Ultrasonics 55 (2015) 133–140

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Ultrasonics

journal homepage: www.elsevier .com/locate /ul t ras

Using the pressure transmission coefficient of a transmitted waveto evaluate some of the mechanical properties of refractory metals

http://dx.doi.org/10.1016/j.ultras.2014.07.0010041-624X/� 2014 Elsevier B.V. All rights reserved.

⇑ Corresponding author at: Department of Mechanical and Materials Engineering,Faculty of Engineering and Built Environment, University Kebangsaan Malaysia,UKM, 43600 UKM Bangi, Selangor Darul Ehsan, Malaysia. Tel.: +60 108916964; fax:+60 389259659.

E-mail address: arshad_ald@yahoo.com (A.A. Mohammed).

Arshed Abdulhamed Mohammed a,b,⇑, Sallehuddin Mohamed Haris b, Mohd Zaki Nuawi b

a Electronic Department, College of Engineering, Diyala University, Iraqb Department of Mechanical and Materials Engineering, Faculty of Engineering and Built Environment, University Kebangsaan Malaysia, UKM, 43600 UKM Bangi,Selangor Darul Ehsan, Malaysia

a r t i c l e i n f o

Article history:Received 10 December 2013Received in revised form 30 June 2014Accepted 1 July 2014Available online 17 July 2014

Keywords:Elastic modulusPressure transmission coefficientRefractory metals and their alloys

a b s t r a c t

Refractory metals have attracted increasing interest in recent years because of their use in many high-temperature applications. However, the characteristics of these metals calculated using loaded tests(such as tensile strength tests) differ considerably from those calculated using one of the most famousmethods in NDT which is called time of flying of the wave (TOF).The present study presents two solutionsbased on calculating the pressure transmission coefficient (PTC) of the transmitted wave between the testsample and magnesium metal. The first is based on the development of a highly accurate algorithm thatlowers the cost by determining the acoustic impedance of the test specimen to calculating mechanicalproperties. Up to 26 theoretical tests were done (10 of these tests for refractory materials) accordingto their known mechanical properties to verify the accuracy of the algorithm. The convergence in resultsranged from 92% to 99%. The second solution was designed to solve the same problem for specimens witha thickness of less than 1 mm. Eight experimental tests were done (five using refractory materials) to ver-ify the accuracy of the second solution, with the convergence in the results ranging from 94% to 97%. Therelationships of the Vrms measured from the oscilloscope with the PTC and with the Fourier transformspectrum were derived. The results of this research were closer to the standard mechanical propertiesfor refractory metals compared with several recent acoustic tests.

� 2014 Elsevier B.V. All rights reserved.

1. Introduction

The role of refractory materials in applications such as X-raytargets, electrical contacts [1], and the nuclear power industry [2]is becoming increasingly difficult to ignore. However, the calcu-lated values of some mechanical properties (for refractory metalsand their alloys) from TOF usually differ considerably from theactual values in load testing (tensile strength test), with differencesranging from 1% to about 50%.

The thickness of the specimen also limits acoustic testing, i.e.,the pulse-echo technique [3] is only accurate when testing speci-mens at least 12.5 mm thick, and this hinders the using of this testin test samples such as the metal sheets, and plates. In addition thethickness (12.5 mm) means cost especially for the expensive met-als and alloys.

Elastic modulus (E) is one of the important characteristics ofmetals and alloys. Therefore, this research focuses on the test, thenthe use of elastic modulus for highly accurately determining theother mechanical properties of refractory metals. Hancock et al.[4] one of the first researchers who referred to the classificationthe Young modulus (E) into static and the dynamic modulus ofelasticity. Since 2003, [5] recommends defining elastic moduliusing two values. The first is the static elastic modulus (Es; orloaded elastic modulus) and the other type is the dynamicelastic modulus (ED; or unloaded elastic modulus), which can bedetermined through unloading tests such as TOF [6,7]. Physics con-version relationships are widely used in testing the properties ofmaterials, e.g., evaluating the time of flight (TOF) wave to deter-mine the acoustic longitudinal wave speed (CL) and transversewave speed (Cs), and then converting them to ETOF using the fol-lowing formula:

ETOF ¼C2

Lqð1þ mÞð1� 2mÞð1� mÞ ð1Þ

where ETOF is the modulus of elasticity calculated using the TOFmethod (in MPa), q is the density (kg/m3), and m is the Poisson’s

134 A.A. Mohammed et al. / Ultrasonics 55 (2015) 133–140

ratio m ¼ 1�2ðCL=CsÞ2

2�2ðCL=CsÞ2. CL and Cs are in meters per second. This method is

explained in detail by Ref. [3]. In fact the controversy revolvesaround the classification of the calculated modulus of elasticity cal-culated from loading and non-loading tests. This section explainedside of this controversy as below.

