An ultrasonic measurement of stress in steel without calibration: the angled shear wave identity
AA N ULTRASONIC MEASUREMENT OF STRESS IN STEEL WITHOUTCALIBRATION : THE ANGLED SHEAR WAVE IDENTITY
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Guo-Yang Li
Harvard Medical School and Wellman Center for PhotomedicineMassachusetts General HospitalBoston, MA 02114, USA [email protected]
Artur L. Gower ∗ Department of Mechanical EngineeringUniversity of SheffieldSheffield, UK [email protected]
Michel Destrade
School of Mathematics, Statistics and Applied MathematicsNUI GalwayGalway, Ireland [email protected]
August 4, 2020 A BSTRACT
Measuring stress levels in loaded structures is crucial to assess and monitor their health, and to predictthe length of their remaining structural life. However, measuring stress non-destructively has provedquite challenging. Many ultrasonic methods are able to accurately predict in-plane stresses in acontrolled laboratory environment, but struggle to be robust outside, in a real world setting. That isbecause they rely either on knowing beforehand the material constants (which are difficult to acquire)or they require significant calibration for each specimen. Here we present a simple ultrasonic methodto evaluate the in-plane stress in situ directly, without knowing any material constants. This methodonly requires measuring the speed of two angled shear waves. It is based on a formula which is exactfor incompressible solids, such as soft gels or tissues, and is approximately true for compressible“hard” solids, such as steel and other metals. We validate the formula against virtual experimentsusing Finite Element simulations, and find it displays excellent accuracy, with a small error of theorder of 1%.
Railroad rails in the real world can degrade greatly because of high levels of built-up mechanical stresses. During coldwinter nights steel rails contract and the resulting tensile stress then promotes fatigue cracks. On hot summer days railsexpand and the resulting compressive stress can trigger catastrophic buckling.Rail steel is one of many examples where stress variations and/or high stresses lead to wear and failure [1]. Otherexamples include stress in steel pipes [2], pressure vessels, and common machine components [3] such as bearingraceways. Accurately evaluating the stress within steel plates and bars would substantially improve our estimates onstructure life [4], and help us to efficiently schedule maintenance and improve safety. Figure 1 shows two examples ofsteel structures that have failed due to a build up of compressive stress: the left shows a buckled steel column and theright a buckled track. Other than buckling, stress also causes material fatigue and crack growth [5].Stress measurement methods based on ultrasonic elastic wave propagation have long been used as they are relativelycheap, safe, quick and non-destructive [1, 8]. The basic setup is shown in Figure 2. In this work we focus on structures ∗ Webpage: arturgower.github.io. a r X i v : . [ phy s i c s . c l a ss - ph ] A ug PREPRINT - A
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4, 2020Figure 1: Examples of failure due to compressive stresses. On the left: A steel column which has buckled (in thelab) due to applied compressive stresses [6]. On the right: a rail track which has buckled due to compressive stressfollowing an Earthquake [7].made of a hard solid, like steel, where we have access to at least one relatively flat surface, such as the web rail. Themethods to measure stress with ultrasonic waves in other scenarios, such as strands and cables [9], can be quite different.One of the simplest methods to measure stress, in a scenario similar to Figure 2, is called ultrasonic birefringence . Itrequires measuring the speed of two shear waves travelling directly across the plate. Then, using the coordinate systemshown in Figure 2, assuming only a uni-axial stress along x , and symmetry along x , we find that [10, 11, 12] ρ ( v − v ) = (cid:18) n µ (cid:19) σ , (Birefringence identity) (1.1)where v ( v , respectively) is the speed of a shear wave propagating in the x − direction and polarised in the x − direction ( x − direction, respectively), and