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

Considere uma estrutura de treliça de telhado simétrica compreendendo os membros verticais, diagonais e horizontais.
Um diagrama de corpo livre é desenhado para analisar as forças nos membros DC e HC usando o método de seção.
Aqui, as cargas e os comprimentos dos membros horizontais e verticais são os parâmetros conhecidos.
Primeiro, somando os momentos sobre o ponto A, a força de reação em E é calculada.
Além disso, usando a condição de equilíbrio de força para a treliça, a força de reação em A é determinada.
A simetria da treliça garante que ambas as forças de reação sejam iguais.
Agora, um corte é feito ao longo de um plano que cruza os membros DC, HC e HG, e um diagrama de corpo livre da seção menor é desenhado.
Tomando a soma dos momentos sobre H dá a força ao longo de DC. O sinal positivo indica a força de tração.
A força ao longo do HC é resolvida em seus componentes seno e cosseno, e a trigonometria é usada para estimar o ângulo.
A condição de equilíbrio de momento em E produz a força em HC, com o sinal negativo indicando a força de compressão.

<ol>
	<li>R. C., Hibbeler Engenharia Mecânica Estática, Pearson. Pág. 291-293</li>
	<li>F.P. Beer, E.R. Johnston, D.F. Mazurek, P.J. Cornwell, B.P. Self, Mecânica Vetorial para Engenheiros Estática e Dinâmica Mecânica de Engenharia Estática, Mc Graw-Hill Education. Pág. 217</li>
</ol>

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método das seções: resolução de problemas i

Considere uma estrutura de treliça simétrica do telhado, composta por membros verticais, diagonais e horizontais. O comprimento de cada membro horizontal é de 4 m. Os comprimentos dos membros verticais FB e HD são de 4 m, enquanto o comprimento do membro GC é de 6 m. As cargas atuando nas juntas F, G e H são de 2 kN, enquanto as nas juntas A e E são de 1 kN.

<img alt="Figure 1" src="/files/ftp_upload/14286/14286_PT_Fig_1v2.svg" style="text-align: center; height: 350px;" />

O método das seções é empregado para calcular as forças atuando nos membros DC e HC. A condição de equilíbrio de momento é aplicada ao ponto A, e os valores conhecidos de forças e distâncias são substituídos na equação de momento.

<img alt="Equation 1" src="/files/ftp_upload/14286/14286_PT_Equation_1v2.svg" style="text-align: center; height: 13px;" />

Isso resulta em uma força de reação estimada de 4 kN no ponto E. Posteriormente, a condição de equilíbrio de força vertical no ponto A revela que a força de reação em A também é de 4 kN.

<img alt="Equation 2" src="/files/ftp_upload/14286/14286_PT_Equation_2v2.svg" style="text-align: center; height: 13px;" />

Devido à simetria da treliça, as forças de reação nos pontos A e E são iguais.

As forças atuando nos membros DC e HC podem ser obtidas ao determinar as forças de reação. Um corte seccional é feito ao longo de um plano que intersecta os membros DC, HC e HG, e um diagrama de corpo livre da seção menor é considerado.

<img alt="Figure 2" src="/files/ftp_upload/14286/14286_PT_Fig_2v2.svg" style="text-align: center; height: 350px;" />

Ao somar os momentos em torno do ponto H, a força ao longo de DC é calculada como sendo positiva 3 kN, indicando uma força de tração.

<img alt="Equation 3" src="/files/ftp_upload/14286/14286_PT_Equation_3v2.svg" style="text-align: center; height: 13px;" />

A força ao longo de CH é resolvida em suas componentes seno e cosseno. A trigonometria é usada para encontrar o ângulo entre o membro CH e o eixo horizontal, que é de 45 graus. Finalmente, aplicando a condição de equilíbrio de momento no ponto E, obtém-se uma força de -1,41 kN para FCH.

<img alt="Equation 4" src="/files/ftp_upload/14286/14286_PT_Equation_4v2.svg" style="text-align: center; height: 13px;" />

Neste caso, o resultado negativo significa uma força compressiva atuando no membro CH.

Método das Seções: Resolução de Problemas I

Considere uma estrutura de treliça de telhado simétrica, composta por barras verticais, diagonais e horizontais. O comprimento de cada membro horizontal é de 4 m. O ...

Education

JoVE Core

Mechanical Engineering

Engenharia Mecânica

Unit 1

Unidade 1

Structural Analysis

Análise Estrutural

KEY TERMS AND CONCEPTS

TERMOS-CHAVE E CONCEITOS

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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method of joints: problem solving i

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method of joints: problem solving ii

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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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Space Truss Forces and Stability

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frames: problem solving ii

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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

O JoVE Core Mechanical Engineering oferece vídeos animados concisos e de fácil compreensão que abrangem tópicos-chave como equilíbrio, forças, análise de treliças e mecânica estrutural. Explore suas aplicações no mundo real e observe esses conceitos em ação por meio de experimentos conduzidos por cientistas.

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Truss Member Force Calculation by Sections

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