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

Considera una struttura a traliccio con giunti senza attrito fissati a una parete e supporto a rulli a terra. Se viene applicata una forza di 150 N al giunto A, determinare le forze in ciascun membro della trave reticolare.
Le forze su ciascun giunto possono essere determinate utilizzando il metodo dei giunti.
In primo luogo, l'angolo tra le aste AC e AB può essere determinato come l'inverso tan del rapporto tra la lunghezza di BC e AB.
In secondo luogo, disegna un diagramma di corpo libero del giunto A e risolvi le forze nelle loro componenti. Applicando le equazioni di equilibrio, le forze incognite possono essere determinate.
Successivamente, disegnando i diagrammi di corpo libero delle joint B e C e utilizzando le equazioni di equilibrio, è possibile determinare le forze incognite alle joint B e C.
Infine, disegna un diagramma a corpo libero delle articolazioni D ed E. Per i membri CD ed EF, risolvere le forze nei componenti orizzontale e verticale. Le forze incognite alle articolazioni D ed E possono essere determinate utilizzando le equazioni di equilibrio.

<ol>
	<li>Hibbeler, R.C. (2016). Ingegneria Meccanica \u2012 Statica e Dinamica. Hoboken, New Jersey: Pearson Prentice Hall. pagine 276-277.</li>
</ol>

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metodo dei giunti: risoluzione del problema ii

Considera una struttura a travata con giunti privi di attrito fissati a una parete e supporto a rulli. Se una forza di 150 N viene applicata al giunto A, le forze in ciascun elemento della travata possono essere determinate utilizzando il metodo dei giunti.

<img alt="Figure 1" src="/files/ftp_upload/14283/14283_PT_Fig_1v2.svg" style="text-align: center; height: 350px;" />

L'angolo tra gli elementi AC e AB può essere trovato prendendo la tangente inversa del rapporto tra le lunghezze di BC e AB, il risultato ottenuto è di 67.38°. Viene considerato un diagramma di corpo libero del giunto A e tutte le forze sconosciute vengono risolte nelle loro rispettive componenti.

La condizione di equilibrio delle forze verticali, può quindi essere utilizzata per calcolare la forza agente sull'elemento CA, che è uguale a 173.08 N.

<img alt="Equation 1" src="/files/ftp_upload/14283/14283_PT_Equation_1v2.svg" style="text-align: center; height: 15px;" />

Allo stesso modo, la condizione di equilibrio delle forze orizzontali, può essere utilizzata per trovare la forza FAB pari a 161.54 N.

<img alt="Equation 2" src="/files/ftp_upload/14283/14283_PT_Equation_2v2.svg" style="text-align: center; height: 15px;" />

Sono considerati i diagrammi di corpo libero dei giunti B e C, e vengono applicate le condizioni di equilibrio delle forze verticali e orizzontali per calcolare le forze sconosciute FBD, FCD e FEC. I valori risultanti per FBD, FCD e FEC sono rispettivamente 161.53 N, 173.08 N e 207.70 N.

Infine, sono considerati i diagrammi di corpo libero dei giunti D ed E, e vengono applicate le condizioni di equilibrio delle forze verticali e orizzontali. Questo fornisce i valori di FED e FDF pari a 138.46 N e 265.39 N, rispettivamente.

Metodo dei giunti: Risoluzione del problema II

Si consideri una struttura a traliccio con giunti privi di attrito fissati a una parete e un supporto a rulli. Se al giunto A viene applicata una forza di 150 N, le forze in ogni ...

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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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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...

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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

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 Force Analysis at Joints A to E

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