ANALYSIS OF NANOBRIDGE TESTS

Acta Mechanica Solida Sinica, Vol. 23, No. 4, August, 2010ISSN 0894-9166Published by AMSS Press, Wuhan, ChinaANALYSIS OF NA NOBRIDGE TESTS**Wing Kin ChanJianrong LiYong WangShengyao ZhangTongyi Zhang*(Department of Mechanical Engineering, Hong Kong University of Science and Technology,Clear Water Bay, Kowloon, Hong Kong, China)Received 20 January 2010, revision received 30 June 2010ABSTRACT This paper analyzes nanobridge tests with consideration of adhesive contact defor-mation, which occurs between a probe tip and a tested nanobeam, and deformation of a substrateor template that supports the tested nanobeam. Analytical displacement- load relation, includingadhesive contact deformation and substrate deformation, is presented here for small deformationof bending. The analytic results are confirmed by finite element analysis. If adhesive contact de-formation and substrate deformation are not considered in the analysis of nanobridge test data,they might lead to lower values of Young's modulus of tested nanobeams.KEY WORDS nanobridge tests, size-dependency, adhesion, contact compliance, surface effect,substrate effect, finite elementI. INTRODUCTIONNanowires are one-dimensional materials with lateral dimension less than 100 nm. Different fromtheir bulk counterparts, nanowires have unique material properties and important applications innanotechnologies1. Reliably characterizing mechanical properties of nanowires and understanding theirbehaviors are of academic merits and applicational importance to nanoscience and nanotechnologies.Nanobridge tests are widely used to evaluate the Young s modulus of nanowiresl2 -12), in which nanowiresare cross trenches or holes and clamped on substrates or templates. An atomic force microscope (AFM)is usually used as a testing machine to bend the double- clamped nanowire vertically. Figure 1(a) is aschematic depiction of the nanobridge test.During the nanobridge test, load and displacement of the probe tip are recorded. In current practice,displacement of the probe tip is regarded as the deflection of bending and classical Euler's beamtheory under small deformation is used to analyze experimental data. Euler's beam theory under smalldeformation gives a linear relation between deflection and load. The bending compliance is defined asthe slope of deflection versus load. When a load is applied on the center of a nanobridge and deflectionis measured at the loading point, the compliance based on Euler's beam theory with clamped ends isgiven by3Ceuler = 192EwwI(1where ENw is the Young's modulus of the nanobeam in the axial direction, I is the second momentof the cross-sectional area, and L is the length of the nanobridge. Rewriting Eq.(1), one calculates theYoung's modulus by ENw = KEuler/CEuler with KEuler = L*/(192I). In the nanobridge test, however,★Corresponding author. Tel: (852) 2358-7192, Fax: (852) 2358 1543, Email: mezhangt@ust.hk** Project supported by an Earmarked Research Grant (No. 622506) from the Research GrantsKong Special Administrative Region, China. The authors would like to thank Mr. Zis helpful discussion.中国煤化工fHCNM HG284.ACTA MECHANICA SOLIDA SINICA2010AFM tip'ln/≤CmorCme SuyMI@伞SsQCEuler(a(bFig. 1. (a) Schematic depiction of nanobridge tests where an AFM tip loads a nanowire vertically; (b) Modeling adhesivecontact deformation and substrate deformation with springs.adhesion contact deformation inevitably occurs between a probe tip and a tested nanobeam and substratedeformation also happens unavoidably. Adhesion contact deformation and substrate deformation arecalled extrinsic deformations in the present work, which are not considered in the current practice ofnanobridge tests. Young s modulus of a nanowire determined from the nanobridge test is lower due toextrinsic deformations, if one does not consider adhesion contact deformation and substrate deformationin the analysis of test data. There are many reports showing that the Young's modulus of a nanowiremeasured by the nanobridge test is smaller than that of its bulk counterpart. For example, the measuredYoung' s modulus of ZnS nanobelts exhibits a 10%-40% reduction compared with the bulk valuel6l. Themeasured Young's modulus of ZnO nanobelts ranges from 32 to 57 GPal10), which is much smaller thanthe bulk value of 140 GPa[13]. The reported Young's modulus of GaN nanowires has an average valueof 43.9 GPa, much smaller than the bulk value of 293 GPal8l. There might be different mechanismsfor the reported lower values of Young's modulus determined from the nanobridge test. The presentanalysis will demonstrate that if adhesion deformation and substrate deformation are not considered inthe data analysis, they will decrease the