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ES240 Solid Mechanics Z. Suo Waves Yet another play of two actors: inertia and elasticity References. H. Kolsky, Stress Waves in Solids, Dover Publications, New York. K.F. Graff, Wave Motion in Elastic Solids, Dover Publications, New York. J.D. Achenbach, Wave propagation in elastic solids. North-Holland, Amsterdam. Also see Achenbach’s acceptance speech of the Timoshenko Medal (http://imechanica.org/node/185). Light and sound. A comparison between light and sound is instructive. What’s vibrating? Vibrating electric field and magnetic field transport light. Vibrating stress field and displacement field transport sound. Media. Light can propagate in vacuum, as well as in certain materials. Sound wave must propagate in elastic media. Spring of air, liquid, solid. When a noisy watch is suspended in a glass jar with a thread. In the beginning, you can see the watch and hear its noise. When the air is sucked out of the jar, you can still see the watch, but cannot hear it. Speeds. The light wave speed is 3 108 m/s in vacuum. The sound speed is about 340 m/s in air, 1500 m/s in water, and 5000 m/s in steel. Frequencies and wavelengths. f c/ . Visible electromagnetic waves. Red: 0.7 m, f = 4.3 1014 Hz. Violet: 0.4 m, f = 7.5 1014 Hz. Audible sound waves. 20 Hz to 20 kHz. In air, they correspond to 17 m and 17 mm. In water, they correspond to 75 m and 7.5 cm. Ultrasound. Ultrasound has frequencies too high to be detected by the human ear. Ultrasound can be generated using piezoelectric materials, which convert an electric field to a stress field. High frequencies correspond to short wavelengths. Ultrasound imaging: see baby inside mother. Non-destructive evaluation (NDE): detect flaws inside materials. Surface Acoustic Wave (SAW) devices for wireless applications. t=0 5mm 10mm x t = 0.5 s 5mm 10mm 15mm t = 1 s 5mm 10mm 15mm x x t = 2 s 5mm 5/31/16 t 15mm 10mm Waves-1 15mm x ES240 Solid Mechanics Z. Suo Longitudinal wave in a rod. The speed of sound in steel is ~ 5 km/s. A frame-by-frame “movie” shows the stress profiles at several times. Before time zero, these is no stress in a steel rod. At t = 0, a hammer hits the end of the rod for 1 s. At t = 0.5 s, 2.5 mm of rod is under compression, but the rest of the rod is stress-free. The delay is caused by the inertia of the matter. At t = 1 s, 5 mm of rod is under compression, but the rest of the rod is stress-free. At t = 2 s, the same 5 mm wave packet travels for 5 mm. The rod behind and ahead of the packet is stress-free. The D'Alambert solution. So far we have appealed to our daily experience about waves. What do the equations say? The equation of motion of a rod is 2u 2u E 2 2 . x t Let c be the wave speed. Define a composite variable, x ct . Let f be an arbitrary function. The function ux, t f represents a fixed displacement profile traveling at speed c to the right. At t = 0, the displacement profile is ux, 0 f x . At time t1 , the displacement profile is ux, t1 f x ct1 . The profile has the same shape as at time t = 0, but moves toward the right by a distance ct1 . We suspect that any function f will satisfy the wave solution. To ascertain this, we need to insert ux, t f into the equation of motion. Using the chain rule in the differential calculus, we obtain that u df df x d x d and u df df c t d t d Insert into the equation of motion, and we have d2 f d2 f E 2 c 2 2 d d The equation is satisfied by any function f provided that 1/ 2 c E / . This relates the speed of the longitudinal elastic wave, c, to Young’s modulus and density. Similarly, gx ct represents a fixed displacement profile traveling at speed c to the left. Let x ct and x ct . Any functions f and g satisfy the equation of motion. The general solution for the displacement is ux,t f x ct gx ct . Denote F df / d and G dg/ d . Velocity of a material particle is v u / t . The general expression for the velocity is vx, t c F x ct Gx ct . The stress field is given by Eu / x . The general expression for the stress is 5/31/16 Waves-2 ES240 Solid Mechanics Z. Suo x, t EF x ct Gx ct . Initial value problems. The time-dependent field ux, t is governed by A PDE (i.e., the equation of motion) Boundary conditions (i.e., the displacement and the stress values at the two ends of the rod) Initial conditions (i.e., the displacement field and velocity field in the rod at time zero.) In reaching the D'Alambert solution, we have used the PDE, but not the boundary conditions and the initial conditions. Here is an example to illustrate how the initial conditions come into play. Imagine a long steel rod. Pull a rod with grips at two points in the rod. Held the forces constant before time zero. Release the grips at time zero. We’d like to find out the waves generated in the rod afterward. x x Before the grips are released. The rod has a static stress field, sx . When the grips are released, at time zero, the stress field is still the same as the static field: EF x Gx sx . At time zero, there is no velocity, so that c F x Gx 0 . A combination of the two equations gives that 1 F x G x sx . 