Chapter 2 Simple models for structures

Description of both fluid flows and elastic structures require infinite degrees of freedom along each of the three spatial dimensions. Doing so may be necessary in some circumstances, but hinders uncovering of basic stability principles. Where possible, it is prudent to judiciously retain only a few degrees of freedom as part of the stability analysis. In this chapter, we will revise some simple models to represent structures.

In many cases, the instability arises from the fluid flow forcing the structure to excite one or few of the natural modes of oscillation of the structure. Under such a circumstance, it is prudent to represent the state of the structure by a few state variables that are excited during the oscillations of those modes. The simplest of such variables is an amplitude of the natural mode, which behaves like a (possibly damped) harmonic oscillator, with equivalent lumped parameters. In this chapter, a generic fluid dynamic force on these degrees of freedom is assumed.

Making this approximation is not entirely without merit, as will be shown in some of the cases below using the method of projection. In doing so, we will practice the method of linearization and the calculation of normal modes, which will also prepare us for such analysis in case of coupled fluid-structure interaction. In addition, we will also consider some structural stability problems in their own right, using the method of linear stability analysis outlined in chapter 1.

2.1 Lumped mass-spring-damper

Sometimes, a structure may be approximated simply as a rigid point mass m attached to a spring with stiffness k and a damper with coefficient b, as shown in fig. 2.1.

Figure 2.1: Schematic of a mass-spring-damper lumped mass system.

A generic fluid dynamic force on the mass is denoted f. The displacement x(t) is the sole state variable describing this system. It obeys

md2xdt2+bdxdt+kx=f. (2.1)

Note that this equation is already linear, so a linearization of this equation around any steady state will not alter it. Also, note that an energy equation can be constructed by multiplying this equation by dx/dt as

ddt[12m(dxdt)2+12kx2]=b(dxdt)2+fdxdt. (2.2)

The rate of energy removed by the damper is b(dx/dt)2 and the work done by the fluid is fdx/dt.

Question 2.1.

Determine the motion of the mass in the absence of the fluid dynamic force and starting from an arbitrary initial condition.

Answer 2.1.

The natural frequence of a free undamped harmonic oscillator (i.e. f=0 and b=0) is given by ω0=k/m. Based on this, the underdamped motion of a free damped oscillator (i.e. f=0 but b0) may be written in terms of the dimensionless variables ω0t=t~ and β=b/2km. The equation of motion for the mass then becomes

d2xdt~2+2βdxdt~+x=0. (2.3)

Since this equation is already linear, we can follow the prescription of section 1.9 and try a solution of the form x(t)est~, to get the following eigenvalue equation for s,

s2+2βs+1=0,which has rootss=β+i1β2. (2.4)

(Note that the nature of the eigenvalue equation is quadratic in this case.) This yields the motion to be

x(t)=eβt~(a1cosωt~+a2sinωt~),whereω2=1β2. (2.5)

Here the condition for the motion being underdamped is β<1.

Question 2.2.

Possibly nonlinear spring: Suppose that the spring is not a Hookean or linear spring, and the restoring force it produces is kxqx3. Write the equation governing the state variable x. Also derive any equation for the energy of the system.

Answer 2.2.

The restoring force appears in the governing equations as

md2xdt2+bdxdt+kx+qx3=f. (2.6)

The energy equation for this case is

ddt[12m(dxdt)2+12kx2+14qx4]=b(dxdt)2+fdxdt. (2.7)

The spring is strain-stiffening if q>0 and strain-softening if q<0.

The spring in the damped mass-spring system can be nonlinear, such as in question 2.2, and so can be the damper.

Question 2.3.

The nonlinear mass-spring system is subject to a constant external fluid dynamic force f. Determine the steady state of the system in response to this force. Linearly perturb this state to determine the equation governing the evolution of the perturbation. Determine whether the perturbation would grow or decay, especially for the strain-softening case q<0.

Answer 2.3.

