2. Żebro |
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2.1. Obciążenia stałe |
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hciężar własny żebra |
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współczynnik obciążenia |
ciężar objętościowy w stanie powietrznosuchym |
przekrój |
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fck |
fcd |
fctk |
fctd |
fctm |
Ecm |
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[Mpa] |
[Mpa] |
[Mpa] |
[Mpa] |
[Mpa] |
[Mpa] |
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16 |
10,6 |
1,3 |
0,87 |
1,9 |
27500 |
[-] |
[kN/m3] |
[m2] |
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fyk |
fyd |
ftk |
nominalna średnica prętów [mm] |
gfp |
gp |
bż*(hż-hpł) |
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[Mpa] |
[Mpa] |
[Mpa] |
1,1 |
25 |
0,4*(0,6-0,12) = |
0,192 |
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355 |
310 |
410 |
6 y 32 |
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Gż = bż*hż*gp*gfp |
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9,31 |
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Gżk = 0,192 m2*25 kN/m3 = |
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4,8 |
kN/m |
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15,21 |
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Gżd = 0,192 m2*25 kN/m3*1,1 = |
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5,28 |
kN/m |
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15,21 |
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hobciążenia stałe płyty działające na żebro |
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Gpł = Eo.st.*(l1+l2)/2 |
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Gpłk = 9,31kN/m2*(3,02 m+3,02 m)/2 = |
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28,12 |
kN/m |
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Gpłd = 10,8kN/m2*(3,02 m+3,02 m)/2 = |
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32,62 |
kN/m |
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Suma obciążeń stałych |
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Eo.st.k = Gżk+Gpłk |
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Eo.st.k = |
32,92 |
kN/m |
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Eo.st.d = Gżd+Gpłd |
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Eo.st.d = |
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kN/m |
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2.2. Obciążenia zmienne |
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hobciążenia zmienne płyty działające na żebro |
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Gpł = Eo.zm.*(l1+l2)/2 |
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Gpł = 2,72kN/m2*(3,02 m+3,02 m)/2= |
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8,21 |
kN/m |
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Gpł = 3,41kN/m2*(3,02 m+3,02 m)/2= |
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10,30 |
kN/m |
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Suma obciążeń zmiennych |
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Eo.zm.k = |
8,21 |
kN/m2 |
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Eo.zm.d = |
10,30 |
kN/m2 |
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2.3. Obliczanie żebra |
PRZĘSŁO |
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MAX |
MIN |
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0,077 |
0,077 |
0,393 |
0,393 |
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0,1 |
-0,023 |
0,446 |
-0,036 |
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-0,107 |
-0,107 |
-0,607 |
-0,607 |
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-0,121 |
-0,004 |
-0,62 |
0,013 |
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0,036 |
0,036 |
0,536 |
0,536 |
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0,081 |
-0,045 |
0,603 |
-0,067 |
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-0,071 |
-0,071 |
-0,464 |
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-0,107 |
0,018 |
-0,571 |
0,085 |
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0,464 |
0,464 |
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0,571 |
-0,085 |
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2.3.1. Wyznaczenie sił wewnętrznych |
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l0XY = 6,50m*1,025 = |
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(l0XY +l0YY)/2 = |
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6,83125 |
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l0YY = L = |
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m |
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Mmax = a*q*lo2 + B*p*lo2 |
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Mmin = a*q*lo2 + g*p*lo2 |
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Współczynniki a, B, g odczytano z Tablic Winklera |
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Podpora skrajna |
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Mmax = 0 |
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Mmin =0 |
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Przęsło skrajne |
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Mmax = 0,077*37,90*6,6632+0,1*10,30*6,6632 = |
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175,239 |
kNm |
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Mmin = 0,077*37,90*6,6632-0,023*10,30*6,6632 = |
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119,013 |
kNm |
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Podpora przedskrajna |
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Mmax = -0,107*37,90*6,83132-0,121*10,30*6,83132 = |
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-247,374204456086 |
kNm |
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Mmin = -0,107*37,90*6,83132-0,004*10,30*6,83132 = |
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-191,146863955281 |
kNm |
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Przęsło przedskrajne |
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Mmax = 0,036*37,90*7,02+0,081*10,30*7,02 = |
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107,722 |
kNm |
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Mmin = 0,036*37,90*7,02-0,045*10,30*72 = |
