EME 231 Engineering Statics: - DOC - DOC by sZ73y0

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									EGR 231 Engineering Statics:                                                             Fall 2011
Lecture 07: Equilibrium in 2-D


Today:
 Homework Questions?
 Newton’s 2nd Law
 2-D equilibrium problems.
                                                                          500N
Homework Problem Assignment 07:
Problem 2.44
Two cables are tied together at C and are loaded
 as shown. Determine the tension
a) in cable AC, b) in cable BC.

                                                                 y                       FB

Problem 2:51
A welded connection is in equilibrium under the action
of the four forces shown. Knowing that F A=8 kN and
FB = 16 kN, determine the magnitudes of the other two                                              FC
forces.                                                                                                 x

                                                                          5      3
                                                                          4
                                                    FA


                                                                     FD

Problem 2.67:
A 600 lb crate is supported by several rope
and pulley arrangements as shown.
Determine for each arrangement the                 T                                                             T
tension in the rope.                                                                 T
                                                                T                                  T




                                          (a              (b            (c                    (d            (e
                                          )               )             )                     )             )
                                          T               T             T                     T             T
Following Today’s Lecture you should be able to:
1) Write Newton’s 2nd law for the case of static equilibrium of particles.

2) Set up and find the forces needed to give static equilibrium of a particle.
Particle Equilibrium in 2-Dimensions:

A particle is in equilibrium if it is at rest, or if it has a constant
velocity. When in equilibrium, the sum of all forces acting on a
body is zero.
           F  0
Free Body Diagram:
A diagram of a body which shows all forces that at upon it and
replaces all connections to the outside with other forces.

Step 1: Isolate the body from the surroundings.

Step 2: Show all forces. Include pure forces and all reactive
forces due to cut or removed constraints.

Step 3: Label know forces with values. Label unknown forces
with variables.

The equation    F  0 is a vector equation. In 2-dimensions this
can be broken down into two scalar equations...
      one in the x- and one in the y- directions.
                    F  0
            F1  F2  F3  0
           ˆ                 ˆ                ˆ
       F1x i  F1 y ˆ  F2 x i  F2 y ˆ  F3x i  F3 y ˆ  0
                    j                 j                j
                           ˆ
       ( F1x  F2 x  F3x )i  ( F1 y  F2 y  F3 y ) ˆ  0
                                                      j
           F   x   i   Fy ˆ  0
                    ˆ        j
or breaking it down into two scalar equations:
  x-dir:    F      x   0       and             y-dir:    F  y   0
                                                            y

Example Problem 1:                                                        F

Find the force F required
                                               FA=150 lbf             θ
to provide static equilibrium
                                                                              x
of the green particle.                                          O


                                                                30o
                                                                              FB=200 lb



--------------------------------------------------------------------------------
Given: FA  150 ˆ + 0ˆ lb
                      i     j
          F  200sin 30 ˆ  200cos30 ˆ lb
            B                i               j
               100 ˆ  173.2ˆ lb
                    i        j
          F  F cosθ ˆ  F sinθˆ
                      i         j

Find the force, F that makes the system in equilibrium:
     F  0
      FA  FB  F  0
     (150 ˆ + 0ˆ)  (100 ˆ  173.2 ˆ)  ( F cos θ ˆ  F sin θˆ )  0
            i   j         i         j              i          j
      (150  100+ F cos θ) ˆ  ( 0  173.2  F sin θ)ˆ  0
                             i                          j

i:    0  150  100+ F cos θ         j:    173.2  F sin θ  0
       F cosθ=50                              F sin θ  173.2

To solve divide the 2nd equation by the 1st.
          F sin θ   173.2
tan θ                   =3.464                  θ  73.9o
          F cos θ    50

Therefore
       50     50
F=                   180.3 lb                  so F =180.3 lb at 73.9
                                                                        o
      cos θ cos 73.9
Example Problem 2
                                                               B                                  C
Find the maximum load value                                                          5       3
which does not exceed a cable                                        60o   A
                                                                                         4
tension of 780 lbf in either cable.
------------------------------------------------------------------

Given:                                                                     W
                4      3                                                                 y       Draw FBD
            FC  FC ˆ + FC ˆ
                    i      j
                5      5                                                       FB                         5
            FB   FB cos60 ˆ  FB sin 60 ˆ                                                      FC           3
                            i             j
                                                                                                          4
            W  0 ˆ  Wˆ
                   i    j                                                                    θ
                                                                               60o
                                                                                                      x
For Equilibrium:                                                                     A

    F  0
    FC  FB  W  0                                        W

     4        3
   ( FC ˆ + FC ˆ )  (  FB cos 60 ˆ  FB sin 60 ˆ )  (  W ˆ )  0
          i        j                 i            j          j
     5        5
        4                                3
i: 0  FC  FB cos 60             j: 0  FC  FB sin 60  W
        5                                5
                                3
  FC  0.625 FB             W  FC  FB sin 60
                                5
At this point we can see that FC < FB, so let FB = 780 lbf and then
calculate the other forces.
 FB  780 lb      FC  0.625 FB  0.625(780)  487.5 lb f
         3                   3
    W  FC  FB sin 60  (487.5)  780sin 60  968lb f
         5                   5
This clearly shows the rope holding the weight exceeds 780 lb, so
redo the calculations letting W = 780 lb.
          3                   3
   780  FC  FB sin 60  (0.625FB )  FB sin 60  1.241FB
          5                   5
        780
  FB         628.5 lb and FC  0.625(628.5)  392.8 lb
       1.241
Example 3
The 40 N collar A can slide on a frictionless vertical rod and is
attached as shown to a spring. The spring is unstretched when
h = 300 mm. Knowing that the constant of the spring is 560 N/m,
determine the value of h for which the system is in equilibrium.
-------------------------------------------------------------------------------
                                                                                  B



