kratownica-mechanika stosowania

mihai · Post autor: **mihai** » 6 cze 2010, o 22:40

sprawdź narazie \(\displaystyle{ \sum_{}^{} Fi_{y} = 0}\)
bo na tym drugim na razie myślę a wyslalem fix przez przypadek

solmech · Post autor: **solmech** » 6 cze 2010, o 22:58

To sa jakies chinskie obliczenia.

\(\displaystyle{ S8 = - \frac{ \sqrt{2} }{2} \cdot S10}\)

Wystarczy podstawic wartosc i wyjdzie -5

mihai · Post autor: **mihai** » 7 cze 2010, o 08:11

Wezel 6 .

\(\displaystyle{ \sum_{}^{} Fi_{y} = 0}\)
\(\displaystyle{ S10 \cdot \frac{ \sqrt{2} }{2} +S8=0}\)
\(\displaystyle{ S8=-S10 \cdot \frac{ \sqrt{2} }{2}}\)
\(\displaystyle{ S8=-7,07 \cdot \frac{ \sqrt{2} }{2}}\)
\(\displaystyle{ S8=\cdot \frac{ -7,07\sqrt{2} }{2}}\)
\(\displaystyle{ S8= \frac{10}{2}}\)
\(\displaystyle{ S8=-5}\)

\(\displaystyle{ \sum_{}^{} Fi_{x} = 0}\)
\(\displaystyle{ -S10\cdot \frac{ \sqrt{2} }{2}-S11+S5=0}\)

\(\displaystyle{ S5=S10\cdot \frac{ \sqrt{2} }{2}S11 / \cdot 2}\)
\(\displaystyle{ S5 \cdot 2=7,07 \cdot 1 -5 \cdot 2/2}\)
\(\displaystyle{ S5=7,07-10}\)
\(\displaystyle{ S5=-2,93}\)

strzałka jest w lewą stronę to pewnie będzie minus Myślę że to tak powinno byc -- 8 cze 2010, o 09:35 --Żeby nie wyjść z wprawy poprawiłem węzeł 8 mam nadzieje że dobrze i jeszcze zrobiłem węzeł 4 Mam nadzieje ze tym razem bedzie ok
WĘZEŁ 8

\(\displaystyle{ \sum_{}^{} Fi_{y} = 0}\)
\(\displaystyle{ -S1 \cdot \frac{ \sqrt{2} }{2} + B_{y} = 0 / \cdot (-1)}\)
\(\displaystyle{ S1 \cdot \frac{ \sqrt{2} }{2} - B_{y} = 0}\)
\(\displaystyle{ -S1 \cdot \frac{ \sqrt{2} }{2} = B_{y} / \cdot 2}\)
\(\displaystyle{ S1 \cdot 1 = 5 \cdot 2}\)

\(\displaystyle{ S1=10}\)

\(\displaystyle{ \sum_{}^{} Fi_{x} = 0}\)

\(\displaystyle{ B_{x}=0}\)

W4

\(\displaystyle{ \sum_{}^{} Fi_{y} = 0}\)
\(\displaystyle{ -S1 \cdot \frac{ \sqrt{2} }{2}-S2\cdot \frac{ \sqrt{2} }{2}=0}\)
\(\displaystyle{ S2\cdot \frac{ \sqrt{2} }{2}=-S1 \cdot \frac{ \sqrt{2} }{2}/ \cdot \frac{2}{\sqrt{2}}}\)
\(\displaystyle{ S2\cdot \frac{ \sqrt{2} }{2} \cdot \frac{2}{\sqrt{2}}=-S1 \cdot \frac{ \sqrt{2} }{2}\cdot \frac{2}{\sqrt{2}}}\)
\(\displaystyle{ S2 \cdot 1=-10 \cdot 1}\)

\(\displaystyle{ S2 =-10}\)

\(\displaystyle{ \sum_{}^{} Fi_{x} = 0}\)
\(\displaystyle{ -S3-S2 \cdot \frac{ \sqrt{2} }{2} -S1\cdot \frac{ \sqrt{2} }{2}=0}\)
\(\displaystyle{ S3=-S2 \cdot \frac{ \sqrt{2} }{2} -S1\cdot \frac{ \sqrt{2} }{2}/ \cdot 2}\)
\(\displaystyle{ S3 \cdot 2=-(-10) \cdot 1 -10\cdot 1/2}\)
\(\displaystyle{ S3 =5 -5}\)
\(\displaystyle{ S3 =0}\)

