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EN0175-21

EN0175-21 - EN0175 11 16 06 Continue on the problem of...

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Unformatted text preview: EN0175 11 / 16 / 06 Continue on the problem of circular hole under uniaxial tension (remote). Stress concentration occurs at a r = , 2 π θ = . T T 3 a Governing equation is: 2 2 = ∇ ∇ φ The stress components in polar coordinates are: 2 2 2 1 1 θ φ φ σ ∂ ∂ + ∂ ∂ = r r r rr ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ∂ ∂ ∂ ∂ − = θ φ σ θ r r r 1 2 2 r ∂ ∂ = φ σ θθ Boundary conditions are: @ a r = , = = θ σ σ r rr @ ∞ = r , 1 1 e e T v v ⊗ = σ ( ) θ σ σ 2 cos 1 2 + = ⋅ = T e e r r rr v v ( ) θ σ θθ 2 cos 1 2 − = T θ σ θ 2 sin 2 T r − = Therefore, the boundary condition at infinity can be decomposed into two parts. Part I: 1 EN0175 11 / 16 / 06 @ ∞ = r , 2 T rr = = θθ σ σ . For this part, we have previously obtained the solution as ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ − = 2 2 1 2 r a T rr σ ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ + = 2 2 1 2 r a T θθ σ = θ σ r Part II: @ ∞ = r , θ σ 2 cos 2 T rr = , θ σ θθ 2 cos 2 T − = , θ σ θ 2 sin 2 T r − = These expressions suggests ( ) θ φ 2 cos r f = . Inserting it into gives 2 2 = ∇ ∇ φ ( ) 4 d d 1 d d 4 d d 1 d d 2 2 2 2 2 2 = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ − + ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ − + r f r r r r r r r r Assume: ( ) ( ) ( ) ( ) 4 , 2 , 2 , 2 2 4 − = ⇒ = + − − ⇒ = λ λ λ λ λ λ r r f ( ) 4 2 3 4 2 2 1 C r C r C r C r f + + + = − Using boundary conditions: @ a r = , = = θ σ σ r rr @ ∞ = r , θ σ 2 cos 2 T rr = , θ σ θ 2 sin 2 T r − = We can determine the constant coefficients as 2 EN0175 11 / 16 / 06 T C 4 1 1 − = , , 2 = C T a C 4 3 4 1 − = , T a C 2 4 2 1 = Adding the solution to part I, the complete solutions of stress components are: θ σ 2...
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EN0175-21 - EN0175 11 16 06 Continue on the problem of...

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