measured Young' s modulus of a nanowire from the nanobridgetest. The effects of adhesion deformation and substrate deformation on determined Young's modulus,which are called extrinsic factors in the present work, depend on the size of a tested nanowire, therebyleading to size-dependent apparent Young s modulus.Nanowires may exhibit size-dependent Young's modulus due to intrinsic mechanism such as thesurface related intrinsic effect. In the nanobridge test on a nanowire, the intrinsic and extrinsic factorsfunction jointly. If extrinsic factors are not considered in the data analysis, nanowires of the same materialmay show the smaller-the stiffer behavior, but the Young 's modulus is still lower than that of the bulkcounterpart, as reported in the literaturel7,11,121. Clearly, the extrinsic factors must be considered inthe analysis of nanobridge test data in order to have intrinsic mechanical properties. The present studycontact deformation and substrate deformation. The following reported analysis results allow one toanalyze nanobridge test data with the consideration of extrinsic deformations so that Young s modulusof a tested nanowire can be accurately determined from the nanobridge test.In macroscopic scales, adhesive force between two solid bodies in contact is generally small due toin microbridge testsl4 16]. In nanometer scales, however, adhesion occurs always between two solidbodies in. contactl17,18l. There are several approaches to include adhesion into the analysis of contactproblemsl19 21), among which the JKR theory is easy to be analytically manipulated. In the presentstudy, we luse the JKR theory to analyze adhesive contact deformation between a probe tip and ananobeam. The effects of substrate deformation have been studied in micro/ nanobending testsl9.22-251In practice, an extra length can be added to the real beam length to include the infuence of substratedeformation, as Baker and Nixl22] did in microcantilever tests and Wong et al.l23)] in nanocantilevertests. In a review article, Zhangl16l systematically described how to model substrate deformation inmicrobridge tests by using two coupled springs for small deformation and three coupled springs for largedeformation. Following the approachl6l, we use two coupled springs to madel substrateformation in中国煤化工MHCNM HGVol. 23, No.4Wing Kin Chan et al: Analysis of Nanobridge Testsnanobridge tests for small deformation. With these approaches, we are able to study nanobridge testsanalytically and clarify quantitatively adhesion deformation and substrate deformation.II. ADHESION IN NANOBRIDGE TESTS2.1. Analytical Contact Model in Nanobridge TestsAn AFM tip is assumed to have a half spherical shape with a radius, Rtip, and a tested nanowireis assumed to be a circular cylinder with a radius of RNw. When the tip contacts with the nanowire,the contact area is elliptical. The principal relative radi of curvature of surfaces, R' and R", take theformsR'=(-+1) =Rup, R”=(1+1(2)、Rtip+ Rvw)To use the JKR model19,26], we assume that the radi of the elliptical contact area is much smallerthan Rtip and RNw. We set a local Cartesian coordinate system with its origin at the ellipse centerand the coordinate axes, x and y, parallel with semi-axes a and b of the llipse, respectively. Accordingto the JKR theory, the pressure exerted by the AFM tip on the nanowire is given by/2p(x,) =p[-()" -(%)]”+1m[l+()-()]“(3)where the first term with Po in the right side of Eq.(3) corresponds to the pressure that gives a uniformdisplacement in the z-direction. The second term with P1 corresponds to the Hertz pressure distribution, .which is exerted between two frictionless solids in contact. Based on the elasticity theory, the pressurefield of Eq.(3) produces the following displacementl26Uz :2pobK(e)」p1b[k(e)- K(e)- E(e)z2_ E(e)/(1-e2)- K(e),21(4)E*e2a2 .e2a2where E* is the reduced modulus defined by1_ 1-喝+ 1-呢wE- EtipENwVtip and UNw, Etip and ENw are Poisson's ratios and Young's moduli of the probe tip and the testednanobeam, respectively; e =√1- b2/a2 is the eccentricity; and K(e) and E(e) are respectively theelliptic integrals of the first and second kinds,K(e)= (1-esin20)- 1/2d0， E(e)= (1 - esin2 0)}/2d0The geometric relation with the two kinds of elliptic integrals is given by Johnson[26]K(e)- E(e)R(7)a2E(e)/b2-K(e)= Rwith the definition of eccentricitye=√1- b2/a2, Eq.(7) gives the relationship between a and b.The elastic displacement can also be expressed by the principal relative radi of curvature asx2(8)2R- 2R"where δ is the displacement of the probe tip. Comparing Eq.(4) with Eq.(8) yields_E*{e2a2K(e) + 28R'[E(e)- K(e)}(9a)4bR'K(e)[E(e)一K(e)]E* e2a2(9b)PI = 26R'[K(e)- E(e)中国煤化工MHCNM HG286 .ACTA MECHANICA SOLIDA SINICA2010Substituting Eqs.