Jak mam żle wynik to sprawdzis czy dobrze rozpisałem?
Dziękuje za wytrwałość

solmech · Post autor: **solmech** » 8 cze 2010, o 22:55

Przepraszam, nie widzialem ze napisales wiadomosci. Mogles mi wyslac prywatna wiadomosc. Jutro na spokojnie sprawdze.
Pozdrawiam
Tomek

mihai · Post autor: **mihai** » 8 cze 2010, o 23:12

Spoko Myślałem że kajs wyjechales Bo nawe Online niewidziałem ciebie

solmech · Post autor: **solmech** » 10 cze 2010, o 14:26

S8 sie zgadza, S5 nie. Popraw.
Wezel 8 jest zle obliczyony, zapomniales o Bx.

mihai · Post autor: **mihai** » 10 cze 2010, o 14:46

W6

\(\displaystyle{ \sum_{}^{} Fi_{y} = 0}\)
\(\displaystyle{ S10 \cdot \frac{ \sqrt{2} }{2} +S8=0}\)
\(\displaystyle{ S8=-S10 \cdot \frac{ \sqrt{2} }{2}}\)
\(\displaystyle{ S8=-7,07 \cdot \frac{ \sqrt{2} }{2}}\)
\(\displaystyle{ S8=\cdot \frac{ -7,07\sqrt{2} }{2}}\)
\(\displaystyle{ S8= \frac{10}{2}}\)
\(\displaystyle{ S8=-5}\)

\(\displaystyle{ \sum_{}^{} Fi_{x} = 0}\)
\(\displaystyle{ -S10\cdot \frac{ \sqrt{2} }{2}-S11+S5=0}\)

\(\displaystyle{ S5=S10\cdot \frac{ \sqrt{2} }{2}S11 / \cdot 2}\)

\(\displaystyle{ S5 \cdot 2=7,07 \cdot 1 -5 \cdot 2/2}\)

\(\displaystyle{ S5 = \frac{7,07}{2} \frac{-5}{2}}\)

\(\displaystyle{ S5=3,53-5}\)
\(\displaystyle{ S5=-1,46}\)

teraz powinno byc dobrze -- 10 cze 2010, o 15:18 --WĘZEŁ 8
\(\displaystyle{ B _{x}=5}\)
\(\displaystyle{ B_{y}=5}\)
Nie wiem jak to zrobić ale spróbuje

\(\displaystyle{ \sum_{}^{} Fi_{y} = 0}\)
\(\displaystyle{ S1 \cdot \frac{ \sqrt{2} }{2} + B_{y} = 0}\)
\(\displaystyle{ S1 \cdot \frac{ \sqrt{2} }{2} = -B_{y}}\)
\(\displaystyle{ S1 \cdot \frac{ \sqrt{2} }{2} =-B_{y} / \cdot 2}\)
\(\displaystyle{ S1 \cdot 1 = -5 \cdot 2}\)

\(\displaystyle{ S1=-10}\)

\(\displaystyle{ \sum_{}^{} Fi_{x} = 0}\)
\(\displaystyle{ -S1 \cdot \frac{ \sqrt{2} }{2} + B_{x} = 0}\)
\(\displaystyle{ S1 \cdot \frac{ \sqrt{2} }{2} = B_{x}}\)
\(\displaystyle{ S1 \cdot \frac{ \sqrt{2} }{2} =B_{x} / \cdot 2}\)
\(\displaystyle{ S1 \cdot 1 = 5 \cdot 2}\)

\(\displaystyle{ S1=10}\)

solmech · Post autor: **solmech** » 10 cze 2010, o 18:48

Nadal S5 zle. Wystarczy podstawic obliczone przez nas sily.

mihai · Post autor: **mihai** » 10 cze 2010, o 19:35

\(\displaystyle{ \sum_{}^{} Fi_{y} = 0}\)
\(\displaystyle{ S10 \cdot \frac{ \sqrt{2} }{2} +S8=0}\)
\(\displaystyle{ S8=-S10 \cdot \frac{ \sqrt{2} }{2}}\)
\(\displaystyle{ S8=-7,07 \cdot \frac{ \sqrt{2} }{2}}\)
\(\displaystyle{ S8=\cdot \frac{ -7,07\sqrt{2} }{2}}\)
\(\displaystyle{ S8= \frac{10}{2}}\)
\(\displaystyle{ S8=-5}\)

\(\displaystyle{ \sum_{}^{} Fi_{x} = 0}\)

\(\displaystyle{ -S10-S11+S5=0}\)

\(\displaystyle{ S5=S10+S11}\)

\(\displaystyle{ S5=7,07+(-5)}\)

\(\displaystyle{ S5=2,07}\)

aaa 8 węzeł dobry?? zapomniałeś sprawdzić

solmech · Post autor: **solmech** » 10 cze 2010, o 20:47

Nie sprawdze dopoki nie obliczysz S5 poprawnie. Jak ma byc dobrze skoro raz masz 2.07 dla S5 a raz cos innego

mihai · Post autor: **mihai** » 10 cze 2010, o 21:53

\(\displaystyle{ \sum_{}^{} Fi_{x} = 0}\)
\(\displaystyle{ -S10\cdot \frac{ \sqrt{2} }{2}-S11+S5=0}\)

