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Ćwiczenie projektowe nr 1 |
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Zaprojektować słup drewniany ściskany osiowo , o przekroju złożonym, przy nastepujacych założeniach |
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1.1 Dane: |
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wstaw dane: |
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obliczeniowa siła ściskająca |
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N = |
105 |
kN |
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długość teoretyczna słupa |
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l = |
5,4 |
m |
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współczynnik długości wyboczeniowej: |
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my = |
1,5 |
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mz = |
1,0 |
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klasa użytkowania |
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2 |
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klasa trwania obciążenia: |
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k mod = |
0,7 |
wstaw odpowiednią wartość z tabeli |
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stałe |
100 |
1 |
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100 |
% |
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klasa drewna |
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C 35 |
wstaw odpowiednie wartość z normy |
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cechy mechaniczne drewna: |
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wytrz. na zginanie |
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f m,k = |
35 |
Mpa |
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wytrz. na ściskanie |
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f e,o,u = |
25 |
Mpa |
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średni moduł spręzystości |
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E0,mean = |
13 |
Mpa |
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gęstość charak. Drewna |
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ρ= |
400 |
kg/m3 |
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połączenie na gwoździe |
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Wartości obliczeniowe |
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f m,d = |
f m,k * |
k mod |
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35 |
0,7 |
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18,85 |
Mpa |
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1,885 |
kN/cm2 |
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g M |
1,3 |
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g M = |
1,3 |
wsp.materiałowy dla materiałów drewnopochodnych |
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k mod = |
0,7 |
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f c,o,d = |
f c,o,k * |
k mod |
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25 |
0,7 |
= |
13,46 |
Mpa |
= |
1,346 |
kN/cm2 |
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g M |
1,3 |
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1.2 Dobór przekroju |
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Wstępne ustalenie napreżeń w elemencie ściskanym osiowo |
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δc,o,d = |
N |
< |
f c,o,d |
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k c * |
Ac |
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k c = |
0,4 - 0,5 |
wsp. wyboczeniowy |
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k c = |
0,4 |
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Ac - |
przekrój obliczeniowy |
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Ac > |
N |
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105 |
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195,00 |
cm2 |
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k c * |
f c,o,d |
0,40 |
1,35 |
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Przyjeto przekrój : |
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z |
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7,5 |
10 |
7,5 |
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25 |
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5 |
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15 |
y |
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5 |
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25 |
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Ac = |
475 |
cm2 |
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2. Przyjęcie wielkość i rozstaw gwoździ |
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- średnica gwoździa |
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t min = |
5 |
cm |
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d = |
0,45 |
0,83 |
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przyjeto |
d = |
5 |
mm |
tabela PN Z-7.4.1-2 |
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d = |
6x125 |
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7 * d |
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35 |
mm |
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t max = |
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13 d - 30 |
*ρ |
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35 |
mm |
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400 |
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- rozstaw gwoździ |
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Minimalne rozstawy i odległości gwoździ odczytano z tablicy PN 7.4.1.1 |
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- odległość miedzy gwoździami |
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< |
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dla |
d |
> |
5,0 |
mm |
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> |
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7,00 |
a1 = |
5 |
