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. 6= + +Sg RFig.10.1: A simple example of the decomposition of a two dimensional distortion of a lattice intoan expansion (˜), a shear (£), and a rotation (R).x2x1Fig.10.2: Shear in two dimensions.The displacement of points in a solid undergoing pure shearis the vector field ¾(x) given by Eq.(10.5) with Sji replaced by £ji: ¾j = £jixi = £j1x1 + £j2y.The integral curves of this vector field are plotted in this figure.The figure is drawn using principalaxes, which are Cartesian, so £12 = £21 = 0, £11 = -£22, which means that ¾1 = £11x1, and¾2 = -£11y.The integral curves of this simple vector field are the hyperbolae shown in the figure.Note that the displacement increases linearly with distance from the origin.The expansion, therefore, is simply the fractional change in volume of a small region of thesolid; see Figure 10.1 for a simple example.The shear tensor £ produces the shearing displacements illustrated in Figures 10.1 and10.2.As it has zero trace, there is no volume change when a body undergoes a pure sheardeformation.The shear tensor has five independent components.However, by rotating ourCartesian coordinates appropriately, we can transform away all the off diagonal elements,leaving the three diagonal elements, which must sum to zero.This is known as a principalaxis transformation.The components of the shear tensor in a cartesian coordinate systemcan be written down immediately from Eq.(10.6) by substituting the Kronecker delta ´ij forthe metric tensor gij and treating all derivatives as partial derivatives:2 "¾x 1 "¾y "¾z 1 "¾x "¾y£xx = - + , £xy = - , (10.10)3 "x 3 "y "z 2 "y "xand similarly for the other components.The analogous equations in spherical and cylindricalcoordinates will be described in the next section.The third term in Eq.(10.7) describes a pure rotation which does not deform the solid. 7To verify this, write ¾ = Æ × x where Æ is a small rotation of magnitude Æ about an axisparallel to the direction of Æ.Using cartesian coordinates in three dimensional Euclideanspace, we can demonstrate by direct calculation that the symmetric part of S vanishes, i.e.,˜ = £ = 0 and that1Rij = - Æk , Æi = - Rjk.(10.11)ijk ijk2Therefore the elements of the tensor R in a cartesian coordinate system just involve the angleÆ.Note that expression (10.11) for Æ and expression (10.7) for Rij imply that Æ is half thecurl of the displacement vector,1Æ = " × ¾.(10.12)2A simple example of rotation is shown in the last picture in Figure 10.1.Let us consider some examples of strains that can arise in physical systems.(i) Understanding how materials deform under various loads is central to mechanical,civil and structural engineering.As we have already remarked, in an elastic solid, thedeformation (i.e.strain ) is proportional to the applied stress.If, for example, wehave some structure of negligible weight and it supports a load, then the amount ofstrain will increase everywhere in proportion to this load.However this law will onlybe obeyed as long as the strain is sufficiently small that the material out of which thestructure is constructed behaves elastically.At a large enough strain, plastic flow willset in and the solid will not return to its original shape after the stress is removed.Thepoint where this happens is known as the elastic limit.For a ductile substance likepolycrystalline copper with a relatively low elastic limit, this occurs at strains [ Pobierz caÅ‚ość w formacie PDF ]
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. 6= + +Sg RFig.10.1: A simple example of the decomposition of a two dimensional distortion of a lattice intoan expansion (˜), a shear (£), and a rotation (R).x2x1Fig.10.2: Shear in two dimensions.The displacement of points in a solid undergoing pure shearis the vector field ¾(x) given by Eq.(10.5) with Sji replaced by £ji: ¾j = £jixi = £j1x1 + £j2y.The integral curves of this vector field are plotted in this figure.The figure is drawn using principalaxes, which are Cartesian, so £12 = £21 = 0, £11 = -£22, which means that ¾1 = £11x1, and¾2 = -£11y.The integral curves of this simple vector field are the hyperbolae shown in the figure.Note that the displacement increases linearly with distance from the origin.The expansion, therefore, is simply the fractional change in volume of a small region of thesolid; see Figure 10.1 for a simple example.The shear tensor £ produces the shearing displacements illustrated in Figures 10.1 and10.2.As it has zero trace, there is no volume change when a body undergoes a pure sheardeformation.The shear tensor has five independent components.However, by rotating ourCartesian coordinates appropriately, we can transform away all the off diagonal elements,leaving the three diagonal elements, which must sum to zero.This is known as a principalaxis transformation.The components of the shear tensor in a cartesian coordinate systemcan be written down immediately from Eq.(10.6) by substituting the Kronecker delta ´ij forthe metric tensor gij and treating all derivatives as partial derivatives:2 "¾x 1 "¾y "¾z 1 "¾x "¾y£xx = - + , £xy = - , (10.10)3 "x 3 "y "z 2 "y "xand similarly for the other components.The analogous equations in spherical and cylindricalcoordinates will be described in the next section.The third term in Eq.(10.7) describes a pure rotation which does not deform the solid. 7To verify this, write ¾ = Æ × x where Æ is a small rotation of magnitude Æ about an axisparallel to the direction of Æ.Using cartesian coordinates in three dimensional Euclideanspace, we can demonstrate by direct calculation that the symmetric part of S vanishes, i.e.,˜ = £ = 0 and that1Rij = - Æk , Æi = - Rjk.(10.11)ijk ijk2Therefore the elements of the tensor R in a cartesian coordinate system just involve the angleÆ.Note that expression (10.11) for Æ and expression (10.7) for Rij imply that Æ is half thecurl of the displacement vector,1Æ = " × ¾.(10.12)2A simple example of rotation is shown in the last picture in Figure 10.1.Let us consider some examples of strains that can arise in physical systems.(i) Understanding how materials deform under various loads is central to mechanical,civil and structural engineering.As we have already remarked, in an elastic solid, thedeformation (i.e.strain ) is proportional to the applied stress.If, for example, wehave some structure of negligible weight and it supports a load, then the amount ofstrain will increase everywhere in proportion to this load.However this law will onlybe obeyed as long as the strain is sufficiently small that the material out of which thestructure is constructed behaves elastically.At a large enough strain, plastic flow willset in and the solid will not return to its original shape after the stress is removed.Thepoint where this happens is known as the elastic limit.For a ductile substance likepolycrystalline copper with a relatively low elastic limit, this occurs at strains [ Pobierz caÅ‚ość w formacie PDF ]