The new steady state displacement of the mass, x0, balances the spring restoring force with the applied fluid dynamic force.

kx0+qx03=f. (2.8)

A closed form expression for x0 in terms of k, q and f may not be possible. We perturb this state as x=x0+x, to yield the equation for the perturbation as

md2xdt2+bdxdt+kx+q(x0+x)3qx03=fkx0qx03=0, (2.9)

where we have used eq. 2.8 to simplify the right-hand side of eq. 2.9. Further, taking the perturbation x to be infinitesimal, the nonlinear term of eq. 2.9 may be linearized as q(x0+x)3qx033qx02x. Thus, the linear equation governing the perturbation becomes

md2xdt2+bdxdt+(k+3qx02)x=0. (2.10)

Equation 2.10 is a linear damped mass-spring equation in its own right, where the equivalent spring stiffness of k+3qx02. If the nonlinear spring is strain-softening, q<0, the equivalent spring is softer. If k+3qx02 becomes negative, the spring does not produce a restoring force, but instead provides a positive feedback for the perturbation to grow. The threshold condition is k+3qx02=0, which combined with eq. 2.8, yields the criteria on the critical f to be fcritical=233k3/2(q)1/2.

Note that the nonlinear spring force produces a restoring force for any displacement less than x0=k/q, but stably holding the extension at any value greater than k/3q=x0/3 is impossible due to sensitivity to perturbations.

Determining the scales for x and t, as well as the dimensionless parameters from eq. 2.6 is a worthy exercise in dimensional analysis. Performing this analysis and following it with the linear stability analysis in dimensionless terms is left for the astute reader.

Figure 2.2: Schematic for the model of buckling. (a) A mass suspended by four springs in equilibrium. (n) Free body diagram for the mass when it is displaced along the x direction.
Question 2.4.

Consider a mass m suspended symmetrically between four springs. Two are along the x axis of stiffness kx and two are along the y axis of stiffness ky. The springs along y have a rest length of Ly. The mass is at the origin when all the springs are at their equilibrium length. The springs along y are then compressed by a length y0. A schematic of this setup is shown in fig. 2.2(a). Examine the stability of the mass under translation along x.

Answer 2.4.

Here x, the displacement of the mass m is the state variable. Free body diagram of the mass m in the displaced state is shown in fig. 2.2(b). The two springs along x exert a restoring force of strength kxx. The extension and orientation of the springs along y must also be determined to examine their influence on the mass. These springs make an angle θ=tan1(xLyy0). The compression of the spring is e=Ly(Lyy0)2+x2, and the force along the length of the spring is kye. The equation governing the state, therefore, is

md2xdt2=2kxx+2kyesinθ. (2.11)

Since the equilibrium to investigate is x=0, the perturbation x=x obeys the same equation as x. When the perturbation is infinitesimal in size, we have

sinθθxLyy0,andey0 (2.12)

The linear equation governing the infinitesimal perturbation is

md2xdt2=2(kx+kyy0Lyy0)x. (2.13)

An examination of the dimensions suggest defining a dimensionless time as t~=tky/m and the follwing two parameters e~=y0/L and k~=kxky. The variable x may be non-dimensionalized by any length. The dimensionless equation governing the perturbation is

d2xdt~2=2(k~+e~1e~)x. (2.14)

The right hand side is equivalent to a modified spring with dimensionless stiffness k~e~/(1e~), which forebodes instability if it becomes negative. Therefore, the criteria for the onset on instability is e~>k~1+k~.

The mechanism is evident from fig. 2.2(b). This is a simplified model of buckling of a structure.

2.2 Pitch and heave

The next simplest structure has two degrees of freedom. An example is depicted schematically in fig. 2.3, where they are the rotational degree called pitch, θ, and the translational degree of freedom called heave, h. While fig. 2.3 shows a symmetric airfoil, the object could be of arbitrary shape. It has mass m and moment of inertia about the centre of mass I=mRg2, where Rg is the radius of gyration. The object is supported at the centre of support, which is a distance xs away from the centre of mass, as shown in the figure. A translational spring with linear stiffness k, and a torsional spring with rotational stiffness κ support the object at the centre of support. And the object also experiences an aerodynamic force of lift, L, in the direction of heave at the centre of pressure, which is a distance xa from the centre of mass. We ignore any damping that may be present in this structure for this analysis.