7 |
cos a |
5 |
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60,0 |
mm |
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przyjeto |
a1 = |
80 |
mm |
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- rozstaw miedzy gwoździami |
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a2 = |
5*d |
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25 |
mm |
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- odległość od końca nie obciążonego |
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a3 = |
10 |
5,0 |
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50,0 |
mm |
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- odległość od krawędzi nieobciążonej |
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a4 = |
5 |
5 |
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25 |
mm |
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3. Wyboczenie słipa w płaszczyznie prostopadłej do szwów: ( oś y ) |
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3.1 Współczynnik redukcyjny gi |
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Jy = |
suma Jyi + |
gy * |
suma Ai * |
eyi2 |
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gy = |
1 |
= |
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1 |
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= |
0,983 |
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1 + |
ρ2 |
* E 0,05 |
* A iy |
* S1 |
1 |
3,14 |
800 |
125,00 |
4 |
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k l 2 |
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773,1 |
540 |
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S1 = |
a1 |
= |
80 |
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40 |
mm |
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2 |
2 |
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S1 = |
4 |
cm |
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E 0,05 = |
8 Gpa |
= |
8000 |
Mpa = |
800 |
kN/cm2 |
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A iy = |
25 |
5 |
= |
125 |
cm2 |
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k = |
2/3 kser |
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k - |
moduł podatności złacza |
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kser - |
moduł podatności gwoździa na 1 cięcie wg 7.2 |
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kser = |
ρu1.5 * |
d 0.8 |
= |
400 |
5 |
= |
1159,6 |
N/mm |
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25 |
25 |
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k = |
2 |
1159,6 |
= |
773,1 |
N/mm |
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3 |
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3.2 Efektywny moment bezwładności |
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Jefy = |
suma Jyi + |
gy * |
suma Ai * |
eyi2 |
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Jefy = |
2 |
7,5 |
15 |
2 |
25 |
5 |
2 |
0,98 |
125 |
10 |
= |
29309 |
cm4 |
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12 |
12 |
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3.3 Smukłość efektywna |
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lef,y = |
my * |
l |
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29739,5833333333 |
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ief,y |
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my = |
1,5 |
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l = |
540 |
cm |
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ief,y = |
Jefy |
= |
29309 |
= |
7,86 |
cm |
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Ac |
475 |
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lef,y = |
1,5 |
540 |
= |
103,12 |
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7,86 |
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3.4 Współczynnik wyboczeniowy |
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naprężenie krytyczne |
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sc,cvity = |
p2 |
* E 0,05 |
= |
3,14 |
8000 |
= |
7,42 |
Mpa |
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lef,y2 |
103,12 |
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smukłość sprawdzona |
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lred,y = |
f e,o,u |
= |
25 |
= |
1,84 |
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sc,cvity |
7,42 |
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współczynnik ky |
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ky = |
0,5[1+bc(lred.y - 0,5) + lerd.y2] |
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bc - wsp dotyczacy prostoliniowości elementów |
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bc = |
0,2 |
dla drewna litego |
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ky = |
2,32 |
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współczynnik wyboczeniowy kc.y |
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kc.y = |
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1 |
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ky + |
ky2 - |
lred2,y |
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kc.y = |
|
1 |
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= |
0,27 |
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2,32 |
2,32 |
1,84 |
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3.5 Sprawdzenie napręzeń w przekroju |
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sc,o,d = |
N |
< |
fc,o,d = |
1,346 |
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kc.y * |
Ac * |
f c,o,d |
Mpa |
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sc,o,d = |
105 |
= |
0,83 |
< |
1,35 |
Mpa |
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0,27 |
475,00 |
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4. Wyboczenie słipa w płaszczyznie równoległej do szwów: ( oś z ) |
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4.1 Współczynnik redukcyjny gi |
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Jz = |
suma Jzi + |
gz * |
suma Ai * |
