Force System Resultant by mikesanye

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									    CHAPTER 4
Force System Resultant
4.1 Moment of a Force - - - Scalar Formulation
      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.


                     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
  3.Resultant Moment of Coplanar Force System

               M RO
                do1           F3
                do2     do3
4.2 Cross Product
    1. Definition

     (1) magnitude of

     (2)Direction of
            perpendicular to the plane containing A & B
2. Law of operation



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
      (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 .

           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

        Point O on axis aa’

                              Ma Ө
                                       Mo= r x F

                                       A           F

                     axis               Axis of projection
(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

  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.


    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
 (1) The couple moment is equivalent to the sum of the moment
 of both couple forces about any arbitrary point 0 in space.

    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

        F         B
                                     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
                                B -F                  F

  5. Resultant couple moment
    Apply couple moment at any point p on a body and add
    them vectorially.

                                    B                     M2

                  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

        A                                    A       equivalent             A
                    equivalent                                         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
Original system
                 Force on Point A
                =Force on point O + couple moment on any point p.

                      F                    F
  o                                   o                    A

      Point O is on the line of action of the force

                 F                        F
 o                   A                o        X   P       A
                                           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

  2. Procedures for Analysis

   (1)Force summations

   (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
                 FR MR0 or FR*MR0 = 0
          (2)Force system
          A. Concurrent Force system
                 F2                        F1                 FR

                           P                      =
                          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=
                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                                  b       FR
         M//           a a
                                              b                    M//
    b                                                     o
               o p                                            p
                                 b                                       b

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