AIRY STRESS FUNCTION FOR TWO DIMENSIONAL INCLUSION ...

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u in θ rS( 1 + κ 1)sin 2θ = − . 4( κ µ + µ ) 1 1 30 (3.3.30) The above equations can be verified by substituting them into the equilibrium equations i.e. (2.1.7). Substitute the constants into equations (3.3.8) and (3.3.9) to obtain the final stress and displacement equations for the part surrounding the circular inclusion i.e. 2 2 µ κ µ out S a S a S 1 σ rr = − 2 + 2 + 2 2r ( 2µ − µ + κ µ ) 2r ( 2µ − µ + κ µ ) 1 1 1 1 2 2 a Sµ 1 a Sκ µ 1 1 2 − 2 + S cos2θ − 2r ( 2µ − µ + κ µ ) 2r ( 2µ − µ + κ µ ) 2 1 1 1 1 4 2 4 2 3a Sµ cos2θ 2a Sµ cos2θ 3a Sµ 1 cos2θ 2a Sµ 1 cos2θ 4 + 2 + 4 − 2 2r ( κ µ + µ ) r ( κ µ + µ ) 2r ( κ µ + µ ) r ( κ µ µ ) , + 1 1 1 1 1 1 2 2 2 µ κ µ out S a S a S a S 1 σ θθ = + 2 − 2 − 2 2 2r 2r ( 2µ − µ + κ µ ) 2r ( 2µ − µ + κ µ ) 1 1 4 4 1 3a Sµ cos2θ 3a Sµ 1 cos2θ − S cos2θ + 4 − 4 2 2r ( κ µ + µ ) 2r ( κ µ + µ ) , 1 1 1 1 4 2 out 1 3a Sµ sin 2θ a Sµ 2θ τ rθ = − S sin2θ − 4 2 + 2 2r ( κ µ + µ ) r κ µ + µ _ sin ( ) 1 1 4 2 3a Sµ 1 sin 2θ a Sµ 1 sin 2θ 4 − 2 2r ( κ µ + µ ) r ( κ µ + µ ) , 1 1 1 1 1 1 1 1 1 1 2 2 2 a S rS rSκ a S a S out 1 µ κ 1µ ur = − + − − + 4rµ 8µ 8µ 4rµ ( 2µ − µ + κ µ ) 4rµ ( 2µ − µ + κ µ ) 1 1 1 1 1 1 1 1 1 4 2 2 rS cos2θ a S cos2θ a S cos2θ a Sκ 1 cos2θ − 3 + + + 4µ 4r ( κ µ + µ ) 4r( κ µ + µ ) 4r( κ µ + µ ) 1 1 1 1 1 1 1 1 1 1 4 2 2 a Sµ cos2θ a Sµ cos2θ a Sκ 1µ cos2θ 3 − 4r µ ( κ µ + µ ) 4rµ ( κ µ µ ) 4rµ ( κ µ µ ) , − + + 1 1 1 1 1 1 (3.3.31)
u out θ 4 2 2 4 S( a ( − µ + µ 1) + a r ( − 1 + κ 1)( − µ + µ 1) + r ( κ 1µ + µ 1)) sin 2θ = − 3 . 4r µ ( κ µ + µ ) 1 1 1 31 (3.3.32) The above equations can be verified by substituting them into the equilibrium equations (2.1.7). Example Problem: Circular Inclusion Problem Consider finding the stresses for a thin Infinite plate with a hole made of Iron, having a two-dimensional circular inclusion made of carbon in the center, subjected to far field tensile loading of 1000N/mm 2 problem. we know that, for Carbon, Poisson’s Ratio, ν = 0.24, Shear Modulus µ= 12.4GPa, Parameter κ for the plane strain from (2.2.12)= 3 - 4ν = 2.04. for Iron Poissons Ratio, ν1 = 0.29, Shear Modulus µ1= 77.5GPa, Parameter κ1 for the plane strain from (2.2.12)= = 3 - 4ν = 1.836. Let, above, we get the radius of the central disc be, a= 100mm, Substituting the above values in the stress and displacement components derived
Page 1 and 2: AIRY STRESS FUNCTION FOR TWO DIMENS
Page 3 and 4: Supervising Professor: Dr. Seiichi
Page 5 and 6: 3.2 Infinite Plate with a Hole Subj
Page 7 and 8: CHAPTER 1 INTRODUCTION 1.1 Overview
Page 9 and 10: our best knowledge. This thesis add
Page 11 and 12: • A finite plate with a hole subj
Page 13 and 14: epresentation It is observed that t
Page 15 and 16: 2 2 2 2 ⎛ ⎞ ⎛ ⎞ 4 ∂ 1 ∂
Page 17 and 18: where λ is called the Lame’s con
Page 19 and 20: σ + σ = σ + σ r θ x y 2iθ σ
Page 21 and 22: γ ( z) = − ψ ( z) = n ∑ k = 1
Page 23 and 24: CHAPTER 3 ADVANCED APPLICATIONS OF
Page 25 and 26: σ σ x y ( ± l, y) = S, ( x, ± c
Page 27 and 28: Sz b b χ ( z) = (log z) b . z z
Page 29 and 30: from which the stress components ca
Page 31 and 32: in σ = 2c + 2c r cosθ − d cos2
Page 33 and 34: 1 2a 4 4 4 2 ( a S + a S cos2θ + 2
Page 35: a b c 1 1 0 2 a d 2 a2 κ 1a2 − 2
Page 39 and 40: τ τ in xy out xy = 0, 6 10 6 2 6
Page 41 and 42: -50 -100 -150 0 τ 100 150 200 alon
Page 43 and 44: CHAPTER 4 CONCLUSIONS AND SUGGESTIO
Page 45 and 46: APPENDIX A MATHEMATICA CODE 39
Page 47 and 48: χ = Integrate[ϕ,z] χ1 = Integrat
Page 49 and 50: σθθ1 =FullSimplify@HD@φ1, 8r,2
Page 51 and 52: THE SIMULTANEOUS EQUATIONS THAT HAV
Page 53 and 54: aS aSκ1 aa1 − − 8 µ1 8 µ1 2
Page 55 and 56: PLATE τrθ1Final = ExpandAτrθ1In
Page 57 and 58: VERIFICATION OF THE EQUILIBRIUM EQU
Page 59: BIOGRAPHICAL INFORMATION Dharshini,

AIRY STRESS FUNCTION FOR TWO DIMENSIONAL INCLUSION ...

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