东南大学：《弹性力学》课程教学课件（英文讲稿）10 Torsion of Prismatic Bars

Torsion of Prismatic Bars

Torsion of Prismatic Bars

OutlineElastic Cylinders with End LoadingTorsion of Cylinders: General formulationStress-Function FormulationDisplacementFormulationMembrane Analogy Solution: Boundary Equation Scheme Solution: Fourier Method - Rectangular SectionMultiplyConnectedCross-SectionsHollow Sections2

Outline • Elastic Cylinders with End Loading • Torsion of Cylinders: General formulation • Stress-Function Formulation • Displacement Formulation • Membrane Analogy • Solution: Boundary Equation Scheme • Solution: Fourier Method – Rectangular Section • Multiply Connected Cross-Sections • Hollow Sections 2

Elastic Cylinders Subjected to End Loadings. Semi-Inverse Method· Zero lateral forces: O, =o, = Tx = 0.: Let us guess the most general form for stressesagdaatx10SayQXatot,doxy0OzayOTyEota0x=0OzaxayRat=0=T=TOzaty=0=ty=ty (x,y)OzMaa=0az3

0. σστ x y xy = = = x xy x y ∂σ ∂τ + ∂ ∂ xz F x z ∂τ + + ∂ 0 xy y x y τ σ = ∂ ∂ + ∂ ∂ yz F y z ∂τ + + ∂ 0 xz yz z F z xyz τ τ σ = ∂ ∂ ∂ +++ ∂∂∂ ( ) ( ) 2 2 0 0 , 0 , 0 xz xz xz y z z yz yz x y z x z z y τ τ τ τ τ σ τ          =   ∂ = ⇒=  ∂  ∂ ⇒ =⇒ ∂ = ∂  = ∂     Elastic Cylinders Subjected to End Loadings • Semi-Inverse Method • Zero lateral forces: • Let us guess the most general form for stresses 3

Elastic Cylinders Subjected to End LoadingsBeltrami-Michell EquationsaFaFOH27ato+0.)=ax(0+2%(ato, +aaaFaF=0yaxaaxaxoyaFaF.aFQOF27satisfiedazOzOzayox+Va0aga?ga)山o.=C,x+Cy+Cz+Cxz+Csyz+Cax?Oz2ayaxoya2a2OFa2aF1C4ax20x02axOy+yz1+vVa2aFaOFCsOyo2O2dyax2Oy1+vA

2 ∇ σ x 2 2 1 1 x y x σ σ ν ∂ + + + ∂ ( ) 2 1 x x y z z F F F F xyz x ν σ ν   ∂ ∂ ∂ ∂ + =− + + −   −∂ ∂ ∂ ∂   2 2 2 0 z y x σ ∂ σ ⇒ = ∂ ∇ 2 2 1 1 x y y σ σ ν ∂ + + + ∂ ( ) 2 1 x y y z z F F F F xyz y ν σ ν   ∂ ∂ ∂ ∂ + =− + + −   −∂ ∂ ∂ ∂   2 2 2 0 xy z y τ σ = ∂ ∇ ∂ ⇒ 2 1 1 x y x y σ σ ν ∂ + + + ∂∂ ( ) x y z F F y x σ   ∂ ∂ + =− +     ∂ ∂ 2 2 2 2 0 1 1 , z z x y z x y σ σσ ν σ ∂ ∇+ + + ∂ ∂ ⇒ = ∂ ∂ ( ) 2 1 x y z z z F F F F xyz z ν σ ν   ∂ ∂ ∂ ∂ + =− + + −   −∂ ∂ ∂ ∂   satisfied                • Beltrami-Michell Equations 2222 222 12 34 5 6 0 zzzz z C x C y C z C xz C yz C x y z xy σσσσ σ ∂∂∂∂ = = = =⇒ = + + + + + ∂ ∂ ∂ ∂∂ Elastic Cylinders Subjected to End Loadings 4 2 2 1 1 xz x y x z τ σσ ν ∂ ∇+ + + ∂∂ ( ) x z z F F z x σ   ∂ ∂ + =− +     ∂ ∂ 2 2 4 2 2 2 2 1 1 1 xz xz yz x y C x y y z τ τ ν τ σσ ν ∂ ∂ ⇒ + =− ∂∂ + ∂ ∇+ + + ∂∂ ( ) y z z F F z y σ   ∂ ∂ + =− +     ∂ ∂ 2 2 5 2 2 1 yz yz C x y τ τ ν      ∂ ∂ ⇒ + =−  ∂∂ + 

