Dynamics of Structures:

Theory and Analysis

Steen Krenk

Technical University of Denmark

1. Free vibrations

2. Forced vibrations

3. Transient response

4. Damping mechanisms

5. Modal analysis I: Basic idea and matrix formulation

6. Modal analysis II: Implementation and system reduction

7. Damping and tuned mass dampers

8. Time integration by Newmark methods

9. Structural response to earthquakes

10. Vibration of cables, bars, etc.

11. Vibration of beams

12. Finite element formulation for bars, beams etc.

13. Course summary

Books

Daniel J. Inman, Engineering Vibration, 2nd Edition, Prentice Hall Interna-tional, Upper Saddle River, N.J., 2001.

J.W. Tedesco, W.G. McDougal and C.A. Ross, Structural Dynamics, Addison-Wesley, Menlo Park, Ca, 1999.

A.K. Chopra, Dynamics of Structures. Theory and application to EarthquakeEngineering, Prentice-Hall, Upper Saddle River, N.J., 2001.

Geradin, M. and Rixen, D., Mechanical Vibrations, Theory and Applicationsto Structural Dynamics, 2’nd ed., Wiley, Chichester, 1997.

I. Langen and R. Sigbjørnsson, Dynamisk Analyse av Konstruktioner, Tapir,Trondheim, 1979.

Journals

Earthquake Engineering and Structural Dynamics.

Journal of Sound and Vibration

Journal of Engineering Structures.

Journal of Structural Engineering.

Journal of Engineering Mechanics.

International Journal for Numerical Methods in Engineering.

Computers and Structures.

General Introduction

Offshore platforms

Tyra South – STAR platform Harald Field (www.maersk.com)

Wind Turbines

Nysted Wind Farm (www.nystedwindfarm.com)

Cable stayed bridge

Erasmus Bridge, Rotterdam (www.h2olland.nl)

Rotating pedestrian bridge

Gateshead Millennium Bridge, Gateshead (www.gateshead.gov.uk/bridge)

Pedestrian bridge

Solferino Bridge, Paris (S. Krenk)

Pedestrian suspension bridge

Millennium Bridge, London (www.galinsky.com/buildings/millenniumbridge/ )

See also the detailed presentation at www.arup.com/millenniumbridge/

Lecture 1: Free Vibrations

• Undamped vibrations

• Damped vibrations

• Logarithmic decrement

• Energy balance

• State variables and state-space

• Discrete time increments

Undamped vibrations

Equation of motion by balance of inertial force mx and elastic force kx withmass m and stiffness k,

mx + k x = 0 (1.1)

Obtain normalized equation by division by m

x + ω2

0 x = 0 (1.2)

where the natural angular frequency ω0 is defined by

ω0 =

√

k

m(1.3)

Note that the (angular) frequency is determined by

ω2

0 =‘stiffness′

‘mass′

In many problems ‘stiffness’ and ‘mass’ can be estimated by use of energymethods.

Solution

Solution requires initial conditions in terms of initial position x0 and initialvelocity v0 = x0.

x(t) = x0 cos(ω0t) +v0

ω0

sin(ω0t) (1.4)

ω0 t

x

ω0 t

v / ω0

Figure 1.1: a) displacement x(t), and b) velocity x(t) as function of time t.

The period T requires the phase ω0t to increase by 2π, Note that the (angular)frequency is determined by

T =2π

ω0

(1.5)

The natural frequency f is the number of oscillations per time unit, whereby

f =1

T=

ω0

2π(1.6)

Damped vibrations

Include energy dissipation through damping force cx,

mx(t) + c x(t) + k x(t) = 0 (1.7)

Note, that all three parameters m, c, k are positive, when the forces arerestoring.

Normalize equation by division by m

x + 2ζω0 x + ω2

0 x = 0 (1.8)

using the natural angular frequency ω0,

ω0 =

√

k

m(1.9)

and the damping ratio ζ,

ζ =c

2√km

(1.10)

The characteristics of the free vibration behavior depends on the magnitudeof the damping ratio ζ.

Solutions

The free vibration solution is of exponential type and is found by substitutionof the partial solution x(t) = Aeλt into the homogeneous equation of motion,yielding the characteristic equation

λ2 + 2ζω0λ + ω2

0 = 0 (1.11)

There are three different cases of damped vibrations, depending on the mag-nitude of the damping ratio ζ. The solutions are given below for initialconditions (x, x)t=0 = (x0, v0).

