A Summary Of Linear Continuum Mechanics 1. a b = a i b i = a b cos θ. det A = ɛ ijk A 1i A 2j A 3k. e 1 e 2 e 3 a b = ɛ ijk a j b k e i =

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1 A Summary Of Linear Continuum Mechanics 1 Vector Analysis Inner / Scalar / Dot product Norm of a vector Kronecker delta a b = a i b i = a b cos θ a = [a a] 1/2 = a 1 a 1 + a 2 a 2 + a 3 a 3 δ ij = I = e i e j = { 1 i = j 0 i j Levi-Civita / Permutation symbol 0 if 2 or 3 indices are equal ɛ ijk = 1 if (ijk) is an even permutation of (123) 1 if (ijk) is an odd permutation of (123) Important properties det A = ɛ ijk A 1i A 2j A 3k ɛ ijk ɛ imn = δ jm δ kn δ jn δ km ɛ ijk ɛ ijm = 2δ km Outer/cross product Tensor / Dyadic product e 1 e 2 e 3 a b = ɛ ijk a j b k e i = a 1 a 2 a 3 b 1 b 2 b 3 a b = a b sin θ A = a b A ij = a i b j or A = a i b j e i e j Some basic operations on tensors & vectors A a = A ij a j e i A a = ɛ jkl A ik a l e i e j a A = a i A ij e j a A = ɛ ikl a k A lj e i e j Coordinate transformation from (x 1, x 2, x 3 ) to ( x 1, x 2, x 3 ) ā i = Q ij a j, Ā ij = Q ik Q jl A kl : Q ij = ē i e j Nabla operator e i = e i i x i

2 2 A Summary Of Linear Continuum Mechanics Gradient operator Divergence operator Curl operator grad(φ) φ = φ x i e i = φ,i e i grad(a) a = a i x j e i e j = a i,j e i e j div(a) a = a i x i = a i,i div(a) A = A ij x j e i = A ij,j e i Important identities curl(a) a = ɛ ijk a j x k e i curl(a) A = ɛ jmn A im x n e i e j curl(grad(φ)) = 0 div(curl(a)) = 0 Laplacian operator φ div (grad (φ)) = i 2 φ x i = φ,ii Stress and Equilibrium Equations Symmetry of stress σ = σ T ; σ ij = σ ji Cauchy's formula σ n = t ; σ ij n j = t i Equilibrium equation σ + f = 0 Equilibrium in cylindrical coordinates e r : σ rr,r + 1 r (σ rr σ θθ ) + 1 r σ rθ,θ + σ rz,z + f r = 0 e θ : σ θr,r + 1 r σ θθ,θ + σ θz,z + 2 r σ θr + f θ = 0 e z : σ zr,r + 1 r σ zr + 1 r σ zθ,θ + σ zz,z + f z = 0

3 A Summary Of Linear Continuum Mechanics 3 Strain and Compatibility Displacement eld Innitesimal deformation u = x X dx dx + u dx Symmetric & antisymmetric decomposition in which u = ε + ω dx = dx + } ε {{ dx} + ω }{{ dx} dx e dx r strain tensor: ε = 1 2 (u + u) with ε rotation tensor: ω = 1 2 (u u) with ω = εt = ωt Rotation vector Stretch of an innitesimal line segment w = 1 2 ε ijkω jk e i = 1 ω 23 + ω 32 ω 13 ω 31 = u ω 12 + ω 21 l = dx dx = dx ε dx dx Change in the angle between two directions Strain compatibility condition ϕ = 2e 1 ε e 2 : e 1 = dx 1 dx 1, e 2 = dx 2 dx 2 Inc (ε) = ε = 0 Calculation of displacement eld from its gradient Michell-Cesaro formula x u (x) = u (x 0 ) + u ds x 0 x u (x) = u (x 0 ) + ω (x 0 ) [x x 0 ] + d s [ ε + ( ε ) ( x x) ] (0.1) x 0

4 4 A Summary Of Linear Continuum Mechanics Linear Elasticity Hooke's law for linear elastic isotropic material σ = λ tr (ε) I + 2µε or σ ij = λε kk δ ij + 2µε ij Volumetric-Deviatoric decomposition σ = σ vol + σ dev : ε = ε vol + ε dev : { σ vol = 1 3tr (σ) I σ dev = σ 1 3tr (σ) I { ε vol = 1 3tr (ε) I ε dev = ε 1 3tr (ε) I Some relations between material constants Alternative forms of Hooke's law Plane strain ε ij = 0 for i = 3 or j = 3 σ vol = 3Kε vol, σ dev = 2µε dev (0.2) K = λ µ µ (3λ + 2µ) E = λ + µ λ ν = 2 (λ + µ) ε = 1 2µ σ λ 2µ (3λ + 2µ) tr(σ) I ε = 1 + ν E σ ν E tr(σ) I σ = E 1 + ν ε νe (1 + ν) (1 2ν) tr(ε) I σ 11 = λ (ε 11 + ε 22 ) + 2µε 11, σ 22 = λ (ε 11 + ε 22 ) + 2µε 22, σ 33 = λ (ε 11 + ε 22 ) ; σ 12 = 2µε 12, σ 13 = 0, σ 23 = 0 Plane stress σ ij = 0 for i = 3 or j = 3 where λ = 2λµ λ+2µ σ 11 = λ (ε 11 + ε 22 ) + 2µε 11, σ 22 = λ (ε 11 + ε 22 ) + 2µε 22, Fundamental equations of elastostatics σ 33 = 0 ; σ 12 = 2µε 12, σ 13 = 0, σ 23 = 0 ε = 1 ( u + u ) 2 σ + b = 0 σ = λtr(ε) I + 2µε (0.3)

