Final Project Report
ESAM 311-2
Christopher Carhart, Dooyoul Lee, Yoke Peng Leong
March 12, 2012
Introduction
Coating is a covering that is applied to the surface of an object, usually referred
to as the substrate. In many cases coatings are applied to improve surface
properties of the substrate such as appearance, adhesion, wetability, corrosion
resistance, wear resistance and scratch resistance. For example, turbine blades
of a turbojet engine are covered with thermal barrier coating which protect the
surface of the blade from heat and oxidation.
To assure the quality of the coating, thickness measurement is used, and
ultrasonic testing is the most popular nondestructive evaluation method for it.
The principle is simple. The coating and the substrate have different acoustic
properties such as the velocity of sound which cause reflection and transmission of ultrasound at the boundary between the two, and the thickness can be
measured by counting the traveling time of reflected ultrasound.
Although the basic idea in simple, there exists uncertainties at the boundary. For example, it can have discontinuities such as void which reflect ultrasound 100%, or surface roughness which scatter ultrasound. It is very difficult to
experiment those conditions. So computer simulation is widely used to research
the response of ultrasound with various boundary conditions.
This project explores the use of the one-dimensional wave equation for
application in non-destructive inspection. There are many ways of posing this
problem, but we will focus on a scenario where a brief impulse wave is sent into
a material, interacts with a defect, and returns a signal to the surface where the
response can be measured. The basic setup for the problem is shown in Figure
1.
The nature of the response measured at the accelerometer depends on how
1
Figure 1: Problem setup
the defect line is treated with respect to wave interaction. We chose to explore
this problem when the defect is fully reflective and when the defect reflects half
of the wave impulse and transmits the other half. Each problem will be explored
separately.
Problem 1: Fully Reflective Defect
The reflectivity of a defect can depend on a variety of things. However, a fully
reflective defect can indicate a void in the material which would be incapable of
transmitting any wave energy. This problem may be expressed mathematically
as the following:

2

0 < x < 2L, t > 0
utt = c uxx
u (x, 0) = f (x) , ut (x, 0) = 0 0 ≤ x ≤ 2L, t > 0


ux (0, t) = 0, ux (L, t) = 0
t>0
where
(
cos (5x) if x ≤ π/10
f (x) =
0
if x > π/10
The differential equation used is the standard, one-dimensional wave equation with a wave speed of c, but the interpretation of u(x, t) in this problem
warrants some explanation. A wave that travels internally to the material may
2
be thought of as a moving density fluctuation. This means that u(x, t) may be
thought of as the deflection of the particles at location, x and time, t from their
original position. Therefore, measuring u(x, t) at x = 0 is measuring the deflection of the material’s boundary. However, in an experimental setting, it may not
be possible to measure u(0, t) directly, but it is possible to use an accelerometer
and process the raw data through double integration to get u(0, t).
Neumann boundary conditions, ux (0, t) = ux (L, t) = 0, were chosen for
this problem because the Neumann boundary conditions allow waves to reflect
and allow the deflection at boundaries, which should match physical intuition.
The physical meaning of Neumann boundary conditions in this case can be
expressed by saying that the particle next to the material’s boundary (in the
x direction) has the same amount of deflection as a neighboring particle on
the material’s boundary. By contrast, Dirichlet boundary conditions would not
allow deflection at boundaries, which is a problem if one is trying to measure
wave response at those same boundaries. Having no boundary conditions would
not allow any reflection at all and only allow full transmission of wave energy
across the defect region.
The initial condition describes a single wave peak at x = 0 with zero initial
velocity. The function f (x) was chosen to make visual analysis of the wave
behavior easy to observe graphically (refer to Figure 2).
1.0
0.8
f HxL
0.6
0.4
0.2
0.0
0.0
0.2
0.4
0.6
0.8
1.0
x
Figure 2: Initial condition, f (x), for problem 1
This problem has a homogenous boundary condition and does not possess
an external forcing function. Separation of variables may be used to solve it:
u (x, t) = X (x) T (t)
3
1 T 00
X 00
=
= −λ
2
c T
X
Eigenvalue problem in x:
(
X 00 + λX = 0
ux (0) = ux (L) = 0
n
√ √ o
X = sin
λx , cos
λx
X 0 (0) = C1 = 0
√
X 0 (L) = −C2 λ
nπ 2
λn =
, n = 0, 1, 2, . . .
L
It is known from solving similar problems that no eigenvalue for λ < 0 exists.
So,
nπx Xn = cos
L
Now solve for t:
(
2
T 00 + nπ
T =0
L
0
T (0) = 0
nπ o
n
nπ t , cos
t
T = sin
L
L
T 0 (0) = C1 = 0
Thus,
Tn = cos
nπ t
L
The full series solution is
∞
X
u (x, t) =
an cos
n=0
nπ nπ t cos
x
L
L
which can also be written as
∞
nπ
nπ
X
1
nπ 1
nπ u (x, t) =
an cos
x−
t + cos
x+
t
2
L
L
2
L
L
n=0
Use initial condition to get an
u (x, 0) =
a0 =
an =
∞
X
an cos
n=0
L
1
L
Z
2
L
Z
nπ x = f (x)
L
f (x) dx
0
L
cos
0
nπ L
4
f (x) dx
1.0
1.0
0.8
0.8
0.8
0.6
0.6
0.6
0.4
0.4
0.2
0.2
0.0
0.0
-0.2
uHx,tL
1.0
uHx,tL
uHx,tL
At this point, the first problem has been solved. The behavior of the wave using
N = 20, c = 10 and L = 10 is shown graphically in Figure 3. The graphs
generated are from a simulation where u(x, t) has been approximated. Instead
of u(x, t) being a sum from n = 0 to n = ∞, ∞ is approximated by N . This
approximation is the cause of the small ripples seen on either side of the wave
impulse.
0.2
0.0
-0.2
0
2
4
6
8
10
-0.2
0
2
4
6
x
10
0
1.0
0.8
0.8
0.6
0.6
0.6
0.4
0.4
0.2
0.2
0.0
0.0
-0.2
uHx,tL
1.0
0.8
4
6
8
10
2
4
6
8
10
0
0.8
0.6
0.6
0.6
0.4
0.0
-0.2
uHx,tL
1.0
0.8
uHx,tL
1.0
0.8
4
6
8
10
10
8
10
0.4
0.0
-0.2
0
2
4
6
x
(g) t = 1.50
6
0.2
-0.2
2
4
(f) t = 1.25
1.0
0
2
x
(e) t = 1.00
0.0
8
0.0
(d) t = 0.75
0.2
10
0.4
x
0.2
8
-0.2
0
x
0.4
6
0.2
-0.2
2
4
(c) t = 0.50
1.0
0
2
x
(b) t = 0.25
uHx,tL
uHx,tL
8
x
(a) t = 0
uHx,tL
0.4
x
(h) t = 1.75
8
10
0
2
4
6
x
(i) t = 2.00
Figure 3: Behavior of the wave using N = 20, c = 10 and L = 10 when the
defect is fully reflective at x = L
If one wanted to construct an experiment to verify the theoretical results,
one could create a setup similar to that shown in Figure 4 where the defect is
created by a physical gap between two sections of material. The inducer sends
an impulse wave into the first section of material which will not be transmitted
into the second section of material. The returning wave may be measured using
an accelerometer placed at x = 0. The resulting u(x, t) at x = 0 is shown in
Figure 5.
5
Figure 4: Experimental setup for problem 1
0.6
uH0,tL
0.4
0.2
0.0
0.0
0.5
1.0
1.5
2.0
t
Figure 5: u(x, t) at x = 0 when defect is fully reflective
Problem 2: 50% Reflective Defect
A defect in material does not necessarily reflect 100 percent of wave energy.
For example, if instead of a void, a defect was composed of a thin crack, or a
6
different material, some of the wave energy would be reflected and some would
be transmitted. While the ratio of transmitted energy vs. reflected energy
depends greatly on the type of defect, we will assume a 50-50 split in wave
energy distribution.
If we change the nature of the defect region to allow 50 percent reflection
and 50 percent transmission of the wave energy, we must solve the problem in
two phases. Phase 1 may be expressed as the following:

2

0 < x < 2L, 0 < t < L/c
utt = c uxx
u (x, 0) = f (x) , ut (x, 0) = 0 0 ≤ x ≤ 2L, 0 < t < L/c


ux (0, t) = 0, ux (2L, t) = 0
0 < t < L/c
where
(
cos (5x) if x ≤ π/10
f (x) =
0
if x > π/10
The solution may be expressed as:
u (x, t) =
∞
X
an cos
n=0
nπ nπ t cos
x
2L
2L
where an is defined as
Z 2L
1
f (x) dx
2L 0
Z
nπ 1 2L
an =
cos
f (x) dx
L 0
2L
a0 =
Note two key differences in these expressions from those of problem 1. Here, t
goes from 0 to Lc and the second Neumann boundary condition has been shifted
from the defect line to the other end of the material at x = 2L. The physical
meaning of these changes is that the initial wave starts out the same and travels
from x = 0 to x = L without any recognition of the defect line. Phase 1 of
this problem may be solved using the identical method outlined in problem 1
while taking the shifted boundary condition into account. However, phase 1
ends when t = Lc which is when the wave peak is centered over the defect line
and this marks the end of phase 1 and the start of phase 2 (refer to Figure 6).
7
1.0
0.8
uHx,tL
0.6
0.4
0.2
0.0
-0.2
0
2
4
6
8
10
x
Figure 6: Wave is centered at x = 5 where 2L = 10 when t =
L
c
Phase 2 may be expressed as the following:

2

0 < x < 2L, t > L/c
utt = c uxx
u (x, 0) = f (x) , ut (x, 0) = 0 0 ≤ x ≤ 2L, t > L/c


ux (0, t) = 0, ux (2L, t) = 0
t > L/c
where
f (x) =


0
1
2 sin (L (x


0
if x < L − π/10
− L + π/10)) if x ≤ L + π/10
if x > π/10
Separation of variables may again be used to solve phase 2 in a manner similar
to that outlined in problem 1. The solution may be expressed as:
u (x, t) =
∞
X
n=0
an cos
nπ (t − L/c) cos
x
2L
2L
nπ
where an is defined as
Z 2L
1
f (x) dx
2L 0
Z
nπ 1 2L
an =
cos
f (x) dx
L 0
2L
a0 =
8
Note that the f (x) term for phase 2 (refer to Figure 7) is identical to the wave
form u(x, Lc ) in phase 1 in Figure 6. In other words, the final condition at t = Lc
from phase 1 is being substituted as the initial condition of phase 2. f (x) is
the exact solution of the initial wave after it is being propagated from x = 0 to
x = L. u(x, Lc ) is the approximation of the exact solution when the peak of the
wave is exactly at x = L.
0.5
0.4
f HxL
0.3
0.2
0.1
0.0
0
2
4
6
8
10
x
Figure 7: f (x) for phase 2 of problem 2
The solution to phase 2 describes a wave that splits at x = L with half of
its energy traveling towards the left and half of its energy traveling towards the
right. This may be seen graphically in Figure 8d.
In this situation, there will be two signals observed at x = 0. The magnitude
of the signals will be half of that observed in problem 1. The first signal will
be observed at approximately t = 2L
c and the second signal will be observed at
approximately t = 4L
(refer
to
Figure
8).
c
It should be noted that this solution only takes the first interaction with
the defect and will not split the wave each successive time a wave crosses the
defect line. This is a limitation of this solution and could be area of future
research.
If one was to construct an experiment to verify theoretical results, one could
create a setup similar to that shown in Figure 9. Figure 9 is similar to Figure 4
except that the gap between the two sections of material has been filled with a
9
1.0
0.8
0.8
0.6
0.6
0.6
0.4
0.4
0.2
0.2
0.0
0.0
-0.2
uHx,tL
1.0
0.8
uHx,tL
uHx,tL
1.0
0.2
0.0
-0.2
0
2
4
6
8
10
-0.2
0
2
4
x
6
8
10
0
(b) t = 0.25
0.8
0.6
0.6
0.6
0.4
0.2
0.2
0.0
0.0
-0.2
uHx,tL
1.0
0.8
uHx,tL
1.0
0.4
6
8
x
10
10
8
10
0.4
0.0
-0.2
0
2
4
6
8
10
0
2
x
(d) t = 0.75
8
0.2
-0.2
4
6
(c) t = 0.50
0.8
2
4
x
1.0
0
2
x
(a) t = 0
uHx,tL
0.4
(e) t = 1.00
4
6
x
(f) t = 1.25
Figure 8: Behavior of the wave using N = 20, c = 10 and L = 5 when the defect
is 50% reflective at x = L
second material or gel. This will allow partial transmission of wave energy. The
experimental results here will likely differ from our theoretical solution due to
the limitations of our solution as stated above. However, two relatively strong
4L
signals should still be observable at t = 2L
c and t = c and the depth of the
defect will still be measurable. The resulting u(x, t) at x = 0 is shown in Figure
10.
Figure 9: Exprimental setup for problem 2
10
0.6
uH0,tL
0.4
0.2
