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

Considere una grúa giratoria con una carga externa de la polea.
Si se conocen las dimensiones de los miembros de la grúa, ¿cuáles son las fuerzas de reacción en las uniones de los pasadores, teniendo en cuenta que las poleas no tienen fricción?
El sistema es una estructura de marco, que consta de un miembro BD de dos fuerzas y un miembro ABC de múltiples fuerzas.
Considerando un diagrama de cuerpo libre y aplicando las condiciones de equilibrio de fuerzas para la sección inferior de la polea, se puede obtener la tensión en el cable.
Teniendo en cuenta el punto C, la tensión en el cable vertical se dirige hacia abajo, mientras que para el cable horizontal, se dirige hacia la junta A.
En el miembro DB, la fuerza FBD se puede expresar mediante un triángulo de pendiente.
La condición de equilibrio de momento en la unión A da la fuerza a lo largo de BD.
Aplicando las condiciones de equilibrio de fuerzas, se calculan las fuerzas de reacción horizontal y vertical en la junta A.
Ahora, considerando el diagrama de cuerpo libre para el miembro BD, las condiciones de equilibrio de fuerzas se pueden aplicar en la unión D para obtener las fuerzas de reacción horizontal y vertical en D.

<ol>
	<li>Hibbeler, R.C. (2016). Ingeniería Mecánica \u2012 Estática y Dinámica. Hoboken, Nueva Jersey: Pearson Prentice Hall. Págs. 305.</li>
	<li>Cerveza, F.P.; Johnston, E.R.; Mazurek, D.F.; Cromwell, P.J. y Self, B.P. (2019). Mecánica vectorial para ingenieros: estática y dinámica. Nueva York: McGraw-Hill. págs. 330</li>
	<li>Meriam, J.L., Kraige, L.G., y Bolton, J.N. (2020). Ingeniería Mecánica \u2012 Estática. Hoboken, Nueva Jersey: John Wiley. págs. 200</li>
</ol>

No se han declarado conflictos de interés.

marcos: resolución de problemas i

Frames: Problem Solving I

Considera una grúa horca con una carga externa suspendida del polipasto. Las dimensiones de los componentes de la grúa se muestran en la figura. Se requiere un análisis sistemático de la estructura para determinar las fuerzas de reacción en los puntos de articulación, asumiendo que los polipastos no tienen fricción. <img alt="Figura 1" src="/files/ftp_upload/14291/14291_PT_Fig_1v2.svg" style="text-align: center; height: 350px;"> El sistema tiene dos componentes estructurales principales: un componente de dos fuerzas BD y un componente de fuerzas múltiples ABC. El componente de dos fuerzas BD se refiere a un elemento recto solo sometido a fuerzas en sus dos extremos, B y D, sin fuerzas adicionales a lo largo de su longitud. Estas fuerzas tienen la misma magnitud pero dirección opuesta, lo que hace que el componente esté en pura tensión o compresión. Por otro lado, el componente de fuerzas múltiples ABC está sometido a más de dos fuerzas distribuidas a lo largo de su longitud. Estas fuerzas pueden incluir cargas externas, fuerzas de reacción en los puntos de articulación y la fuerza ejercida por el cable. Debido a las múltiples fuerzas que actúan en el componente ABC, experimenta una distribución de tensiones más compleja en comparación con el componente más simple de dos fuerzas BD. <img alt="Figura 2" src="/files/ftp_upload/14291/14291_PT_Fig_2v2.svg" style="text-align: center; height: 350px;"> Teniendo en cuenta la sección de la polea inferior, el peso de la carga equilibra la tensión en los cables, lo que resulta en una tensión hacia arriba de 10 kN para cada cable. Ahora, teniendo en cuenta la sección de la polea superior, la tensión T en el cable vertical se dirige hacia abajo, mientras que apunta hacia la unión A para el cable horizontal. La tensión en el cable vertical también es de 10 kN, ya que forma parte del mismo sistema continuo de cable. En el componente DB, la fuerza FBD se puede descomponer en sus componentes horizontales y verticales utilizando un triángulo de pendiente. La condición de equilibrio de momentos en la unión A da FBD como 50 kN. La condición de equilibrio de fuerzas horizontales se puede aplicar en la unión A. <img alt="Ecuación 1" src="/files/ftp_upload/14291/14291_PT_Equation_1v2.svg" style="text-align: center; height: 15px;"> Sustituyendo los valores de la longitud AB, AC y el radio de la polea C, se obtiene la fuerza FBD como 50 kN. La condición de equilibrio de fuerzas horizontales da la fuerza de reacción en A como 40 kN. <img alt="Ecuación 2" src="/files/ftp_upload/14291/14291_PT_Equation_2v2.svg" style="text-align: center; height: 15px;"> De manera similar, utilizando la condición de equilibrio de fuerzas verticales, se estima que la fuerza de reacción vertical en A es de -20 kN. <img alt="Ecuación 3" src="/files/ftp_upload/14291/14291_PT_Equation_3v2.svg" style="text-align: center; height: 15px;"> Las condiciones de equilibrio de fuerzas se pueden aplicar en la unión D para obtener las fuerzas de reacción horizontales y verticales en D. <img alt="Ecuación 4" src="/files/ftp_upload/14291/14291_PT_Equation_4v2.svg" style="text-align: center; height: 15px;"> <img alt="Ecuación 5" src="/files/ftp_upload/14291/14291_PT_Equation_5v2.svg" style="text-align: center; height: 15px;"> Los resultados obtenidos indican que las fuerzas de reacción horizontales y verticales en el punto D son de -30 kN y 40 kN, respectivamente.

Marcos: Resolución de problemas I

Considere una grúa giratoria con una carga externa suspendida de la polea. Las dimensiones de los miembros de la grúa se muestran en la figura. Un análisis sistemático ...

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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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