4 2D Elastostatic Problems In Polar Coordinates

4 2D ElastostaticProblems in PolarCoordinatesMany problems are most conveniently cast in terms of polar coordinates. To this end,first the governing differential equations discussed in Chapter 1 are expressed in terms ofpolar coordinates. Then a number of important problems involving polar coordinates aresolved.53

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Section 4.14.1 Cylindrical and Polar Coordinates4.1.1Geometrical AxisymmetryA large number of practical engineering problems involve geometrical features whichhave a natural axis of symmetry, such as the solid cylinder, shown in Fig. 4.1.1. Theaxis of symmetry is an axis of revolution; the feature which possesses axisymmetry(axial symmetry) can be generated by revolving a surface (or line) about this axis.create cylinder byrevolving a surfaceabout the axis ofsymmetryaxis ofsymmetryFigure 4.1.1: a cylinderSome other axisymmetric geometries are illustrated Fig. 4.1.2; a frustum, a disk on a shaftand a sphere.Figure 4.1.2: axisymmetric geometriesSome features are not only axisymmetric – they can be represented by a plane, which issimilar to other planes right through the axis of symmetry. The hollow cylinder shown inFig. 4.1.3 is an example of this plane axisymmetry.Solid Mechanics Part II55Kelly

Section 4.1axisymmetric planerepresentative offeatureFigure 4.1.3: a plane axisymmetric geometriesAxially Non-Symmetric GeometriesAxially non-symmetric geometries are ones which have a natural axis associated withthem, but which are not completely symmetric. Some examples of this type of feature,the curved beam and the half-space, are shown in Fig. 4.1.4; the half-space extends to“infinity” in the axial direction and in the radial direction “below” the surface – it can bethought of as a solid half-cylinder of infinite radius. One can also have plane axially nonsymmetric features; in fact, both of these are examples of such features; a slice throughthe objects perpendicular to the axis of symmetry will be representative of the wholeobject.Figure 4.1.4: a plane axisymmetric geometries4.1.2Cylindrical and Polar CoordinatesThe above features are best described using cylindrical coordinates, and the planeversions can be described using polar coordinates. These coordinates systems aredescribed next.Stresses and Strains in Cylindrical CoordinatesUsing cylindrical coordinates, any point on a feature will have specific (r ,θ , z )coordinates, Fig. 4.1.5:Solid Mechanics Part II56Kelly

Section 4.1rθz– the radial direction (“out” from the axis)– the circumferential or tangential direction (“around” the axis –counterclockwise when viewed from the positive z side of the z 0 plane)– the axial direction (“along” the axis)zθzrrz 0 planeFigure 4.1.5: cylindrical coordinatesThe displacement of a material point can be described by the three components in theradial, tangential and axial directions. These are often denoted byu u r , v uθ and w u zrespectively; they are shown in Fig. 4.1.6. Note that the displacement v is positive in thepositive θ direction, i.e. the direction of increasing θ .wvuFigure 4.1.6: displacements in cylindrical coordinatesThe stresses acting on a small element of material in the cylindrical coordinate system areas shown in Fig. 4.1.7 (the normal stresses on the left, the shear stresses on the right).Solid Mechanics Part II57Kelly

Section 4.1σ zzσ zrσθ zσ θθσ rrσ rrσ θθσ rθσ zzFigure 4.1.7: stresses in cylindrical coordinatesThe normal strains ε rr , ε θθ and ε zz are a measure of the elongation/shortening ofmaterial, per unit length, in the radial, tangential and axial directions respectively; theshear strains ε rθ , ε θ z and ε zr represent (half) the change in the right angles between lineelements along the coordinate directions. The physical meaning of these strains isillustrated in Fig. 4.1.8.zstrain at point oCε rr unit elongation of oAε θθ unit elongation of oBθε zz unit elongation of oCε rθ ½ change in angle AoBBε θ z ½ change in angle BoCoε zr ½ change in angle AoCArFigure 4.1.8: strains in cylindrical coordinatesPlane Problems and Polar CoordinatesThe stresses in any particular plane of an axisymmetric body can be described using thetwo-dimensional polar coordinates (r ,θ ) shown in Fig. 4.1.9.Solid Mechanics Part II58Kelly

Section 4.1θrFigure 4.1.9: polar coordinatesThere are three stress components acting in the plane z 0 : the radial stress σ rr , thecircumferential (tangential) stress σ θθ and the shear stress σ rθ , as shown in Fig. 4.1.10.Note the direction of the (positive) shear stress – it is conventional to take the z axis out ofthe page and so the θ direction is counterclockwise. The three stress components whichdo not act in this plane, but which act on this plane ( σ zz , σ θ z and σ zr ), may or may notbe zero, depending on the particular problem (see later).σ θθσ rrσ rθσ rrσ rθσ θθFigure 4.1.10: stresses in polar coordinatesSolid Mechanics Part II59Kelly

