Problem 4.4 Derive the ramp response by nding the particular and homogeneous solutions of the ordinary di. equation.

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1 Homework #4 Name: HyoSoo Kim UFID: Date:/6/6 Problem 4.4 Derive the ramp response by nding the particular and homogeneous solutions of the ordinary di. equation. The unit ramp fuction is r(t) = tu(t) () from t = For a viscously damped single-degree-of-freedom system with unit ramp function. rewritting() mx + c _x + kx = r(t) () x + &w n _x + wnx = t u(t) (3) m q where, & = c w ; w k nm n = m To nd the soulttion of the nd order di erential equation where, xh(t) :homogeneous solution xp(t) :particular soultion -For Homogeneous soultion let x(t) = e t The characteristic equation is x(t) = x h (t) + x p (t) (4) x + &w n _x + w nx = (5) + &wn + w n = ; = &wn iwd (6) x (t) = e ( &wn+i)t = e &wnt e it = e &wnt (cos t + i sin t) (7)

2 x (t) = e ( &wn i)t = e &wnt e it = e &wnt (cos t i sin t) (8) To make real dimension solution (x (t) + x (t)) = e &wnt cos t (9) i (x (t) x (t)) = e &wnt sin t () The Homogeneous solution with integral constant is To nd particular soution x h (t) = e &wnt (C sin t + C cos t) () from the method of undetermined coe cient, x + &w n _x + wnx = t u(t) () m x p = K t + K (3) _x p = K (4) inserting (3,4,)into() x = (5) &w n K + w n(k t + K ) = t m (6) w nk t + (&w n K + w nk ) = t m K = mw n (7) (8) the particular solution is K = &w nk w n = & mw 3 n (9) the solution is ) x p = t mw n & mw 3 n = k (t & w n ) () ) x(t) = e &wnt (C sin t + C cos t) + t mw n & mw 3 n ()

3 For the ramp function with initial conditon zeros x() = ; v() = x (t) = ( &w n ) e &wnt (C sin t+c cos t)+e &wnt (C cos t C sin t)+ mw n () so the ramp response is & x() = C mwn 3 = ) C = & mw 3 n x () = ( &w n ) C + C + mwn = ) C = & mw n x(t) = k ft & w n + e &wnt [ & sin t + & w n cos t]gu (t) (3) Comparing eq..3 x (t) = e &wnt ( &w nx + v sin t + x cos t) x = & w n k (4) v = k (5) 3

4 Problem 4.6 Derive the response of a viscously damped single-degree-of-freedom system shown in Fig. 4.3 Using the superposition principle to express ramp fucntion for the triangular pulse F (t) = F T [r(t) r(t T ) + r(t T )] (6) where, r(t) is the ramp function The response in terms of time-shifted ramp response is x(t) = F T [r(t) r(t T ) + r(t T)] (7) where, r(t) is the ramp response 4

5 Problem 4.7 Derive the response of a viscously damped single-degree-of-freedom system shown in Fig Using the superposition principle to express ramp fucntion for the trapezoidal pulse F (t) = F T [r(t) r(t T ) r(t 3T ) + r(t T )] (8) where, r(t) is the ramp function The response in terms of time-shifted ramp response is x(t) = F T [r(t) r(t T ) r(t 3T ) + r(t T)] (9) where, r(t) is the ramp response 5

6 Problem 4.9 Derive the response of a visously damped single-degree-of-freedom system to the force by means of the convolution integral plot the response given conditions: m = kg c = 4N s=m k = 4; 8N=m F = N = From the convolution integral, the respose is x (t) = F (t) = F e t u(t) (3) F (t)g(t )d = F (t )g(t)d (3) For underdamped single-degree-of-freedom, the impulse response is, inserting (3,3) into (??) x (t) = F m g(t) = m e &wnt sin tu(t) (3) e (t ) u(t )e &wn sin u()d (33) = F e t m e ( &wn) sin d = F e t im e ( &wn) [e i e i ]d = F e t im [ e [( &wn)+i] d e [( &wn) i] d &wn)+iwd] = F e t [ e[( j t im ( &w n ) + i e [( &wn) i] ( &w n ) i j t ] = F e t [ e[( &wn)+iwd]t e [( &wn) iwd]t + im ( &w n ) + i ( &w n ) + i ( &w n ) i ( &w n ) i = F e t im [ e[( &wn)+iwd]t ( &w n ) + i e [( &wn) iwd]t ( &w n ) i 6

