Name: truss member force calculation by sections
Uploaded: 2025-05-22T09:06:03.000Z
Duration: 1 min 27 s
Channel: JoVE

Рассмотрим симметричную стропильную конструкцию кровли, состоящую из вертикальных, диагональных и горизонтальных стержней.
Для анализа сил, действующих на стержни постоянного и постоянного тока, с помощью метода сечения строится диаграмма свободного тела.
Здесь нагрузки и длины горизонтальных и вертикальных стержней являются известными параметрами.
Во-первых, суммируя моменты около точки А, вычисляется сила реакции в точке Е.
Далее, используя условие равновесия сил для фермы, определяется сила реакции в точке А.
Симметрия фермы гарантирует, что обе силы реакции равны.
Теперь делается разрез вдоль плоскости, пересекающей стержни DC, HC и HG, и рисуется диаграмма свободного тела меньшего сечения.
Суммирование моментов вокруг H дает силу вдоль постоянного тока. Положительный знак указывает на растягивающее усилие.
Сила вдоль HC распадается на синусоидальную и косинусную составляющие, а тригонометрия используется для оценки угла тета.
Состояние равновесия момента в точке E дает силу на HC, где отрицательный знак указывает на сжимающую силу.

<ol>
	<li>Р. К., Хиббелер Инженерная механика Статика, Пирсон. стр. 291-293</li>
	<li>Ф.. Бир, Э.Р. Джонстон, Д.Ф. Мазурек,.Д. Корнуэлл, Б.. Селф, Векторная механика для инженеров Статика и динамика Инженерная механика Статика, Mc Graw-Hill Education. С. 217</li>
</ol>

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Метод сечений: решение задачи i

Рассмотрим симметричную крышевую ферму, состоящую из вертикальных, диагональных и горизонтальных элементов. Длина каждого горизонтального элемента составляет 4 м. Длины вертикальных элементов FB и HD равны 4 м, а длина элемента GC составляет 6 м. Нагрузки, действующие в узлах F, G и H, равны 2 кН, а в узлах A и E - 1 кН.

<img alt="Figure 1" src="/files/ftp_upload/14286/14286_PT_Fig_1v2.svg" style="text-align: center; height: 350px;" />

Применяется метод сечений для расчета сил, действующих на элементы DC и HC. Условие равновесия моментов применяется к точке A, а известные значения сил и расстояний подставляются в уравнение момента.

<img alt="Equation 1" src="/files/ftp_upload/14286/14286_PT_Equation_1v2.svg" style="text-align: center; height: 13px;" />

Это приводит к оценочной реакционной силе 4 кН в точке E. Впоследствии условие вертикального равновесия сил в точке A показывает, что реакционная сила в точке A также равна 4 кН.

<img alt="Equation 2" src="/files/ftp_upload/14286/14286_PT_Equation_2v2.svg" style="text-align: center; height: 13px;" />

Из-за симметрии фермы реакционные силы в точках A и E равны.

Силы, действующие на элементы DC и HC, могут быть получены после определения реакционных сил. Выполняется сечение плоскостью, пересекающей элементы DC, HC и HG, и рассматривается схема свободных тел меньшей секции.

<img alt="Figure 2" src="/files/ftp_upload/14286/14286_PT_Fig_2v2.svg" style="text-align: center; height: 350px;" />

Путем суммирования моментов относительно точки HH сила вдоль DC рассчитывается как положительная 3 кН, что указывает на растягивающую силу.

<img alt="Equation 3" src="/files/ftp_upload/14286/14286_PT_Equation_3v2.svg" style="text-align: center; height: 13px;" />

Сила вдоль CH разлагается на синусовые и косинусовые компоненты. Тригонометрия используется для определения угла между элементом CH и горизонтальной осью, который составляет 45°. Наконец, применение условия равновесия моментов в точке E дает силу -1.41 кН для FCH.

<img alt="Equation 4" src="/files/ftp_upload/14286/14286_PT_Equation_4v2.svg" style="text-align: center; height: 13px;" />

В этом случае отрицательный результат означает сжимающую силу, действующую на элемент CH.

Метод сечений: решение задачи I

Рассмотрим симметричную конструкцию стропильной фермы, состоящую из вертикальных, диагональных и горизонтальных элементов. Длина каждого горизонтального стержня составляет 4 м. Длина ...

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Mechanical Engineering

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Unit 1

Раздел 1

Structural Analysis

Структурный анализ

KEY TERMS AND CONCEPTS

ОСНОВНЫЕ ТЕРМИНЫ И ПОНЯТИЯ

Method of Sections: Problem Solving I

truss member force calculation by sections

Truss member forces can be found with the method of sections in a symmetrical roof truss. The structure here includes vertical, diagonal, and horizontal members. Each horizontal member is 4 m long. The vertical members FB and HD are 4 m long, and member GC is 6 m long.

The roof truss carries loads at several joints. Joints F, G, and H each have a 2 kN load. Joints A and E each have a 1 kN load. To start the analysis, moment equilibrium is used at point A. The force and distance values are placed into the moment equation, which gives a reaction force of 4 kN at point E.

Next, vertical force equilibrium at point A gives the reaction force at A as 4 kN. Because the truss is symmetrical, the reactions at A and E are equal. Once the reactions are known, the forces in members DC and HC can be found. A section is cut through members DC, HC, and HG, and a free-body diagram is drawn for the smaller part.

Moments about point H are then summed to solve for the force in DC. The result is 3 kN, which is positive and shows tension. The force in CH is handled with trigonometry. The angle between member CH and the horizontal axis is found to be 45°. Using moment equilibrium at point E gives FCH = -1.41 kN. The negative sign means member CH is in compression.

Truss Member Force Calculation by Sections

truss-member-force-calculation-by-sections

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Truss Member Force Analysis by Sections

Truss member force analysis by sections finds the force in selected members of a truss. In this problem, forces F1 and F2 act at joints B and D. The goal is to determine the forces in members EF, DC, and DF. The method uses the idea that a truss in equilibrium also has each of its sections in equilibrium. A free-body diagram of the whole truss is drawn first. The moment equation about joint A is then used to estimate the support reaction at point E. Next, a cut is made along a sectional plane...

method of sections: problem solving i

Truss member forces can be found with the method of sections in a symmetrical roof truss. The structure here includes vertical, diagonal, and horizontal members. Each horizontal member is 4 m long. The vertical members FB and HD are 4 m long, and member GC is 6 m long. The roof truss carries loads at several joints. Joints F, G, and H each have a 2 kN load. Joints A and E each have a 1 kN load. To start the analysis, moment equilibrium is used at point A. The force and distance values are...

method of sections: problem solving ii

Truss member force analysis by sections finds the forces in selected members of a wall-mounted truss. The structure here includes diagonal, vertical, and horizontal members. The known loads and member lengths are used to solve for the forces in CB, GB, and GH. A cut is made through members CB, GB, and GH. The right side of the truss is then isolated as a free-body diagram. Next, the moment equilibrium equation is taken about point G. This step gives the force in member CB as 4 kN, and the...

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key terms and concepts

Learn how the method of sections is used on a symmetrical roof truss to find support reactions and member forces, including tension and compression.

Truss Member Force Calculation by Sections | 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.

mechanical engineering

Truss Member Force Calculation by Sections

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