ρ is the current mass density. x x stress σ directangled θ θ Figure 2: To measure the uni-axial stress σ we can send two ultrasonic waves either directly across the steel plate, or atan angle. In the former method (birefringence) we need an a priori knowledge of the elastic constants to deduce thestress; by contrast, the angled method (this paper) gives instant access to the stress.Although these two wave speeds are quite easy to measure, the accuracy of the resulting measurement of σ is governedby the accuracy of the term in brackets on the right-hand side–called the birefringence constant –, which involves µ , asecond-order Lamé elastic constant, and n , a third-order Murnaghan elastic constant. Birefringence constants are verydifficult to measure in situ , and instead are measured in the lab, often on pristine steel samples, which obviously canbe very different from their in-service and worn counterparts (for instance, these constants change with temperatureand wear [13].) In addition, whereas the mass density ρ and the second-order constants λ , µ can be measured quiteconsistently from one steel sample to another, the measurements of the third-order constants vary widely, as can bechecked in Table 1. Even with a perfect measurement of the shear wave speeds, we can expect this variation in theconstants to lead to an error of , as shown in Figure 3, where the birefringence constant of ten samples of steel isreported.Another notable ultrasonic method which use changes in ultrasonic speeds to measure stress below the material surfaceis the ‘longitudinal critically refracted method’ [4, 14]. It is mostly used for plates, as its penetration depth is limited toone wavelength. There are further alternative methods for measuring stress along the surface of the material, includingusing Rayleigh or Lamb waves [15]. These usually require specific surface conditions, and often that the surface ofthe material is the same as the bulk. Similar to Birefringence, these methods require prior knowledge of the materialconstants or significant calibration. Designing a method to measure stress, without knowing the third order constantshas long been a goal worthy of investigation and applications [16].2 PREPRINT - A
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REX 535Hecla 138AHecla 17Hecla 37Rail 2Rail 1 - - - - - B i r e f r i ngen c e c on s t an t: + n / ( μ ) Nickel -s teel N i c k e l -s t e e l S / N V T
Figure 3: Range of values for the birefringence constant of 10 samples of steel. Note that the constants were allmeasured in a highly controlled laboratory environments. Depending on the type of steel, this constant varies between10% (Nickel-steel) and 40% (Rail steel). The references and values of the constants are given in Table 1.In this paper, we propose a method to measure stress directly based on the following approximate identity ρ v − v cos(2 θ ) (cid:39) σ − σ , when θ ± θ = 90 ◦ , (Angled shear-wave identity) (1.2)where ρ is the mass density before applying stress, and v , v are the speeds of the quasi-shear waves polarised inthe ( x , x ) plane and traveling in the directions along (cos θ , sin θ , and (cos θ , sin θ , , respectively. Figure 2shows examples of these angled shear waves.Note that the identity above requires no prior knowledge on the material’s elastic constants and, as we shall see, it isaccurate for a wide range of stresses in steel, with an error typically around 1%. The problem of a small-amplitude body shear wave travelling in a homogeneously stressed, initially isotropic, incom-pressible solid is quite straightforward to solve.Consider such a solid, subject to a static uniform Cauchy stress σ . Call ( x , x , x ) the axes along the principaldirections of stress, σ , σ , σ the corresponding principal stresses, and λ , λ , λ the principal stretch ratios, whichoccur along those directions. Then consider wave propagation in the ( x , x ) principal plane, along n = (cos θ, sin θ, ,say.Because of incompressibility, only pure shear waves can propagate: one (out-of-plane) pure shear wave polarised along x , and another (in-plane) pure shear wave polarised along ( − sin θ, cos θ, and travelling with speed v