
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Przekrój pojedyńczo zbrojony |
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As=Msd/((d-0,5xeff)*fyd) |
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As=298,4/((6-0,5*0,304)*35) |
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As2= |
1,59418741318517 |
(cm2) |
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Przyjmuję 6 f 6 o As1= |
|
1,68 |
(cm2) |
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Przęsło 2 |
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6) |
M2max= |
221,38 |
(kNcm) |
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SM=0 |
a*fcd*b*x*(d1-0,5x)-Msd=0 |
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SM=0 |
1*1,33*100*x*(6-0,5x)-Msd=0 |
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x1=11,98>d1 |
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x2=0,3 |
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xeff=0,8*x=0,8*0,38= |
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0,24 |
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eeff=xeff/d=0,04<eeff.lim=0,53 |
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Przekrój pojedyńczo zbrojony |
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As=Msd/((d-0,5xeff)*fyd) |
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As=221,38/((6-0,5*0,24)*35) |
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As2= |
1,1756771109931 |
(cm2) |
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Przyjmuję 4 f 6 o As1= |
|
1,12 |
(cm2) |
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Przęsło 3 |
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7) |
M3max= |
270,29 |
(kNcm) |
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SM=0 |
a*fcd*b*x*(d1-0,5x)-Msd=0 |
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SM=0 |
1*1,33*100*x*(6-0,5x)-Msd=0 |
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x1=11,92>d1 |
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x2=0,36 |
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xeff=0,8*x=0,8*0,36= |
|
0,285 |
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eeff=xeff/d=0,0475<eeff.lim=0,53 |
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Przekrój pojedyńczo zbrojony |
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As=Msd/((d-0,5xeff)*fyd) |
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As=270,29/((6-0,5*0,285)*35) |
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As2= |
1,441450569962 |
(cm2) |
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Przyjmuję 5 f 6 o As1= |
|
1,4 |
(cm2) |
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Sprawdzenie warunku na ścinanie dla Vsdmax |
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Dla Vsd= |
13,628 |
kN |
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VRd1=2,2*tRd*bw*d>=Vsd |
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VRd1=2,2*0,026*100*6=34,32 |
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VRd1>=Vsd |
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Nie należy zbroić na ścinanie |
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Sprawdzenie SGU |
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a) |
Sprawdzenie ugięć dla przęsła 1 |
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rL=As1/b*d |
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rL=1,68/100*6 |
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rL= |
0,305454545454545 |
[%] |
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ssobl=Msd/As*z= |
|
298,4/(1,68*(6-0,5*0,304)) |
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ssobl= |
33,212237774691 |
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leff/d= |
160/6 = |
29,0909090909091 |
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(leff/d)dop= |
28 |
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(250/ssobl)*(leff/d)dop>=leff/d |
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|
210,765683646113 |
>= |
26,67 |
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Warunek spełniony |
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b) |
Sprawdzenie ugięć dla przęsła 2 |
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rL=As1/b*d |
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rL=1,12/100*6 |
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rL= |
0,203636363636364 |
[%] |
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ssobl=Msd/As*z= |
|
221,38/(1,12*(6-0,5*0,24)) |
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ssobl= |
36,7399097185343 |
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leff/d= |
160/6 = |
33,0909090909091 |
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(leff/d)dop= |
28 |
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(250/ssobl)*(leff/d)dop>=leff/d |
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|
190,52850302647 |
>= |
30,33 |
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Warunek spełniony |
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c) |
Sprawdzenie ugięć dla przęsła 3 |
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rL=As1/b*d |
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rL=(1,4/100*6)*100% |
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rL= |
0,254545454545454 |
[%] |
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ssobl=Msd/As*z= |
|
270,29/(1,4*(6-0,5*0,36)) |
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ssobl= |
36,0362642490501 |
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leff/d= |
160/6 = |
33,0909090909091 |
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(leff/d)dop= |
35 |
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(250/ssobl)*(leff/d)dop>=leff/d |
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|
242,810962299752 |
>= |
30,33 |
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Warunek spełniony |
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Sprawdzenie stanu granicznego zarysowania |
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wlim=0,3 |
(mm) |
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ss1=Msd/As*d*z |
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ss1=298,4/(1,68*6*0,9) |
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ss1= |
35,8826358826359 |
(kN/cm2) = |
328,92 |
(kN/m2) |
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rL=As1/b*d |
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rL=1,68/100*6 |
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rL= |
0,305454545454545 |
[%] |
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fmax=f 8 |
>= |
frzecz=f 6 |
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Warunek spełniony |
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2,1 |
Zestawienie obciążeń (kN/mb) |
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Obciążenia stałe |
|
|
Obciążenie charakterys. |
|
|
Obciążenie obl. |
|
|
Lastriko |
(0,02*22*1,82) |
|
|
0,801 |
1,3 |
|
1,041 |
|
Gładź cementowa |
|
(0,03*21*1,82) |
|
1,15 |
1,3 |
|
1,495 |
|
Izolacja styropianem (0,03*0,45*1,82) |
|
|
|
0,025 |
1,3 |
|
0,033 |
|
Ciężar własny płyty (0,08*25*1,82) |
|
|
|
3,64 |
1,1 |
|
4,004 |
|
Tynk cem-wap. (0,015*19*1,82) |
|
|
|
0,519 |
1,3 |
|
0,675 |
|
Ciężar własny żebra (0,2*0,32*25) |
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1,6 |
1,1 |
|
1,76 |
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Razem= |
gk= |
7,735 |
|
go= |
9,008 |
|
Obciążenie użytkowe (8*1,82)= |
|
|
pk= |
14,56 |
1,2 |
p0= |
17,47 |
|
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Razem= |
|
22,295 |
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|
26,478 |
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Obciążenie zastępcze (kN/mb) |
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gp=g0+0,25*p0=9,008+0,25*17,47 |
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gp= |
13,3755 |
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pp=0,75*p0=0,75*17,47 |
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pp= |
13,1025 |
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2,2 |
Schemat statyczny i rozpiętości obliczeniowe |
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Schemat belki pięcioprzesłowej |
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1
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5,4 |
5,4 |
|
5,4 |
5,4 |
|
5,4 |
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|
Dobór materiałów (kN/cm2) |
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|
Klasa betonu - B25 |
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fcd= |
1,33 |
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trd= |
0,026 |
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|
Stal klasy A-III |
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fyd= |
35 |
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xeff,lim= 0,53 |
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2,4 |
Wymiarowanie żebra na zginanie (cm) |
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Dla przęsła nr 1 |
|
|
Dla przęsła nr 2,3 |
|
|
Nad podporą |
|
|
beff= |
111,8 |
(cm) |
beff= |
95,6 |
(cm) |
h1= |
45 |
(cm) |
beff1= |
45,9 |
(cm) |
beff1= |
37,8 |
(cm) |
d1= |
42 |
(cm) |
beff2= |
45,9 |
(cm) |
beff2= |
37,8 |
(cm) |
a= |
3 |
(cm) |
hf= |
8 |
(cm) |
hf= |
8 |
(cm) |
bw(p)= |
30 |
(cm) |
bw= |
20 |
(cm) |
bw= |
20 |
(cm) |
|
|
|
H= |
40 |
(cm) |
H= |
40 |
(cm) |
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d= |
37 |
(cm) |
d= |
37 |
(cm) |
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beff |
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MBmax |
MBmax' |
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hf |
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h1 |
d1 |
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H |
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beff1 |
beff2 |
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bw |
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ZGINANIE-obliczenia |
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Przęsło nr 1 |
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Przyjmuję beff=111,8cm |
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Sprawdzam czy przekrój pozornie czy rzeczywiście teowy |
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Mht=a*beff*ht*(d-0,5ht) |
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Mht=1*111,8*8*(37-0,5*8) |