(9a) and (9b) into Eq.(3) gives the pressure distribution. Integrating the pressure overthe contact area yields the applied force:E* πa{68R'E(e)- K(e)]-a2e2K(e)}P= | p(x, y)drdy =6R[K(e)- E()]K(e)(10)In the nanobridge test, a probe tip penetrates into a tested beam for some distance, which is identicalto the above discussed adhesion contact displacement δ at the tip. In addition, the tested nanobeamdeflects and has bending deflection Ub at the loading point. Therefore, the total displacement of theprobe tip is given byu= ub+δ(11a)The bending deflection ub can be calculated using the beam bending theory[14 16]. Under linear smalldeflection, the bending deflection is linearly proportional to the load, i.e., up= C.P, where C representsthe bending compliance with or without the consideration of substrate deformation, which will be clarifiedwhen needed. Thus, the total displacement of the probe tip given by Eq.(11a) is rewritten asu=C.P+δ(11b)Equation (11b) provides a relationship among the three variables, u, P and δ. As indicated in Eq.(10),the applied load is related to semi-major axis a of the contact area. To have the relationship betweendisplacement u and load P, which are measurable in the nanobridge test, we must determine a as afunction of u. Thus, we express P and 8 in terms of u and a, and then determine a by energy minimizationbased on the JKR theory. With Eqs.(11b), (10) and (4), we haveK(e){CE*. πa3e2 + 6uR'[K(e)- E()]}6(u,a)=- 6R([K(e) 一Ele)lCE*ra+ K(e)[(12a)P(u, a) =E*ra{6uR'[K(e)- E(e)] -a2e2K()}(12b)6R'[K(e)- E(e)][CE*πa+ K(e)]The elastic strain energy only due to adhesion contact is calculated to beUeln=j | p(x,y)u.drdy = πaE*( a4e2[R" + R'(1-e2)]8δa2e2δa2[R"+R'(1-e2)]]l 40(R)2[K(e)- E(e)R"T 2K(e)~ 12R'[K(e)- E(e)]12F'Ke) 2} (a3)The elastic strain energy due to bending is given byUo2= 2CP2(14)The surface energy Ug is the product of the contact surface area and the adhesive surface energy perunit area r, i.e,U。= γπab= γπa2√1-e2(15)Combining Eqs.(13), (14) and (15) yields the total free energy:Utotal=Ue1 + Ue2- Us(16)which is a function of a and u. For a given u, minimization of Utotal with respect to a determines thevalue of a, meaning thataUtotal |=0(17a)82Utotal 1一>0(17b)8a2中国煤化工MHCNM HGVol. 23, No.4Wing Kin Chan et al: Analysis of Nanobridge Tests150 Equations (17a) and (17b) determine the valule of a as a fumctionot，, with which E2a) and (12b)give the probe tip displacement due to adhesion contact and the load as functions of u.When no adhesion exists between a probe tip and a tested nanobeam, the derivation is reduced to thatfor Hertz contact. Appendix A describes in detail the analysis of nanobridge tests with Hertz contact.Furthermore, the AFM probe may be pressed on a flat lattice pl60 of a nanowire or a nanobelt, whennanobridge tests are conducted on nanobelts or nanowires with rectangular or hexagonal CrOSS- sections.In such a case, the contact area will be circular. The above described analysis of nanobridge tests withelliptical adhesion contact can be reduced to that for circular adhesion contact, which is described inAppendix B for detail.1400i002.2. Results from Analytical Contact ModelSince substrate deformation can be separately considered from adhesion contact, we first analyzethe role of adhesion contact in the nanobridge test without considering substrate deformation. Withdiferent values of γ and the input data: ENw = 200 GPa, Rtip 1290 nm, RNw = 50 nm, and L = 1000nm, we plot the load P US. tip displacement u in Fig.2(a) for 0.1 nm≤u≤1 nm and in Fig.2(b)for 1 nm≤u≤10 nm to illustrate clearly the effect of adhesion contact. Due to adhesion, there arenegative forces in the P-u plots, indicating that the probe tip and the tested nanobeam attract eachother and this attraction becomes stronger when the adhesjon enrsy 7 is higher, as shown in Fig.2(a).550 As expected, the P-u curve with adhesion approaches tothat y4the Hertz solution when the valueof γ is reduced to zero. Nevertheless, the relation between P and u is nonlinear if adhesion contactor Hertz contact is considered, especially, in the small tip displacement region, as shown in Fig.2(a).The nonlinearity raises a question that is how to take the slope of load vUS. tip displacement curve andhow much error will be induced if the slope is taken from fergtregions of the P-u curves. With the .input data, the compliance of an Euler nanobridge is 5.30hich is called the Euler complianceand serves here as a reference. From the Hertz solution, the compliances by linear ftting of P Us. ufortheregionsof0.1nm≤u≤1nm,1nm≤u≤10nmand10nm≤u≤30nmare7.32m/N,6.43 m/N and 6.16 m/N, respectively, implying that the effect of contact compliance decreases slowly磊0 during the loading. When the tip displacement is suficienEy larf6OOthe compliance with Hertz contactwill approach the Euler compliance. The inset of Fig.2(b) shows that the P-u curve of an Euler beam,where u is linearly proportional to P and the slope of u Us. P gives the Euler compliance, is nearlyparallel to that of a nanobeam with Hertz contact deformation.