Vibrations of Single Degree of Freedom Systems

Vibrations of Single Degree of Freedom Systems CEE 541. Structural Dynamics Department of Civil and Environmental Engineering Duke University Henri P. Gavin Fall, 16 This document describes free and forced dynamic responses of single degree of freedom (SDOF) systems. The prototype single degree of freedom system is a spring-mass-damper system in which the spring has no damping or mass, the mass has no stiffness or damping, the damper has no stiffness or mass. Furthermore, the mass is allowed to move in only one direction. The horizontal vibrations of a single-story building can be conveniently modeled as a single degree of freedom system. Part 1 of this document describes some useful trigonometric identities. Part shows how damped SDOF systems vibrate freely after being released from an initial displacement with some initial velocity. Part 3 covers the resposne of damped SDOF systems to persistent sinusoidal forcing. Consider the structural system shown in Figure 1, where: f(t) = external excitation force x(t) = displacement of the center of mass of the moving object m = mass of the moving object, f I = d dt (mẋ(t)) = mẍ(t) c = linear viscous damping coefficient, f D = cẋ(t) k = linear elastic stiffness coefficient, f S = kx(t) x(t) m k, c 11111111111 11111111111 f(t) c k m f(t) x(t) Figure 1. The proto-typical single degree of freedom oscillator.

CEE 541. Structural Dynamics Duke University Fall 16 H.P. Gavin The kinetic energy T (x, ẋ), the potential energy, V (x), and the external forcing and dissipative forces, p(x, ẋ), are T (x, ẋ) = 1 m(ẋ(t)) (1) V (x) = 1 k(x(t)) () p(x, ẋ) = cẋ(t) + f(t) (3) The general form of the differential equation describing a SDOF oscillator follows directly from Lagrange s equation, d T (x, ẋ) T (x, ẋ) V (x) + p(x, ẋ) =, (4) dt ẋ x x or from simply balancing the forces on the mass, F = : fi + f D + f S = f(t). (5) Either way, the equation of motion is: mẍ(t) + cẋ(t) + kx(t) = f(t), x() = d o, ẋ() = v o (6) where the initial displacement is d o, and the initial velocity is v o. The solution to equation (6) is the sum of a homogeneous part (free response) and a particular part (forced response). This document describes free responses of all types and forced responses to simple-harmonic forcing. 1 Trigonometric and Complex Exponential Expressions for Oscillations 1.1 Constant Amplitude An oscillation, x(t), with amplitude X and frequency ω can be described by sinusoidal functions. These sinusoidal functions may be equivalently written in terms of complex exponentials e ±iωt with complex coefficients, X = A + ib and X = A ib. (The complex constant X is called the complex conjugate of X.) x(t) = X cos(ωt + θ) (7) = a cos(ωt) + b sin(ωt) (8) = X e +iωt + X e iωt (9)

Dynamics of Single Degree of Freedom Systems 3 To relate equations (7) and (8), recall the cosine of a sum of angles, X cos(ωt + θ) = X cos(θ) cos(ωt) X sin(θ) sin(ωt) (1) Comparing equations (1) and (8), we see that a = X cos(θ), b = X sin(θ), and a + b = X. (11) Also, the ratio b/a provides an equation for the phase shift, θ, tan(θ) = b a (1) 6 period, T response, x(t) 4 - amplitude, -X -4-6 4 6 8 1 time, t, sec Figure. A constant-amplitude oscillation. To relate equations (8) and (9), recall the expression for a complex exponent in terms of sines and cosines, X e +iωt + X e iωt = (A + ib) (cos(ωt) + i sin(ωt)) + (A ib) (cos(ωt) i sin(ωt)) (13) = A cos(ωt) B sin(ωt) + ia sin(ωt) + ib cos(ωt) + A cos(ωt) B sin(ωt) ia sin(ωt) ib cos(ωt) = A cos(ωt) B sin(ωt) (14)

4 CEE 541. Structural Dynamics Duke University Fall 16 H.P. Gavin Comparing equations (14) and (8), we see that a = A, b = B, and tan(θ) = B A. (15) Any sinusoidal oscillation x(t) can be expressed equivalently in terms of equations (7), (8), or (9); the choice depends on the application, and the problem to be solved. Equations (7) and (8) are easier to interpret as describing a sinusoidal oscillation, but equation (9) can be much easier to work with mathematically. These notes make use of all three forms. One way to interpret the compex exponential notation is as the sum of complex conjugates, and Xe +iωt = [A cos(ωt) B sin(ωt)] + i[a sin(ωt) + B cos(ωt)] X e iωt = [A cos(ωt) B sin(ωt)] i[a sin(ωt) + B cos(ωt)] as shown in Figure 3. The sum of complex conjugate pairs is real, since the imaginary parts cancel out. Im A sin ωt + B cos ωt +ω A + B ω X exp(+i t) A cos ωt B sin ωt ω A + B Re A sin ωt B cos ωt X* exp( i ω t) Figure 3. Complex conjugate oscillations. The amplitude, X, of the oscillation x(t) can be found by finding the the sum of the complex amplitudes X and X. X = X + X = A + B = a + b (16)