given by[17, 18] ρ v = α cos θ + δ cos θ sin θ + γ sin θ. (2.1)Here ρ is the (constant) mass density of the solid, δ is a complicated function of the derivatives of the strain energydensity W and the stretch ratios (but it is not needed), and α = σ − σ λ − λ λ , γ = σ − σ λ − λ λ . (2.2)Then use Equation (2.1) to compute in turn the squared waves speeds v and v of the waves travelling in the θ = θ and in the θ = π/ − θ directions, respectively. Then, by subtraction, the δ term disappears and we are left with thefollowing exact formula , ρ ( v − v ) = ( σ − σ ) cos 2 θ. (2.3)Note that this formula is independent of W and gives direct access to the stress difference σ − σ once the speeds andthe mass density are measured. It is valid irrespective of the magnitude of the strains and stresses, can prove usefulfor precise stress assessment in soft, isotropic, incompressible matter, such as gels [19, 20] or some biological tissues[21, 22]. 3 PREPRINT - A
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The exact formula (2.3) does not translate exactly to a stressed compressible solid–such as steel–for two reasons. First,because the mass density now changes with the stresses. The second reason is more subtle, but nonetheless known [23]:the in-plane shear wave is no longer a pure shear wave, but instead is a quasi-shear wave . In fact, there are now twoin-plane body waves travelling in the n − direction: the quasi-shear wave and a quasi-longitudinal wave.To find their characteristics, we write down and solve the equations governing small-amplitude motion in a homogeneoussolid subject to a uniform stress [24, 25], ρ ∂ u j ∂t = A ijkl ∂ u l ∂x k ∂x i , (2.4)where u = u ( x , t ) is the mechanical displacement, and the A ijkl are the so-called instantaneous elastic moduli [26].Looking for in-plane quasi-shear waves of the form u = ( U , U , i k ( x cos θ + x sin θ − vt ) where k is the wavenumber,the equation of motion (2.4) becomes M (cid:20) U U (cid:21) = with M = (cid:20) A cos θ + A sin θ − ρv ( A + A ) cos θ sin θ ( A + A ) cos θ sin θ A cos θ + A sin θ − ρv (cid:21) . (2.5)We then find the wave speeds by solving det M = 0 , (2.6)which is a bi-quadratic in v .The instantaneous moduli depend on the stresses σ i and on the elastic constants, and can be calculated in a straight-forward (albeit long-winded) manner, see details in Appendix A. As steel can sustain only infinitesimal strains in theelastic regime, we adopt the so-called third-order elasticity model for its strain-energy density.As seen in the appendix, the moduli can be computed explicitly at the level of approximation afforded by third-orderelasticity, as A = λ + 2 µ + (2 c + 2 d + 1) σ + ( a + 2 b )( σ + σ ) , A = λ + 2 µ + (2 c + 2 d + 1) σ + ( a + 2 b )( σ + σ ) , A = µ + (2 b + d )( σ + σ ) , A = λ + ( a + c )( σ + σ ) , A = A + σ , A = A + σ , (2.7)where the non-dimensional coefficients a , b , c , d are defined in the appendix (they are constants of order 1, varyingin value from -3.6 to + 0.7 for rail steel, see Table 2). A further justification for using the form (2.7) is that there issignificant evidence [27] that for steel the moduli have a linear dependence on the Cauchy stress, even beyond smalllevels of stress.Staying within the same level of approximation, we may now solve (2.6) for v , choose the solution corresponding to aquasi-shear wave, and then expand the result. We obtain: ρv = µ + ( b + d/ σ + σ ) + σ cos θ + σ sin θ. (2.8)Having found the wave speed, we may return to (2.5) and determine the direction of polarisation. Instead of being alongthe π/ θ direction, as it would be for a pure shear wave, it is along the π/ θ + θ (cid:63) direction for the quasi-shear wave,where θ (cid:63) is the offset. Hence, calling U the amplitude of the wave, we have [ U , U ] = U [ − sin( θ + θ (cid:63) ) , cos( θ + θ (cid:63) )] .Substituting into the first line of the homogeneous system (2.5), we conclude that tan( θ + θ (cid:63) ) = M /M , or tan( θ + θ (cid:63) ) = (cid:20) − c + dλ + µ ( σ − σ ) (cid:21) tan θ. (2.9)Now we go back to (2.8), the equation giving the wave speed. Calling v and v the speeds of the waves travelling inthe θ = θ and θ = π/ − θ directions, respectively, and