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44,141 |
kNm |
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Podpora środkowa |
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Mmax = -0,071*37,90*72-0,107*10,30*7,02 = |
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-185,834 |
kNm |
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Mmin = -0,071*37,60*7,02+0,018*10,30*7,02 = |
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-122,757 |
kNm |
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2.3.2. Wyznaczenie sił poprzecznych |
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Vmax = a*q*lo + B*p*lo |
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Vmin = a*q*lo + g*p*lo |
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7,00 |
6,663 |
Podpora skrajna |
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32,92 |
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8,21 |
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Vmax = 0,393*37,90*6,663+0,446*10,30*6,663 = |
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129,826 |
kN |
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Vmin = 0,393*37,90*6,663-0,036*10,30*6,663 = |
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96,755 |
kN |
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Podpora przedskrajna L |
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Vmax = -0,607*37,90*6,663-0,62*10,30*6,663 = |
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-195,796 |
kN |
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Vmin = -0,607*37,90*6,663+0,013*10,30*6,663 = |
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-152,365 |
kN |
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Podpora przedskrajna P |
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Vmax = 0,536*37,90*7,0+0,603*10,30*7,0 = |
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185,654 |
kN |
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Vmin = 0,536*37,90*7,0-0,067*10,30*7,0 = |
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137,356 |
kN |
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Podpora środkowa L |
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Vmax = -0,464*37,90*7-0,571*10,30*7 = |
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-164,248 |
kN |
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Vmin = -0,464*37,90*7+0,085*10,30*7 = |
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-116,959 |
kN |
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Podpora środkowa P |
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Vmax = 0,464*37,90*7+0,571*10,30*7 = |
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164,248 |
kN |
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Vmin = 0,464*37,90*7-0,085*10,30*7 = |
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116,959 |
kN |
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2.3.3. Wymiarowanie żebra - obliczenie potrzebnego zbrojenia |
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Warunki: |
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M' = 0,33*(Mprzęsłamin+Mpodporymin) = 0,33* |
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1) Płyta monolitycznie połączona z żebrem |
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2) t/h >=0,05 |
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0,12/0,60 = |
0,20 |
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3)Płyta po stronie ściskanej |
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c |
fi |
d |
h |
b |
alfa |
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Wszystkie powyższe warunki są spełnione |
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Założenia: |
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2 |
0,6 |
9,2 |
12 |
100 |
0,85 |
założono, że przekrój pracuje jako pozornie teowy, więc można go obliczać jak przekrój prostokątny |
dp |
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15 |
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Dane: |
d = hż-gr.otul = |
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0,585 |
m |
gr. otuliny a = 0,015m |
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bż |
hż |
l |
ld = 0,8 l |
b'p = 0,15ld |
b'd = 2b'p+bż |
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[m] |
[m] |
[m] |
[m] |
[m] |
[m] |
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0,4 |
0,6 |
6,5 |
5,2 |
0,78 |
1,96 |
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hpł |
Rb |
z=hż - gr.otul - hpł/2 |
Nb = b'd+hpł*Rb |
Mpt = Nb*z |
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[m] |
MPa |
[m] |
kN |
kNm |
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0,12 |
14,3 |
0,525 |
3363,36 |
1765,764 |
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Mmax <= Mpt |
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175,239<1765,764 |
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Wymiarowanie zbrojenia: |
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0,85 |
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10,6 |
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31 |
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Przęsło skrajne |
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msc = M/(b'd*d2*a*fcd) = 175,239/(1,96*0,5852*0,85*10,6*103) = |
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0,029 |
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z tablic odczytano z = |
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1 |
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Pole przekroju zbrojenia wyznaczono ze wzoru: |
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As1 = M/(z*d*fyd) = |
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9,66 |
cm2 |
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Przyjęto 5 prętów f 16 mm o łącznym polu przekroju 10,05 cm2 |
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bż |
hż |
l |
ld = 0,8 l |
b'p = 0,15ld |
b'd = 2b'p+bż |
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[m] |
[m] |
[m] |
[m] |
[m] |
[m] |
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0,4 |
0,6 |
8 |
6,4 |
0,96 |
2,32 |
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hpł |
Rb |
z=hż - gr.otul - hpł/2 |
Nb = b'd+hpł*Rb |
