Dynamics of structures

Transcription

1 Dynamics of structures 1.2 Viscous damping Luc St-Pierre October 30, / 22

2 Summary so far We analysed the spring-mass system and found that its motion is governed by: mẍ(t) + kx(t) = 0 k y m x x With solution: A x(t) = A sin(ω n t + φ) t -A T (And x(t) = ae λt is an alternative solution.) 2 / 22

3 Viscous dampers Viscous dampers are often used in mechanical systems; here is an example of a car damper. In this course, we assume that dampers/dashpots have a linear behaviour in the form: f c (t) = cẋ(t) where the constant c has units of Ns/m. 3 / 22

4 Damped system Free-body diagram k x( t) f k = k x x( t) c m Friction-free surface fc = c x m The equation of motion for the damped system shown here is given by: f k f c = mẍ kx cẋ = mẍ mẍ + cẋ + kx = 0 4 / 22

5 Characteristic equation Let s assume the following solution: x(t) = ae λt ẋ(t) = aλe λt ẍ(t) = aλ 2 e λt with initial values x(0) = x 0 and ẋ(t) = v 0. Substituting in the equation motion gives: mẍ(t) + cẋ(t) + kx(t) = 0 maλ 2 e λt + caλe λt + kae λt = 0 (mλ 2 + cλ + k)e λt = 0 mλ 2 + cλ + k = 0 This is the characteristic equation and its roots are: λ 1,2 = c ± c 2 4mk 2m 5 / 22

6 Regimes of behaviour The roots of the characteristic equation are: λ 1,2 = c ± c 2 4mk 2m which gives three types of motion: Overdamped: c 2 4mk > 0 λ 1 and λ 2 are real. Critically damped: c 2 4mk = 0 λ 1 = λ 2 is real. Underdamped: c 2 4mk < 0 λ 1 and λ 2 are complex. This is clear, but it can be made more elegant by defining new variables. 6 / 22

7 Critical damping coefficient and damping ratio By definition, critically damped motion implies that: c 2 4mk = 0 c cr 2 km = 2mω n where c cr is the critical damping coefficient and ω n = k/m is the natural angular frequency. Let s use c cr to define the damping ratio: ζ c c cr = c 2mω n = c 2 km λ 1,2 = c ± c 2 4mk 2m = c ω n ± 2 ( ) km c 2 2mω n 2m 2 1 km λ 1,2 = ζω n ± ω n ζ / 22

8 Characteristic root equation revisited The characteristic root equation can be re-written as: λ 1,2 = c ± c 2 4mk 2m = ζω n ± ω n ζ 2 1 and again, we have three types of motion: Overdamped: ζ > 1 Critically damped: ζ = 1 Underdamped: 0 < ζ < 1 λ 1 and λ 2 are real. λ 1 = λ 2 is real. λ 1 and λ 2 are complex. Remember that we are interested in λ 1 and λ 2 because they are part of our solution: x(t) = a 1 e λ 1t + a 2 e λ 2t 8 / 22

9 Underdamped motion 0 < ζ < 1 The underdamped complex solutions are: λ 1,2 = ζω n ± ω n ζ 2 1 = ζω n ± ω n j 1 ζ 2 = ζω n ± ω d j where ω d = ω n 1 ζ 2 is the damped natural angular frequency. Our solution can be written: x(t) = a 1 e λ 1t + a 2 e λ 2t = e ζωnt ( a 1 e jω dt + a 2 e jω dt ) Using Euler relations e θj = cos θ + j sin θ and e θj = cos θ j sin θ, x(t) becomes: x(t) = e ζωnt [(a 1 + a 2 ) cos ω d t + j(a 1 a 2 ) sin ω d t] At the moment, x(t) is complex. How can we ensure that we have a real solution? 9 / 22

10 Underdamped motion 0 < ζ < 1 Our complex underdamped solution is: x(t) = e ζωnt [(a 1 + a 2 ) cos ω d t + j(a 1 a 2 ) sin ω d t] where a 1 and a 2 are complex numbers. We can make x(t) real by forcing a 1 and a 2 to be conjugate pairs, meaning that: A consequence is that: a 1 = b + dj and a 2 = b dj A 2 a 1 + a 2 = 2b and A 1 j(a 1 a 2 ) = 2d where A 1 and A 2 are both real. Substituting in x(t) gives: x(t) = e ζωnt [A 1 sin ω d t + A 2 cos ω d t] = Ae ζωnt sin(ω d t + φ) where A = A A2 2 and φ = arctan(a 2/A 1 ). 10 / 22

11 Underdamped motion 0 < ζ < 1 Our (real) underdamped solution now is: x(t) = Ae ζωnt sin(ω d t + φ) and next we have to find A and φ. With initial conditions x(0) = x 0 and ẋ(0) = v 0, we obtain: x(0) = x 0 = A sin φ and ẋ(0) = v 0 = A [ ζω n sin φ + ω d cos φ] Solving this system of two equations and two unknowns returns: (v 0 + x 0 ζω n ) A = 2 + (x 0 ω d ) 2 ( ) x0 ω d and φ = arctan v 0 + x 0 ζω n ω 2 d 11 / 22

