9 34 plane stress problems in many problems of

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Unformatted text preview: ess condition is one of plane stress. The basic definition of this state of stress has already been given in Sec. 1.8. In this section we shall present the governing equations for the solution of plane stress problems. To exemplify the case of plane stress, consider a thin plate, as in Fig. 3.3, wherein the loading is uniformly distributed over the thickness, parallel to the plane of the plate. This geometry contrasts with that of the long prism previously discussed, which is in a state of plane strain. To arrive at tentative conclusions with regard to the stress within the plate, consider the fact that z, xz, and yz are zero 3.4 Plane Stress Problems 99 ch03.qxd 12/20/02 7:20 AM Page 100 FIGURE 3.3. Thin plate under plane stress. on both faces of the plate. Because the plate is thin, the stress distribution may be very closely approximated by assuming that the foregoing is likewise true throughout the plate. We shall, as a condition of the problem, take the body force Fz = 0 and Fx and Fy each to be functions of x and y only. As a consequence of the preceding, the stress is specified by x, z = y, xy = xz yz (a) =0 The nonzero stress components remain constant over the thickness of the plate and are functions of x and y only. This situation describes a state of plane stress. Equations (1.11) and (1.41), together with this combination of stress, again reduce to the forms found in Sec. 3.3. Thus, Eqs. (3.6) and (3.7) describe the equations of equilibrium and the boundary conditions in this case, as in the case of plane strain. Substitution of Eq. (a) into Eq. (2.28) yields the following stress–strain relations for plane stress: x = 1 1 E y = 1 1 E xy = x - y2 y - x2 z =- (3.10) xy G and xz = = 0, yz E 1 x + y2 (3.11a) Solving for x + y from the sum of the first two of Eqs. (3.10) and inserting the result into Eq. (3.11a), we obtain z 100 =- 1- 1 x + Chapter 3 y2 (3.11b) Two-Dimensional Problems in Elasticity ch03.qxd 12/20/02 7:20 AM Page 101 Equations (3.11) define the out-of-plane principal strain in terms of the in-plane stresses 1 x, y2 or strains 1 x, y2. Because z is not contained in the other governing expressions for plane stress, it can be obtained independently from Eqs. (3.11); then z = 0w/0z may be applied to yield w. That is, only u and v are considered as independent variables in the governing equations. In the case of plane stress, therefore, the basic strain–displacement relations are again given by Eqs. (3.1). Exclusion from Eq. (2.3) of z = 0w/0z makes the plane stress equations approximate, as is demonstrated in the section that follows. The governing equations of plane stress will now be reduced, as in the case of plane strain, to three equations involving stress components only. Since Eqs. (3.1) apply to plane strain and plane stress, the compatibility condition represented by Eq. (3.8) applies in both cases. The latter expression may be written as follows, substituting strains from Eqs. (3.10) and employing Eqs. (3.6): ¢ 02 02 + 2 ≤1 0x2 0y x + y2 = - 11 + 2 ¢ 0Fy 0Fx + ≤ 0x 0y (3.12) This equation of compatibility, together with the equations of equilibrium, represents a useful form of the governing equations for problems of plane stress. To summarize the two-dimensional situations discussed, the equations of equilibrium [Eqs. (3.6)], together with those of compatibility [Eq. (3.9) for plane strain and Eq. (3.12) for plane stress] and the boundary conditions [Eqs. (3.7)], provide a system of equations sufficient for determination of the complete stress distribution. It can be shown that a solution satisfying all these equations is, for a given problem, unique [Ref. 3.1]. That is, it is the only solution to the problem. In the absence of body forces or in the case of constant body forces, the compatibility equations for plane strain and plane stress are the same. In these cases, the equations governing the distribution of stress do not contain the elastic constants. Given identical geometry and loading, a bar of steel and one of Lucite should thus display identical stress distributions. This characteristic is important in that any convenient isotropic material may be used to substitute for the actual material, as, for example, in photoelastic studies. It is of interest to note that by comparing Eqs. (3.5) with Eqs. (3.10) we can form Table 3.1, which facilitates the conversion of a plane stress solution into a plane strain solution, and vice versa. For instance, conditions of plane stress and plane strain prevail in a narrow beam and a very wide beam, respectively. Hence, in TABLE 3.1 Solution Plane stress Plane strain Plane strain 3.4 To Convert to: Plane stress Plane Stress Problems E is Replaced by: E 1- 2 1+2 E 11 + 22 is Replaced by: 11+ 101 ch03.qxd 12/20/02 7:20 AM Page 102 a result pertaining to a thin beam, EI would become EI/11 - 22 for the case of a wide beam. The stiffness in the latter case is, for = 0.3, about 10% greater owing to the prevention of sidewise displacement (Secs. 5.2 and 13.3)....
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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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