Two-Dimensional Problemsin Cartesian Coordinates
Two-Dimensional Problems in Cartesian Coordinates
OutlineIntroductionPolynomial SolutionsUniaxial Tension of a BarPure Bending of a BeamBeam under Uniform Transverse LoadingRiver DamFourier MethodsBeam under Sinusoidal Loading Rectangular Domain with Arbitrary Symmetric TractionLoads2
Outline • Introduction • Polynomial Solutions • Uniaxial Tension of a Bar • Pure Bending of a Beam • Beam under Uniform Transverse Loading • River Dam • Fourier Methods • Beam under Sinusoidal Loading • Rectangular Domain with Arbitrary Symmetric Traction Loads 2
Two-Dimensional Plane Elasticity. Using the Airy Stress Function approach, it was shownthat the plane elasticity formulation with harmonic bodyforce potential reduces to a single governing biharmonicequation.atyQQU72Tax1+KayaaxayaxoyBoundary conditions need to be satisfied to complete asolutionInverse or Semi-Inverse Method is typically applied3
Two-Dimensional Plane Elasticity • Using the Airy Stress Function approach, it was shown that the plane elasticity formulation with harmonic body force potential reduces to a single governing biharmonic equation. • Boundary conditions need to be satisfied to complete a solution. • Inverse or Semi-Inverse Method is typically applied. ( ) 4 44 4 2 4 22 4 22 2 2 2 2 1 2 1 , , x y xy V x xy y V V y x x y ψ ψψ κ ψ κ ψψ ψ σ στ ∂ ∂∂ − + + =∇ = ∇ ∂ ∂∂ ∂ + ∂∂ ∂ = + = + =− ∂ ∂ ∂ ∂ 3
Polvnomial Solutions: In Cartesian coordinates we choose Airy stress function solution ofXpolynomial formV(x,y)=ZZAm"y"m=0 n=0 Noted that the three lowest order terms with m + n ≤ 1 do notcontribute to the stresses and will therefore be droppedSecond order terms will produce a constant stress fieldy(x, y)= A2ox? + Aixy+ Ao22ay2 Ao2, 0, = 2A20, Tx, =-AOv2. Third-order terms will give a linear distribution of stress, and so onfor higher-order polynomials.Terms with m + n ≤ 3 will automatically satisfy the biharmonicequation for any choice of constants A,mn4
Polynomial Solutions • In Cartesian coordinates we choose Airy stress function solution of polynomial form 0 0 (, ) m n mn m n ψ xy A x y ∞ ∞ = = = ∑∑ • Noted that the three lowest order terms with m + n ≤ 1 do not contribute to the stresses and will therefore be dropped. • Second order terms will produce a constant stress field • Third-order terms will give a linear distribution of stress, and so on for higher-order polynomials. • Terms with m + n ≤ 3 will automatically satisfy the biharmonic equation for any choice of constants Amn. 2 2 20 11 02 2 2 02 20 11 (, ) 2, 2, x y xy x y A x A xy A y AAA y ψ ψ σ στ = ++ ∂ = = = = − ∂ 4
Polynomial Solutions. For m + n ≥ 4 , constants Amn will have to be related to have thepolynomial satisfy the biharmonic equation. (Specifying additionalequations on Amn.)a4a4a422A.."y"0=axayaxa1m=0n=0=ZZm(m-1)(m-2)(m-3) Am*"-y" +2ZZm(m-1) n(n-1) Am*"-~ y"-2m=4n=0m=2 n=2n(n-1)(n-2)(n-3) A.mx"y"-4m=0n=4(m+2)(m+1)m(m-1) Am+2,n-2 +2m(m-1) n(n-1) AmZ+(n+2)(n+1)n(n-1) Am=-2,n+2(m+2)(m+1)m(m-1) Am+2,n-2 +2m(m-1)n(n-1) Am+(n +2)(n+1)n(n-1) Am-2,n+2 = 05
• For m + n ≥ 4 , constants Amn will have to be related to have the polynomial satisfy the biharmonic equation. (Specifying additional equations on Amn.) ( )( )( ) ( ) ( ) ( )( )( ) ( )( ) ( ) ( ) ( ) ( )( ) ( ) 4 44 4 4 22 4 0 0 4 2 2 4 0 2 2 4 0 4 2 2 2 2 0 2 123 2 1 1 123 21 1 2 1 1 21 1 m n mn m n m n m n mn mn m n m n m n mn m n m n mn m n Axy x xy y mm m m A x y mm nn A x y nn n n A xy m m mm A mm nn A n n nn A ψ ∞ ∞ = = ∞ ∞ ∞ ∞ − − − = = = = ∞ ∞ − = = + − − + ∂ ∂∂ =∇ = + + ∂ ∂∂ ∂ = −− − + − − + −− − ++ − + − − = ++ + − ∑∑ ∑∑ ∑∑ ∑∑ , , ( )( ) ( ) ( ) ( ) ( )( ) ( ) 2 2 2 2 2 2 2 2 21 1 2 1 1 21 1 0 m n m n m n mn m n x y m m mm A mm nn A n n nn A ∞ ∞ − − = = + − − + ++ − + − − ⇒ ++ + − = ∑∑ , , Polynomial Solutions 5