Spoiny część a:) |
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Przekrój: |
HEB 1000 |
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M = |
0,65 |
x MR |
kNcm |
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V = |
0,6 |
x VR |
kN |
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STAL |
St3S |
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Fd = |
215 |
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wartości: |
jednostki: |
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36 |
t
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Dane: |
h |
1000 |
mm |
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tw |
19 |
mm |
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b |
300 |
mm |
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t |
36 |
mm |
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tw |
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r |
30 |
mm |
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A |
400 |
cm2 |
1000 |
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19 |
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m |
- |
kg/m |
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Ix |
644700 |
cm4 |
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Iy |
- |
cm4 |
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Wx |
12890 |
cm3 |
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Wy |
- |
cm3 |
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ix |
- |
cm |
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iy |
- |
cm |
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0 |
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66 |
78 |
105 |
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140,5 |
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9 |
10 |
14 |
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b |
300 |
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164,92 |
235,08 |
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0,4123 |
0,5877 |
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164,92 |
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1 |
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Wyznaczenie wysokości urzytecznej środnika |
hśr |
868,00 |
mm |
hśr = h -2 x t - 2 x r |
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Klasa przekroju |
E |
1,00 |
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ŚRODKI: |
klasa: |
I |
II |
III |
IV |
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b |
45,68 |
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Jeleli III lub IV klasa przekroju przejdz do pkt 1.2.
JEŻELI: |
III |
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t |
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IV |
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PAS: |
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b |
3,90 |
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Jeleli III lub IV klasa przekroju przejdz do pkt 1.2.
JEŻELI: |
III |
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t |
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IV |
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Pkt 1.1 |
I |
II |
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Nośność obliczeniowa przy jednokierunkowym zginaniu - MR dla klasy 1 i 2 zgodnie z PN-90/B-03200 pkt 4.5.2 : MR=AlfaPWfd |
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Alfa |
1,07 |
Izwyk Alfap=1,0
IHEB Alfap=1,05
IPE Alfap=1,07
wybierz: |
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MR = |
296534,45 |
kN/cm |
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Nośność obliczeniowa przy ścinaniu VR zgodnie z PN-90/B-03200 pkt 4.5.2 ; VR = 0,58Avfd |
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VR = |
2056,55 |
kN |
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OSTATECZNIE PRZYJĘTO |
MR = |
192747,39 |
kN/cm |
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VR = |
1233,93 |
kN |
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Pkt 1.2 |
III |
IV |
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Nośność obliczeniowa przy zginaniu wzór 43 |
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MR= Ψ x W x fd |
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Stan krytyczny |
Ψ = φp |
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Smukłość: |
λ = |
b |
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t |
56 |
215 |
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0 |
1 |
3 |
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λ = |
0,326315789473684 |
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φp = |
1,960 |
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MR = |
543088,858646979 |
kNcm |
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Nośność obliczeniowa przy ścinaniu |
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VR = φp x 0,58 x Av x fd |
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λpv = |
b |
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t |
56 |
215 |
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φpv = |
1 |
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λpv |
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φpv = |
2,903 |
mniejsze niż 1 |
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λpv = |
0,344 |
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VR = |
5970,636 |
kN |
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Ostatecznie przyjęto: |
M = |
353007,76 |
kNcm |
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V = |
3582,3816 |
kN |
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Połaczenie Spawane |
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Przekrój: |
|
HEB 1000 |
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Stal: |
St3S |
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fd = |
215 |
Mpa |
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Dane: |
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[Miara] |
[Jednostki] |
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36 |
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h |
1000 |
mm |
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tw |
19 |
mm |
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b |
300 |
mm |
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t |
36 |
mm |
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r |
30 |
mm |
1000 |
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19 |
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A |
400 |
cm2 |
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m |
- |
kg |
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Ix |
644700 |
cm4 |
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Iy |
- |
cm4 |
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Wx |
12890 |
cm3 |
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Wy |
- |
cm3 |
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300 |
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ix |
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cm |
