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 SaintVenant’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 rigidbody 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 = − ⇒ = +−