Force resultant Lecture by alicejenny

VIEWS: 14 PAGES: 45

									                                                     University of Palestine
                                   College of Engineering & Urban Planning


            ENGINEERING MECHANICS
              STATICS & DYNAMICS




               Instructor: Eng. Eman Al.Swaity
                                                                               1
Lecture 5                                        Chapter 4: Force resultant
                               Chapter(4)
                       Force System Resultants
    Chapter Objectives
    •   To discuss the concept of the moment of a force and show
        how to calculate it in two and three dimensions.
    •   To provide a method for finding the moment of a force
        about a specified axis.
    •   To define the moment of a couple.
    •   To present methods for determining the resultants of
        nonconcurrent force systems.
    •   To indicate how to reduce a simple distributed loading to
     a resultant force having a specified location.


                                                                           2
Lecture 5                                             Chapter 4: Force resultant
 Today’s Objectives :
 Students will be able to:
 a) understand and define moment, and,
 b) determine moments of a force in 2-D
    and 3-D cases.


   The moment (Torque) of a force about a point or axis provides a
   measure of the tendency of the force to cause a body to rotate
   about the point or axis. For example, consider the horizontal
   force Fx, which acts perpendicular to the handle of the wrench
   and is located a distance dy from point 0, is seen that this force
   tends to cause the pipe to turn about the Z axis.
                                                                        3
Lecture 5                                         Chapter 4: Force resultant
    Magnitude: The magnitude of Mo is




   where d is referred to as the moment arm or perpendicular
   distance from the axis at point 0 to the line of action of the force.
   Units of moment magnitude consist of force times distance, e.g.,
   N. m or lb. ft.
                                                                           4
Lecture 5                                           Chapter 4: Force resultant
     Direction:
     The direction of Mo will be specified
     by using the "right-hand rule." To do
     this, the fingers of the right hand are
     curled such that they follow the sense
     of rotation, which would occur if the
     force could rotate about point 0, The
     thumb then points along the moment
     axis so that it gives the direction and
     sense of the moment vector, which
     is upward and perpendicular to the
     shaded plane containing F and d.

                                                                    5
Lecture 5                                      Chapter 4: Force resultant
                            READING
                                                         F = 10 N
   1. What is the moment of the 10 N force
      about point A (MA)?
      A) 10 N·m       B) 30 N·m    C) 13 N·m
                                                        d=3m
      D) (10/3) N·m   E) 7 N·m                   • A




                                                                    6
Lecture 5                                      Chapter 4: Force resultant
            APPLICATIONS (continued)




                    What is the effect of the 30 N
                    force on the lug nut?




                                                      7
Lecture 5                        Chapter 4: Force resultant
                         MOMENT IN 2-D




   The moment of a force about a point provides a measure of the
   tendency for rotation (sometimes called a torque).


                                                                     8
Lecture 5                                       Chapter 4: Force resultant
                          MOMENT IN 2-D (continued)

     In the 2-D case, the magnitude of the moment is
     Mo = F d




    As shown, d is the perpendicular distance from point O to the
    line of action of the force.

    In 2-D, the direction of MO is either clockwise or
    counter-clockwise depending on the tendency for rotation.
                                                                       9
Lecture 5                                         Chapter 4: Force resultant
                          MOMENT IN 2-D (continued)
                      F
            a
   b                            For example, MO = F d and the
    O                           direction is counter-clockwise.
     d
     Often it is easier to determine MO by using the components of F
     as shown.                   Fy       F

                                      Fx
                  b        a
                     O
   Using this approach, MO = (FY a) – (FX b). Note the different
   signs on the terms! The typical sign convention for a moment in
   2-D is that counter-clockwise is considered positive. We can
   determine the direction of rotation by imagining the body pinned
   at O and deciding which way the body would rotate because of
   the force.
                                                                    10
Lecture 5                                       Chapter 4: Force resultant
                              EXAMPLE 1
                                      Given: A 400 N force is
                                             applied to the frame
                                             and  = 20°.
                                      Find: The moment of the
                                            force at A.


                                      Plan:
     1) Resolve the force along x and y axes.
     2) Determine MA using scalar analysis.