Ciccotti and Mulargia [8] used Eq. (1) to calculate ETOF andregarded ETOF is ED and then they compared it with the Es for seism-ogenic rocks (in the Italian Apennines); they found that the ED was10% greater than the Es. Furthermore, the difference between the Es

and ED increases with the density of the specimens [9].Recently several published papers studied this difference [10–

15]. On the other hand, many published studies that used acoustictests without referring to this difference and regarded the two (ED

and Es) as the same and mentioned them as the modulus of elastic-ity (E) [16–27].

This research offers new insights into improving the accuracy ofthe readings and lowering the minimum dimensions of the speci-mens that can be tested, by calculating the pressure transmissioncoefficient of the transmitted wave between test specimens of var-ious metals, especially refractory metals and magnesium, which isusually used as the control metal in all tests.

2. Theoretical part

The available data were collected from authorized sources of 26materials [28–30]. These data include the specifications of eightmetals and three alloys for refractory materials, as shown in bluein Table 1, whereas the other metals are non-refractory metals.In Table 1, Es represents the elastic modulus calculated from thetensile test, whereas ETOF is the elastic modulus calculated using

Table 1Mechanical properties of some metals and alloys.

Metal Name

υ CL ρ (kg/m3)

Es

(GPa)ETOF

(Gpa)RT

%

Magnesium 0.35 5740 1738 45 35.67 79.

Aluminum 0.35 6350 2699 70 67.8 96.

Beryllium 0.075 12800 1850 287 299.41 95.

Ti-6Al-4V 0.342 5800 4430 114 96.04 84.

Titanium 0.345 6100 4450 120 105.4 87.

Zirconium 0.38 4262 6506 97 63.13 65.

Zinc 0.249 4170 7133 104 103.55 99.

Niobium 0.397 3480 8570 104 49.53 47.

Vanadium 0.365 6000 6160 127.6 128.7 99.

Brass 0.34 4430 8520 100 108.63 91.

Silver 0.367 3640 10500 82.7 79.91 96.

Copper 0.343 4660 8941 130 124.62 95.

TitaniumCarbide

0.182 8270 5150 310 323.69 95.

Steel 4340 0.28 5850 7800 206 208.8 98.

Iron 0.29 5900 7800 196 207.19 94.

Chromium 0.21 6850 7190 279 299.7 92.

Cobalt 0.32 5730 8900 211 204.2 96.

Hafnium 0.26 3840 13310 141 160.4 86.

Nickel 0.312 5810 8902 199.5 215.46 91.

Tantalum 0.342 3400 16654 185.7 124.07 66.

Uranium 0.25 3370 18950 175 179.34 97.

Molybdenum 0.293 6370 10220 325 313.98 96.

Rhodium 0.26 6190 12410 379 388.62 97.

Ruthenium 0.25 6530 12370 432 439.55 98.

W Ni Fe 0.29 5040 17700 345 343.09 99.

Tungsten 0.28 5180 19300 411 405.08 98.