2E The stress field afterward is x, t EF x ct Gx ct 1 sx ct sx ct 2 After the grips are released, the static stress profile splits into two waves, one traveling to the right, and the other to the left. Reflection from a free end. We now give an illustration of the boundary condition. An incident wave packet in the rod, hits the free end of the rod, and reflects back into the rod. Let’s say the rod lies on the axis, x < 0, and the free end is x 0 . The incident wave travels from the left to the right. The boundary condition: Stress vanishes at all time at x 0 . An incident compressive wave, after reflection, becomes a tensile wave. Dynamic fracture. 5/31/16 Waves-3 ES240 Solid Mechanics Z. Suo How do we obtain the solution by doing algebra? We know the shape of the incident wave, x, t EF x ct . That is, we know the function F . The reflected wave travels from the right to the left. It must be of the form x, t EGx ct . Its shape G is to be determined. The stress in the rod is the sum of the incident and the reflected wave: x, t EF x ct EGx ct . How to determine the function G ? The boundary condition: at the free end x 0 the stress is zero at all time, namely, 0,t 0 . Put this boundary condition to the general expression for the stress, and we have 0 EF ct EGct . Denote ct by Q. We have GQ F Q . That is, we have expressed the function G in terms of the known function F. The independent variable can be anything. The stress in the rod is given by x, t EF x ct EF x ct . The first term is the incident wave, running towards the positive x-direction. The second term is the reflected wave, running towards the negative x-direction. At the free end, x = 0, the stress indeed vanishes at all time. If an incident wave induces a compressive stress in the rod, the reflected wave induces a tnesile stress in the rod. t1 c Free end c t2 t3 Standing waves. A normal mode is a standing wave. For example, consider a rod fixed at one end and free to move on the other end. From the two boundary conditions, the standing wave of the longest wavelength has 4L . Only a quarter of this wavelength is inside the rod. The frequency of the fundamental mode is given by f c / , which recovers the result we obtained before. You can understand other normal modes in the similar way. Acoustic impedance. An alternative form of the D’Alambert solution is x x u x, t f t g t . c c The particle velocity v u / t is 5/31/16 Waves-4 ES240 Solid Mechanics The axial force P AEu / x is Z. Suo x x v x, t F t G t . c c x x P x, t RF t RG t , c c where R A E . Note that P Rv for a wave propagates in the positive x direction, and P Rv for a wave propagates in the negative x direction. These relations may remind you of Hooke’s law, except that force is proportional to the velocity, rather than elongation. The quantity R is called the acoustic impedance, and has the unit of force per unit velocity. Reflection and transmission at the interface of two materials. Now consider two rods 1/ 2 joined at x = 0. The rod on the left-hand side has properties A1 , E1 , 1 , and c1 E1 / 1 . The rod on the right-hand side has properties A2 , E2 , 2 , and c2 E2 / 2 . An incident wave comes from rod 1 towards the interface. Upon hitting the interface, the wave is partly reflected back to rod 1, and partly transmitted into rod 2. 1/ 2 Let the incident wave be x u x, t f t . c1 The arbitrary function f represents the form of the incident wave, and is known. Because the displacement and force are continuous across the junction at all time, the reflected wave in rod 1 should take the same wave form, but run in the opposite direction, and have a different amplitude: x u x, t af t . c1 Everything about the reflected wave is known, except for the amplitude a. The net field in rod 1 is the superposition of the incident and reflected waves: x x u1 x, t f t af t . c1 c1 The axial force in rod 1 takes the form x x P1 x, t R1 F t R1aF t , c1 c1 where F df / d . The transmitted wave in rod 2 takes the form x u2 x, t bf t . c2 x P2 x, t R2 bF t