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iy |
- |
cm |
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1 |
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Wyznaczenie wysokości urzytecznej środnika |
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hśr = h -2 x t - 2 x r |
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hśr |
= |
868,00 |
mm |
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Klasa Przekroju |
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Środnik: |
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- |
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I klasa |
II klasa |
III klasa |
IV klasa |
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+ |
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66ε |
78ε |
105ε |
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66,00 |
78,00 |
105,00 |
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ε |
= |
215 |
= |
215 |
= |
1,00 |
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fd |
21,5 |
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zał: |
ε |
< |
1 |
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b |
= |
868,00 |
= |
45,68 |
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t |
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19 |
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45,68 |
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< |
66,00 |
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I klasa przekroju |
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Pas: |
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- |
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I klasa |
II klasa |
III klasa |
IV klasa |
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9ε |
10ε |
14ε |
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b |
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9,00 |
10,00 |
14,00 |
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b |
= |
1/2 b - tw |
= |
140,5 |
= |
3,90 |
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t |
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t |
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36 |
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3,90 |
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< |
9,00 |
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I klasa przekroju |
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2 |
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Nośność obliczeniowa przy jednokierunkowym zginaniu - MR dla klasy 1 i 2 zgodnie z PN-90/B-03200 pkt 4.5.2 : MR=αpWfd |
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αp - zalezy od rodzaju przekroju. Jeżeli Izwyk αp=1,0 ; IHEB αp=1,05; IPE αp=1,07 |
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αp |
= |
1,07 |
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Wx |
= |
12890 |
cm3 |
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fd |
= |
21,5 |
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Mr = |
1,07 |
x |
12890 |
x |
21,5 |
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Mr |
= |
296534,45 |
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kN/cm |
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Ostatecznie: |
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M = |
0,65 |
x |
296534,45 |
= |
192747,39 |
kN/cm |
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Nośność obliczeniowa przy ścinaniu VR zgodnie z PN-90/B-03200 pkt 4.5.2 ; VR = 0,58Avfd |
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Av |
= |
hśr |
x |
tw |
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Av |
= |
86,8 |
x |
1,9 |
= |
164,92 |
cm2 |
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fd |
= |
21,5 |
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VR |
= |
0,58 |
x |
164,92 |
x |
21,5 |
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VR |
= |
2056,55 |
kN |
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Ostatecznie: |
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V = |
0,6 |
x |
2056,55 |
= |
1233,93 |
kN |
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3 |
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Dobór geometryczny przekładek: |
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An - pole nakłądek |
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Ap - pole przekładek |
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An =Ap |
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Ac - pole 2 teownika |
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Ac = A - Av |
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Av - pole użyteczne przekroju |
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An +Ap =Ac |
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Ac = |
400,00 |
cm 2 |
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0,65 |
x |
Ac |
= |
260,00 |
cm 2 |
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Nakładka górna : |
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b1 = |
bf - 4cm |
= |
26 |
cm |
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An1 = |
b x t |
= |
108 |
cm 2 |
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t1 = |
An1 / b1 |
= |
4,15 |
cm |
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Przyjęto |
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b1 = |
26 |
cm |
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t1 = |
2,8 |
cm |
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Nakładka dolna : |
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b2 = |
bf + 4cm |
= |
34 |
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An2 = |
b x t |
= |
108 |
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t2 = |
An2 / b2 |
= |
3,18 |
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Przyjęto |
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b2 = |
34 |
cm |
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t2 = |
2,2 |
cm |
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4 |
|
Przekładki: |
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b-tw |
= |
140,5 |
mm |
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2 |
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|
tg30' |
= |
0,577 |
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Xmin |
= |
tg30' |
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|
140,5 |
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Xmin |
= |
81,07 |
mm |
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hg max = h - 2•t - 2•xmin |
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hg max |
= |
69,39 |
cm |
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grubość: |
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2 x gp > gsr |
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gp |
= |
1,15 |
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Przyjęto |
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|
Proporcje: |
[%] |
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|
Przekładka |
40,02 |
|
|
hg = |