Eq. (1). RTOF represents the percentage of convergence between Es

and ETOF (as shown Table 1), where:

RTOF ¼ 100�

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiEs � ETOF

Es

� �2s

� 100 ð2Þ

Z in Table 1 represents acoustic impedance, where Z = q � CL.From calculated values of RTOF in Table 1, it is so clear that ES is

so close to ETOF for non-refractory metals like AL, Zn, Brass and Ag.This means that TOF is suitable for non-refractory metals. While, inthe same table, the values of RTOF showed mismatches or a big dif-ference between ES and ETOF for refractory metals and their alloys,especial for Ti, Zr, Nb, and Ta and their alloys. To find out the realreasons behind these results, it must refer to the origin of Eq. (1)which means the development equation of motion of wave in iso-tropic media which is Navier government equation as shown inbelow [31]:

ðkþlÞuj;ijþluj;ijþqfi ¼q€ui ði; j¼1;2;3Þ ðNavier government equationÞ

where k, l are Lame constants and u is the displacement of parti-cles. Eq. (1) was derived base on neglected the term of q � f i whichrepresents the body force [32], where the body force is forces thatacts throughout the volume of a body which are gravity, magnetic,electrostatic attraction [33]. In fact body forces can be neglected fornormal materials [32] such as AL, Zn, brass and Ag, therefore it canbe seen the acceptable correspond between ES and ETOF for thesemetals in Table 1. While this simplification is un-acceptable forthe materials have electrostatic force like Ti, Zr, Nb, and Ta. Theelectrostatic force (F) can be determined from the electric potentialenergy, this energy associated with the configuration of a particularinside the metal and this energy depend on two things first, the

OF Z (Kg/m2 s)

*106

ES× ρ GPa. kg/m3

PTC ES1

(GPa)RS %

28 9.97 78210 1.73 41.79 92.88

87 17.13 188930 1.35 69.27 98.95

67 23.68 530950 1.11 278.61 97.07

24 25.69 505020 1.05 112.9 99.03

83 27.14 534000 1.01 122.68 97.76

08 27.72 631082 1.01 89.7 92.47

57 29.74 741832 0.95 114.46 89.93

62 29.82 891280 0.95 103.93 99.93

13 36.96 786016 0.81 132.36 96.26

36 37.74 852000 0.8 102.35 97.64

63 38.22 868350 0.79 86.622 95.25

86 41.66 1162330 0.74 136.85 94.72

58 42.59 1596500 0.73 255.37 82.37

63 45.63 1606800 0.69 207.57 99.23

28 46.02 1528800 0.69 212.51 91.57

57 49.25 2006010 0.65 273.35 97.97

77 50.99 1877900 0.63 200.12 94.84

23 51.11 1876710 0.63 136.77 97

99 51.72 1775949 0.62 226.43 86.49

81 56.62 3092647 0.58 173.78 93.58

51 63.86 3316250 0.52 178.94 97.74

61 65.1 3321500 0.51 339.67 95.48

45 76.81 4703390 0.45 371.98 98.14

25 80.77 5343840 0.43 415.66 96.21

44 89.2 6106500 0.39 359.14 95.9

56 99.97 7932300 0.35 407.46 99.13

A.A. Mohammed et al. / Ultrasonics 55 (2015) 133–140 135

amount of electric charge (Q) that this particle has and its relativeposition to other electrically charged objects (electric field (EF)),therefore F = Q � E [34]. Hussain et al. [35] pointed out the intensityand values of the electrostatic of Zr-Silicate for different amounts ofthis material. Also [36] referred to the electrostatic properties ofTa2O5, TiO2, ZrO2. As well [37,38] proved this properties for Ti, like-wise [39] for Nb. In addition, it is well known that the piezoelectricmaterials are materials have ability to convert the mechanicalenergy to electric energy [40,41] and that means the particles ofpiezoelectric elements have an electrostatic force [42,43], and it iswell known that Ti and Zr are the essential elements in PZT (Leadzirconium titanate with the chemical formula Pb[ZrxTi1�x]O3

0 6 x 6 1), which is regarded one of the most famous materials inproducing piezoelectric elements. Ta is also regarded one of the ele-ments entering in manufacturing piezoelectric elements [44,45].Therefore the results of ES and ETOF are not identical for Ti, Zr, Nb,and Ta as shown in Table 1.

2.1. The theoretical test for proposed method

The main idea of this research is to calculate the pressure trans-mission coefficient (PTC) between the test specimen and Mg metal.Mg is mainly used because its acoustic impedance and E � q arethe lowest among metals, as shown in Table 1. This accounts forthe seemingly exponential relationship between the PTC (betweenMg and other metals or alloys) and the corresponding E � q forthese metals and alloys. To calculate the theoretical PTC, we useda setup with a Mg base that immerses the test sample in water,as shown in Fig. 1. where ZW, ZSP, and ZMg are the acousticimpedance of water, the test specimen, and Mg, respectively.PTC1 is the transmission coefficient between the water and the testspecimen, and PTC2 is transmission coefficient between the testspecimen and Mg.