Dynamics of Single Degree of Freedom Systems 5 Note, again, that equations (7), (8), and (9) are all equivalent using the relations among (a, b), (A, B), X, and θ given in equations (11), (1), (15), and (16). 1. Decaying Amplitude The dynamic response of damped systems decays over time. Note that damping may be introduced into a structure through diverse mechanisms, including linear viscous damping, nonlinear viscous damping, visco-elastic damping, friction damping, and plastic deformation. ll but linear viscous damping are somewhat complicated to analyze, so we will restrict our attention to linear viscous damping, in which the damping force f D is proportional to the velocity, f D = cẋ. To describe an oscillation which decays with time, we can multiply the expression for a constant amplitude oscillation by a positive-valued function which decays with time. Here we will use a real exponential, e σt, where σ <. 6 4 -X e σt period, T response, x(t) - -4-6 4 6 8 1 time, t, sec Figure 4. A decaying oscillation.

6 CEE 541. Structural Dynamics Duke University Fall 16 H.P. Gavin Multiplying equations (7) through (9) by e σt, x(t) = e σt X (cos(ωt + θ)) (17) = e σt (a cos(ωt) + b sin(ωt)) (18) = e σt (Xe iωt + X e iωt ) (19) = Xe (σ+iω)t + X e (σ iω)t () = Xe λt + X e λ t (1) Again, note that all of the above equations are exactly equivalent. The exponent λ is complex, λ = σ + iω and λ = σ iω. If σ is negative, then these equations describe an oscillation with exponentially decreasing amplitudes. Note that in equation (18) the unknown constants are σ, ω, a, and b. Angular frequencies, ω, have units of radians per second. Circular frequencies, f = ω/(π) have units of cycles per second, or Hertz. Periods, T = π/ω, have units of seconds. In the next section we will find that for an un-forced vibration, σ and ω are determined from the mass, damping, and stiffness of the system. We will see that the constant a equals the initial displacement d o, but that the constant b depends on the initial displacement and velocity, as well mass, damping, and stiffness.

Dynamics of Single Degree of Freedom Systems 7 Free response of systems with mass, stiffness and damping Using equation (1) to describe the free response of a single degree of freedom system, we will set f(t) = and will substitute x(t) = Xe λt into equation (6). mẍ(t) + cẋ(t) + kx(t) =, x() = d o, ẋ() = v o, () mλ Xe λt + cλxe λt + kxe λt =, (3) (mλ + cλ + k)xe λt =, (4) Note that m, c, k, λ and X do not depend on time. For equation (4) to be true for all time, (mλ + cλ + k)x =. (5) Equation (5) is trivially satisfied if X =. The non-trivial solution is mλ + cλ + k =. This is a quadratic equation in λ which has the roots, λ 1, = c m ± ( c m ) k m. (6) The solution to a homogeneous second order ordinary differential equation requires two independent initial conditions: an initial displacement and an initial velocity. These two independent initial conditions are used to determine the coefficients, X and X (or A and B, or a and b) of the two linearly independent solutions corresponding to λ 1 and λ. The amount of damping, c, qualitatively affects the quadratic roots, λ 1,, and the free response solutions. Case 1 c = undamped If the system has no damping, c =, and λ 1, = ±i k/m = ±iω n. (7) This is called the natural frequency of the system. Undamped systems oscillate freely at their natural frequency, ω n. The solution in this case is x(t) = Xe iω nt + X e iω nt = a cos ω n t + b sin ω n t, (8)

8 CEE 541. Structural Dynamics Duke University Fall 16 H.P. Gavin which is a real-valued function. The amplitudes depend on the initial displacement, d o, and the initial velocity, v o. Case c = c c critically damped If (c/(m)) = k/m, or, equivalently, if c = mk, then the discriminant of equation (6) is zero, This special value of damping is called the critical damping rate, c c, c c = mk. (9) The ratio of the actual damping rate to the critical damping rate is called the damping ratio, ζ. ζ = c c c. (3) The two roots of the quadratic equation are real and are repeated at λ 1 = λ = c/(m) = c c /(m) = mk/(m) = ω n, (31) and the two basic solutions are equal to each other, e λ 1t = e λ t. In order to admit solutions for arbitrary initial displacements and velocities, the solution in this case is x(t) = x 1 e ω nt + x t e ω nt. (3) where the real constants x 1 and x are determined from the initial displacement, d o, and the initial velocity, v o. Details regarding this special case are in the next section. Case 3 c > c c over-damped If the damping is greater than the critical damping, then the roots, λ 1 and λ are distinct and real. If the system is over-damped it will not oscillate freely. The solution is x(t) = x 1 e λ 1t + x e λ t, (33) which can also be expressed using hyperbolic sine and hyperbolic cosine functions. The real constants x 1 and x are determined from the initial displacement, d o, and the initial velocity, v o.

