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CHAPTER 4 Force System Resultant 4.1 Moment of a Force - - - Scalar Formulation 1.Moment A measure of the tendency of the force to cause a body to rotate about the point or axis. • Torque (T) 扭力 • Bending moment (M) 彎曲力矩 T M P M 2. Vector quantity (1) Magnitude ( N-m or lb-ft) Mo = Fd d = moment arm or perpendicular distance from point O to the line of action of force. d o Lime of action (sliding vector) (2) Direction Right-Hard rule A. Sense of rotation ( Force rotates about Pt.O) Curled fingers B. Direction and sense of moment Thumb 3.Resultant Moment of Coplanar Force System Fn M RO don do1 F3 do2 do3 F1 F2 4.2 Cross Product 1. Definition (1) magnitude of (2)Direction of perpendicular to the plane containing A & B 2. Law of operation (1) (2) (3) 3. Cartesian Vector Formulation (1) Cross product of Cartesian unit vectors. i i j j k k 0 i j i j sin 90 k k j j k i ; k i j i (2) Cross product of vector A & B in Cartesian vector form A Ax i Ay j Az k B Bx i B y j Bz k A B ( Ax i Ay j Az k ) ( Bx i B y j Bz k ) Ax B y i j Ax Bz j k Ay Bx i j Ay Bz k j Az B y k i Az B y k j (Ay Bz -Az B y )i ( Az Bx Az Bx ) j (Ax B y Ay Bx )k A Ax i Ay j Az k i j k A B Ax Ay Az Bx By Bz Ay Az Ax Az Ax Ay i j k By Bz Bx Bz Bx By 4.3 Moment of a Force – Vector Formulation 1. Moment of a force F about pt. O Mo = r x F where r = A position vector from pt. O to any pt. on the line of action of force F . F d o (1) Magnitude Mo=|Mo|=| r x F | =| r|| F | sinθ=F r sinθ =F d (2) Direction Curl the right-hand fingers from r toward F (r cross F ) and the thumb is perpendicular to the plane containing r and F. 4.4 Principle of moments Varignon’s theorem The moment of a force about a point is equal to the sum of the moment of the force’s components about the point . F F1 r F2 o Mo=r x F Mo= r x (F1+F2) F = F1+F2 = r x F1+ r x F2 = MO1+MO2 4.5 Moment of a force about a specified Axis 1. Objective Find the component of this moment along a specified axis passes through the point about which the moment of a force is computed. 2. Scalar analysis (See textbook) 3. Vector analysis a Point O on axis aa’ b Ma Ө Mo= r x F O b’ A F Moment axis Axis of projection a’ (1) Moment of a force F about point 0 Mo = r × F Here, we assume that bb’ axis is the moment axis of Mo (2) Component of Mo onto aa´ axis Ma = Ma ua Ma=Mo cosθ =Mo ●ua=( r × F ) ● ua =trip scalar product Here Ma=magnitude of Ma ua= unit vector define the direction of aa´ axis If U a U ax i U ay j U azk r rx i ry j rz k F Fx i Fy j Fz k then M a Ua ( r F ) U ax U ay U az rx ry rz Fx Fy Fz 4.Method of Finding Moment about a specific axis (1) Find the moment of the force about point O Mo = r x F (2) Resolving the moment along the specific axis Ma = Maua = (Mo•ua) ua =[ua •( r x F )]ua 4.6 moment of a couple 1. Definition ( couple) 偶力矩 Two parallel forces have the same magnitude, opposite distances, and are separated by a perpendicular distance d. d 2. Scalar Formulation (1) Magnitude M=Fd (2) Direction & sense (Right-hand rule) •Thumb indicates the direction •Curled fingers indicates the sense of rotation 3. Vector Formulation M= r x F |M|=M=|r x F |=r F sinθ θ d =F d F F r Remark: (1) The couple moment is equivalent to the sum of the moment of both couple forces about any arbitrary point 0 in space. -F Mo= rAx( -F )+ rB x F =(-rA+rB) x F r B A =r x F= M F rB r o A (2) Couple moment is a free vector which can act at any point in space. F B -F r A Mo=Mo’= r x F=M o o’ 4. Equivalent Couples The forces of equal couples lie either in the same plane or in planes parallel to one another. A F F d -F plane A // plane B -F d B -F F 5. Resultant couple moment Apply couple moment at any point p on a body and add them vectorially. M2 M1 A B M2 M1 MR=ΣM=Σ r x F 4.7 Equivalent system 1. Equivalent system When the force and couple moment system produce the same “external” effects of translation and rotation of the body as their resultant , these two sets of loadings are said to be equivalent. 2. Principle of transmissibility The external effects on a rigid body remain unchanged,when a force, acting a given point on the body, is applied to another point lying on line of action of the force. line of action Same external effect F A P Internal effect ? P F Internal stresses are different. 3. Point O is on the line of action of the force F A A equivalent A equivalent F F o o -F o Original system Sliding vector 4. Point O is not on the line of action of the force F Couple moment F Mc= r x F M=r x F A F A r P A o line of o F o action Original system -F Force on Point A =Force on point O + couple moment on any point p. Example: F F o o A A Point O is on the line of action of the force F F o A o X P A d M= F d (Free vector) Mo= F d Point O is not on the line of action of the force 4.8 Resultant of a force & couple system 1. Objective Simplify a system of force and couple moments to their resultants to study the external effects on the body. 2. Procedures for Analysis (1)Force summations FR=F1+F2+……+ΣF (2)Moment summations MR0= ΣMC+r1o*F1+r2o*F2= ΣMC+ ΣM0 MC:Couple moment in the system Mo:Couple moment about pt.O of the force in the system. 4.9 Further Reduction of a force & couple system 1. Simplification to a single Resultant Force (1)Condition FR MR0 or FR*MR0 = 0 (2)Force system A. Concurrent Force system F2 F1 FR Equivalent P = System Fn no couple moment B.Coplanar Force System y F1,F2,F3 on xy plane F3 M1&M2:z direction MR0=ΣMC+ Σr * F P MR0 F2 x => => d= FR F1 FR=ΣF C. Parallel Force System 1. F1 // F2 //……// Fn 2. MR0 perpendicular to FR , MR0=ΣM+ Σr*F z z FR z F1 r2 F2 MR0 FR= ΣF r1 y = y = y M1 p o r3 F3 x x MR0 x M2 d = -------------- 2. Reduction to a wrench |FR|d=|MR0| FR (1) Condition: FR MR0 MR0=M +M// M = moment component FR M// = moment component // FR (2) Wrench (or Screw) An equivalent system reduces a simple resultant force FR and couple moment MR0 at pt.0 to a collinear force FR and couple moment M// at pt. FR MRo a b o FR b FR M// a a a b M// b o o p p b b a a