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Mht= |
29515,2 |
kNcm >= |
Msd= |
4423 |
kNcm |
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Przekrój pozornie teowy |
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Obliczenia przeprowadzam jak dla belki beff*H=111,8*40cm |
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Mmax= |
4423 |
(kNcm) |
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SM=0 |
a*fcd*b*x*(d-0,5x)-Msd=0 |
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SM=0 |
1*1,33*111,8*x*(37-0,5x)-Msd=0 |
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x1=73,18>d1 |
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x2=0,813 |
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xeff=0,8*x=0,8*0,813= |
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0,65 |
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eeff=xeff/d=0,017<eeff.lim=0,53 |
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Przekrój pojedyńczo zbrojony |
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As=Msd/((d-0,5xeff)*fyd) |
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As=4423/((37-0,5*0,65)*35) |
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As1= |
3,41544401544402 |
(cm2) |
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Przyjmuję 8 f 8 o As1= |
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4 |
(cm2) |
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Przęsło nr 2 |
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Przyjmuję beff=95,6cm |
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Sprawdzam czy przekrój pozornie czy rzeczywiście teowy |
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Mht=a*beff*ht*(d-0,5ht) |
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Mht=1*95,6*8*(37-0,5*8) |
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Mht= |
25238,4 |
kNcm >= |
Msd= |
2387 |
kNcm |
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Przekrój pozornie teowy |
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Obliczenia przeprowadzam jak dla belki beff*H=95,6*40cm |
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Mmax= |
2387 |
(kNcm) |
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SM=0 |
a*fcd*b*x*(d-0,5x)-Msd=0 |
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SM=0 |
1*1,33*95,6*x*(37-0,5x)-Msd=0 |
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x1=73,49>d1 |
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x2=0,511 |
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xeff=0,8*x=0,8*0,511= |
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0,408 |
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eeff=xeff/d=0,011<eeff.lim=0,53 |
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Przekrój pojedyńczo zbrojony |
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As=Msd/((d-0,5xeff)*fyd) |
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As=2387/((37-0,5*0,408)*35) |
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As1= |
1,85346233286227 |
(cm2) |
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Przyjmuję 4 f 8 o As1= |
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2 |
(cm2) |
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Przęsło nr 3 |
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Przyjmuję beff=95,6cm |
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Sprawdzam czy przekrój pozornie czy rzeczywiście teowy |
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Mht=a*beff*ht*(d-0,5ht) |
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Mht=1*95,6*8*(37-0,5*8) |
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Mht= |
25238,4 |
kNcm >= |
Msd= |
2995 |
kNcm |
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Przekrój pozornie teowy |
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Obliczenia przeprowadzam jak dla belki beff*H=95,6*40cm |
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Mmax= |
2995 |
(kNcm) |
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SM=0 |
a*fcd*b*x*(d-0,5x)-Msd=0 |
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SM=0 |
1*1,33*95,6*x*(37-0,5x)-Msd=0 |
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x1=73,36>d1 |
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x2=0,64 |
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xeff=0,8*x=0,8*0,64= |
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0,51 |
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eeff=xeff/d=0,014<eeff.lim=0,53 |
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Przekrój pojedyńczo zbrojony |
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As=Msd/((d-0,5xeff)*fyd) |
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As=2995/((37-0,5*0,51)*35) |
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As1= |
2,32879108916665 |
(cm2) |
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Przyjmuję 5 f 8 o As1= |
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2,5 |
(cm2) |
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Podpora B |
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1) |
MBmax= |
5781 |
(kNcm) |
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SM=0 |
a*fcd*b*x*(d-0,5x)-Msd=0 |
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SM=0 |
1*1,33*20*x*(42-0,5x)-Msd=0 |
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x1=78,46>d1 |
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x2=5,54 |
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xeff=0,8*x=0,8*5,54= |
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4,3 |
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eeff=xeff/d=0,1<eeff.lim=0,53 |
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Przekrój pojedyńczo zbrojony |
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As=Msd/((d-0,5xeff)*fyd) |
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As=5781/((42-0,5*4,3)*35) |
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As1= |
4,1448288223696 |
(cm2) |
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Przyjmuję 9 f 8 o As1= |
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2) |
MBmax'= |
MBmax-Qpmin*b/2-(gp+pp)*b/2*b/4 |
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MBmax'= |
5781-36,15*30/2-(13,375+13,1025)*30/2*30/4 |
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MBmax'= |
2259,975 |
(kNcm) |
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SM=0 |
a*fcd*b*x*(d-0,5x)-Msd=0 |
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SM=0 |
1*1,33*20*x*(37-0,5x)-Msd=0 |
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x1=71,64>d |
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x2=2,36 |
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xeff=0,8*x=0,8*2,36= |
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1,89 |
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eeff=xeff/d=0,051<eeff.lim=0,53 |
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Przekrój pojedyńczo zbrojony |
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As=Msd/((d-0,5xeff)*fyd) |
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As=2259,975/((37-0,5*3,54)*35) |
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As2= |
1,79089486300691 |
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Przyjmuję 4 f 8 o As1= |
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2 |
(cm2) |
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Podpora C |
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1) |
MCmax= |
4634 |
(kNcm) |
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SM=0 |
a*fcd*b*x*(d-0,5x)-Msd=0 |
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SM=0 |
1*1,33*20*x*(42-0,5x)-Msd=0 |
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x1=79,6>d1 |
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x2=4,38 |
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xeff=0,8*x=0,8*4,38= |
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3,5 |
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eeff=xeff/d=0,095<eeff.lim=0,53 |
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Przekrój pojedyńczo zbrojony |
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As=Msd/((d-0,5xeff)*fyd) |
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As=4634/((42-0,5*3,5)*35) |
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As1= |
3,28944099378882 |
(cm2) |
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Przyjmuję 8 f 8 o As1= |
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4 |
(cm2) |
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2) |
MCmax'= |
MCmax-QCmin*b/2-(gp+pp)*b/2*b/4 |
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MCmax'= |
4634-31,55*30/2-(13,375+13,1025)*30/2*30/4 |
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MCmax'= |
1112,975 |
(kNcm) |
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SM=0 |
a*fcd*b*x*(d-0,5x)-Msd=0 |
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SM=0 |
1*1,33*20*x*(37-0,5x)-Msd=0 |
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x1=72,85>d |
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x2=1,15 |
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xeff=0,8*x=0,8*1,15= |
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0,92 |
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eeff=xeff/d=0,025<eeff.lim=0,53 |
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Przekrój pojedyńczo zbrojony |
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As=Msd/((d-0,5xeff)*fyd) |
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As=1112,975/((37-0,5*1,15)*35) |
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As2= |
0,870259598092111 |
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Przyjmuję 2 f 8 o As1= |
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1,01 |
(cm2) |
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Sprawdzenie warunku na ścinanie dla Vsdmax |
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Podpora A |
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Dla Vsd= |
40,1 |
kN |
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Sprawdzenie nośności na rozciąganie przekroju betonowego |
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VRd1=[k*trd*(1,2+40*rl)+0,15*scp]*bw*d |
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k=1,6-d = |
1,23 |
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rl=Asl/bw*d=4/20*37 |