69 |
cm |
|
Nakłądka |
59,98 |
|
|
gp = |
1,4 |
cm |
|
Pas |
58,77 |
|
|
Sprawdzenie przekładek |
|
Środnik |
41,23 |
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|
Wp = |
2 x gp x hp2 |
= |
2246,74735958867 |
cm3 |
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|
6 |
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Wś = |
gs x h sr 2 |
x |
0,65 |
= |
1550,79773333333 |
cm3 |
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|
6 |
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Warunek spełniony |
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Wp |
> |
Wś |
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|
2246,74735958867 |
> |
1550,79773333333 |
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5 |
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Ip = |
2 x gp x hp3 |
= |
77946,7431583136 |
cm4 |
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12 |
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In = |
b1 x t1 3 |
+ |
An1xY12 |
+ |
b1 x t1 3 |
+ |
An2xY22 |
|
12 |
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|
12 |
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In = |
382209,55 |
cm4 |
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In+p = |
= |
460156,30 |
cm4 |
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Wn |
= |
In |
= |
7245,15 |
cm3 |
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Ymax |
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Warunek spełniony |
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|
Wn+Wp |
> |
Ws x |
0,65 |
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|
9491,89784172759 |
> |
8378,5 |
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|
Rozkłąd momentów |
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|
Mr = |
192747,39 |
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|
Mn = |
Mr x In |
= |
160097,548101909 |
kN/cm |
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|
I p+n |
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|
Mp = |
Mr x Ip |
= |
32649,8443980907 |
kN/cm |
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|
I p+n |
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Siły działające na nakładki |
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α|| = |
0,8 |
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N = |
Mn |
= |
1600,97548101909 |
kN |
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h |
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Dobór spoiny |
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0,2t1< |
a1 |
<0,7t2 |
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|
t1 < t2 |
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|
0,55 |
< a1 < |
2,52 |
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a1 |
= |
0,8 |
cm |
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|
Długość spoiny |
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|
Lmin = |
N |
= |
58,17 |
cm |
10a < |
l |
<100a |
|
a α|| fd |
8 |
< l < |
80 |
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|
Przyjęto |
|
L = |
59 |
|
l = |
4cm |
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l = |
b |
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6 |
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|
Wyznaczenie spoiny przykładek |
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|
0,2t1< |
a2 |
<0,7t2 |
|
|
t1 < t2 |
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|
0,28 |
< a2 < |
1,33 |
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|
a2 |
= |
0,8 |
cm |
|
Długość przekładki |
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d = |
21 |
cm |
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69 |
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0,8 |
21 |
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|
Środek ciężkości |
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|
e = |
ΣSi |
= |
hg x a2 x 1/2d+1/2 a2 |
|
|
|
Σ A |
hg x a2 + 2 x d x a2 |
|
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|
e = |
6,54 |
cm |
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|
x = |
e + 1/2 d |
= |
17,04 |
cm |
|
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|
y = |
1/2d +1/2a2 |
= |
35,09 |
cm |
|
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|
|
Ix= |
2*(d*a23) |
+ |
y2*a2*d |
+ |
a2*hp3 |
|
|
|
12 |
12 |
|
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|
Iy= |
2*(a2*d3) |
+ |
x2*a2*d |
+ |
hp*a23 |
+ |
(1/2*a+(d-x)^2)*hp*a |
12 |
12 |
|
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|
Ix= |
63651,6703898285 |
cm 4 |
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|
Iy= |
12049,6475781689 |
cm 4 |
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|
Io = |
Ix |
+ |
Iy |
= |
75701,32 |
cm 4 |
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|
7 |
|
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|
Mo = |
Mp + V * x |
= |
26838,5151796036 |
kNcm |
|
|
2 |
|
|
|
|
|
|
|
|
|
|
|
Vo = |
V |
= |
616,96572 |
kN |
|
|
2 |
|
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|
|
τp = |
Vo |
= |
6,9237163816376 |
|
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|
|
Σal |
|
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|
|
τy = |
Mo X |
= |
6,04150542825775 |
|
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|
|
Io |
|
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|
τx = |
Mo y |
= |
12,4416333064805 |
|
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|
Io |
|
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|
Warunek spełniony |
|
|
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|
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|
|
τ = |
17,97 |
< |
19,35 |
αfd |
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8 |
Połaczenie Spawane |
|
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|
|
Przekrój: |
|
HEB 1000 |
|
|
|
|
|
|
|
|
|
|
|
|
Stal: |
St3S |
|
|
Dane: |
|
[Miara] |
[Jednostki] |
|
fd = |
215 |
Mpa |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
36 |
|
|
h |
1000 |
mm |
|
|
|
|
|
|
tw |
19 |
mm |
|
|
|
|
|
|
b |
300 |
mm |
|
|
|
|
|
|
t |
36 |
mm |
|
|
|
|
|
|
r |
30 |
mm |
1000 |
|
19 |
|
|
|
|
|
|
|
|
|
|
|
|
A |
400 |
cm2 |
|
|
|
|
|
|
m |
- |
kg |
|
|
|
|
|
|
Ix |
644700 |
cm4 |
|
|
|
|
|
|
Iy |
- |
cm4 |
|
|
|
|
|
|
Wx |
12890 |
cm3 |
|
|
|
|
|
|
Wy |
- |
cm3 |
|
|
300 |
|
|
|
ix |
- |
cm |
|
|
|
|
|
|
iy |
- |
cm |
|
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|
1 |
|
|
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|
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|
|
Wyznaczenie wysokości urzytecznej środnika |
|
|
|
|
|
|
|
|
|
|
|
hśr = h -2 x t - 2 x r |
|
|
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|
|
|
|
|
|
|
|
|
|
hśr |
= |
868,00 |
mm |
|
|
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|
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|
|
|
|
Klasa Przekroju |
|
|
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|
|
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|
|