ezi2 |
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gz = |
1 |
= |
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1 |
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= |
1,02 |
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1 + |
p2 |
* E 0,05 |
* A iż |
* S1 |
1 |
3,14 |
800 |
112,5 |
4 |
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k l 2 |
|
773,1 |
540 |
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S1 = |
a1 |
= |
80 |
= |
40 |
mm |
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2 |
2 |
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S1 = |
4 |
cm |
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E 0,05 = |
8 Gpa |
= |
8000 |
Mpa = |
800 |
kN/cm2 |
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A iz = |
7,5 |
15 |
= |
112,5 |
cm2 |
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k = |
2/3 kser |
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k - |
moduł podatności złacza |
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kser - |
moduł podatności gwoździa na 1 cięcie wg 7.2 |
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kser = |
ru1.5 * |
d 0.8 |
= |
400 |
5 |
= |
1159,6 |
N/mm |
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25 |
25 |
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k = |
2 |
1159,6 |
= |
773,1 |
N/mm |
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3 |
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4.2 Efektywny moment bezwładności |
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Jefz = |
suma Jzi + |
gz * |
suma Ai * |
ezi2 |
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Jefy = |
2 |
15 |
7,5 |
2 |
112,5 |
8,75 |
5 |
25 |
2 |
1,02 |
= |
31573 |
cm4 |
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12 |
12 |
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4.3 Smukłość efektywna |
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31302,0833333333 |
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lef,z = |
mz * |
l |
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ief,z |
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mz = |
1,0 |
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l = |
540 |
cm |
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ief,z = |
Jefz |
= |
31573 |
= |
8,15 |
cm |
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Ac |
475 |
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lef,z = |
1,0 |
540 |
= |
66,23 |
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8,15 |
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4.4 Współczynnik wyboczeniowy |
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|
naprężenie krytyczne |
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sc,cvitz = |
p2 |
* E 0,05 |
= |
3,14 |
8000 |
= |
17,98 |
Mpa |
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lef,z2 |
66,23 |
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smukłość sprawdzona |
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lred,z = |
f e,o,u |
= |
25 |
= |
1,18 |
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sc,cvitz |
17,98 |
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współczynnik kz |
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kz = |
0,5[1+bc(lred.z - 0,5) + lerd.z2] |
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bc - wsp dotyczacy prostoliniowości elementów |
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bc = |
0,2 |
dla drewna litego |
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ky = |
1,26 |
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współczynnik wyboczeniowy kc.z |
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kc.z = |
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1 |
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kz + |
kz2 - |
lred2,z |
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kc.z = |
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1 |
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= |
0,58 |
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1,26 |
1,26 |
1,18 |
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3.5 Sprawdzenie napręzeń w przekroju |
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sc,o,d = |
N |
< |
fc,o,d = |
1,346 |
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kc.z * |
Ac |
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sc,o,d = |
105 |
= |
0,38 |
< |
1,35 |
Mpa |
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0,58 |
475,00 |
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nośność połączeń |
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Vdz= |
3,00 |
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S1z= |
984,37 |
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beta= |
1 |
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Fdz= |
0,38 |
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t2= |
66,5 |
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t1= |
50 |
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Rd1= |
2724,44 |
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fh,1,k= |
20,24 |
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Rd2= |
3623,51 |
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fh,1,d= |
10,90 |
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Rd3= |
1337,16 |
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d= |
5 |
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Rd4= |
1474,98 |
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Myd= |
9091,83 |
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Rd5= |
1189,86 |
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Rd6= |
1094,93 |
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Rd,min= |
1094,93 |
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Fdz= |
380,40 |
< |
Rd,min= |
1094,93 |
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Vdy= |
6,54 |
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S1y= |
1250 |
