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By Martin H. Sadd

Workouts came upon on the finish of every bankruptcy are a big factor of the textual content as they supply homework for scholar engagement, difficulties for examinations, and will be utilized in classification to demonstrate different gains of the subject material. This suggestions guide is meant to help the teachers of their personal specific use of the workouts. assessment of the options can help you be sure which difficulties could top serve the pursuits of homework, assessments or be utilized in category.

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

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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.

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