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

Considera una struttura reticolare simmetrica del tetto che comprende gli elementi verticali, diagonali e orizzontali.
Viene disegnato un diagramma a corpo libero per analizzare le forze sulle aste DC e HC utilizzando il metodo della sezione.
Qui, i carichi e le lunghezze degli elementi orizzontali e verticali sono i parametri noti.
Innanzitutto, sommando i momenti intorno al punto A, viene calcolata la forza di reazione in E.
Inoltre, utilizzando la condizione di equilibrio della forza per il traliccio, viene determinata la forza di reazione in A.
La simmetria del traliccio assicura che entrambe le forze di reazione siano uguali.
A questo punto, viene eseguito un taglio lungo un piano che interseca gli elementi DC, HC e HG e viene disegnato un diagramma a corpo libero della sezione più piccola.
Prendendo la somma dei momenti intorno a H si ottiene la forza lungo DC. Il segno positivo indica la forza di trazione.
La forza lungo HC viene risolta nelle sue componenti seno e coseno e la trigonometria viene utilizzata per stimare l'angolo theta.
La condizione di equilibrio del momento in E produce la forza su HC, con il segno negativo che indica la forza di compressione.

<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 i

Considera la struttura di un tetto simmetrico, composta da elementi verticali, diagonali ed orizzontali. La lunghezza di ogni elemento orizzontale è di 4 m. Le lunghezze degli elementi verticali FB e HD sono 4 m, mentre la lunghezza dell'elemento GC è di 6 m. I carichi che agiscono sui nodi F, G e H sono di 2 kN, mentre quelli sui nodi A ed E sono di 1 kN.

<img alt="Figure 1" src="/files/ftp_upload/14286/14286_PT_Fig_1v2.svg" style="text-align: center; height: 350px;" />

Il metodo delle sezioni viene utilizzato per calcolare le forze che agiscono sugli elementi DC ed HC. La condizione di equilibrio del momento viene applicata al punto A e i valori noti delle forze e delle distanze vengono sostituiti nell'equazione del momento.

<img alt="Equation 1" src="/files/ftp_upload/14286/14286_PT_Equation_1v2.svg" style="text-align: center; height: 13px;" />

Ciò porta ad una forza di reazione stimata di 4 kN nel punto E. Successivamente, la condizione di equilibrio delle forze verticali nel punto A, rivela che anche la forza di reazione nel punto A è di 4 kN.

<img alt="Equation 2" src="/files/ftp_upload/14286/14286_PT_Equation_2v2.svg" style="text-align: center; height: 13px;" />

A causa della simmetria del travetto, le forze di reazione nei punti A ed E sono uguali.

Le forze che agiscono sugli elementi DC ed HC possono essere ottenute determinando le forze di reazione. Viene effettuato un taglio sezione lungo un piano che interseca gli elementi DC, HC e HG, e viene considerato il diagramma di corpo libero della sezione più piccola.

<img alt="Figure 2" src="/files/ftp_upload/14286/14286_PT_Fig_2v2.svg" style="text-align: center; height: 350px;" />

Sommando i momenti rispetto al punto H, si calcola che la forza lungo DC è di 3 kN, indicando una forza di trazione.

<img alt="Equation 3" src="/files/ftp_upload/14286/14286_PT_Equation_3v2.svg" style="text-align: center; height: 13px;" />

La forza lungo CH viene risolta nei suoi componenti seno e coseno. Si utilizza la trigonometria per trovare l'angolo tra l'elemento CH e l'asse orizzontale, che risulta essere di 45°. Infine, applicando la condizione di equilibrio del momento al punto E, si ottiene una forza di -1,41 kN per FCH.

<img alt="Equation 4" src="/files/ftp_upload/14286/14286_PT_Equation_4v2.svg" style="text-align: center; height: 13px;" />

In questo caso, il risultato negativo indica che sull'elemento CH agisce una forza di compressione.

Metodo delle sezioni: Risoluzione dei problemi I

Considera una struttura a capriate simmetrica del tetto, composta da elementi verticali, diagonali e orizzontali. La lunghezza di ogni elemento orizzontale è di 4 m. Il ...

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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 Analysis by Sections

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method of sections: problem solving i

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...

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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

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 Calculation by Sections

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