CHAPTER 4 Force System Resultant 4 1 Moment

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 do 1 do 2 don do 3 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 j

(2) Cross product of vector A & B in Cartesian vector form

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 1 r o Mo=r x F F = F 1+F 2 Mo= r x (F 1+F 2) = r x F 1+ r x F 2 = MO 1+MO 2

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 O rx. F A b’ Moment axis Mo= a’ Axis of projection

(1) Moment of a force F about point 0 (2) Mo = r × F (3) Here, we assume that bb’ axis is the moment axis of Mo (2) Component of Mo onto aa´ axis M a = M a 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 then 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 M a = Ma ua = (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 =F d F θ d r F

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= r. Ax( -F )+ r. B x F =(-r. A+r. B) x F =r x F= M r B A r. B F o r. A (2) Couple moment is a free vector which can act at any point in space. B F -F r o A Mo=Mo’= r x F=M o’

4. Equivalent Couples The forces of equal couples lie either in the same plane or in planes parallel to one another. F d F A -F -F d plane A // plane B F B -F 5. Resultant couple moment Apply couple moment at any point p on a body and add them vectorially. M 2 M 1 A B M 2 M 1 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 P Internal effect ? F A P Internal stresses are different.

3. Point O is on the line of action of the force F A equivalent o A F equivalent o o -F F A Original system Sliding vector 4. Point O is not on the line of action of the force F Couple moment F A F o Original system line of action r Mc= r x F A M=r x F P F o A o -F Force on Point A =Force on point O + couple moment on any point p.

Example: F o A o F A Point O is on the line of action of the force F o d Mo= F d F A o X P A M= F d (Free vector) 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=F 1+F 2+……+ΣF (2)Moment summations MR 0= ΣMC+r 1 o*F 1+r 2 o*F 2= ΣMC+ ΣM 0 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 MR 0 or FR*MR 0 = 0 (2)Force system A. Concurrent Force system F 2 F 1 FR Equivalent P = System Fn no couple moment B. Coplanar Force System y F 1, F 2, F 3 on xy plane F 3 F 2 M 1&M 2: z direction x => MR 0=ΣMC+ Σr * F => P MR 0 d= FR F 1 FR=ΣF

C. Parallel Force System 1. F 1 // F 2 //……// Fn 2. MR 0 perpendicular to FR , z F 1 MR 0=ΣM+ Σr*F z r 2 F 2 r 1 MR 0 y FR FR= ΣF = y M 1 p r 3 x F 3 z x o x MR 0 d = ------- M 2 |FR|d=|MR 0| 2. Reduction to a wrench (1) Condition: FR MR 0=M +M// M = moment component FR M// = moment component // FR FR

(2) Wrench (or Screw) An equivalent system reduces a simple resultant force FR and couple moment MR 0 at pt. 0 to a collinear force FR and couple moment M// at pt. FR MRo a b o FR M// b a a a b o p FR b a o M// p b