Środnik: |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
- |
|
|
I klasa |
II klasa |
III klasa |
IV klasa |
|
|
|
+ |
|
66ε |
78ε |
105ε |
|
|
|
|
|
66,00 |
78,00 |
105,00 |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
ε |
= |
215 |
= |
215 |
= |
1,00 |
|
|
fd |
21,5 |
|
|
|
|
|
|
|
|
|
|
|
zał: |
ε |
< |
1 |
|
|
|
|
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|
|
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|
|
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|
|
|
|
|
|
|
|
|
|
|
b |
= |
868,00 |
= |
45,68 |
|
|
|
|
t |
19 |
|
|
|
|
|
|
|
|
|
|
|
|
|
105,00 |
< |
45,68 |
|
IV klasa przekroju |
|
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|
Pas: |
|
|
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|
|
|
|
|
|
|
|
|
|
|
|
|
- |
|
I klasa |
II klasa |
III klasa |
IV klasa |
|
|
|
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|
9ε |
10ε |
14ε |
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b |
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9,00 |
10,00 |
14,00 |
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b |
= |
1/2 b - tw |
= |
140,5 |
= |
3,90 |
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t |
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t |
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36 |
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3,90 |
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< |
14,00 |
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III klasa przekroju |
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2 |
Nośność obliczeniowa przy zginaniu wzór 43 |
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MR= Ψ x W x fd |
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Stan krytyczny |
Ψ = φp |
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Smukłość: |
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λ = |
b |
K = |
0,4 |
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t |
56 |
215 |
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K - współczynik podparcia i obciązenia ścianki wg PN -90/B-03200 tab 8 |
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φp - Współczynik nie stateczniści przyjmować należy od smukłości względnej wg tab.9 |
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φp |
= |
1,960 |
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Wx |
= |
12890 |
cm3 |
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fd |
= |
21,5 |
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Mr = |
1,960 |
x |
12890 |
x |
21,5 |
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Mr |
= |
543088,858646979 |
kN/cm |
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Ostatecznie: |
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M = |
0,65 |
x |
543088,858646979 |
= |
192747,39 |
kN/cm |
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Nośność obliczeniowa przy ścinaniu |
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VR = φp x 0,58 x Av x fd |
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λpv = |
b |
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K = |
0,8 |
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t |
56 |
215 |
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λpv = |
0,3444 |
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φpv = |
1 |
= |
2,903 |
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λpv |
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Av = |
hśr |
x |
tw |
= |
164,92 |
cm2 |
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VR = |
2,903 |
x |
0,58 |
x |
164,92 |
x |
21,5 |
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VR = |
5970,636 |
kN |
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Ostatecznie |
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V = |
0,6 |
x |
5970,636 |
= |
3582,38 |
kN |
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3 |
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Dobór geometryczny przekładek: |
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An - pole nakłądek |
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Ap - pole przekładek |
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An =Ap |
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Ac - pole 2 teownika |
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Ac = A - Av |
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Av - pole użyteczne przekroju |
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An +Ap =Ac |
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Ac = |
400 |
cm 2 |
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0,65 |
x |
Ac |
= |
260,00 |
cm 2 |
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Nakładka górna : |
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b1 = |
bf - 2cm |
= |
28 |
cm |
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An1 = |
b x t |
= |
108 |
cm 2 |
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t1 = |
An1 / b1 |
= |
3,86 |
cm |
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Przyjęto |
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b1 = |
28 |
cm |
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t1 = |
4,0 |
cm |
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Nakładka dolna : |
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b2 = |
bf + 2cm |
= |
32 |
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An2 = |
b x t |
= |
108 |
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t2 = |
An2 / b2 |
= |
3,38 |
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Przyjęto |
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b2 = |
32 |
cm |
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t2 = |
3,4 |
cm |
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Proporcje: |
[%] |
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Przekładka |
39,5245992260918 |
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Nakłądka |
60,4754007739082 |
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2Teownik |
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Pas |
41,23 |
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Środnik |
58,77 |
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4 |
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Przekładki: |
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b-tw |
= |
140,5 |
mm |
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2 |
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tg30' |
= |
0,577 |
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Xmin |
= |
tg30' |
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140,5 |
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Xmin |
= |
81,07 |
mm |
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hg max = h - 2•t - 2•xmin |
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hg max |
= |
70,59 |
cm |
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grubość: |
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2 x gp > gsr |
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gp |
= |
1,15 |
