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 crosssection 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 straindisplacement 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