Mpt = Nb*z |
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[m] |
MPa |
[m] |
kN |
kNm |
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0,12 |
14,3 |
0,525 |
3981,12 |
2090,088 |
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Mmax <= Mpt |
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107,722<1942,8552 |
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Przęsło przedskrajne |
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msc = M/(b'd*d2*a*fcd) = 107,722/(2,32*0,5852*0,85*10,6*103) = |
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0,015 |
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z tablic odczytano z = |
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1 |
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Pole przekroju zbrojenia wyznaczono ze wzoru: |
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As1 = M/(z*d*fyd) = |
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5,94 |
cm2 |
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Przyjęto 3 pręty f 16 mm o łącznym polu przekroju 6,03 cm2 |
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Zasięg momentu ujemnego w przęśle skrajnym |
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Zbrojenie nad podporami |
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a = ((p+q)*l)/(8*(p+q/4)) = |
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Podpora przedskrajna |
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48,19 |
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Mkr = Mpmax+Vmax*x-(p+q)*x2/2 |
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x = bż*0,5 = |
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0,2 |
m |
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Mkr = -247,37+185,654*0,2 - 48,19*0,2*0,2/2 = |
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-211,207 |
kNm |
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msc = M/(b'd*d2*a*fcd) = 229,773/(2,32*0,5852*0,85*10,6*103) = |
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0,030 |
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z tablic odczytano z = |
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1 |
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Pole przekroju zbrojenia wyznaczono ze wzoru: |
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As1 = M/(z*d*fyd) = |
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11,65 |
cm2 |
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Przyjęto 6 prętów f 16 mm o łącznym polu przekroju 12,06 cm2 |
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Podpora środkowa |
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0,00 |
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Mkr = Mpmax+Vmax*x-(p+q)*x2/2 |
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x = bż*0,5 = |
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0,2 |
m |
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Mkr = -85,834+164,248*0,2 - 48,19*0,2*0,2/2 = |
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-153,948 |
kNm |
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msc = M/(b'd*d2*a*fcd) = 153,948/(2,32*0,5852*0,85*10,6*103) = |
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0,022 |
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z tablic odczytano z = |
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1 |
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Pole przekroju zbrojenia wyznaczono ze wzoru: |
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As1 = M/(z*d*fyd) = |
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8,49 |
cm2 |
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Przyjęto 5 prętów f 16 mm o łącznym polu przekroju 10,05 cm2 |
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Zbrojenie na ścinanie (podpora skrajna) |
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Maksymalna siła działająca na podporę Vmax= |
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129,826 |
kN |
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W przypadku obciążenia równomiernie rozłożonego do obliczeń można przyjąć siłę tnącą w odległości od lica podpory z poniższego wzoru: |
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Vsd = Vmax - (0,6-0,015-0,015*0,5+0,4)*(p+q) = |
82,716 |
kN |
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Przyjęto, że do podpory zostaną doprowadzone dołem minimum 4 pręty f 16 o łącznym polu przekroju 8,04 cm2 |
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Wyznaczenie siły VRD1 |
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VRD1 = [1,4*k*tRD*(1,2+40*r1)+0,15*Nsd/Ac]*bż*dż = |
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119,1305752 |
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dż = 0,6-0,015-0,015*0,5 = |
0,5775 |
m |
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k = 1,6-dż = |
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1,0225 |
m |
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tRD = |
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0,26 |
MPa |
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r1 = As1/bż*dż = |
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0,0046 |
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Nie potrzebne jest zbrojenie na ścinanie gdyż spełniony został warunek VSD<VRD1 |
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Zbrojenie na ścinanie (podpora przedskrajna) |
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Maksymalna siła działająca na podporę Vmax= |
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195,796 |
kN |
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W przypadku obciążenia równomiernie rozłożonego do obliczeń można przyjąć siłę tnącą w odległości od lica podpory z poniższego wzoru: |
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Vsd = Vmax - (0,6-0,015-0,015*0,5+0,4)*(p+q) = |
148,686 |
kN |
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Przyjęto, że do podpory zostaną doprowadzone 4 pręty f 16 o łącznym polu przekroju 8,04 cm2 |
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Wyznaczenie siły VRD1 |
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VRD1 = [1,4*k*tRD*(1,2+40*r1)+0,15*Nsd/Ac]*bż*dż = |
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119,1305752 |