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Przyjęto |
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hg = |
110 |
cm |
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gp = |
1,3 |
cm |
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Sprawdzenie przekładek |
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Wp = |
2 x gp x hp2 |
= |
5243,33333333333 |
cm3 |
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6 |
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Wś = |
gs x h sr 2 |
x |
0,65 |
= |
1550,79773333333 |
cm3 |
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6 |
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Warunek spełniony |
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Wp |
> |
Wś |
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5243,33 |
> |
1550,79773333333 |
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5 |
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Ip = |
2 x gp x hp3 |
= |
288383,333333333 |
cm4 |
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12 |
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In = |
b1 x t1 3 |
+ |
An1xY12 |
+ |
b1 x t1 3 |
+ |
An2xY22 |
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12 |
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12 |
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In = |
588135,97 |
cm4 |
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In+p = |
= |
876519,31 |
cm4 |
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Wn |
= |
In |
= |
10900,06 |
cm3 |
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Ymax |
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Warunek spełniony |
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Wn+Wp |
> |
Ws x |
0,65 |
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16143,39 |
> |
1550,80 |
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Rozkłąd momentów |
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Mr = |
353007,76 |
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Mn = |
Mr x In |
= |
236864,79 |
kN/cm |
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I p+n |
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Mp = |
Mr x Ip |
= |
116142,97 |
kN/cm |
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I p+n |
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Siły działające na nakładki |
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α|| = |
0,8 |
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N = |
Mn |
= |
2368,65 |
kN |
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h |
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Dobór spoiny |
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0,2t1< |
a1 |
<0,7t2 |
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t1 < t2 |
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0,79 |
< a1 < |
2,52 |
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a1 |
= |
0,9 |
cm |
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Długość spoiny |
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Lmin = |
N |
= |
76,51 |
cm |
10a < |
l |
<100a |
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a α|| fd |
9 |
< l < |
90 |
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Przyjęto |
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L = |
78 |
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l = |
4cm |
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l = |
b |
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6 |
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Wyznaczenie spoiny przykładek |
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0,2t1< |
a2 |
<0,7t2 |
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t1 < t2 |
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0,26 |
< a2 < |
1,33 |
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a2 |
= |
0,5 |
cm |
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Długość przekładki |
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d = |
50 |
cm |
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110 |
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0,5 |
50 |
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Środek ciężkości |
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e = |
ΣSi |
= |
hg x a2 x 1/2d+1/2 a2 |
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Σ A |
hg x a2 + 2 x d x a2 |
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e = |
13,10 |
cm |
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x = |
e + 1/2 d |
= |
38,10 |
cm |
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y = |
1/2d +1/2a2 |
= |
55,25 |
cm |
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Ix= |
2*(d*a2^3) |
+ |
y^2*a2*d |
+ |
a2*hp^3 |
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12 |
12 |
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Iy= |
2*(a*d^3) |
+ |
x^2*a*d |
+ |
hp*a^3 |
+ |
(1/2*a+(d-x)^2)*hp*a |
12 |
12 |
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Ix= |
208087,5 |
cm 4 |
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Iy= |
91105,7738095238 |
cm 4 |
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Io = |
Ix |
+ |
Iy |
= |
299193,27 |
cm 4 |
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7 |
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Mo = |
Mp + V * x |
= |
126307,324182864 |
kNcm |
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2 |
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Vo = |
V |
= |
1791,1908 |
kN |
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2 |
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τp = |
Vo |
= |
17,05896 |
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Σal |
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τy = |
Mo X |
= |
16,0822719262797 |
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Io |
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τx = |
Mo y |
= |
23,3243200030826 |
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Io |
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Warunek spełniony |
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τ = |
40,53 |
< |
19,35 |
αfd |
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8 |
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ŚRODEK CIĘŻKOŚCI |
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2TEO |
|
1/2TEO |
|
|
Wymiary teownika |
|
[cm] |
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Stosunek pól |
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|
grubość półki |
gp = |
1,20 |
|
Pole półki |
36,00 |
cm2 |
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pas |
0,48 |
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h |
360 |
|
180 |
mm |
|
szerokość półki |
sp = |
30,00 |
|
75,42 |
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szer |
170 |
|
170 |
mm |
|
grubość środnika |
gs = |
0,90 |
|
Pole środnika |
39,42 |
cm2 |
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śr |
0,52 |
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gs |
8 |
|
8 |
mm |
|
wysokość środnika |
