Name: jib crane reaction forces in frames
Uploaded: 2023-09-22T11:49:25.000Z
Duration: 1 min 24 s
Channel: JoVE

Prenons l’exemple d’une potence avec une charge externe de la poulie.
Si les dimensions des éléments de la grue sont connues, quelles sont les forces de réaction au niveau des joints des axes, étant donné que les poulies sont sans frottement ?
Le système est une structure de cadre, composée d’un membre BD à deux forces et d’un membre ABC à plusieurs forces.
En considérant un schéma à corps libre et en appliquant les conditions d’équilibre des forces pour la section inférieure de la poulie, la tension dans le câble peut être obtenue.
En ce qui concerne le point C, la tension dans le câble vertical est dirigée vers le bas, tandis que pour le câble horizontal, elle est dirigée vers le joint A.
Dans le membre DB, la force FBD peut être exprimée à l’aide d’un triangle de pente.
La condition d’équilibre de moment à l’articulation A donne la force le long de BD.
En appliquant les conditions d’équilibre des forces, les forces de réaction horizontales et verticales à l’articulation A sont calculées.
Maintenant, si l’on considère le diagramme à corps libre de l’élément BD, les conditions d’équilibre des forces peuvent être appliquées à l’articulation D pour obtenir les forces de réaction horizontales et verticales à D.

<ol>
	<li>Hibbeler, R.C. (2016). Génie mécanique \u2012 Statique et dynamique. Hoboken, New Jersey : Pearson Prentice Hall. p. 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 i

Frames: Problem Solving I

Considérez une grue à flèche avec une charge externe suspendue à la poulie. Les dimensions des éléments de la grue sont indiquées dans la figure. Une analyse systématique de la structure du cadre est requise pour déterminer les forces de réaction au niveau des articulations, en supposant que les poulies sont sans friction. <img alt="Figure 1" src="/files/ftp_upload/14291/14291_PT_Fig_1v2.svg" style="text-align: center; height: 350px;"> Le système a deux composantes structurelles principales : un élément à deux forces BD et un élément à forces multiples ABC. L'élément à deux forces BD fait référence à un élément droit soumis uniquement à des forces à ses deux extrémités, B et D, sans forces supplémentaires agissant le long de sa longueur. Ces forces sont égales en magnitude mais opposées en direction, ce qui fait que l'élément est soit en tension pure, soit en compression pure. En revanche, l'élément à forces multiples ABC est soumis à plus de deux forces réparties le long de sa longueur. Ces forces peuvent inclure des charges externes, des forces de réaction aux articulations et la force exercée par le câble. En raison des forces multiples agissant sur l'élément ABC, il subit une distribution de contraintes plus complexe par rapport à l'élément à deux forces plus simple BD. <img alt="Figure 2" src="/files/ftp_upload/14291/14291_PT_Fig_2v2.svg" style="text-align: center; height: 350px;"> En ce qui concerne la section de la poulie inférieure, le poids de la charge équilibre la tension dans les câbles, ce qui entraîne une tension ascendante de 10 kN pour chaque câble. Maintenant, en ce qui concerne la section de la poulie supérieure, la tension T dans le câble vertical est dirigée vers le bas, tandis qu'elle indique le point A pour le câble horizontal. La tension dans le câble vertical est également de 10 kN car il fait partie du même système de câbles continu. Dans l'élément DB, la force FBD peut être résolue en ses composantes horizontales et verticales en utilisant un triangle de pente. L'équilibre momentané à l'articulation A donne FBD comme étant égal à 50 kN. La condition d'équilibre des forces horizontales peut être appliquée à l'articulation A. <img alt="Equation 1" src="/files/ftp_upload/14291/14291_PT_Equation_1v2.svg" style="text-align: center; height: 15px;"> En remplaçant les valeurs des longueurs AB, AC et le rayon de la poulie C, la force FBD est obtenue comme étant de 50 kN. La condition d'équilibre des forces horizontales donne la force de réaction à A comme étant de 40 kN. <img alt="Equation 2" src="/files/ftp_upload/14291/14291_PT_Equation_2v2.svg" style="text-align: center; height: 15px;"> De même, en utilisant la condition d'équilibre des forces verticales, la force de réaction verticale à A est estimée à -20 kN. <img alt="Equation 3" src="/files/ftp_upload/14291/14291_PT_Equation_3v2.svg" style="text-align: center; height: 15px;"> Les conditions d'équilibre des forces peuvent être appliquées à l'articulation D pour obtenir les forces de réaction horizontales et verticales à D. <img alt="Equation 4" src="/files/ftp_upload/14291/14291_PT_Equation_4v2.svg" style="text-align: center; height: 15px;"> <img alt="Equation 5" src="/files/ftp_upload/14291/14291_PT_Equation_5v2.svg" style="text-align: center; height: 15px;"> Les résultats obtenus indiquent que les forces de réaction horizontales et verticales au point D sont de -30 kN et 40 kN, respectivement.

Cadres : Résolution de problèmes I

Prenons l’exemple d’une potence avec une charge externe suspendue à la poulie. Les dimensions des éléments de la grue sont indiquées sur la figure. Une analyse systématique ...

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jib crane reaction forces in frames

Jib crane reaction forces in frames are found by analyzing the pin joints, the cable tension, and the load carried by the pulley. The structure in this problem has an external load suspended from the pulley, and the member sizes are shown in the figure. Because the pulleys are frictionless, the force balance can be worked out step by step.

The frame has two main parts. Member BD is a two-force member, which means it is a straight part loaded only at its two ends, D and B. Those end forces are equal in size and opposite in direction, so the member is in either tension or compression. Member ABC is a multi-force member, so it carries more than two forces along its length. These forces include the pin reactions, the external load, and the cable force, which makes its stress state more complex.

The lower pulley section shows that the load weight is balanced by the cable tension. That gives an upward tension of 10 kN in each cable. In the upper pulley section, the tension T in the vertical cable points downward, while the horizontal cable tension points toward joint A. The vertical cable tension is also 10 kN because it is part of the same continuous cable system.

To solve member DB, the force FBD is split into horizontal and vertical parts with a slope triangle. A moment balance at joint A gives FBD as 50 kN. The horizontal force balance at A then uses the lengths AB and AC and the pulley radius at C to confirm this value. From that balance, the reaction at A is 40 kN in the horizontal direction, and the vertical reaction at A is -20 kN. The force balance at joint D gives the reaction forces there as -30 kN horizontally and 40 kN vertically.

Jib Crane Reaction Forces in Frames

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frames: problem solving i

Jib crane reaction forces in frames are found by analyzing the pin joints, the cable tension, and the load carried by the pulley. The structure in this problem has an external load suspended from the pulley, and the member sizes are shown in the figure. Because the pulleys are frictionless, the force balance can be worked out step by step. The frame has two main parts. Member BD is a two-force member, which means it is a straight part loaded only at its two ends, D and B. Those end forces are...

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key terms and concepts

Analyze a jib crane frame with a frictionless pulley, two-force member BD, and multi-force member ABC to find reaction forces at joints A and D.

Jib Crane Reaction Forces in Frames | JoVE K12

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Jib Crane Reaction Forces in Frames

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