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

Si consideri una struttura reticolare arbitraria composta da elementi diagonali, verticali e orizzontali fissati al muro.
Il metodo delle sezioni può essere utilizzato per calcolare la forza che agisce sulle aste CB, GB e GH.
Qui, i carichi e le lunghezze degli elementi orizzontali e verticali sono i parametri noti.
Viene eseguito un taglio lungo un piano che interseca le aste CB, GB e GH e viene disegnato un diagramma a corpo libero della sezione laterale destra.
L'equazione del momento di equilibrio attorno al punto G fornisce la forza lungo CB. Il segno positivo indica tensione nell'asta. 
Ora, FGB può essere espresso utilizzando un triangolo di pendenza in BCG, mentre FGH può essere espresso utilizzando un triangolo di pendenza in CEG.
Considerando la somma delle forze verticali e orizzontali, si possono scrivere le equazioni di equilibrio delle forze.
Il valore di FCB è sostituito. Le equazioni di equilibrio delle forze sono risolte simultaneamente per ottenere la forza lungo le aste GH e GB.
Il segno negativo di FGH indica che la forza è di compressione, mentre il segno positivo di FGB indica la forza di trazione.

<ol>
	<li>R. C., Hibbeler Engineering Mechanics Statics, Pearson. pp. 291-293</li>
	<li>F.P. Beer, E.R. Johnston, D.F. Mazurek, P.J. Cornwell, B.P. Self, Meccanica vettoriale per ingegneri Statica e dinamica Ingegneria meccanica Statica, Mc Graw-Hill Education. Pagine 217</li>
</ol>

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metodo delle sezioni: risoluzione dei problemi ii

Considera una struttura a capriata arbitraria, composta da membri diagonali, verticali e orizzontali fissati al muro. Per calcolare la forza agente sui membri CB, GB e GH, può essere utilizzato il metodo delle sezioni. I carichi e le lunghezze dei membri orizzontali e verticali sono parametri noti, come mostrato nella figura.

<img alt="Figure 1" src="/files/ftp_upload/14287/14287_PT_Fig_1v2.svg" style="text-align: center; height: 350px;" />

Per iniziare, viene effettuato un taglio lungo un piano che interseca i membri CB, GB e GH, e viene disegnato un diagramma di corpo libero della sezione destra.

<img alt="Figure 2" src="/files/ftp_upload/14287/14287_PT_Fig_2v2.svg" style="text-align: center; height: 350px;" />

Viene applicata l'equazione di equilibrio dei momenti attorno al punto G.

<img alt="Equation 1" src="/files/ftp_upload/14287/14287_PT_Equation_1v2.svg" style="text-align: center; height: 13px;" />

Il risultato fornisce la forza lungo il membro CB pari a 4 kN, con un segno positivo che indica tensione nel membro.

Ora, FCB può essere espressa utilizzando un triangolo di pendenza in BCG, mentre FGH può essere espressa utilizzando un triangolo di pendenza in CEG. Considerando la somma delle forze verticali e orizzontali, le equazioni di equilibrio delle forze possono essere scritte come segue:

<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;" />

Il valore di FCB viene sostituito, e le equazioni di equilibrio delle forze vengono risolte simultaneamente. Il risultato fornisce la forza FGH di -7.454 kN e FGB di 3.772 kN. Il segno negativo di FGH indica che la forza è compressiva, mentre il segno positivo per FGB indica una forza di trazione.

Metodo delle sezioni: Risoluzione dei problemi II

Si consideri una struttura reticolare arbitraria composta da elementi diagonali, verticali e orizzontali fissati al muro. Per calcolare la forza che agisce sulle aste ...

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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

truss-force-analysis-using-method-of-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

JoVE Core Mechanical Engineering offers concise, easy-to-understand animated videos covering key topics such as equilibrium, forces, truss analysis, and structural mechanics. Explore their real-world applications, and witness these concepts in action through scientist-led experiments.

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Truss Member Force Analysis by Sections

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