hs = |
43,80 |
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tf |
12,7 |
|
12,7 |
mm |
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r |
18 |
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Wymiary blach |
|
[cm] |
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|
76,00 |
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Napr |
21,5 |
A |
116 |
|
58 |
mm |
|
grubość nakładki |
tn = |
1,00 |
|
Pole nakładki |
30,00 |
cm2 |
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|
n |
0,39 |
|
21,3601608386871 |
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|
szerokość nakładki |
sn = |
30,00 |
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grubość przykładki |
tp = |
1,00 |
|
Pole przykładek |
46,00 |
cm2 |
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|
p |
0,61 |
|
21,1691350100484 |
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|
wysokość przykładki |
hp = |
23,00 |
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|
IPE500 wymiary w mm h=500, s szer=200, g=10,2, t szer stopki 16, r=21, A=116 Ix=48200, Iy=2140, Wx=1930, Wy=214, ix=20,4 , iy=4,31 |
|
odległość przykł-półka |
d = |
8,05 |
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|
wysokość robocza przykł |
|
35,75 |
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Środek ciężkości |
|
[cm] |
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teownika |
Yt = |
13,36 |
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układu blach |
Yb = |
13,36 |
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21,3601608386871 |
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|
Środek ciężkości teownika Yt musi być w przybliżeniu równy środkowi ciężkości układu blach (nakładki i przykładek) Yb. |
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|
21,1691350100484 |
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|
Przekrój |
1/2 IKS 900 |
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0,9 |
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N = |
0,91 |
x A x fd |
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a1 |
d |
a2 |
d |
a1 |
|
14,55 |
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30 |
20 |
50 |
20 |
30 |
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Wyznaczenie sił w przekroju: |
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|
290 |
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Stal: |
St4W |
|
fd |
215 |
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|
Śruby: |
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Przekładki i Nakładki |
|
Rm = |
410 |
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Klasa: |
4.6 |
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Stal: |
St4W |
|
Re = |
255 |
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Rm= |
400 |
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Nr = A x fd |
Re = |
240 |
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a1 |
30 |
220 |
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d |
20 |
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a2 |
50 |
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Nr= |
1621,53 |
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d |
20 |
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a1 |
30 |
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N= |
1475,5923 |
kN |
|
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|
39,45 |
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|
240 |
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|
Rozkład sił na nakładke i przekładki |
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Nsr = |
N |
Asr |
kN |
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Ac |
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Środnik |
Nsr = |
771,25 |
kN |
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|
Sprawdzenie: |
|
|
|
1475,5923 |
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|
Np = N - Nsr |
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|
N= |
1475,5923 |
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Nsr+Np.= |
1475,59 |
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Pas |
Np = |
704,34 |
kN |
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NAKŁADKI : |
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Fr = n x n x Srmin |
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n - |
Liczba śrub |
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Wytrzymałość trzpienia na ścinanie |
|
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|
0 |
7 |
0 |
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8 |
14 |
1 |
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Srv = 0,45 x Rm x Av x m |
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16 |
24 |
2 |
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27 |
45 |
3 |
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Av - |
pole powieszchni trzpienia |
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m - |
Liczba powieszchni scinania |
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Delta do D |
20 |
|
ż |
|
-0,4 |
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Średnica śruby |
20 |
|
m= |
1 |
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Av= |
3,14 |
cm2 |
n= |
16 |
szt |
Wstaw: |
2 |
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Srv = |
56,55 |
kN |
1 |
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Fr= |
904,75 |
kN |
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nakładka |
582,470644736842 |
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Przekładka |
893,121655263158 |
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Osłabienie przekroju: |
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Liczba śrub w przekroju |
4 |
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An = |
21,2 |
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At = |
30 |
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Aψ = |
27,27 |
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ψ = |
0,91 |
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Sigma |
19,42 |
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Sigma el |
21,36 |
< |
21,5 |
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Przekładki: |
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Fr = n x n x Srmin |
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n - |
Liczba śrub |
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Wytrzymałość trzpienia na ścinanie |
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Srv = 0,45 x Rm x Av x m |
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0 |
7 |
0 |
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8 |
14 |
1 |
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Av - |
pole powieszchni trzpienia |
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16 |
24 |
2 |
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m - |
Liczba powieszchni scinania |
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27 |
45 |
3 |
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Średnica śruby |