Underdamped, 0 < ζ < 1 :

Two complex roots of characteristic equation, λ = ω0(−ζ ± i√

1− ζ2).

Imaginary part is expressed by defining the damped natural frequency

ωd = ω0

√

1− ζ2 (1.12)

This leads to complex partial solutions of the form

x(t) = A e−ζω0t

︸ ︷︷ ︸

amplitude

e±iωdt

︸ ︷︷ ︸

oscillation

(1.13)

The last factor can be expressed in terms of sin(ωdt) and cos(ωdt).

x(t) = x0 e−ζω0t [ cos(ωdt) +

ζω0

ωdsin(ωdt) ]

+v0

ωde−ζω0t sin(ωdt)

(1.14)

ω0 t

x

ω0 t

v / ω0

Figure 1.2: a) displacement x(t), and b) velocity x(t) for ζ = 0.05.

The period of the oscillatory factor in damped free oscillations is Td = 2π/ωd.

Critically damped, ζ = 1 :

The characteristic equation has the real double root, λ = −ω0, and in thiscase the amplitude A is replaced by a linear function A+Bt.

x(t) = [ x0 + (ω0x0 + v0) t ] e−ω0t (1.15)

Overdamped, 1 < ζ :

The characteristic equation has two real roots λ = ω0(−ζ ±√

ζ2 − 1). In-troducing the parameter ωd = ω0

√

ζ2 − 1 < ζω0 the solution is expressed interms of hyperbolic functions as

x(t) = x0 e−ζω0t [ cosh(ωdt) +

ζω0

ωdsinh(ωdt) ]

+v0

ωde−ζω0t sinh(ωdt)

(1.16)

Logarithmic decrement

ω0 t

x,aa

j aj+1

Figure 1.3: Displacement record with maxima aj and aj+1 for ζ = 0.05.

Underdamping can also be characterized by the decrease in amplitude fromone maximum to the next. The ratio between any two maximum valuesfollowing each other is constant, and the logarithmic decrement is definedas

δ = ln

(ajaj+1

)

= ln(eζω0Td

)=

2π ζ√

1− ζ2(1.17)

or for lightly damped systems

δ ' 2π ζ for ζ << 1 (1.18)

NOTE: Lightly damped structures may have ζ as low as 0.001 .

Energy balance

Form the rate of work by multiplication of the equation of motion with x,

x(t)

(

mx(t) + c x(t) + k x(t)

)

= 0 (1.19)

Rewrite as time derivative,

d

dt

(

1

2mx2 + 1

2k x2

)

= −c x2 ≤ 0 (1.20)

This defines the mechanical energy E as

E = Ekin + Eel (1.21)

with kinetic energy

Ekin = 1

2mx2 (1.22)

and elastic energy

Eel = 1

2k x2 (1.23)

The mechanical energy of undamped free vibrations is constant.

When the motion can be described (estimated) by a single degree of freedom,the equation of undamped motion can be determined from the time derivativeof the mechanical energy.

State vector and phase-plane

Time histories of displacement x(t) and velocity v(t) = x(t) against time t.

ω0 t

x

ω0 t

v / ω0

Figure 1.4: a) displacement x(t), and b) velocity x(t) as function of time t.

Combination into a single three-dimensional graph as shown in Fig. , withtime t along the first axis.

ω0 t

x

v / ω0

v / ω0

x

Figure 1.5: a) response path and b) phase-plane diagram.

Projection of the state vector (x(t), x(t)) on the phase-plane.

Use of normalized coordinates x, x/ω0 reduces to ‘near-circle’.

Discrete time increment - Undamped system

Undamped equation of free vibrations, ζ = 0,

x + ω2

0 x = f(t)

Free oscillations described entirely in terms of initial conditions,

x = x0 cos(ω0t) + x0ω−1

0sin(ω0t)

x = −x0ω0 sin(ω0t) + x0 cos(ω0t)

State vector (x, x)1 at time t = ∆t in terms of initial state vector (x, x)0.

Recurrence relation for any pair of state vectors with time separation ∆t.

