The Pennsylvania State University

The Graduate School

Eberly College of Science

DYNAMIC BUCKLING AND TENSION OF AN ELASTIC BEAM

UNDER IMPACT OF A PROJECTILE

A Thesis inMathematics

byQichuan Bai

Submitted in Partial Fulfillmentof the Requirements

for the Degree of

Master of Arts

August 2013

The thesis of Qichuan Bai was reviewed and approved* by the following:

Andrew BelmonteProfessor of MathematicsThesis Adviser

Qiang DuProfessor of Mathematics

Xiantao LiAssociate Professor of Mathematics

Yuxi ZhengProfessor of MathematicsHead of the Department of Mathematics

*Signatures are on file in the Graduate School.

Abstract

Our problem is to consider a thin elastic beam subject to the impact of a projectile on one

end, with the other end fixed. This thin beam would have normal (lateral) displacement x

and tangential (compressional) displacement y at the same time. Different ways to employ

the tension T give different models. Most traditional ones in earlier works assume tension

is uniform, decided by some pre-selected modes or other parameters. In Saint-Venant

approach, the tension T is estimated in terms of the velocity of the projectile U0 and the

property of the beam, such as the Young’s modulus E, the cross section area of the beam

A. Later many models incorporate a non-uniform tension, where a constitutive equation

of tension solely based on the compression of the tangential displacement y. Most recently,

there arises a new way to construct a Poisson equation for tension T when studying the

fluid-fiber interactions.

Our model derives a Poisson-type equation for tension without the interaction with

fluids. We borrow the idea of taking the constraint of arc length parameter to get an

equation for tension. For boundary conditions, we use the contact boundary idea of Saint-

Venant. We incorporate the derivation of energy law to get a set of new boundary conditions.

We have performed an extensive numerical study of this model. The coupled system

would be one hyperbolic equation plus a elliptic-type problem. Our numerical scheme is

a combination of semi-implicit finite difference in time and Chebyshev spectral method in

space. This combination of finite difference and spectral method appears often literature.

We investigate the dynamic buckling under impact with our model on contact boundary

conditions. We find that the larger hammer mass will result in larger growth in the buckling

amplitude, which agrees with experimental results. We observe a hook developed in the

iii

simulation results which has not yet been observed in the physics experiments, which might

lead to a different scenario of experiment setting.

iv

Contents

List of Figures vi

List of Symbols vii

Chapter 1 Background 1

1.1 Euler-Bernoulli beam equation . . . . . . . . . . . . . . . . . . . . . . . . . 1

1.1.1 Euler buckling . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2

1.1.2 Euler-Bernoulli beam . . . . . . . . . . . . . . . . . . . . . . . . . . 2

1.2 Saint-Venant approach to impact problems . . . . . . . . . . . . . . . . . . 3

Chapter 2 Derivation of buckling tension system 4

2.1 Tension equation due to inextensibility in literature . . . . . . . . . . . . . . 4

2.2 Poisson-type tension equation . . . . . . . . . . . . . . . . . . . . . . . . . . 5

Chapter 3 Energy law for the beam and hammer 7

3.1 Energy law for the rod . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

3.2 Energy law for the rod and hammer system . . . . . . . . . . . . . . . . . . 9

3.2.1 Proposed new boundary conditions and corresponding energy laws . 10

3.2.2 Conjecture of a new boundary condition . . . . . . . . . . . . . . . . 11

Chapter 4 Numerical algorithm and linear stability analysis 13

4.1 Numerical scheme . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13

4.2 Von Neumann linear stability analysis . . . . . . . . . . . . . . . . . . . . . 15

Chapter 5 Numerical results and discussion 19

v

5.1 Hammer constrained in x direction . . . . . . . . . . . . . . . . . . . . . . . 19

5.1.1 Test of different hammer weight MH . . . . . . . . . . . . . . . . . . 20

5.1.2 The hook under larger initial velocity . . . . . . . . . . . . . . . . . 20

5.1.3 v0 = 0.025, lower ε = 1×10−3, from no hook to slight hook for a long

time MH ∈ (0.01, 0.08) . . . . . . . . . . . . . . . . . . . . . . . . . 22

5.2 Hammer allowed 2D motion with full projection for tension . . . . . . . . . 24

Chapter 6 Conclusions and future work 29

Bibliography 29

vi

List of Figures

5.1 Tension projection, velocity (left) and rod shape (right), MH = 0.1, time =

9.922 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

5.2 Tension projection, velocity (left) and rod shape (right), MH = 0.3, time =

26.62 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

5.3 Tension projection, rod shape in long time, MH = 0.5, time = 30 (left),

time = 49.5 (right) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22

5.4 T projection, rod shape, MH = 0.3, time = 22.5 (left), time = 30 (right) . . 22

5.5 T projection, velocity (left) and impact end angle (right), MH = 0.3, time =

33.24, the blue horizontal line is θ = π/2 . . . . . . . . . . . . . . . . . . . 23

5.6 Velocity in x-axis (left) and angle θ, MH = 0.04, time = 13.3, the blue

horizontal line is θ = π/2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24