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Fdy= |
1,10 |
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Fdy= |
1095,85 |
> |
Rd,min= |
1094,93 |
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Ćwiczenie projektowe nr 2 |
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przekrój α-α |
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Nośność belki stropowej |
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25 |
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1. Schemat statyczny |
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α |
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7,5 |
10 |
7,5 |
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h1= |
5 |
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L= |
540 |
cm |
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α |
e1= |
10 |
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h2= |
15 |
25 |
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h1= |
5 |
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2. Określenie cech geometrycznych i parametrów mechanicznych drewna |
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- pole całkowite |
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Ac= |
475,0 |
cm2 |
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- moduł podatności gwoździa na 1 cięcie |
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kser= |
1159,65 |
N/mm |
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- moduł podatności złącza |
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k= |
773,1 |
N/mm |
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- wytrzymałość na rozciąganie |
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ft,o,d= |
1,29 |
kN/cm2 |
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1,29230769230769 |
- klasa drewna |
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C35 |
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- efektywny moment bezwładności |
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Jef,y= |
29739 |
cm4 |
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- współczynnik redukcyjny |
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γy= |
0,983 |
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- wytrzymałość na zginanie |
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fm,k= |
35 |
MPa |
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- obliczeniowa wytrzymałość na zginanie |
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fm,d= |
1,885 |
kN/cm2 |
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- gęstość charakterystyczna drewna |
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ρ= |
400 |
kg/m3 |
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- obliczeniowa wytrzymałość na ściskanie |
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fc,o,d= |
1,346 |
kN,cm2 |
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3. Nośność belki ze względu na zginanie |
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3.1 Naprężenia krawędziowe środnika |
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σh=(Mo*(h2/2))/Jef,y<fm,d σh= |
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7472,88 |
15 |
= |
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29739 |
*2 |
1,885 |
kN/cm2 = |
18,85 |
MPa |
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Mo=(fm,d*Jef,y)/(h2/2) |
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Mo= |
1,885 |
29739 |
= |
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15 |
/2 |
7472,88 |
kNcm = |
74,73 |
kNm |
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Mo=(ql2)/8=>qo=(8*Mo)/l2 Mo= |
8 * |
7472,88 |
= |
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540 |
*2 |
0,2050 |
kN/cm = |
20,50 |
kN/m |
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3.2 Naprężenia krawędziowe półki |
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σf=σ1+σmf< fm,d |
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3.3 Naprężenia ściskające w osi pasów |
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σ1=(Mo*e1)/Jef,y*γ<fc,o,d σ1= |
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4073,43 |
10,000 |
= |
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29739 |
0,983 |
1,346 |
kN/cm2 = |
13,46 |
MPa |
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Mo=(fc,o,d*Jef,y)/(e1*γ) |
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Mo= |
1,346 |
29739 |
= |
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10 |
0,983 |
4073,43 |
kNcm = |
40,73 |
kNm |
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Mo=(ql2)/8=>qo=(8*Mo)/l2 Mo= |
8 * |
4073,43 |
= |
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540 |
*2 |
0,1118 |
kN/cm = |
11,18 |
kN/m |
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3.4 Naprężenia rozciągające w osi pasów |
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σ1=(Mo*e1)/Jef,y*γ<ft,o,d σ1= |
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3910,50 |
10,000 |
= |
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29739 |
0,983 |
1,292 |
kN/cm2 = |
12,92 |
MPa |
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Mo=(fc,o,d*Jef,y)/(e1*γ) |
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Mo= |
1,292 |
29739 |
= |
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10 |
0,983 |
3910,50 |
kNcm = |
39,10 |
kNm |
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Mo=(ql2)/8=>qo=(8*Mo)/l2 Mo= |
8 * |
3910,50 |
= |
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540 |
*2 |
0,1073 |
kN/cm = |
10,73 |
kN/m |
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3.5 Naprężenia krawędziowe od zginania |
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σmf=(Mo*h1)/Jef,y*2<fm,d σmf= |
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11405,61 |
5,000 |
= |
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29739 |
*2 |
0,959 |
kN/cm2 = |
9,59 |
MPa |
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Mo=(fm,d*(2*Jef,y))/(h1*γ) |
