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

Stellen Sie sich eine symmetrische Dachstuhlstruktur vor, die aus vertikalen, diagonalen und horizontalen Elementen besteht.
Es wird ein Freikörperdiagramm gezeichnet, um die Kräfte auf DC- und HC-Stäbe mit der Schnittmethode zu analysieren.
Dabei sind die Lasten und die Längen der horizontalen und vertikalen Stäbe die bekannten Parameter.
Zuerst wird durch Summierung der Momente um den Punkt A die Reaktionskraft bei E berechnet.
Des Weiteren wird unter Verwendung der Kraftgleichgewichtsbedingung für das Fachwerk die Reaktionskraft bei A bestimmt.
Die Symmetrie des Fachwerks sorgt dafür, dass beide Reaktionskräfte gleich sind.
Nun wird ein Schnitt entlang einer Ebene vorgenommen, die DC-, HC- und HG-Stäbe schneidet, und es wird ein Freikörperdiagramm des kleineren Schnitts gezeichnet.
Nimmt man die Summe der Momente um H, erhält man die Kraft entlang DC. Das Pluszeichen gibt die Zugkraft an.
Die Kraft entlang von HC wird in ihre Sinus- und Kosinuskomponenten aufgelöst, und die Trigonometrie wird verwendet, um den Winkeltheta zu schätzen.
Der Momentengleichgewichtszustand bei E ergibt die Kraft auf HC, wobei das negative Vorzeichen die Druckkraft anzeigt.

<ol>
	<li>R. C., Hibbeler Technische Mechanik Statik, Pearson. S. 291-293</li>
	<li>F.P. Beer, E.R. Johnston, D.F. Mazurek, P.J. Cornwell, B.P. Self, Vektormechanik für Ingenieure Statik und Dynamik Technische Mechanik Statik, Mc Graw-Hill Bildung. S. 217</li>
</ol>

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methode der schnitte: problemlösung i

Betrachten Sie eine symmetrische Dachträgerstruktur, bestehend aus senkrechten, diagonalen und horizontalen Elementen. Die Länge jedes horizontalen Elements beträgt 4 m. Die Längen der senkrechten Elemente FB und HD betragen 4 m, während die Länge des Elements GC 6 m beträgt. Die Lasten, die an den Knoten F, G, und H wirken, betragen jeweils 2 kN, während diejenigen an den Knoten A und E 1 kN betragen. <img alt="Figure 1" src="/files/ftp_upload/14286/14286_PT_Fig_1v2.svg" style="text-align: center; height: 350px;"> Die Methode der Schnitte wird angewendet, um die auf die Elemente DC und HC wirkenden Kräfte zu berechnen. Die Momentengleichgewichtsbedingung wird am Punkt A angewendet und die bekannten Werte von Kräften und Abständen in die Momentengleichung eingesetzt. <img alt="Equation 1" src="/files/ftp_upload/14286/14286_PT_Equation_1v2.svg" style="text-align: center; height: 13px;"> Dies führt zu einer geschätzten Reaktionskraft von 4 kN am Punkt E. Anschließend zeigt die vertikale Kraftgleichgewichtsbedingung am Punkt A, dass die Reaktionskraft an A ebenfalls 4 kN beträgt. <img alt="Equation 2" src="/files/ftp_upload/14286/14286_PT_Equation_2v2.svg" style="text-align: center; height: 13px;"> Aufgrund der Symmetrie des Trägers sind die Reaktionskräfte an den Punkten A und E gleich. Die auf die Elemente DC und HC wirkenden Kräfte können ermittelt werden, sobald die Reaktionskräfte bestimmt sind. Ein Schnitt wird entlang einer Ebene gemacht, die die Elemente DC, HC, und HG, schneidet, und es wird ein Freikörperdiagramm des kleineren Abschnitts betrachtet. <img alt="Figure 2" src="/files/ftp_upload/14286/14286_PT_Fig_2v2.svg" style="text-align: center; height: 350px;"> Durch Summierung der Momente um Punkt H wird die Kraft entlang DC berechnet und beträgt 3 kN, was auf eine Zugkraft hinweist. <img alt="Equation 3" src="/files/ftp_upload/14286/14286_PT_Equation_3v2.svg" style="text-align: center; height: 13px;"> Die Kraft entlang CH wird in ihre Sinus- und Kosinuskomponenten aufgelöst. Trigonometrie wird verwendet, um den Winkel zwischen dem Element CH und der horizontalen Achse zu finden, der 45° beträgt. Schließlich ergibt die Anwendung der Momentengleichgewichtsbedingung am Punkt E eine Kraft von -1,41 kN für FCH. <img alt="Equation 4" src="/files/ftp_upload/14286/14286_PT_Equation_4v2.svg" style="text-align: center; height: 13px;"> In diesem Fall deutet das negative Ergebnis auf eine Druckkraft hin, die auf das Element CH wirkt.

Methode der Schnitte: Problemlösung I

Stellen Sie sich eine symmetrische Dachstuhlstruktur vor, die aus vertikalen, diagonalen und horizontalen Elementen besteht. Die Länge jedes horizontalen Elements beträgt 4 m. Die Länge ...

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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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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

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Truss Member Force Calculation by Sections

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