chap3_0130473928

# At a distance equal to the depth of the beam they are

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Unformatted text preview: n. 3.11 STRESS CONCENTRATION FACTORS The discussion of Sec. 3.9 shows that, for situations in which the cross section of a load-carrying member varies gradually, reasonably accurate results can be expected if we apply equations derived on the basis of constant section. On the other hand, where abrupt changes in the cross section exist, the mechanics of materials approach cannot predict the high values of stress that actually exist. The condition referred to occurs in such frequently encountered configurations as holes, notches, and fillets. While the stresses in these regions can in some cases (for example, Table 3.2) be analyzed by applying the theory of elasticity, it is more usual to rely on experimental techniques and, in particular, photoelastic methods. The finite element method (Chapter 7) is very efficient for this purpose. It is to be noted that irregularities in stress distribution associated with abrupt changes in cross section are of practical importance in the design of machine elements subject to variable external forces and stress reversal. Under the action of stress reversal, progressive cracks (Sec. 4.4) are likely to start at certain points at which the stress is far above the average value. The majority of fractures in machine elements in service can be attributed to such progressive cracks. It is usual to specify the high local stresses owing to geometrical irregularities in terms of a stress concentration factor, k. That is, k= maximum stress nominal stress (3.51) Clearly, the nominal stress is the stress that would exist in the section in question in the absence of the geometric feature causing the stress concentration. The technical 3.11 Stress Concentration Factors 123 ch03.qxd 12/20/02 7:20 AM Page 124 FIGURE 3.13. Example 3.4. Circular hole in a plate subjected to uniaxial tension: (a) tangential stress distribution for = ; /2; (b) tangential stress distribution along periphery of the hole. literature contains an abundance of specialized information on stress concentration factors in the form of graphs, tables, and formulas.* EXAMPLE 3.4 A large, thin plate containing a small circular hole of radius a is subjected to simple tension (Fig. 3.13a). Determine the field of stress and compare with those of Example 3.3. Solution The boundary conditions appropriate to the circumference of the hole are r = r = 0, r=a (a) For large distances away from the origin, we set r, , and r equal to the values found for a solid plate in Example 3.3. Thus, from Eq. (3.39), for r = q , r = = 1 2 o 11 1 2 o11 + cos 2 2 - cos 2 2, r =- 1 2o sin 2 (b) For this case, we assume a stress function analogous to Eq. (h) of Example 3.3, £ = f11r2 + f21r2 cos 2 (c) in which f1 and f2 are yet to be determined. Substituting Eq. (c) into the biharmonic equation (3.37) and noting the validity of the resulting expression for all , we have *See, for example, Refs. 3.8 through 3.11. 124 Chapter 3 Two-Dimensional Problems in Elasticity ch03.qxd 12/20/02 7:20 AM Page 125 ¢ ¢ d2f1 d2 1d 1 df1 + ≤¢ 2 + ≤=0 2 r dr r dr dr dr (d) 4f2 d2f2 d2 1d 4 1 df2 + - 2≤ ¢ 2 + - 2≤ =0 2 r dr r dr dr r dr r (e) The solutions of Eqs. (d) and (e) are (Prob. 3.35) f1 = c1r2 ln r + c2r2 + c3 ln r + c4 f2 = c5r2 + c6r4 + (f) c7 + c8 r2 (g) where the c’s are the constants of integration. The stress function is then obtained by introducing Eqs. (f) and (g) into (c). By substituting £ into Eq. (3.29), the stresses are found to be r = c111 + 2 ln r2 + 2c2 + = c113 + 2 ln r2 + 2c2 - r = ¢ 2c5 + 6c6r2 - c3 6c7 4c8 - ¢ 2c5 + 4 + 2 ≤ cos 2 2 r r r c3 6c7 + ¢ 2c5 + 12c6r2 + 4 ≤ cos 2 r2 r (h) 6c7 2c8 - 2 ≤ sin 2 4 r r The absence of c4 indicates that it has no influence on the solution. According to the boundary conditions (b), c1 = c6 = 0 in Eq. (h), because as r : q the stresses must assume finite values. Then, according to the conditions (a), the equations (h) yield 2c2 + c3 = 0, a2 2c5 + 6c7 4c8 + 2 = 0, 4 a a 2c5 - 6c7 2c8 - 2 =0 4 a a Also, from Eqs. (b) and (h) we have o = - 4c5, o = 4c2 Solving the preceding five expressions, we obtain c2 = o/4, c3 = - a2 o/2, c5 = - o/4, c7 = - a4 o/4, and c8 = a2 o/2. The determination of the stress distribution in a large plate containing a small circular hole is completed by substituting these constants into Eq. (h): r 3.11 1 2o = r = 1 2o =- B ¢1 B ¢1 + 1 2o ¢1 - Stress Concentration Factors a2 3a4 4a2 ≤ + ¢ 1 + 4 - 2 ≤ cos 2 R r2 r r a2 3a4 ≤ - ¢ 1 + 4 ≤ cos 2 R r2 r 3a4 2a2 + 2 ≤ sin 2 r4 r (3.52a) (3.52b) (3.52c) 125 ch03.qxd 12/20/02 7:20 AM Page 126 FIGURE 3.14. Example 3.4. Graph of tangential and radial stresses for = /2 versus the distance from the center of the plate shown in Fig. 3.13a. The tangential stress distribution along the edge of the hole, r = a, is shown in Fig. 3.13b using Eq. (3.52b). We observe from the figure that 1 2max = 3 o, 1 2min = - o, = ; /2 = 0, =; The latter indicates that there exists a small area experiencing compressive stress. On the other hand, from Eq. (3.39) for = ; /2, 1 2max = o. The stress c...
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