Dynamics of Single Degree of Freedom Systems 9 Case 4 < c < c c under-damped If the damping rate is positive, but less than the critical damping rate, the system will oscillate freely from some initial displacement and velocity. The roots are complex conjugates, λ 1 = λ, and the solution is x(t) = X e λt + X e λ t, (34) where the complex amplitude depends on the initial displacement, d o, and the initial velocity, v o. We can re-write the dynamic equations of motion using the new dynamic variables for natural frequency, ω n, and damping ratio, ζ. Note that c k m = c 1 c k c = = k m m k m m k km m = ζω n. (35) mẍ(t) + cẋ(t) + kx(t) = f(t), (36) ẍ(t) + c mẋ(t) + k m x(t) = 1 f(t), (37) m ẍ(t) + ζω n ẋ(t) + ωn x(t) = 1 f(t), (38) m The expression for the roots λ 1,, can also be written in terms of ω n and ζ. λ 1, = c m ± = ζω n ± ( c ) k m, (39) m (ζω n ) ωn, (4) = ζω n ± ω n ζ 1. (41)

1 CEE 541. Structural Dynamics Duke University Fall 16 H.P. Gavin Some useful facts about the roots λ 1 and λ are: ω = Imλ λ 1 + λ = ζω n λ 1 λ = ω n ζ 1 λ 1 x ω n ω d ω n = 1 4 (λ 1 + λ ) 1 4 (λ 1 λ ) ω n = λ 1 λ ζω n σ = Re λ ζ = (λ 1 + λ )/(ω n ) λ x ω d.1 Free response of critically-damped systems The solution to a homogeneous second order ordinary differential equation requires two independent initial conditions: an initial displacement and an initial velocity. These two initial conditions are used to determine the coefficients of the two linearly independent solutions corresponding to λ 1 and λ. If λ 1 = λ, then the solutions e λ1t and e λt are not independent. In fact, they are identical. In such a case, a new trial solution can be determined as follows. Assume the second solution has the form x(t) = u(t)x e λ t, (4) ẋ(t) = u(t)x e λ t + u(t)λ x e λ t, (43) ẍ(t) = substitute these expressions into ü(t)x e λ t + u(t)λ x e λ t + u(t)λ x e λ t ẍ(t) + ζω n ẋ(t) + ω nx(t) =, collect terms, and divide by x e λ t, to get ü(t) + ω n (ζ 1) u(t) + ω n(1 ζ)u(t) = (44) or ü(t) = (since ζ = 1). If the acceleration of u(t) is zero then the velocity of u(t) must be constant, u(t) = C, and u(t) = Ct, from which the new trial solution is found. x(t) = u(t)x e λ t = x t e λ t.

Dynamics of Single Degree of Freedom Systems 11 So, using the complete trial solution x(t) = x 1 e λt + x te λt, and incorporating initial conditions x() = d o and ẋ() = v o, the free response of a criticallydamped system is:. Free response of underdamped systems x(t) = d o e ω nt + (v o + ω n d o ) t e ω nt. (45) If the system is under-damped, then ζ < 1, ζ 1 is imaginary, and λ 1, = ζω n ± iω n ζ 1 = σ ± iω. (46) The frequency ω n ζ 1 is called the damped natural frequency, ω d, ω d = ω n ζ 1. (47) It is the frequency at which under-damped SDOF systems oscillate freely, With these new dynamic variables (ζ, ω n, and ω d ) we can re-write the solution to the damped free response, x(t) = e ζω nt (a cos ω d t + b sin ω d t), (48) = Xe λt + X e λ t. (49) Now we can solve for X, (or, equivalently, A and B) in terms of the initial conditions. At the initial point in time, t =, the position of the mass is x() = d o and the velocity of the mass is ẋ() = v o. x() = d o = Xe λ + X e λ (5) = X + X (51) = (A + ib) + (A ib) = A = a. (5) ẋ() = v o = λxe λ + λ X e λ, (53) = λx + λ X, (54) = (σ + iω d )(A + ib) + (σ iω d )(A ib), (55) = σa + iω d A + iσb ω d B + σa iω d A iσb ω d B, (56) = σa ωb (57) = ζω n d o ω d B, (58)

1 CEE 541. Structural Dynamics Duke University Fall 16 H.P. Gavin from which we can solve for B and b, B = v o + ζω n d o ω d and b = v o + ζω n d o ω d. (59) Putting this all together, the free response of an underdamped system to an arbitrary initial condition, x() = d o, ẋ() = v o is x(t) = e ζω nt ( d o cos ω d t + v ) o + ζω n d o sin ω d t. (6) ω d 6 v o d o 4 -X e -ζ ω n t damped natural period, T d response, x(t) - -4-6 4 6 8 1 time, t, sec Figure 5. Free response of an under-damped oscillator to an initial displacement and velocity.