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Mo= |
1,885 |
29739 |
= |
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5 |
0,983 |
11405,61 |
kNcm = |
114,06 |
kNm |
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Mo=(ql2)/8=>qo=(8*Mo)/l2 Mo= |
8 * |
11405,61 |
= |
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540 |
*2 |
0,3129 |
kN/cm = |
31,29 |
kN/m |
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3.6 Naprężenia łączne na krawędzi pasa |
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σf=σmf+σ1<fm,d |
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qol2/8*h1+ |
qol2/8*e1 |
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2*Jef,y |
Jef,y |
γ < fm,d= |
1,885 |
/ * Jef,y |
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qol2 |
* h1 + |
qol2 |
* e1 *γ = |
1,885 |
29739 |
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16 |
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8 |
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qo* |
540 |
2 |
540 |
2 |
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16 |
5 |
+ go 8 |
10 |
0,983 |
= |
56046,5769230769 |
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go * |
449351,708660604 |
= |
56046,5769230769 |
/ |
449351,708660604 |
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go= |
0,12472763726689 |
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Ćwiczenie projektowe nr 3 |
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Dźwigar klejony |
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1. Dane |
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- rozpiętość dźwigara |
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L= |
13 |
m |
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- klasa użytkowania |
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2 |
kmod= |
0,8 |
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γf= |
1,3 |
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- klasa drewna |
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C30 |
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- sprężystość |
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Eo,mean= |
12 |
Gpa = |
1200 |
kN/cm2 |
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Gmean= |
0,75 |
Gpa = |
75 |
kN/cm2 |
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E0,05= |
8 |
Gpa = |
800 |
kN/cm2 |
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- wytrzymałość na zginanie |
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fm,k= |
30 |
MPa = |
3 |
kN/cm2 |
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- wytrzymałość na ściskanie w poprzek włókien |
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fc,90,k= |
5,7 |
MPa = |
0,57 |
kN/cm2 |
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- wytrzymałość na ścinanie |
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fv,k= |
3 |
MPa = |
0,3 |
kN/cm2 |
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- wytrzymałość na rozciąganie w poprzek włókien |
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ft,90,k= |
0,4 |
MPa = |
0,04 |
kN/cm2 |
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2. Obliczeniowe wytrzymałości |
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- na zginanie |
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fm,g,d= |
(fm,k*kmod) |
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3 |
0,8 |
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γf |
= |
1,3 |
= |
1,85 |
kN/cm2 |
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- na ściskanie w poprzek włókien |
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fc,90,g,d= |
(fc,90,k*kmod) |
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0,57 |
0,8 |
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γf |
= |
1,3 |
= |
0,351 |
kN/cm2 |
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- na ścinanie |
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fv,g,d= |
(fv,k*kmod) |
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0,3 |
0,8 |
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γf |
= |
1,3 |
= |
0,185 |
kN/cm2 |
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- na rozciąganie w poprzek włókien |
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ft,90,g,d= |
(ft,90,k*kmod) |
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0,04 |
0,8 |
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γf |
= |
1,3 |
= |
0,025 |
kN/cm2 |
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3. Wstępny dobór przekroju |
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5 |
o |
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h= |
1,10 |
m |
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hp= |
0,53 |
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0,002916666666667 |
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L= |
13 |
m |
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h=(1/12 - 1/17)*L |
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h= |
1,08333333333333 |
0,76 |
przyjęto |
h= |
1,1 |
m |
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b> |
8 cm |
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h/b<10 |
0,08 |
m = |
8,5 |
cm |
przyjęto |
b= |
10 |
cm |
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hp=h-(0,5*L*tgα) |
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hp= |
0,53 |
m = |
53 |
cm |
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tgα= |
0,0875 |
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GL 30 |
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0,2h |
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C30 |
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0,6h |
C30 |
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h= |
110 |
cm |
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0,2h |
C30 |
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10 |
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4. Ustalenie przekroju obliczeniowego |