\(\displaystyle{ S5=S10\cdot \frac{ \sqrt{2} }{2}+S11}\)

\(\displaystyle{ S5=7,07\cdot \frac{ \sqrt{2} }{2}+(-5)}\)

\(\displaystyle{ S5=5-5}\)
\(\displaystyle{ S5=-0}\)

solmech · Post autor: **solmech** » 10 cze 2010, o 23:35

Gdybys podstawil dokladna wartosc, czyli \(\displaystyle{ 5 \cdot \sqrt{2}}\) wyszlo by dokladnie 0, i tak ma wlasnie byc.

mihai · Post autor: **mihai** » 11 cze 2010, o 09:45

Poprawiłem W6

\(\displaystyle{ \sum_{}^{} Fi_{y} = 0}\)
\(\displaystyle{ S10 \cdot \frac{ \sqrt{2} }{2} +S8=0}\)
\(\displaystyle{ S8=-S10 \cdot \frac{ \sqrt{2} }{2}}\)
\(\displaystyle{ S8=-7,07 \cdot \frac{ \sqrt{2} }{2}}\)
\(\displaystyle{ S8=\cdot \frac{ -7,07\sqrt{2} }{2}}\)
\(\displaystyle{ S8= \frac{10}{2}}\)

\(\displaystyle{ \Rightarrow S8=-5}\)

\(\displaystyle{ \sum_{}^{} Fi_{x} = 0}\)
\(\displaystyle{ -S10\cdot \frac{ \sqrt{2} }{2}-S11+S5=0}\)
\(\displaystyle{ S5=S10\cdot \frac{ \sqrt{2} }{2}+S11}\)
\(\displaystyle{ S5=7,07\cdot \frac{ \sqrt{2} }{2}+(-5)}\)
\(\displaystyle{ S5=5-5}\)

\(\displaystyle{ \Rightarrow S5=0}\)

-- 11 cze 2010, o 09:55 --

\(\displaystyle{ \sum_{}^{} Fi_{y} = 0}\)
\(\displaystyle{ S1 \cdot \frac{ \sqrt{2} }{2} + B_{y} = 0}\)

\(\displaystyle{ S1 \cdot \frac{ \sqrt{2} }{2}=-B_{y} / \cdot \frac{2}{\sqrt{2}}}\)
\(\displaystyle{ S1 \cdot \frac{ \sqrt{2} }{2} \cdot \frac{2}{\sqrt{2}} = - \frac{2}{\sqrt{2}} \cdot 5}\)
\(\displaystyle{ S1 \cdot 1= 5 \sqrt{2} \approx -7,07}\)

\(\displaystyle{ \Rightarrow S1 = -7,07}\)

\(\displaystyle{ \sum_{}^{} Fi_{x} = 0}\)

\(\displaystyle{ -S1 \cdot \frac{ \sqrt{2} }{2} + B_{x} = 0}\)

\(\displaystyle{ S1 \cdot \frac{ \sqrt{2} }{2}=B_{x} / \cdot \frac{2}{\sqrt{2}}}\)
\(\displaystyle{ S1 \cdot \frac{ \sqrt{2} }{2} \cdot \frac{2}{\sqrt{2}} = \frac{2}{\sqrt{2}} \cdot 5}\)
\(\displaystyle{ S1 \cdot 1= 5 \sqrt{2} \approx 7,07}\)

\(\displaystyle{ \Rightarrow S1 = 7,07}\)

-- 11 cze 2010, o 10:11 --

Poprawiłem także węzeł 4 w węźle 4 podstawiłem \(\displaystyle{ S1 = -7,07}\) czy mam podstawic na osi \(\displaystyle{ Y}\) \(\displaystyle{ S1=-7,07}\) a na osi \(\displaystyle{ X}\) \(\displaystyle{ S1=7,07}\) co?? bo nie wiem

\(\displaystyle{ \sum_{}^{} Fi_{y} = 0}\)
\(\displaystyle{ -S1 \cdot \frac{ \sqrt{2} }{2}-S2\cdot \frac{ \sqrt{2} }{2}=0}\)
\(\displaystyle{ S2\cdot \frac{ \sqrt{2} }{2}=-S1 \cdot \frac{ \sqrt{2} }{2}/ \cdot \frac{2}{\sqrt{2}}}\)
\(\displaystyle{ S2\cdot \frac{ \sqrt{2} }{2} \cdot \frac{2}{\sqrt{2}}=-S1 \cdot \frac{ \sqrt{2} }{2}\cdot \frac{2}{\sqrt{2}}}\)
\(\displaystyle{ S2 \cdot 1=-7,07 \cdot 1}\)