5.7 Shape of rod (left) and y v.s. s (right), MH = 0.04, time = 13.3 . . . . . . 24

5.8 Velocity in x-axis (left) and angle θ, MH = 0.05, time = 35.1, the blue

horizontal line is θ = π/2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25

5.9 Shape of rod before blow up (left) MH = 0.05, time = 34.5, (right) MH =

0.07, time = 33 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25

5.10 Velocity in x-axis (left) and y-axis (right), MH = 0.1, time = 9.92 . . . . . 26

5.11 Shape of rod (left) and y v.s. s (right), MH = 0.1, time = 9.92 . . . . . . . 26

5.12 Velocity in x-axis (left) and y-axis (right), MH = 0.2, time = 15.1 . . . . . 27

5.13 Shape of rod (left) and y v.s. s (right), MH = 0.2, time = 15.1 . . . . . . . 27

5.14 Rod Shape (left) and y (right), MH = 0.5, time = 60 . . . . . . . . . . . . . 28

5.15 Impact end angle for MH = 0.3 (left) and MH = 0.5 (right), the blue hori-

zontal line is θ = π/2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28

vii

List of Symbols

ε Dimensionless Young’s modulus

µ Dimensionless Hooke’s constant

MH Mass of hammer

A Cross section area of the rod

ρ Linear density of the rod

dL Dimensionless aspect ratio of the beam

vo Initial velocity of hammer

H(t) Position of hammer

T Tension along the elastic beam

β Penalty parameter for length constraint

N Mesh size for computation domain

A0 Amplitude of random initial imperfection

viii

Chapter 1

Background

If we apply some axial load onto a simply supported beam, it would tend to bend if the

load is large enough. This phenomena is called buckling. The earliest model considers only

the deflection variable u in the direction perpendicular to the neutral position of a straight

beam.

There have been intensive studies on the buckling beam problem during centuries earliest

dating back to Euler’s equation around 1750 where the beam is in quasi-static status, see

[6]. The buckling of a beam could describe the failure of some mechanic structures, thus it

has received broad attention in the engineering field.

There is the celebrated Euler-Bernoulli model which provides a means of calculating

the deflection characteristics of beams with the load varying slowly with position and time.

This is so-called classical beam theory or just simply beam theory.

1.1 Euler-Bernoulli beam equation

When a beam is subject to some load with appropriate value, the beam will experience

transverse deformation. The deformation shape could be described as a minimal curve

u(x, t) to the following functional:

S =

∫ T

0

∫ L

0

1

2ρA(

∂u

∂t)2 − 1

2EI(

∂2u

∂x2)2 + q(x, t)u(x, t) dxdt. (1)

1

2

We use u(x) to denote the lateral displacement of the beam while it has length L, linear

density ρ and area A. E is the Young’s modulus and I is the moment of inertia, q(x, t) is

the transverse load on the beam. The corresponding Euler-Lagrangian equation would be

ρA∂2u

∂t2= q(x, t)− EI ∂

4u

∂x4. (2)

1.1.1 Euler buckling

Imagine that we increase the compressional force on one end of the beam very slowly while

beam is in a quasi-static state, then the beam will tend to bend once the force exceeds

a critical value Fcritical. This bifurcation problem was first studied by Euler which is the

starting point of this research field. However, we may just consider Euler’s model to be a

special case of the beam equation. Dropping the acceleration term, we recover the celebrated

Euler’s buckling equationd2

dx2(EI

d2u

dx2) = q(x). (3)

In the compressional case, often the effect of transverse load q(x) is considered to be the

constant compressional force projecting on the normal direction,namely q(x) = Puxx,thus

take y = uxx

EId2y

dx2= −Fy

Suppose that the static beam is pinned at both ends, y(0) = y(1) = 0, the shape of lateral

displacement would be a sine function. The corresponding critical load which corresponds

to the lowest wave number possible would be

Fcritical =π2EI

L2.

1.1.2 Euler-Bernoulli beam

For the case where tension is decided by compressional force in [13], both lateral displace-

ment u(x, t) and tangential displacement v(x, t) will be considered. Besides the bending

moment term, the gradient of the tension would also contribute to the total force, which

3

gives rise to the term ∂x(P (x, t)ux), where P (x, t) is the axial force caused by compression.

The system would be ρAutt = ∂x(P (x, t)ux)− EIuxxxx

ρAvtt = ∂x(P (x, t))(4)

Under the assumption that the material has linear elasticity, the axial force P satisfies

P (x, t) = K∂v

∂x, (5)

where K is the bulk compressional modulus. Thus both v and P obey the wave equation

with speed c =√E/ρA.