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dż = 0,6-0,015-0,015*0,5 = |
0,5775 |
m |
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k = 1,6-dż = |
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1,0225 |
m |
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tRD = |
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0,26 |
MPa |
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r1 = As1/bż*dż = |
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0,0046 |
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Konieczne jest zbrojenie na ścinanie gdyż nie spełniony został warunek VSD<VRD1 |
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Długość odcinka drugiego rodzaju |
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lt = (VSD-VRD1)/(p+q) = |
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0,61 |
m |
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ctg q = |
2,00 |
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Rozmieszczenie prętów odgiętych |
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Maksymalny rozstaw prętów odgiętych |
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a2 = c+f1+0,5*f+Dh = |
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3,4 |
cm |
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Dh |
c |
f1 |
f |
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[cm] |
[cm] |
[cm] |
[cm] |
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0,5 |
1,5 |
0,6 |
1,600 |
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Przyjęto odgięcie prętów pod kątem a = 45o |
ctg 45o = |
1,00 |
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s2max = (dż-a2)*ctg45o+0,2*hż = |
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66 |
cm |
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liczba płaszczyzn odgięcia n=lt/s2max= |
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0,9 |
przyjęto n = |
1 |
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Przyjęty odstęp prętów odgiętych |
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s2 = lt/n = |
61,3 |
cm |
< =s2max |
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Siła poprzeczna VRD3,2 przenoszona przez pręty odgięte f 16 |
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As2 |
fyd |
z = 0,9*dż |
sin 45o |
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[cm2] |
[kN/cm2] |
[cm] |
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2,01 |
31 |
51,975 |
0,707 |
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VR3,2 =( AS2*fyd*z*(ctgq+ctga)*sina)/s2 = |
112,008 |
kN |
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Do obliczenia strzemion należy więc przyjąć siłę VRD3,1 = VSD -VR3,2= |
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36,678 |
kN |
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Wymagany odstęp strzemion dwuramiennych f 6 |
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As1 |
fyd |
z = 0,9*dż |
cos a |
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[cm2] |
[kN/cm2] |
[cm] |
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0,56 |
31 |
51,975 |
0,707 |
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s1 = AS1*fyd*z*ctgq/VRD3,1 = |
49,2 |
cm |
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Graniczna siła poprzeczna VRD2 ze względu na ukośne ścinanie |
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n = 0,7-fck/200 = |
0,620 |
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fck = |
16 |
MPa |
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1,06 |
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DVRD2 = AS2*fyd*z*cosa/s2 = |
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37,336 |
kN |
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n*fcd*bż*z*(ctgq/(1+ctg2q))*(ctga/(2*ctgq+ctga)) = |
109,306 |
kN |
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Przyjęto DVRD = |
109,306 |
kN |
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VRD2 = n*fcd*bż*z*(ctgq/(1+ctg2q))+DVRD = |
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655,834 |
kN |
>VSD = 195,796 |
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Rozmieszczenie strzemion |
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Maksymalny dopuszczalny odstęp strzemion smax wzdłuż osi belki |
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Ponieważ spełniona została nierówność 0,2VRD2<VSD<0,667VRD2 |
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131,167 |
<195,795< |
437,441 |
więc |
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smax=0,6*dż= |
34,65 |
cm |
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Maksymalny rozstaw strzemion smax ze względu na wymagany minimalny stopień zbrojenia na ścinanie |
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r1 = |
0,0009 |
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smax = As1/r1*bż = |
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15,56 |
cm |
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Uwzględniając powyższe wyniki przyjęto na odcinkach drugiego rodzaju strzemiona dwuramienne f 6 co 15 cm, natomiast na pozostałej części przęsła co 30 cm |
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Zbrojenie na ścinanie (podpora środkowa) |
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Maksymalna siła działająca na podporę Vmax= |
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164,248 |
kN |
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W przypadku obciążenia równomiernie rozłożonego do obliczeń można przyjąć siłę tnącą w odległości od lica podpory z poniższego wzoru: |
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Vsd = Vmax - (0,6-0,015-0,015*0,5+0,4)*(p+q) = |
117,138 |
kN |
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Przyjęto, że do podpory zostaną doprowadzone 3 pręty f 16 o łącznym polu przekroju 6,03 cm2 |
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Wyznaczenie siły VRD1 |
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VRD1 = [1,4*k*tRD*(1,2+40*r1)+0,15*Nsd/Ac]*bż*dż = |
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119,1305752 |
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dż = 0,6-0,03-0,03*0,5 = |
0,5775 |
m |
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k = 1,6-dż = |
