Name: hydraulic hoist force analysis
Uploaded: 2023-09-22T11:49:25.000Z
Duration: 1 min 26 s
Channel: JoVE

Weźmy pod uwagę wciągnik hydrauliczny podtrzymujący ładunek.
Zakładając uproszczone schematyczne przedstawienie tej konstrukcji ramy, jaka siła działa na pręty BD i BF, pod warunkiem, że znane są wymiary pręta i ciężar obciążenia?
Tutaj BF i BD są dwoma elementami siłowymi, podczas gdy EFG i EDC są elementami o wielu siłach.
Rysowany jest wykres swobodnego ciała dla pręta EFG wskazujący wszystkie siły.
Składowe pionowe i poziome siły FBF można wyrazić za pomocą trójkąta nachylenia.
Warunek równowagi momentu w złączu E daje siłę F BF .
Następnie, stosując warunek równowagi sił poziomych w złączu E, ocenia się poziomą siłę reakcji w E.
Podobnie, warunek równowagi sił pionowych daje pionową siłę reakcji na złączu E.
Teraz, rysując wykres swobodnego ciała dla pręta EDC, składowe siły wzdłuż BD można wyrazić za pomocą innego trójkąta nachylenia.
Stosując warunek równowagi momentów w punkcie C, oblicza się siłę F BD.

<ol>
	<li>Hibbeler, R.C. (2016). Mechanika inżynierska \u2012 Statyka i dynamika. Hoboken, New Jersey: Pearson Prentice Hall. Strony 305.</li>
	<li>Piwo, F.P.; Johnston, E.R.; Mazurek, D.F.; Cromwell, PJ i Self, BP (2019). Mechanika wektorowa dla inżynierów \u2012 Statyka i dynamika. Nowy Jork: McGraw-Hill. Strona 330</li>
	<li>Meriam, J.L., Kraige, L.G. i Bolton, J.N. (2020). Mechanika inżynierska \u2012 Statyka. Hoboken, New Jersey: John Wiley. Strona 200</li>
</ol>

ramy: rozwiązywanie problemów ii

Rozważ podnośnik hydrauliczny wspierający obciążenie 1 kN. Przyjmując uproszczoną schematyczną reprezentację tej struktury ramowej, można określić siłę działającą na elementy BD i BF.

<img src="/files/ftp_upload/14292/14292_PT_Fig_1v2.svg" style="text-align: center; height: 350px;" />

Wymiary elementów i masa obciążenia są znane. Ta struktura może zostać uznana za ramę, w której BF i BD działają jako elementy dwusiłowe, podczas gdy EFG i EDC są elementami wielosiłowymi.

Rozważa się diagram swobodnego ciała dla elementu EFG. Siłę pochyłą FBF można rozłożyć na składowe pionową i poziomą. Warunek równowagi momentu stosuje się do węzła E.

<img alt="Equation 1" src="/files/ftp_upload/14292/14292_PT_Equation_1v2.svg" style="text-align: center; height: 15px;" />

Siła FBF jest obliczana jako 1,546 kN. Następnie, do węzła E stosuje się warunek równowagi siły poziomej.

<img alt="Equation 2" src="/files/ftp_upload/14292/14292_PT_Equation_2v2.svg" style="text-align: center; height: 15px;" />

Podstawiając wartość składowej poziomej FBF do równania równowagi, oblicza się siłę reakcji poziomej w E jako 0,375 kN. Analogicznie postępuje się przy zastosowaniu warunku równowagi siły pionowej.

<img alt="Equation 3" src="/files/ftp_upload/14292/14292_PT_Equation_3v2.svg" style="text-align: center; height: 15px;" />

Obliczona siła reakcji pionowej w węźle E wynosi 0,500 kN.

Teraz rozważa się diagram swobodnego ciała dla elementu EDC. Składowe pozioma i pionowa FBD mogą być wyrażone za pomocą trójkąta nachylenia. Warunek równowagi momentu w punkcie C jest stosowany.

<img alt="Equation 4" src="/files/ftp_upload/14292/14292_PT_Equation_4v2.svg" style="text-align: center; height: 15px;" />

Siła FBD jest obliczana jako 1,677 kN.

Ramy: Rozwiązywanie problemów II

Weźmy pod uwagę wciągnik hydrauliczny przenoszący obciążenie 1 kN. Zakładając uproszczone schematyczne przedstawienie tej konstrukcji ramy, siła działająca na BD i ...

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Frames: Problem Solving II

hydraulic hoist force analysis

Hydraulic hoist force analysis shows how to find the forces in key members of a simple frame. The load is 1 kN, and the structure is represented with known member dimensions. In this setup, BF and BD act as two-force members. EFG and EDC are multi-force members, so each one is analyzed with a free-body diagram.

The first step is to study member EFG. The inclined force in BF is split into vertical and horizontal parts. Then the moment equilibrium condition is applied at joint E. Using this step, the force in BF is found to be 1.546 kN.

Next, the horizontal force equilibrium at joint E is used. The horizontal part of the BF force is substituted into the equation. This gives the horizontal reaction force at E as 0.375 kN. The vertical force equilibrium at the same joint is also applied. The vertical reaction force at E is calculated as 0.500 kN.

A free-body diagram for member EDC is then considered. The horizontal and vertical parts of FBD are written using a slope triangle. Finally, the moment equilibrium condition at point C is applied. This calculation gives the force in BD as 1.677 kN.

Hydraulic Hoist Force Analysis

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key terms and concepts

Analyze a hydraulic hoist frame with free-body diagrams, equilibrium equations, and two-force members to find forces in BF, BD, and reactions at E.

Hydraulic Hoist Force Analysis | JoVE K12

JoVE Core Mechanical Engineering offers concise, easy-to-understand animated videos covering key topics such as equilibrium, forces, truss analysis, and structural mechanics. Explore their real-world applications, and witness these concepts in action through scientist-led experiments.

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Hydraulic Hoist Force Analysis

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