东南大学：《弹性力学》课程教学课件（英文讲稿）08 Two-Dimensional Problems in Cartesian Coordinates

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

Polvnomial SolutionsThis method produces polynomial stress distributions, and thuswould not satisfy general boundary conditions However, we can modify such boundary conditions using SaintVenant's principle and replace a non-polynomial condition with astatically equivalent loadingThe solution to the modified problem would then be accurate atpoints sufficiently far away from the boundary where adjustmentswere madeThis formulation is most useful for problems with rectangulardomains in which one dimension is much larger than the otherThis would include a variety of beam problems6

• This method produces polynomial stress distributions, and thus would not satisfy general boundary conditions. • However, we can modify such boundary conditions using Saint￾Venant’s principle and replace a non-polynomial condition with a statically equivalent loading. • The solution to the modified problem would then be accurate at points sufficiently far away from the boundary where adjustments were made. • This formulation is most useful for problems with rectangular domains in which one dimension is much larger than the other. This would include a variety of beam problems. Polynomial Solutions 6

Example: Uniaxial Tension of a Bar4y+2cx→21Boundary Conditions: ,(±l,y)= T, t(±l,y)=O,),,(x,±c)=0, Tx(x,±c)=0Since the boundary conditions specify constant stresses on all boundaries, try asecond-orderstressfunctionoftheform=Ao2J2 ,=2A02 ,,=Tx =0The first boundary condition implies that Ao2 = T/2, and all other boundaryconditions are identically satisfied.Thereforethe stress field solutionis givenbyO,=T,O,=Tx=0.7

Example: Uniaxial Tension of a Bar • Boundary Conditions: ( , ) , ( , ) 0; ( , ) 0, ( , ) 0 . x xy y xy στ σ τ ± = ± = ±= ±= ly T ly x c x c • Since the boundary conditions specify constant stresses on all boundaries, try a second-order stress function of the form 2 02 02 2, 0 ψ σ στ = ⇒ = == Ay A x y xy • The first boundary condition implies that A02 = T/2, and all other boundary conditions are identically satisfied. Therefore the stress field solution is given by , 0. σ στ x = = = T y xy 7

Example: Uniaxial Tension of a BarDisplacement Field (Plane Stress)ouT1Tu1EaxEEav2T1y+g(x)1.:E[ayEEOuOv= 0 = f'(y)+g(x) =02gayaxuf(y)= -O.y+ u。.. . Rigid-Body Motion一(g(x) = 0,x +v。They do not contribute to the strain or stress fields. Recall that, thedisplacements are determined from the strain field up to an arbitrary rigid-body motion.“Fixity conditions" needed to determine these terms explicitly, i.eu(0,0) =v(0,0) = 0,(0,0) = 0 = f(y)= g(x)= 08

Example: Uniaxial Tension of a Bar • Displacement Field (Plane Stress) 1 ( ) ( ) 1 ( ) ( ) 2 0 () () 0 ( ) ( ) x xy y yx xy xy o o o o u T T u x fy xE E E v T T v y gx yE E E u v f y gx y x fy y u gx x v ε σ νσ ε σ νσ ν ν τ ε µ ω ω ∂  == − = = + ∂      ⇒ ∂  = = − =− =− + ∂  ∂ ∂ + = = =⇒ + = ′ ′ ∂ ∂  =− + ⇒   = + . . . Rigid-Body Motion • They do not contribute to the strain or stress fields. Recall that, the displacements are determined from the strain field up to an arbitrary rigid￾body motion. u(0,0) = v(0,0) = ωz (0,0) = 0 ⇒ f ( y) = g(x) = 0 • “Fixity conditions” needed to determine these terms explicitly, i.e. 8

Example: Pure Bending of a Beamty1M2cMx+21Boundary Conditions: ,(x,±c)= 0 , Tx,(x,±c)=↑x(±l,y)= 0[o (±l,y)dy=0, ,(l,y)ydy = -MExpecting a linear bending stress distribution, try third-order stress function ofthe form= Ao3J3 → ,=6Ao3J,,=T, = 0Moment boundary condition implies that Ao3 = -M/4c3, and all other boundaryconditions are identically satisfied.Thus the stressfield is3My,,=Tx=0a2c3.9

Example: Pure Bending of a Beam • Boundary Conditions: • Expecting a linear bending stress distribution, try third-order stress function of the form 3 03 03 6, 0 ψ σ στ = ⇒ = == Ay Ay x y xy • Moment boundary condition implies that A03 = -M/4c3, and all other boundary conditions are identically satisfied. Thus the stress field is ∫− ∫− σ ± = σ ± = − σ ± = τ ± = τ ± = c c x c c x y xy xy l y dy l y ydy M x c x c l y ( , ) 0 , ( , ) ( , ) 0 , ( , ) ( , ) 0 3 3 , 0 2 x y xy M y c σ στ =− = = 9

Example: Pure Bending of a BeamDisplacementField (Plane Stress)ou3M3M1voxy+ f(y)2Ec3yuaxE2Ec3av13M3Mvvo.)二+ g(x)2Ec3J2ayE4Ec3MOuavx+ f'(y)+g(x)=002Ec3ayaxf(y)=-oy+u3M40*+0++g(x)“Fixity conditions" to determine rigid-body motion terms, i.e. a simplysupported beam3Mu:2Ec3v(±l, 0) = 0 and u(-l, 0) = 03M?3M=u。=0。=0, V。14Ec34Ec310

• “Fixity conditions” to determine rigid-body motion terms, i.e. a simply supported beam 3 3 2 3 3 3 2 3 1 3 3 ( ) ( ) 2 2 1 3 3 ( ) ( ) 2 4 3 0 () () 0 2 ( ) 3 ( ) 4 x xy y yx o o o o u M M y u xy f y x E Ec Ec v M M y v y gx y E Ec Ec uv M x f y gx y x Ec fy y u M gx x x v Ec ε σ νσ ν ε σ νσ ν ω ω ∂  = = − =− =− + ∂      ⇒ ∂  == − = = + ∂  ∂ ∂ + = ⇒− + + = ′ ′ ∂ ∂  =− +  ⇒  = ++   2 3 ( ,0) 0 and ( ,0) 0 3 0, 4 oo o vl ul Ml u v Ec ω ±= −= ⇒ = = =− • Displacement Field (Plane Stress) Example: Pure Bending of a Beam 10 ( ) 3 2 22 3 3 2 3 4 M u xy Ec M v yxl Ec ν  = −  ⇒   = +− 