Dynamics of Single Degree of Freedom Systems 13.3 Free response of over-damped systems If the system is over-damped, then ζ > 1, and ζ 1 is real, and the roots are both real and negative λ 1, = ζω n ± ω n ζ 1 = σ ± ω d. (61) Substituting the initial conditions x() = d o and ẋ() = v o into the solution (equation (33)), and solving for the coefficients results in x 1 = v o + d o (ζω n + ω d ) ω d, (6) x = d o x 1. (63) Substituting the hyperbolic sine and hyperbolic cosine expressions for the exponentials results in x(t) = e ζω nt ( d o cosh ω d t + v o + ζω n d o ω d ) sinh ω d t. (64) 7 v o 6 d o 5 response, x(t) 4 3 1 critically damped ζ=1.5 over damped ζ=5. ζ=. -1.5 1 1.5.5 3 3.5 4 time, t, sec Figure 6. Free response of critically-damped (yellow) and over-damped (violet) oscillators to an initial displacement and velocity.

14 CEE 541. Structural Dynamics Duke University Fall 16 H.P. Gavin The undamped free response can be found as a special case of the underdamped free response. While special solutions exist for the critically damped response, this response can also be found as limiting cases of the underdamped or over-damped responses..4 Finding the natural frequency from self-weight displacement Consider a spring-mass system in which the mass is loaded by gravity, g. The static displacement D st is related to the natural frequency by the constant of gravitational acceleration. D st = mg/k = g/ω n (65).5 Finding the damping ratio from free response Consider the value of two peaks of the free response of an under-damped system, separated by n cycles of motion x 1 = x(t 1 ) = e ζω nt 1 (A) (66) x 1+n = x(t 1+n ) = e ζω nt 1+n (A) = e ζω n(t 1 +nπ/ω d ) (A) (67) The ratio of these amplitudes is x 1 x 1+n = e ζω nt 1 e ζω n(t 1 +nπ/ω d ) = e ζωnt1 e ζω nt 1e nπζω n /ω d = e nπζ/ 1 ζ, (68) which is independent of ω n and ω d. Defining the log decrement δ(ζ) as ln(x 1 /x 1+n )/n, δ(ζ) = πζ 1 ζ (69) and, inversely, ζ(δ) = δ 4π + δ δ π (7) where the approximation is accurate to within 3% for ζ <. and is accurate to within 1.5% for ζ <.1.

Dynamics of Single Degree of Freedom Systems 15.6 Summary To review, some of the important expressions relating to the free response of a single degree of freedom oscillator are: X cos(ωt + θ) = a cos(ωt) + b sin(ωt) = Xe +iωt + X e iωt X = a + b ; tan(θ) = b/a; X = A + ib; A = a/; B = b/; mẍ(t) + cẋ(t) + kx(t) = ẍ(t) + ζω n ẋ(t) + ω nx(t) = x() = d o, ẋ() = v o ω n = k m ζ = c c c = c mk ω d = ω n ζ 1 x(t) = e ζω nt ( d o cos ω d t + v ) o + ζω n d o sin ω d t ω d ( ζ < 1) δ = 1 n ln ( ) x1 x 1+n ζ(δ) = δ 4π + δ δ π

16 CEE 541. Structural Dynamics Duke University Fall 16 H.P. Gavin 3 Response of systems with mass, stiffness, and damping to sinusoidal forcing When subject to simple harmonic forcing with a forcing frequency ω, dynamic systems initially respond with a combination of a transient response at a frequency ω d and a steady-state response at a frequency ω. The transient response at frequency ω d decays with time, leaving the steady state response at a frequency equal to the forcing frequency, ω. This section examines three ways of applying forcing: forcing applied directly to the mass, inertial forcing applied through motion of the base, and forcing from a rotating eccentric mass. 3.1 Direct Force Excitation If the SDOF system is dynamically forced with a sinusoidal forcing function, then f(t) = F cos(ωt), where ω is the frequency of the forcing, in radians per second. If f(t) is persistent, then after several cycles the system will respond only at the frequency of the external forcing, ω. Let s suppose that this steady-state response is described by the function then and x(t) = a cos ωt + b sin ωt, (71) ẋ(t) = ω( a sin ωt + b cos ωt), (7) ẍ(t) = ω ( a cos ωt b sin ωt). (73) Substituting this trial solution into equation (6), we obtain mω ( a cos ωt b sin ωt) + cω ( a sin ωt + b cos ωt) + k (a cos ωt + b sin ωt) = F cos ωt. (74) Equating the sine terms and the cosine terms ( mω a + cωb + ka) cos ωt = F cos ωt (75) ( mω b cωa + kb) sin ωt =, (76)