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rl= |
0,005405405405405 |
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scp=0 |
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VRd1=[1,23*0,026*(1,2+40*0,0054)]*20*37 |
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VRd1= |
33,51504 |
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a= |
0,5*t |
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t= |
0,45 |
m |
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fck= |
20 |
Mpa |
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Vsdkr=Vsd-q*a=40,1-(13,3755+13,1025)*0,45/2 |
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Vsdkr= |
36,1283 |
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Vsdkr |
>= |
VRd1 |
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Konieczne jest zbrojenie na ścinanie |
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Sprawdzenie nośności ściskanych krzyżulców betonowych |
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VRd2=n*fcd*b*z*cotq/(1+cotq) |
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z=0,9*d=0,9*37 |
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z= |
33,3 |
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n=0,7-fck/200=0,7-20/200 |
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n= |
0,6 |
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VRd2=0,6*1,33*20*33,3*0,5 |
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VRd2= |
265,734 |
>= |
Vsdkr |
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Nośność krzyżulców betonowych jest zapewniona |
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Vrd1=Vsdkr-q*c |
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c=(Vsdkr-Vrd1)/q=(36,13-33,51504)/(13,3755+13,1025) |
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c= |
0,098695520809729 |
m = |
9,8695520809729 |
cm < 3*d |
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Przyjmuję strzemiona 2 cięte f 8 |
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Asw1= |
1 |
cm2 |
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Stal klasy A-l |
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fywd= |
21 |
kN/cm2 |
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Vsd2=Vsdkr-q*c=36,13-(13,3755+13,0125)*0,01*9,869 |
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Vsd2= |
33,51504 |
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s1<=(Asw1*z*fywd*cotq)/Vsd2=(1*33,3*21*1)/33,51504 |
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s= |
20,865259298512 |
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Przyjmuję rozstaw strzemion: |
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Na odcinku c co 20 cm |
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Podpora BL |
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Dla Vsd= |
59,78 |
kN |
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Sprawdzenie nośności na rozciąganie przekroju betonowego |
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VRd1=[k*trd*(1,2+40*rl)+0,15*scp]*bw*d |
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k=1,6-d = |
1,23 |
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rl=Asl/bw*d=4,5/20*37 |
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rl= |
0,006081081081081 |
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scp=0 |
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VRd1=[1,23*0,026*(1,2+40*0,0054)]*20*37 |
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VRd1= |
34,15464 |
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a= |
0,5*t |
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t= |
0,3 |
m |
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fck= |
20 |
Mpa |
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Vsdkr=Vsd-q*a=59,78-(13,3755+13,1025)*0,3/2 |
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Vsdkr= |
55,8083 |
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Vsdkr |
>= |
VRd1 |
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Konieczne jest zbrojenie na ścinanie |
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Sprawdzenie nośności ściskanych krzyżulców betonowych |
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VRd2=n*fcd*b*z*cotq/(1+cotq) |
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z=0,9*d=0,9*37 |
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z= |
33,3 |
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n=0,7-fck/200=0,7-20/200 |
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n= |
0,6 |
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VRd2=0,6*1,33*20*33,3*0,5 |
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VRd2= |
265,734 |
>= |
Vsdkr |
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Nośność krzyżulców betonowych jest zapewniona |
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Vrd1=Vsdkr-q*c |
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c=(Vsdkr-Vrd1)/q=(55,8083-33,51504)/(13,3755+13,1025) |
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c= |
0,817798172067377 |
m = |
81,7798172067377 |
cm < 3*d |
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Przyjmuję strzemiona 2 cięte f 8 |
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Asw1= |
1 |
cm2 |
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Stal klasy A-l |
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fywd= |
21 |
kN/cm2 |
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Vsd2=Vsdkr-q*c=55,8083-(13,3755+13,0125)*0,01*81,77982 |
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Vsd2= |
34,15464 |
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s<=(Asw1*z*fywd*cotq)/Vsd2=(1*33,3*21*1)/34,15464 |
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s= |
20,4745241056559 |
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Przyjmuję rozstaw strzemion: |
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Na odcinku c co20 cm |
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Podpora BP |
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Dla Vsd= |
53,51 |
kN |
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Sprawdzenie nośności na rozciąganie przekroju betonowego |
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VRd1=[k*trd*(1,2+40*rl)+0,15*scp]*bw*d |
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k=1,6-d = |
1,23 |
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rl=Asl/bw*d=4,5/20*37 |
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rl= |
0,006081081081081 |
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scp=0 |
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VRd1=[1,23*0,026*(1,2+40*0,0061)]*20*37 |
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VRd1= |
34,15464 |
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a= |
0,5*t |
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t= |
0,3 |
m |
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fck= |
20 |
Mpa |
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Vsdkr=Vsd-q*a=53,51-(13,3755+13,1025)*0,3/2 |
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Vsdkr= |
49,5383 |
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Vsdkr |
>= |
VRd1 |
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Konieczne jest zbrojenie na ścinanie |
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Sprawdzenie nośności ściskanych krzyżulców betonowych |
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VRd2=n*fcd*b*z*cotq/(1+cotq) |
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z=0,9*d=0,9*37 |
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z= |
33,3 |
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n=0,7-fck/200=0,7-20/200 |
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n= |
0,6 |
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VRd2=0,6*1,33*20*33,3*0,5 |
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VRd2= |
265,734 |
>= |
Vsdkr |
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Nośność krzyżulców betonowych jest zapewniona |
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Vrd1=Vsdkr-q*c |
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c=(Vsdkr-Vrd1)/q=(49,5383-34,154)/(13,3755+13,1025) |
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c= |
0,580997809502228 |
m = |
58,0997809502228 |
cm < 3*d |
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Przyjmuję strzemiona 2 cięte f 8 |
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Asw1= |
1 |
cm2 |
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Stal klasy A-l |
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fywd= |
21 |
kN/cm2 |
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Vsd2=Vsdkr-q*c=49,5383-(13,3755+13,0125)*0,01*58,099 |
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Vsd2= |
34,15464 |
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s<=(Asw1*z*fywd*cotq)/Vsd2=(1*33,3*21*1)/34,15464 |
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s= |
20,4745241056559 |
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Przyjmuję rozstaw strzemion: |
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Na odcinku c co20 cm |
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Podpora CL |
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Dla Vsd= |
49,16 |
kN |
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|
|
Sprawdzenie nośności na rozciąganie przekroju betonowego |
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VRd1=[k*trd*(1,2+40*rl)+0,15*scp]*bw*d |
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k=1,6-d = |
1,23 |
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rl=Asl/bw*d=4/20*37 |
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rl= |
0,005405405405405 |
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scp=0 |
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VRd1=[1,23*0,026*(1,2+40*0,0054)]*20*37 |
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VRd1= |
33,51504 |
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a= |
0,5*t |
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t= |
0,3 |
m |
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fck= |
20 |
Mpa |
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Vsdkr=Vsd-q*a=49,16-(13,3755+13,1025)*0,3/2 |
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Vsdkr= |
45,1883 |
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Vsdkr |