subtracting the corresponding equations (2.8), we obtain ρ ( v − v ) = ( σ − σ ) cos(2 θ ) . (2.10)Finally, we recall that the current mass density ρ , which is usually not known, can be related to ρ , the initial massdensity (measured before applying stress). From (A.3) in the appendix we see that, within the context of third-orderelasticity, (2.10) is equivalent to (1.2), at the same order of approximation.Note that using (2.8) to compute the difference of the squared wave speeds of any two shear waves travelling in differentdirections would also produce a formula giving direct access to the stress without involving the material constants.4 PREPRINT - A
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Practically, there are a number of ways to generate the angled shear waves travelling in the θ and θ directions. Forexample, we could apply a shear force on the boundary of a wedge with inclination θ , as shown in Figure 2. However,because they are quasi-shear waves, their direction of propagation is not exactly orthogonal to their polarisation, whichis to say it is not exactly orthogonal to the wedge face.In the previous section we saw that the quasi-shear wave propagating in the θ direction is polarised in the π/ θ + θ (cid:63) direction. It follows that the quasi-shear wave polarised in the π/ θ direction (launched by a wedge with inclination θ ) is propagating in the θ − θ (cid:63) direction, where θ (cid:63) is computed from (2.9). Calling v the speed of the wave in the θ = θ + θ (cid:63) direction, and v that of the wave in the θ = π/ − θ direction, we find again that ρ ( v − v ) =( σ − σ ) cos(2 θ ) , because the correction due to the offset is of higher order and negligible in third-order elasticity.Hence we conclude that the perturbation in the propagation direction due to the quasi-shear wave character of the wavedoes not affect the prediction of the stress based on (2.3).This result is useful, for example, when using finite element software to conduct a virtual experiment, where it is easierto specify the polarisation than the wave direction. To use the angled shear-wave identity (1.2) and measure the stress robustly, we need to determine how sensitive it is toexperimental errors, including those coming from misaligned wedges, imprecise wave speeds, varying stress levels,among others. After investigating a number of scenarios, we found that the largest source of imprecision comes fromerrors in the direction of propagation, such as those arising in the case of misaligned wedges. o o o o o o o o o o o Wave direction θ W a v e d i r e c t i o n e rr o r δ θ Stress error % % % % % % % % % Figure 4: Relative error in the predicted stress σ − σ due to an error in the direction of propagation ( δθ ) of one ofthe shear waves. For example, when one of the waves travels in the direction θ = 34 ◦ ± δθ (while the other wavetravels in the direction θ = 56 . ◦ ), we expect: a error in predicting the stress when δθ (cid:39) ◦ , and a error inpredicting the stress when δθ (cid:39) . ◦ .Assume one of the angled shear waves has a slight error in its direction of propagation, so that instead of travelling inthe θ direction it is travelling in the θ + δθ direction. Call v its speed and v the speed of the shear wave propagatingat the angle θ = π/ − θ . Then by perturbation of (2.8), we find that the difference between the squared wave speedsis given by ρ ( v − v ) = ( σ − σ ) [cos(2 θ ) − δθ sin(2 θ )] . (2.11)From this equation we deduce that the relative error in the predicted stress is equal to δθ tan(2 θ ) , which we evaluatefor various directions θ in Figure 4. We find that it is small when the precision of the angle θ is of the order of 1 ◦ and θ is close to 35 ◦ .In the next section, we validate the identity (1.2) further by conducting virtual experiments.5 PREPRINT - A
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To validate the method, we use Finite Element (FE) analysis (Abaqus 6.13, Dassault Systèmes R (cid:13) ) to conduct virtualexperiments. We built a plane strain model and created a user-subroutine UHYPER to implement a third-order elasticmodel, such as (A.2). An illustration of the FE simulation is shown in Figure 5(a).