Dynamics of Single Degree of Freedom Systems 17 which is a set of two equations for the two unknown constants, a and b, for which the solution is k mω cω cω k mω a(ω) = b(ω) = a b = F, (77) k mω (k mω ) + (cω) F (78) cω (k mω ) + (cω) F. (79) Referring to equations (7) and (1) in section 1.1, the forced vibration solution (equation (71)) may be written x(t) = a(ω) cos ωt + b(ω) sin ωt = X(ω) cos (ωt + θ(ω)). (8) The angle θ is the phase between the force f(t) and the response x(t), and tan(θ(ω)) = b(ω) a(ω) = cω (81) k mω Note that π < θ(ω) < for all positive values of ω, meaning that the displacement response, x(t), always lags the external forcing, F cos(ωt). The ratio of the response amplitude X(ω) to the forcing amplitude F is X(ω) F = a (ω) + b (ω) F = 1 (k mω ) + (cω). (8) This equation shows how the response amplitude X depends on the amplitude of the forcing F and the frequency of the forcing ω, and has units of flexibility. Let s re-derive this expression using complex exponential notation! The equations of motion are mẍ(t) + cẋ(t) + kx(t) = F cos ωt = F (ω)e iωt + F (ω)e iωt. (83) In a solution of the form, x(t) = X(ω)e iωt + X (ω)e iωt, the coefficient X(ω) corresponds to the positive exponents (positive frequencies), and X (ω) corresponds to negative exponents (negative frequencies). Positive exponent coefficients and negative exponent coefficients are independent and may be found

18 CEE 541. Structural Dynamics Duke University Fall 16 H.P. Gavin separately. Considering the positive exponent solution, the forcing is expressed as F (ω)e iωt and the partial solution X(ω)e iωt is substituted into the forced equations of motion, resulting in from which ( mω + ciω + k) X(ω) e iωt = F (ω) e iωt, (84) X(ω) F (ω) = 1 (k mω ) + i(cω), (85) which is complex-valued. This complex function has a magnitude X(ω) F (ω) = 1 (k mω ) + (cω), (86) the same as equation (8) but derived using e iωt in just three simple lines. Equation (85) may be written in terms of the dynamic variables, ω n and ζ. Dividing the numerator and the denominator of equation (8) by k, and noting that F/k is a static displacement, x st, we obtain X(ω) F (ω) = X(ω) = X(Ω) x st = X(Ω) x st = 1/k ( 1 m k ω) + i ( c, k ω) (87) F (ω)/k ( ( ) ) ), 1 ω ω n + i (ζ ωωn (88) 1 (1 Ω ) + i (ζω), (89) 1 (1 Ω ) + (ζω), (9) where the frequency ratio Ω is the ratio of the forcing frequency to the natural frequency, Ω = ω/ω n. This equation is called the dynamic amplification factor. It is the factor by which displacement responses are amplified due to the fact that the external forcing is dynamic, not static. See Figure 7.

Dynamics of Single Degree of Freedom Systems 19 1 8 frequency response: F to X ζ=.5 X / X st 6.1 4..5 1..5 1 1.5.5 3 ζ=.5 phase, degrees -45-9 -135 ζ=1. -18.5 1 1.5.5 3 frequency ratio, Ω = ω / ω n Figure 7. The dynamic amplification factor for external forcing, X/xst, equation (89). To summarize, the steady state response of a simple oscillator directly excited by a harmonic force, f(t) = F cos ωt, may be expressed in the form of equation (7) x(t) = F /k ζω cos(ωt + θ), tan θ = (91) (1 Ω ) + (ζω) 1 Ω or, equivalently, in the form of equation (8) x(t) = where Ω = ω/ω n. F /k (1 Ω ) + (ζω) [ (1 Ω ) cos ωt + (ζω) sin ωt ], (9)

CEE 541. Structural Dynamics Duke University Fall 16 H.P. Gavin 3. Support Acceleration Excitation When the dynamic loads are caused by motion of the supports (or the ground, as in an earthquake) the forcing on the structure is the inertial force resisting the ground acceleration, which equals the mass of the structure times the ground acceleration, f(t) = m z(t). m k, c 11 z(t) x(t) z(t) c k m x(t) Figure 8. The proto-typical SDOF oscillator subjected to base motions, z(t) m ( ẍ(t) + z(t) ) + cẋ(t) + kx(t) = (93) mẍ(t) + cẋ(t) + kx(t) = m z(t) (94) ẍ(t) + ζω n ẋ(t) + ω nx(t) = z(t) (95) Note that equation (95) is independent of mass. Systems of different masses but with the same natural frequency and damping ratio have the same behavior and respond in exactly the same way to the same support motion. If the ground displacements are sinusoidal z(t) = Z cos ωt, then the ground accelerations are z(t) = Zω cos ωt, and f(t) = m Zω cos ωt. Using the complex exponential formulation, we can find the steady state response as a function of the frequency of the ground motion, ω. mẍ(t) + cẋ(t) + kx(t) = m Zω cos ωt = mz(ω)ω e iωt + mz (ω)ω e iωt (96) The steady-state response can be expressed as the sum of independent complex exponentials, x(t) = X(ω)e iωt +X (ω)e iωt. The positive exponent parts are independent of the negative exponent parts and can be analyzed separately.