c2 Everything about the reflected wave is known, except for the amplitude b. 5/31/16 Waves-5 ES240 Solid Mechanics Z. Suo To determine the amplitudes of the reflected and transmitted waves, a and b, we invoke the boundary conditions: at x = 0 the force in rod 1 equals that in rod 2, and the particle velocity in rod 1 equals that in rod 2. Thus, 1 a b R1 1 a R2b Solving for the two amplitudes, we find that R R2 2R1 . a 1 , b R1 R2 R1 R2 We note several special cases: When the two rods have the same impedance, R1 R2 , the two amplitudes become a 0 and b 0 . Upon hitting the joint, the wave does not reflect, but fully transmits into rod 2. The two rods are said to have matched impedance. When rod 2 has much lower impedance than rod 2, R2 / R1 1 , the two amplitudes become a 1 and b 2 . The force transmitted into rod 2 is vanishingly small, so that all the energy of the incident wave in rod 1 will be reflected back into rod 1. The reflected wave changes the sign of the axial force. When rod 2 has much higher impedance than rod 2, R2 / R1 1 , the two amplitudes become a 1 and b 0 , all the energy of the incident wave in rod 1 will be reflected back into rod 1. The reflected wave keeps the sign of the axial force as that of the incident wave. Bending wave in a beam. The equation of motion of a beam is 4w 2w EI 4 A 2 . x t This equation cannot be satisfied by an arbitrary function wx, t f x ct Instead, we look for solution of a more special form: wx, t a sin kx t , where k is the wave number, and the frequency. Inserting the above form into the equation of motion, we obtain EI 2 k . A Phase velocity. The velocity of a pure sinusoidal wave is known as the phase velocity, written as cp / k . The phase velocity of the pure sinusoidal wave in the beam is EI cp k. A The phase velocity of this wave depends on the wave number. Consequently, sinusoidal waves of different wave numbers propagate in a beam at different phase velocities. Such a wave is 5/31/16 Waves-6 ES240 Solid Mechanics Z. Suo known as a dispersive wave. By contrast, the longitudinal wave in a rod is nondispersive, so that a wave of an arbitrary shape can propagate without change in the rod. Note that the velocity increases with the wave number, and becomes infinite for very short wavelengths. This is clearly an artifact of our model. The equation of motion is based on the classical theory of beams, a theory that breaks down when the wavelength becomes too small. Group velocity. Another useful description of a dispersive wave is the group velocity, defined by cg d / dk . For the wave in the beam, the group velocity is EI cg 2 k. A The definition of the group velocity is motivated as follows. Consider two sinusoidal waves with the same amplitude but slightly different wave numbers: a sin k1x 1t , a sin k2 x 2t . The overall response is the superposition of the two waves: w x, t a sin k1 x 1t a sin k2 x 2t 2 k1 k2 2 k k 2a cos 1 2 x 1 t sin x 1 t 2 2 2 2 The combined wave is a wave of the average wave number k1 k2 / 2 , modulated by a wave of a smaller wave number k1 k2 / 2 (i.e., a longer wave length). The former wave is called the carrier, and the latter the group. The group propagates at the velocity 1 2 d . k1 k2 dk Sketch a carrier wave modulated by a group wave. Plane waves in a 3D, isotropic elastic medium. When the length scale of a disturbance is small compared to size of the body to all three directions, a material particle is undisturbed before the disturbance arrives. We may as well regarded the body as an infinite body. A plane wave is a wave that propagates in a given direction, and the displacement field is fixed in another direction, and the amplitude of the displacement in any plane normal to the direction of propagation is invariant. Without any calculation, we may make the following remarks: Because the problem has no length scale, a plane wave in a 3D medium is nondispersive. Because the medium is isotropic, plane waves in all directions are identical. Because the medium is linearly elastic, we can superimpose plane waves to obtain any complex waves. Two kinds of plane waves exist in an isotropic elastic solid. A longitudinal wave has the displacement in the direction of the wave propagation. A transverse wave has the displacement normal to the direction of wave propagation. We consider the longitudinal wave in this lecture, and leave the transverse wave as a homework problem. Deformation geometry. Because all directions are equivalent, we only need to consider the wave propagate in one direction, denoted by x. For a longitudinal wave, the only nonzero displacement component is ux,t . Consequently, the only nonzero strain component is 5/31/16 Waves-7 ES240 Solid Mechanics x Z. Suo u . x That is, the solid is in a uniaxial strain state. Material law. Because of Poisson’s effect, this strain induces normal stresses in all three directions. Symmetry dictates that y z . Writing Hooke’s law in the y-direction, we obtain that 1 y 0 y z x , E so that y z x . 