PTC1 ¼2ZSP

ZW þ ZSPð3Þ

PTC2 ¼2ZMg

ZSP þ ZMgð4Þ

PTC ¼ PTC1 � PTC2 ð5Þ

Through the compensation equation (3) and (4) in (5)

PTC ¼ 4ZSP � ZMg

ZW þ ZSPð Þ ZSP þ ZMg� � ð6Þ

Incident wave

Reflected wave

Transmissionwave

WaterSpecimen

TestMg

FirstReflectedSecondReflected

PTC1

PTC2

ZW

ZSP

ZMg

Mg

Fig. 1. Transmission wave through water–test specimen–magnesium.

where ZW = 1.5 � 106 N s/m3 and ZMg = 9.97612 � 106 N s/m3

So PTC ¼ 39:9ZSP

1:5þ ZSPð Þ ZSP þ 9:97612ð Þ ð7Þ

Note that the value of PTC, in Eq. (7) depends on the value of ZSP.Using the acoustic impedances of the metals and alloys in Table 1,calculated using TOF for CL only, the values of PTC for all these met-als and alloys were calculated as shown in Table 1. For measuringCL only, It can be use the method which was explained in [3] or anyother methods of TOF.

The data was then arranged from the highest PTC value (magne-sium), which equals 1.73, to the lowest PTC value (tungsten), whichequals 0.35. Then, the relationship between E � q and PTC wasplotted, as shown in Fig. 2. By curve fitting all these points, the totalrelationship between E � q and PTC can be calculated. However, wedivided the curve into three sections to increase the accuracy and toreduce the order of polynomials to reduce the cost, thereby simpli-fying the corresponding electronic logic circuit.

The following equations were derived from Fig. 2:

1. The selected region in the dotted line (A1) is represented by thefollowing equation:

ES1¼�5:66211þ73:0387T�275:856T2þ425:253T3�234:885T4

q�107

ð8:aÞ

This equation can be used for materials that have an acousticimpedance of Z P 51 � 106 kg/m2 s [from cobalt to materials withhigher Z, as shown in Table 1].

1. For region A2 and for materials with a Z range of 51 �1066 Z 6 29 � 106 kg/m2 s [from niobium to chromium in

Table 1], the equation is as follows:

ES1 ¼�6:86293þ 58:1587T � 102:833T2 þ 52:8154T3

q� 106

ð8:bÞ

3. For region A3 and for materials with Z 6 29 � 106 kg/m2 s [frommagnesium to zinc in Table 1], the equation is as follows:

ES1¼2:55909�8:25426Tþ9:88962T2�5:196T3þ1:00751T4

q�108

ð8:cÞ

where ES1 is the modulus of elasticity calculated from the set ofEq. (8). Table 1 explains in detail the ES1 values from Eq. (8),whereas RS represents the percentage of convergence between Es

and ES1, where RS equals:

RS ¼ 100�

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiEs � ES1

Es

� �2s

� 100 ð9Þ

Knowing the CL of any material would then allow calculation ofits density using Archimedes law, and facilitates accurately calcu-lating ES1 directly using Eqs. (7) and (8), especially for refractorymetals and alloys. This method reduces the risk of error andreduces the cost of measuring devices.

However, Eqs. (7) and (8) enabled us to calculate the elasticmodulus accurately but the limitation on the dimensions of thespecimens test still exists. This issue was addressed in theexperimental part.

3. Experimental part

To achieve the second objective of this research, a practicalmodel was built based on the theoretical design [Fig. 1], as shown

Fig. 2. Relationship of E � q with T for refractory and non-refractory materials.

Fig. 3. (a) Testing the system using different samples; (b) magnified view of the setup (The mechanism test) with details.

