Name: beam shear force diagram
Uploaded: 2025-05-22T09:06:03.000Z
Duration: 1 min 27 s
Channel: JoVE

せん断図は、ビームの長さに沿ったせん断力の分布をグラフィカルに表現したものです。
両端で支持され、垂直な荷重を受けるビームABを考えてみましょう。そのせん断図を作成します。
ビームのフリーボディ線図を描画します。力とモーメントの平衡方程式を使用して、各支持体の反力を計算できます。
梁に断面の方法を適用します。
次に、領域 AP 内の点 A からの任意の距離を考えます。この断面の自由体図を描画します。平衡条件下では、断面内のせん断力は支持体の反力と等しくなります。
同様に、領域PQ内の任意の点について考えます。自由体図を描画し、平衡方程式を使用することで、断面内のせん断を決定できます。
最後に、任意の点はセクションQB内で考慮されます。自由体図を描画し、平衡方程式を思い出すことにより、断面のせん断を決定することができ、これはビームの終わりまで一定に保たれ、最終的にはゼロになり、平衡を維持します。

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	<li>ヒベラー、RC(2016)。エンジニアリングメカニクス \u2012 静力学と力学。ニュージャージー州ホーボーケン:ピアソンプレンティスホール。361ページ。</li>
	<li>ビール、FP;ジョンストン、ER;マズレック、D.F;クロムウェル、PJおよびセルフ、BP(2019)。エンジニアのためのベクトル力学 \u2012 静力学と動力学。ニューヨーク:マグロウヒル。381ページ。</li>
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せん断図

梁力学の研究において、せん断図は梁の長さに沿ったせん断力の分布を理解する上で重要な役割を果たします。両端で支持され、直角の荷重を受ける一本の梁ABを考えてみましょう。

まず、梁の自由体図を描き、梁に作用するすべての外部力と内部反力を表します。力とモーメントの平衡方程式を用いて、各支持点での反力を計算することができます。与えられた梁では、点Aと点Bの鉛直成分の反力はそれぞれ24 kNと18 kNです。

<img alt="図1" src="/files/ftp_upload/14301/14301_PT_Figure_1v2.svg" style="text-align: center; height: 300px;" />

せん断図は、梁の長さに沿ったせん断力の分布を示すグラフィカルな表現です。この図を作成するために、断面法が梁に適用されます。次に、領域AP内の点Aからの任意の距離を考えます。この断面の自由体図を描いて、その上に作用する力を分析します。平衡条件のもとでは、断面のせん断力は支持点での反力と等しくなります。ここではV1=24 kNです。

同様に、領域PQ内の任意の点を考えます。この断面の自由体図を描き、平衡方程式を使ってせん断力V2を決定することができます。

<img alt="式1" src="/files/ftp_upload/14301/14301_Equation_1v2.svg" style="text-align: center; height: 13px;" />

最後に、領域QB内の任意の点を考えます。この断面の自由体図を描き、平衡方程式を再度適用して、断面内のせん断力V3を求めます。

<img alt="式2" src="/files/ftp_upload/14301/14301_Equation_2v2.svg" style="text-align: center; height: 13px;" />

このせん断力は、梁の終端まで一定の値を保ち続け、最終的にゼロになります。これにより平衡が保たれます。

ビーム力学の研究では、せん断図はビームの長さに沿ったせん断力の分布を理解する上で重要な役割を果たします。考える。。。

Education

JoVE Core

Mechanical Engineering

機械工学

Unit 1

ユニット1

Internal Forces

内力

KEY TERMS AND CONCEPTS

主要用語と概念

Shear Diagram

beam shear force diagram

Beam shear force diagrams show how shear changes along the length of a supported beam. In beam mechanics, they help students track the internal shear force caused by perpendicular loads. The beam AB in this example is supported at both ends, so the first step is to draw a free-body diagram that shows the external forces and internal reactions.

The reaction forces at supports A and B are found with equilibrium equations for force and moment. For the beam shown, the vertical reaction at A is 24 kN and the vertical reaction at B is 18 kN. These values become the starting point for building the shear diagram.

To form the shear diagram, the method of sections is used. A point is chosen at an arbitrary distance from A in the region AP, and a free-body diagram of that section is drawn. From equilibrium, the shear force in that section is V1 = 24 kN.

The same process is repeated in the region PQ. An arbitrary point is selected, the section is isolated, and the equilibrium equation is applied to find V2. A third section is then taken in the region QB. Its free-body diagram gives V3, which stays constant until the end of the beam and then goes to zero so the beam remains in equilibrium.

Beam Shear Force Diagram

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key terms and concepts

Learn how to build a beam shear force diagram using free-body diagrams, reaction forces, and the method of sections for a supported beam.

Beam Shear Force Diagram | JoVE K12

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mechanical engineering

Beam Shear Force Diagram

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