>= |
VRd1 |
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Konieczne jest zbrojenie na ścinanie |
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Sprawdzenie nośności ściskanych krzyżulców betonowych |
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VRd2=n*fcd*b*z*cotq/(1+cotq) |
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z=0,9*d=0,9*37 |
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z= |
33,3 |
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n=0,7-fck/200=0,7-20/200 |
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n= |
0,6 |
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VRd2=0,6*1,33*20*33,3*0,5 |
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VRd2= |
265,734 |
>= |
Vsdkr |
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|
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Nośność krzyżulców betonowych jest zapewniona |
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Vrd1=Vsdkr-q*c |
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c=(Vsdkr-Vrd1)/q=(45,1883-33,515)/(13,3755+13,1025) |
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c= |
0,440866379635924 |
m = |
44,0866379635924 |
cm < 3*d |
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Przyjmuję strzemiona 2 cięte f 8 |
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Asw1= |
1 |
cm2 |
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Stal klasy A-l |
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fywd= |
21 |
kN/cm2 |
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Vsd2=Vsdkr-q*c=45,1883-(13,3755+13,0125)*0,01*44,08664 |
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Vsd2= |
33,51504 |
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s1<=(Asw1*z*fywd*cotq)/Vsd2=(1*33,3*21*1)/33,515 |
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s= |
20,865259298512 |
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Przyjmuję rozstaw strzemion: |
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Na odcinku c co20 cm |
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Podpora CP |
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Dla Vsd= |
51,43 |
kN |
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|
|
Sprawdzenie nośności na rozciąganie przekroju betonowego |
|
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|
|
VRd1=[k*trd*(1,2+40*rl)+0,15*scp]*bw*d |
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k=1,6-d = |
1,23 |
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rl=Asl/bw*d=4/20*37 |
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rl= |
0,005405405405405 |
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scp=0 |
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|
VRd1=[1,23*0,026*(1,2+40*0,0054)]*20*37 |
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VRd1= |
33,51504 |
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a= |
0,5*t |
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t= |
0,3 |
m |
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fck= |
20 |
Mpa |
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Vsdkr=Vsd-q*a=51,43-(13,3755+13,1025)*0,3/2 |
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Vsdkr= |
47,4583 |
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Vsdkr |
>= |
VRd1 |
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|
|
Konieczne jest zbrojenie na ścinanie |
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|
|
Sprawdzenie nośności ściskanych krzyżulców betonowych |
|
|
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|
|
VRd2=n*fcd*b*z*cotq/(1+cotq) |
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|
z=0,9*d=0,9*37 |
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z= |
33,3 |
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n=0,7-fck/200=0,7-20/200 |
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n= |
0,6 |
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VRd2=0,6*1,33*20*33,3*0,5 |
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|
VRd2= |
265,734 |
>= |
Vsdkr |
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|
|
Nośność krzyżulców betonowych jest zapewniona |
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|
|
Vrd1=Vsdkr-q*c |
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|
c=(Vsdkr-Vrd1)/q=(47,4583-33,515)/(13,3755+13,1025) |
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c= |
0,526597930357278 |
m = |
52,6597930357278 |
cm < 3*d |
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|
|
Przyjmuję strzemiona 2 cięte f 8 |
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|
|
Asw1= |
1 |
cm2 |
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|
|
Stal klasy A-l |
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|
fywd= |
21 |
kN/cm2 |
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|
|
Vsd2=Vsdkr-q*c=47,4583-(13,3755+13,1025)*0,01*52,65979 |
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Vsd2= |
33,51504 |
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s<=(Asw1*z*fywd*cotq)/Vsd2=(1*33,3*21*1)/33,515 |
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s= |
20,865259298512 |
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|
|
Przyjmuję rozstaw strzemion: |
|
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|
Na odcinku c co20 cm |
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Sprawdzenie SGU |
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a) |
Sprawdzenie ugięć dla przęsła 1 |
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rL=As1/b*d |
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rl=(Asl/bw*d)*100%=(4/20*37)*100% |
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rl= |
0,540540540540541 |
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ssobl=Msd/As*z= |
|
4423/(4*(37-0,5*0,65)) |
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ssobl= |
30,1499659168371 |
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leff/d= |
540/37= |
14,5945945945946 |
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(leff/d)dop= |
28 |
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(250/ssobl)*(leff/d)dop>=leff/d |
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|
232,172733438842 |
>= |
26,67 |
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|
|
Warunek spełniony |
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|
b) |
Sprawdzenie ugięć dla przęsła 2 |
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rL=As1/b*d |
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rL=2/20*37 |
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|
rL= |
0,27027027027027 |
[%] |
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|
ssobl=Msd/As*z= |
|
2387/(2*(37-0,5*0,408)) |
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ssobl= |
32,4355908250897 |
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|
leff/d= |
540/37 = |
14,5945945945946 |
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(leff/d)dop= |
35 |
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|
|
(250/ssobl)*(leff/d)dop>=leff/d |
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|
|
269,765395894428 |
>= |
30,33 |
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|
|
Warunek spełniony |
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|
|
c) |
Sprawdzenie ugięć dla przęsła 3 |
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|
rL=As1/b*d |
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rL=2,5/37*20 |
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|
rL= |
0,337837837837838 |
[%] |
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|
|
ssobl=Msd/As*z= |
|
2995/(2,5*(37-0,5*0,51)) |
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|
ssobl= |
32,6030752483331 |
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|
|
leff/d= |
540/37 |
14,5945945945946 |
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|
(leff/d)dop= |
35 |
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|
|
(250/ssobl)*(leff/d)dop>=leff/d |
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|
|
268,379590984975 |
>= |
30,33 |
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|
|
Warunek spełniony |
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|
Sprawdzenie stanu granicznego zarysowania |
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|
wlim=0,3 |
(mm) |
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|
|
ss1=Msdmax/As*d*z |
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|
|
ss1=4423/(4*37*0,9) |
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|
|
ss1= |
33,2057057057057 |
(kN/cm2)= |
336,26 |
(Mpa) |
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|
|
rL=As1/b*d |
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|
rL=4/20*37 |
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|
rL= |
0,540540540540541 |
[%] |
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|
|
fmax=f 10 |
>= |
frzecz=f 8 |
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|
|
Warunek spełniony |
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|
3,1 |
Zestawienie obciążeń (kN/mb) |
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Obciążenia stałe |
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Obciążenie charakterys. |
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Obciążenie obl. |
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Lastriko |
(0,02*22*1,82*5,4) |
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4,32 |
1,3 |
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5,616 |
Gładź cementowa |
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(0,03*21*1,82*5,4) |
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6,21 |
1,3 |
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8,073 |
Izolacja styropianem (0,03*0,45*1,82*5,4) |
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0,135 |
1,3 |
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0,175 |
Ciężar własny płyty (0,08*25*1,82*5,4) |
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19,66 |
1,1 |
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21,63 |
Tynk cem-wap. (0,015*19*1,82*5,4) |
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2,8 |
1,3 |
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3,64 |
Ciężar własny podciągu (0,3*0,42*25*5,4) |
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17,01 |
1,1 |
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18,71 |