一400150]rtz、1800]| Euler beam1600 I001400]Euler beam120050γ= 0.4N/m00 ]-50800]75600 ]号65400]|γ= 0.1N/n--50200]。FEwihHertrontootHertz contact35404550-100Y= 1.0N/ml-20(0.6-0.4-0.20.00.20.40.60.81.0Tip displacement u (nm)TIp dsplacement u (nm)a)b)Fig. 2. Tip displacement-load curves for adhesive contact with different values of r and for Hertz contact, (a)0.1nm≤u≤1| nm and (b) 1nm≤u≤10 nm and the inset for 30nm≤u≤50 nm, where200=20nm,RNw=50nmandL=1000nm.-0.6tQe4oQ12odpbiabnle wftk thle konideraltibon of adhesion contact iscalled the Jkr4omplignce Cj&R,10 12which is related to 7, ENw, Rtip, RNw and L. As discussed above for Hertz contact, the average slope ofu-P varies with the linear fitting region of u-P curves. Therefore, we take the JK R compliance from tworegions, 0.1 nm≤u≤1 nm and 1 nm≤u≤10 nm, and normalize it by the Euler compliance. Firstly,we study the effects of 7 and ENw, which represent, respectively, the intrinsic strenoth of adhesion and中国煤化工Tipspcertr u(nn)fYHCN MHGMACTA MECHANICA SOLIDA SINICA2010the elastic property of a tested nanobeam. We use geometric dimensions of Rtip= 20 nm, RNw = 50nm, and L = 1000 nm to plot curves of CJkR/CEuler US. γ for different values of ENw in Fig.3. The ratio1.9Crkr/CEuler decreases with increasing r monotonically for each given value of ENw, while for a givenvalue of 7, CJkR/CEuler increases with increasing ENw. This behavior occurs in both tip displacementregions,asshowninFig.3(a)for0.1nm≤u≤1nmandinFig.3(b)for1nm≤u≤10nm.Theeffect of geometric parameters Rtip, RNw and L on the JKRftompliance may be studied with twodimensionless variables of RNw/L and Rtip/RNw. For example, we use Rtip/RNw = 2.5, ENw = 6001.8GPa and γ = 0.4 N/m to calculate CJkr/CEuler as a function of L, ranging from 1000 nm to 1500nm, for RNw = 25, 27.5, 50, 62.5 and 75 nm, which are plotted in Fig.4(a) for 0.1 nm≤u≤1 nmand in Fig.4(b) for 1 nm≤u≤10 nm. As expected, a longer beam with a smaller radius makes theclassical Euler beam theory more appropriate. On the other hand, a larger RNw and/or a shorter Lincrease the contact deformation, as shown in Fig.4. Although this trend is the same in both fitting1.7regions, the value of CJKR fitted in the large tip displacementytegion of 1 nm≤u≤10 nm is muchcloser to CEuler than that fitted from the short tip displacement Tegion of0.1 nm≤u≤1 nm, as shownby Figs. 4(a) and 4(b). For a given radius of nanobeams, as mentioned above, a longer beam is moreappropriate when using the classical Euler beam theory. Figure 5(a) clearly illustrates this behavior byplotting CJkr/CEuler vS. L/Rww for RNw = 50 nm, and Btip = 10, 20 and 30 nm. Figure 5(a) showsalso that a large tip radius reduces the effect of adhesive contact deformation. This is because for a名1.6given load, a probe tip with a lager radius will penetrate jless into the tested nanobeam. Figure 5(b)shows CJkr/CEuler us. L/Rtip for Rtip= 20 nm, and RNw = 3040 and 50 nm, indicating again thata longer and thinner beam is more appropriate when using the classical Euler beam theory.|1.122ENw= 800 GPa占1.51.32-Enw= 600 GPaENw =.800 GPa1.311.30-1.121 t1 291Ew = 000 GPeζ 1.120ENw= 400 GPa198。1.41入ENw= 400 GPa .1.275ENw= 200 GPa.119 tErw=200GPa.1.25L0.0 0.2 0.4 0.6 0.8 1.011. 0.2 0.4 0.6 0.8 1.0γ (N/m)r (N/m)(a)0.1nm≤u≤ 1mm(b) 1nm≤u≤10 omCFig. 3. CjkR/CEuler as a function of r with diferent Eww values,CrRjis Arearly fted in regions, where Rtip = 20 nm,RNw=50nmandL=1000nm.1.21.9-1.30Rww=775 nm1.25RNw= 75 nm1.11.5FRw-62.5m l高1.20 ~ Rrw-62.5 nm=25 nm!1.15Rww=50m号13-Rw-5m5 1.10RNw=37.5 mm1.2 RNw-37.5 n1.05RNw=25 nm1.00仁BNw=25 n1.(10000 1100 1200 1300 1400 15001000 1100 1200 1300 1400 1500L (nm)(a)0.1nm≤u≤1 om(b) 1nm≤u≤10 nm100 500012010140000 R500* CIKR slime fte 11010200二0014000 15002.3. Finite Element Analysis with Adhesion ContactIn this section, we implement finite element (FE) analysis of nanobridge tests with adhesion contact to中国煤化工YHCNM HG(ao.1nmk钆≤1nm(b)1m<饥≤10 nmVol. 