136 A.A. Mohammed et al. / Ultrasonics 55 (2015) 133–140

in Fig. 3. This system included a pulse generator 10 MHz TGP110 toprovide the system with a square pulses. An Agilent 3000 Seriesoscilloscope, which can generate results as images and Excel files,was used in this study. The oscilloscope was controlled by connect-ing it to a computer or laptop, as shown in Fig. 3. The setup used inthe test was made of Mg and was manufactured in the Germany–Malaysian institute. The main reason for using Mg was explainedin the theoretical section; furthermore, the low Z of Mg allowsthe setup to maintain the intensity of the wave as it passes throughit. Two piezoelectric transducer disks (SM211) with a frequencyf = 7.2 MHz were used as the emitter and the receiver, as shownin Fig. 3a. An SM211 piezoelectric transducer was selected because

it has a high frequency, damping, and electromechanical couplingcoefficient; these advantages allow SM211 transducers to obtainhigh axial resolutions [32]. In addition, SM211 is low cost and iswidely available in the market [46]. For all these reasons togetherand because the response of SM211 very disciplined, therefore [41]built the mathematical model of this type of piezoelectrictransducer and compare it with the standard test of piezoelectrictransducer (sending and receiving) and got very good identitybetween the theoretical and experimental response. The emitterwas connected to the pulse generator whereas the receiver wasconnected to the oscilloscope. The dimensions of the two transducerswere 22 mm diameter (D) and 0.25 mm thickness. Each transducer

Fig. 4. Response of the system in Fig. 3a using tantalum as the test sample.

-0.00 -0.00 0.00 0.00 0.00

Time History

-0.02

-0.01

0.00

0.01

0.02

Vol

tage

(V

olte

)

Brass StTa

ALTi6AL4V

TiZr Nb

Fig. 5. Experimental time–voltage curve for all test samples.

A.A. Mohammed et al. / Ultrasonics 55 (2015) 133–140 137

has two silver electrodes. These transducers were affixed usingsilver epoxy. Silver epoxy offers the following advantages after1:1 mixing: (a) cold soldering has strong adhesion after it dries,(b) high printability on electrical conductors because its electricresistance is equal to 1.7 � 10�4 O m. The dimensions of Part A inFig. 3b was designed within the near field (N) of the emitter, whichwas equal to 226 mm, where the near field was calculated fromlow (N = D2f /4CLT), where CLT is the propagation velocity alongthe thickness of the transducer, which is equal to 3840 m/s.

The dimensions of the three parts (Part A–B and the frame) areillustrated in Fig. 3b.

Eight specimens test with dimensions 25 mm � 25 mm � 1 mmwere used. Four of the eight specimens were imported from theAdvent Research Materials Ltd. (Oakfield industrial Estate): zirco-nium, titanium, niobium, and tantalum. The alloys Ti6AL4V, steel4340, and brass were bought from Malaysian markets. The thicknessof the test specimen is the important dimension in these tests,where all of the test specimens have a thickness of 1 mm to avoidthe effect of the attenuation. The attenuation diagram [page 577in the appendix of [47] is illustrated on the right side of the diagram:for materials with thickness of 1 mm and less, the effect of the atten-uation is very small and almost the same for most materials. There-fore, we excluded the effect of the attenuation on the test specimens.

The test starts when the specimen, AL as an example, wasplaced in the small pool filled with water. Then, part A in Fig. 3bwas turned back to the original position, as shown in Fig. 3a, andthe reading was taken after pausing the picture for reading. Thistest was repeated for all specimens.

4. Results and discussion

Fig. 4 represents the response of the system in Fig. 3a using tan-talum as the test sample. During the test, the receiver responded

Table 2Experimental output voltage for each material.

Type of voltage (V) AL Ti6AL4V Titanium Zirco

Vpp 3.26E�02 3.12E�02 3.12E�02 3.0Vmax 1.88E�02 1.78E�02 1.82E�02 1.8Vmin �1.38E�02 �1.34E�02 �1.30E�02 �1.2Vavg �1.88E�04 �3.08E�04 �3.07E�04 �3.9Vamp 3.26E�02 3.12E�02 3.12E�02 3.0Vtop 1.88E�02 1.78E�02 1.82E�02 1.8Vbase �1.38E�02 �1.34E�02 �1.30E�02 �1.2Vrms 1.08E�03 1.06E�03 1.04E�03 1.0Vover (%) 60.70 140.40 141.70 144.7Vpre (%) 77.90 61.50 67.30 73.7

with yellow, whereas the right corner of the squire pulse fromthe pulse generator connected directly to second channel of theoscilloscope (in second channel of the pulse generator) respondedwith green.

