Question 17.32: The static deflection of an automobile on its springs is 80 ...

The static deflection of an automobile on its springs is 80 mm. Find the critical speed when the automobile is travelling on a road which can be approximated by a sine wave of amplitude 60 mm and a wave length 12 m (Fig.17.56). Assume the damping to be 0.05. Also calculate the amplitude of vibration at 60 km/h.

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\omega_{n}=\sqrt{\frac{g}{\delta_{s t}}}=\sqrt{\frac{9.81}{80 \times 10^{-3}}}=11.07 rad / s .

For dynamic magnification to be maximum,

\beta=\sqrt{1-2 \zeta^{2}} .

=\sqrt{1-2 \times(0.05)^{2}}=0.9975 .

\text { Excitation frequency, } \omega=\beta \omega_{n} .

=0.9975 \times 11.07=11.04 rad / s .

\text { Critical velocity of vehicle, } v=f \lambda=\frac{\omega \lambda}{2 \pi} .

\frac{11.04 \times 12}{2 \lambda}=21.09 m / s .

=75.92 km / h .

Excitation frequency, at 60 km/h,

\omega=\frac{2 \pi \times 60 \times 10^{3}}{3600 \times 12}=8.827 rad / s .

\beta=\frac{\omega}{\omega_{n}}=\frac{8.827}{11.07}=0.7883 .

\text { Amplitude of vehicle, } X=\frac{y \sqrt{1+(2 \zeta \beta)^{2}}}{\sqrt{\left(1-\beta^{2}\right)^{2}+(2 \zeta \beta)^{2}}} .

=\frac{60 \sqrt{1+(2 \times 0.05 \times 0.7883)^{2}}}{\sqrt{\left\{1-(0.07883)^{2}\right\}^{2}+(2 \times 0.05 \times 0.7883)^{2}}}=0.7883 .

=155.65 mm .

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