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Multibody systems simulation software 103 02 x z y I Action-only force x z y J 03 Fig. 3.24 Action-only force 02 z x y I Torque acting on the I marker x Reaction torque acting on the J marker z y J 03 Fig. 3.25 Action–reaction torque Rotational forces may be defined to be action–reaction or action-only. In either case the torque produced is assumed to act on the I marker. For an action-only torque it is again the z-axis of the J marker that is used, in this case to define the axis of rotation. If the torque is action–reaction the z-axes of the I and J marker must be parallel and point in the same direction as shown in Figure 3.25. Considering next the definition of translational spring elements we can start with a definition that is linear and introduce the use of system variables for the formulation of a force. As can be seen in Figure 3.26 the formulation of the spring force will be dependent on the length of the spring. This is made available through a system variable defined here as DM(I, J) which represents the scalar magnitude of the displacement between the I and the J marker at any point in time during the simulation. The spring force F S is initially defined here to be linear using F S k(DM(I, J) L) (3.45) where k is the spring stiffness and L is the free length of the spring, at zero force. The equation used in (3.45) to determine F S follows the required convention that the scalar value of force produced is positive when the spring is in compression, zero when it is at its free length and negative when it is in tension.
104 Mutibody Systems Approach to Vehicle Dynamics I I I DM (I, J) L DM (I, J) J J J Fig. 3.26 COMPRESSION Positive force AT FREE LENGTH Zero force Spring in compression, at free length and in tension TENSION Negative force F S DM (I, J) L F S k(L DM (I, J)) Compression Tension Fig. 3.27 Formulation of a linear spring force This should not be confused with the components of any reaction force recovered at the I or the J marker. The sign of these will be dependent not only on the state of the spring but also on the orientation of the spring force line of action and the reference frame in which the components are being resolved. The formulation of the spring force F S is shown graphically in Figure 3.27. Note that the formulation used here produces a force that is positive in compression and negative in tension. This is opposite to the convention used for stresses in stress analysis and finite element programs. An example of the syntax that could be used to formulate this using a SFORCE statement, would be SFORCE/0509,I0205,J0409,TRANS,FUNCTION 40*(250-DM(0205,0409)) where using the units that are consistent throughout this text we would have: FUNCTION the spring force F S (N) The spring stiffness k 40 N/mm The free length L 250 mm DM(0205,0409) the magnitude of the displacement between I and J (mm)
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104 Mutibody Systems Approach to Vehicle Dynamics<br />
I<br />
I<br />
I<br />
DM (I, J)<br />
L<br />
DM (I, J)<br />
J<br />
J<br />
J<br />
Fig. 3.26<br />
COMPRESSION<br />
Positive force<br />
AT FREE LENGTH<br />
Zero force<br />
Spring in compression, at free length and in tension<br />
TENSION<br />
Negative force<br />
F S<br />
DM (I, J)<br />
L<br />
F S k(L DM (I, J))<br />
Compression<br />
Tension<br />
Fig. 3.27<br />
Formulation of a linear spring force<br />
This should not be confused with the components of any reaction force recovered<br />
at the I or the J marker. The sign of these will be dependent not only on<br />
the state of the spring but also on the orientation of the spring force line of<br />
action and the reference frame in which the components are being resolved.<br />
The formulation of the spring force F S is shown graphically in Figure 3.27.<br />
Note that the formulation used here produces a force that is positive in compression<br />
and negative in tension. This is opposite to the convention used for<br />
stresses in stress analysis and finite element programs. An example of the syntax<br />
that could be used to formulate this using a SFORCE statement, would be<br />
SFORCE/0509,I0205,J0409,TRANS,FUNCTION<br />
40*(250-DM(0205,0409))<br />
where using the units that are consistent throughout this text we would have:<br />
FUNCTION the spring force F S (N)<br />
The spring stiffness k 40 N/mm<br />
The free length L 250 mm<br />
DM(0205,0409) the magnitude of the displacement between I and J<br />
(mm)