5 A Summary Of Linear Continuum Mechanics 5 Navier equation (λ + µ) u + µ u + b = 0 (0.4) Beltrami-Michell equation σ ν tr(σ) + ν b I + b + b = 0 (0.5) 1 ν Remark 1. The equilibrium equation σ + b = 0 still has to be satised together with Beltrami-Michell equation as it is not included in it. Displacement Functions Homogeneous Navier equation (no body force) Helmholtz theorem suggests (λ + µ) u + µ u + b = 0 u = φ + ψ where φ and ψ are scalar and vector potential functions. The result of substitution is Special solutions are of the form (λ + 2µ) φ + µ ψ = 0 (0.6) φ = constant, ψ = constant or more specialized φ = 0, ψ = 0 (0.7) Galerkin vector v oers a solution to Navier equation as u = k v v Special choice of gives the following form k = λ + 2µ λ + µ = 2 (1 ν) v = 0 (0.8) Love's strain function L is obtained by a special choice of Galerkin vector as v 1 = v 2 = 0, v 3 = L Then the Navier's problem is simplied to biharmonic equation L = 0 (0.9)

6 6 A Summary Of Linear Continuum Mechanics And the displacement eld is given immediately u 1 = 2 L x 1 x 3, u 2 = 2 L x 2 x 3, u 3 = 2 (1 ν) L 2 L x 2 3 (0.10) Papkovich-Neuber approach gives a completely general solution to Navier's equation that is based on harmonic functions, with displacement eld presented by [ u = A B + A x ] 4 (1 ν) in which A and B are vector and scalar elds, and x is the position vector. Then substitution in Navier equation yields µ A (λ + 2µ) B λ + µ (x A) = 0 2 which is satised if A = 0 B = 0 (0.11) Papkovich-Neuber functions are related to Galerkin vector by A = 2 (1 ν) v B = v A x 4 (1 ν) (0.12) Some Useful Mathematics Dierential operators in cylindrical coordinates (r, θ, z) e r r + e 1 θ r θ + e z z a = 1 r r (ra r) + 1 a θ r θ + a z z φ = 1 ( r φ ) φ r r r r 2 θ φ z 2 = 2 φ r φ r r φ r 2 θ φ z [ 2 1 a z a = r θ a ] [ θ ar e r + z z a ] z e θ + 1 [ r r r (ra θ) a ] r e z θ

7 A Summary Of Linear Continuum Mechanics 7 Dierential operators in spherical coordinates (r, θ, φ) e r r + e 1 θ r θ + e 1 φ r sin θ φ a = 1 ( r 2 ) 1 r 2 a r + r r sin θ θ (sin θa θ) + 1 a φ r sin θ φ ξ = 2 ξ r ξ r r + cos θ ξ r 2 sin θ θ ξ r 2 θ r 2 sin 2 θ a = 1 [ r sin θ θ (sin θa φ) a ] θ e r + 1 [ 1 φ r sin θ + 1 [ r r (ra θ) a ] r e φ θ Cylindrical Harmonics (Bessel's Dierential Equation) 2 ξ φ 2 a r φ r (ra φ) The cylindrical part of Laplace's equation takes the form of Bessel's equation of n th order (n R) as There are three kinds of functions satisfying this equation. x 2 d2 y dx 2 + x dy dx + ( x 2 n 2) y = 0 (0.13) ] e θ Bessel's functions of the rst kind J n (x) = 1 π π which are non-singular at x = 0. [see g. 1] 0 cos(nt x sin t) dt Fig. 1: Bessel's functions of the rst kind for n = 0, 1, 2, 3.

8 8 A Summary Of Linear Continuum Mechanics Bessel's functions of the second kind (Neumann Functions) which are singular at x = 0. [see g. 2] Y n (x) = J n(x) cos(nπ) J n (x) sin(nπ) Fig. 2: Bessel's functions of the second kind for n = 0, 1, 2, 3. Bessel's functions of the third kind (Hankel Functions) are the complex solutions to the Bessel's equation. They are linear combination of the two previous kinds. H ± n (x) = J n (x) ± iy n (x) Dierent cases of Bessel's equation and their solutions 1. For integer values of n the solution is of the form y(x) = C 1 J n (x) + C 2 Y n (x) 2. For non-integral and real values of n y(x) = C 1 J n (x) + C 2 J n (x) 3. For general complex solution y(x) = C 1 H + n (x) + C 2 H n (x)

9 A Summary Of Linear Continuum Mechanics 9 Spherical Harmonics The angular part of Laplace's equation in spherical coordinates after substitution of the ansatz y = Φ(φ) Θ(θ) gives sin θ Θ ( d sin θ dθ ) + l (l + 1) sin 2 θ = m 2 dθ dθ 1 d 2 Φ Φ dφ 2 = m2 ; m R (0.14) While the real valued solution to the second equation is very well known as Φ(φ) = Asin(mφ) + Bcos(mφ) solution of the rst equation is given for m { l,..., 0,..., l 1, l} as Θ(θ) = P m l (cos θ) with P m l (x) being the Legendre polynomial. Finally the real valued normalized spherical harmonics take the form Y ms l (θ, φ) = Y mc l (θ, φ) = 2l + 1 (l m)! 2π (l + m)! Pm l (cos θ) sin(mφ) 2l + 1 (l m)! 2π (l + m)! Pm l (cos θ) cos(mφ) The above functions have important orthogonality property (0.15) 2π π 0 0 Y m l (θ, φ) Y m l (θ, φ) sin θ dθ dφ = δ mm δ ll (0.16)