1 Writing Hooke’s law in the x-direction, we obtain that 1 x x y z , E or E 1 x . 1 1 2 x The coefficient is the stiffness under the uniaxial strain conditions. Newton’s second law reduces to x 2u 2 . x t Putting the three ingredients together, we obtain the equation of motion: E 1 2u 2u . 1 1 2 x2 t 2 Except for the coefficient on the left-hand side, this equation of motion is identical to that for the rod. Thus, the same solution procedure applies. The longitudinal wave speed is E 1 cl . 1 1 2 By comparison, for a transverse wave, the only nonzero displacement is normal to the direction of propagation. We may write vx, t . Following the same procedure, we find that the transverse wave speed is E ct . 2 1 Take a representative value of Poisson’s ratio, 0.3 , and we obtain that cl 1.16 E / , ct 0.62 E / . Recall that the wave speed in a rod is E / . The general form of plane waves in an isotropic elastic medium. Let s be a unit vector in the direction along which a plane wave propagates. For a longitudinal wave, the displacement field takes the form sx ux, t sf t , cl 5/31/16 Waves-8 ES240 Solid Mechanics Z. Suo where f is the profile of the displacement field. For a transverse wave, the displacement can be in any direction normal to the direction of propagation, and takes the form sx ux, t ag t , ct where a is a unit vector normal to s. The general form of a plane wave propagates in direction s in an isotropic material is the superposition of three waves: sx sx sx ux, t sf t ag t bh t , cl ct ct where a and b are unit vectors normal to s. Vectors a and b are also normal to each other. Plane waves in 3D, anisotropic elastic medium. A combination of the momentum balance equation ij 2u 2i x j t and the stress-strain relation u ij Cijkl k xl leads to the the equation of motion 2u k 2u Cijkl 2i . xl x j t Because these is no length scale in the problem, the wave is nondispersive. Let the plane wave propagate in a direction specified by a unit vector normal to the plane, s. The displacement field takes the form sx ui x, t ai f t . c where a is a unit vector in the direction of the displacement, and f is the profile of the displacement. Insert this expression into the equation of motion, and we obtain that Cijkl sl s j ak c 2ai . Consequently, the wave velocity is determined by an eigenvalue of the matrix Cijkl nl n j . Because this matrix is symmetric and positive-definite, each direction of propagation will have three distinct plane waves. Associated with each eigenvalue is an eigenvector, ai , which is determined up to a scalar. This eigenvector gives the direction of the displacement of the plane wave. The three eigenvectors associated with the three velocities are orthogonal to one another. However, none of them need be longitudinal or transverse to the direction of propagation. Let the three wave velocities be c1 , c2 , c3 , and the associated eigenvectors be a1 , a 2 , a3 . These quantities depend on the direction of the propagation, s. The general form of a plane wave takes the form sx sx sx ux, t a1 f1 t a2 f 2 t a3 f3 t . c1 c2 c3 5/31/16 Waves-9 ES240 Solid Mechanics Z. Suo Direction of reflection or refraction. Snell’s law. Two materials are bonded on a planar interface. An incident plane wave propagates in direction s and velocity c. We would like to determine the directions of reflected and refracted waves. Let one of such waves propagate in sx t . direction s at velocity c . The field of the incident wave is a function of the argument c s x t . The continuity at the interface The field of the other wave is a function of the argument c must be maintained at all times. Consequently, the two arguments must be equal for all time and for all x on the interface, namely, s x s x , c c or s s x 0 c c s s for all x on the interface. That is, the vector is normal to the interface. Thus, the vector c c s is in the plane spanned by s and the normal vector of the interface, known as the plane of incidence. Let be the angle from the normal vector of the interface to s, and be the angle from s s is normal to the interface, in the direction the normal vector to s . Because