The Agilent oscilloscope enabled providing the details of anyresponse, such as the wave in Fig. 4. These responses includedthe details of the voltages: Peak to peak (Vpp), Maximum (Vmax),Minimum (Vmin), Average (Vavg), Amplitude (Vamp), Top (Vtop), Base(Vbase), Vrms, Overshoot (Vover), and Pre-shoot (Vpre) as shown inTable 2.

To clarify the other side of the comparison, all these responseshave been put together to determine how the response changeswith the type of metal tested, as shown in Fig. 5.

Practically all these types of the voltage suffers from isolationduring the repeated readings (for the same test to achieve higherconfidence values) because of the slight vibration of the emergingwave except Vrms, which is commonly more stable than any othertype of voltage. Vrms was computed as follows:

V rms ¼

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiPni¼1x2

i

n

s;

where xi = value at ith point and n = number of points [48].Therefore, this research took this voltage as a base to distin-

guish between the metals.From the data in Tables 1 and 2, the relationship between Vrms

and E � q and the relationship between Vrms and PTC were plotted,as shown Fig. 4a and b.

nium Niobium Brass Steel Tantalum

4E�02 3.10E�02 3.04E�02 2.90E�02 2.66E�020E�02 1.84E�02 1.80E�02 1.72E�02 1.56E�024E�02 �1.26E�02 �1.24E�02 �1.18E�02 �1.10E�027E�04 �2.28E�04 �2.38E�04 �2.32E�04 �3.55E�044E�02 3.10E�02 3.04E�02 2.90E�02 2.66E�020E�02 1.84E�02 1.80E�02 1.72E�02 1.56E�024E�02 �1.26E�02 �1.24E�02 �1.18E�02 �1.10E�024E�03 1.00E�03 1.02E�03 9.59E�04 9.38E�040 143.20 144.70 70.30 60.200 66.50 68.40 77.20 75.90

0.92 0.94 0.96 0.98 1 1.02 1.04 1.06 1.08 1.10

0.5

1

1.5

2

2.5

3

3.5x 106

Vrms (mVolte)Mod

ulus

of

Ela

stic

ity*D

ensi

ty (

GPa

*kg/

m3 )

Ta

St

AL

Ti6AL4VTi

ZrBrass

Nb

a1=2.96e+006b1=0.9317c1=0.02712a2=8.732e+005b2=1.004c2=0.06301

Y=a1*exp(-((x-b1)/c1)2+a2*exp(-((x-b2)/c2)2)

(a) (b)

Fig. 6. Experimental relationship of Vrms (a) with elastic modulus � Density; (b) relationship with transmission coefficient (PTC).

0.00 200.00 400.00 600.00Number of Samples

0.00

0.05

0.10

0.15

0.20

0.25

Four

ier

Spec

trum

AL

Ti6AL4VTiZr

NbBrass

StTa

Fig. 7. Experimental comparison among the Fourier transform spectrum of allspecimen tests.

Vrms (mVolte)

11.50

12.00

12.50

13.00

13.50

14.00

Four

ier

Spec

trum

Ta

St

Brass

Nb

Zr

Ti

Ti6AL4V

AL

0.92 0.96 1.00 1.04 1.08 1.12

(a)

Fig. 8. Relationship of total Fourier transform spectrum w

138 A.A. Mohammed et al. / Ultrasonics 55 (2015) 133–140

The two relationships in Fig. 6a and b proves the sensitivity ofVrms to changes in material properties. Fig. 6a illustrates the rela-tionship of Vrms caused by changes in the materials (from specimentest to Mg) with the magnitude of E � q for these specimen tests.

The experimental results of these tests prove the direct rela-tionship between Vrms and E � q, which equals the following:forrange of Vrms (0.9 6 Vrms 6 1.08) mV

ES2 � q ¼ a1e� Vrms�b1

c1

n o2

þ a2e� Vrms�b2

c2

n o2

ð10Þ

where ES2 modulus of elasticity calculated from experimental part,a1 = 2.96 � 106, b1 = 0.9317, c1 = 0.02712, a2 = 8.732 � 105,b2 = 1.004, and c2 = 0.06301.

