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

考虑一个由竖直杆、斜杆和水平杆组成的对称屋顶桁架结构。
绘制受力图，以使用截面法分析 DC 和 HC 构件上的受力情况。
此处，水平杆件和竖直杆件的载荷及其长度均为已知参数。
首先，对点 A 求力矩和，计算出点 E 处的支座反力。
此外，利用桁架的力平衡条件，可确定 A 点的支座反力。
桁架的对称性确保了两个支座反力相等。
现在，沿一个相交平面进行切割 直流电 HC 和 HG 成员，并绘制较小部分的受力分析图。
对点 H 取力矩和，可得沿 DC 方向的力。正值表示该力为拉力。
沿 HC 的力被分解为正弦和余弦分量，并利用三角学估算角度 theta。
在 E 点的力矩平衡条件可得出 HC 上的力，负号表示该力为压力。

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	<li>R. C.，Hibbeler Engineering Mechanics Statics，皮尔逊，第 291-293 页</li>
	<li>F.P. Beer， E.R. Johnston， D.F. Mazurek， P.J. Cornwell， B.P. Self， 工程师的矢量力学 静力学和动力学 工程力学静力学， Mc Graw-Hill Education.第 217 页</li>
</ol>

剖析法：问题求解 i

假设有一个对称的屋顶桁架结构，该结构由垂直、对角和水平支撑的构件组成。每个水平支撑构件的长度为4m。垂直支撑的构件FB和HD的长度为4m，而支撑构件GC的长度为6m。在连接点F、G和H处的荷载为2kN，而在连接点A和E处的荷载为1kN。 <img alt="Figure 1" src="/files/ftp_upload/14286/14286_PT_Fig_1v2.svg" style="text-align: center; height: 350px;"> 采用截面法来计算支撑构件DC和HC上的作用力。将力矩平衡条件应用于A点，并且将已知的力和距离值代入到力矩方程中。 <img alt="Equation 1" src="/files/ftp_upload/14286/14286_PT_Equation_1v2.svg" style="text-align: center; height: 13px;"> 通过以上条件可以得出在点 E 处的反作用力为4kN。随后，位于A点处竖直力的平衡条件揭示了A点处的反作用力也是4kN。 <img alt="Equation 2" src="/files/ftp_upload/14286/14286_PT_Equation_2v2.svg" style="text-align: center; height: 13px;"> 由于桁架具有对称性，因此位于A点和E点处的反作用力是相等的。 在确定了反作用力后，可以得到用来支撑构件DC和HC上的作用力。沿着一个与支撑构件DC、HC和HG相交的平面来进行截面切割，则可以绘制出较小截面的受力分析图。 <img alt="Figure 2" src="/files/ftp_upload/14286/14286_PT_Fig_2v2.svg" style="text-align: center; height: 350px;"> 通过对点H来进行力矩求和，可以计算出沿着DC方向的力为3kN，并以此来表示拉力。 <img alt="Equation 3" src="/files/ftp_upload/14286/14286_PT_Equation_3v2.svg" style="text-align: center; height: 13px;"> CH上的力被分解为其正弦分量和余弦分量。通过三角形法则可以得出CH和水平轴之间的角度为45度。最后，通过点E处的力矩平衡条件可以得出位于FCH的力为-1.41kN。 <img alt="Equation 4" src="/files/ftp_upload/14286/14286_PT_Equation_4v2.svg" style="text-align: center; height: 13px;"> 在这种情况下，负结果能够用来表示作用于CH的压缩力。

剖析法：问题求解 I

考虑一个由竖直、斜向和水平杆件组成的对称屋顶桁架结构。每个水平杆件的长度为 4 m。

Education

教育

JoVE Core

Mechanical Engineering

机械工程

Unit 1

第一单元

Structural Analysis

结构分析

KEY TERMS AND CONCEPTS

关键术语与概念

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.

mechanical engineering

Truss Member Force Calculation by Sections

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