20 |
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m= |
2 |
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Delta do D |
20 |
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Av= |
3,142 |
cm2 |
n= |
15 |
szt |
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Srv = |
60,788 |
kN |
n = |
1 |
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Wstaw: |
2 |
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|
1,33333333333333 |
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28,2666666666667 |
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Fr= |
911,820 |
kN |
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0,942222222222222 |
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Osłabienie przekroju: |
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Liczba śrub w przekroju |
3 |
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An = |
16,4 |
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At = |
23 |
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Aψ = |
21,09 |
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ψ |
0,92 |
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Sigma |
19,42 |
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Sigma el |
21,17 |
< |
21,5 |
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|
POŁACZENIE ŚRUBOWE |
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|
Przekrój |
1/2 IKS 900 |
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|
|
Stosunek pól [%] |
h = |
45,00 |
cm |
Środnik |
0,52 |
b = |
30,00 |
cm |
Pas |
0,48 |
gs= |
0,90 |
cm |
Przekładka |
0,61 |
gp= |
1,20 |
cm |
Nakładka |
0,39 |
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|
Wymiary blach |
grubość nakładki |
tn = |
1,00 |
szerokość nakładki |
sn = |
30,00 |
grubość przykładki |
tp = |
1,00 |
|
Środek ciężkości |
wysokość przykładki |
hp = |
23,00 |
|
1/2 teo |
13,36 |
cm |
odległość przykł-półka |
d = |
8,05 |
|
blachy |
13,36 |
cm |
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N = |
0,91 |
x A x fd |
kN |
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Rodzaj stali: |
St4W |
|
fd = |
215 |
[MPa] |
|
Przekładki i Nakładki |
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|
Stal: |
St4W |
|
Rm = |
410 |
[MPa] |
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Re = |
255 |
[MPa] |
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|
Klasa śrób |
4.6 |
|
Rm = |
400 |
[MPa] |
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Re = |
240 |
[MPa] |
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|
Nośność obliczeniowa przy osiowym rozciąganiu zgodnie z PN-90/B-03200 pkt 4.3.2 ; Nr = A fd |
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|
Pole przekroju |
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|
A= |
75,42 |
cm |
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Siła |
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Nr = |
75,42 |
x |
21,5 |
= |
1621,53 |
kN |
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|
siła obliczeniowa |
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|
N = |
0,91 |
x |
1621,53 |
= |
1475,5923 |
kN |
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1 |
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Rozkład sił miedzy nakładke i przekładki |
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Siła przypadająca na środnik |
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|
Nsr = |
N |
x |
Aśr |
|
kN |
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|
A |
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|
Nsr = |
1475,5923 |
x |
39,42 |
|
kN |
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|
75,42 |
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|
Nsr = |
771,25 |
|
kN |
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Siła przypadająca na pas |
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Np = |
N |
- |
Nsr |
|
kN |
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Np = |
1475,5923 |
- |
771,25 |
|
kN |
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Np = |
704,34 |
|
kN |
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|
NAKŁADKA |
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|
Zgodnie z PN-90/B-03200 pkt 6.2.4.2. Połączenie Zakładkowe, Nośność połaczenia zakładkowego: |
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Fr = n x n x Sr |
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n - |
liczba łaczników |
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η = |
1 |
- |
1 - 15d |
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|
200d |
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Sr - |
miarodajna nośność łacznika |
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2 |
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Wytrzymałość trzpienia na ścinanie |
|
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|
Srv = 0,45 x Rm x Av x m |
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Av - |
pole powieszchni trzpienia |
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m - |
Liczba powieszchni scinania |
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Średnica śruby |
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d = |
20 |
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Pole trzpienia |
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Av = |
π * d2 |
= |
3,14 |
cm2 |
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|
4 |
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Liczba powieszchni scinania |
|
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m = |
1 |
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Nośność trzpienia |
|
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Srv = |
56,55 |
kN |
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Nośność układu śrub |
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Liczba śrub |
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n = |
16 |
szt |
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η = |
1 |
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Fr = |
904,75 |
kN |
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Tabela 14 |
PN -90/B-03200 |
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Średnica śruby |
8<d<14 |
16<d<24 |
27<d<45 |
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Rodzaj otworu |
Mkasymalne średnice otworów d0 [mm] |
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Δ = |
1 |
Δ = |
2 |
Δ = |
3 |
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PN-90/B-03200 str24 tab 15 |
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a1 = |
1,5 x d |
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a2 = |
2,5 x d |
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a1 |
d |
a2 |
d |
a1 |
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30 |
20 |
50 |
20 |
30 |
[mm] |
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3 |
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Osłabienie przekroju |
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Liczba śrub w przekroju |