Fig. 4b proves the positive relationship between Vrms and PTC.This relationship is represented by the following equation:forrange of Vrms (0.9 6 Vrms 6 1.08) mV

PTC ¼ a1 sinðb1 � V rms þ c1Þ þ a2 sinðb2 � V rms þ c2Þþ a3sinðb3 � V rms þ c3Þ ð11Þ

where a1 = 9.643, b1 = 3.17, c1 = 15.76, a2 = 1.877, b1 = 15.96,c1 = 24.83, a3 = 0.0922, b3 = 120.4, and c1 = 30.11.

0.40 0.60 0.80 1.00 1.20 1.40

Transmission Coefficient (T)

11.50

12.00

12.50

13.00

13.50

14.00

Four

ier

Spec

trum

Ta

St

Brass

Nb

Zr

Ti

Ti6AL4V

AL(b)

ith (a) Vrms and with (b) transmission coefficient (T).

0

10

20

30

40

50

60

70

80

Ti-6Al-4V Titanium Zirconium Hafnium Tantalum

G1 in (Gpa)

Gs in (Gpa)

G2 in (Gpa)

Fig. 9. The difference between G1 and G2 is proportional to the value of Gs.

A.A. Mohammed et al. / Ultrasonics 55 (2015) 133–140 139

Fig. 7 was plotted to determine the amount of energy transmit-ted in each test and to perform a comparison among them.Thesmall colored lines on the Y-axis (Fourier transform spectra) inFig. 7 represent the spectra that were plotted on this axis. Fig. 7gives an impression of the behavior of the energy transition, butit does not provide a clear comparison between these metals.The total energy (total Fourier transform spectrum) of each metalseparately provides a distinguishing relationship of the Fouriertransform spectrum with Vrms and with PTC, as shown in Fig. 8aand b.

Calculating ES1 using the algorithm in Eq. (8) or using ES2 in thenew algorithm in Fig. 8 (for thin specimens) has been proven the-oretically and practically. The other mechanical properties werecalculated by deriving Eq. (1) to calculate the Poisson’s ratio:

m ¼ðE� KÞ þ

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiðE� KÞ2 þ 8kðk� EÞ

q4K

ð12Þ

where k ¼ C2L � q.

Using Eqs. (12) and (8) or (10) (depending on dimensions of thespecimen), several mechanical properties were accurately calcu-lated (for refractory metal) using physics conversion formulas.These properties include transverse waves, shear elastic modulus(G), bulk modulus, Lamé’s first parameter, and P-wave modulus.Table 1 demonstrates the difference in accuracy among ES, ETOF,

and ES1, whereas Fig. 9 illustrates the difference in accuracy (forsome of refractory metals and their alloys) among shear modulus(G1) calculated using the method explained by Ref. [3] and shearelastic modulus (G2), which was calculated using Eqs. (8) and(12), whereas Gs was calculated using a normal shearing test [28].

5. Conclusion

We drew the following conclusions based on the results

� The mechanical properties of refractory metals and their alloyscalculated using conventional tests (loaded tests) differ signifi-cantly from those using recent acoustic tests (TOF).� This research addresses the discrepancies in the elastic moduli

of refractory metals and their alloys by developing a new algo-rithm that calculates the pressure transmission coefficient of atransmitted wave. It also proposes a method determining theproperties of materials with thickness of 1 mm.� The new method is less costly and less prone to error because it

depends on the use of piezoelectric transducers that generatelongitudinal waves and do not require the use of shearingvelocity.

� Vrms is the most stable voltage parameters measured duringthese acoustic tests, which depends on the output voltage.� Vrms can be considered an indicator for Fourier transform spec-

trum in these types of tests.� Transmission coefficient (PTC) is directly proportional to both

Vrms and Fourier transform spectrum.� The E � q in the specimen tests exhibits an exponential rela-

tionship with Vrms.

Acknowledgements

Authors would like to thank Prof. Prakash Thamburaja, atUniversity Kebangsaan Malaysia (UKM), for the important discussionwith him.

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