23, No. 4Wing Kin Chan et al: Analysis of Nanobridge Tests289.1.4151.4-13RNw-50 nm1.3-20m1.2-号1.2-tLmm导40mm.1.1Rip= 30 nm ==--------===== .10元25301.03050L/RwL/ Rupb)Fig. 5. C.kr/CEuler against L/Rww for diferent Rtip values, where CJKR is linearly fitted in regions 0.1 nm≤u≤1nm and 1 nm≤u≤10 nm and plotted by solid and dashed lines, respectively, where ENw = 200 GPa and γ= 0.4 N/mare input data in both (a) and (b), and RNw = 50 nm and Rtip = 10, 20 and 30 nm in (a), whereas Rtip = 20 nm,RNw=30,40and50nmin(b).14confirm the analytic results. The commercial code, ABAQUS, is adopted here. We use the Lennard-Jones(LJ) potential to describe adhesion, in which the contact pressure is given by p(a) = ψ[(do/a)]0 -(do/a)41/6, where do is the equilibrium distance, ψ is the adhesion stiffness per unit area, a is theseparation between two interaction points. The adhesive energy γ is defined as the work done per unitarea to separate two unit faces from the equilibrium position and is calculatedby γ= Jdo p(a)da, which13 gives γ = rbdo/27. In this study, do = 0.2 nm is used. The contact pressure force given by the LJ potentialis programmed through the user subroutine, UINTER. The load-tip displacement relation is governedby five variables ψ, ENw, Rtip, RNw and L. For simplicity, we fixed rPtip at 20 nm because this is a typicalvalue for a sharp AFM tip. RNw and L are examined from 20 nm to 50 nm and from 800 nm to 1500 nm,respectively, as listed in Table 1. Two valuesofψ= 13.5 GPa(γ= 0.1N/m) andψ= 1.35 GPa(γ= 0.01N/m) and various values of ENw, from 200 GPa to 800 GPa, are used in the FE analysis. The probe tip isloaded at the middle of a tested nanobridge and only a quarter of the studied symmetric system is used in2the FE analysis. Figure 6 shows the used FE mesh, whose density is high at and near the contact region.The clamped boundary condition is adopted atthe ends of the nanobridge without consideringsubstrate deformation. The AFM tip is treated asrigid and its vertical movement is prescribed. Thetip displacement u is zero when the tip is 0.2 nmaway from the tested nanobeam. The total forceacting on the tip is calculated for each prescribedtip displacement from 0.0 nm to 1 nm. To compare .the FE results with the analytic results, we plottwo FE load-tip displacement curves in Fig.2(a)，Fig.6 Finite element meshes showing high density near thecontact region.Table 1. Ratio of Cr to C]KRL/Rww104034.328.6 二242016L (nm)10-_ γ(N/m) Eww (GPa)1000 8001500 1200 1$0100012008002000.9796 0.9869 0.9879 0.9899 1 .0095 1 .0039 1 .0238 1 .0355 1 .08854000.98220.9896 0.99150.9955l .0176. 1.0074 1.0332 1.0417 1.0967).16050A820 0.216 0.9400.99971.02230 1.0094 50388 1.04400 1.1011 900.9832 0.9933 0.9958 1.0028 1.0264 1.0107 1.0429 1.0471 1.10390.9915 1.0027 1.0032 1.0037 1.0220 10235 1.0349 1.0550 1.11670.9974 1.0084 1.0070 1.0072 1.0252 1.0268 1.0384 1.0588 1.12040.016001.0007 1.0116 1.00881.00881 .02681.0283 1.0402 1.0609 1.12271.00301.0137 1.00991.00999 1 .0279 1. 0293 1 .04131. ( )623 1. 1243L/RW中国煤化工THCNM HG290 .ACTA MECHANICA SOLIDA SINICA2010showing that the FE results are close to the analytical results. Again, the average total compliance iscalculated from linear fitting ofu vs. P curve in the region0.1 nm≤u≤1 nm. The FE results showthat the dependence of the FE compliance CRr; on parameters ψ (r), ENw, Rtip, RNw and L are allcorrespondingly the same as the dependence of the analytic compliance CJKR. Table 1 lists the ratio ofCFRR to CJKR, indicating that the difference between them is below 6.3% except for the L/RNw= 16case, in which the shear deformation is noticeable and has an effect on CjRR. The good agreementbetween the FE results and the analytic results verifies the analytic study.2.4. Contact Spring Compliancewe may model it as an additional spring attached between the probe tip and the tested nanobeam, asillustrated in Fig.1(b), and the spring compliance for adhesion contact Cac is calculated fromCac = CJkR - Ceuler(18)Based on the dimension analysis, we deduce an expression for adhesive contact compliance asA( ENw RNw( RNw”(RNw)Cac = EwwRww ()"(Reip)(Rirw)(19a)orKac(ENw)Cac= -ENw(19b)where A, m, n, P, and q are fitting parameters and Umax is the upper limit of the tip displacement inthe ftting region. Note that Kiac might still be a weak function of ENw. To determine the values ofA, m, n, p and q, we compute as many values of Cac as possible with various values of ENw, RNw,Rtip, L and 7, and for diferent values of UImax. The analytical approach is adopted here because of its .analytic nature. Thus, Rtip = 10, 20, 30 and 40 nm; RNw = 20, 30, 40 and 50 nm; L = 800, 1000, 1200,1400, 1600, 1800 and 2000 nm; ENw = 200, 400, 600, 800 and 1000 GPa; γ = 0.1, 0.4 and 1 N/m;do= 0.2 nm; and the tip displacement regions fromu= 1 nm to umax =5, 10 and 20 nm are used inthe computation, which leads to 5040 combinations. Fitting the 5040 cases gives0.08180.34890.74220.2549( 0.1538Evw RNw'Kac =( Rww)(19c)RNw(Rw)RNw ;Umax ;with correlation coefficient 0.90.Similarly, we calculate the total compliance of the nanobeam including Hertz contact CHertz. Thespring compliance for the Hertz contact Che is calculated fromChe = CHertz - CEuler(20)The adhesive surface energy per unit area γ is not involved in Hertz contact so that the dimensionanalysis result is reduced toChe( Rww(21a)ENwRNw \ Rtip)"(点)”(n)Umax,Khe_Che=(21b)Then, fitting the numerical results calculated from the analytic approach with the same parameters asmentioned above for the JKR contact, we have0.27810.95660.3230( 0.1814/Rww\'RNwRww) RtipRw(21c)Equations (19) and (21) allow us to estimate the additional compliances induced by adhesive contactand Hertz contact, respectively.