Ciężar własny żebra (0,2*0,32*25*5,4) |
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8,64 |
1,1 |
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10,01 |
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Razem = |
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gk= |
58,775 |
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go= |
67,854 |
Obciążenie użytkowe (8*1,82*5,4) = |
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pk= |
78,62 |
1,2 |
p0= |
94,35 |
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Razem= |
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137,395 |
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162,204 |
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3,2 |
Schemat statyczny i rozpiętości obliczeniowe |
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7,06 |
7,28 |
7,28 |
7,28 |
7,06 |
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Dobór materiałów (kN/cm2) |
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Klasa betonu - B25 |
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fcd= |
1,33 |
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trd= |
0,026 |
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Stal klasy A-III |
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fyd= |
35 |
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xeff,lim= |
0,53 |
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3,4 |
Wymiarowanie podciągu na zginanie (cm) |
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hf= |
8 |
(cm) |
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bw= |
30 |
(cm) |
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H= |
50 |
(cm) |
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d= |
47 |
(cm) |
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h1= |
56,6666666666667 |
(cm) |
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d1= |
53,6666666666667 |
(cm) |
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a= |
3 |
(cm) |
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bw(s)= |
40 |
(cm) |
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MBmax |
MBmax' |
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d |
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H |
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h1 |
d1 |
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bw |
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ZGINANIE-obliczenia |
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Przęsło nr 1 |
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Mmax= |
41132 |
(kNcm) |
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SM=0 |
a*fcd*b*x*(d-0,5x)-Msd=0 |
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SM=0 |
1*1,33*30*x*(47-0,5x)-Msd=0 |
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x1=59,1>d1 |
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x2=34,86 |
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xeff=0,8*x=0,8*34,86= |
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27,89 |
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zeff=xeff/d=0,59>zeff.lim=0,53 |
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Przekrój podwójnie zbrojony |
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Określenie nośności (M1) strefy ściskanej |
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xeff,lim=zeff,lim*d = 0,53*47= |
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24,91 |
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SMs => a*fcd*bw*xeff.lim*(d-0,5xeff.lim) = M1 |
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M1=1*1,33*30*24,91*(47-24,91) |
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M1= |
21955,44981 |
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Określenie przekroju zbrojenia rozciąganego(As11) |
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As11=M1/((d-0,5xeff,lim)*fyd) |
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As=21955,45/((47-0,5*24,91)*35) |
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As1= |
18,1588816326531 |
(cm2) |
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dM = Msd-M1 |
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dM=41132-21955,45= |
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19176,55019 |
kNcm |
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As12=dM/((d-a2)*fyd)=19176,55/((47-3)*35) |
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As12= |
12,4523053181818 |
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Zatem : |
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Zbrojenie w strefie rozciąganej: |
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As1 = As11+As12 = 18,1588 + 12,45 = |
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30,6111869508349 |
cm2 |
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Przyjmuję 10 f 20 o As1= |
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31,42 |
(cm2) |
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As2=As12= |
12,45 |
(cm2) |
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Przyjmuję 4 f 20 o As1= |
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12,57 |
(cm2) |
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Przęsło nr 2 |
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Mmax= |
30289 |
(kNcm) |
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SM=0 |
a*fcd*b*x*(d-0,5x)-Msd=0 |
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SM=0 |
1*1,33*30*x*(47-0,5x)-Msd=0 |
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x1=73,28>d1 |
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x2=20,7 |
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xeff=0,8*x=0,8*20,7= |
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16,57 |
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zeff=xeff/d=0,35<zeff.lim=0,53 |
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Przekrój pojedyńczo zbrojony |
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As=Msd/((d-0,5xeff)*fyd) |
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As=30289/((47-0,5*16,57)*35) |
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As1= |
22,3530931163632 |
(cm2) |
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Przyjmuję 8 f 20 o As1= |
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25,13 |
(cm2) |
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Przęsło nr 3 |
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Mmax= |
33928 |
(kNcm) |
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SM=0 |
a*fcd*b*x*(d-0,5x)-Msd=0 |
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SM=0 |
1*1,33*30*x*(47-0,5x)-Msd=0 |
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x1=69,54>d1 |
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x2=24,1 |
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xeff=0,8*x=0,8*15,24= |
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19,27 |
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zeff=xeff/d=0,41<zeff.lim=0,53 |
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Przekrój pojedyńczo zbrojony |
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As=Msd/((d-0,5xeff)*fyd) |
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As=33928/((47-0,5*19,27)*35) |
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As1= |
25,943300644224 |
(cm2) |
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Przyjmuję 9 f 20 o As1= |
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28,27 |
(cm2) |
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Podpora B |
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1) |
MBmax= |
32622 |
(kNcm) |
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SM=0 |
a*fcd*b*x*(d-0,5x)-Msd=0 |
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SM=0 |
1*1,33*30*x*(53,667-0,5x)-Msd=0 |
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x1=88,9>d1 |
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x2=18,38 |
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xeff=0,8*x=0,8*18,38= |
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14,7 |
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zeff=xeff/d=0,27<zeff.lim=0,53 |
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Przekrój pojedyńczo zbrojony |
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As=Msd/((d-0,5xeff)*fyd) |
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As=32622/((53,667-0,5*14,7)*35) |
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As1= |
23,5071158349847 |
(cm2) |
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Przyjmuję 8 f 20 o As1= |
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25,13 |
(cm2) |
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2) |
MBmax'= |
MBmax-Qpmin*b/2-(g0+p0)*b/2*b/4 |
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MBmax'= |
326,22-253,94*0,4/2-162,204*0,4/2*0,4/4 |
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MBmax'= |
272,18792 |
(kNm) = |
27218,792 |
(kNcm) |
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SM=0 |
a*fcd*b*x*(d-0,5x)-Msd=0 |
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SM=0 |
1*1,33*30*x*(47-0,5x)-Msd=0 |
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x1=76>d |
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x2=17,94 |
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xeff=0,8*x=0,8*17,94= |
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14,35 |
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zeff=xeff/d=0,3<zeff.lim=0,53 |
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Przekrój pojedyńczo zbrojony |
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As=Msd/((d-0,5xeff)*fyd) |
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As=27218,79/((47-0,5*14,35)*35) |
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As2= |
19,5274267778675 |
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Przyjmuję 7 f 20 o As1= |
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21,99 |
(cm2) |
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Podpora C |
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1) |
MCmax= |
29502 |
(kNcm) |
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SM=0 |
a*fcd*b*x*(d-0,5x)-Msd=0 |
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SM=0 |
1*1,33*30*x*(53,667-0,5x)-Msd=0 |
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x1=91,1>d1 |