                                                                    11
Lecture 5                                       Chapter 4: Force resultant
                            EXAMPLE 1 (continued)




                                 Solution
       +  Fy = -400 cos 20° N
            +  Fx = -400 sin 20° N
            + MA = {(400 cos 20°)(2) + (400 sin 20°)(3)} N·m
                  = 1160 N·m

                                                                    12
Lecture 5                                       Chapter 4: Force resultant
                             EXAMPLE 2
        Find    the moment of F1 at point o

                               F1=100kN         F2=200kN

                     +                          O

                                       4.5m


                     M = F.d
                     M = 100  4.5 = 450 kN.m

            Find the moment of F2 at point o
                    M = F.d
                    M = 200  0 = 0 kN.m
                                                                    13
Lecture 5                                       Chapter 4: Force resultant
           Example 3
             Fine the moment of all forces at point A
                             50N

                        1m         100N             100N
                50N
                        1m                 A               50N
                                                                        -
         Don’t forget
          the signs
                              3m     3m        2m    2m


                 Moment at A = F1d1 + F2d2 + F3d3 +…. =  Fi .d i
                 Five forces means five moments
                   MA= (-)500 + (-)1002 + 1003 + (-)506 + 501
                   MA= -150 N.m

                                                                        14
Lecture 5                                           Chapter 4: Force resultant
           Moment theorem (Varignon’s theorem)
             The moment of force F at point O is
               equal to the sum of moments of the
               force components at the same point O
             Example 4
               Fined the magnitude of the moment
               about point O in four deferent ways
               Technique I
                  the moment arm d
                       d = 4cos40o + 2sin40o
                       d = 4.35m
                  the moment = Fd                                     -
                       M = -6004.35 = -2610 N.m
                                                                      15
Lecture 5                                         Chapter 4: Force resultant
     Example 4
       Technique II
         Replace the force by its components
              F1 = 600cos40o = 460N                                       -
              F2 = 600sin40o = 386N
         Moment theorem
              M0= (-)4604 + (-)3862
              M0 = -2610N.m
       Technique III
         the principle of transmissibility
              d1 = 4 + 2tan40o = 5.68m
              M0= (-)4605.68 + 3860                             -
              M0 = -2610N.m
                                                                      16
Lecture 5                                      Chapter 4: Force resultant
     Example 4
       Technique IV
          the principle of transmissibility
               d1 = 2 + 4tan50o = 6.77m
               M0= 4600 + (-)3866.77 = -2610N.m




                                                        -



                                                                 17
Lecture 5                                    Chapter 4: Force resultant
                   GROUP PROBLEM SOLVING
                                    Given: A 40 N force is
                                           applied to the wrench.
                                     Find: The moment of the
                                           force at O.
                                    Plan: 1) Resolve the force
                                          along x and y axes.
                                           2) Determine MO using
                                           scalar analysis.
       Solution: +  Fy = - 40 cos 20° N
                 +  Fx = - 40 sin 20° N
       + MO = {-(40 cos 20°)(200) + (40 sin 20°)(30)}N·mm
            = -7107 N·mm = - 7.11 N·m
                                                                  18
Lecture 5                                     Chapter 4: Force resultant
   For each case illustrated in Fig. 4-4, determine the moment of the
   force about point O.




                                                                      19
Lecture 5                                         Chapter 4: Force resultant
                                20
Lecture 5   Chapter 4: Force resultant
    In general, the cross product of two vectors A and B results in
    another vector C , i.e., C = A  B. The magnitude and direction
    of the resulting vector can be written as
                     C = A  B = A B sin  UC
    Here UC is the unit vector perpendicular to both A and B
    vectors as shown (or to the plane containing the
    A and B vectors).                                               21
Lecture 5                                        Chapter 4: Force resultant
                            CROSS PRODUCT


   The right hand rule is a useful tool for determining the direction of
   the vector resulting from a cross product.
   For example: i  j = k
   Note that a vector crossed into itself is zero, e.g., i  i = 0




                                                                       22
Lecture 5                                          Chapter 4: Force resultant
                   CROSS PRODUCT (continued)
   Of even more utility, the cross product can be written as




   Each component can be determined using 2  2 determinants.




                                                                     23
Lecture 5                                        Chapter 4: Force resultant
  The moment of a force F about point 0, or actually about the
  moment axis passing through 0 and perpendicular to the plane
  containing 0 and F, Fig. 4-12a, can be expressed using the vector
  cross product, namely,




                                                                      24
Lecture 5                                        Chapter 4: Force resultant
                                25
Lecture 5   Chapter 4: Force resultant
                                26
Lecture 5   Chapter 4: Force resultant
           Moment
            M=rF
              r = xi + yj + zk
              F = FXi + FYj + FZk

                  i       j      k
             M  rx      ry      rz
                 FX      FY      FZ

            M = (ryFz – rzFy)i + (rzFx – rxFz)j + (rxFy – ryFx)k
              Mx = ryFz – rzFy
              My = rzFx – rxFz
              Mz = rxFy – ryFx

                                                                                   27
Lecture 5                                                      Chapter 4: Force resultant
                           EXAMPLE 2

                           Given: a = 3 in, b = 6 in and c = 2 in.
                           Find: Moment of F about point O.
            o              Plan:

                             1) Find rOA.
                             2) Determine MO = rOA  F .

