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

Prenons l’exemple d’un palan hydraulique supportant une charge.
En supposant une représentation schématique simplifiée de cette structure de cadre, quelle force agit sur les éléments BD et BF, à condition que les dimensions de l’élément et le poids de la charge soient connus ?
Ici, BF et BD sont deux membres de la force, tandis que EFG et EDC sont des membres de plusieurs forces.
Un schéma de corps libre pour l’élément EFG indiquant toutes les forces est dessiné.
Les composantes verticales et horizontales de la force FBF peuvent être exprimées à l’aide d’un triangle de pente.
La condition d’équilibre des moments à l’articulation E donne la force F BF .
Ensuite, en appliquant la condition d’équilibre de la force horizontale à l’articulation E, la force de réaction horizontale à l’articulation E est évaluée.
De même, la condition d’équilibre de la force verticale donne la force de réaction verticale à l’articulation E.
Maintenant, en dessinant un diagramme de corps libre pour le membre EDC, les composantes de la force le long de BD peuvent être exprimées à l’aide d’un autre triangle de pente.
En appliquant la condition d’équilibre de moment au point C, la force F BD est calculée.

<ol>
	<li>Hibbeler, R.C. (2016). Génie mécanique \u2012 Statique et dynamique. Hoboken, New Jersey : Pearson Prentice Hall. Pages 305.</li>
	<li>Bière, F.P. ; Johnston, E.R. ; Mazurek, D.F ; Cromwell, P.J. et Self, B.P. (2019). Mécanique vectorielle pour les ingénieurs \u2012 Statique et dynamique. New York : McGraw-Hill. pp 330</li>
	<li>Meriam, J.L., Kraige, L.G. et Bolton, J.N. (2020). Génie mécanique \u2012 Statique. Hoboken, New Jersey : John Wiley. page 200</li>
</ol>

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cadres : résolution de problèmes ii

Considérons un élévateur hydraulique supportant une charge de 1 kN. En supposant une représentation simplifiée de ce cadre, la force agissant sur les membres BD et BF peut être déterminée. <img alt="Figure 1" src="/files/ftp_upload/14292/14292_PT_Fig_1v2.svg" style="text-align: center; height: 350px;"> Les dimensions des membres et le poids de la charge sont des paramètres connus. Cette structure peut être considérée comme un cadre avec BF et BD agissant en tant que membres à deux forces, tandis que EFG et EDC sont des membres à forces multiples. Un schéma cinématique du membre EFG est considéré. La force inclinée FBF peut être résolue en composantes verticales et horizontales. La condition d’équilibre des moments est appliquée à l’articulation E. <img alt="Equation 1" src="/files/ftp_upload/14292/14292_PT_Equation_1v2.svg" style="text-align: center; height: 15px;"> La force FBF est calculée pour être de 1,546 kN. Ensuite, la condition d’équilibre des forces horizontales est appliquée à l’articulation E. <img alt="Equation 2" src="/files/ftp_upload/14292/14292_PT_Equation_2v2.svg" style="text-align: center; height: 15px;"> En substituant la valeur de la composante horizontale de FBF dans l’équation d’équilibre, la force de réaction horizontale à E est calculée à 0,375 kN. De même, la condition d’équilibre des forces verticales est appliquée. <img alt="Equation 3" src="/files/ftp_upload/14292/14292_PT_Equation_3v2.svg" style="text-align: center; height: 15px;"> La force de réaction verticale calculée à l’articulation E est de 0,500 kN. Maintenant, un schéma cinématique du membre EDC est considéré. Les composantes horizontales et verticales de FBD peuvent être exprimées à l’aide d’un triangle de pente. La condition d’équilibre des moments au point C est appliquée. <img alt="Equation 4" src="/files/ftp_upload/14292/14292_PT_Equation_4v2.svg" style="text-align: center; height: 15px;"> La force FBD est calculée comme étant de 1,677 kN.

Cadres : Résolution de problèmes II

Considérons un palan hydraulique supportant une charge de 1 kN. En supposant une représentation schématique simplifiée de cette structure de cadre, la force agissant sur BD et ...

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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

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Hydraulic Hoist Force Analysis

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