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

Rozważmy symetryczną konstrukcję wiązary dachowej składającą się z prętów pionowych, ukośnych i poziomych.
Rysowany jest wykres swobodnego ciała w celu analizy sił działających na pręty DC i HC przy użyciu metody przekroju.
W tym przypadku znanymi parametrami są obciążenia i długości prętów poziomych i pionowych.
Najpierw, sumując momenty wokół punktu A, oblicza się siłę reakcji w punkcie E.
Ponadto, korzystając z warunku równowagi sił dla kratownicy, określa się siłę reakcji w A.
Symetria kratownicy zapewnia, że obie siły reakcji są równe.
Teraz wykonuje się cięcie wzdłuż płaszczyzny przecinającej pręty DC, HC i HG i rysuje się schemat swobodnego ciała mniejszego przekroju.
Biorąc pod uwagę sumę momentów wokół H, otrzymujemy siłę wzdłuż DC. Znak dodatni wskazuje siłę rozciągającą.
Siła wzdłuż HC rozkłada się na składowe sinus i cosinus, a do oszacowania kąta theta stosuje się trygonometrię.
Warunek równowagi momentu w E daje siłę działającą na HC, przy czym znak ujemny wskazuje siłę ściskającą.

<ol>
	<li>R. C., Hibbeler Inżynieria Mechanika Statyka, Pearson. Str. 291-293</li>
	<li>F.P. Beer, E.R. Johnston, D.F. Mazurek, P.J. Cornwell, B.P. Self, Mechanika wektorowa dla inżynierów, statyka i dynamika, mechanika inżynierska, Edukacja Mc Graw-Hill. Str. 217</li>
</ol>

metoda przekrojów: rozwiązywanie problemów i

Rozważ symetryczną konstrukcję kratownicowego dachu, składającą się z elementów pionowych, skośnych i poziomych. Długość każdego elementu poziomego wynosi 4 m. Długości elementów pionowych FB i HD wynoszą 4 m, natomiast długość elementu GC wynosi 6 m. Obciążenia działające w węzłach F, G i H wynoszą 2 kN, a w węzłach A i E wynoszą 1 kN.

<img alt="Figure 1" src="/files/ftp_upload/14286/14286_PT_Fig_1v2.svg" style="text-align: center; height: 350px;" />

Metoda przekrojów jest stosowana do obliczenia sił działających na elementy DC i HC. Warunek równowagi momentu jest stosowany w punkcie A, a znane wartości sił i odległości są podstawiane do równania momentu.

<img alt="Equation 1" src="/files/ftp_upload/14286/14286_PT_Equation_1v2.svg" style="text-align: center; height: 13px;" />

W wyniku tego szacuje się siłę reakcji równą 4 kN w punkcie E. Następnie przez równoważenie sił pionowych w punkcie A stwierdza się, że siła reakcji w A wynosi również 4 kN.

<img alt="Equation 2" src="/files/ftp_upload/14286/14286_PT_Equation_2v2.svg" style="text-align: center; height: 13px;" />

Ze względu na symetrię kratownicy, siły reakcji w punktach A i E są równe.

Siły działające na elementy DC i HC można uzyskać po ustaleniu sił reakcji. Wykonuje się cięcie przekroju wzdłuż płaszczyzny przecinającej elementy DC, HC i HG i rozważa się siłę wewnętrzną mniejszego przekroju na wykresie swobodnego ciała.

<img alt="Figure 2" src="/files/ftp_upload/14286/14286_PT_Fig_2v2.svg" style="text-align: center; height: 350px;" />

Poprzez sumowanie momentów wokół punktu H oblicza się siłę wzdłuż DC równą dodatniemu 3 kN, co wskazuje na siłę rozciągającą.

<img alt="Equation 3" src="/files/ftp_upload/14286/14286_PT_Equation_3v2.svg" style="text-align: center; height: 13px;" />

Siła wzdłuż CH jest rozkładana na składowe sinus i cosinus. Trygonometria jest używana do obliczenia kąta między elementem CH a osią poziomą, który wynosi 45°. Ostatecznie, zastosowanie warunku równowagi momentu w punkcie E daje siłę -1,41 kN dla FCH.

<img alt="Equation 4" src="/files/ftp_upload/14286/14286_PT_Equation_4v2.svg" style="text-align: center; height: 13px;" />

W tym przypadku wynik ujemny oznacza siłę ściskającą działającą na element CH.

Metoda przekrojów: Rozwiązywanie problemów I

Rozważmy symetryczną konstrukcję wiązary dachowej, składającą się z prętów pionowych, ukośnych i poziomych. Długość każdego pręta poziomego wynosi 4 m. Długość ...

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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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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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