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x2=16,23 |
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xeff=0,8*x=0,8*16,23= |
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12,98 |
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zeff=xeff/d=0,25<zeff.lim=0,53 |
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Przekrój pojedyńczo zbrojony |
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As=Msd/((d-0,5xeff)*fyd) |
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As=29502/((53,667-0,5*16,25)*35) |
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As1= |
17,867186159421 |
(cm2) |
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Przyjmuję 6 f 20 o As1= |
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18,85 |
(cm2) |
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2) |
MCmax'= |
MCmax-QCmin*b/2-(gp+pp)*b/2*b/4 |
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MCmax'= |
29502-232,68*0,4/2-162,204*0,4/2*0,4/4 |
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MCmax'= |
245,23992 |
(kNm) = |
24523,992 |
(kNcm) |
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SM=0 |
a*fcd*b*x*(d-0,5x)-Msd=0 |
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SM=0 |
1*1,33*30*x*(47-0,5x)-Msd=0 |
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x1=78,3>d |
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x2=15,2 |
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xeff=0,8*x=0,8*15,2= |
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12,16 |
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eeff=xeff/d=0,25<eeff.lim=0,53 |
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Przekrój pojedyńczo zbrojony |
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As=Msd/((d-0,5xeff)*fyd) |
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As=24523,99/((47-0,5*12,16)*35) |
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As2= |
17,1233012149141 |
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Przyjmuję 6 f 20 o As1= |
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18,85 |
(cm2) |
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Sprawdzenie warunku na ścinanie |
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Podpora A |
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Vsd= |
206,59 |
kN |
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Sprawdzenie nośności na rozciąganie przekroju betonowego |
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VRd1=[k*trd*(1,2+40*rl)+0,15*scp]*bw*d |
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k=1,6-d = |
1,13 |
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rl=Asl/bw*d=31,42/30*47 |
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rl= |
0,022283687943263 |
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scp=0 |
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VRd1=[1,13*0,026*(1,2+40*0,0222)]*30*47 |
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VRd1= |
86,635744 |
<= |
Vsd= |
206,59 |
kN |
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Konieczne jest zbrojenie na ścinanie |
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fck= |
20 |
Mpa |
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Sprawdzenie nośności ściskanych krzyżulców betonowych |
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VRd2=n*fcd*b*z*cotq/(1+cotq) |
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z=0,9*d=0,9*47 |
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z= |
42,3 |
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n=0,7-fck/200=0,7-20/200 |
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n= |
0,6 |
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VRd2=0,6*1,33*30*42,3*0,5 |
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VRd2= |
506,331 |
>= |
Vsd |
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Nośność krzyżulców betonowych jest zapewniona |
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Przyjmuję strzemiona 4 cięte f 12 |
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Asw1= |
4,52 |
cm2 |
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Stal klasy A-l |
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fywd= |
21 |
kN/cm2 |
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s<=(Asw1*z*fywd*cotq)/Vsd=(4,52*42,3*21*1)/206,59 |
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s= |
19,4351904738855 |
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Przyjmuję rozstaw strzemion co 19 cm |
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Podpora BL |
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Dla Vsd= |
280,44 |
kN |
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Sprawdzenie nośności na rozciąganie przekroju betonowego |
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VRd1=[k*trd*(1,2+40*rl)+0,15*scp]*bw*d |
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k=1,6-d = |
1,13 |
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rl=Asl/bw*d=25,13/30*47 |
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rl= |
0,017822695035461 |
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scp=0 |
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VRd1=[1,13*0,026*(1,2+40*0,0178)]*30*47 |
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VRd1= |
79,243736 |
<= |
Vsd= |
280,44 |
kN |
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Konieczne jest zbrojenie na ścinanie |
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fck= |
20 |
Mpa |
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Sprawdzenie nośności ściskanych krzyżulców betonowych |
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VRd2=n*fcd*b*z*cotq/(1+cotq) |
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z=0,9*d=0,9*47 |
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z= |
42,3 |
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n=0,7-fck/200=0,7-20/200 |
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n= |
0,6 |
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VRd2=0,6*1,33*30*42,3*0,5 |
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VRd2= |
506,331 |
>= |
Vsd |
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Nośność krzyżulców betonowych jest zapewniona |
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Przyjmuję strzemiona 4 cięte f 12 |
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Asw1= |
4,52 |
cm2 |
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Stal klasy A-l |
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fywd= |
21 |
kN/cm2 |
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s<=(Asw1*z*fywd*cotq)/Vsd=(4,52*42,3*21*1)/280,44 |
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s= |
14,3172015404365 |
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Przyjmuję rozstaw strzemion co 14 cm |
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Podpora Bp |
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Dla Vsd= |
253,94 |
kN |
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|
Sprawdzenie nośności na rozciąganie przekroju betonowego |
|
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VRd1=[k*trd*(1,2+40*rl)+0,15*scp]*bw*d |
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k=1,6-d = |
1,13 |
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rl=Asl/bw*d=25,13/30*47 |
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rl= |
0,017822695035461 |
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scp=0 |
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VRd1=[1,13*0,026*(1,2+40*0,0178)]*30*47 |
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VRd1= |
79,243736 |
<= |
Vsd= |
253,95 |
kN |
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|
Konieczne jest zbrojenie na ścinanie |
|
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fck= |
20 |
Mpa |
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|
|
Sprawdzenie nośności ściskanych krzyżulców betonowych |
|
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VRd2=n*fcd*b*z*cotq/(1+cotq) |
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z=0,9*d=0,9*47 |
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z= |
42,3 |
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n=0,7-fck/200=0,7-20/200 |
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n= |
0,6 |
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VRd2=0,6*1,33*30*42,3*0,5 |
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|
VRd2= |
506,331 |
>= |
Vsd |
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|
|
Nośność krzyżulców betonowych jest zapewniona |
|
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|
|
Przyjmuję strzemiona 4 cięte f 12 |
|
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|
|
Asw1= |
4,52 |
cm2 |
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|
|
Stal klasy A-l |
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|
fywd= |
21 |
kN/cm2 |
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|
|
s<=(Asw1*z*fywd*cotq)/Vsd=(4,52*42,3*21*1)/253,95 |
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s= |
15,8112782547058 |
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|
Przyjmuję rozstaw strzemion co 15 cm |
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Podpora CL |
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Dla Vsd= |
232,68 |
kN |
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|
|
Sprawdzenie nośności na rozciąganie przekroju betonowego |
|
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|
|
VRd1=[k*trd*(1,2+40*rl)+0,15*scp]*bw*d |
|
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|
k=1,6-d = |
1,13 |
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rl=Asl/bw*d=18,85/30*47 |
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rl= |
0,013368794326241 |
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scp=0 |
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|
VRd1=[1,13*0,026*(1,2+40*0,0133)]*30*47 |
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|
|
VRd1= |
71,86348 |
<= |
Vsd= |
232,68 |
kN |
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|
|
Konieczne jest zbrojenie na ścinanie |
|
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|
fck= |
20 |
Mpa |
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Sprawdzenie nośności ściskanych krzyżulców betonowych |
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VRd2=n*fcd*b*z*cotq/(1+cotq) |