4 |
szt |
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Δ = |
2 |
Nn = |
N x (An/An+p)= |
582,47 |
kN |
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Przekrój netto |
z otworami |
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An = |
tn x (sn -(d+Δ)) |
cm2 |
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An = |
21,2 |
cm2 |
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Przekrój bruto |
bez otworów |
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At = |
tn x sn |
cm2 |
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At = |
30 |
cm2 |
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Sprawdzenie pola przekroju |
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Aψ = |
An |
0,8 x Rm |
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Re |
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Aψ = |
27,27 |
cm2 |
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ψ = |
Aψ |
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At |
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ψ = |
0,9 |
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Naprezenia w przekroju bruto |
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σ = |
Nn |
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MPa |
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At |
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σ = |
19,42 |
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MPa |
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Napreżenia osłabionego elementu |
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σel = |
σ |
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MPa |
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ψ |
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σel = |
21,36 |
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MPa |
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Warunek spełniony |
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σel |
< |
fd |
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21,36 |
< |
21,5 |
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4 |
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PRZEKŁADKI |
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Zgodnie z PN-90/B-03200 pkt 6.2.4.2. Połączenie Zakładkowe, Nośność połaczenia zakładkowego: |
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Fr = n x n x Sr |
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n - |
liczba łaczników |
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η = |
1 |
- |
1 - 15d |
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200d |
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Sr - |
miarodajna nośność łacznika |
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Wytrzymałość trzpienia na ścinanie |
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Srv = 0,45 x Rm x Av x m |
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Av - |
pole powieszchni trzpienia |
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m - |
Liczba powieszchni scinania |
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Średnica śruby |
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d = |
20 |
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Pole trzpienia |
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Av = |
π * d2 |
= |
3,14 |
cm2 |
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4 |
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Liczba powieszchni scinania |
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m = |
2 |
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Nośność trzpienia |
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Srv = |
60,79 |
kN |
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5 |
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Nośność układu śrub |
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Liczba śrub |
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n = |
15 |
szt |
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η = |
1 |
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Fr = |
911,82 |
kN |
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Tabela 14 |
PN -90/B-03200 |
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|
Średnica śruby |
8<d<14 |
16<d<24 |
27<d<45 |
|
Rodzaj otworu |
Mkasymalne średnice otworów d0 [mm] |
|
|
|
Δ = |
1 |
Δ = |
2 |
Δ = |
3 |
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Osłabienie przekroju |
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Liczba śrub w przekroju |
3 |
szt |
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Δ = |
2 |
Npk = |
N x (Ap/An+p)= |
893,12 |
kN |
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Przekrój netto |
z otworami |
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An = |
tn x (sn -(d+Δ)) |
cm2 |
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An = |
16,4 |
cm2 |
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Przekrój bruto |
bez otworów |
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At = |
tn x sn |
cm2 |
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At = |
23 |
cm2 |
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Sprawdzenie pola przekroju |
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Aψ = |
An |
0,8 x Rm |
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Re |
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Aψ = |
21,09 |
cm2 |
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ψ = |
Aψ |
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At |
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ψ = |
0,9 |
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6 |
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Naprezenia w przekroju bruto |
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σ = |
Npk |
|
MPa |
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At |
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σ = |
19,42 |
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MPa |
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Napreżenia osłabionego elementu |
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σel = |
σ |
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MPa |
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ψ |
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σel = |
21,17 |
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MPa |
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Warunek spełniony |
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σel |
< |
fd |
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21,17 |
< |
21,5 |
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a1 |
30 |
[mm] |
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d |
20 |
[mm] |
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a2 |
50 |
[mm] |
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d |
20 |
[mm] |
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a1 |
30 |
[mm] |
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7 |