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

Stellen Sie sich ein hydraulisches Hebezeug vor, das eine Last trägt.
Welche Kraft wirkt bei vereinfachter schematischer Darstellung dieser Rahmenstruktur auf BD- und BF-Stäbe, sofern die Stababmessungen und das Gewicht der Last bekannt sind?
Hier sind BF und BD zwei Truppenmitglieder, während EFG und EDC Mehrtruppenmitglieder sind.
Es wird ein Freikörperdiagramm für das Bauteil EFG gezeichnet, in dem alle Kräfte angegeben sind.
Die vertikalen und horizontalen Komponenten der Kraft FBF können mit Hilfe eines Steigungsdreiecks ausgedrückt werden.
Der Momentengleichgewichtszustand am Gelenk E ergibt die Kraft F BF .
Wenn dann die horizontale Kraftgleichgewichtsbedingung an Gelenk E angewendet wird, wird die horizontale Reaktionskraft bei E ausgewertet.
In ähnlicher Weise ergibt die Gleichgewichtsbedingung der vertikalen Kraft die vertikale Reaktionskraft am Gelenk E.
Wenn Sie nun ein Freikörperdiagramm für den Stab EDC zeichnen, können die Komponenten der Kraft entlang BD mit Hilfe eines weiteren Steigungsdreiecks ausgedrückt werden.
Unter Anwendung der Momentengleichgewichtsbedingung am Punkt C wird die Kraft F BD berechnet.

<ol>
	<li>Hibbeler, R.C. (2016). Technische Mechanik \u2012 Statik und Dynamik. Hoboken, New Jersey: Pearson Prentice Hall. Seite 305.</li>
	<li>Bier, F.P.; Johnston, E.R.; Mazurek, D.F.; Cromwell, P.J. und Self, B.P. (2019). Vektormechanik für Ingenieure \u2012 Statik und Dynamik. New York: McGraw-Hill. Seite 330</li>
	<li>Meriam, J.L., Kraige, L.G. und Bolton, J.N. (2020). Technische Mechanik \u2012 Statik. Hoboken, New Jersey: John Wiley. Seite 200</li>
</ol>

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rahmen: problem lösung ii

Berücksichtigen Sie einen hydraulischen Aufzug mit einer Last von 1 kN. Unter der Annahme einer vereinfachten schematischen Darstellung dieser Rahmeneinheit kann die auf die Elemente BD und BF wirkende Kraft bestimmt werden.
<img alt="Figure 1" src="/files/ftp_upload/14292/14292_PT_Fig_1v2.svg" style="text-align: center; height: 350px;" />
Die Abmessungen der Elemente und das Gewicht der Last sind bekannte Parameter. Diese Struktur kann als Rahmen betrachtet werden, bei dem BF und BD als Zweikraft-Elemente wirken, während EFG und EDC Mehrfachkraft-Elemente sind.
Ein Freikörperdiagramm für das Element EFG wird erstellt. Die geneigte Kraft FBF kann in die vertikale und horizontale Komponente aufgelöst werden. Die Momentengleichgewichtsbedingung wird auf die Verbindung E angewendet.
<img alt="Equation 1" src="/files/ftp_upload/14292/14292_PT_Equation_1v2.svg" style="text-align: center; height: 15px;" />
Die Kraft FBF wird als 1,546 kN berechnet. Dann wird die horizontale Kräftgleichgewichtsbedingung an der Verbindung E. angewendet.
<img alt="Equation 2" src="/files/ftp_upload/14292/14292_PT_Equation_2v2.svg" style="text-align: center; height: 15px;" />
Der in die Gleichgewichtsgleichung eingesetzte Wert der horizontalen Komponente von FBF ergibt eine horizontale Reaktionskraft von 0,375 kN an E. Ebenso wird die vertikale Kräftgleichgewichtsbedingung angewendet.
<img alt="Equation 3" src="/files/ftp_upload/14292/14292_PT_Equation_3v2.svg" style="text-align: center; height: 15px;" />
Die berechnete vertikale Reaktionskraft an der Verbindung E beträgt 0,500 kN.
Nun wird ein Freikörperdiagramm für das Element EDC betrachtet. Die horizontalen und vertikalen Komponenten von FBD können mit einem Steigungsdreieck ausgedrückt werden. Die Momentengleichgewichtsbedingung am Punkt C wird angewendet.
<img alt="Equation 4" src="/files/ftp_upload/14292/14292_PT_Equation_4v2.svg" style="text-align: center; height: 15px;" />
Die Kraft FBD wird als 1,677 kN berechnet.

Rahmen: Problem Lösung II

Stellen Sie sich ein hydraulisches Hebezeug vor, das eine Last von 1 kN trägt. Geht man von einer vereinfachten schematischen Darstellung dieser Rahmenstruktur aus, so zeigt sich die Kraft, die auf BD und ...

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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

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Hydraulic Hoist Force Analysis

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