Time ( μ s) Par ti cle velocity (arbitrary units) P a r t i c l e v e l o c i t y ( a r b i t r a r y un i t s ) d σ θ (a) (b) (c) θ -1 10 y x σ Figure 5: FE simulation of the quasi-shear wave propagation. (a) Schematic of the FE model. The stress is determinedby σ and the out-of-plane strain. The wave is induced by a body force defined in the local coordinate systems. (b)Snapshot of the shear wave propagation. Six points at equal distance d are used to measure the wave speed. (c) Timeprofiles of the particle velocities at these points shown in (b).For the material parameters, we choose the values for the sample ‘Rail steel 1’ in Table 1. We then apply a uni-axialstress along the horizontal direction accompanied by some out-of-plane strain. By prescribing the out-of-plane strain,we can achieve different stress levels and keep the displacement induced by shear wave motion to be zero in thisdirection. The top and bottom surfaces are stress free.To describe our setup, let us introduce a local coordinate system ( x, y ) as shown in Figure 5(a), where the x axis isoriented at a θ angle from the horizontal. To imitate a real experiment [28], we generate shear waves by applying animpulse force in the y direction, which would be like applying a shearing force to the face of the wedges shown inFigure 2. We expect this impulse to generate a quasi-shear wave polarised in the y direction, and propagating in adirection that is not exactly along x . We will however assume that the quasi-shear wave does approximately propagatein the x direction. As discussed in Section 2.3, we expect this assumption to introduce a very small error. In a realexperiment, assuming that the quasi-shear wave propagates in the x direction would simplify the experimental setup. Sothe success of these FE simulations give confidence that this experimental setup would successfully predict the stress.The impulse we use is a body force F ( x, y ; t ) defined by F ( x, y ; t ) = F e − x a y b e − ( t − t ) τ sin(2 πf t ) , (3.1)with F = 0 . N/mm , a = 1 mm, b = 60 mm, t = 6 . µ s, τ = 2 . µ s, and f = 0 . MHz. We point out that wechoose to use a rather low frequency f for convenience. We checked that taking higher frequencies leads to the sameresults, but are much more time-consuming to simulate. The mesh size is approximately / of the shear wavelengthto ensure the convergence of the simulation.We use six points spaced with equal distance d along the x axis, as shown in Figure 5(b), to measure the shear wavespeed. Here d denotes the distance in the stressed configuration, which is approximately equal to mm in the initialconfiguration. In Figure 5(c) we plot the time profiles of the particle velocities at these six points. The phase differencesbetween every two adjacent points were measured in the Fourier domain. We then calculate the shear wave speed v from the averaged phase difference ∆ φ by [29] v = 2 πf d ∆ φ . (3.2)6 PREPRINT - A
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4, 2020In the next section we use our FE model to investigate which angle θ leads the most accurate prediction of the stress,when using our identity (1.2). We then validate our predictions over a wider range of stress in Section 3.2. Here we study how the angle of propagation θ influences the stress predicted by the identity (1.2) when using the wavespeeds v simulated by our FE model. For this section, we take the uni-axial stress as σ = 300 MPa and no out-of-planestrain: λ = 1 .As we can see in Figure 6(a), the shear wave speed v decreases with θ . The wave speeds corresponding to θ = θ < ◦ and θ = 90 ◦ − θ are v and v in the identity (1.2), which then yield the stress σ , as shown in Figure 6(b). For thefour different cases θ = 27 ◦ , . ◦ , ◦ , ◦ , the stress identified by (1.2) agrees well with the stress σ used in theFE model, with a very small difference ( ≤ ).