This leads to the coupled system: ρAutt = K∂x(uxvx)− EIuxxxx

ρAvtt = Kvxx(6)

1.2 Saint-Venant approach to impact problems

To explicitly set the force in Euler-Bernoulli equation, Saint-Venant in 1883 introduced a

contact boundary condition, which was derived by applying Newton’s third law in [4] for

the hammer and the rod,

MHvtt(t) = −EAvx, (7)

where all the values are taken at the impact end. This expression gave a convenient way

to characterize the stress profile P (x, t) in a short time frame without actually solving the

wave equation

P (x, t) = EAUocexp− x

Lpexp ct

Lp, (8)

where U0 is the initial velocity of the hammer and c =√E/ρ is the speed of sound in

the rod. There is detailed discussion of the physical interpretation and its significance for

parameter Lp in [1], where an approximation of uniform tension is later introduced.

Chapter 2

Derivation of buckling tension

system

2.1 Tension equation due to inextensibility in literature

The tension plays a vital role in the entire buckling model and many other PDE systems.

If the reference variable of a curve always stays as arc length parameter, which means the

moving curve has an invariance of infinitesimal material length.

‖ ~Xs(s, t)‖ = 1, ∀s ∈ Ω, t ∈ [0, time] (9)

In an earlier work of self-knotting chain in [2] and a later work on interaction of fiber and

fluid in [3], (9) was used to derive a Poisson equation for tension.

In both cases, the tension variable acts like the Lagrangian multiplier in the entire

system, similar to the role of pressure in the Navier-Stokes system. In [3] the governing

equation for the displacement is parabolic, which has a damping effect on the whole system.

Furthermore as the fiber is immersed in a fluid which slows down the motion, the boundary

condition for tension will simply be homogeneous Dirichlet boundary condition. Whereas

in [2] there is also a third order damping term.

In our dynamic buckling model, however, we don’t have an explicit damping term or

4

5

smoothing term and our governing equation for displacement is hyperbolic. Our boundary

condition for tension will be a constitutional equation which is based on Hooke’s law. All of

these facts give an extra challenge for our numerical simulation as well as the corresponding

analytical discussion.

In [3], to enforce numerical stability, a penalty term β(1 − | ~Xs|) was artificially added

into the tension equation, where β is the penalty coefficient. We will also adopt this in our

model to get better numerical stability.

2.2 Poisson-type tension equation

For our dynamical buckling case where we have a small projectile hitting on the top of the

beam with high speed, then buckling would happen, namely the beam would be compressed

and experience tangential and transverse displacements at the same time. To denote that

we want to have arc length parameter, we use new notation: lateral displacement x(s, t)

and tangential displacement y(s, t), and the tension along the rod T (s, t).

The balance of momentum would reads as following after appropriate non-dimensionalization

process:

~Xtt = (T ~Xs)s − ε ~Xssss, s ∈ [π/2, π/2], (10)

where T is the tension in the rod. If we start with s as the arc length parameter (in

real computation, we use a reference variable), ‖ ~Xs(s, 0)‖ = 1, then we again have the

inextensibility condition (9). We differentiate (10) with respect to s on both sides, to get

~Xtts = Tss ~Xs + 2Ts ~Xss + T ~Xsss − ε ~X5s.

Taking into account the equalities below, obtained by differentiating (9),

~Xs · ~Xss = 0, ~Xss · ~Xss + ~Xs · ~Xsss = 0,

3 ~Xss · ~Xsss + ~Xs · ~Xssss = 0, 3‖ ~X3s‖2 + 4 ~Xss · ~Xssss + ~Xs · ~X5s = 0,

~Xs · ~Xst = 0, ~Xst · ~Xst + ~Xs · ~Xstt = 0, (11)

6

where ~X5s = ~Xsssss, we will have an elliptic tension equation

Tss − ‖ ~Xss‖2T = −‖ ~Xts‖2 − ε(

3‖ ~X3s‖2 + 4 ~Xss · ~Xssss

). (12)

Now we have the coupled PDE system consisting of (10),(12).

~Xtt = (T ~Xs)s − ε ~X4s.

‖ ~Xs‖2Tss − ‖ ~Xss‖2T = −‖ ~Xts‖2 − ε(

3‖ ~Xss‖2 + 4 ~Xss · ~Xssss

).

(13)

Chapter 3

Energy law for the beam and

hammer

3.1 Energy law for the rod

Our original equation is

~Xtt = (T ~Xs∗)s∗ − ε ~Xs∗s∗s∗s∗ , s∗ ∈ Ω = (0, L(t)) (14)

where s∗ is arc length variable and L(t) is the total length of the rod, ~X = (x, y · dL).

Now if I use s ∈ (−π/2, π/2) as a reference variable, we have

s∗(s, t) =

∫ s

−π/2| ~Xs(s, t)|ds (15)

L(t) =

∫ π/2

−π/2| ~Xs(s, t)|ds (16)

Now we will derive an energy law for the buckling beam system.