                                                      h
                                                                                      k=560 N/m

                                                                A



                                                                    300 mm        C
    Example 3 Solution:
    The 40 N collar A can slide on a frictionless vertical rod and is
    attached as shown to a spring. The spring is unstretched when
    h = 300 mm. Knowing that the constant of the spring is 560 N/m,
    determine the value of h for which the system is in equilibrium.
    -------------------------------------------------------------------------------
                                                                                      B
    Geometry                       FBD         Fs
               300mm
                                                            h
                                        θ                                                 k=560 N/m
    300        L1
    mm                         N                                      A
h                                   A

                    L2                                                    300 mm      C
                                        W = 40 N
     d     θ



    Vector Forces:
               FS  Fs sinθ ˆ + Fs cosθ ˆ
                            i           j
               N   N ˆ  0ˆ
                         i    j
               W  0 ˆ  40ˆ N
                     i     j

    Spring Equation:             FS  k L  k ( L2  L1 )  560( L2  L1 )

    Equilibrium:            F  0
                        FS  N  W  0
                        Fs sin θ ˆ + Fs cosθ ˆ  N ˆ  40ˆ  0
                                 i           j     i     j
    i: Fs sin θ  N  0                 j: Fs cos θ  40  0
          N  Fs sin θ                          Fs cos θ  40
Geometry relationship:                                                        300mm

L1  0.3002  0.3002  0.4243m
                                                                    300       L1
L2  0.300  (0.300  d )
                2                 2
                                                                    mm
                                                             h
              0.3  d
      cos θ                                                                       L2
                L2
                                                                     d    θ
              0.3
      sin θ 
               L2

So now start combining the equations:

       Fs cos θ  40                        FS  562( L2  L1 )
become
           0.3  d
      Fs            40                      FS  562( L2  0.4243)
             L2
and combining these two give

     40  ( L  0.4243)(0.3  d )  0.4243 
         2                      1      (0.3  d )
    562            L2                 L1 

Since L2 and d are related by
      L2  0.3002  (0.300  d )2

we finally get a single equation with a single variable, but it's not fun to solve being
nonlinear in form.
                       0.4243                       
  0.071174  1                                      (0.3  d )
                 0.3002  (0.300  d )2             
                                                    
Rewrite as a function equation to zero.
                        0.4243                   
 f (d )  0  1                                  (0.3  d )  0.071174
                  0.3002  (0.300  d )2         
                                                 
This can be solved by plotting the function and locating the root.
                          0.4243                    
   f (d )  0  1                                   (0.3  d )  0.071174
                    0.3002  (0.300  d )2          
                                                    




Matlab code used to generate graph:

 L1=0.4243
 k=562
 s=0.3
 W=40
 d=linspace(0.11,0.12)
 x=(1-(L1./sqrt(s.^2+(s+d).^2))).*(s+d)-W/k
 h1=plot(d,x,'b')
 set(h1,'linewidth',2.0)
 xlabel('d')
 ylabel('residual')
 grid on


Therefore:        d = 0.115      or    h = d + 0.3 = 0.415 m

------------------------------------------------------------------------------------------------------

An alternative method is to use Mathematica (available at LANDesk Management
download site) to solve the equation. Using the Mathematica Solve command:
Example 4:                                                            y             FB
An irregularly shaped machine component
is held in the position shown by three clamps.
Knowing that Fa=940 N, determine the
magnitudes of three forces Fb and Fc exerted
by the other two clamps                               FA                      50o
                                                                                    x
                                                                              70o


--------------------------------------------------------------------------------
Given: FA  940 ˆ + 0ˆ N
                                                                         FC
                    i     j
          F   F cos50 ˆ  F sin50 ˆ
             B        B       i        B     j
                  0.6428FB ˆ  0.7660FB ˆ
                             i             j
           FC   FC cos70 ˆ  FC sin 70 ˆ
                           i             j
                  0.3420F ˆ  0.9397F ˆ
                              i  C          j     C


Find the forces, FB and FC that give the system equilibrium:
     F  0
     FA  FB  FC  0
 ( 940 ˆ + 0ˆ )  (  0.6428FB ˆ  0.7660FB ˆ )  (  0.3420FC ˆ  0.9397 FC ˆ )  0
       i    j                  i            j                  i             j
     ( 940  0.6428FB  0.3420FC ) ˆ +( 0  0.7660FB  0.9397FC )ˆ  0
                                   i                             j


i:     0  940  0.6428FB  0.3420 FC            j: 0  0.7660 FB  0.9397 FC  0
              940  0.6428FB                                      0.7660 FB
         FC                                                 FC 
                   0.3420                                           0.9397
          FC  2749  1.8795FB                              FC  0.8152 FB

To solve set the two expressions for FC equal:
                                                                2749
 2749  1.8795FB  0.8152 FB                     FB                       1020 N
                                                           0.8152  1.8795
then
                 FC  0.8152 FB  0.8152(1020)  832 N
Workout Problem:
Two cables are tied together at C and are loaded
 as shown. Determine the tension
a) in cable AC, b) in cable BC.

								
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