15 25 35 45 55 65 75 θ ( o ) W a v e s p ee d ( m / s ) I d e n t i fi e d s t r e ss ( M P a ) θ ( o ) (a) (b) Figure 6: Shear wave speed obtained by the FE simulation and the stress predicited by (1.2). (a) Shows the how thewave speed depends on the angle θ . (b) Shows the stress predicted by using the FE results fed into (1.2). The horizontalline denotes the value of σ actually applied to the FE model. We now fix the angle shown in Figure 5 by choosing θ = 35 ◦ , θ = 55 ◦ and study the accuracy of the angled shearwave identity (1.2) for different stress levels. The stress levels we investigate are σ =
100 to 1,500 MPa, which cover allpossible stress levels in rail, as standard rail has a maximum tensile strength between 700MPa and 1100MPa [30, 31].As shown in Figure 7, we find that the predicted stresses agree well with the stresses actually applied, with a relativeerror less than 1.5%, and an error less than 1% when below the maximum tensile strength 1100 MPa.One difference between the FE model and the identity (1.2) is that our FE model imposes an out-of-plane stress σ ,where as (1.2) has zero out-of-plane stress. To confirm that this difference has no impact on our predictions we conducta FE simulation with plane-stress (PS) and then another simulation under plane-strain (PE).To achieve PS, we adapt the stretch ratio λ so we can maintain σ = 0 when applying a uniaxial stress. To achieve PEwe fix λ = 1 . and allow σ to change. Figure 8(a) shows the stress along the horizontal σ against the wave speeds v and v . The wave speeds obtained from the PS model are (slightly) smaller than those obtained from the PE model,and this difference is significant when it comes to predicting the stress. However, when using these wave speeds topredict the stress with (1.2) (see Figure 8(b)), we find that both the PE and PS models predict the same levels of stress.This indicates that the identity (1.2) is in practice independent from a small amount of out-of-plane stress, which wethen confirmed analytically. This independence is highly convenient, because in practice the deformation state mayneither be plane stress nor plane strain. One significant drawback of most ultrasonic methods to measure stress is that they need significant calibration for eachspecimen. This requires either prior knowledge on the material constants, or complicated measurement systems.7
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Applied stress (MPa) I d e n t i fi e d s t r e ss ( M P a ) Applied stress (MPa) R e l a t i v e e rr o r (a) (b) Figure 7: Comparison between the stress applied in the FE simulation and the stress identified by the identity (1.2).Here the wedges angle, as shown in Figure 5, for the two waves are θ = 35 ◦ , θ = 55 ◦ . (a) Applied stress vs predictedstress. (b) The relative error of the predicted stress. Note that the maximum tensile strength of standard rail steel ismaximum tensile strength 1100 MPa. Applied stress (MPa) W a v e s p ee d ( m / s ) Iden ti fied stress (MPa), PS I d e n t i fi e d s t r e ss ( M P a ) , P E v v PS PE(a) (b)
Figure 8: Comparison between plane stress (PS) and plane strain (PE) simulations when the waves travel along θ = 35 ◦ and θ = 55 ◦ . (a) The speeds v and v vs the applied stress σ . (b) Comparison of the identified stress obtained fromthe plane stress and plane strain simulations.In this paper we propose a solution to these issues by introducing a method that can be used to measure in-plane stresswithout any calibration. The method requires two angled shear wave speed measurements. At its heart is a formulawhich is exact for incompressible solids, and applies to compressible solids within the context of initially isotropicthird-order elasticity.The main advantage of our proposed method is that it does not require prior knowledge on the material constants. Onechallenge in using this method includes fabricating wedges with a precisely chosen inclination: an error of 1% in theinclination typically results in a error of 4% when predicting the stress, see Figure 4. Other aspects that still needinvestigation include how errors in measuring the wave speeds result in errors in predicting the stress, though this willdepend on the specific experimental realisation. Another issue that needs to be considered is how texture anisotropyaffects the predictionsof this method [16].We validated the method against Finite Element simulations and found its precision to be excellent, and the predictionsto be robust in terms of out-of-plane stress. It remains to be demonstrated if this new method is able to accuratelypredict stresses in the lab and on the ground. 8 PREPRINT - A
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Acknowledgements
Artur Gower is grateful to Prof. Rob Dwyer-Joyce and Prof. Roger Lewis for introducing this problem, and for theircollaboration. Artur Gower and Michel Destrade are grateful for partial support from the UKAN grant EPSRC (grantno. EP/R005001/1).