The state variables h and θ describe the dynamical state of the object. The evolution of the state is governed by force and torque balance around the centre of mass and satisfies the following equations

mh¨+kh+kxssinθ =L, (2.15)
Iθ¨+kxshcosθ+κθ+kxs2sinθcosθ =Lxacosθ, (2.16)

where the double-dot decoration above h and θ means the second derivative with time t. For what follows, we will ignore the external force of lift, i.e. take L=0, and examine the structural dynamics. For this purpose, we determine the steady state to be h=0 and θ=0. Perturbation about the steady state are made as h=0+h and θ=0+θ. Assuming the perturbations to be small (i.e. infinitesimal), they satisfy

mh¨+kh+kxsθ =0, (2.17a)
Iθ¨+kxsh+κθ =0, (2.17b)

where κ=κ+kxs2.

Figure 2.3: Schematic showing a structure with two degrees of freedom – pitch θ and heave h.

These equations may also be expressed in terms of the state vector 𝐱=[h,θ]T as

𝑴𝐱¨+𝑲𝐱=0,𝑴=[m00I],𝑲=[kkxskxsκ] (2.18)

where 𝑴 is the mass matrix and 𝑲 is the stiffness matrix. Note that both are symmetric and positive definite.

An equation for energy can be constructed by multiplying eq. 2.17a with h˙=dh/dt and eq. 2.17b with θ˙=dθ/dt and adding them to yield

ddt[12mh˙2+12Iθ˙2+12kh2+kxshθ+12κθ2]=0, (2.19)

or in matrix notation, where 𝐱˙=d𝐱/dt,

ddt[12𝐱˙T𝑴𝐱˙+12𝐱T𝑲𝐱]=0. (2.20)

The energy equation guides the rescaling towards a dimensionless formulation with the rescaling

t=t~mk,h=Rgh~,θ=θ~,δ=2xsRg,andωa2=κkRg2. (2.21)

The dimensionless equations governing the perturbations are

d2h~dt~2+h~+δ2θ~=0, (2.22)
d2θ~dt~2+δ2h~+ωa2θ~=0. (2.23)

The parameter δ serves as the coupling between the two degrees of freedom. If δ=0, pitch and heave are uncoupled and each have a natural frequency of ωa and 1, respectively. Because energy neither grows nor decays in this system, we expect pure oscillations for the two degrees of freedom, even when they are coupled. Since these equations are linear, we may seek exponentially growing solutions eiωt~, so that s=iω satisfies

[1δ/2δ/2ωa2][h~θ~]=ω2[h~θ~]. (2.24)

Solving the eigenvalue equation yields for the two values of ω2

(1ω2)δ/2=δ/2(ωa2ω2)ω1,22=1+ωa2±δ2+(1ωa2)22. (2.25)

The first property we note is that ω2 is real, so the growth rate is purely imaginary and thus there is neither growth nor decay of the perturbations. When δ=0, the frequencies agree with our expectation for the uncoupled degrees of freedom. A real physical system will show decay of these perturbations owing to some damping.

The corresponding eigenvectors are

𝐱^1,2=[h~θ~]=[δ/2ω1,221]=c[ω1,22ωa2δ/2], (2.26)

where c is a constant, which ensures that the lengths of the two forms of the eigenvector are equal to each other. It can be readily seen from eq. 2.26 that when δ=0, the eigenvectors have either the h component or θ, but not both, thus verifying that in this case the two modes are indeed decoupled. In this case, naturally the dot product between the two independent eigenvectors is zero, so they are orthogonal. The orthogonality between the two eigenvectors for the case δ0 can be easily verified using the forms in eq. 2.26 and the property that ω12+ω22=1+ωa2. This nifty property is left for the reader to discover.

2.3 A general structure with a finite number of degrees of freedom

In general, an abstract structure with a finite degrees of freedom may be represented using the state variable vector 𝐱, where the vector has as many components as the degrees of freedom. The state vector can obey an equations for its evolution written as eq. 1.3 from section 1.4. Let us consider the case where the undeformed state is given by 𝐱=0 is the steady state, where there are no external forces, and all displacements and internal forces vanish. However, in the absence of damping, the dynamics must conserve energy, and therefore, the following form for the evolution emerges for the perturbation 𝐱 when linearized about the undeformed steady state.

