Solids Mechanics, Including Elastic Wave Propagation. Professor - PowerPoint PPT Presentation

Solids Mechanics, Including Elastic Wave Propagation. Professor Julius Kaplunov – Typeset by Foil T EX –

Part 1. Elementary introduction. 1D rod 1. Equilibrium Equation (”Statics”) ∫ x +∆ x A { σ ( x + ∆ x ) − σ ( x ) } = Aρ u tt ( ξ ) dξ x Next, as it follows from the Mean Value theorem σ ( x + ∆ x ) − σ ( x ) = ρ ∆ xu tt ( x + θ ∆ x ) , 0 ≤ θ ≤ 1 Let us ∆ x → 0 σ ( x + ∆ x ) − σ ( x ) lim = ρ lim ∆ x → 0 u tt ( x + θ ∆ x ) ∆ x ∆ x → 0 – Typeset by Foil T EX – 1

σ x = ρu tt 2.Strain (”Geometry”) u ( x + ∆ x ) − u ( x ) lim = u x = ϵ ∆ x ∆ x → 0 3. Constitutive relations (”Physics”) A. Elasticity (i) σ = Eϵ - Hook’s Law for a Linearly elastic rod strains are small ϵ ≪ 1 √ Now σ x = Eu xx and u xx − 1 E c 2 u tt = 0 with c = ρ (ii) Finite elastic deformations ϵ ∼ 1 – Typeset by Foil T EX – 2

Say, now σ = Eϵ + ϵ 3 which is an example of physical non-linearity. Then, ) 2 ( du σ x = Eu xx + 3 η u xx dx and u xx − 1 c 2 u tt + 3 η E u 2 x u xx = 0 If η is small, then it is a room for asymptotics. B. Plasticity (i) Perfect plasticity (ii) Plasticity with hardening / softening – Typeset by Foil T EX – 3

C. Elastic - plastic materials (i) Elastic - perfectly plastic material (ii) Elastic - hardening plastic material D. Time - dependent materials Small deformations σ = Eϵ + µϵ t - Voight material σ x = Eu xx + µu xxt u xx − 1 c 2 u tt + µ Eu xxt = 0 Often we get a small µ finalising with a singular perturbed problem. – Typeset by Foil T EX – 4

Part 2. Linear Isotropic Elasticity. 2.1. Stress and Strain tensors and constitutive relations. 1. Stress tensor σ ij Equilibrium eqns σ ij,i = ρu j,tt (2 . 1) with j = 1 , 2 , 3 and Einstein convention is assumed Symmetry : σ ij = σ ji 2. Strain tensor ϵ ij = 1 2( u i,j + u j,k ) (2 . 2) i, j = 1 , 2 , 3 – Typeset by Foil T EX – 5

1 → x | x 1 2 → y | x 2 3 → z | x 3 3. Constitutive relation σ ij = λϵ kk δ ij + 2 µϵ ij (2 . 3) λ and µ denote Lame constants Young modulus E = µ (3 λ + 2 µ ) λ + µ Poisson ratio λ ν = 2( λ + µ ) usually (0 < ν < 0 . 5) – Typeset by Foil T EX – 6

ϵ ij = 1 + ν σ ij − ν Eσ kk δ ij E 2.2. Equation of Motion in terms of displacements. Clear from (2.2) and (2.3) σ ij,i = λδ ij ϵ kk,i + µ ( u i,ji + u j,ii ) On substituting into (2.1) ( λ + µ ) ϵ kk,j + µ ∆ u j = ρu j,tt Finally, ( λ + µ ) graddiv − → u + µ ∆ − → u − ρ − → u tt = 0 (2 . 4) where − → u = ( u 1 , u 2 , u 3 ) – Typeset by Foil T EX – 7

2.3. Shear and Dilatation waves. u = gradφ + curl − → → − ψ (2 . 5) It follows from (2.5) div − → u = ∆ φ On substituting (2.5) into (2.4) and taking into account the last formula 2 ∆ − → ψ − − → grad ( c 2 1 ∆ φ − φ tt ) + curl ( c 2 ψ tt ) = 0 1 = λ +2 µ 2 = µ where c 2 , c 2 ρ denote the speeds of the dilatation and β shear waves. Thus, c 2 1 ∆ φ − φ tt = 0 and – Typeset by Foil T EX – 8

