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Extra resources for Elasticity: Theory, Applications and Numerics, Second Edition: Solutions Manual

Example text

All rights reserved. 3-25. 14) or exercise 1 - 16, ∇ ⋅ u = 1 ∂ 1 ∂u θ ∂u z ( rur ) + + r ∂r r ∂θ ∂z ∂Tr 1 1 ∂Tθ ∂Tz + Tr + + r ∂θ ∂r r ∂z ∂ 1 = (σ r e r + τ rθ eθ + τ rz e z ) + (σ r e r + τ rθ eθ + τ rz e z ) r ∂r 1 ∂ ∂ + ( τ rθ e r + σ θ eθ + τ θz e z ) + (τ rz e r + τ θz eθ + σ z e z ) r ∂θ ∂z ∂τ ∂σ ∂τ 1 = r e r + rθ eθ + rz e z + ( σ r e r + τ rθ eθ + τ rz e z ) r ∂r ∂r ∂r ∂σ ∂τ 1  ∂τ  +  rθ e r + τ rθ eθ + θ eθ − σ θ e r + θz e z  r  ∂θ ∂θ ∂θ  ∂τ ∂τ ∂σ + rz e r + θz e θ + z e z ∂z ∂z ∂z Collecting terms in each coordinate direction and adding the appropriated body force gives ⇒ ∇⋅σ = ∂σ r 1 ∂τ rθ ∂τ rz 1 + + + ( σ r − σ θ ) + Fr = 0 r ∂r r ∂θ ∂z ∂τ rθ 1 ∂σ θ ∂τ θz 2 + + + τ rθ + Fθ = 0 r ∂θ r ∂r ∂z ∂τ rz 1 ∂τ θz ∂σ z 1 + + + τ rz + Fz = 0 ∂r ∂z r ∂θ r Copyright © 2009, Elsevier Inc.

Ij = Cijkl ekl = (αδij δ kl + βδ ik δ jl + γδil δ jk )ekl = αekk δij + β eij + γe ji = αekk δij + (β + γ )eij letting α = λ and (β + γ ) = 2μ, σij = λekk δij + 2μeij 4-4. 7) ⇒ α = λ and β = μ , γ = μ ⇒ Cijkl = λδ ij δ kl + μ(δil δ jk + δik δ jl ) 2 2 Since λ = k − μ ⇒ Cijkl = μ(δil δ jk + δik δ jl ) + (k − μ)δij δ kl 3 3 E Eν Since μ = and λ = ⇒ 2(1 + ν) (1 + ν)(1 − 2ν) Eν E Cijkl = (δil δ jk + δik δ jl ) δij δ kl + (1 + ν)(1 − 2ν) 2(1 + ν) 4-5. 10)providing and 2μ 2μ(3λ + 2μ) E E μ(3λ + 2μ) λ and this will be true if E = and ν = (λ + μ ) 2(λ + μ) σij = λekk δij + 2μeij ⇒ σ kk = (3λ + 2μ)ekk ⇒ ekk = 4-6.

3-23. σ x = σ x ( x) , σ y = σ z = τ xy = τ yz = τ xz = 0 , Fx = ρg , Fy = Fz = 0 x ∂σ x ∂τ yx ∂τ zx ∂σ + + + Fx = 0 ⇒ x + ρg = 0 ⇒ σ x = − ∫ ρgdx = ρg (l − x) l ∂x ∂y ∂z ∂x ∂τ xy ∂σ y ∂τ zy + + + Fy = 0 ⇒ 0 = 0 ∂x ∂y ∂z ∂τ xz ∂τ yz ∂σ z + + + Fz = 0 ⇒ 0 = 0 ∂x ∂y ∂z 3-24. 0  − p 0  σ ij = − pδ ij = 0 − p 0    0 − p   0 σ ij , j + Fi = 0 ⇒ − p, j δ ij + Fi = 0 ⇒ p,i = Fi or ∇p = F Using a scalar approach ∂σ x ∂τ yx ∂τ zx ∂p + + + Fx = 0 ⇒ = Fx ∂x ∂y ∂z ∂x ∂τ xy ∂σ y ∂τ zy ∂p + + + Fy = 0 ⇒ = Fy ∂x ∂y ∂z ∂y ∂τ xz ∂τ yz ∂σ z ∂p + + + Fz = 0 ⇒ = Fz ∂x ∂y ∂z ∂z ∴ ∇p = F Copyright © 2009, Elsevier Inc.