Name: truss member force analysis by sections
Uploaded: 2025-05-22T09:06:03.000Z
Duration: 1 min 30 s
Channel: JoVE

Considérons une structure de treillis arbitraire composée d’éléments diagonaux, verticaux et horizontaux fixés au mur.
La méthode des sections peut être utilisée pour calculer la force agissant sur les barres CB, GB et GH.
Ici, les charges et les longueurs des éléments horizontaux et verticaux sont les paramètres connus.
Une coupe est faite le long d’un plan coupant les membres CB, GB et GH et un schéma de corps libre de la section du côté droit est dessiné.
L’équation du moment d’équilibre autour du point G donne la force le long de CB. Le signe positif indique une tension dans le membre. 
Maintenant, FGB peut être exprimé à l’aide d’un triangle de pente dans BCG, tandis que FGH peut être exprimé à l’aide d’un triangle de pente dans CEG.
En considérant la somme des forces verticales et horizontales, les équations d’équilibre des forces peuvent être écrites.
La valeur de FCB est substituée. Les équations d’équilibre des forces sont résolues simultanément pour obtenir la force le long des membres GH et GB.
Le signe négatif de FGH indique que la force est de compression, tandis que le signe positif de FGB indique la force de traction.

<ol>
	<li>R. C., Hibbeler Engineering Mechanics Statics, Pearson, p. 291 à 293</li>
	<li>F.P. Beer, E.R. Johnston, D.F. Mazurek, P.J. Cornwell, B.P. Self, Mécanique vectorielle pour les ingénieurs Statique et dynamique Génie mécanique Statique, Mc Graw-Hill Éducation. pp. 217</li>
</ol>

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méthode des sections : résolution de problème ii

Considérez une structure de treillis arbitraire composée de membres diagonaux, verticaux et horizontaux fixés au mur. Pour calculer la force agissant sur les membres CB, GB et GH, la méthode des sections peut être utilisée. Les charges et les longueurs des membres horizontaux et verticaux sont des paramètres connus, comme le montre la figure. <img alt="Figure 1" src="/files/ftp_upload/14287/14287_PT_Fig_1v2.svg" style="text-align: center; height: 350px;"> Pour commencer, une coupe est réalisée le long d'un plan intersectant les membres CB, GB et GH, et un diagramme de corps libre de la section du côté droit est dessiné. <img alt="Figure 2" src="/files/ftp_upload/14287/14287_PT_Fig_2v2.svg" style="text-align: center; height: 350px;"> L'équation d'équilibre des moments autour du point G est appliquée. <img alt="Equation 1" src="/files/ftp_upload/14287/14287_PT_Equation_1v2.svg" style="text-align: center; height: 13px;"> Le résultat donne la force le long du membre CB comme étant de 4 kN, avec un signe positif indiquant une tension dans le membre. Maintenant, FCB peut être exprimé à l'aide d'un triangle de pente dans BCG, tandis que FGH peut être exprimé à l'aide d'un triangle de pente dans CEG. En considérant la sommation des forces verticales et horizontales, les équations d'équilibre des forces peuvent être écrites comme suit : <img alt="Equation 2" src="/files/ftp_upload/14287/14287_PT_Equation_2v2.svg" style="text-align: center; height: 15px;"> <img alt="Equation 3" src="/files/ftp_upload/14287/14287_PT_Equation_3v2.svg" style="text-align: center; height: 13px;"> La valeur de FCB est substituée, et les équations d'équilibre des forces sont résolues simultanément. Le résultat donne la force FGH comme étant -7,454 kN et FGB comme étant 3,772 kN. Le signe négatif de FGH indique que la force est compressive, tandis que le signe positif de FGB indique la force de traction.

Méthode des sections : résolution de problème II

Considérons une structure de treillis arbitraire composée d’éléments diagonaux, verticaux et horizontaux fixés au mur. Pour calculer la force agissant sur les membres ...

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Method of Sections: Problem Solving II

truss member force analysis by sections

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 positive sign shows that CB is in tension.

The force in CB is then written with a slope triangle in triangle BCG. The force in GH is written with a slope triangle in triangle CEG. After that, the sums of vertical and horizontal forces are set equal to zero. The equation for CB is substituted, and the remaining equations are solved at the same time.

The final values are FGH = -7.454 kN and FGB = 3.772 kN. The negative sign for GH means the member is in compression. The positive sign for GB means the member is in tension.

Truss Member Force Analysis by Sections

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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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Learn how the method of sections isolates a truss cut, uses moment and force balance, and identifies tension or compression in members CB, GB, and GH.

Truss Member Force Analysis by Sections | JoVE K12

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Truss Member Force Analysis by Sections

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