     Solution rOA = {3 i + 6 j – 0 k} in
        MO =     i j k
                3 6 0    = [{6(-1) – 0(2)} i – {3(-1) – 0(3)} j +
                3 2 -1      {3(2) – 6(3)} k] lb·in
Lecture 5                = {-6 i + 3 j – 12 k} lb·in 4: Force resultant
                                            Chapter
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Lecture 5   Chapter 4: Force resultant
                                30
Lecture 5   Chapter 4: Force resultant
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Lecture 5   Chapter 4: Force resultant
                                32
Lecture 5   Chapter 4: Force resultant
                                33
Lecture 5   Chapter 4: Force resultant
       Recall that when the moment of a force is computed about a point, the
       moment and its axis are always perpendicular to the plane containing
       the force and the moment arm.
       In some problems it is important to find the component of this moment
       along a specified axis that passes through the point. To solve this problem
       either a scalar or vector analysis can be used.




                                                                                     34
Lecture 5                                                  Chapter 4: Force resultant
                          APPLICATIONS




    With the force F, a person is
    creating the moment MA.
    What portion of MA is used in
    turning the socket?

                                                             35
Lecture 5                                Chapter 4: Force resultant
                                 SCALAR ANALYSIS
  Recall that the moment of a force about any point A is MA= F dA
  where dA is the perpendicular (or shortest) distance from the point
  to the force’s line of action. This concept can be extended to find
  the moment of a force about an axis.
   In the figure above, the
   moment about the y-axis
   would be My= 20 (0.3) =
   6 N·m. However this
   calculation is not always
   trivial and vector analysis
   may be preferable.




                                                                        36
Lecture 5                                          Chapter 4: Force resultant
            VECTOR ANALYSIS




                                                  37
Lecture 5                     Chapter 4: Force resultant
                                38
Lecture 5   Chapter 4: Force resultant
                                39
Lecture 5   Chapter 4: Force resultant
                                40
Lecture 5   Chapter 4: Force resultant
       Example
       Determine the moment MZ of T about z-axis passing O
          Technique I
                           AB      (12  0)i  (0  15) j  (9  0)k
            T  T .n F  T     10
                           AB        (12  0) 2  (0  15) 2  (9  0) 2
            T  10  0.566i  0.707 j  0.424k 

            T  5.66i  7.07 j  4.24k
             MO  r F
             M O  15 j  (5.66i  7.07 j  4.24k )

             M O  84.9k  63.6i

             M Z  84.9kN .m
               Minus sign indicate that the Mz is
               in the negative Z direction
                                                                             41
Lecture 5                                                Chapter 4: Force resultant
             Technique II
               The force T resolved into two components
                 Tz & Txy
                                  9
                 X Y  tan 1             25.1O
                               122  152
                 T XY  T cos xy  9.06kN
                 T Z  T sin  xy  4.24kN
             TZ is parallel to the Z-axis. That’s means no
             moment comes from TZ about Z-axis.
             MZ is due only to TXY
                 M Z  T XY .d
                          12
                 d                  .15  9.37
                       12  15
                          2      2


                  M Z  9.06  9.37  84.9kN .m
                                                                                 42
Lecture 5                                                    Chapter 4: Force resultant
                           EXAMPLE

                                      Given: A force is applied to
          A                           the tool to open a gas valve.
                                      Find: The magnitude of the
                 B                    moment of this force about
                                      the z axis of the value.


                                      Plan:
1) We need to use Mz = u • (r  F).
2) Note that u = 1 k.
3) The vector r is the position vector from A to B.
4) Force F is already given in Cartesian vector form.
                                                                      43
                             EXAMPLE (continued)

            A                   rAB = {0.25 sin 30° i + 0.25 cos30° j} m

                  B
                                    = {0.125 i + 0.2165 j} m
                                F = {-60 i + 20 j + 15 k} N
                                Mz = (rAB  F)


                            0       0         K1
                   = with the 0.2165 SmartDraw 5.
                MzCreated0.125 Trial Edition of0
                          -60      20         15

                   = 1{0.125(20) – 0.2165(-60)} N·m
                   = 15.5 N·m

                                                                        44
Lecture 5                                           Chapter 4: Force resultant
                                          University of Palestine
                        College of Engineering & Urban Planning




            ENGINEERING MECHANICS
              STATICS & DYNAMICS




                                                                    45
Lecture 5                             Chapter 4: Force resultant

								
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