{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

example_IV_2_02

# example_IV_2_02 - Example IV.2.2 Find the response of the...

This preview shows pages 1–4. Sign up to view the full content.

1 Example IV.2.2 Find the response of the single DOF system shown below. z y(z) y 0 cos( π z/2a) x k m a a L L L v c Equation of Motion T = 1 2 m ˙ x 2 R = 1 2 c ˙ x 2 U = 1 2 k x y ( ) 2 Applying Lagrange’s equations: m ˙ ˙ x + c ˙ x + k x = ky t ( ) = f t ( ) Fourier series for excitation The road roughness is periodic in the space variable z; that is, y(z) = y(z+L). Specifically we can write for -L/2 < z < L/2: f z ( ) = 0 ; L/ 2 < z < a = ky 0 cos π z / 2a ( ) a < z < a = a < z < Using z = vt, we can write the road roughness as a function of time t as:

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
2 f t ( ) = 0 ; L /2v < t < a / v = ky 0 cos ω t a/ v < t < a / v = a / v < t < where ω = π v/2a . Also, y(t) is an even periodic function in time with a fundamental period and fundamental frequency of: T = L v Ω = 2 π T = 2 π v L From this we can write: f 0 = k T y t ( ) dt T/2 = k T y t ( ) a/v a/v y t ( ) 0 for a / v < t < = k T y 0 cos ω t dt a/v = k T y 0 sin ω t ω t = a/v t = = k T y 0 ω a / v ( ) −ω a / v ( ) ω = kv L y 0 2sin π / 2 ( ) π v / 2a = 4 ka y 0 π L
3 f cj =

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 4

example_IV_2_02 - Example IV.2.2 Find the response of the...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online