Name: truss force analysis at joints a to e
Uploaded: 2023-09-22T11:49:24.000Z
Duration: 1 min 30 s
Channel: JoVE

Considere uma estrutura de treliça com juntas sem atrito fixadas a uma parede e suporte de rolos no solo. Se uma força de 150 N for aplicada à junta A, determine as forças em cada membro da treliça.
As forças em cada junta podem ser determinadas usando o método das juntas.
Em primeiro lugar, o ângulo entre os membros AC e AB pode ser determinado como o inverso da razão entre o comprimento de BC e AB.
Em segundo lugar, desenhe um diagrama de corpo livre da articulação A e resolva as forças em seus componentes. Aplicando as equações de equilíbrio, as forças desconhecidas podem ser determinadas.
Em seguida, desenhando os diagramas de corpo livre das articulações B e C e usando equações de equilíbrio, as forças desconhecidas nas articulações B e C podem ser determinadas.
Finalmente, desenhe um diagrama de corpo livre das articulações D e E. Para barras CD e EF, resolva as forças nos componentes horizontal e vertical. As forças desconhecidas nas juntas D e E podem ser determinadas usando as equações de equilíbrio.

<ol>
	<li>Hibbeler, R.C. (2016). Engenharia Mecânica \u2012 Estática e Dinâmica. Hoboken, Nova Jersey: Pearson Prentice Hall. pág. 276-277.</li>
</ol>

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método dos nós: resolução de problemas ii

Considere uma estrutura de treliça com nós sem atrito fixados em uma parede e com suporte de rolos. Se uma força de 150 N é aplicada no nó A, as forças em cada membro da treliça podem ser determinadas usando o método dos nós.

<img alt="Figure 1" src="/files/ftp_upload/14283/14283_PT_Fig_1v2.svg" style="text-align: center; height: 350px;" />

O ângulo entre os membros AC e AB pode ser encontrado ao tomar a tangente inversa da razão dos comprimentos de BC e AB, e o resultado obtido é 67.38 °. Um diagrama de corpo livre do nó A é considerado, e todas as forças desconhecidas são resolvidas em seus respectivos componentes.

A condição de equilíbrio das forças verticais pode então ser usada para calcular a força atuando no membro CA, que é igual a 173.08 N.

<img alt="Equation 1" src="/files/ftp_upload/14283/14283_PT_Equation_1v2.svg" style="text-align: center; height: 15px;" />

Da mesma forma, a condição de equilíbrio das forças horizontais pode ser usada para encontrar a força F_AB como 161.54 N.

<img alt="Equation 2" src="/files/ftp_upload/14283/14283_PT_Equation_2v2.svg" style="text-align: center; height: 15px;" />

Diagramas de corpo livre dos nós B e C são considerados, e as condições de equilíbrio das forças verticais e horizontais são aplicadas para calcular as forças desconhecidas F_BD, F_CD e F_EC. Os valores resultantes para F_BD, F_CD e F_EC são, respectivamente, 161.53 N, 173.08 N e 207.70 N.

Diagramas de corpo livre dos nós D e E são finalmente considerados, e as condições de equilíbrio das forças verticais e horizontais são aplicadas. Isso fornece os valores de F_ED e F_DF como 138.46 N e 265.39 N, respectivamente.

Método dos Nós: Resolução de problemas II

Considere uma estrutura de treliça com juntas sem atrito fixadas a uma parede e suporte de rolos. Se uma força de 150 N for aplicada à junta A, as forças em cada ...

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 Joints: Problem Solving II

truss force analysis at joints a to e

Truss force analysis at joints A to E uses the method of joints to find the force in each member. The truss has frictionless joints, a wall support, and a roller support. A 150 N force is applied at joint A, and the member forces are solved from joint to joint.

First, the angle between members AC and AB is found using the inverse tangent of the ratio of BC to AB. This gives an angle of 67.38°. A free-body diagram of joint A is then drawn, and each unknown force is split into vertical and horizontal components.

The vertical force equilibrium equation is used to find the force in member CA, which is 173.08 N. The horizontal force equilibrium equation gives the force in member AB as 161.54 N. Free-body diagrams for joints B and C are then used to continue the calculation.

At joints B and C, the vertical and horizontal equilibrium conditions are applied again. This gives FBD = 161.53 N, FCD = 173.08 N, and FEC = 207.70 N. Free-body diagrams of joints D and E are used last, and the same equilibrium steps give FED = 138.46 N and FDF = 265.39 N.

Truss Force Analysis at Joints A to E

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Truss force analysis at joints A to E uses the method of joints to find the force in each member. The truss has frictionless joints, a wall support, and a roller support. A 150 N force is applied at joint A, and the member forces are solved from joint to joint. First, the angle between members AC and AB is found using the inverse tangent of the ratio of BC to AB. This gives an angle of 67.38°. A free-body diagram of joint A is then drawn, and each unknown force is split into vertical and...

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key terms and concepts

Use the method of joints to solve truss forces at joints A to E. Find member forces from free-body diagrams, angles, and equilibrium equations.

Truss Force Analysis at Joints A to E | JoVE K12

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Truss Force Analysis at Joints A to E

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