Time separation only appears in the form of the non-dimensional parameter

α = ω0 ∆t

Non-dimensional form, when x is replaced with ∆tx.[

x

∆t x

]

i+1

=

[

cosα α−1 sinα

−α sinα cosα

][

x

∆t x

]

i

Discrete time increment - Damped system

Normalized equation for free vibrations

x + 2ζ ω0 x + ω2

0 x = 0

Damped natural angular frequency,

ωd = ω0

√

1− ζ2

Free damped oscillations given in terms of the initial conditions

x = x0 e−ζω0t

(

cos(ωdt) +ζω0

ωdsin(ωdt)

)

+x0

ωde−ζω0t sin(ωdt)

x = −x0

ω20

ωde−ζω0t sin(ωdt) + x0e

−ζω0t

(

cos(ωdt)−ζω0

ωdsin(ωdt)

)

Two non-dimensional time scales, conveniently defined as

α = ωd∆t , β = ζ ω0 ∆t

Recurrence relations for time increment ∆t,

[

x

∆t x

]

i+1

=e−β

α

[

α cosα + β sinα sinα

−(α2 + β2) sinα α cosα− β sinα

][

x

∆t x

]

i

Structure of time-stepping algorithm

The exact solution (xn, xn) can be obtained at discrete times 0,∆t, 2∆t, · · ·by starting at the initial conditions (x0, x0) and multiplying with the matrixA, found above, in each step.

y0 = (x0,∆t x0)T

for i = 0 : n− 1

yi+1 = Ayi

(x,∆t x) = yT

Table 1.1: Direct time-stepping algorithm.

NOTE: The simplicity of the algorithm is obtained because the naturalfrequency and damping ratio is known. For multi-degree-of-freedom systems,this procedure requires a modal analysis, described later.

Exercise 1.1

The natural frequency of a single-degree-of-freedom-system depends on theratio of ‘stiffness’ to ‘mass’. For a simple mass–spring system the relation isω2

0 = k/m.

a) For the beam shown in the figure the force-displacement relation for atransverse force at the end is

F = 3EI

l3u

What is the natural frequency for transverse vibrations of a heavy massM fixed to the end of the beam, when the mass of the beam is neglected.

b) Let the mass M = 1000 kg be supported by a beam of length l = 10m. What is the bending stiffness EI necessary to give the frequencyf = 2.0 Hz.

Figure 1.6: Beam supporting a mass M .

Exercise 1.2

Consider the massm supported by a spring with stiffness k. The motion of themass m is described by x. The mass of the spring ms is small relative to themass m, and therefore the motion of the spring is quasi-static. This impliesthat the spring extends uniformly. Thus, the left end point is a rest, whilethe right end point moves x. A non-dimensional coordinate ξ is introducedsuch that the spring is described by 0 ≤ ξ ≤ 1. This implies that the motionof a point described by the position ξ is ξx.

a) Find the total kinetic energy of the mass m and the spring in terms of x.

b) Use energy balance to find an expression for the natural angular fre-quency ω0.

c) The contribution from the spring can be included as an extra contribu-tion to the effective mass meff ,

ω2

0 = k/meff with meff = m + (?)ms

Find the coefficient (?) of the spring mass.

Figure 1.7: Beam supporting a mass M .

Exercise 1.3

The figure shows a water column of total length l and cross-section area A.The mass density is ρ. When the water surface at the left side is displacedthe distance x downward, the water surface at the right side is lifted the samedistance x and conversely. The system is exposed to downward gravitationwith acceleration constant g.

a) Express the potential energy Epot and the kinetic energy Ekin as functionof x and x, respectively.

b) Use an energy balance argument to find an expression of the naturalangular frequency ω0.

c) The angular frequency ω0 is independent of the mass density ρ. Why?

Figure 1.8: Beam supporting a mass M .

The motion of the fluid may be constrained, whereby damping is introduced.This device can be used as damper of ship roll motion.

Summary

• Examples have been given of structures that often exhibit dynamic be-havior and must be analyzed for dynamic effects.

• Free vibrations of a single degree-of-freedom system constitute an ex-change between potential and kinetic energy. The time scale is charac-terized by the natural angular frequency ω0.

• The square of the natural (angular) frequency is determined by the ratio‘stifness’/‘mass’, i.e. ω2

0 = k/m.

• Damping is charactized by the non-dimensional damping ratio ζ, de-scribing attenuation per vibration cycle. Vibrations with ζ < 1 are‘underdamped’ and may have damping ratio as low as 0.001.

• Damping may be measured from attenuation of free vibration responsein terms of the logarithmic decrement δ = 2πζ.

• The equation of motion may be considered as the time derivative ofan energy balance equation. For simple systems with distributed massor stiffness the equation of motion may be obtained from the energybalance relation.

• The displacement and velocity may be combined into a state vector

(x, x) describing the response of the system. Initial conditions and re-sponse are conveniently represented in a phase-plane.