12 Underdamped motion 0 < ζ < 1 Underdamped solution is: x(t) = Ae ζωnt sin(ω d t+φ) Most mechanical systems vibrate with an underdamped response. Displacement D. J. Inman Time (sec) 12 / 22

13 Overdamped motion ζ > 1 The overderdamped solution is given by: x(t) = a 3 e ( ζωn+ω d)t + a 4 e ( ζωn ω d)t Here, a 3 and a 4 are real. I have changed the subscripts to differentiate with the underdamped case. Initial conditions prescribe: x(0) = x 0 = a 3 + a 4 and ẋ(0) = v 0 = ζω n (a 3 + a 4 ) + ω d (a 3 a 4 ) and solving this system of two equations and two unknowns returns: a 3 = v 0 + (ζω n + ω d )x 0 2ω d and a 4 = (ω d ζω n )x 0 v 0 2ω d 13 / 22

14 Overdamped motion ζ > 1 Underdamped solution: x(t) = a 3 e ( ζωn+ω d)t + a 4 e ( ζωn ω d)t The overdamped response does not oscillate: x(t) returns to rest exponentially. Displacement (mm) k=225n/m m=100kg and ζ=2 0.6 x 0 =0.4mm v 0 =1mm/s 0.5 x 0 =0.4mm v 0 =0mm/s x =0.4mm v 0 =-1mm/s D. J. Inman Time (sec) 14 / 22

15 Critically damped motion ζ = 1 Since λ 1,2 = ζω n = ω n, the critically damped solution becomes: x(t) = a 5 e ωnt + a 6 te ωnt Again, a 5 and a 6 are real. The initial conditions prescribe: x(0) = x 0 = a 5 a 5 = x 0 ẋ(0) = v 0 = a 5 ω n + a 6 a 6 = v 0 + ω n x 0 With these, our solution becomes: x(t) = [x 0 + (v 0 + ω n x 0 )t] e ωnt 15 / 22

16 Critically damped motion ζ = 1 Critically damped solution: x(t) = [x 0 + (v 0 + ω n x 0 )t] e ωnt Displacement (mm) k=225n/m m=100kg and ζ=1 x 0 =0.4mm v 0 =1mm/s x 0 =0.4mm v 0 =0mm/s x 0 =0.4mm v 0 =-1mm/s 0 D. J. Inman Time (sec) The critically damped response represents the smallest value of ζ that produces a nonoscillatory motion. It provides the fastest return to zero without oscillations. 16 / 22

17 Example: damped system Consider a damped system with mass m = kg, spring k = N/m and dashpot c = 0.11 Ns/m. Evaluate: the natural angular frequency ω n, the frequency f n and the period T of the oscillations, and the damping ratio ζ. Is the system under/over/critically damped? 17 / 22

18 Solution: damped system The natural angular frequency is given by: k ω n = m = 132 rad/s The period and frequency are: T = 2π = 2π ω n s and f n = 1 T = ω n 2π = Hz, 2π respectively. Finally, the damping ratio is: ζ = c c cr = c 2 km = The system is underdamped because ζ < / 22

19 Example: human knee In knee-lock position, the human leg has a natural frequency f n = 20 Hz and a damping ratio ζ = Jumping from a height of 18 mm corresponds to applying an initial velocity v 0 = 0.6 m and zero initial displacement. For these conditions: Calculate and plot the response of the system x(t). Evaluate the maximum acceleration experienced by the leg if damping is neglected. 19 / 22

20 Solution: human knee The natural angular frequency is: ω n = 2πf n = 2π rad/s while the damped natural angular frequency is: ω d = ω n 1 ζ 2 = rad/s Since ζ < 1, the system is underdamped and its solution is given by: x(t) = Ae ζωnt sin(ω d t + φ) where (v 0 + x 0 ζω n ) A = 2 + (x 0 ω d ) ωd 2 = m and ( ) ( ) x0 ω d 0 φ = arctan = arctan = 0 rad v 0 + x 0 ζω n / 22

21 Solution: human knee Therefore, the solution for the system is: x(t) = Ae ζωnt sin(ω d t + φ) = e 28.15t sin(122.47t) and this expression is plotted below. x (mm) Ae ζωnt x t = Ae ζωnt sin ω d t + φ 1 0 Time (s) / 22

22 Solution: human knee Otherwise, if damping is neglected then the solution is given by: x(t) = 1 ω ω nx v2 0 sin(ω nt + φ) n and differentiating gives the corresponding acceleration: ẍ(t) = ω n ω 2 nx v2 0 sin(ω nt + φ) Therefore, the maximum acceleration is obtained when sin(ω n t + φ) = ±1 giving: max(ẍ(t)) = ω n ω 2 nx v2 0 = m/s g 22 / 22