Extension of CvlindersAssumptionsLoad P, is applied at centroid of cross-sectionsonobendingeffectsUsing Saint-Venant Principle, exact endtractions are replaced by staticallyequivalent uniform loadingThus assume stress ,isuniformoverany cross-section throughout the solidP=0TUaTAUsing stress results into Hooke's law and combining with the strain-displacement relations givesvP.uIntegrating andP.AEvP.vP.ouavOwdropping rigid-bodyVP.axAEayAEOzAEmotion terms suchAEouavOvowouaw0that displacementsP.ayOxOzOzdyaxWvanish at originAE5

Assumptions • Load Pz is applied at centroid of cross￾section so no bending effects • Using Saint-Venant Principle, exact end tractions are replaced by statically equivalent uniform loading • Thus assume stress σz is uniform over any cross-section throughout the solid , 0 z z xz yz P A ⇒ = == σ ττ • Using stress results into Hooke’s law and combining with the strain￾displacement relations gives 0, 0, 0 , , = ∂ ∂ + ∂ ∂ = ∂ ∂ + ∂ ∂ = ∂ ∂ + ∂ ∂ = ∂ ν ∂ = − ∂ ν ∂ = − ∂ ∂ z u x w y w z v x v y u AE P z w AE P y v AE P x u z z z Integrating and dropping rigid-body motion terms such that displacements vanish at origin z AE P w y AE P v x AE P u z z z = ν = − ν = − Extension of Cylinders x y z Pz ℓ S R 5

Torsion of CylindersGuided by Observationsfrom Strength of Materials:. Projection of each section on x-yplane rotates as rigid-body aboutcentral axis: Amount of projected sectionrotation is linear function of axialcoordinate: Plane cross-sections will not remainplane after deformation thus leadingto a warping displacement6

x y z T ℓ S R Guided by Observations from Strength of Materials: Torsion of Cylinders 6 • Projection of each section on x-y plane rotates as rigid-body about central axis • Amount of projected section rotation is linear function of axial coordinate • Plane cross-sections will not remain plane after deformation thus leading to a warping displacement

Torsional Deformation. In-plane / projected displacementsu=-rβsin=-βy, =rβcos=βxS· Angleoftwist: β=αzOX. The warping displacement is assumed to beRa function of only the in-plane coordinates=u=-αyz, v=αxz, w=w(x,y): Now must show assumed displacement formwill satisfy all elasticity field equations7

u r y vr x =− =− = = β θ β β θβ sin , cos β α= z. ⇒ =− = = u yz v xz w w x y α α , , ( , ). Torsional Deformation • In-plane / projected displacements • Angle of twist: • The warping displacement is assumed to be a function of only the in-plane coordinates 7 • Now must show assumed displacement form will satisfy all elasticity field equations

Stress Function Formulation. Strain and stress field=0=6,=8,,=0Ta8,=8,=0axyVLu=-αyzowow1.V=αxzL2axOxw=w(x,y)ow1ow=G8ax+Tax+V2V22ayay: Two stress components satisfy. Equilibrium equations result inatyeotyagatot,-2GαVY=0ayaxaxOzdyayayayay2GαUTax?ay2Xzdyax: Stress compatibility isy=y(x,y) Prandtl Stress Functionautomatically satisfied8