the vector c c s s parallel to the interface, the two vectors and must have c c an equal magnitude and be in opposite directions. n Consequently, s sin sin . c c s This is known as Snell’s law. These equations determine the directions of the reflected and refracted waves. The arguments we have used are general, so that Snell’s law is applicable to any waves, e.g., elastic waves and electromagnetic waves. In general, an incident plane wave will generate three refracted reflected waves and three refracted waves. These waves propagate in directions in the plane of incidence, with the angles determined by Snell’s law. For an isotropic material, the wave speed is known, so that Snell’s law determines the incident direction of the reflected or refracted wave. For an anisotropic material, c itself is a function of , and the function c ( ) is determined by the eigenvalue problem. reflected The combination of Snell’s law and the function c ( ) determines both the wave speed c and the angle . 5/31/16 Waves-10 ES240 Solid Mechanics Z. Suo We can also give explicit equations for the unit vector in the direction of a reelected or refracted wave. Let n be the unit vector normal to the interface, pointing away from the half space that contains the incident wave. Thus, for the reflected wave, s s cos cos n c c c c For a refracted wave s s cos cos n . c c c c Amplitude of reflection. Consider a half space, and a known wave sx ux, t af t. c is incident upon the surface of the half space. Here, a is the unit vector in the direction of the displacement, s is the unit vector pointing in the direction of the propagation, c is the wave speed, and f is the wave form. Three waves reflected back. The net displacement field in the body is the superposition of the four waves: sx s x s x s x ux, t af t Aaf t Aaf t Aaf t c c c c Except for the amplitudes A, A, A , everything else about the reflected waves has been determined. To determine the amplitudes, we use the traction free boundary conditions. In particular, when x is on the surface of the half space, all the arguments of the functions are equal: sx s x s x s x t t t t c c c c Let n be the unit vector normal to the surface of the half space. The traction vector of the surface is u Aap sq Aap sq Aapsq df a s ti n j Cijpq p n j Cijpq p q . xq c c c d c The traction free condition ti 0 gives 3 linear algebraic equations for the three amplitudes: Aap sq Aap sq Aapsq a s 0. n j Cijpq p q c c c c Reflection from a free surface of an isotropic material. If the incident wave is a transverse wave with the displacement vector perpendicular to the plane of incidence, the wave is entirely reflected as a wave of the same kind, and the amplitude is 1. If the incident wave is a transverse wave with the displacement vector in the plane of incidence, the reflected consists of a transverse wave of the same kind and a longitudinal wave. The amplitudes of the two reflected waves are 2 2cl / ct sin 2 cos 2 sin 2l sin 2 cl / ct cos 2 2 , . At A l 2 2 sin 2l sin 2 cl / ct cos 2 2 sin 2l tan 2 cl / ct cos 2 2 5/31/16 Waves-11 ES240 Solid Mechanics Z. Suo If the incident wave is a longitudinal wave, the reflected wave consists of a longitudinal wave and a transverse wave with the displacement vector in the plane of incidence. The amplitudes of the two reflected waves are 2 sin 2 t sin 2 cl / ct cos 2 2 t 2cl / ct sin 2 cos 2t , At . Al 2 2 2 sin 2t sin 2 cl / ct cos 2 2t sin 2 t sin 2 cl / ct cos 2 t Exercise. A wave incident upon a bimaterial interface generates three reflected waves and three refracted waves. The amplitudes of the six waves are determined by the continuity of the displacement vector and the traction vector. Establish a set of six equations that determine the six amplitudes. Stroh representation. (Stroh, A.N., 1962. Steady state problems in anisotropic elasticity. Math. Phys. 41, 77-103.) A body is in a state of stress independent of the coordinate x3, and the source of stress is moving through the body at a constant speed v in the x1 direction. Let ( x, y ) be a coordinate system moving at the source speed, relating to the fixed coordinates as x x1 vt , y x2 . To an observer moving at speed v, the steady state field is time-independent. That is, the displacements are functions of x, y . Consider a specific steady state displacement field, ui Ai f ( x py ) , where A1 , A2 , A3 and p are constants, and f ( ) is an arbitrary, one-variable function. The displacement field will satisfy the equation of motion provided Ci1k1 v 2 ik pCi1k 2 pCi 2k1 p 2Ci 2k 2 Ak 0 . This is a