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

考虑一个铰接节点无摩擦的桁架结构，一端固定在墙上，另一端由地面上的滚轴支座支撑。若在节点 A 施加一个 150 N 的力，则确定桁架各杆件中的受力情况。
可以使用节点法确定每个节点上的受力。
首先，杆件 AC 与 AB 之间的夹角可通过 BC 与 AB 长度之比的反正切值确定。
其次，绘制节点 A 的受力分析图，并将力分解为各个分量。通过应用平衡方程，可求解出未知力。
接下来，通过绘制节点 B 和 C 的受力分析图，并利用静力学平衡方程，可确定节点 B 和 C 处的未知力。
最后，绘制结点 D 和 E 的受力图。对于构件 CD 和 EF，将力分解为水平和垂直分量。利用平衡方程可确定结点 D 和 E 处的未知力。

<ol>
	<li>希伯勒，RC（2016 年）。工程力学 \u2012 静力学和动力学。新泽西州霍博肯:Pearson Prentice Hall。第 276-277 页。</li>
</ol>

节点法：问题求解ii

假设有一个无摩擦接头固定在墙壁和滚轮支撑上的桁架结构。如果对节点A施加150N的力，则可以使用节点法来确定桁架在每个构件中的力。 <img alt="Figure 1" src="/files/ftp_upload/14283/14283_PT_Fig_1v2.svg" style="text-align: center; height: 350px;"> 构件AC和AB之间的夹角可以通过取BC和AB长度比的反正切来得到，计算得出的结果为67.38°。绘制节点A的受力分析图，并将所有未知的力分解为各自的分力。 然后可以利用竖直力的平衡条件来计算作用在构件CA上的力，结果为173.08N。 <img alt="Equation 1" src="/files/ftp_upload/14283/14283_PT_Equation_1v2.svg" style="text-align: center; height: 15px;"> 类似地，可以利用水平力的平衡条件来找到力FAB，得出的结果为161.54N。 <img alt="Equation 2" src="/files/ftp_upload/14283/14283_PT_Equation_2v2.svg" style="text-align: center; height: 15px;">

绘制节点B和节点C的受力分析图，并通过使用竖直方向和水平方向的力的平衡条件来计算其中未知的力FBD、FCD和FEC。计算的结果分别为161.53N、173.08N和207.70N。

最后绘制节点D和节点E的受力分析图，并应用竖直和水平力的平衡条件。这给出了FED和FDF的值分别为138.46N和265.39N。

节点法：问题求解II

考虑一个由无摩擦铰接节点组成的桁架结构，其一端固定于墙面，另一端由滚轴支座支撑。若在节点A施加150 N的力，则可求解各杆件中的内力……

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Mechanical Engineering

机械工程

Unit 1

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Structural Analysis

结构分析

KEY TERMS AND CONCEPTS

关键术语与概念

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

introduction to structures

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Truss Joint Force Analysis by Equilibrium

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method of joints: problem solving ii

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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method of sections: problem solving i

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Space Truss Forces and Stability

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frames

Frame Analysis in Structural Systems

Frames are important parts of many mechanical and structural systems used every day. They are valued for stability and for their ability to bear heavy loads. A frame is built from two-force members and multi-force members, and the parts are connected with pin joints. Trusses are different because they are made entirely of two-force members. Frames appear in many practical settings. They can support beams and columns in buildings. They are also used in automobile chassis construction. In the...

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frames: problem solving ii

Hydraulic Hoist Force Analysis

Hydraulic hoist force analysis shows how to find the forces in key members of a simple frame. The load is 1 kN, and the structure is represented with known member dimensions. In this setup, BF and BD act as two-force members. EFG and EDC are multi-force members, so each one is analyzed with a free-body diagram. The first step is to study member EFG. The inclined force in BF is split into vertical and horizontal parts. Then the moment equilibrium condition is applied at joint E. Using this...

machines

Cutting Pliers as a Force-Multiplying Machine

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machines: problem solving i

Toggle Clamp Force Analysis

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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.

mechanical engineering

Truss Force Analysis at Joints A to E

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