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z=0,9*d=0,9*47 |
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z= |
42,3 |
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n=0,7-fck/200=0,7-20/200 |
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n= |
0,6 |
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VRd2=0,6*1,33*30*42,3*0,5 |
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VRd2= |
506,331 |
>= |
Vsd |
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Nośność krzyżulców betonowych jest zapewniona |
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Przyjmuję strzemiona 4 cięte f 12 |
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Asw1= |
4,52 |
cm2 |
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Stal klasy A-l |
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fywd= |
21 |
kN/cm2 |
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s<=(Asw1*z*fywd*cotq)/Vsd=(4,52*42,3*21*1)/232,68 |
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s1= |
17,2559566787004 |
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Przyjmuję rozstaw strzemion co 17 cm |
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Podpora Cp |
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Dla Vsd= |
251 |
kN |
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Sprawdzenie nośności na rozciąganie przekroju betonowego |
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VRd1=[k*trd*(1,2+40*rl)+0,15*scp]*bw*d |
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k=1,6-d = |
1,13 |
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rl=Asl/bw*d=18,85/30*47 |
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rl= |
0,013368794326241 |
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scp=0 |
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VRd1=[1,13*0,026*(1,2+40*0,0133)]*30*47 |
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VRd1= |
71,86348 |
<= |
Vsd= |
337,66 |
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Konieczne jest zbrojenie na ścinanie |
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fck= |
20 |
Mpa |
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Sprawdzenie nośności ściskanych krzyżulców betonowych |
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VRd2=n*fcd*b*z*cotq/(1+cotq) |
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z=0,9*d=0,9*47 |
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kN |
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z= |
42,3 |
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n=0,7-fck/200=0,7-20/200 |
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n= |
0,6 |
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VRd2=0,6*1,33*30*42,3*0,5 |
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VRd2= |
506,331 |
>= |
Vsd |
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Nośność krzyżulców betonowych jest zapewniona |
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Przyjmuję strzemiona 4 cięte f 12 |
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Asw1= |
4,52 |
cm2 |
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Stal klasy A-l |
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fywd= |
21 |
kN/cm2 |
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s<=(Asw1*z*fywd*cotq)/Vsd2=(4,52*42,3*21*1)/251 |
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s= |
15,9964780876494 |
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Przyjmuję rozstaw strzemion co 15 cm |
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Połączenie żebra z podciągiem |
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1) |
Podpora B |
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Vsd <= 2*fctd*b*d |
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280,44 <= 2*0,1*30*47=282 |
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2) |
Podpora C |
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Vsd <= 2*fctd*b*d |
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232,68 <= 2*0,1*30*47=282 |
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Sprawdzenie stanów granicznych użytkowania |
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Sprawdzenie ugięć dla Msdmax |
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Msdmax= |
34749 |
kNcm |
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a(oo,to)=ak*Msd*leff2/B(oo,to)<=alim |
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ak=5/48*(1-(MA+MB)/10Mo=5/48*(1-(0+278,04)/10*347,49)= |
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0,095831894443006 |
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Sprawdzenie czy przekrój jest zarysowany |
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Dla B25 Ecm = |
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2900 |
kNcm |
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Es= |
20000 |
kNcm |
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Mcr=fctm*Wc |
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fctm= |
0,22 |
kNcm2 |
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Wc=b*h2/6=30*502/6 |
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Wc= |
12500 |
cm3 |
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Mcr=0,22*12500 = |
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2750 |
kNcm |
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Mcr <= |
Msdmax |
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Przekrój jest zarysowany |
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tab 3 PN |
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2*Ac./u =2*300*500/(2*300+2*500) = |
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18,75 |
cm
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f(oo,to) = |
2,5 |
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Eceff = Ecm/1+f(oo,to) = 2,9000/1+2,5 = |
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8285,71428571429 |
Mpa = |
828,571428571429 |
kNcm |
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(ssr/ss)=(Mcr/Msd)=2750/34749= |
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0,079138968027857 |
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b1= |
1,0 |
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b2= |
0,5 |
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ae=Es/Eceff=20000/828,5714= |
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24,1379310344828 |
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II=bw*h3/12+As1*ae*(0,5*h*a)2 |
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II = 30*503/12+25,13*24,13793*(0,5*50-3)2 = |
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679572,275862069 |
cm4 |
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xeff = fyd*As1/(a*fcd*bw)=35*31,42/(1*1,33*30) |
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xeff = |
27,5614035087719 |
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III=bw*xeff3/12+bw*xeff*(0,5*h-0,5xeff)2+As1*ae*(0,5h-a)2 |
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III=30*27,563/12+30*27,56*(0,5*50-0,5*27,56)2+31,42*24,13793*(0,5*50-3)2 |
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III= |
523490,324114433 |
cm4 |
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B(oo,to)=Eceff*III/[1-b1*b2*(ssr/ss)*(1-III/II)] |
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B(oo,to)=828,5714*523490,3/[1-1*0,5*(2750/34749)*(1-27,596/679572)] |
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B(oo,to)= |
4377272745,48441 |
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a(oo.to)=ak*Msd*leff2/B(oo,fo)=0,09*34749*7062/4377272745= |
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0,379191137669516 |
cm |
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a(oo.to) < alim=3cm |
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Warunek spełniony |
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Sprawdzenie szerokości rozwarcia rys prostopadłych |
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wk=b*Srm*esm<wlim=0,3 |
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Srm=50+0,25*k1*k2*f/rr |
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k1= |
0,8 |
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k2= |
0,5 |
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hteff=2,5*(h-d)=2,5*(50-47) |
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hteff= |
7,5 |
cm |
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Acteff=bw*hteff= 30*7,5 |
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Acteff= |
225 |
cm2 |
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rr=As/Acteff =25,13/225 |
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rr= |
0,139644444444444 |
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Srm=50+0,25*0,8*0,5*20/0,111688 |
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Srm= |
64,3220878421388 |
mm = |
6,43220878421388 |
cm |
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esm-średnie odkształcenie zbrojenia |
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esm=ss/Es*[1-b1*b2*(ssr/ss)2] |
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b1= |
1 |
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b2= |
0,5 |
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fck= |
2 |
kN/cm2 |
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Msd=0,8*fck*x*bw*(d-0,8*x/2) |
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x=1,25*d-(1,5625*d2-3,125*Msd/(fck*bw))0,5 |
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x=1,25*47-(1,5625*472-3,125*32946/(2*30))0,5 |
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x= |
10,8298259069692 |
cm |
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z=d-x/2 =47-10,2084/2 = |
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41,5850870465154 |
cm |
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ss=Msd/z*As1 =32946/41,89578*25,13 |
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ss= |
26,5949094186789 |
kN/cm |
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(ssr/ss)=(Mcr/Msd)=2750/32946 = |
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0,079138968027857 |
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esm = 31,29247/20000*[1-1*0,5*0,083472] = |
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0,001325581388775 |
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wk=b*Srm*esm<wlim=0,3 |
