chap3_0130473928

28 2r 0r f 0 r 0r in the absence of body forces eqs

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Unformatted text preview: n of the infinitesimal element abcd, denoting the r and displacements by u and v , respectively. The general deformation experienced by an element may be regarded as composed of (1) a change in length of the sides, as in Figs. 3.8a and b, and (2) rotation of the sides, as in Figs. 3.8c and d. In the analysis that follows, the small angle approximation sin L is employed, and arcs ab and cd are regarded as straight lines. Referring to Fig. 3.8a, it is observed that a u displacement of side ab results in both radial and tangential strain. The radial strain r, the deformation per unit length of side ad, is associated only with the u displacement: r 3.8 Basic Relations in Polar Coordinates = 0u 0r (3.30a) 113 ch03.qxd 12/20/02 7:20 AM Page 114 FIGURE 3.8. Deformation and displacement of an element in polar coordinates. The tangential strain owing to u, the deformation per unit length of ab, is 1 2u = 1r + u2 d - r d u = r rd (d) Clearly, a v displacement of element abcd (Fig. 3.8b) also produces a tangential strain, 1 2v = 10v/0 2 d 1 0v = r0 rd (e) since the increase in length of ab is 10v/0 2d . The resultant tangential strain, combining Eqs. (d) and (e), is = 1 0v u + r0 r (3.30b) Figure 3.8c shows the angle of rotation eb ¿ f of side a ¿ b ¿ due to a u displacement. The associated strain is 1 r 10u/0 2 d 1 0u = r0 rd 2u = (f) The rotation of side bc associated with a v displacement alone is shown in Fig. 3.8d. Since an initial rotation of b – through an angle v /r has occurred, the relative rotation gb – h of side bc is 1 r 2v = v 0v r 0r (g) The sum of Eqs. (f) and (g) provides the total shearing strain 114 Chapter 3 Two-Dimensional Problems in Elasticity ch03.qxd 12/20/02 7:20 AM Page 115 1 0u v 0v + r0 r 0r = r (3.30c) The strain–displacement relationships in polar coordinates are thus given by Eqs. (3.30). Hooke’s Law To write Hooke’s law in polar coordinates, we need only replace subscripts x by r and y by in the appropriate Cartesian equations. In the case of plane stress, from Eqs. (3.10) we have r = 1 1 E = r 1 1 E = 1 G - 2 - r r2 (3.31) r For plane strain, Eqs. (3.5) lead to 1+ [11 - 2 E 1+ [11 - 2 E = r = = r 1 G - - r r] (3.32) r Transformation Equations Replacement of the subscripts x ¿ by r and y ¿ by r = =1 = r cos2 x x y 2 sin + x2 sin + y y sin2 +2 cos + 2 cos xy -2 xy y = =1 r = r cos2 r 2 sin + sin2 2 sin cos + 2 cos -2 + +2 xy r, r r r sin cos 1cos2 xy We can also express x, xy, and y in terms of placing with - in Eqs. (1.13). Thus, x in Eqs. (1.13) results in - sin2 2 sin cos r , and (Problem 3.26) by re- sin cos 1cos2 - sin2 2 Basic Relations in Polar Coordinates (3.34) sin cos Similar transformation equations may also be written for the strains 3.8 (3.33) r, r , and . 115 ch03.qxd 12/20/02 7:20 AM Page 116 Compatibility Equation It can be shown that Eqs. (3.30) result in the following form of the equation of compatibility: 02 1 02 +2 r0 0r2 r 2 + 20 10r 1 02 r 1 0r = +2 r 0r r 0r r 0r 0 r0 (3.35) To arrive at a compatibility equation expressed in terms of the stress function £ , it is necessary to evaluate the partial derivatives 02 £ /0x2 and 02 £ /0y2 in terms of r and by means of the chain rule together with Eqs. (a). These derivatives lead to the Laplacian operator: §2 £ = 1 0£ 1 02 £ 02 £ 02 £ 02 £ + = + +2 2 2 2 r 0r r 02 0x 0y 0r (3.36) The equation of compatibility in alternative form is thus §4 £ = ¢ 10 1 02 02 + + 2 2 ≤ 1 §2 £ 2 = 0 r 0r r0 0r2 (3.37) For the axisymmetrical, zero body force case, the compatibility equation is, from Eq. (3.9) [referring to (3.36)], §21 r + 2= d21 r + dr 2 2 + 1 d1 r + r dr 2 =0 (3.38) The remaining relationships appropriate to two-dimensional elasticity are found in a manner similar to that outlined in the foregoing discussion. EXAMPLE 3.3 A large thin plate is subjected to uniform tensile stress o at its ends, as shown in Fig. 3.9. Determine the field of stress existing within the plate. Solution For purposes of this analysis, it will prove convenient to locate the origin of coordinate axes at the center of the plate as shown. The state of stress in the plate is expressed by x = o, y = xy =0 2 The stress function, £ = oy /2, satisfies the biharmonic equation, Eq. (3.14). The geometry suggests polar form. The stress function £ may be transformed by substituting y = r sin , with the following result: FIGURE 3.9. Example 3.3. A plate in uniaxial tension. 116 Chapter 3 Two-Dimensional Problems in Elasticity ch03.qxd 12/20/02 7:20 AM Page 117 £= 1 4 or 11 - cos 2 2 2 (h) The stresses in the plate now follow from Eqs. (h) and (3.29): r = = r 1 2 1 2 =- o11 o11 1 2o + cos 2 2 - cos 2 2 (3.39) sin 2 Clearly, substitution of y = xy = 0 could have led directly to the foregoing result, using the transformation expressions of stress, Eqs. (3.33). Part B—Stress Concentrations 3.9 STRESSES DUE TO CONCENTRATED LOADS Let us now consider a concentrated force P or F acting at the vertex of a very large or semi-infinite wedge (Fig. 3.10). The load distribution alon...
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This note was uploaded on 01/09/2011 for the course ME 201 taught by Professor Yok during the Spring '08 term at Boğaziçi University.

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