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

Envisagez une structure symétrique en treillis de toit comprenant les éléments verticaux, diagonaux et horizontaux.
Un schéma à corps libre est dessiné pour analyser les forces sur les éléments DC et HC à l’aide de la méthode de section.
Ici, les charges et les longueurs des éléments horizontaux et verticaux sont les paramètres connus.
Tout d’abord, en additionnant les moments autour du point A, on calcule la force de réaction en E.
De plus, en utilisant la condition d’équilibre de force pour le treillis, la force de réaction en A est déterminée.
La symétrie de la poutre garantit que les deux forces de réaction sont égales.
Maintenant, une coupe est faite le long d’un plan coupant les membres DC, HC et HG, et un diagramme de corps libre de la plus petite section est dessiné.
En prenant la somme des moments autour de H, on obtient la force le long de DC. Le signe positif indique la force de traction.
La force le long de HC est résolue en ses composantes sinus et cosinus, et la trigonométrie est utilisée pour estimer l’angle thêta.
La condition d’équilibre du moment en E donne la force sur HC, le signe négatif indiquant la force de compression.

<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 problèmes i

Considérez une structure de fermes de toit symétrique, composée de membres verticaux, diagonaux et horizontaux. La longueur de chaque membre horizontal est de 4 m. Les longueurs des membres verticaux FB et HD sont de 4 m, tandis que la longueur du membre GC est de 6 m. Les charges agissant aux points F, G et H sont de 2 kN, tandis que celles aux points A et E sont de 1 kN. <img alt="Figure 1" src="/files/ftp_upload/14286/14286_PT_Fig_1v2.svg" style="text-align: center; height: 350px;"> La méthode des sections est utilisée pour calculer les forces agissant sur les membres DC et HC. La condition d'équilibre du moment est appliquée au point A, et les valeurs connues des forces et des distances sont substituées dans l'équation du moment. <img alt="Equation 1" src="/files/ftp_upload/14286/14286_PT_Equation_1v2.svg" style="text-align: center; height: 13px;"> Cela donne une force de réaction estimée de 4 kN au point E. Ensuite, la condition d'équilibre des forces verticales au point A révèle que la force de réaction au point A est également de 4 kN. <img alt="Equation 2" src="/files/ftp_upload/14286/14286_PT_Equation_2v2.svg" style="text-align: center; height: 13px;"> En raison de la symétrie de la ferme, les forces de réaction aux points A et E sont égales. Les forces agissant sur les membres DC et HC peuvent être obtenues une fois que les forces de réaction sont déterminées. Une coupe sectionnelle est réalisée le long d'un plan intersectant les membres DC, HC et HG, et un diagramme de corps libre de la section plus petite est considéré. <img alt="Figure 2" src="/files/ftp_upload/14286/14286_PT_Fig_2v2.svg" style="text-align: center; height: 350px;"> En additionnant les moments autour du point H, la force le long de DC est calculée comme étant de 3 kN positive, indiquant une force de traction. <img alt="Equation 3" src="/files/ftp_upload/14286/14286_PT_Equation_3v2.svg" style="text-align: center; height: 13px;"> La force le long de CH est résolue en ses composantes sinus et cosinus. La trigonométrie est utilisée pour trouver l'angle entre le membre CH et l'axe horizontal, qui est de 45°. Enfin, en appliquant la condition d'équilibre du moment au point E, on obtient une force de -1.41 kN pour FCH. <img alt="Equation 4" src="/files/ftp_upload/14286/14286_PT_Equation_4v2.svg" style="text-align: center; height: 13px;"> Dans ce cas, le résultat négatif signifie une force de compression agissant sur le membre CH.

Méthode des sections : Résolution de problèmes I

Considérez une structure symétrique en treillis de toit, composée d’éléments verticaux, diagonaux et horizontaux. La longueur de chaque élément horizontal est de 4 m. Le ...

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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 Calculation by Sections | JoVE K12

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Truss Member Force Calculation by Sections

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