2 ∆ − → ψ − − → c 2 ψ tt = 0 (2 . 6) 4. Rayleigh and Love waves. 4.1. Plane and antiplane strain. ∂ For both of them ∂x 3 ≡ 0 Plane strain u i = u i ( x 1 , x 2 ) , i = 1 , 2 and u 3 = 0 Antiplane strain u i = 0 , u 3 = u 3 ( x 1 , x 2 ) 4.2. Rayleigh waves. Consider plane strain of a half - space x 2 ≥ 0 Traction free surface x 2 = 0 ( σ 22 = σ 21 = 0) (2 . 7) – Typeset by Foil T EX – 9

( σ 23 ≡ 0) for plane strain. Traveling wave solutions φ = Ae − αx 2 + iq ( x 1 − ct ) (2 . 8) ψ 3 = Be − βx 2 + iq ( x 1 − ct ) ; ψ i = 0( i = 1 , 2) Conditions : α > 0 , β > 0 correspond to surface waves (decay as x 2 → ∞ ) Here c - phase velocity. On substituting (2.8) into (2.6) and (2.8) into (2.5) we get √ √ 1 − c 2 1 − c 2 α = q β = q 1 , c 2 c 2 2 u 1 = φ , 1 + ψ 3 , 2 = iq ( Ae − αx 2 − βBe − βx 2 ) – Typeset by Foil T EX – 10

u 2 = φ , 2 − ψ 3 , 1 = ( − αAe − αx 2 − iqBe − βx 2 ) (2 . 9) Here and below we omit factor exp [ iq ( x − ct )] . On making use of constitutive relations (2.3) in (2.7) and taking into account geometric relations (2.2) and formulae (2.9) we arrive at homogeneous equations in A and B. They are √ 2 − c 2 ( ) 1 − c 2 A + 2 i B = 0 c 2 c 2 2 2 √ 2 − c 2 1 − c 2 ( ) − 2 i A + B = 0 (2 . 10) c 2 c 2 2 2 Solvability of (2.10) yields R ( γ ) = (2 − γ 2 ) 2 − 4 √ (1 − γ 2 )(1 − κ 2 γ 2 ) = 0 – Typeset by Foil T EX – 11

With √ 1 − 2 γ √ γ = c r κ = c 2 µ = 2 − γ = (2 . 11) , λ + 2 µ c 2 c 1 R - Rayleigh denominator, c = c r - Rayleigh wave speed. We will prove that there exist a root γ < 1 of R ( γ ) = 0 at 0 ≤ ν < 1 2 . This root is unique for given ν . 4.3. Love waves. c 2 < c ∗ 2 Antiplane problem ∂ 2 u 3 + ∂ 2 u 3 ∂ 2 u 3 = 1 ∂x 2 ∂x 2 c 2 ∂t 2 1 2 2 ∂ 2 u ∗ + ∂ 2 u ∗ ∂ 2 u ∗ = 1 3 3 3 (2 . 12) ∂x 2 ∂x 2 c ∗ 2 ∂t 2 1 2 2 – Typeset by Foil T EX – 12

Traction free surface x 2 = − H , i.e. ⇒ ∂u 3 σ 23 = 0 ⇐ = 0 (2 . 13) ∂x 2 Contact conditions at x 2 = 0 , u 3 = u ∗ 3 , σ 23 = σ ∗ 23 or ∂u ∗ µ∂u 3 3 = µ ∗ (2 . 14) ∂x 2 ∂x 2 Let us u 3 = f ( x 2 ) e iq ( x 1 − ct ) (2 . 15) 3 = f ∗ ( x 2 ) e iq ( x 1 − ct ) u ∗ – Typeset by Foil T EX – 13

On substituting (2.15) into (2.12) we get   √ ∂ 2 f c 2 + q 2 α 2 f = 0 ,  α = − 1  ∂x 2 c 2 2 2   √ ∂ 2 f ∗ 1 − c 2 − q 2 β 2 f ∗ = 0 ,  β = (2 . 15)  ∂x 2 c ∗ 2 2 2 Thus, we have a decaing at x 2 → ∞ wave (Love wave): f ( x 2 ) = A sin( αqx 2 ) + B cos( αqx 2 ) f ∗ ( x 2 ) = Ce − βqx 2 (2 . 16) It follows from contact conditions (2.14) A = − µ ∗ β B = C, µα C – Typeset by Foil T EX – 14

