美国麻省理工大学：《结构力学》（英文版） Unit 20 Solutions for Single Spring-Mass Systems

MT-1620 al.2002 Unit 20 Solutions for Single Spring-Mass Systems Paul A Lagace Ph D Professor of aeronautics Astronautics and engineering systems Paul A Lagace @2001

MIT - 16.20 Fall, 2002 Unit 20 Solutions for Single Spring-Mass Systems Paul A. Lagace, Ph.D. Professor of Aeronautics & Astronautics and Engineering Systems Paul A. Lagace © 2001

MT-1620 al.2002 Return to the simplest system the single spring-mass This is a one degree-of-freedom system with the governing equation mg t q F First consider Free Vibration Set f=o resulting in mq+kq=0 The solution to this is the homogeneous solution to the general equation For an Ordinary Differential Equation of this form, know that the solution is of the form g() mppt kepl=o =mp+k=0(in order to hold for all t Paul A Lagace @2001 Unit 20-2

qt MIT - 16.20 Fall, 2002 Return to the simplest system: the single spring-mass… This is a one degree-of-freedom system with the governing equation: m q˙˙ + k q = F First consider… Free Vibration ⇒ Set F = 0 resulting in: m q˙˙ + k q = 0 The solution to this is the homogeneous solution to the general equation. For an Ordinary Differential Equation of this form, know that the solution is of the form: () = e pt ⇒ m p2e pt + k e pt = 0 ⇒ 2 mp + k = 0 (in order to hold for all t) Paul A. Lagace © 2001 Unit 20 - 2

MT-1620 al.2002 k →P m Where a natural frequency of single m degree-of-freedom system [rad/sec stiffness Important concept that natural frequency mass So have the equation q()=C +I0 t +ce Paul A Lagace @2001 Unit 20-3

qt it i t MIT - 16.20 Fall, 2002 2 k ⇒ p = − m ⇒ p = ± i k m where: i = − 1 k m = ω = natural frequency of single degree-of-freedom system [rad/sec] Important concept that natural frequency = stiffness mass So have the equation: () = C1 e+ ω + C2 e− ω Paul A. Lagace © 2001 Unit 20 - 3

MT-1620 Fall 2002 from mathematics know this is (0)=C sinat C cost general solution Now use the initial conditions 0q=9→C q g0 This results in g(t)= q sinat go cost ith This is the basic, unforced response of the system So if one gives the system an initial displacement a and then lets go Paul A Lagace @2001 Unit 20-4

qt MIT - 16.20 Fall, 2002 from mathematics, know this is: q t() = C1 ′sinω t + C2 ′ cosω t general solution Now use the Initial Conditions: @ t = 0 q = q0 ⇒ C2 ′ = q0 @ t = 0 q˙ = q˙0 ⇒ C1 ′ = q ˙ 0 ω This results in: () = q ˙ 0 sinω t + q0 cosω t ω with: ω = k m This is the basic, unforced response of the system So if one gives the system an initial displacement A and then lets go: q0 = A q ˙ 0 = 0 Paul A. Lagace © 2001 Unit 20 - 4

MT-1620 al.2002 The response is q(t)=Acosot Figure 20.1 Basic unforced dynamic response of single spring-mass system A e 七 But, generally systems have a force, so need to consider Paul A Lagace @2001 Unit 20-5

qt MIT - 16.20 Fall, 2002 The response is: () = A cosω t Figure 20.1 Basic unforced dynamic response of single spring-mass system But, generally systems have a force, so need to consider: Paul A. Lagace © 2001 Unit 20 - 5

MT-1620 Fall 2002 Forced vibration The homogeneous solution is still valid, but must add a particular solution The simplest case here is a constant load with time Figure 20.2 Representation of constant applied load with time (think of the load applied suddenly step function at t= O) The governing equation is mq kq= F The particular solution has no time dependence since the force has no time dependence Paul A Lagace @2001 Unit 20-6

MIT - 16.20 Fall, 2002 Forced Vibration The homogeneous solution is still valid, but must add a particular solution The simplest case here is a constant load with time… Figure 20.2 Representation of constant applied load with time (think of the load applied suddenly ⇒ step function at t = 0) The governing equation is: m q˙˙ + k q = F0 The particular solution has no time dependence since the force has no time dependence: F0 qparticular = k Paul A. Lagace © 2001 Unit 20 - 6

MT-1620 al.2002 Now use the homogeneous solution with this to get the total solution qt)=C snot C2cosot k The Initial Conditions are q 0)=0 F6 q(0)=0→C1=0 So the final solution is F6 q coS ot k with Plotting this Paul A Lagace @2001 Unit 20-7

qt MIT - 16.20 Fall, 2002 Now use the homogeneous solution with this to get the total solution: F () = C1 sinω t + C2 cosω t + 0 k The Initial Conditions are: @ t = 0 q = 0 q˙ = 0 q () 0 = 0 ⇒ C = − F0 2 k q ˙() 0 = 0 ⇒ C1 = 0 So the final solution is: F q = 0 (1 − cosω t) k with ω = k m Plotting this: Paul A. Lagace © 2001 Unit 20 - 7

MT-1620 Fall 2002 Figure 20.3 Ideal dynamic response of single spring-mass system to constant force Dynamic response Static response 七 Note that Dynamic response= 2 X static response dynamic magnification factor'-will be larger when considering stresses over their static values Know this doesn't really happen(i. e response does not continue forever) What has been left out? DAMPING Paul A Lagace @2001 Unit 20-8

MIT - 16.20 Fall, 2002 Figure 20.3 Ideal dynamic response of single spring-mass system to constant force Dynamic response Static response Note that: Dynamic response = 2 x static response “dynamic magnification factor” - will be larger when considering stresses over their static values Know this doesn’t really happen (i.e. response does not continue forever) What has been left out? DAMPING Paul A. Lagace © 2001 Unit 20 - 8

MT-1620 al.2002 Actual behavior would be Figure 20.4 Actual dynamic response with damping of single spring mass system to constant force rate of damping dependent on magnitude of damper(c) Have considered a simple case. But, in general forces are not simple steps. Consider the next level The Unit Impulse Paul A Lagace @2001 Unit 20-9

MIT - 16.20 Fall, 2002 Actual behavior would be… Figure 20.4 Actual dynamic response (with damping) of single spring￾mass system to constant force rate of damping dependent on magnitude of damper (c) Have considered a simple case. But, in general forces are not simple steps. Consider the next “level”… The Unit Impulse Paul A. Lagace © 2001 Unit 20 - 9

MT-1620 al.2002 An impulse occurs at time t=t. Such a force is represented by the Dirac delta function: F(0=8(t Where t≠τ and ∫(-)d Recall: force x time impulse Hence g()6(t-)t =0 everywhere but att=t )」。0(- now a constant with regard to time Paul A Lagace @2001 Unit 20-10

MIT - 16.20 Fall, 2002 An impulse occurs at time t = τ. Such a force is represented by the Dirac delta function: F t() = δ (t − τ ) where: δ (t − τ ) = 0 @ t ≠ τ δ (t − τ ) → ∞ @ t = τ and: ∞ −∞ δ (t − τ ) dt = 1 ∫ Recall: force x time = impulse Hence, ∞ −∞ g t ∫ ( ) δ (t − τ ) dt = 0 everywhere but at t = τ ∞ = g( ) τ ∫−∞ δ (t − τ ) dt now a constant = 1 with regard to time Paul A. Lagace © 2001 Unit 20 - 10