set of linear algebraic equations for Ak . To represent a nontrivial displacement field, Ak cannot be all zero. Consequently, the determinant of the above equations must vanish, which leads to a polynomial equation of degree six in p. Denote the six roots by p , labeling 1, 2, 3 so that, if complex roots occur, p , p are complex conjugates, and giving the positive to the root with positive imaginary part. The labeling for real roots will be specified later. Following Stroh, we will only consider the case that the p are all distinct; equal roots may be regarded as the limiting case of distinct roots. For each p , we can determine a column Ak up to a scaling factor. Make Ak real when p is real, and Ak , Ak , complex conjugates when p is complex. For any six arbitrary functions f ( ) , the linear combination ui 3 A f z 1 i satisfies the equation of motion. Here z x p y . Summation over a Greek suffix will always be indicated explicitly. Any steady state solution can be represented in this form; the six Stroh functions f are to be determined by boundary conditions. Make f real when p is real, and f , f complex conjugate when p is complex, so that the displacements will always be real-valued. 5/31/16 Waves-12 ES240 Solid Mechanics Z. Suo One can expresses the stresses in terms of the Stroh functions: i2 3 L f z , i 1 i1 Li p v 2 Ai f z , 3 1 with Li Ci 2 k1 p Ci 2 k 2 Ak . We use ( ) to indicate the differentiation of any one-variable function. The above equationshold for any steady source speed whether greater or less than the sonic speeds of the solid. When v 0 , all the p are complex. When v is sufficiently large, all the p are real. There are three critical speeds, V3 V2 V1 . When v passes V , a pair of roots p change from complex to real. If then v V , the roots p are complex, and the functions f are complex analytic functions. If v V , the two roots p are real, and the equations, x p y constant , are the characteristic lines, along which the real functions f have constant values. For a half space, the anti-plane and the in-plane deformation decouple. We will only consider the in-plane deformation, for which the determinant has four roots: p 21 v 2 cl2 1 , p2 2 v 2 ct2 1 . The longitudinal and shear wave speeds are given by 12 12 2(1 ) , ct , cl (1 2 ) It is evident that cl and cs are also the critical speeds. When the crack speed v surpasses cl or cs, a pair of roots change from complex to real. The related matrices are all 2 by 2, as given by 2 p1 1 p 22 1 p2 , L A , 2 1 2 p2 p1 1 p 2 For a subsonic crack, v ct , all roots are imaginary numbers: p1 i 1 v 2 cl2 i l , Thus, p2 i 1 v 2 ct2 it . z1 x i l y, z2 x i t y . Rayleigh wave. A wave can propagate undiminished along the surface of a half space. However, the wave is localized near the surface, and the amplitude decays beneath the surface. There is no intrinsic length scale in this problem, so that the surface wave is nondispersive. We will only consider isotropic materials, although the method below is applicable to anisotropic materials. Say we seek a surface wave with the velocity v below the transverse body wave, so that the eigenvalues are both complex. Let the complex potential be f1 z1 , f 2 z2 . The displacement field is 2 2 1 1 ui Ai f z Ai f z . 5/31/16 Waves-13 ES240 Solid Mechanics Z. Suo The traction vector is 2 2 1 1 i 2 Li f z Li f z . On the surface of the half space, y 0 , the traction vanishes, so that Lf x Lf x 0 . The first term is analytic in the lower half plane, and the second term is analytic in the upper half plane. Consequently, Lf z 0 for any z. To have nonvanishing field, we must require that det L 0 . Consequently, the speed of surface wave is determined by 2 16 1 v 2 cl2 1 v 2 ct2 1 v 2 ct2 . This is an algebraic equation for v. An approximate solution of the Rayleigh wave speed is 0.87 1.12 vR . 1 Waves in laminates. Let us now look at waves propagating in a direction parallel to the interfaces of a laminate of different materials. An example is the Love wave, propagating in a laminate of a layer bonded to a half space. The thickness of each layer provides a length scale, so that the waves are dispersive. Each material has its own set of complex functions. We look for solutions of the form f z a exp ikz b exp ikz . Such a function ensures that the x dependence is sinusoidal, with the wave number k. The wave velocity v and the constants a and b are determined by requiring continuity of traction and displacements, which leads to an eigenvalue problem. The algebra is messy but straightforward. 5/31/16 Waves-14