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 transmis- sion 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 bound- ary. For example, it can have discontinuities such as void which reflect ultra- sound 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
Transcript
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 referredto as the substrate. In many cases coatings are applied to improve surfaceproperties of the substrate such as appearance, adhesion, wetability, corrosionresistance, wear resistance and scratch resistance. For example, turbine bladesof a turbojet engine are covered with thermal barrier coating which protect thesurface of the blade from heat and oxidation.
To assure the quality of the coating, thickness measurement is used, andultrasonic testing is the most popular nondestructive evaluation method for it.The principle is simple. The coating and the substrate have different acousticproperties such as the velocity of sound which cause reflection and transmis-sion of ultrasound at the boundary between the two, and the thickness can bemeasured by counting the traveling time of reflected ultrasound.
Although the basic idea in simple, there exists uncertainties at the bound-ary. For example, it can have discontinuities such as void which reflect ultra-sound 100%, or surface roughness which scatter ultrasound. It is very difficult toexperiment those conditions. So computer simulation is widely used to researchthe response of ultrasound with various boundary conditions.
This project explores the use of the one-dimensional wave equation forapplication in non-destructive inspection. There are many ways of posing thisproblem, but we will focus on a scenario where a brief impulse wave is sent intoa material, interacts with a defect, and returns a signal to the surface where theresponse can be measured. The basic setup for the problem is shown in Figure1.
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 explorethis problem when the defect is fully reflective and when the defect reflects halfof the wave impulse and transmits the other half. Each problem will be exploredseparately.
Problem 1: Fully Reflective Defect
The reflectivity of a defect can depend on a variety of things. However, a fullyreflective defect can indicate a void in the material which would be incapable oftransmitting any wave energy. This problem may be expressed mathematicallyas the following:
utt = c2uxx 0 < x < 2L, t > 0
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
f (x) =
{cos (5x) if x ≤ π/10
0 if x > π/10
The differential equation used is the standard, one-dimensional wave equa-tion with a wave speed of c, but the interpretation of u(x, t) in this problemwarrants some explanation. A wave that travels internally to the material may
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be thought of as a moving density fluctuation. This means that u(x, t) may bethought of as the deflection of the particles at location, x and time, t from theiroriginal position. Therefore, measuring u(x, t) at x = 0 is measuring the deflec-tion of the material’s boundary. However, in an experimental setting, it may notbe possible to measure u(0, t) directly, but it is possible to use an accelerometerand process the raw data through double integration to get u(0, t).
Neumann boundary conditions, ux(0, t) = ux(L, t) = 0, were chosen forthis problem because the Neumann boundary conditions allow waves to reflectand allow the deflection at boundaries, which should match physical intuition.The physical meaning of Neumann boundary conditions in this case can beexpressed by saying that the particle next to the material’s boundary (in thex direction) has the same amount of deflection as a neighboring particle onthe material’s boundary. By contrast, Dirichlet boundary conditions would notallow deflection at boundaries, which is a problem if one is trying to measurewave response at those same boundaries. Having no boundary conditions wouldnot allow any reflection at all and only allow full transmission of wave energyacross the defect region.
The initial condition describes a single wave peak at x = 0 with zero initialvelocity. The function f(x) was chosen to make visual analysis of the wavebehavior easy to observe graphically (refer to Figure 2).
0.0 0.2 0.4 0.6 0.8 1.00.0
0.2
0.4
0.6
0.8
1.0
x
fHxL
Figure 2: Initial condition, f(x), for problem 1
This problem has a homogenous boundary condition and does not possessan external forcing function. Separation of variables may be used to solve it:
u (x, t) = X (x)T (t)
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1
c2T ′′
T=X ′′
X= −λ
Eigenvalue problem in x: {X ′′ + λX = 0
ux (0) = ux (L) = 0
X ={sin(√
λx), cos
(√λx)}
X ′ (0) = C1 = 0
X ′(L) = −C2
√λ
λn =(nπL
)2, n = 0, 1, 2, . . .
It is known from solving similar problems that no eigenvalue for λ < 0 exists.So,
Xn = cos(nπxL
)Now solve for t: {
T ′′ +(nπL
)2T = 0
T ′ (0) = 0
T ={sin(nπLt), cos
(nπLt)}
T ′ (0) = C1 = 0
Thus,
Tn = cos(nπLt)
The full series solution is
u (x, t) =
∞∑n=0
ancos(nπLt)cos(nπLx)
which can also be written as
u (x, t) =
∞∑n=0
an
[1
2cos(nπLx− nπ
Lt)
+1
2cos(nπLx+
nπ
Lt)]
Use initial condition to get an
u (x, 0) =
∞∑n=0
ancos(nπLx)
= f (x)
a0 =1
L
∫ L
0
f (x) dx
an =2
L
∫ L
0
cos(nπL
)f (x) dx
4
At this point, the first problem has been solved. The behavior of the wave usingN = 20, c = 10 and L = 10 is shown graphically in Figure 3. The graphsgenerated are from a simulation where u(x, t) has been approximated. Insteadof u(x, t) being a sum from n = 0 to n = ∞, ∞ is approximated by N . Thisapproximation is the cause of the small ripples seen on either side of the waveimpulse.