Multiply (14) by ~Xt, we get

~Xtt · ~Xt = (T ~Xs∗)s∗ · ~Xt − ε ~Xs∗s∗s∗s∗ · ~Xt, s∗ ∈ (0, L(t)), (17)

7

8

To validate this equation we need an inertial coordinate system to avoid additiona fictitious

force. Thus we choose s∗ = 0 as the fixed end and s∗ = L(t) as the impact end. Integrating

both sides along the entire rod from 0 to L(t), we have

∫ L(t)

0

d

dt

‖ ~Xt‖2

2ds∗ =

∫ L(t)

0(T ~Xs∗)s∗ ~Xtds

∗ − ε∫ L(t)

0

~Xs∗s∗s∗s∗~Xtds

∗ (18)

= −∫ L(t)

0T ~Xs∗

~Xts∗ds∗ + T ~Xs∗

~Xt|∂Ω

+ ε

(− ~Xs∗s∗s∗

~Xt|∂Ω + ~Xs∗s∗~Xts∗ |∂Ω −

∫ L(t)

0

d

dt

‖ ~Xs∗s∗‖2

2ds∗

).

Now we need to consider boundary conditions. We have for the rod,

~Xt(0, t) = 0, ~Xs∗s∗ |∂Ω = 0. (19)

Also keep in mind for the arc length parameter, we have

‖ ~Xs∗‖2(s∗, t) = 1, ~Xs∗~Xs∗t = 0.

Rearranging terms in (18), we have

∫ L(t)

0

d

dt

(‖ ~Xt‖2

2+ ε‖ ~Xs∗s∗‖2

2

)ds∗ =

(T ~X∗s ~Xt − ε ~Xs∗s∗s∗

~Xt

)|s∗=L(t). (20)

Let’s define

f(s∗, t) =‖ ~Xt‖2

2+ ε‖ ~Xs∗s∗‖2

2.

Recall from fundamental calculus, we have for any continuous function g(s, t) with gt(s, t)

also continuous

d

dt

∫ b(t)

a(t)g(s, t)ds = g(b(t), t)b′(t)− g(a(t), t)a′(t) +

∫ b(t)

a(t)gt(s, t)ds.

9

Thus

d

dt

∫ L(t)

0

‖ ~Xt‖2

2+ ε‖ ~Xs∗s∗‖2

2ds∗ = f(L(t), t)L′(t)− f(0, t) · 0 +

∫ L(t)

0

d

dtf(s∗, t)ds∗ (21)

= f(L(t), t)L′(t) +

∫ L(t)

0

d

dt

(‖ ~Xt‖2

2+‖ ~Xs∗s∗‖2

2

)ds∗

=‖ ~Xt(L(t), t)‖2

2L′(t) +

∫ L(t)

0

d

dt

(‖ ~Xt‖2

2+‖ ~Xs∗s∗‖2

2

)ds∗.

Combining these equations, we find our first energy law for the rod:

d

dt

∫ L(t)

0

(‖ ~Xt‖2

2+ ε‖ ~Xs∗s∗‖2

2

)ds∗ =

‖ ~Xt(L(t), t)‖2

2L′(t) + (T ~Xs∗

~Xt − ε ~Xs∗s∗s∗~Xt)|s∗=0.

(22)

Now define the energy of the rod to be

E1(t) =

∫ L(t)

0

(‖ ~Xt‖2

2+ ε‖ ~Xs∗s∗‖2

2

)ds∗. (23)

The energy law (22) tells us that the change of total energy of the rod is purely decided

by what happens on the impact end. However, the right hand side does not have a definite

sign, we can not easily tell whether the energy E1(t) is going up or down.

3.2 Energy law for the rod and hammer system

We haven’t talked about the hammer (projectile) yet. Ignoring gravity, the total force

exerted on the hammer will be the tension exerted by the rod which is along the tangential

direction. If we only consider the hammer to move in the x axis, we have the governing

ODE of hammer to be as follows in our current coordinates:

MHH′′(t) = −T (0, t) cos θ, cos θ =

xs

| ~Xs|= xs∗ ,

H ′(0) = −v0 (24)

10

where s is the reference variable. One can simply check for the sign: if the rod gets

compressed, then T < 0 and the hammer should slow down, as H ′(0) is negative. H ′(t)

should later get closer to 0. Thus H ′′(t) is positive, which makes a negative sign necessarily.

The current contact boundary condition for the rod/hammer means

H(t) = x(0, t). (25)

It is easy to see that

d

dt

MH |H ′(t)|2

2=

∫ t

0Txs∗H

′(t)dt =

∫ t

0Txs∗xtdt. (26)

Note that there is obvious similarity between this RHS and the first term in the RHS of

(22), the former is just the x component of the latter.

We get only one component as we confine our hammer to move only along the x axis,

considering the position of hammer is always (H(t), 0). To be consistent, our impact end

will also have no displacement along y axis.

3.2.1 Proposed new boundary conditions and corresponding energy laws

Here we propose two new boundary conditions and derive corresponding energy laws.

