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

Рассмотрим гидравлический подъемник, поддерживающий груз.
Предполагая упрощенное схематическое изображение этой конструкции рамы, какая сила действует на стержни BD и BF, если размеры стержня и вес нагрузки известны?
Здесь BF и BD являются двумя силовыми членами, в то время как EFG и EDC являются многосиловыми членами.
Строится диаграмма свободного тела для EFG стержня с указанием всех сил.
Вертикальная и горизонтальная составляющие силы FBF могут быть выражены с помощью треугольника наклона.
Условие равновесия момента в соединении E дает силу F BF .
Затем, применяя условие горизонтального равновесия сил в соединении E, оценивается сила горизонтальной реакции в точке E.
Аналогично, условие равновесия вертикальной силы дает вертикальную силу реакции в соединении E.
Теперь, рисуя диаграмму свободного тела для стержня EDC, компоненты силы вдоль BD можно выразить с помощью другого треугольника наклона.
Применяя условие равновесия момента в точке C, вычисляется сила F BD.

<ol>
	<li>Хиббелер, Р.К. (2016). Инженерная механика \u2012 Статика и динамика. Хобокен, Нью-Джерси: Пирсон Прентис Холл. С. 305.</li>
	<li>Бир, Ф..; Джонстон, Э.Р.; Мазурек, Д.Ф; Кромвель,.Д. и Селф, Б.. (2019). Векторная механика для инженеров \u2012 Статика и динамика. Нью-Йорк: Макгроу-Хилл. С. 330</li>
	<li>Мериам, Д.Л., Крейдж, Л.Г., и Болтон, Д.Н. (2020). Инженерная механика \u2012 Статика. Хобокен, Нью-Джерси: Джон Уайли. С. 200</li>
</ol>

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Рамы: Решение задачи ii

Рассмотрим гидравлическую подъемную платформу, поддерживающую нагрузку в 1 кН. При условии упрощенного схематического представления этой конструкции, можно определить силу, действующую на элементы BD и BF.

<img alt="Figure 1" src="/files/ftp_upload/14292/14292_PT_Fig_1v2.svg" style="text-align: center; height: 350px;" />

Известны размеры элементов и вес нагрузки. Эта конструкция может быть рассмотрена как фрейм, при этом BF и BD являются элементами с двумя силами, а EFG и EDC - элементами с несколькими силами.

Выполняется диаграмма сил для элемента EFG. Наклонная сила FBF может быть разложена на вертикальную и горизонтальную составляющие. К месту E применяется условие моментального равновесия.

<img alt="Equation 1" src="/files/ftp_upload/14292/14292_PT_Equation_1v2.svg" style="text-align: center; height: 15px;" />

Сила FBF рассчитывается как 1,546 кН. Затем, горизонтальное условие равновесия сил применяется в точке E.

<img alt="Equation 2" src="/files/ftp_upload/14292/14292_PT_Equation_2v2.svg" style="text-align: center; height: 15px;" />

Подставляя значение горизонтальной составляющей FBF в уравнение равновесия, горизонтальная реакция в точке E вычисляется как 0,375 кН. Аналогично применяется условие вертикального равновесия сил.

<img alt="Equation 3" src="/files/ftp_upload/14292/14292_PT_Equation_3v2.svg" style="text-align: center; height: 15px;" />

Вычисленная вертикальная реакция в точке E составляет 0,500 кН.

Теперь рассмотрим диаграмму сил для элемента EDC. Горизонтальные и вертикальные составляющие FBD могут быть выражены с использованием наклонного треугольника. Применяется условие моментального равновесия в точке C.

<img alt="Equation 4" src="/files/ftp_upload/14292/14292_PT_Equation_4v2.svg" style="text-align: center; height: 15px;" />

Сила FBD рассчитывается как 1,677 кН.

Рамы: Решение задачи II

Рассмотрим гидравлическую тали, поддерживающую нагрузку 1 кН. Предполагая упрощенное схематическое изображение этой структуры рамы, сила, действующая на BD и ...

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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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