7 = F e t im [ [( &w n) i ][e [( &wn)+iwd]t ] [( &w n ) + i ][e [( &wn) iwd]t ] ( &w n ) + w d () [( &w n ) i ][e [( &wn)+i]t ] = [( &w n ) i ]e [( &wn)+i]t [( &w n ) i ] = ( &w n )e [( &wn)+i]t i e [( &wn)+i]t ( &w n ) + i () [( &w n ) + i ][e [( &wn) i]t ] = [( &w n ) + i ]e [( &wn) i]t ( &w n ) i = ( &w n )e [( &wn) i]t + i e [( &wn) i]t ( &w n ) i () () = ( &w n )[e [( &wn)+i]t e [( &wn) i]t i [e [( &wn)+i]t +e [( &wn) i]t ]+i = ( &w n )[e t e &wnt e i e t e &wnt e i i [e t e &wnt e i +e t e &wnt e i ]+i = ( &w n )e t e &wnt [e it e it ] i e t e &wnt [e it + e it ] + i x (t) = F e t im [ ( &w n)e t e &wnt [e iwdt e iwdt ] i e t e &wnt [e iwdt + e iwdt ] + i ( &w n ) + w d (34) 7

8 F ( + &w n )e &wnt e iwdt e iwdt = m[( &w n ) + wd ][ i The response is e &wnt eiwdt + e it +e t ] (35) x(t) = F t m[( &w n ) + wd ][e e &wnt (cos t + (&w n ) sin t)]u(t) (36)

9 Problem 4. Derive the response of an undamped single-degree of freedom system to the force of the triangular pulse The ramp function F (t) = F T [r(t) r(t T ) r(t T )] (37) The impulse response for undamped system g(t) = mw n sin w n t (38) for t < T x (t) = mw n ( F T ) sin w n (t ) d (39) = F mw nt (t w n sin w n t) (4) for T < t < T x (t) = mw n [ Z T= ( F T ) sin w n (t ) d + F ( T= T sin w n (t ) d] (4) = F mw nt f t T + w n T [ sin w n(t T ) sin w nt]g (4) for t > T x(t) = mw n [ Z T= ( F T ) sin w n (t ) d + F ( T= T sin w n (t ) d] (43) = F mw 3 nt [ sin w n(t T ) sin w nt sin w n (t )] (44) 9

10 Problem 4.6 Derive the response of a viscously damped single-degree-of freedom system to the force F (t) = F e t u(t) by means of Laplace Transformation method using Laplace transformation mx + c _x + kx = F e t (45) x = x(ms + cs + k) = F s + F ms + cs + k = As + B ms + cs + k + C s + (46) (47) F = (s + )(As + B) + C(ms + cs + k) (48) F = (A + Cm)s + (A + B + cc)s + B + Ck (49) m A 4 C5 4B5 = 4 5 (5) k C F F m A = m c + k = F c m + k = m B = F (m c) m c + k = F ( &w n ) &w n + wn F &w n + w n (5) (5) C = m F c + k = m F &w n + w n (53) where & = x = c mw n ; w n = k m x = ( m )( F &w n+w s + F( n (s + &w n ) + wd m (As + B) (s + &w n ) + wd + C s + &wn) &w n+w ) n + ( m ) F &w n+wn s + (54) (55) = ( m )( Fs+F F &w n) &w n+wn (s + &w n ) + wd + ( m ) F &w n+wn s + (56) &w n + w n = ( &w n ) + w d (57)

11 F (s + &w n ) x = m[( &w n ) + wd ][ (s + &w n ) + wd + F s + ] (58) = F m[( &w n ) + w d ][ (s + &w n ) (s + &w n ) + w d + (&w n ) (s + &w n ) + wd + s + ] = m[( F &w n ) + w d ][e &wnt cos t (&w n ) e &wnt sin( t)+e t ] (59) so, the solution is ) x(t) = F t m[( &w n ) + wd ]fe e &wnt [cos t + (&w n ) sin( t)]gu(t) (6)