中国煤化工MHCNM HGVol. 23, No. 4Wing Kin Chan et al: Analysis of Nanobridge Tests291III. SUBSTRATE DEFORMATION IN NANOBRIDGE TESTSZhang et al. developed a theoretical approach to include substrate deformation into the bendinganalysis for microbridge tests under large deformationl15, 16,27] and for microcantilever testsl16,27]. Undersmall deformation, longitudinal stretching can be ignored and thus the number of coupled springsis reduced to two, which correspond to the rotational and vertical displacement, respectively. Theconstitutive equation of the two coupled springs is thus given byl14, 16,28]:[ SQq Sqm(22){}=ISMq SMM.]{%}where Sij(i,j= Q, M) rept; the substrate compliances reflecting the substrate deformation undera unit load and/or a moment; Q and M are the vertical load and moment, respectively, and δ and θare the vertical displacement and rotation angle of the substrate, respectively, at the joint location ofsubstrate with a tested nanowire. Based on the Betti-Rayleigh reciprocal theorem, SqM is equal toSMQ. Incorporating substrate deformation into the boundary conditions at nanobridge ends, we derivethe total compliance for an Euler beam with substrate deformation asL3[SMq- 12/(8EtI)]2. SQQ .CSD= 48ENwI 2[SMM+ 1(2E1)]十2The substrate compliances Sij are calculated out from FE analysis. In general, the substrate compliancesSij are the function of Young's modulus ENw and Poisson's ratio Nw of a tested nanobeam, Young'smodulus Es and Poisson s ratio vs of substrate and nanowire radius RNw. In the present work, VNw andvs are assumed to be constants, independent of RNw. Based on dimensional analysis, we can expressSij asaMM( ENwbMMaQQ_ ( ENw，( Eww1QSMM =SqEs(RNw)( EsEsRw Es,SMQ = Es(Rw)( Eswhere aij and bij (i, j = Q, M) are parameters to be determined from ftting finite element analysisresults. In the present analysis, the substrate is taken to be silicon with a Reuss average Young s modulusof 159 GPa and a Poson's ratio of 0.25. A nanowire is adhered on the substrate by Van de Waalsforce firmly, which is elucidated in Appendix C. The used FE model is schematically shown in Fig.7.substrate through a rigid cylinder. The rigid cylinder has the same cross -section as the nanowire, andits dimension along the beam length is infinitesimal. Under a given load or moment, we calculate thedisplacement and rotation of the rigid cylinder to determine the values of Sij. The rigid cylinder makesthe deformation uniformly cross the connection edge of the substrate so that the substrate deformationappropriately links with the beam deformation. The values of Sij are calculated for a range of ENw/Esfrom 0.05 to 50 and RNw ranging from 8 nm to 48 nm. Then, ftting the calculated values of Sij withEq.(24) yields the values of aij and bij; which are tabulated in Table 2.Q,8|一RNwM, 0Rigid block勇ViewAL wfixedL-wfixed.ViewB”37.5RNwZ,50RtNwz, wView B.Fig. 7. Schematic diagram used in the finite element calculations of spring complia中国煤化工YHCNM HG50RNW么，12Vi ewAVi evB292ACTA MECHANICA SOLIDA SINICA2010Table 2. Empirical parameters for the spring compliances, SijSMM SQq SMqaj 1.253.360.74bi; -0.729 - - 0.300- 0.294After examining the value of CsD for a range of L, RNw and ENw, we found that SMM plays themost important role in substrate deformation. Thus, simplifying Eq.(23) by taking SQq = SMQ = 0,we have[4(25a)CsD= 48ENwI- 64EwI(2SMMENw1 + D)orKsD(ENw)(25b)ENwNote that k'SD is still a function of ENw.IV. SIZE-DEPENDENT APPARENT YOUNG'S MODULUSDuring the nanobridge test, adhesive contact deformation and substrate deformation occur simul-taneously. Thus, the total compliance should beCsD,+C%e = ^'SD十Iac(26)quation (26) paves the way for us to calculate a correct value of ENw in the nanobridge test. Lettingw denote the apparent Young 's modulus evaluated from total compliance by using Eq.(1), we calculatethe ratio of E'w to Evw .ENw__ KEuler(27)ENw KsD + KacAs described above, the effects of adhesive con-tact deformation and substrate deformation on.4+、Cs- 300N/muthe apparent Young's modulus relate to the di-:2Cs = 200 N/mTneter of tested nanowires. For a given length,he larger the diameter is, the greater the effects当10ill be. Therefore, adhesive contact deformation__ 200N/m Cg--100N/mand substrate deformation lead to size-dependent-- 300N/tapparent Young s modulus. To illustrate the size-145 10152025303540dependent apparent Young's modulus, we plot theRNw (nm)ratio ENw/Ew vS. nanowire diameter RNw in Fig. 8 Intrinsic efct and extrinsic efet as a function of theFig.8, where the used data are E. = 473 GPa, nanowire radius, where E.