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1,0225 |
m |
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tRD = |
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0,26 |
MPa |
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r1 = As1/bż*dż = |
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0,0046 |
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Nie potrzebne jest zbrojenie na ścinanie gdyż spełniony został warunek VSD<VRD1 |
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Sprawdzenie stanu granicznego użytkowania |
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Wyznaczenie rozwarcia rysy ukośnej na odcinku z pretami odgiętymi |
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wk = 4t2l/rw*Es*fck |
t = VSD/b*d |
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rw = rw1+rw2 |
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rw1 = As1/b*s1 |
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l = 1/(3*(rw1/B1*F1+rw2/b2*F2)) |
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rw2 = As2/b*sina*s2 |
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VSD max |
Es |
fck |
b1 = b2 |
s1 |
s2 |
b |
sina |
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131,1 |
200000 |
30 |
1 |
0,071 |
0,382 |
0,3 |
0,707 |
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As1 |
As2 |
F1 |
F2 |
d |
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5,6E-05 |
0,000314 |
0,006 |
0,020 |
0,348 |
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t = |
1256 |
kN/m2 = |
1,256 |
MPa |
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rw1 = |
0,002629107981221 |
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rw = |
0,006504589022219 |
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rw2 = |
0,003875481040999 |
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l = |
527,460615878789 |
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wk = |
0,085282160772812 |
< w lim = 0,3 mm |
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Na odcinkach zbrojonych strzemionami i prętami odgiętymi szerokość rys jest mniejsza |
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od dopuszczalnej |
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Wyznaczenie rozwarcia rysy ukośnej na odcinku bez pretów odgiętych |
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t = |
919 |
kN/m2 = |
0,919 |
MPa |
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VSD max |
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rw1 = |
0,001964912280702 |
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rw = |
0,001964912280702 |
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95,9 |
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l = |
1017,85714285714 |
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wk = |
0,291664401466837 |
< w lim = 0,3 mm |
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Na odcinkach zbrojonych strzemionami szerokość rys jest mniejsza od dopuszczalnej |
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rw = |
0,002 |
> rmin = 0,0013 |
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Stan graniczny ugięcia |
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Mmax = |
95,8 |
kNm |
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b= |
0,3 |
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Miarodajny wymiar przekroju |
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h= |
0,4 |
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d= |
0,348 |
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171,429 |
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mm |
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Końcowy współczynnik pełzania |
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Ec,eff = Ecm/1+F(oo,la) |
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=32/1+2,5= |
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9,14 |
GPa |
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aef = Es/Eceff =200/9,14 |
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21,875 |
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r =9,426*10-4/0,4*0,3= |
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0,009028735632184 |
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Charakterystyka przekroju niezarysowanego, położenie osi obojętnej |
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Acs= |
0,140619375 |
m2 |
X1= |
0,221701614731256 |
m |
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Scs= |
0,0311755425 |
m3 |
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Ics= |
0,001985420677627 |
m4 |
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Wcs= |
0,015720079662162 |
m3 |
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Moment rysujący |
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Mcr = Wcs*fctm=0,01572*2900 = |
45,588 |
< Mmax = 95,8 |
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Przekrój pracuje jako zarysowany |
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Moment bezwładnosci przekroju po zarysowaniu |
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0,3*0,5*XII*XII-21,87*9,426*0,0001*(0,348-XII)=0 |
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Xll= |
0,160531 |
m |
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Icsll= |
0,000478042111533 |
m4 |
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ss= |
821,820324291446 |
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esm= |
0,00364384630033 |
Średnie odkształcenia w stali |
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Act,eff= |
239,5 |
cm2 |
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rr= |
0,039356993736952 |
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Srm= |
100,8168894547 |
mm |
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Wk= |
0,477569624545318 |
> w lim = 0,3 mm |
Nie spełnione |
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Ugięcie belki |
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Bh= |
5,27765797790609 |
MNm2 |
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a= |
76,7241169495641 |
>alim= |
30 |
mm |
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Warynek ugięcia nie spełniony |
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