Dynamics of Single Degree of Freedom Systems 1 Assuming persistent excitation and ignoring the transient resposne (the particular part of the solution), the response will be harmonic. Considering the positive exponent part of the forcing mz(ω)ω e iωt, the positive exponent part of the steady-state response will have a form Xe iωt. Substituting these expressions into the differential equation (96), collecting terms, and factoring out the exponential e iωt, the frequency response function is X(ω) Z(ω) = = mω (k mω ) + i(cω), Ω (1 Ω ) + i(ζω) where Ω = ω/ω n (the forcing frequency over the natural frequency), and See Figure 9. (97) X(Ω) Z(Ω) = Ω (98) (1 Ω ) + (ζω) Finally, let s consider the motion of the mass with respect to a fixed point. This is called the total motion and is the sum of the base motion plus the motion relative to the base, z(t) + x(t). and X + Z Z X + Z Z = X Z + 1 = (1 Ω ) + i(ζω) + Ω (1 Ω ) + i(ζω) 1 + i(ζω) = (1 Ω ) + i(ζω) (99) 1 + (ζω) = = Tr(Ω, ζ). (1) (1 Ω ) + (ζω) This function is called the transmissibility ratio, Tr(Ω, ζ). It determines the ratio between the total response amplitude X + Z and the base motion Z. See figure 1. For systems that have a longer natural period (lower natural frequency) than the period (frequency) of the support motion, (i.e., Ω > ), the transmissibility ratio is less than 1 especialy for low values of damping ζ. In such systems the motion of the mass is less than the motion of the supports and we say that the mass is isolated from motion of the supports.

CEE 541. Structural Dynamics Duke University Fall 16 H.P. Gavin 1 8 ζ=.5 frequency response: Z to X X / Z 6.1 4..5 1..5 1 1.5.5 3 ζ=.5 phase, degrees -45-9 -135 ζ=1. -18.5 1 1.5.5 3 frequency ratio, Ω = ω / ω n Figure 9. The dynamic amplification factor for base-excitation, X/ Z, equation (97). transmissibility: Z to (X+Z) 1 ζ=.5 8 (X+Z) / Z 6.1 4..5 ζ=1..5 1 1.5.5 3 phase, degrees -45 ζ=1. -9-135 ζ=.5-18.5 1 1.5.5 3 frequency ratio, Ω = ω / ω n Figure 1. The transmissibility ratio (X + Z)/Z = Tr(Ω, ζ), equation (99).

Dynamics of Single Degree of Freedom Systems 3 3.3 Eccentric-Mass Excitation Another type of sinusoidal forcing which is important to machine vibration arises from the rotation of an eccentric mass. Consider the system shown in Figure 11 in which a mass µm is attached to the primary mass m via a rigid link of length r and rotates at an angular velocity ω. In this single degree of freedom analysis, the motion of the primary mass is constrained to lie along the x coordinate and the forcing of interest is the x-component of the reactive centrifugal force. This component is µmrω cos(ωt) where the angle ωt is the counter-clockwise angle from the x coordinate. The equation of motion with m k, c x(t) ω 11111111111 11111111111 Figure 11. The proto-typical SDOF oscillator subjected to eccentric-mass excitation. this forcing is c k m r ω x(t) µm mẍ(t) + cẋ(t) + kx(t) = µmrω cos(ωt) (11) This expression is simply analogous to equation (83) in which F = µ m r ω. With a few substitutions, the frequency response function is found to be X r = µ Ω (1 Ω ) + i (ζω), (1) which is completely analogous to equation (97). The plot of the frequency response function of equation (1) is simply proportional to the function plotted in Figure 9. The magnitude of the dynamic force transmitted between a machine supported on dampened springs and the base, f T, is the sum of the forces in the springs and the dampers. The ratio of the transmitted force to the force generated by the eccentric mass, µmrω, is the transmissibility ratio, (1). f T = Tr(Ω, ζ) (13) µmrω

4 CEE 541. Structural Dynamics Duke University Fall 16 H.P. Gavin Note, here, though, that the denomonator, µmrω, increases with ω. The transmitted force amplitude increases with µω Tr(Ω, ζ). Multiplying both sides of equation (13) by Ω we obtain the transmission ratio. f T = Ω Tr(Ω, ζ) (14) µmrωn Unlike the transmissibility ratio, which asymptotically approaches with increasing Ω, the vibratory force transmitted from an eccentric mass excitation is when Ω = but increases with Ω for Ω >. This increasing effect is significant for ζ >., as shown in Figure 1. For high frequency Ω (X+Z) / Z 1 8 6 4 transmission: Ω (Z to (X+Z)) ζ=.5.1 ζ=1...5.5 1 1.5.5 3 phase, degrees -45 ζ=1. -9-135 ζ=.5-18.5 1 1.5.5 3 frequency ratio, Ω = ω / ω n Figure 1. The transmission ratio Ω Tr(Ω, ζ), equation (14). ratios (ω n < ω/) and low damping ratios (ζ <.), the force transmitted from the rotating machinery to the floor ( f T ) is less than half of the force generated by the machinery (µmrω ). Further, for ζ., the value of the transmitted force is rougly independent of the forcing frequency.

