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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
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