A The instantaneous elastic moduli
The theory and practice of acousto-elasticity has a long and distinguished history. Classic works on the topic includecontributions by Brillouin [32], Hugues and Kelly [33], Toupin and Bernstein [10], Hayes and Rivlin [34], or Biot [35].Later, Ogden [26] formalised its equations in a compact and elegant manner within the framework of exact non-linearelasticity. Hence they obtained the following expressions for the instantaneous elastic moduli in (2.5) A = λ ∂σ ∂λ , A = λ ∂σ ∂λ , A = λ ∂σ ∂λ + σ , A = σ − σ λ − λ λ , A = σ − σ λ − λ λ , A = σ λ − σ λ λ − λ . (A.1)Now to make progress, we must specify the strain energy density W so that we can compute the Cauchy stresscomponents σ i = λ i ∂W/∂λ i ( i = 1 , , , no sum) and their derivatives with respect to the stretch ratios.As we are focusing on steel, which deforms very little elastically, it is sufficient to use the general strain energy ofthird-order weakly nonlinear elasticity. We take it in its Murnaghan [36] form, W = ( λ + 2 µ ) I − µI + ( (cid:96) + 2 m ) I − mI I + nI , (A.2)where λ, µ are the Lamé constants, (cid:96), m, n are the Murnaghan constants, and I , I , I are the first three principalinvariants of the Green-Lagrange strain tensor E . We align the coordinate axis so that E is diagonal in the ( x , x , x ) coordinate system, with components E i = ( λ i − / ( i = 1 , , ).The current mass density ρ which is usually not known, is related to ρ , the initial mass density (measured beforeapplying stress) through [33]: ρ = ρ / (1 + 2 I + 4 I + 8 I ) . (A.3)Another commonly used form is the Landau strain energy that uses the (third-order) Landau constants A , B , C . Theyare connected to the Murnaghan constants m , n , (cid:96) through (cid:96) = B + C, m = A B, n = A. (A.4)Then the stress components and eventually, the instantaneous moduli are straight-forward to compute with a ComputerAlgebra System.By assuming small strains, the instantaneous moduli for third-order elastic materials can be written explicitly in terms ofthe stresses, see Gower et al. [37] and Tanuma and Man [38], resulting in (2.7), where the non-dimensional coefficients a , b , c , d are defined in terms of Murnaghan constants as a = 2 (cid:96)µ − λ + ( n − m − λ )( λ + µ ) µ (3 λ + 2 µ ) , c = 2 λ + 2 m − n µ , (A.5) b = − λn + µ (4 λ + 2 µ − m + n )2 µ (3 λ + 2 µ ) , d = 2 + n µ . (A.6)For (2.7) to be asymptotically equivalent to classical third-order elastic models, such as (A.2), the constants b , c , d needto be linked through [37, 38]: λ (1 + 3 b + d ) + µ (1 + 2 b − c ) = 0 , (A.7)as can be checked easily.For the material parameters of steel, we use the data collected in Table 1. Whereas Table 2 gives the values of theconstants a , b , c , d for the Rail steel 1 and Rail steel 2 specimens.9 PREPRINT - A
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4, 2020 ρ λ µ (cid:96) m n Rail steel 1 [39] 7800 115.8 79.9 -248 -623 -714Rail steel 2 [40] 7777 112.9 80.8 -88.9 -591.6 -903.8Nickel steel [41] 909 78.0 -46 -590 -730Nickel-steel S/NVT [42] 109.0 81.7 -56 -671 -785Hecla 37 carbon steel [43] 7823 111 82.1 -461 -636 -708Hecla 17 carbon steel [43] 7825 110.5 82.0 -328 -595 -668Hecla 138A steel [43] 7843 109 81.9 -426.5 -619 -708REX 535 nickel steel [43] 7065 109 81.8 -327.5 -578 -676Table 1: Material parameters for several samples of steel. The density ρ is in kg/m and the elastic constants are inGPa ( λ , µ : second-order Lamé constants; (cid:96) , m , n : third-order Murnaghan constants.) λ µ a b c d Rail steel 1 [39] 115.8 79.9 0.701 -0.118 -1.88 -2.47Rail steel 2 [40] 112.9 80.8 0.127 0.370 -0.332 -3.59Table 2: λ and µ are reported in GPa; a , b , c , and d are non-dimensional constants. References [1] M. Hirao and H. Ogi.
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