𝑴d2𝐱dt2+𝑲𝐱=0, (2.27)

where 𝑴 is the mass matrix and 𝑲 the stiffness matrix. Both 𝑲 and 𝑴 are symmetric and positive-definite matrices. These matrices in general may not be diagonal or sparse. The system conserves the energy E defined as

E=12𝐱˙T𝑴𝐱˙+12𝐱T𝑲𝐱. (2.28)

For this reason, any perturbation made around the steady state must oscillate, say with frequency ω. The frequency satisfies the eigenvalue equation

𝑲𝐱^=ω2𝑴𝐱^, (2.29)

where the state is assumed to evolve as 𝐱=𝐱^eiωt. Solution of this equation yields both the many natural frequencies ωi and the corresponcing mode shapes 𝐱^i, for i=1,2,,n, where n is the number of degrees of freedom. Because of the symmetry and positive-definiteness of 𝑲 and 𝑴, we are guaranteed that the frequencies are real and that the various modes are mutually orthogonal, i.e.

𝐱^jT𝑴𝐱^i=0whenji. (2.30)

Using the properties of 𝐱^i and ωi, we may construct the general evolution of the perturbation in the presence of an external forcing, say from the surrounding fluid. If the infinitesimal forcing on the structure is written as 𝐟, it responds according to

𝑴d2𝐱dt2+𝑲𝐱=𝐟. (2.31)

It is possible to look for a solution to eq. 2.31 of the form

𝐱(t)=a1(t)𝐱^1+a2(t)𝐱^2++an(t)𝐱^n, (2.32)

where a1(t), a2(t), …, an(t) are amplitudes of the various modes, which are to be determined. (Here we assume that the mode shapes 𝐱^1, 𝐱^2, …, 𝐱^n have been precomputed.) Substituting eq. 2.32 into eq. 2.31, then yields

i=1n(ai¨+ωi2ai)𝑴𝐱^i=𝐟, (2.33)

where we have used eq. 2.29 and the double-dot decoration denoted second derivative in t. Now, exploit orthogonality of the eigenvectors from eq. 2.30, and take a dot product with one of the eigenvectors 𝐱^j. Noting that all the terms in the sum vanish, except for the i=j, then yields the equation for the amplitude ai(t) as

ai¨+ωi2ai=𝐱^iT𝐟𝐱^iT𝑴𝐱^i. (2.34)

In this manner, the response of a structure to an external forcing can be determined by projection on the natural modes of oscillation of the structure.

2.4 A stretched string or a one-dimensional membrane

A string or a wire that is stretched along it length (say, in the direction of the x-axis) presents a simple model for a continuum elastic structure. The wire length is L, tension in the wire is T and it has a mass μ per unit length, as shown in fig. 2.4. The point on this string at the coordinate x displaces in a direction perpendicular to the length (the Y-direction) by a distance y(x,t). If an external force f per unit length is applied to the string along the y-direction, the displatement obeys

μ2yt2=T2yx2+f. (2.35)

The displacement of the string is zero at the two ends x=0 and x=L.

Figure 2.4: A string or a membrane stretched between x=0 and x=L, which has a transverse displacement y(x,t).

Figure 2.4 also shows a membrane stretched along x with tension T per unit length. The tension perpendicular to plane of the page (i.e. along the z direction) does not enter the formulation because z=0. In this case, μ is the mass of the membrane per unit area, and f is the external force per unit area perpendicular to the membrane. The membrane displacement y(x,t) is also governed by eq. 2.35. Because of the analogy between the one-dimensional membrane and a stretched string, we will only consider the string from now on.

The stretched string satisfies the energy equations

t0L[μ2(yt)2+T2(yx)2]dx=0Lfytdx. (2.36)

Here the integral proportional to μ is the kinetic energy, the one proportional to T is the elastic potential energy, and the right-hand side is the work done by the external force. For the stretched membrane a trivial integral along z may also be included.

Let us first determine the natural modes of vibrations of the string (or the membrane, as the case may be). For the natural modes, we set f=0. The equilibrium shape of the string in this case is y=0. Perturbing the displacement from the steady state as y=0+y yields the equation for the perturbation

μ2yt2=T2yx2. (2.37)

This equation is already linear so no further approximation for an infinitesimal perturbation needs to be made.

