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					Method of Superposition

Sometimes the number external forces and inertial forces acting on a mechanism is too much
for graphical solution. In this case we apply the method of superposition. Using superposition
the entire system is broken up into (n) problems, where n is the number of forces, by
considering the external and inertial forces of each link individually. Response of a linear
system to several forces acting simultaneously is equal to the sum of responses of the system
to the forces individually. This approach is useful because it can be performed by graphically.
Example: Onto crank CD of the mechanism shown, a CCW torque of 100 N.m is acting.
Onto the slider of link 6 a 50 N force is acting horizontally rightward. Calculate the amount of
torque required on crank AB to keep the mechanism in static equilibrium using the method of
“superposition”.




      6                 50 N
            F
                                      5                             C

                         3
                                          E




                                                                            80
                    B
                                                                        4
      T                  2
                                                          100 N.m
                60°




     A
                                                                        D
                1


AB = BE=CE = 7 cm.             CD = EF = 6 cm.             AD = 10 cm.
e = 8 cm.                      θ =120 o

Solution:
This problem can not be solved graphically. We have to use the method of superposition.
   1- Assume, only 50 N force is acting onto mechanism. Then,
Link 6, 3 are three force member.
Link 5, 4 are two force member.


In this example we are beginning with link 5. Since link 5 is two force member, this requires
that FF and FE have equal magnitudes and opposite directions. We can now select the link 6
from which the force analysis can be started. Then, the force of FF can be found from force
polygon as:
v
FF = 59.8 N ∠147 o


Link 4 is two force member, so line of applications of forces acting on this link is known.
Then , we can analyze the link 3 that is three force member.

               N                                FF
                          50 N
                                                      F
              F               FF
                                                                             5
                   50 N

                                                                                     FE
                                       N
          FF =59,8 N




                                                     FE                          C
                                                                             FC           FC
                                   3
                               B                          E                                    C

                                       FB
                          B

              T               2                                                                     4
  d=37,6 mm                                                   FB=49 N

    A
                                            FE =59,8 N                  FC

                                                                                               FD
            N                             F
                                          F
                     50
                     N                          F
           F          F
                     F

                                                                    5
                50
                N
                                                                        F
                                N                                       E

       FF =59,8 N




From the force polygon;
 v
FB = 59.8 N ∠170 o
 v
FC = N ∠94 o
From the free body diagram of Link 2, torque T becomes:
v             37,6
T1 = d * FB =      * 59.8 = N .m CCW      ANS
              1000

   2- Assume, only 100 N.m torque is acting onto mechanism. Then,
Link 3 is two force member.
There is no force and torque on link 5 and 6.
                     F

            F
                                                 5                                FC
                                                         E
                                                                              C

                                           3
                              B                      E
                         FB
                                                                  FC              C


                                                             d3=58 mm
                                          FB                                           4
                                  B

                     T2                                                 100 N.m
                                      2
   d2 =26.7 mm

                 A                                                 FD
       FA

From link 4;
         100
FC =          = 1724,14 N
        0.058
From link 3;
FC = FB
Then,
v               26.7
T2 = d 2 * FB =      * 1724,14 =48,275 N .m CW
                1000
v v v
T = T1 + T2 = 48,275 − 2,784 =45,492 N.m CW
Example: Find the magnitude and direction of the moment that must be applied to link 2 to
drive the linkage against the forces shown using the method of “superposition”. Sketch free-
body diagrams of each link and show all forces acting.



                                                 C     100 N


                                   3
                                                                          D
                                                                          200 N
                                                               F
                 2 B
                                                                   4
                 30°




     A                                                    E
                 1

AB = 2 cm              BC = 7 cm.            BD = 9 cm.             ED = 5 cm.
AE = 7 cm              EF = 3,5 cm.          CB = 4 cm.

Solution:
We have to use the method of superposition.
   3- Assume, only 200 N force is acting onto mechanism. Then,
Link 3 is two force member.
Link 4 is three force member.


In this example we are beginning with link 4. Since link 3 is two force member, this requires
that FB and FD have equal magnitudes and opposite directions. So, line of applications of
forces acting on this link is known. Then , we can analyze the link 4 that is three force
member. Then, the force of FD can be found from force polygon as:

v
FD = 204 N ∠18 o
                                            C

                                3                                       FD
                                                                    D

                                                                             FD         D
                                                                                        200 N
                  B                                                          F
         FD=FB
                                                                                  4
                  2 B      FB
                                d                               E
             T1
    FA                                                         FE

                                                                    200 N
                                                    FE= 63 N
                                                                             FD=204 N




From the free body diagrams;
 v
FB = 204 N ∠18 o
From the free body diagram of Link 2, torque T becomes:
v              4,3
T1 = d * FB =      * 204 = 0,877 N .m CCW    ANS
              1000

   4- Assume, only 100 N.m torque is acting onto mechanism. Then,
Link 4 is two force member.
Link 3 is three force member.
                                            C   100 N

                                3                                   FD
                                                                D



                 B                                 100 N
       FB                                                                    FD
                                    FD= 105 N
                                                        FB=188 N
                                                                             D

                                                                         4
                 2       FB d
            T1
                     B                                      E
                                                   FE
  FA


From force polygon of the link 3;
 v
FD = 105 N ∠49 o
 v
FB = 188 N ∠25 o
Then,
v                1,6
T2 = d 2 * FB =      * 188 = 0,3 N .m CW
                1000
v
T = T1 + T2 = 0,877 + 0,3 = 1,177 N .m CW

				
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posted:11/27/2012
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