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

Stellen Sie sich eine Fachwerkstruktur mit reibungsfreien Verbindungen vor, die an einer Wand befestigt sind, und einer Rollenstütze auf dem Boden. Wenn eine Kraft von 150 N auf die Verbindung A aufgebracht wird, dann bestimmen Sie die Kräfte in jedem Element des Fachwerkbinders.
Die Kräfte an jedem Gelenk können mit der Methode der Gelenke bestimmt werden.
Zunächst kann der Winkel zwischen den Stäben AC und AB als Tan-Kehrwert des Verhältnisses der Länge von BC und AB bestimmt werden.
Zeichnen Sie zweitens ein Freikörperdiagramm von Gelenk A und lösen Sie die Kräfte in ihre Bestandteile auf. Durch die Anwendung der Gleichgewichtsgleichungen können die unbekannten Kräfte bestimmt werden.
Als nächstes können durch das Zeichnen der Freikörperdiagramme der Gelenke B und C und unter Verwendung von Gleichgewichtsgleichungen die unbekannten Kräfte an den Gelenken B und C bestimmt werden.
Zeichnen Sie abschließend ein Freikörperdiagramm der Gelenke D und E. Für die Stäbe CD und EF lösen Sie die Kräfte in die horizontalen und vertikalen Komponenten auf. Die unbekannten Kräfte an den Gelenken D und E können mit Hilfe der Gleichgewichtsgleichungen bestimmt werden.

<ol>
	<li>Hibbeler, R.C. (2016). Technische Mechanik \u2012 Statik und Dynamik. Hoboken, New Jersey: Pearson Prentice Hall. Seiten 276-277.</li>
</ol>

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methode der gelenke: problemlösung ii

Betrachten Sie eine Fachwerkstruktur mit reibungsfreien Gelenken, die an einer Wand und einer Rollenstütze befestigt sind. Wenn eine Kraft von 150 N auf das Gelenk A angewendet wird, können die Kräfte in jedem Mitglied des Fachwerks mit der Methode der Gelenke bestimmt werden. <img alt="Figure 1" src="/files/ftp_upload/14283/14283_PT_Fig_1v2.svg" style="text-align: center; height: 350px;"> Der Winkel zwischen den Mitgliedern AC und AB kann durch die Berechnung des inversen Tangens des Verhältnisses der Längen von BC und AB gefunden werden, und das Ergebnis ist 67,38°. Ein Freikörperdiagramm des Gelenks A wird betrachtet, und alle unbekannten Kräfte werden in ihre jeweiligen Komponenten zerlegt. Die Bedingung für das Gleichgewicht der vertikalen Kräfte kann dann verwendet werden, um die Kraft zu berechnen, die auf das Mitglied CA wirkt, welche 173,08 N beträgt. <img alt="Equation 1" src="/files/ftp_upload/14283/14283_PT_Equation_1v2.svg" style="text-align: center; height: 15px;"> Ähnlich kann die Bedingung für das Gleichgewicht der horizontalen Kräfte verwendet werden, um die Kraft FAB als 161,54 N zu finden. <img alt="Equation 2" src="/files/ftp_upload/14283/14283_PT_Equation_2v2.svg" style="text-align: center; height: 15px;"> Freikörperdiagramme der Gelenke B und C werden betrachtet, und die Bedingungen für das Gleichgewicht der vertikalen und horizontalen Kräfte werden angewendet, um die unbekannten Kräfte FBD, FCD, und FEC zu berechnen. Die resultierenden Werte für FBD, FCD, und FEC sind jeweils 161,53 N, 173,08 N und 207,70 N. Freikörperdiagramme der Gelenke D und E werden schließlich betrachtet, und die Bedingungen für das Gleichgewicht der vertikalen und horizontalen Kräfte werden angewendet. Dies ergibt die Werte von FED und FDF als 138,46 N bzw. 265,39 N.

Methode der Gelenke: Problemlösung II

Stellen Sie sich eine Fachwerkstruktur mit reibungsfreien Verbindungen vor, die an einer Wand und einer Rollenstütze befestigt sind. Wird eine Kraft von 150 N auf das Gelenk A aufgebracht, so werden die Kräfte in jedem Gelenk ...

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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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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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