Then, the substitution of (2.16) into boundary condition (2.13) at free surface yields tan( αqH ) = µ ∗ β (2 . 17) µα which is the dispersion relation for Love waves. It determines the phase speed versus wave number, i.e. c = c ( qH ) . There are infinitely many Love waves. Part 3. Lamb (Rayleigh - Lamb)waves. Consider an infinite layer of thickness 2 h with fraction free faces Recall the equations of motion in plane strain 2(1 + ν ) ∗ (1 − 2 ν ) graddivu − ρ∂ 2 u E E 2(1 + ν )∆ u + ∂t 2 = 0 , (3 . 1) – Typeset by Foil T EX – 15

where u = ( u 1 , u 2 , 0) is the displacement vector whose components do not depend on x 3 ( u k = u k ( x 1 , x 2 , t ) , k = 1 , 2); ∆ is Laplacian. The ”displacements - stresses” formulae are ( ∂u 1 ) E ν ∂u 2 σ 11 = + , 2(1 + ν ) κ 2 1 − ν ∂x 1 ∂x 2 ( ∂u 1 ) Eν + ∂u 2 σ 33 = , (3 . 2) 2(1 − ν 2 ) κ 2 ∂x 1 ∂x 2 ( ) E ν ∂u 1 + ∂u 2 σ 22 = , 2(1 + ν ) κ 2 1 − ν ∂x 1 ∂x 2 E ( ∂u 2 + ∂u 1 ) σ 21 = , 2(1 + ν ) κ 2 ∂x 1 ∂x 2 σ 13 = σ 23 = 0 . – Typeset by Foil T EX – 16

We impose homogeneous boundary conditions on faces x 2 = ± h σ 21 = σ 22 = 0 (3 . 3) We specify displacement as before in (2.9) u 1 = ∂φ + ∂ψ 3 , u 2 = ∂φ − ∂ψ 3 (3 . 4) ∂x 1 ∂x 2 ∂x 2 ∂x 1 where ϕ and ψ 3 are potentials. Substituting (3.4) into (3.1) we obtain two equations ∂ 2 φ ∂ 2 ψ 3 ∆ 2 φ − 1 ∆ 2 ψ 3 − 1 ∂t 2 = 0 , (3 . 5) ∂t 2 , c 2 c 2 1 2 where ∆ 2 = ∂ 2 + ∂ 2 . ∂x 2 ∂x 2 1 2 – Typeset by Foil T EX – 17

Let us introduce the dimensionless coordinates ξ 1 = x 1 h , ζ = x 2 h and τ = tc 2 h and seek the solution to equations (3.5) in the form φ = f ( ζ ) exp[ i ( Kξ 1 − Ω τ )] , ψ 3 = g ( ζ ) exp[ i ( Kξ 1 − Ω τ )] (3 . 6) Inserting the latter into (3.5) we have ∂ 2 f ∂ζ 2 − α 2 f = 0 , (3 . 7) ∂ 2 g ∂ζ 2 − β 2 g = 0 , (3 . 8) where α 2 = K 2 − κ 2 Ω 2 , β 2 = K 2 − Ω 2 . The vibration modes corresponding to the above equations are separated into two groups. The modes of the first group are – Typeset by Foil T EX – 18

symmetric with respect to the midplane of the layer ζ = 0 and those of the second group are antisymmetric. First, examine the symmetric modes . For them the displacement u 1 and the stresses σ 11 , σ 22 , σ 33 are even with respect to the thickness variable ζ and the displacement u 2 and the stress σ 21 are odd. The solutions to the equations (3.7), (3.8) are given by f = A cosh( αζ ) , g = B sinh( βζ ) (3 . 9) where A and B are arbitrary constants. Because of the symmetry it is sufficient to obey the boundary conditions only on one of the faces. The boundary conditions on the other face are satisfied automatically. Expressing the stresses entering into the boundary conditions (3.3) in terms of the functions f and g defined by formulae (3.9) we obtain a system of two linear equations: AKiα sinh α + Bγ 2 sinh β = 0 – Typeset by Foil T EX – 19

Aγ 2 cosh α − BKiβ cosh β = 0 (3 . 10) where γ 2 = K 2 − 1 2Ω 2 . Equating the determinant of this system to zero we obtain the Rayleigh-Lamb dispersion equation [classical works by Lord Rayleigh (1889) and Lamb (1889)] γ 4 cosh α sinh β − α 2 K 2 sinh α cosh β = 0 . (3 . 11) β α Displacements and stresses are expressed as γ 2 sinh β cosh( αζ ) − αβ sinh α cosh( βζ ) ( ) u 1 = RKi , γ 2 sinh β sinh( αζ ) − K 2 sinh α sinh( βζ ) ( ) u 2 = Rα , – Typeset by Foil T EX – 20