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Xo= |
l*hp |
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2*h |
= |
314 |
cm |
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hxo=hp+Xo*tgα |
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hxo= |
81 |
cm |
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- Wskaźnik wytrzymałości |
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Wxo= |
b*h2 |
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10 |
110 |
2 |
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6 |
= |
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6 |
= |
20166,6666666667 |
cm3 |
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- Moment bezwładności |
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Jxo= |
b*h3 |
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10 |
110 |
3 |
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12 |
= |
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12 |
= |
1109166,66666667 |
cm4 |
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5. Obciążenia obliczeniowe i charakterystyczne |
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Rodzaj obciążenia |
Obciążenie charakterystyczne |
Współczynnik obciążeniowy |
Obciążenie obliczeniowe |
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[-] |
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1) pokrycie dachowe |
0,233 |
1,1 |
0,257 |
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- płyta TRIMOTERM SNV 100 |
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2) płatwie (założono przekrój 120x500, ciężar drewna wg tabl. Z1-1 lp.6 w PN-82/B-02001) |
0,08 |
1,1 |
0,088 |
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3) ciężar własny dźwigara (przyjęto na podstawie p.4.2 – wzór Z2-1 z załącznika 2 PN-82/B-02001) |
0,18 |
1,1 |
0,2 |
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Σ gi |
0,495 |
1,1 |
0,545 |
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1) obciążenie śniegiem wg. |
0,72 |
1,5 |
1,08 |
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(PN80/B-02010) |
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Strefa śniegowa II Qk= 0,9 kN/m2 |
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2) obciążenie wiatrem zgodnie z normą PN77/B-2011 pominięto |
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Obciążenie całkowite |
1,215 |
1,34 |
1,625 |
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- obciążenie obliczeniowe |
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q= |
0,5 |
kN/m |
p= |
0,7 |
kN/m |
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qo=q*1,1+p*1,5 |
|
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qo= |
1,625 |
kN/m |
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|
- obciążenie charakterystyczne |
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q |
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1,2 |
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qk= |
1,1 |
+ p qk= |
1,1 + |
0,7 |
= |
1,82 |
kN/m |
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6. Siły w przekroju obliczeniowym |
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Qxo=qo*(0,5*L - Xo) |
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Qxo= |
1,625 |
* 0,5 * |
13 |
3,14 |
= |
5,46 |
kN |
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Mxo= |
qo*xo |
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1,625 |
3,14 |
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2 |
* (L-xo) |
Mxo= |
2 |
* |
13 |
3,14 |
= |
25,1533591260658 |
kNm |
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|
7. Naprężenia normalne w przekroju obliczeniowym |
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- włókna dolne |
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σm,o,d=(1+4tg2α)* |
|
Mxo |
=(1+4* |
0,09 |
2 * |
2515,33591260658 |
= |
0,13 |
kN/cm2 |
< |
fm,d= |
1,85 |
kN/cm2 |
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Wxo |
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|
20166,6666666667 |
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- włókna górne |
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sinα |
0,087155742747658 |
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fm,α,d= |
fm,d |
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|
1,85 |
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cosα |
0,996194698091746 |
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fm,d |
*sinα2+cosα2 |
|
fm,α,d = |
1,85 |
0,087155742747658 |
0,996194698091746 |
= |
1,79 |
kN/cm2 |
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fc,90,d |
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|
0,351 |
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σm,α,d=(1+4tg2α)* |
|
Mxo |
=(1+4* |
0,09 |
2 * |
2515,33591260658 |
= |
0,13 |
kN/cm2 |
< |
fm,α,d= |
1,79 |
kN/cm2 |
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Wxo |
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|
20166,6666666667 |
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8. Stateczność giętna przy zginaniu w przegroju obliczeniowym |
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σm,0,d<kcrit*fm,d |
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- smukłość sprawdzona |
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ld*hx*fm,d |
|
Eo, mean |
|
1200 |
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|
λred,m= |
π*b2*Eo,oo5 * |
|
G mean |
λred,m= |
3,14 |
10 |
800 |
75 |
= |
0,68 |
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gdzie: |
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rozstaw płatwi |
|
ld= |
300 |
cm |
kcrit= |
1,56-0,75*λred,m |
|
kcrit= |
1,00 |
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σm,0,d= |
0,13 |
< |
1,85 |
= |
1,00 |
1,85 |
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9. Naprężenia w strefie kalenicowej |
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h1=h-(0,5*h)*tgα |
|
h1= |
110 |
- 0,5 * |
110 |
0,0875 |