Dynamics of Single Degree of Freedom Systems 5 3.4 Finding the damping from the peak of the frequency response function For lightly damped systems, the frequency ratio of the resonant peak, the amplification of the resonant peak, and the width of the resonant peak are functions to of the damping ratio only. Consider two frequency ratios Ω 1 and Ω such that H(Ω 1, ζ) = H(Ω, ζ) = H peak/ where H(Ω, ζ) is one of the frequency response functions described in earlier sections. The frequency ratio corresponding to the peak of these functions Ω peak, and the value of the peak of these functions, H peak are given in Table 1. Note that the peak coordinate depends only upon the damping ratio, ζ. Since Ω Ω 1 = (Ω Ω 1 )(Ω + Ω 1 ) and since Ω + Ω 1, ζ Ω Ω 1 (15) which is called the half-power bandwidth formula for damping. For the first, second, and fourth frequency response functions listed in Table 1 the approximation is accurate to within 5% for ζ <. and is accurate to within 1% for ζ <.1. Table 1. Peak coordinates for various frequency response functions. H(Ω, ζ) Ω peak H peak Ω Ω 1 1 (1 Ω )+i(ζω) 1 ζ 1 4ζ (1 ζ ) 4ζ 1 ζ iω 1 (1 Ω 1 )+i(ζω) 4ζ 4ζ 1 + ζ Ω (1 Ω )+i(ζω) 1 1 ζ 1 4ζ (1 ζ ) 4ζ 1 ζ 1 8ζ (1 ζ ) 1+i(ζΩ) (1 Ω )+i(ζω) ((1+8ζ ) 1/ 1) 1/ ζ 8ζ 4 8ζ 4 4ζ 1+ 1+8ζ ouch.

6 CEE 541. Structural Dynamics Duke University Fall 16 H.P. Gavin 4 Real and Imaginary. Even and Odd. Magnitude and Phase. Using the rules of complex division, it is not hard to show that R[H(Ω, ζ)] = R[H( Ω, ζ)] (16) I[H(Ω, ζ)] = I[H( Ω, ζ)]. (17) That is, R[H(Ω)] is an even function and I[H(Ω)] is an odd function. This fact is true for any dynamical system for which the inputs and outputs are real-valued. For any frequency response function, the magnitude H(Ω, ζ) and phase θ(ω, ζ) may be found from H(Ω, ζ) = (R[H(Ω, ζ)]) + (I[H(Ω, ζ)]) (18) I[H(Ω, ζ)] tan θ(ω, ζ) = (19) R[H(Ω, ζ)] Expressions for the magnitude of various frequency response functions are given by equations (89), (98), and (99). The magnitude and phase lag of these functions are plotted in Figures 7, 9, and 1. Real and imaginary parts of H(Ω, ζ) are plotted in Figure 13. Note the following: The real and imaginary parts are even and odd, respectively. The real part of X/x st (force to response displacement) is zero at Ω = 1. The phase at Ω = 1 is 9 degrees. The real part of X/Z (support motion to response motion) is zero at Ω = 1. The phase at Ω = 1 is 9 degrees. The imaginary part of iωx/x st (force to response velocity) is zero at Ω = 1. The phase at Ω = 1 is 9 degrees. The real part of iωx/x st (force to response velocity) is maximum at Ω = 1. The real part of iωx/x st (force to response velocity) is positive for all values of Ω.

Dynamics of Single Degree of Freedom Systems 7 6 X / X st 4 - (a) -4-6 - -1.5-1 -.5.5 frequency ratio, Ω = ω / ω n 1 1.5 6 4 (b) i Ω X / X st - -4 - -1.5-1 -.5.5 frequency ratio, Ω = ω / ω n 1 1.5 6 4 (c) X / Z - -4-6 - -1.5-1 -.5.5 1 1.5 frequency ratio, Ω = ω / ω n 6 4 (d) (X+Z) / Z - -4-6 - -1.5-1 -.5.5 1 1.5 frequency ratio, Ω = ω / ω n Figure 13. The real (even) and imaginary (odd) parts of frequency response functions, ζ =.1.