The first one is to extend the motion of hammer (also the impact end) into the full

two dimensional plane. It might not exactly resemble our physical intuition. The previous

restriction of motion has its experimental roots, see ??. As the hammer is hitting the

rod with such a very high speed that the deviation along the lateral y direction might

not be easily observable. That is the reason we see the motion of hammer is very much

centralized around the neutral position of the straight rod. However small the motion in

the y direction is, we can extend our model include the motion on the second axis. The

usual contact boundary condition is a special case of our current assumption. Thus we can

reset the ODE for the hammer, which effectively change our boundary condition for the

11

impact end. Now, ~H(t) is a vector representing the position of hammer. We have

MH~H ′′(t) = −T ~Xs∗(0, t), (27)

and contact boundary conditions reads

~H(t) = ~X(0, t). (28)

Similarly, we can get the energy law for the system of the rod and the hammer. The

interaction term involving T ~Xs~Xt will be canceled and we find

d

dt

(∫ L(t)

0

(‖ ~Xt‖2

2+ ε‖ ~Xs∗s∗‖2

2

)ds∗ +

MH |H ′(t)|2

2

)=‖ ~Xt(L(t), t)‖2

2L′(t)−ε ~Xs∗s∗s∗

~Xt(0, t)

(29)

Denote

E2(t) =MH |H ′(t)|2

2, (30)

which is the kinetic energy of the hammer. Then the second energy law (29) states that the

boundary term will decide the behavior of the total energy E = E1 + E2, which includes

the kinetic energy of the hammer and the rod plus the elastic energy of the rod.

3.2.2 Conjecture of a new boundary condition

At this moment, we seem to know how the total energy will depends on the boundary term.

However, is it possible to eliminate the second term of RHS in (29)? Here is a conjecture

of a new boundary condition for the hammer which might suffice:

MH~H ′′(t) = −T ~Xs∗(0, t) + ε ~Xs∗s∗s∗ , (31)

and contact boundary condition again reads

~H(t) = ~X(0, t). (32)

12

Following the same procedure, now we will have another energy law:

d

dt

(∫ L(t)

0

(‖ ~Xt‖2

2+ ε‖ ~Xs∗s∗‖2

2

)ds∗ +

MH |H ′(t)|2

2

)=‖ ~Xt(L(t), t)‖2

2L′(t). (33)

Compared with (33), apparently the RHS is one term less, which can be interpreted as the

interaction between the rod and hammer will cancel out each other under current boundary

condition. This cancellation can be viewed as Newton’s third law in terms of energy. More

importantly the sign of the RHS solely depends on L′(t). If we have fixed total rod length,

then we have the total energy to be conserved. Even when L′(t) 6= 0, (33) tells us that the

tendency of change in the total energy is associated only with the change of total length at

that moment.

Chapter 4

Numerical algorithm and linear

stability analysis

4.1 Numerical scheme

Now we need to solve the following coupled PDE system,

~Xtt = (T ~Xs)s − ε ~X4s.

‖ ~Xs‖2Tss − ‖ ~Xss‖2T = −‖ ~Xts‖2 − ε(

3‖ ~Xss‖2 + 4 ~Xss · ~Xssss

).

(34)

along with boundary conditions:

~Xt(0, t) = 0, ~Xs∗s∗ |∂Ω = 0. (35)

For other boundary conditions we will have several options to choose from.

Our algorithm works as follows:

• Initialize the whole system, with random initial input ~X0, calculate the initial tension

T 0.

• Use explicit (Euler) scheme to generate ~X1, then T 1.

13

14

• For time level n ≥ 2, use semi-implicit scheme to first solve

~Xn+1tt = (Tn ~Xn

s )s − ε ~Xn+14s ,

then use ~Xn+1 to solve Tn+1.

Note here that we have the option to do a fixed point iteration on each time level. Here we

give the description of the core algorithm for the time evolution part:

~Xn+1tt = (Tn ~Xn

s )s − ε ~Xn+14s .

To incorporate our boundary conditions on Xss, we introduce w = xss, v = yss. Then the

problem becomes

xtt = Tw − εwss

xss = w

We have many different choices at this moment in terms of numerical scheme. A naive

one is to use finite difference scheme for both time and spatial variables. We tried explicit

or implicit scheme in time and central difference scheme in space. The results are not very

satisfactory. After some numerical tests on benchmark problems for a similar system, we

find that the error in the high order derivatives spreads out very fast.

To best eliminate the error due to spatial differential operator ∂s, we choose the spectral

methods which is known for its exponential accuracy. The combination of finite difference

and spectral method are also easily found in literature, such as [8]. There are several typical

examples of the spectral method. A popular one is Fourier spectral method, which can be

combined with finite difference scheme such as in [5].

In our problem, however, we don’t have convenient periodic boundary condition, which is

necessary for Fourier spectral method. To work with our non-periodic boundary conditions,

we turn to Chebyshev spectral method, and use a non-uniform Chebyshev grid as our

computational domain. For the finite difference scheme in time, we adopt semi-implicit

15

scheme, to get better stability, while not getting too much non-linearity for the discretized

system.

Our discretized system under semi-implicit central difference scheme in time and spectral

differentiation in space looks as follows:

2I dt2 (−diag(Tn) + εD2)

D2 −I

xn+1

wn+1

=

2xn − xn−1

0

where D2 = D2, D is the spectral differentiation matrix with respect to Chebyshev grid

from [7] after a linear scaling of factor π2 . I is identity matrix and Tn is the explicit tension

from last time level. Let’s define

A =

2I dt2 (−diag(Tn) + εD2)

D2 −I

,which will largely decide the stability of our numerical scheme. Here the D matrix is neither

symmetric nor anti-symmetric. A typical D matrix has following form (choose grid points

N = 3, D is (N + 1)× (N + 1)):

196 −4 4/3 −1/2

1 −1/3 −1 1/3

−1/3 1 1/3 −1

1/2 −4/3 4 −196

.

