10000

Flywheel 1.0234 kg.m2

Equivalent length cm of 7.0 cm dia. solid steel shaft

Stiffness Nm/Rad 0.3907 x 106 1.0916 x 106 1.0916 x 106 ^ 0916 x 106

1.0916 x 106 Torque

1.3205 x 106

Figure 10.34 Typiclal equivalent torsional system for a six-cylinder engine

Equivalent length cm of 7.0 cm dia. solid steel shaft

Stiffness Nm/Rad 0.3907 x 106 1.0916 x 106 1.0916 x 106 ^ 0916 x 106

1.0916 x 106 Torque

1.3205 x 106

Figure 10.34 Typiclal equivalent torsional system for a six-cylinder engine linearly along the shaft length. Then treating the crank inertias as masses and the equivalent length from the flywheel as fixing their position on a beam, a point of balance can be found in a manner analogous to finding the centre of mass. This is done by using the flywheel as the reference point then multiplying the inertia masses by their equivalent length from the flywheel, adding their products and dividing the sum by the total of the masses, including that of the flywheel, to find the point of balance. This approximates to the node with the crank masses vibrating against the flywheel. The approximate natural frequency can then be found by regarding the flywheel as being a simple torsional pendulum on a shaft having the length just determined from eqn (8.3).

0 0

Post a comment