\(\displaystyle{ \Rightarrow S2 =-7,07}\)

\(\displaystyle{ \sum_{}^{} Fi_{x} = 0}\)
\(\displaystyle{ -S3-S2 \cdot \frac{ \sqrt{2} }{2} +S1\cdot \frac{ \sqrt{2} }{2}=0}\)
\(\displaystyle{ S3=-S2 \cdot \frac{ \sqrt{2} }{2} +S1\cdot \frac{ \sqrt{2} }{2}}\)
\(\displaystyle{ S3= -(-7,07) \cdot 0,7 +7,07 \cdot 0,7}\)
\(\displaystyle{ S3=5+5}\)

\(\displaystyle{ \Rightarrow S3=10}\)

solmech · Post autor: **solmech** » 13 cze 2010, o 14:10

\(\displaystyle{ S1=-5 \sqrt{2}}\)

\(\displaystyle{ S2=5 \sqrt{2}}\)

\(\displaystyle{ S3=-10}\)

\(\displaystyle{ S4=-10}\)

\(\displaystyle{ S5=0}\)

\(\displaystyle{ S6=5 \sqrt{2}}\)

\(\displaystyle{ S7=-10}\)

\(\displaystyle{ S8=-5}\)

\(\displaystyle{ S9=-5}\)

\(\displaystyle{ S10=5 \sqrt{2}}\)

\(\displaystyle{ S11=-5}\)

\(\displaystyle{ S12=-5}\)

Powinno sie zgadzac.

mihai · Post autor: **mihai** » 13 cze 2010, o 14:31

W6 mam dobrze
W8 mam dorze

\(\displaystyle{ \sum_{}^{} Fi_{y} = 0}\)
\(\displaystyle{ S1 \cdot \frac{ \sqrt{2} }{2} + B_{y} = 0}\)

\(\displaystyle{ S1 \cdot \frac{ \sqrt{2} }{2}=-B_{y} / \cdot \frac{2}{\sqrt{2}}}\)
\(\displaystyle{ S1 \cdot \frac{ \sqrt{2} }{2} \cdot \frac{2}{\sqrt{2}} = - \frac{2}{\sqrt{2}} \cdot 5}\)
\(\displaystyle{ S1 \cdot 1= -5 \sqrt{2}}\)
\(\displaystyle{ S1= -5 \sqrt{2}}\)

\(\displaystyle{ \sum_{}^{} Fi_{x} = 0}\)

\(\displaystyle{ -S1 \cdot \frac{ \sqrt{2} }{2} + B_{x} = 0}\)

\(\displaystyle{ S1 \cdot \frac{ \sqrt{2} }{2}=B_{x} / \cdot \frac{2}{\sqrt{2}}}\)
\(\displaystyle{ S1 \cdot \frac{ \sqrt{2} }{2} \cdot \frac{2}{\sqrt{2}} = \frac{2}{\sqrt{2}} \cdot 5}\)
\(\displaystyle{ S1 \cdot 1= 5 \sqrt{2}}\)
\(\displaystyle{ S1 = 5 \sqrt{2}}\)

W4 jest tez ok

\(\displaystyle{ \sum_{}^{} Fi_{y} = 0}\)
\(\displaystyle{ -S1 \cdot \frac{ \sqrt{2} }{2}-S2\cdot \frac{ \sqrt{2} }{2}=0}\)
\(\displaystyle{ S2\cdot \frac{ \sqrt{2} }{2}=-S1 \cdot \frac{ \sqrt{2} }{2}/ \cdot \frac{2}{\sqrt{2}}}\)
\(\displaystyle{ S2\cdot \frac{ \sqrt{2} }{2} \cdot \frac{2}{\sqrt{2}}=-S1 \cdot \frac{ \sqrt{2} }{2}\cdot \frac{2}{\sqrt{2}}}\)
\(\displaystyle{ S2 \cdot 1=-(-5 \sqrt{2})}\)
\(\displaystyle{ S2 =5 \sqrt{2}}\)

\(\displaystyle{ \sum_{}^{} Fi_{x} = 0}\)
\(\displaystyle{ -S3-S2 \cdot \frac{ \sqrt{2} }{2} +S1\cdot \frac{ \sqrt{2} }{2}=0}\)
\(\displaystyle{ S3=-S2 \cdot \frac{ \sqrt{2} }{2} +S1\cdot \frac{ \sqrt{2} }{2}}\)
\(\displaystyle{ S3= - 5 \sqrt{2}\cdot \frac{ \sqrt{2} }{2} -5 \sqrt{2}\cdot \frac{ \sqrt{2} }{2}}\)
\(\displaystyle{ S3=-5-5}\)
\(\displaystyle{ S3=-10}\)

Dziękuje bardzo za pomoc przy kratownicach