= 473 GPa, Rtip = 20 nm, andRtip= 20 nm, γ= 0.4 N/m and L = 1000 nm. L= 1000 nm.We take the Young's modulus of SiC in the (111> directions, E, = 473 GPa, which is independent ofthe nanowire diameter, as an example to illustrate the size-dependent Young's modulus. As shown bythe solid curve in Fig.8, adhesive contact deformation and substrate deformation make the apparentYoung's modulus smaller and the ratio ENw/ ENw decrease almost linearly with the increase of thediameter RNw. The results might explain the experimental results that the Young's modulus of ananowire measured by the nanobridge test is smaller than that of its bulk counterpart, as is describedin the Introduction.Many researchers[29- -32] suggest that surface elastic constants are intrinsic properties of nanowiresthat make the apparent Young's modulus of nanowires size-dependent. Zhang et al.(30] proposed a .fundamental energy-based mechanics approach for the study of surface-related intrinsic effects on size-dependent Young's modulus. In the energy-based mechanics approach, a nanowire is treated as acomposite of a bulk core and surfaces (including edges). Let E, andCs be Young's moduli of the coreand surfaces, respectively, the bending stiffness D of a nanowire is given byD= EbI +csIs .中国煤化工MHCNM HGRNW nm)Vol. 23, No. 4Wing Kin Chan et al: Analysis of Nanobridge Tests293.where Is is the second moment of lines and Is = π (RNw)”for circular section nanowiresl31. The .apparent Young s modulus E is calculated by D/I. Without considering the extrinsic effect, E increases(or decreases) for a positive (or negative) Cs, as the diameter decreases. For either a positive or negativeCs, E approaches E when the diameter becomes sufficiently large. In nanobridge tests, intrinsic andextrinsic effects co-exist. Considering the surface related intrinsic effect, we may replace ENw in theabove analysis with E because E is the bending equivalent Young's modulus. Let Eex denote themeasured Young's modulus by using Eq.(1). The normalized Eex/E is plotted in Fig.8 for diferentvalues of Cs. As expected, the curve with Cs = 0, meaning there is no surface-related intrinsic effect,separates the curves for Cs > 0 from those for Cs < 0. For a negative Cs, the value of Eex/ E increasesfrom a lower value to a maximum, which is still smaller than unity, and then decreases as the nanowireradius increases. For a positive Cg， however, the nanowire exhibits the -smaller-the harder behavior,but approaches a value lower than the bulk one due to the extrinsic effects, which might explain theexperimental resut,11,121.1V. CONCLUDING REMARKSIn nanomechanics, a crucial and challenging issue is the size-dependent apparent Young 's modulus.In the present work, we separate factors that cause size-dependent apparent Young s modulus into twocategories, intrinsic and extrinsic factors. The extrinsic factors are also called extrinsic deformations here,which include adhesion contact deformation and substrate deformation. Extrinsic deformations occurduring nanobridge tests, which are, however, often ignored in the literature when Young's modulusof a nanowire is characterized by the nanobridge test. The present work studies systematically therole of extrinsic deformations in the nanobridge test and the results show that the apparent Young'smodulus measured without consideration of the extrinsic deformations is lower than the real value.The apparent Young's modulus might also exhibit the-smaller-the-harder behavior, but the values ofapparent Young s modulus of nanowires with larger diameters are smaller than the Young s modulus ofthe bulk counterpart. Excluding extrinsic factors is the essential step to explore intrinsic size-dependentYoung's modulus in nanomaterials.In the present work, adhesive contact deformation is analyzed based on the JKR model and adhesivecontact-induced compliance is modeled by a spring connecting a probe tip and a tested nanobeam.Under small deflection, the bending deformation of the tested nanobeam is linear, but adhesive contactdeformation is nonlinear. Analytical solution is presented here and verifed by the FE analysis. Thedeveloped analytic approach is powerful in solving adhesion-bending coupling problems. However, theanalytic solution is complex due to the nonlinear deformation of adhesion contact. To be user. friendly,the present work