Statyka - ukł. sił równoległych - zła odp.??

Konstrukcje inżynierskie: kratownice, belki, ramy i inne.
Jake
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Statyka - ukł. sił równoległych - zła odp.??

Post autor: Jake »

Witam,

Zad.7. : Na rysunku 5.14 (pod spodem link) pokazana jest belka obciążona dźwigiem podnoszącym cięzar \(\displaystyle{ P=10 ^{4} N}\). Cięzar dźwigu wynosi \(\displaystyle{ 5*10 ^{4}N}\) i leży na prostej CD. Wyznaczyć reakcje podpór w pktach A i B, pomijając ciężar belki.

Pod spodem link do obrazka, tam także odpowiedzi podane przez skrypt (Engel, Giergiel) Nr zad.: 7.

Męczyłem sie juz nie wiem ile, by otrzymać wynik i nic, a na dobitke podane odpowiedzi sprawdzają sie tylko w 1 z 4 ułożonych przeze mnie równaniach ...



Z góry wielkie dzięki za jakieś objaśnienie, wskazówki!! (mam wielką nadzieje, że pojawi sie tu jakaś odp. ) Pozdr.
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miki999
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Statyka - ukł. sił równoległych - zła odp.??

Post autor: miki999 »

Wymyśliłem coś takiego.

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Możemy zapisać równania równowagi momentów sił względem pkt. \(\displaystyle{ P_1}\) i \(\displaystyle{ P_2}\). Zapomniałem narysować jednego pręta, ale dla momentów nie ma to większego znaczenia.

Mamy 2 niewiadome, 2 równania- możemy wyznaczyć \(\displaystyle{ P_{1_y}\ oraz\ P_{2_y}}\)


Ponadto możemy poniższą belkę rozdzielić na:
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Oznaczenia dobierz sobie sam. Do układu po lewej możemy zapisać 3 równania równowagi i do układu po prawej również.

Nie sprawdzałem wyniku (sposobu zresztą też) i jeżeli to jakieś zad. dom. to lepiej sprawdź, bo nie jestem ekspertem z mechaniki
Ewentualnie jeżeli ktoś bardziej kompetentny to przeczyta, to niech zweryfikuje.


Pozdrawiam.
Jake
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Posty: 77
Rejestracja: 8 gru 2008, o 20:49
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Statyka - ukł. sił równoległych - zła odp.??

Post autor: Jake »

Mom. utwierdzenia zapomniałeś/pominąłeś.

ufff.... poszło , a tyle się męczyłem..

Wynik jest częściowo błędny (wg mnie, moze ktoś jeszcze skomentuje) tzn.
\(\displaystyle{ R _{A}=53750 N, R _{B} =6250N, M _{u} =20,5*10 ^{4}}\)

Dlaczego wydaje mi się, że Twoja metoda jest prawidłowa (oprócz tego, że logiczna)? Bo jak liczyłem dla całości:
\(\displaystyle{ M _{(A)}=M _{u} +12R _{B}-(P+Q _{dźw})4-P*4}\) to po wstawieniu wartości mom. utwierdzenia i reak.B zgadzało sie (liczyłem też to inaczej, wyznaczając odl. wypadkowej Q i P i też było ok), natomiast 2. równanie (kierunek Oy, sił poziomych nie ma):
\(\displaystyle{ R _{A}+R _{B}=Q+P}\) - nie zgadzało sie , czyli prawdopodobnie z tą siłą reakcji A było coś nie tak (powinno z tego wyjść właśnie 53750..

Dzieki wielkie. ALeż na tym czasu straciłem.. szkoda życią.. nast. razem szybciej uderze do Mózgów. Pozdr.

P.S. Wszystkie moje równ. były prawidłowe ( o ile to jest dobrze ) - przynajmniej tyle z tego pożytku..
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