# R r θ 2 x y 1 x x figure p7 27 ˆ z g ρ f e l z fig

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r r θ 2 x y = 1 x x = Figure P7-27 ˆ z g ρ = − f e L z Fig. P7.30
166 SOLUTIONS MANUAL Solution: Using the cylindrical coordinate system shown in the figure above, the bound- ary conditions of the problem can be stated as: At z = 0 , L : σ zz = 0 , σ zr = 0 , σ = 0 , At r = a : u r = 0 , u θ = 0 , u z = 0 . (1) The body force component is f z = - ρg . Neglecting the end effects, we can assume the following form of solution for the antiplane strain problem: u r = 0 , u θ = 0 , u z = U ( r ) . (2) The nonzero strains and stresses are given by 2 ε rz = dU dr , σ zr = μ dU dr . (3) Substitution into the third equilibrium equation yields (the other two equilibrium equa- tions are identically satisfied) zr dr + 1 r σ zr - ρg = 0 1 r d dr r dU dr = ρg μ . (4) The solution of the equation is given by r dU dr = ρg μ r 2 2 + A, or dU dr = ρg μ r 2 + A r , U ( r ) = ρg μ r 2 4 + A log r + B, where A and B are constants of integration that must be determined using the boundary conditions. The boundary conditions associated with the antiplane strain problem are that (a) U ( r ) is finite at r = 0, and (b) U ( a ) = 0. The first condition gives A = 0 and the second one leads to B = - ( ρg/ 4 μ ) a 2 . The first boundary condition can be replaced by the requirement that σ rz = 0 at r = 0. This will lead to the conclusion that dU/dr = 0 at r = 0, from which we arrive at the same result (that is, A = 0). The solution becomes u z ( r ) = U ( r ) = - ρga 2 4 μ 1 - r 2 a 2 . (5) The stress field becomes σ θz = 0 , σ zr = ρg 2 r. (6) Note that the boundary conditions (1) of the 3D problem are not satisfied at z = 0 , L . Hence, it is only an approximate solution. 7.31 An external hydrostatic pressure of magnitude p is applied to the surface of a spherical body of radius b with a concentric rigid spherical inclusion of radius a , as shown in Fig. P7.31. Determine the displacement and stress fields in the spherical body. Using the stress field obtained, determine the stresses at the surface of a rigid inclusion in an infinite elastic medium. a b p Rigid spherical core Spherical shell ( ) , μ λ Figure P7-32 Fig. P7.31
CHAPTER 7: LINEARIZED ELASTICITY 167 Solution: We use the semi-inverse method to solve the problem. Assume that u φ = u θ = 0 and u R = U ( R ). The boundary conditions are: At r = b : σ RR = - p ; At R = a : u R = U ( a ) = 0 (1) The solution of the Navier equations give the displacement [see Eq. (7.3.13)] U ( R ) = AR + 1 R 2 B, σ RR ( R ) = (2 μ + 3 λ ) A - 4 μ R 3 B, (2) where μ and λ are the Lam´ e constants. Using the boundary conditions, we obtain (2 μ + 3 λ = 3 K ) U ( a ) = 0 B = - a 3 A ; σ RR ( R ) = 3 K + 4 a 3 μ R 3 A, (3) σ RR ( b ) = - p A = - p 3 K + 4 a 3 μ b 3 . (4) Hence the displacement u R and stress field are given by u R ( R ) = - b 3 pR 3 Kb 3 + 4 μa 3 1 - a 3 R 3 (5) σ RR ( R ) = - 1 + 2 α ( a 3 /R 3 ) 1 + β p, α = 2 μ 3 K , β = 2 α a 3 b 3 (6) σ θθ ( R ) = σ φφ = - 1 - α ( a 3 /R 3 ) 1 + β p. (7) To obtain the stresses at the surface of a rigid inclusion in an infinite elastic medium, we let b → ∞ and obtain σ RR = - 1 + 4 μa 3 3 KR 3 p, σ θθ = σ φφ = - 1 - 2 μa 3 3 KR 3 p . (8) At the interface of the rigid inclusion and the elastic medium ( R = a ), the stresses are σ RR = - 1 + 4 μ 3 K p, σ θθ ( R ) = σ φφ = - 1 - 2 μ 3 K p . (9) 7.32 Consider the concentric spheres shown in Fig. P7.32. Suppose that the core is elastic
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