To make the formulation dimensionless, we define t~=tLTμ and x~=x/L, so eq. 2.37 becomes

2yt~2=2yx~2,for0<x~<1, (2.38)

and y(x~=0,t~)=y(x~=1,t~)=0.

When f=0, energy is conserved, so we expect perturbations to neither grow nor decay. Hence we seek a solution to eq. 2.38 of the form y=y^(x~)eiωt~. The mode shape y^(x~) satisfies

d2y^dx~2+ω2y^=0 (2.39)

with y^=0 at x~=0 and 1. A non-trivial solution only exists if ω=ωn=nπ for n=1,2,, in which case the mode shape is

y^n(x~)=sin(nπx~). (2.40)

The mode shapes satisfy the orthogonality condition

01y^n(x~)y^m(x~)dx~=0,ifmn, (2.41)

which can be easily verified. (Note that the notion of dot products is replaced by integrals in this case.)

If an external force acts on the string, we can determine the response of the string in the following manner. First, we transform eq. 2.35 to a dimensionless form by using f~=fL2/T to yield

2yt~2=2yx~2=f~, (2.42)

Let us write the perturbation as a linear combination of the modes as

y(x,t)=n=0an(t)y^n(x~). (2.43)

Substituting eq. 2.43 in eq. 2.42, with the intention of determining how the amplitudes an(t) respond to the forcing f~, yields

n=1(a¨n+ωn2an)y^n(x~)=f~(x~,t~), (2.44)

where, as before, the double-dot decoration denotes second derivative with t~. We use orthogonality to decouple the evolution of distinct modes. For doing so, we multiply with y^m(x~) and integrate along x~ as

01y^m(x~)[n=1(a¨n+ωn2an)y^n(x~)]dx~=01y^m(x~)f~(x~,t~)dx~. (2.45)

The orthogonality condition from eq. 2.41 then facilitates the simplification

a¨m+ωm2am=01y^m(x~)f~(x~,t~)dx~01y^m(x~)2dx~. (2.46)

2.5 Euler beam

The stretched string serves as an example that applies the general formalism of normal modes to determine the response of a continuum structure to external forcing. To test their understanding, the reader is invited to apply the formalism to the example of an Euler beam. The state of the beam is again described by the variable y(x,t), which satisfies the governing equation

μ2yt2=By4x4+T2yx2+f. (2.47)

Here μ, T, and f are the same as before, and B is the bending stiffness or bending rigidity of the beam cross section. In what follows, we will take the tension T to be zero. This equations must be supplemented with two boundary conditions at each end, which are of the form

  1. 1.

    A given linear combination of displacement and shear force is specified,

    ayBby3x3=r1(t), (2.48)

    and

  2. 2.

    A given linear combination of the slope and the bending moment is specified,

    cyx+Bd2yx2=r2(t), (2.49)

where a, b, c, d, r1 and r2 are specified constants. Without loss of generality, b and d may be taken to be unity so long as they are non-zero. These boundary conditions may be interpreted as the ends of the beam being supported by linear and torsional springs of stiffness a and c, respectively, and an external force (r1) or torque (r2) being applied there.

Question 2.5.

Derive the principle of conservation of energy for the Euler beam that satisfies eqs. 2.47, 2.48 and 2.49.

Answer 2.5.

Multiply eq. 2.47 by y/t and integrate in x from 0 to L, with a few judicious integrations by parts and application of the boundary conditions, to get

12t{0L[μ(yt)2+B(2yx2)2]dx+[aby2+cd(yx)2]0L}=0Lfytdx+[r1byt+r2dy2xt]0L. (2.50)

The term proportional to μ is the kinetic energy of the beam, the term proportional to B is the bending energy (which is proportional to square of curvature), the terms proportional to a and c are the energies stored in the linear and torsional springs at the end, the term proportional to f on the right hand side is the work done by the distributed external force, and the terms proportional to r1 and r2 are the works done by the external boundary forces and torques. Here we have tacitly assumed that b and d are positive, so they can be taken to be unity. The diligent reader is left the task of deriving the version of this equation when b and d are zero.

Figure 2.5: Shapes of the first few modes of oscillations of the Euler beam.
Question 2.6.

Determine the frequency and shapes of the natural modes of oscillations of a cantilevered beam. (For a cantilevered beam, at one end the displacement and angle are zero, while at the other end the moment and shear force are zero.)

