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

Rozważ konstrukcję kratownicową z beztarciowymi połączeniami przymocowanymi do ściany i podporą rolkową na ziemi. Jeśli do złącza A zostanie przyłożona siła 150 N, określ siły w każdym elemencie kratownicy.
Siły na każdym połączeniu można określić za pomocą metody połączeń.
Po pierwsze, kąt między prętami AC i AB można wyznaczyć jako odwrotność brązu stosunku długości BC i AB.
Po drugie, narysuj diagram swobodnego ciała stawu A i rozłóż siły na ich składowe. Stosując równania równowagi, można wyznaczyć nieznane siły.
Następnie, rysując wykresy ciał swobodnych stawów B i C oraz używając równań równowagi, można wyznaczyć nieznane siły w stawach B i C.
Na koniec narysuj diagram swobodnego ciała stawów D i E. W przypadku prętów CD i EF rozmieść siły na komponenty poziome i pionowe. Nieznane siły w stawach D i E można wyznaczyć za pomocą równań równowagi.

<ol>
	<li>Hibbeler, R.C. (2016). Mechanika inżynierska \u2012 Statyka i dynamika. Hoboken, New Jersey: Pearson Prentice Hall. Strony 276-277.</li>
</ol>

metoda równoważenia węzłów: rozwiązywanie problemów ii

Rozważ konstrukcję kratownicy z węzłami nieruchomymi na ścianie i podporą rolkową. Jeśli siła 150 N jest przyłożona do węzła A, siły w każdym z prętów kratownicy można określić za pomocą metody węzłów.

<img alt="Figure 1" src="/files/ftp_upload/14283/14283_PT_Fig_1v2.svg" style="text-align: center; height: 350px;" />

Kąt pomiędzy prętami AC i AB można znaleźć, obliczając odwrotną tangens stosunku długości BC i AB,, a wynik wynosi 67,38°. Rozważany jest diagram sił dla węzła A, a wszystkie nieznane siły są rozłożone na odpowiednie składowe.

Warunek równowagi sił pionowych może zostać użyty do obliczenia siły działającej na pręt CA, która wynosi 173,08 N.

<img alt="Equation 1" src="/files/ftp_upload/14283/14283_PT_Equation_1v2.svg" style="text-align: center; height: 15px;" />

Podobnie, warunek równowagi sił poziomych może być użyty do znalezienia siły FAB równą 161,54 N.

<img alt="Equation 2" src="/files/ftp_upload/14283/14283_PT_Equation_2v2.svg" style="text-align: center; height: 15px;" />

Diagramy sił dla węzłów B i C są rozważane, a warunki równowagi sił pionowych i poziomych są stosowane do obliczenia nieznanych sił FBD , FCD , i FEC. Wynikowe wartości dla FBD , FCD , i FEC wynoszą odpowiednio 161,53 N, 173,08 N i 207,70 N.

Diagramy sił dla węzłów D i E są ostatecznie rozważane, a warunki równowagi sił pionowych i poziomych są stosowane. Daje to wartości FED i FDF równą odpowiednio 138,46 N i 265,39 N.

Metoda równoważenia węzłów: Rozwiązywanie problemów II

Rozważ konstrukcję kratownicową z beztarciowymi połączeniami przymocowanymi do ściany i wspornika rolkowego. Jeżeli do przegubu A przyłoży się siłę 150 N, siły w każdym ...

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Mechanical Engineering

Inżynieria mechaniczna

Unit 1

Jednostka 1

Structural Analysis

Analiza strukturalna

KEY TERMS AND CONCEPTS

Kluczowe terminy i pojęcia

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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