静力学第2.pptVIP

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2017-03-23 发布于江苏
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静力学第2.ppt

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静力学第2

Chapter 2 Resultants of force systems 2.1 Introduction 2.2 Reduction of Concurrent Force Systems 1.Definition:各力作用线汇交于一点的力系----concurrent forces . 2.Reduction----replacing a system of concurrent forces with a single equivalent force. The equivalent scalar equations are: Method 1 Scalar Method 2 Vector 2.3 Moment of a Force About a Point 1.Definition The moment of a force F about a point O, called the moment center, is defined as Mo= r?F A vector where: r is the vector from the point O to any point on the line of action of F. Unit: N ?m. The magnitude of Mo is where d is the perpendicular distance from O to the line of action of F, called the moment arm of F. Then: Mo=Fd 1.vector solution Writing vector F as following 2.Scalar solution: component of F at point B in vector form: 3. Scalar solution: component of F at point C 2.4 Moment of a Force About a Axis 力对轴之矩是一个力使一个物体绕轴转动趋势的度量 a. Physical characteristic 具体计算时可利用:(a) Principle of momentThe moment of a force about a given axis is equal to the sum of the moments of its components about that axis.(b) Geometric interpretation C. Rectangular components Writing r and F in rectangular representations: r=rxi+ryj+rzk F= Fxi+Fyj+Fzk For MLL=Mo·? ? Mx=Mo·i; My=Mo·j; Mz=Mo·k ? Mo= Mxi+ Myj+ Mzk Sample problem 2.3 Determine the combined moment of P and Q about the x-axis, given that P=80i+120k N and Q=-50k N. Express your answer in vector form. Solution: From principle of moments: + mx(P) = mx(Pz)+mx(Px) = 0.4?120=48 Nm + mx(Q) = -0.4 ? 50 = -20 Nm ? + mx(P+Q) = 48-20 = 28 Nm in vector form: mx(P+Q)= 28 i Nm Sample problem 2.4 Determine the magnitude of the force F given that its moment about an axis directed from point B toward point C is 137.3 lb·ft. Solution: mBC(F) = mB(F)??BC = rBA?F??BC