Answer 2.6.

We use the non-dimensionalization:

t~=tμL4B,x~=xL. (2.51)

It is left as an exercise to the reader to derive the eigenvalue equation for the natural dimensionless frequency ω and the mode shape y^(x~), which is

y^4x~4=ω2y^, (2.52)

with y^=dy^/dx~=0 at x~=0 and d2y^/dx~2=d3y^/dx~3=0 at x~=1. Let k=ω1/2, i.e. the positive square root of ω, because it is convenient to write the expressions in terms of k. The general solution of eq. 2.52 is

y^(x~)=asinhkx~+bsinkx~+ccoshkx~+dcoskx~, (2.53)

where a, b, c and d are constants of integration. Some of these constants are determined by the boundary conditions. For instance, the conditions at x~=0 imply b=a and d=c, so they may be eliminated. The boundary conditions at x~=1 then require

a(sinhk+sink)+c(coshk+cosk)=0, (2.54a)
a(coshk+cosk)+c(sinhksink)=0. (2.54b)

The only solution for general k is a=c=0, but that is the trivial solution to eq. 2.52. A non-trivial solution exists only when the two equations in eq. 2.54 are identical. The condition for a non-trivial solution to exist is

coshkcosk=1, (2.55)

which is satisfied by a countably infinite values of k, say k1, k2,. The first few values are: k1=4.69409113, k2=7.8547, k3=10.9955, k4=14.1372. For large n, the value of kn may be approximated to be

kn=(n+12)π+(1)ncosh(n+12)π

This formula can be seen to be acurate for n=2 upto four decimal places, and gets more and more accurate for larger n. The accuracy of this formula can be determined by evaluating the respective kn and substituting the result in eq. 2.55. The dimensional natural frequency of oscillations may now be written as

Ωn=km2BμL4. (2.56)

The mode shape is given by

y^n(x~)=C(sinhknx~sinknx~sinhkn+sinkncoshknx~cosknx~coshkn+coskn), (2.57)

where C is an arbitrary constant. These mode shapes are visualized in fig. 2.5.

It is left as an exercise to the reader to either verify or prove the the othogonality condition eq. 2.41, which these modes obey. The author does not recommend actually multiplying y^m with y^n and integrating to prove the orthogonality condition.

2.6 Conclusion

In this chapter, we considered simple models for some structures and analyzed their behaviors in terms of perturbations about the steady state. In cases where we expected harmonic oscillations, we were able to apply the formalism of linear theory to determine the mode shape and natural frequency by assuming a time dependence of eiωt, for some real ω. In cases where we expected an instability, we were also able to apply linear theory to determine the exponential growth of the unstable perturbations.

In this process, we found that while the deformation of the structure itself may be most complicated, it may be decomposed into mode shapes, which exhibit a simple harmonic behaviour for their amplitude. We also saw the method of projection, which exploits orthogonality of modes, to decouple the influence of a general external forcing on individual modes.

We also constructed the energy conservation principle for the structures we considered, including the influence of the external forcing on the rate of change of energy. The energy of a perturbation is a good indicator of its growth or decay, and therefore the derivation of an energy equation is instrumental in stability analyses.

The matter in this chapter is only but an introduction to the otherwise rich subject of structures and elasticity, and especially to the topic of simplified descriptions of them. Prime topic in this subject, which we omitted, is the theory of rods, shells and plates, which presents a one- and two-dimensional approximation to three-dimensional objects which are thin along the remaining dimension(s). Furthermore, structural instabilities are a formidable topic in their own right, without the need for the complicated coupling with the fluid flow.

We conclude with the following question on the buckling of elastic beams.

Question 2.7.

Consider an Euler beam governed by eq. 2.47, which is under compression, so T<0, say T=F. Assume simply pinned boundary conditions so that b=r1=c=r2=0, and without loss of generality a=d=1. There are no external forces on the beam, so f=0. The beam buckles when the compressive force F exceeds a threshold. Use the formulation of linear stability analysis to determine this threshold.

Answer 2.7.

(Fcritical=Bπ2L2).

The answer is provided, but the process of arriving at it is entrusted to the keen reader so that they can sharpen their abilities.