provides two empirical equations based on the analytic solution. The two empiricalequations give general expression of adhesion contact compliance and Hertz contact compliance, whichare both valid for the maximum tip displacement up to around 20 nm. Following the previous workl4,16),the substrate deformation is modeled by two coupled springs connecting to each of the nanobeam ends tothe substrate. The spring compliances are calculated for silicon substrate and presented in a generalizedexpression for a large range of nanobeam diameters. In this way, the explicit results presented here allowone to exclude the extrinsic factors and determine Young 's modulus of a nanowire from the nanobridgetest more accurately. The approach developed here can be directly applied to the adhesion-bendingproblem in nanocantilever tests.References1] Xia,Y.N, Yang,P.D., Sun,Y.G, Wu,Y.Y., Mayers,B., Gates,B., Yin,Y.D., Kim,F. and Yan,Y.Q, One-dimensional nanostructures: Synthesis, characterization and applications. Advanced Materials, 2003, 15(5):Salvetat,J.P., Briggs,G.A.D.. Bonard,J.M., Bacsa,R.R., Kulik,A.J.. Stockl,T., Burnham,N.A. and Forro,L.,Elastic and shear moduli of single-walled carbon nanotube ropes. 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Physical Review B,2010, 81(19): 195427.中国煤化工MHCNM HGVol. 23, No. 4Wing Kin Chan et al: Analysis of Nanobridge Tests295APPENDIX A: ANALYTICAL SOLUTION OFHERTZ CONTACT WITH ELLIPTICAL CONTACT AREAIn Hertz contact, two surfaces in contact are frictionless and only normal force is interacted betweentwo bodies. In this case, Eq.(2) is reduced top(x,y)=pI[-()-()](29)This pressure field produces a displacement field:Plb「k(e)- K(e)- E(e)。E(e)/(1-e)-K(e),21(30)Ere2a2Substituting Eq.(9b) into Eq.(30) and letting x= y = 0, we have the tip displacemente2a2K(e)(31)0=2R([K(e)- E(e)Then, the total force is calculated to bee2a3E*π2(32)3R'[K(e)- E(e)Substituting Eqs.(31) and (32) into the tip displacement u =C. P+ 8 yields2CE* π2e2a3 + 3K(e)e2a2(33)6R' [K(e)- E(e)]Equations (33) and (32) give the relations among u, a and P.When the tip is loaded on a nanowire with a rectangular or polygonal cross- section, the contact areawill be circular. For circular Hertz contact, the eccentricity is zero and both elliptical integrals have avalue of π/2. Then, Eqs.(32) and (33) are respectively reduced to4πa3 E*(34)3Rtipa2(4CE*πa + 3)(35)APPENDIX B: ANALYTICAL SOLUTION OFADHESION CONTACT WITH CIRCULAR CONTACT AREAFor circular adhesion contact, the deformation is axisymmetric and local cylindrical coordinates, rand 0, are used here. The vertical displacement for circular adhesion contact is given byUz=δ(1--2(36)2hRtip)/The adhesion contact pressure field is expressed as .1/2/2p= Po(1-))+;(-影)(37)where a is the circular radius of contact. The contact pressure reproduces a local displacement field(38)[0+层(-别中国煤化工MHCNM HG296.ACTA MECHANICA SOLIDA SINICA2010Comparing Eq.(36) and Eq.(38), we havePo= -)(39)E* 2a(40)π RtiBy integrating the contact pressure over the contact area, we obtain the total force2a(38Rtip-a2)E*P= |. σ.2πrdr=(41)3Rtipwith Eq.(41) and u= C. P+δ, we express P and δ in terms of u and a, respectively, as2E*a(3Ruipu-a2)(42)3Rtip(1 + 2aE*C)8_ 3Rtipu+2a3E*C(43)0=3Rtip(1 + 2aE*C)The elastic strain energy only due to adhesion contact is calculated byUel= I p(r)u;drdy= E* ((44)3 RtipEquations (41)-(44) can be deduced from Eqs.(10)-(13), respectively, as the eccentricity e approacheszero. The total energy is then given by(2。_ 28a31 a5CP2Viotal=E* (82a- 3Rip +R2Using Eq.(17a), we have9E* Rpu2 - a(18γπRp) - a2(72γπRp E*C + 18E* Rripu)-a2(72γπRpE*2C2 + 24E*CRipu) + 9a'E* +24E*2a*C+ 16E*3C2a°=0 (46)which yields4E*Ca3(47a)E*If there is no bending deformation, the bending compliance is null, i.e., C = 0. In this case, thedisplacement is induced completely by adhesion contact and Eq.(47a) is reduced toa22γπc(47b)Rtip-V~ E*which is that given by the JKR theory[19,26].APPENDIX C: ADHESION BETWEEN A NANOWIRE AND ITS SUBSTRATEThe lateral contact length l of a nanowire on its substrate may be described by an adhesion angleφ as shown in Fig.9. The vertical distance of thecylindrical surface of the nanowire to the substratesurface is given by h(φ) = ho + RNw(1 - cos q).The total force acting on the cylinder is calculatedby using the Lennard-Jones (LJ) potential and ho .RNWis determined by force equilibrium. The contact isassumed to be lost when the adhesive pressure isThsmaller than one tenth of its maximum. For RNw =16, 32, 48 and 96 nm, we calculated the contactangles to be 16*,119,99 and 6°, respectively. Finite Fig. 9 The lateral contaact length l :nd adhesitangle φ be-element analysis verifies the analysis results.tween a nanowire and the中国煤化工MHCNMH G