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54 Multibody Systems Approach to Vehicle Dynamics<br />

Suspension strut<br />

D<br />

Body 6<br />

Free-body diagram<br />

D<br />

{F D6G4 } 1<br />

G<br />

Body 4<br />

G<br />

{F G4D6 } 1<br />

Fig. 2.24<br />

Vector notation for line of sight forces<br />

Z 1<br />

M Pz<br />

{M P } 1<br />

M Py<br />

Y P<br />

O 1<br />

1<br />

X 1<br />

M Px<br />

Fig. 2.25<br />

Vector representation of a moment<br />

Considering next the equilibrium of the body shown in Figure 2.26 it can<br />

be seen that a force acting at point P will produce a reaction moment at<br />

point Q. In order to simplify the diagram only the components of the applied<br />

force and the resulting moment are shown here. Since we are looking only<br />

at the derivation of the moment at Q the reaction force at Q is also omitted.<br />

The moment generated at point Q, which is the point where Body 2 is fixed to<br />

the non-moving ground part Body 1, is designated as {M Q21 } 1 . Applying the<br />

same principle as used for a force we would read this as the moment acting<br />

at point Q, on Body 2 due to its connection to Body 1, with components<br />

resolved parallel to the axes of reference frame O 1 . For convenience and to<br />

assist with the interpretation of the derivation that follows the components of<br />

the force {F P } 1 , the moment {M Q21 } 1 and the relative position vector {R PQ } 1<br />

have all been set up in Figure 2.26 to have components that are positive.<br />

Using the traditional approach the three equations of moment equilibrium,<br />

M Qx 0, M Qy 0 and M Qz 0 could be transformed to produce the<br />

following three equations:<br />

M Q21x PQ z F Py PQ y F Pz (2.146)<br />

M Q21y PQ z F Px PQ x F Pz (2.147)<br />

M Q21z PQ x F Py PQ y F Px (2.148)

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