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wk = 1,3*6,790688*0,001559 = |
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0,011084341128994 |
cm = |
0,110843411289935 |
mm |
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wk < wlim = 0,3 mm |
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Warunek spełniony |
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Sprawdzenie szerokości rozwarcia rys ukośnych do osi elementu |
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Vsdkr= |
174,72 |
kN |
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t =Vsd/bw*d=174,72/30*47 |
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t = |
0,123914893617021 |
kN/cm2 |
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rw1=Asw1/bw*s=31,42/30*19 |
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rw1= |
0,053888503677935 |
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rw2=Asw2/(bw*s2*sina) = |
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0 |
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rw=rw1 |
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b1= |
1 |
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l = 1/3*[rw1/b1*f1+rw2/b2*f2]=1/3*(0,054/1*0,2) |
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l = |
1,23712224531416 |
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wk =4t 2*l/(rw*Es*fck)=4*0,1242*1,237/(0,054*20000*2) |
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wk = |
0,00004 |
< 0,3 mm |
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Warunek spełniony |
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mm |
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Zestawienie obciążeń na słup |
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a) |
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Pokrycie papowe 0,2*5,4*7,28*1/cos5,71= |
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7,9 |
1,3 |
10,27 |
Wylewka 0,03*21*7,28*5,4*1/cos5,71= |
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24,89 |
1,3 |
32,357 |
Styropian 0,45*0,15*5,4*7,28*1/cos5,71= |
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2,67 |
1,3 |
3,471 |
Płyta żelbetowa 0,06*25*7,28*5,4*1/cos5,71= |
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59,26 |
1,1 |
65,186 |
Tynk cem-wap.0,015*19*7,28*5,4*1/cos5,71= |
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11,26 |
1,3 |
14,638 |
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Razem = |
105,98 |
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125,922 |
Ciężar żebra i podciągu |
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(4*0,15*0,35*5,4*25)+(0,25*0,45*7,28*25) = |
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48,825 |
1,1 |
53,7075 |
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Razem = |
154,805 |
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179,6295 |
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Długość żebra = |
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5,4 |
m |
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Długość podciągu = |
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7,28 |
m |
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Ilość kondygnacji = |
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5 |
m |
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bw(Sł) = |
0,4 |
m |
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Wysokość kondyg.= |
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4 |
m |
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b) Obciążenie śniegiem (strefa I): |
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Sk = qk*c= 0,7*0,8*5,4*7,28 = |
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22,01472 |
1,4 |
30,820608 |
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c) Obciążenie z pozostałych kondygnacji: (162,204*5-1) = |
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810,02 |
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d) Ciężar własny słupa: |
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0,4*0,4*(4,0-0,5)*5*25 = |
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70 |
1,1 |
77 |
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Obciążenie całkowite na poziomie posadzki parteru: |
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Nsd=a+b+c+d=179,629+30,82+810,02+77= |
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1097,470108 |
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Obciążenie długotrwałe: |
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Nsd,lt = Nsd-Sk-(1-yd)*pd*x = 1097,47-30,82-0,2*10*1,2*5,4*7,28 = |
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972,3007 |
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Długość obliczeniowa słupa: |
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lo = 0,7*lcol =0,7*4,0 = |
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2,8 |
m |
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l0/h = 2,8/0,4 = |
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7 |
m |
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Należy uwzględnić smukłość i obc. długotrałe |
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Określenie mimośrodu początkowego |
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eo = ea+ee |
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lcol/600 = 400/600 =
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0,666666666666667 |
mm |
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ea = |
h/30 = 40/30 = |
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1,33333333333333 |
cm |
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10mm |
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ea = |
1,33 |
cm |
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ee = |
0 |
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eo = ea = |
1,13 |
cm |
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Dobór materiałów kN/cm2 |
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Klasa betonu - B25 |
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fcd= |
1,33 |
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trd= |
0,026 |
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Stal klasy A-III |
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fyd= |
35 |
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xeff,lim= 0,53 |
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Ecm = |
2900 |
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Es = |
20000 |
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Określenie nośności słupa |
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As2 |
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As1 |
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fyd*As1*hs*es1+a*fcd*Acc,eff*(0,5*xeff-(es2-a2))-fyd*As2*es2 = 0
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xeff <= xeff.lim
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c= 2*(1-xeff)/(1-x,lim)-1 |
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xeff.lim < xeff <= 1 |
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-1 |
xeff > 1
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Ncrit = (6,4/lo2)*(Ecm*Ic/kit*(0,11/(0,1+eo/h)+0,1)+Es*Is) |
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eo/h >= |
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0,5-0,01*lo/h-0,01*fcd=0,5-0,01*2,8/40-0,01*1,33= |
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0,4167 |
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eo/h = 1,13/40 = |
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0,03325 |
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kit =1+0,5*Nsd,lt/Nsd*foo,to = 1+0,5*972,3007/1097,47*2,0 |
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kit= |
1,88594731912279 |
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Przyjmuję 4 f 16 o As = 2*4,02 = |
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8,04 |
cm2 |
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a1 = a2 = |
3 |
cm |
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d = |
37 |
cm |
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Ncrit = (6,4/2802)*(2900*404/1,88*(0,11/(0,1+0,4167)+0,1)+20000*2*4,02*172) |
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Ncrit = |
104339,219191649 |
kN |
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h = 1/(1-1097,47/104339,219) = |
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1,01063009991346 |
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etot = 1,01*1,33 = |
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1,34750679988462 |
cm |
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es1 = 1,347+0,5*40-3 = |
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18,3475067998846 |
cm |
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es2 = 0,5*40-3-1,347 = |
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15,6524932001154 |
cm |
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cs = 2*(1-xeff)/(1-0,53)-1 = 3,25-4,25*xeff |
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fyd*As1*cs*es1+a*fcd*Acc,eff*(0,5*xeff-(es2-a2))-fyd*As2*es2 = 0
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-35*4,02*(3,25-4,25*xeff)*18,347+0,85*1,33*40*372*xeff*(xeff/2-(15,652-3)/37)-35*4,02*15,652=0 |
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30953,09xeff2+10971,05xeff - 31760,43 = 0 |
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a= |
30953,09 |
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b= |
10971,05 |
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c= |
-31760,43 |
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b2-4ac=10971,052-4*30953,09*(-31760,43)=
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4052697731,0173 |
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x1 < 0 |
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x2=(-10971,05+63660,8022)/(2*30953,09) = |
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0,851122652940209 |
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xeff/d = |
0,85 |
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cs = |
-0,367271274995887 |
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NRd = a*fcd*b*xeff+As2*fyd - c*As1*fyd = 0,85*1,33*40*37*0,85+4,02*35+0,367*4,02*35 |
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Nrd = |
1616,4224239323 |
kN |
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