0 2 4 6 8 10-0.2
0.0
0.2
0.4
0.6
0.8
1.0
x
uHx,
tL
(a) t = 0
0 2 4 6 8 10-0.2
0.0
0.2
0.4
0.6
0.8
1.0
x
uHx,
tL
(b) t = 0.25
0 2 4 6 8 10-0.2
0.0
0.2
0.4
0.6
0.8
1.0
x
uHx,
tL
(c) t = 0.50
0 2 4 6 8 10-0.2
0.0
0.2
0.4
0.6
0.8
1.0
x
uHx,
tL
(d) t = 0.75
0 2 4 6 8 10-0.2
0.0
0.2
0.4
0.6
0.8
1.0
x
uHx,
tL
(e) t = 1.00
0 2 4 6 8 10-0.2
0.0
0.2
0.4
0.6
0.8
1.0
xuH
x,tL
(f) t = 1.25
0 2 4 6 8 10-0.2
0.0
0.2
0.4
0.6
0.8
1.0
x
uHx,
tL
(g) t = 1.50
0 2 4 6 8 10-0.2
0.0
0.2
0.4
0.6
0.8
1.0
x
uHx,
tL
(h) t = 1.75
0 2 4 6 8 10-0.2
0.0
0.2
0.4
0.6
0.8
1.0
x
uHx,
tL
(i) t = 2.00
Figure 3: Behavior of the wave using N = 20, c = 10 and L = 10 when thedefect 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 iscreated by a physical gap between two sections of material. The inducer sendsan impulse wave into the first section of material which will not be transmittedinto the second section of material. The returning wave may be measured usingan accelerometer placed at x = 0. The resulting u(x, t) at x = 0 is shown inFigure 5.
5
Figure 4: Experimental setup for problem 1
0.0 0.5 1.0 1.5 2.0
0.0
0.2
0.4
0.6
t
uH0,
tL
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 wouldbe transmitted. While the ratio of transmitted energy vs. reflected energydepends greatly on the type of defect, we will assume a 50-50 split in waveenergy distribution.
If we change the nature of the defect region to allow 50 percent reflectionand 50 percent transmission of the wave energy, we must solve the problem intwo phases. Phase 1 may be expressed as the following:
utt = c2uxx 0 < x < 2L, 0 < t < L/c
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
f (x) =
{cos (5x) if x ≤ π/10
0 if x > π/10
The solution may be expressed as:
u (x, t) =
∞∑n=0
ancos(nπ
2Lt)cos(nπ
2Lx)
where an is defined as
a0 =1
2L
∫ 2L
0
f (x) dx
an =1
L
∫ 2L
0
cos(nπ
2L
)f (x) dx
Note two key differences in these expressions from those of problem 1. Here, tgoes from 0 to L
c and the second Neumann boundary condition has been shiftedfrom the defect line to the other end of the material at x = 2L. The physicalmeaning of these changes is that the initial wave starts out the same and travelsfrom x = 0 to x = L without any recognition of the defect line. Phase 1 ofthis problem may be solved using the identical method outlined in problem 1while taking the shifted boundary condition into account. However, phase 1ends when t = L
c which is when the wave peak is centered over the defect lineand this marks the end of phase 1 and the start of phase 2 (refer to Figure 6).
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0 2 4 6 8 10-0.2
0.0
0.2
0.4
0.6
0.8
1.0
x
uHx,
tL
Figure 6: Wave is centered at x = 5 where 2L = 10 when t = Lc
Phase 2 may be expressed as the following:utt = c2uxx 0 < x < 2L, t > L/c
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 if x < L− π/1012sin (L (x− L+ π/10)) if x ≤ L+ π/10
0 if x > π/10
Separation of variables may again be used to solve phase 2 in a manner similarto that outlined in problem 1. The solution may be expressed as:
u (x, t) =
∞∑n=0
ancos(nπ
2L(t− L/c)
)cos(nπ
2Lx)
where an is defined as
a0 =1
2L
∫ 2L
0
f (x) dx
an =1
L
∫ 2L
0
cos(nπ
2L
)f (x) dx
8
Note that the f(x) term for phase 2 (refer to Figure 7) is identical to the waveform u(x, Lc ) in phase 1 in Figure 6. In other words, the final condition at t = L
cfrom phase 1 is being substituted as the initial condition of phase 2. f(x) isthe exact solution of the initial wave after it is being propagated from x = 0 tox = L. u(x, Lc ) is the approximation of the exact solution when the peak of thewave is exactly at x = L.
0 2 4 6 8 10
0.0
0.1
0.2
0.3
0.4
0.5
x
fHxL
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 ofits energy traveling towards the left and half of its energy traveling towards theright. This may be seen graphically in Figure 8d.
In this situation, there will be two signals observed at x = 0. The magnitudeof the signals will be half of that observed in problem 1. The first signal willbe observed at approximately t = 2L
c and the second signal will be observed at
approximately t = 4Lc (refer to Figure 8).