Section 4.24.2 Differential Equations in Polar CoordinatesHere, the two-dimensional Cartesian relations of Chapter 1 are re-cast in polarcoordinates.4.2.1Equilibrium equations in Polar CoordinatesOne way of expressing the equations of equilibrium in polar coordinates is to apply achange of coordinates directly to the 2D Cartesian version, Eqns. 1.1.8, as outlined in theAppendix to this section, §4.2.6. Alternatively, the equations can be derived from firstprinciples by considering an element of material subjected to stresses σ rr , σ θθ and σ rθ ,as shown in Fig. 4.2.1. The dimensions of the element are Δr in the radial direction, andrΔθ (inner surface) and (r Δr )Δθ (outer surface) in the tangential direction.σ rθ σ θθ σ rθΔθ θ σ rθΔr r σσ rr rr Δr rσ rθ σ θθΔθ θrσ rrΔθσ rθσ θθFigure 4.2.1: an element of materialSumming the forces in the radial direction leads to Fr σ rr σ rr Δr (r Δr )Δθ σ rr rΔθ r σΔθ Δθ (σ θθ )Δr sin σ θθ θθ Δθ Δr sin2 θ2 σ rθΔθ Δθ (σ rθ )Δr 0 cosΔθ Δr cos σ rθ 2 θ2 (4.2.1)For a small element, sin θ θ , cos θ 1 and so, dividing through by ΔrΔθ , σ rr(r Δr ) σ rr σ θθ Δθ σ θθ σ rθ 0 r2 θ θ(4.2.2)A similar calculation can be carried out for forces in the tangential direction { Problem1}. In the limit as Δr , Δθ 0 , one then has the two-dimensional equilibrium equationsin polar coordinates:Solid Mechanics Part II60Kelly

Section 4.2 σ rr 1 σ rθ 1 (σ rr σ θθ ) 0 r θr r σ rθ 1 σ θθ 2σ rθ 0 r θr r4.2.2Equilibrium Equations (4.2.3)Strain Displacement Relations and Hooke’s LawThe two-dimensional strain-displacement relations can be derived from first principles byconsidering line elements initially lying in the r and θ directions. Alternatively, asdetailed in the Appendix to this section, §4.2.6, they can be derived directly from theCartesian version, Eqns. 1.2.5, u r r1 uθ u r 2-D Strain-Displacement Expressionsr θr1 1 u r uθ uθ rr 2 r θε rr ε θθε rθ(4.2.4)The stress-strain relations in polar coordinates are completely analogous to those inCartesian coordinates – the axes through a small material element are simply labelledwith different letters. Thus Hooke’s law is nowε rr ε rr 4.2.31[σ rr νσ θθ ], ε θθ 1 [σ θθ νσ rr ], ε rθ 1 ν σ rθEEEνε zz (σ rr σ θθ )EHooke’s Law (Plane Stress)(4.2.5a)1 ν[(1 ν )σ rr νσ θθ ], ε θθ 1 ν [ νσ rr (1 ν )σ θθ ], ε rθ 1 ν σ rθEEEHooke’s Law (Plane Strain) (4.2.5b)Stress Function RelationsIn order to solve problems in polar coordinates using the stress function method, Eqns.3.2.1 relating the stress components to the Airy stress function can be transformed usingthe relations in the Appendix to this section, §4.2.6:σ rr 2φ 1 φ 1 φ 1 2φ1 φ 1 2φσσ 2 ,, θθrθr r r θ 2 r r θ r 2 θ r r θ r 2(4.2.6)It can be verified that these equations automatically satisfy the equilibrium equations4.2.3 { Problem 2}.Solid Mechanics Part II61Kelly

Section 4.2The biharmonic equation 3.2.3 becomes 2 1 1 2 2 2r r r θ 2 r4.2.42 φ 0 (4.2.7)The Compatibility RelationThe compatibility relation expressed in polar coordinates is (see the Appendix to thissection, §4.2.6)1 2 ε rr 2 ε θθ 2 2 ε rθ 1 ε rr 2 ε θθ2 ε 2 rθ 0222r r θ r rr r rr θr θ4.2.5(4.2.8)Problems1. Derive the equilibrium equation 4.2.3b2. Verify that the stress function relations 4.2.6 satisfy the equilibrium equations 4.2.3.3. Verify that the strains as given by 4.2.4 satisfy the compatibility relations 4.2.8.4.2.6Appendix to §4.2From Cartesian Coordinates to Polar CoordinatesTo transform equations from Cartesian to polar coordinates, first note the relationsx r cosθ ,y r sin θ(4.2.9)r x 2 y 2 , θ arctan( y / x)Then the Cartesian partial derivatives become r θ x x r x r θ y y r y sin θ cosθ r θ r cosθ sin θ r θ r θ θ(4.2.10)The second partial derivatives are thenSolid Mechanics Part II62Kelly

