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Kinematics and dynamics of rigid bodies 53
equation (2.136) with the forces as drawn in the free-body diagram shown
in Figure 2.23 would be consistent with the scale factors having positive
values. Should the solution yield negative scale factors this would simply
involve a reversal of the sense of the force from that initially assumed in the
free-body diagram.
In this case, since the forces act along the line of the linkages, we equate
forces throughout the structure as follows:
{F A52 } 1 {F A25 } 1 {F B21 } 1 {F B12 } 1 (2.139)
{F A53 } 1 {F A35 } 1 {F C31 } 1 {F C13 } 1 (2.140)
{F A54 } 1 {F A45 } 1 {F D41 } 1 {F D14 } 1 (2.141)
Clearly by solving the unknown forces acting on the pin, Body 5, which
connects to all three linkages the complete force distribution in this system
will be known. Setting up the equation of equilibrium for Body 5 we get
∑{F 5 } 1 {0} 1 (2.142)
{F A } 1 {F A52 } 1 {F A53 } 1 {F A54 } 1 {0} 1 (2.143)
Using the information developed in equations (2.136) to (2.141) we can
now write
{F A } 1 f 2 {R BA } 1 f 3 {R CA } 1 f 4 {R DA } 1 {0} 1 (2.144)
The direction of the position vectors as defined in equation (2.144) should
be carefully noted. These have been selected to maintain the correct positive
sign convention throughout the equation. Expanding equation (2.144)
would lead to
⎡FAx
⎤ ⎡BAx
⎤ ⎡CAx
⎤ ⎡DAx
⎤ ⎡0⎤
⎢
F
⎥
(2.145)
⎢ Ay ⎥
f
⎢
BA
⎥
f
⎢
CA
⎥
f
⎢
DA
⎥ ⎢ ⎥
2 ⎢ y ⎥
3 ⎢ y ⎥
4 ⎢ y ⎥
0
⎢ ⎥
⎣⎢
FAz
⎦⎥
⎣⎢
BAz
⎦⎥
⎣⎢
CAz
⎦⎥
⎣⎢
DAz
⎦⎥
⎣⎢
0⎦⎥
For a given applied force {F A } 1 , and taking the geometry of the points A, B,
C and D to be known, equation (2.145) yields the three equations required
to solve the three unknowns f 2 , f 3 and f 4 and hence solve the force distribution
in this example.
The notation that has been used to define forces in the example shown can
be used where the forces represent reactions generated at a point in a multibody
system. The notation needs to be altered for forces acting along the
line of sight of a force element, such as a spring or damper, connecting two
bodies as shown in Figure 2.24.
The vector representation of a moment is not as straightforward to interpret as
the vector representation of a force. The moment {M P } 1 shown acting about
point P in Figure 2.25 is represented by a vector that is orientated along an
axis about which the moment acts. The length of the vector represents the
magnitude of the moment and the direction of the vector is that which is
consistent with a positive rotation about the axis as shown. The components
of the vector, M Px , M Py and M Pz , are resolved parallel to a reference frame,
in this case O 1 . The double-headed arrows used in Figure 2.25 are intended
to distinguish the moment vector from that of a force.