0 0 1 2 (, ) 1 2 x y z xy x y z xy xz xz yz yz u yz w w v xz y G y x x w wxy w w x G x y y εεεε σσστ α α εα τα εα τα = = = = = = = = = −   ∂ ∂ = ⇒ =−+ ⇒ = −+     ∂ ∂ =   ∂ ∂ =+ =+     ∂ ∂ Stress Function Formulation • Strain and stress field • Equilibrium equations result in yz z xz x y ∂τ ∂τ ∂σ + + ∂ ∂ F z z + ∂ 0 , xz yz y x ψ ψ τ τ = ∂ ∂ ⇒ = =− ∂ ∂ • Two stress components satisfy 2 2 2 2 2 2 2 xz yz G y x G x y τ τ α ψ ψ ψ α ∂ ∂ − =− ∂ ∂ ∂ ∂ ⇒ ∇ = + =− ∂ ∂ • ψ = ψ(x,y) • Prandtl Stress Function 8 • Stress compatibility is automatically satisfied

Stress Function Formulation: BCs.On lateral surfaceT"=gin,+Van,+t=0=0=0T=n,+,n,+t=0=0=0=0=T"=tn+tyn,+o.aydxdyay dy0=0=i.e. set. y = 0axdsdsOy ds On ends: T" =±tx, T" =±ty, T" =o, =0 More interested in satisfying the resultantend-loadingsdyds: 0 = P, = [L, trdxdy,: 0= P, = [, TydxdydxSdydx:0= P,= J,.dxdy,:0=M,=JJ,y9.dxdy1dsds: 0 = M, = [L, xg dxdy, :T = M, = [L(xt,y- - yTx-)dxdyx

n T x x = σ x yx n + τ y zx z n n +τ 0 00 n T y xy τ =⇒= = x y n + σ y zy z n n +τ 0 00 n T nn n z xz x yz y z z ττσ =⇒= =++ i.e. 0 0 0, set: 0. dy dx d y ds x ds ds ψ ψ ψ ψ = ⇒ ∂ ∂    +− − = ⇒ =    ∂ ∂   =  1 :0 , 2 :0 3 :0 x xz y yz A A z z P dxdy P dxdy P τ τ σ = = = = = = ∫∫ ∫∫ , 4 :0 x z A dxdy M y = = σ ∫∫ 5 :0 A y z dxdy = = M xσ ∫∫ , 6: z ( yz xz ) A A dxdy T M x y dxdy = = − τ τ ∫∫ ∫∫ Stress Function Formulation: BCs • On lateral surface • On ends: • More interested in satisfying the resultant end-loadings , 0 nnn TTT x xz y yz z z =± =± = = τ τσ , x y dy dx n n ds ds = = − x y z T ℓ S R n

Stress Function Formulation: BCs. On ends (with a big help from the Green's theorem)- are automatically satisfied.: J, tdxdy= J,% dxdy =-],wdx= J,wn,ds = 0satisfied J,,dxdy = -J, % dxdy = - ], vdy =-J, yn,ds = 0 satisfieda(xy)a(yyayay: T = J,(x - ) dxdy = -JJdxdy :dxdyOxayayaxT=2], ddy= -[(-μdx+ xydy) +2[[ y dxdy =The assumed stress function yields a governing Poisson equationThe stress function vanishes on the lateral boundary.The overall torque is related to the integral of the stress Function. The remaining end conditions are automatically satisfied10

( ) ( ) 3 5 are automatically satisfied. 1 : 0 satisfied 2 : 0 satisfied 6 : xz A A SS yz A A SS yz xz y A x A dxdy dxdy y d dx n ds xdy dxdy d x x T x y dxdy x y dxdy y n ds xy x ψ τ ψ ψ ψ τ ψ ψ ψ ψ ψ τ τ − ∂ = = = = ∂ ∂ =− =− =− = ∂   ∂ ∂ ∂ = − =− + =− +    ∂ ∂ −  ∂ ∫∫ ∫∫ ∫ ∫ ∫∫ ∫∫ ∫ ∫ ∫∫ ∫∫ ( ) ( ) 2 A S y dxdy y y dx x dy ψ ψ ψ ψ   ∂   − ∂   =− − + ∫∫ ∫ 2 2 A A + ⇒= ψ ψ dxdy T dxdy ∫∫ ∫∫ Stress Function Formulation: BCs • On ends (with a big help from the Green’s theorem) • The assumed stress function yields a governing Poisson equation. • The stress function vanishes on the lateral boundary. • The overall torque is related to the integral of the stress Function. • The remaining end conditions are automatically satisfied. 10