It should be noted that this solution only takes the first interaction withthe defect and will not split the wave each successive time a wave crosses thedefect line. This is a limitation of this solution and could be area of futureresearch.
If one was to construct an experiment to verify theoretical results, one couldcreate a setup similar to that shown in Figure 9. Figure 9 is similar to Figure 4except that the gap between the two sections of material has been filled with a
9
0 2 4 6 8 10-0.2
0.0
0.2
0.4
0.6
0.8
1.0
x
uHx,
tL
(a) t = 0
0 2 4 6 8 10-0.2
0.0
0.2
0.4
0.6
0.8
1.0
x
uHx,
tL
(b) t = 0.25
0 2 4 6 8 10-0.2
0.0
0.2
0.4
0.6
0.8
1.0
x
uHx,
tL
(c) t = 0.50
0 2 4 6 8 10-0.2
0.0
0.2
0.4
0.6
0.8
1.0
x
uHx,
tL
(d) t = 0.75
0 2 4 6 8 10-0.2
0.0
0.2
0.4
0.6
0.8
1.0
x
uHx,
tL
(e) t = 1.00
0 2 4 6 8 10-0.2
0.0
0.2
0.4
0.6
0.8
1.0
x
uHx,
tL
(f) t = 1.25
Figure 8: Behavior of the wave using N = 20, c = 10 and L = 5 when the defectis 50% reflective at x = L
second material or gel. This will allow partial transmission of wave energy. Theexperimental results here will likely differ from our theoretical solution due tothe limitations of our solution as stated above. However, two relatively strongsignals should still be observable at t = 2L
c and t = 4Lc and the depth of the
defect will still be measurable. The resulting u(x, t) at x = 0 is shown in Figure10.
Figure 9: Exprimental setup for problem 2
10
0.0 0.5 1.0 1.5 2.0
0.0
0.2
0.4
0.6
t
uH0,
tL
Figure 10: u(x, t) at x = 0 when defect is 50% reflective
11
Appendix A Mathematica Simulation for Problem 1utt c2 uxx for t 0, 0 x Lux0, t 0uxL, t 0ux, 0 gxutx, 0 0
In[1]:= c 10;L 10;
In[8]:= gx_ : PiecewiseCos5 x, x Pi 10, 0, x Pi 10;Plotgx, x, 0, 1, PlotRange Full, FrameLabel Style"x", 14, Style"fx", 14,Frame True, FrameTicks Automatic, None, Automatic, None
Out[9]=
0.0 0.2 0.4 0.6 0.8 1.00.0
0.2
0.4
0.6
0.8
1.0
x
fx
In[10]:= aa0 1 L Integrategx, x, 0, L Chop;aa Table2 L IntegrateCosn Pi x L gx, x, 0, L, n, 1, 20 Chop;uu aa0 Sumaan Cosn Pi c t L Cosn Pi x L, n, 1, 20 N Chop;
Printed by Mathematica for Students
In[13]:= AnimateShowPlotuu . x xx . t tt, xx, 0, L,PlotRange 0, L, 0.3, 1, FrameLabel Style"x", 14, Style"ux,t", 14,Frame True, FrameTicks Automatic, None, Automatic, None,
In[16]:= aaa0 1 L Integratefx, x, 0, L Chop;aaa Table2 L IntegrateCosn Pi x L fx, x, 0, L, n, 1, 20 Chop;uuu aaa0 Sumaaan Cosn Pi c t L Cosn Pi x L, n, 1, 20 N Chop;
Phase 2In[19]:= f2x_ : Piecewise
0, x 5 Pi 10, .5 Sin5 x 5 Pi 10, x 5 Pi 10, 0, x 5 Pi 10;Plotf2x, x, 0, L, PlotRange Full, FrameLabel Style"x", 14, Style"fx", 14,Frame True, FrameTicks Automatic, None, Automatic, None
Out[20]=
0 2 4 6 8 100.0
0.1
0.2
0.3
0.4
0.5
x
fx
In[21]:= aaa02 1 L Integratef2x, x, 0, L Chop;aaa2 Table2 L IntegrateCosn Pi x L f2x, x, 0, L, n, 1, 20 Chop;uuu2 aaa02 Sumaaa2n Cosn Pi c t L Cosn Pi x L, n, 1, 20 N Chop;
simulation.nb 3
Printed by Mathematica for Students
Overall ResultIn[24]:= sound Animate
ShowPiecewisePlotuuu . x xx . t tt, xx, 0, L, PlotRange 0, L, 0.3, 1,FrameLabel Style"x", 14, Style"ux,t", 14, Frame True,FrameTicks Automatic, None, Automatic, None, tt 0.5,
Plotuuu2 . x xx . t tt 0.5, xx, 0, L, PlotRange 0, L, 0.3, 1,FrameLabel Style"x", 14, Style"ux,t", 14, Frame True,FrameTicks Automatic, None, Automatic, None, tt 0.5,