8 CEE 541. Structural Dynamics Duke University Fall 16 H.P. Gavin 5 Vibration Sensors A vibration sensor may be accurately modeled as an inertial mass supported by elements with stiffness and damping (i.e. as a single degree of freedom oscillator). Vibration sensors mounted to a surface ideally measure the velocity or acceleration of the surface with respect to an inertial reference frame. The electrical signals generated by vibration sensors are actually proportional to the velocity of the mass with respect to the sensor s case (for seismometers) or the deformation of the elastic elements of the sensor (for accelerometers). The frequency response functions and sensitivities of seismometers and accelerometers have qualitative differences. 5.1 Seismometers Seismometers transduce the velocity of a magnetic inertial mass to electrical current in a coil. The transduction element of seismometers is therefore based on Ampere s Law, and seismometers are made from sprung magnetic masses that are guided to move within a coil fixed to the instrument housing. The electrical current generated within the coil is proportional to the velocity of the magnetic mass relative to the coil; so the mechanical input is the velocity of the case ż(t) = iωze iωt and the electrical output is proportional to the velocity of the inertial mass with respect to the case v(t) = κẋ(t) = κiωx(ω)e iωt. These variables are related in the frequency domain by equation (97). V iωz = κiωx iωz = κx Z = κω (1 Ω ) + i(ζω) (11) The sensitivity of seismometers approaches zero as Ω approaches zero. In order for seismometers to measure low frequencies, the natural frequency of seismometers is designed to be quite small (less than.5 Hz, and sometimes less than.1 Hz). In comparison to accelerometers, seismometers are heavy, large, delicate, sensitive, high output sensors which do not require external power or amplification. They require frequency-domain calibration.

Dynamics of Single Degree of Freedom Systems 9 5. Accelerometers Accelerometers transduce the deformation of inertially-loaded elastic elements within the sensor to an electrical charge, a voltage, or a current. Accelerometers may be designed with many types of transduction elements, including piezo-electric materials, strain-gages, variable capacitance components, and feed-back stabilization. In all cases, the mechanical input to the accelerometer is the acceleration of the case z(t) = ω Ze iωt, the electrical output is proportional to the deformation of the spring v(t) = κx(t) = κxe iωt, and these variables are related in the frequency domain by V ω Z = κ/ω n (1 Ω ) + i(ζω) (111) So the sensitivity of an accelerometer increases with decreasing natural frequency. Piezo-electric accelerometers have natural frequencies in the range of 1 khz to khz and damping ratios in the range of.5% to 1.%. Because of electrical coupling considerations and the sensitivity of peizo-electric accelerometers to low-frequency temperature transients, the sensitivity and signal-to-noise ratio of piezo-electric accelerometers below 1 Hz can be quite poor. Micro-machined electro-mechanical silicon (MEMS) accelerometers are monolithic with their signal-conditioning micro-circuitry. They typically have natural frequencies in the 1 Hz to 5 Hz range and are accurate down to frequencies of Hz (constant acceleration, e.g., gravity). These sensors are damped to a level of about 7% of critical damping. Force-balance accelerometers utilize feedback circuitry to magnetically stabilize an inertial mass. The force required to stabilize the mass is directly proportional to the acceleration of the sensor. Such sensors can be made to measure accelerations in the µg range down to frequencies of Hz, and have natural frequencies on the order of 1 Hz, also with damping of about 7% of critical damping.

3 CEE 541. Structural Dynamics Duke University Fall 16 H.P. Gavin Accelerometers are typically light, small, and rugged but require electrical power, amplification, and signal conditioning. 5.3 Design Considerations for Accelerometers Accelerometers should have a uniform amplitude spectrum and a linear phase spectrum (minimum phase distortion) over the frequency bandwidth of the application. The phase distortion is the deviation in the phase lag from a linear phase shift. Figure 14 is a close-up of the frequency-response function of equation (111) over the typical frequency band width of accelerometer applications. A linear phase spectrum is equivalent to a constant time delay. Consider a phase-lagged dynamic response in which the phase θ changes linearly with frequency, θ = τω. v(t) = V cos(ωt + τω) = V cos(ω(t + τ)) (11) shifted which is the same as a response that is shifted in time by a constant time increment of τ for all frequencies. As can be seen in Figure 14, a level of damping in the range of.67 to.71 provides an amplitude distortion of less than.5% and a phase distortion of less than.5 degrees for a bandwidth up to 3% of the sensor natural frequency. A sensor damping of ζ = / provides an optimally flat amplitude response and an extremely small phase distortion (<. degrees?) up to 5% of the sensor natural frequency. For frequency ratios in this range, the phase spectrum is [ ] ζω θ(ω) = tan 1 ζω (113) 1 Ω giving a time lag τ of approximately ζ/ω n. For ζ = /, the time lag τ evaluates to τ = /ω n = /(πf n ).3T n.

Dynamics of Single Degree of Freedom Systems 31 1.15 Accelerometer Sensitivity (ω Z to X) 1.1 X / (ω Z) 1.5 1.995.99 ζ =.65 ζ =.675.985 ζ =.77.1..3.4.5 phase distortion, degrees 1.5 -.5 ζ =.65 ζ =.675 ζ =.77-1.1..3.4.5 frequency ratio, Ω = ω / ω n Figure 14. Accelerometer frequency response functions