= |
105 |
cm |
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55 |
55 |
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h= |
110 |
cm |
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h1= |
105 |
cm |
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110 |
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|
- naprężenia normalne wzdłuż włókien |
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Mkmax= |
0,125* |
go * |
L2 |
Mkmax= |
0,125 * |
1,625 |
13 |
2 = |
34,328125 |
kNm |
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|
|
|
|
- wskaźnik wytrzymałości |
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bw*h2 |
|
10 |
110 |
2 |
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Wk= |
6 |
Wk= |
6 |
= |
20166,6666666667 |
cm3 |
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|
|
- naprężenia |
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σm,d= |
(1+1,4* |
0,0875 |
+ 5,4 * |
0,0875 |
3432,8125 |
|
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|
20166,6666666667 |
= |
0,179081999690242 |
kN/cm2 |
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|
0,179081999690242 |
kN/cm2 |
< |
kr= |
1,85 |
kN/cm2 |
|
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|
|
- naprężenia rozciągające prostopadle do włókien w strefie ściskanej |
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gdzie: |
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kdis= |
1,4 |
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|
|
- objętość odnieniona stała |
|
|
Vo= |
0,01 |
m3 |
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|
|
- objętość strefy kalenicowej |
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1,1 |
1,05 |
1,1 |
0,1 |
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V= 2* |
2 |
* |
2 * |
= |
0,12 |
m3 |
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Vb= 2* |
1,1 |
1,05 |
13 |
0,1 |
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2 |
* |
2 * |
= |
1,40 |
m3 |
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V= |
0,12 |
< |
2 |
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3 |
1,40 |
= |
0,932481868526321 |
m3 |
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|
σm,o,d= |
0,13 |
< |
0,132647315164069 |
|
|
!!!!! |
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|
- naprężenia rozciągające dla dźwigara dwutrapezowego |
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|
3432,8125 |
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|
σt,90,d= |
0,2* |
0,0875 |
20166,6666666667 |
= |
0,002978500936464 |
kN/cm2 |
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σt,90,d= |
0,002978500936464 |
< |
0,132647315164069 |
kN/cm2 |
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Warunek spełniony |
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10. Naprężenia ścinające na podporze |
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- siła poprzeczna |
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Qmax= |
1,625 |
13 |
= |
10,5625 |
kN |
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2 |
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|
- moment statyczny |
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Sy= |
10 |
53 |
53 |
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2 |
4 |
= |
3528,81075567342 |
cm3 |
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Jy= |
10 |
53 |
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12 |
|
= |
124996,049447816 |
cm4 |
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|
- naprężenia |
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τv= |
10,5625 |
3528,81075567342 |
|
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|
124996 |
10 |
= |
0,029819393309996 |
kN/cm2 |
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τv= |
0,030 |
< |
0,185 |
kN/cm2 |
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|
Warunek spełniony |
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11. Sprawdzenie ugięć |
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|
dla klasy 2: |
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|
- obciążenie stałe |
|
|
kdef= |
0,8 |
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|
- obciążenie zmienne |
|
|
kdef= |
0,25 |
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|
- ugięcia doraźne ( z uwzględnieniem ścinania) |
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1,1 |
2 |
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1+19,2* |
13 |
|
= |
2,03 |
* UM |
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Uinst= |
UM* |
0,15+0,85* |
|
0,53 |
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1,1 |
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|
- ugięcie od zginania |
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Jk= |
10 |
110 |
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12 |
|
= |
1109166,66666667 |
cm4 |
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|
# ugięcie od obciążenia stałego |
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UgM= |
5* |
1,6 |
13 |
4 |
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384* |
1200 |
1109167 |
= |
0,45 |
cm |
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|
# ugięcie od obciążenia zmiennego |
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UpM= |
5* |
1,2 |
13 |
4 |
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384* |
1200 |
1109167 |
= |
0,34 |
cm |
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|
- ugięcie kalenicowe |
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Ufin= |
2,03 |
0,45 |
(* 1 + |
0,8 |
2,03 |
0,34 |
(* 1+ |
0,25 |
) = |
2,52 |
cm |
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|
Unet,fin= |
L/200 = |
1300 |
/ 200 = |
4,33333333333333 |
cm |
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Ufin= |
2,52 |
cm |
< |
Unet,fin= |
4,33333333333333 |
cm |
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Warunek spełniony |
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