4.2 Von Neumann linear stability analysis

The von Neumann stability analysis or Fourier stability analysis provides necessary and

sufficient condition for stability in the sense of Lax-Richtmeyer [9] when the PDE is linear

and have constant-coefficient with periodic boundary conditions, while the scheme uses no

more than two time levels. Our PDE system is highly non-linear, but we can still get some

insight from the linear stability analysis of approximate linear system to decide our time

16

and spatial step ratio.

For our linear counter part

utt = Tuxx − εuxxxx,

we have the following stability analysis:

Say we use central different scheme (explicit) in time and our spacial differentiations are

accurate, then the round off error errnj satisfy the same discretized equation:

errn+1j + errn−1

j − 2errnj = dt2(Terrnss − εerrnssss)|x=xj .

Assume the growth of error have a typical form

errm(x, t) = eateikmx, errn+1j = ea(t+∆t)eikmx,

where km = m+ π/2, since err = 0 boundary x = ±π/2 Then we have

ea∆t + e−a∆t = 2 + ∆t2(−Tk2m − εk4

m), (36)

namely

(ea∆t)2 −(2 + ∆t2(−Tk2

m − εk4m))ea∆t + 1 = 0.

Solving this quadratic equation, let B = −Tk2m − εk4

m,

ea∆t =(2 + ∆t2B)±

√∆t4B2 + 4∆t2B

2

= 1 +∆t2

2B ± ∆t

2

√∆t2B2 + 4B. (37)

If we require that the numerical error does not accumulate, then we want the amplification

factor |ea∆t| = |r1,2| ≤ 1. While r1r2 = 1, we can only expect |ea∆t| = |r1,2| = 1, which

means we need to have complex roots, or ∆t = 0.

17

If B < 0, which means

k2m >

−Tε,

we have

|ea∆t| = |r1,2| = 1.

The scheme will be stable for any ∆t. Note this will always be true when tension is positive.

While the rod is being stretched, the tension is positive thus the numerical solution under

such scheme is stable, which agrees with the fact that the solution of an unstretched rod is

a stable equilibrium.

Say we use central different scheme (implicit) in time and our spacial differentiations

are accurate, then the round off error errnj satisfy the same discretized equation:

errn+1j + errn−1

j − 2errnj = dt2(Terrn+1ss − εerrn+1

ssss)|x=xj .

Assume the growth of error have a typical form

errm(x, t) = eateikmx, errn+1j = ea(t+∆t)eikmx,

where km = m+ π/2, since err = 0 boundary x = ±π/2 Then we have

ea∆t + e−a∆t = 2 + ∆t2(−Tk2m − εk4

m)ea∆t, (38)

namely

(ea∆t)2(1 + ∆t2(Tk2

m + εk4m))− 2ea∆t + 1 = 0.

Solving this quadratic equation, let B = Tk2m + εk4

m,

ea∆t =(2±

√4− 4(1 + ∆t2B)

2(1 + ∆t2B)

=1±

√−∆t2B

1 + ∆t2B. (39)

18

If B > 0, which means

k2m >

−Tε,

we have

|ea∆t| = |r1,2| =1√

1 + ∆t2B< 1.

The scheme will also be stable for any ∆t. Not surprisingly, the implicit scheme will have

a better performance as the error will shrink to zero with the eigenvalues strictly less than

one. Given the fact that implicit scheme is more expensive to implement, this is a natural

benefit.

As our system is forth order in space and highly non-linear, it is desirable to have a

more stable scheme. That is the reason why we adopt semi-implicit scheme in time.

Chapter 5

Numerical results and discussion

5.1 Hammer constrained in x direction

Now let’s look at the numerical results for contact boundary condition without gravity force

where the hammer is constrained to move only in x direction.

~Xtt = (T ~Xs)s − ε ~Xssss, s ∈ (−π/2, π/2)

Tss + lsTs − ‖ ~Xss‖2T = −‖ ~Xts‖2 − ε(

3‖ ~X3s‖2 + 4 ~Xss · ~Xssss

)+ β(1− | ~Xs|2),

(40)

where l(s, t) = | ~Xs|2 and the additional lsTs term is introduced for stability reasons.

Here we are conducting our computation on a reference domain Ω = (−π/2, π/2). To

maintain the reference variable s to be not too far away from arc length variable s∗, we also

introduce the penalty term β(1− | ~Xs|2) as in [3].

Our contact boundary conditions without gravity force are the following:

For rod position ~X(−π/2, t) = (H(t), 0)

~X(π/2, t) = (π/2, 0)

~Xss|∂Ω = 0

(41)

19

20

For tension T (−π/2, t) = µ(| ~Xs|2 − 1)

Ts(π/2, t) = 0.

