chap3_0130473928

H and 329 r 1 2 1 2 r o11 1 2o o11 339 clearly

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Unformatted text preview: g the thickness (z direction) is uniform. The thickness of the wedge is taken as unity, so P or F is the load per unit thickness. In such situations, it is convenient to use polar coordinates and the semi-inverse method. In actuality, the concentrated load is assumed to be a theoretical line load and will be spread over an area of small finite width. Plastic deformation may occur locally. Thus, the solutions that follow are not valid in the immediate vicinity of the application of load. FIGURE 3.10. Wedge of unit thickness subjected to a concentrated load per unit thickness: (a) knife edge or pivot; (b) wedge cantilever. 3.9 Stresses Due to Concentrated Loads 117 ch03.qxd 12/20/02 7:20 AM Page 118 Compression of a Wedge (Fig. 3.10a). Assume the stress function £ = cPr sin (a) where c is a constant. It can be verified that Eq. (a) satisfies Eq. (3.37) and compatibility is ensured. For equilibrium, the stresses from Eqs. (3.29) are r = 2cP cos , r = 0, r =0 (b) The force resultant acting on a cylindrical surface of small radius, shown by the dashed lines in Fig. 3.10a, must balance P. The boundary conditions are therefore expressed by = 2 L 0 = 0, r 1 r (c) =; cos 2rd = - P (d) Conditions (c) are fulfilled by the last two of Eqs. (b). Substituting the first of Eqs. (b) into condition (d) results in 4cP L 0 cos2 Integrating and solving for c: c = - 1/12 knife edge is therefore r r1 =- P cos 1 2 + d = -P + sin 2 2. The stress distribution in the sin 2 2 , = 0, r =0 (3.40) This solution is due to J. H. Michell [Ref. 3.6]. The distribution of the normal stresses x over any cross section m - n perpendicular to the axis of symmetry of the wedge is not uniform (Fig. 3.10a). Applying Eq. (3.34) and substituting r = L/cos in Eq. (3.40), we have x = r cos2 =- L1 P cos4 + 1 2 sin 2 2 (3.41) The foregoing shows that the stresses increase as L decreases. Observe also that the normal stress is maximum at the center of the cross section 1 = 02 and a minimum at = . The difference between the maximum and minimum stress, ¢ x, is, from Eq. (3.41), ¢ x =- P11 - cos4 2 L1 + 1 2 sin 2 2 (e) = 10°, ¢ x = - 0.172P/L is about 6% of the average normal For instance, if stress calculated from the elementary formula 1 x2elem = - P/A = - P/2L tan = 118 Chapter 3 Two-Dimensional Problems in Elasticity ch03.qxd 12/20/02 7:20 AM Page 119 - 2.836P/L. For larger angles, the difference is greater; the error in the mechanics of materials solution increases (Prob. 3.31). It may be demonstrated that the stress distribution over the cross section approaches uniformity as the taper of the wedge diminishes. Analogous conclusions may also be drawn for a conical bar. Note that Eqs. (3.40) can be applied as well for the uniaxial tension of tapered members by assigning r a positive value. Bending of a Wedge (Fig. 3.10b) We now employ £ = cFr 1 sin 1, with force. The equilibrium condition is 1 /22 + 1 L /22 - 1 12r d r cos 1 measured from the line of action of the = 2cF 1 from which, after integration, c = - 1/12 90° - , we have F cos 1 /22 + cos2 1 L /22 - 1 d 1 = -F - sin 2 2. Thus, by replacing F sin 1 with It is seen that if 1 is larger than /2 the radial stress is positive, that is, tension exists. Because sin = y/r, cos = x/r, and r = 2x2 + y2, the normal and shearing stresses at a point over any cross section m - n, using Eqs. (3.34) and (3.42), may be expressed as r =- r1 - 1 2 1 sin 2 2 x = =- r1 =F y = r sin2 sin 2 2 , = 0, r = 0 (3.42) r1 - 1 sin 2 2 2 x2y sin 2 1x2 + y222 F sin3 =r1 - 1 sin 2 2 2 y3 F 1 2 (3.43) sin 2 1x2 + y222 F sin2 cos = r sin cos = r1 - 1 sin 2 2 2 xy2 F =- 1 sin 2 1x2 + y222 2 =- - xy 1 2 F sin cos2 2 r cos =- - 1 2 Using Eqs. (3.43), it can be shown that (Prob. 3.33) across a transverse section x = L of the wedge: x is a maximum for = ; 30°, y is a maximum for = ; 60°, and xy is a maximum for = ; 45°. To compare the results given by Eqs. (3.43) with the results given by the elementary formulas for stress, consider the series 3.9 Stresses Due to Concentrated Loads 119 ch03.qxd 12/20/02 7:20 AM Page 120 sin 2 =2 - 12 23 12 25 + 3! 5! It follows that, for small angle , we can disregard all but the first two terms of this series to obtain 2 = sin 2 + 12 23 6 (f) By introducing the moment of inertia of the cross I = 2y3 = 2x3 # tan3 , and Eq. (f), we find from Eqs. (3.43) that 3 3 x Fxy tan =B¢ I ≤ cos 3 4 R, xy Fy2 tan =B¢ I section ≤ cos4 R m - n, 3 (g) For small values of , the factor in the bracket is approximately equal to unity. The expression for x then coincides with that given by the flexure formula, - My/I, of the mechanics of materials. In the elementary theory, the lateral stress y given by the second of Eqs. (3.43) is ignored. The maximum shearing stress xy obtained from Eq. (g) is twice as great as the shearing stress calculated from VQ/Ib of the elementary theory and occurs at the extreme fibers (at points m and n) rather than the neutral axis of the rectangular cross section. In the case of loading in both compression and bending, superposition of the effe...
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