0.0
0.0
0.5
1.0
1.5
2.0
t
Figure 10: u(x, t) at x = 0 when defect is 50% reflective
11
Appendix A Mathematica Simulation for Problem 1
u tt  c2 uxx for t  0, 0  x  L
u x 0, t  0
u x L, t  0
ux, 0  gx
u t x, 0  0
In[1]:=
c  10;
L  10;
In[8]:=
gx_ : PiecewiseCos5 x, x  Pi  10, 0, x  Pi  10;
Plotgx, x, 0, 1, PlotRange  Full, FrameLabel  Style"x", 14, Style"fx", 14,
Frame  True, FrameTicks  Automatic, None, Automatic, None
1.0
0.8
fx
0.6
Out[9]=
0.4
0.2
0.0
0.0
0.2
0.4
0.6
0.8
1.0
x
In[10]:=
aa0  1  L Integrategx, x, 0, L  Chop;
aa  Table2  L IntegrateCosn Pi x  L gx, x, 0, L, n, 1, 20  Chop;
uu  aa0  Sumaan Cosn Pi c t  L Cosn Pi x  L, n, 1, 20  N  Chop;
Printed by Mathematica for Students
simulation.nb
In[13]:=
AnimateShowPlotuu . x  xx . t  tt, xx, 0, L,
PlotRange  0, L,  0.3, 1, FrameLabel  Style"x", 14, Style"ux,t", 14,
Frame  True, FrameTicks  Automatic, None, Automatic, None,
GraphicsTextStyle"t  ", 14, 4.5, 0.9,
GraphicsTextStylePaddedFormtt, 6, 2, 14, 5, 0.9,
GraphicsTextStyle"s ", 14, 5.9, 0.9, tt, 0, 2, AnimationRunning  False
tt
1.0
t  0.49 s
0.8
0.6
Out[13]=
ux,t
2
0.4
0.2
0.0
0.2
0
2
4
6
8
10
x
Appendix B Mathematica Simulation for Problem 2
u tt  c2 uxx for t  0, 0  x  L
u x 0, t  0
u x L, t  0
ux, 0  f x
u t x, 0  0
In[5]:=
c  10;
L  10;
Printed by Mathematica for Students
simulation.nb
 Phase 1
In[14]:=
fx_ : PiecewiseCos5 x, x  Pi  10, 0, x  Pi  10;
Plotfx, x, 0, 1, PlotRange  Full, FrameLabel  Style"x", 14, Style"fx", 14,
Frame  True, FrameTicks  Automatic, None, Automatic, None
1.0
0.8
fx
0.6
Out[15]=
0.4
0.2
0.0
0.0
0.2
0.4
0.6
0.8
1.0
x
In[16]:=
aaa0  1  L Integratefx, x, 0, L  Chop;
aaa  Table2  L IntegrateCosn Pi x  L fx, x, 0, L, n, 1, 20  Chop;
uuu  aaa0  Sumaaan Cosn Pi c t  L Cosn Pi x  L, n, 1, 20  N  Chop;
 Phase 2
In[19]:=
f2x_ : Piecewise
0, x  5  Pi  10, .5 Sin5 x  5  Pi  10, x  5  Pi  10, 0, x  5  Pi  10;
Plotf2x, x, 0, L, PlotRange  Full, FrameLabel  Style"x", 14, Style"fx", 14,
Frame  True, FrameTicks  Automatic, None, Automatic, None
0.5
0.4
fx
0.3
Out[20]=
0.2
0.1
0.0
0
2
4
6
8
10
x
In[21]:=
aaa02  1  L Integratef2x, x, 0, L  Chop;
aaa2  Table2  L IntegrateCosn Pi x  L f2x, x, 0, L, n, 1, 20  Chop;
uuu2  aaa02  Sumaaa2n Cosn Pi c t  L Cosn Pi x  L, n, 1, 20  N  Chop;
Printed by Mathematica for Students
3
simulation.nb
 Overall Result
In[24]:=
sound  Animate
ShowPiecewisePlotuuu . x  xx . t  tt, xx, 0, L, PlotRange  0, L,  0.3, 1,
FrameLabel  Style"x", 14, Style"ux,t", 14, Frame  True,
FrameTicks  Automatic, None, Automatic, None, tt  0.5,
Plotuuu2 . x  xx . t  tt  0.5, xx, 0, L, PlotRange  0, L,  0.3, 1,
FrameLabel  Style"x", 14, Style"ux,t", 14, Frame  True,
FrameTicks  Automatic, None, Automatic, None, tt  0.5,
GraphicsTextStyle"t  ", 14, 4.5, 0.9,
GraphicsTextStylePaddedFormtt, 6, 2, 14, 5, 0.9,
GraphicsTextStyle"s ", 14, 5.9, 0.9
, tt, 0, 1.3, AnimationRunning  False
tt
1.0
t  0.75 s
0.8
0.6
Out[24]=
ux,t
4
0.4
0.2
0.0
0.2
0
2
4
6
8
x
Printed by Mathematica for Students
10