Section 4.2 sin θ sin θ 2 cos θ cos θ2r θ r θ r r x sin θ sin θ sin θ sin θ cos θ cos θ cos θ cos θ r θ r θ r θ r r r θ r r 1 1 1 2 1 2 2 2 2 θsin2 cos 2 θ 2 sin 2 θ 2 r r r r θ r θ r r θ (4.2.11)Similarly,1 2 2 222 1 sincosθθ r r r 2 θ 2 y 2 r 2 1 1 2 sin 2θ 2 r θ r r θ 2 1 1 2 2 sin θ cos θ 2 x yr r r 2 θ 2 r 1 1 2 cos 2θ 2 θrrθr (4.2.12)Equilibrium EquationsThe Cartesian stress components can be expressed in terms of polar components using thestress transformation formulae, Part I, Eqns. 3.4.7. Using a negative rotation (see Fig.4.2.2), one hasσ xx σ rr cos 2 θ σ θθ sin 2 θ σ rθ sin 2θσ yy σ rr sin 2 θ σ θθ cos 2 θ σ rθ sin 2θ(4.2.13)σ xy sin θ cos θ (σ rr σ θθ ) σ rθ cos 2θApplying these and 4.2.10 to the 2D Cartesian equilibrium equations 3.1.3a-b lead to1 σ rθ 11 σ θθ 2σ rθ σ σcos θ rr (σ rr σ θθ ) sin θ rθ 0r θrr θr r r1 σ rθ 11 σ θθ 2σ rθ σ σsin θ rr (σ rr σ θθ ) cos θ rθ 0r θrr θr r r(4.2.14)which then give Eqns. 4.2.3.yθrθxFigure 4.2.2: rotation of axesSolid Mechanics Part II63Kelly

Section 4.2The Strain-Displacement RelationsNoting thatu x u r cosθ uθ sin θu y u r sin θ uθ cosθ,(4.2.15)the strains in polar coordinates can be obtained directly from Eqns. 1.2.5: u x x sin θ cos θ (u r cos θ uθ sin θ ) rr θ u1 1 u r uθ uθ 1 uθ u r cos 2 θ r sin 2 θ sin 2θ 2 r θ rr rr r θε xx (4.2.16)One obtains similar expressions for the strains ε yy and ε xy . Substituting the results intothe strain transformation equations Part I, Eqns. 3.8.1,ε rr ε xx cos 2 θ ε yy sin 2 θ ε xy sin 2θε θθ ε xx sin 2 θ ε yy cos 2 θ ε xy sin 2θε rθ sin θ cos θ (ε yy ε xx ) ε xy cos 2θ(4.2.17)then leads to the equations given above, Eqns. 4.2.4.The Stress – Stress Function RelationsThe stresses in polar coordinates are related to the stresses in Cartesian coordinatesthrough the stress transformation equations (this time a positive rotation; compare withEqns. 4.2.13 and Fig. 4.2.2)σ rr σ xx cos 2 θ σ yy sin 2 θ σ xy sin 2θσ θθ σ xx sin 2 θ σ yy cos 2 θ σ xy sin 2θσ rθ sin θ cos θ (σ yy σ xx ) σ xy cos 2θ(4.2.18)Using the Cartesian stress – stress function relations 3.2.1, one hasσ rr 2φ 2φ 2φ22cosθsinθsin 2θ x y y 2 x 2(4.2.19)and similarly for σ θθ , σ rθ . Using 4.2.11-12 then leads to 4.2.6.Solid Mechanics Part II64Kelly

Section 4.2The Compatibility RelationBeginning with the Cartesian relation 1.3.1, each term can be transformed using 4.2.11-12and the strain transformation relations, for example 1 1 2 2 ε xx 21 2 22 1 2 cossinsin2θθθ r r r 2 θ 2 x 2 r 2 r θ r r θ (ε rr cos 2 θ ε θθ sin 2 θ ε rθ sin 2θ )(4.2.20)After some lengthy calculations, one arrives at 4.2.8.Solid Mechanics Part II65Kelly