(42)

For Hammer,

MHHtt(t) = T (−π/2, t) cos θ, cos θ =xs√

x2s + dL2y2

s

(43)

Ht(0) = v0. (44)

5.1.1 Test of different hammer weight MH

First we test a group of different MH , namely the weight of hammer ranging from 0.1 to

0.8, and stop at the moment when velocity of the hammer v(t) arrives at 0.

It can be seen that with fixed initial velocity v0, a heavier hammer will require more time

to stop at v = 0, and the buckling amplitude growths in this process. The following results

have fixed these parameters with same random initial input N = 80, ε = 5× 10−3, µ = 5×

10−4, v0 = 0.025, β = 2, dL = 20, A0 = 5× 10−4, F = rand(8, 1), y = A0∑8

i=1 F (i) sin(is +

π/2)/√i. For small value of MH , the velocity reaches 0 within time < 30. For larger

MH(≥ 0.5), however, the velocity approaches near 0 at long times and suddenly drops

to negative value. The long time behavior of the rod shape is in Figure 5.3. The code

eventually blows up.

5.1.2 The hook under larger initial velocity

Under previous conditions, here we want to see what will happen if v0 increases to 0.035,

with smaller ε and µ. We observe a hook which develops during the process of slowing down

the hammer, in which the rod doubles back at its impact end. The hook continues to grow

in size, see Figure 5.4. The velocity of the hammer can increase after the hook develops,

see Figure 5.5. The hook has not yet been observed in previous experiments.

The dynamic buckling experiments are conducted in a way that there is a horizontal

plane (the cross section of the hammer) at the impact end [1]. However, our model does

not posses such restriction. Our model allow the small portion of rod at the impact end to

21

Figure 5.1: Tension projection, velocity (left) and rod shape (right), MH = 0.1, time = 9.922

Figure 5.2: Tension projection, velocity (left) and rod shape (right), MH = 0.3, time = 26.62

go faster than its neighbors. The possible experiments corresponding to our model is in a

setting where a projectile is always glued to the impact end, with no cross section against

the rod. In this way, there will be no constraint on the angle of the rod. In Figure 5.5 we

observed the velocity to be greater than the initial velocity, which could result from the fact

that the fixed end is providing some energy into the whole system by pushing back on the

projectile. The velocity of the hammer experiences a sharp turn around time = 6. We can

zoom in to find out that the velocity curve is still smooth around that turning point, which

has very large second order derivative w.r.t. spatial variable.

22

Figure 5.3: Tension projection, rod shape in long time, MH = 0.5, time = 30 (left), time =49.5 (right)

Figure 5.4: T projection, rod shape, MH = 0.3, time = 22.5 (left), time = 30 (right)

The following results have fixed these parameters with same random initial input N =

80, ε = 1×10−3, µ = 1×10−4, v0 = 0.035, β = 2, dL = 20, A0 = 5×10−4, F = rand(8, 1), y =

A0∑8

i=1 F (i) sin(is+ π/2)/√i.

5.1.3 v0 = 0.025, lower ε = 1× 10−3, from no hook to slight hook for a long

time MH ∈ (0.01, 0.08)

The following results have fixed these parameters with same random initial input N =

80, ε = 1×10−3, µ = 1×10−4, v0 = 0.025, β = 2, dL = 20, A0 = 5×10−4, F = rand(8, 1), y =

A0∑8

i=1 F (i) sin(is+ π/2)/√i.

23

Figure 5.5: T projection, velocity (left) and impact end angle (right), MH = 0.3, time =33.24, the blue horizontal line is θ = π/2

From MH = 0.01 to MH = 0.1, the former gives no buckling, almost straight solution,

while the latter allows buckling with a slight hook lasting for a long time.

To really understand the transition, we check a gradually growing hammer mass MH ∈

(0.01, 0.08), which gives a gradually increased buckling amplitude, and bridge the gap be-

tween no hook in Figure 5.6 and a slight hook lasting for a long time situation in Figure 5.5.

However, note that at MH = 0.05, in Figure 5.8 no hook is observed during the simulation

time. For MH ≥ 0.06, there is a substantial amount of time during which a slight hook is

observed. We observe from the velocity part in Figure 5.8 that as MH ≥ 0.05, the velocity

didn’t decrease all the way down to 0, but oscillates around some positive value for a long

time.

So it is quite clear that higher hammer mass gives more energy into the system, and

yields larger buckling amplitude and longer motion time and existence of a hook during the

simulation time once the mass of hammer is greater than a threshold.

24

Figure 5.6: Velocity in x-axis (left) and angle θ, MH = 0.04, time = 13.3, the blue horizontalline is θ = π/2

Figure 5.7: Shape of rod (left) and y v.s. s (right), MH = 0.04, time = 13.3

5.2 Hammer allowed 2D motion with full projection for ten-

sion

When should we stop the simulation? We first use the criteria where the velocity in the x

direction of the impact end vx < 1× 10−5.

The following results have fixed these parameters with same random initial input N =

80, ε = 5×10−3, µ = 5×10−4, v0 = 0.025, β = 2, dL = 20, A0 = 5×10−4, F = rand(8, 1), y =

A0∑8

i=1 F (i)sin(is+ π/2)/√i. For small value of MH , the velocity reaches 0 even faster in

25

Figure 5.8: Velocity in x-axis (left) and angle θ, MH = 0.05, time = 35.1, the blue horizontalline is θ = π/2

Figure 5.9: Shape of rod before blow up (left) MH = 0.05, time = 34.5, (right) MH =0.07, time = 33

x direction. However, the velocity in y, although small, is always increasing. Compared to

model with x-axis only projection of tension, it takes slightly shorter (almost the same) to

get zero velocity (in x). For larger MH(≥ 0.6), the code will eventually blow up.

As MH gets larger, both the deviation in y and the total amplitude get larger. Similar

to previous model, as MH = 0.5, there is slight a hook which develops for a very short time

period, see Figure 5.15. No hook is observed during simulaiton time when MH < 0.5

26

Figure 5.10: Velocity in x-axis (left) and y-axis (right), MH = 0.1, time = 9.92

Figure 5.11: Shape of rod (left) and y v.s. s (right), MH = 0.1, time = 9.92

27

Figure 5.12: Velocity in x-axis (left) and y-axis (right), MH = 0.2, time = 15.1

Figure 5.13: Shape of rod (left) and y v.s. s (right), MH = 0.2, time = 15.1

28

Figure 5.14: Rod Shape (left) and y (right), MH = 0.5, time = 60

Figure 5.15: Impact end angle for MH = 0.3 (left) and MH = 0.5 (right), the blue horizontalline is θ = π/2

Chapter 6

Conclusions and future work

We derive a new model for the dynamic buckling of an elastic rod which gives a new

constitutive equation for the non-uniform tension T (s, t). We also derive several energy

laws which give guidance to our choice of boundary conditions. We implement a numerical

algorithm with semi-implicit finite difference scheme in time and spectral method in space

on non-uniform Chebyshev grids. We explore the role of the mass of the hammer in our

system. We can see that larger mass always gives the system more energy thus results in

larger buckling amplitude, which agrees with our physical intuition. Also we discover the

hook phenomenon, which hasn’t been observed in the experiments yet. This is because

our model doesn’t place a constraint on the angle of the rod with the cross section of the

hammer. This can be a prediction for a new experiment where the rod and the impact

projectile are always glued together but otherwise freely rotates around each other. We can

further explore other parameters, as well as improve the existing numerical algorithm. This

is a field that still worth further investigation.

29

Bibliography

[1] J. R. Gladden, N. Z. Handzy , A. Belmonte, and E. Villermaux, Dynamic buckling andfragmentation in brittle rods, Physical Review Letters 94 (3) (2005), p. 035503.

[2] A. Belmonte, M. J. Shelley, S. T. Eldakar, and C. H. Wiggins, Dynamic patterns andself-knotting of a driven hanging chain, Physical Review Letters 87.11 (2001): 114301.

[3] Tornberg, Anna-Karin, and Michael J. Shelley, Simulating the dynamics and interac-tions of flexible fibers in Stokes flows, Journal of Computational Physics 196.1 (2004):8-40.

[4] E. A. H. Love, A Treatise on the Mathematical Theory of Elasticity, Dover Publications,New York, 1944, 4th ed.

[5] Chen, Gui Qiang, Qiang Du, and Eitan Tadmor, Spectral viscosity approximations tomultidimensional scalar conservation laws, Mathematics of Computation 61.204 (1993):629-643.

[6] Zeeman, E. C, Euler buckling, Structural stability, the theory of catastrophes, and ap-plications in the sciences, Springer Berlin Heidelberg, 1976. 373-395.

[7] Trefethen, Lloyd N., Spectral methods in MATLAB, Vol. 10. Society for Industrial andApplied Mathematics, 2000.

[8] Boyd, John Philip, Chebyshev and Fourier spectral methods, Courier Dover Publica-tions, 2001.

[9] Lax, Peter D., and Robert D. Richtmyer, Survey of the stability of linear finite differ-ence equations, Communications on Pure and Applied Mathematics 9.2 (1956): 267-293.

[10] Orszag, Steven A, Spectral methods for problems in complex geometries, Journal ofComputational Physics 37.1 (1980): 70-92.

[11] Lindberg, Herbert E, Little Book of Dynamic Buckling, LCE Science/Software, PennValley CA, USA, 2003.

[12] Kenny, S., N. Pegg, and F. Taheri, Dynamic elastic buckling of a slender beam withgeometric imperfections subject to an axial impulse, Finite elements in analysis anddesign 35.3 (2000): 227-246.

30

31

[13] Waas, Anthony, and Wooseok Ji, The Temporal Evolution of Buckling in a DynamicallyImpacted Column, Journal of Applied Mechanics 1 (2012): 339.

[14] Huang, Wei-Xi, Soo Jai Shin, and Hyung Jin Sung, Simulation of flexible filaments ina uniform flow by the immersed boundary method, Journal of Computational Physics226.2 (2007): 2206-2228.