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FLYWHEELS 7.17
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cos ( /8) = 0.340 7.11 sin ( /8) 1 0.283(48)2(41.9)2 55(48) 0.340 1 + 0.340 32.2(12) 7.11 /8 3( 45.8) 3 = 3660 psi
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(7.42) (7.43)
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Table 7.5 shows the values for points P1, P2 , P3, and P4 , giving additional digits for checking computer programs. Although the stress formulation was based on a spoke of length ra, the maximum tension stress in the spoke is found for the hub radius, 5.2 in. From Eq. (7.37), = 4(55) 3(5.2)2 0.283(48)2(41.9)2 3+ = 2850 psi 6(32.2)(12) 7.11(11) (48)2 (7.44)
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If the spokes are of standard design, an elliptical cross section twice as long in the direction of motion as in the axial direction, the dimensions for an 11-in2 spoke would be 2(11)/ = 2.65 in by 5.29 in. The section modulus is then Zs = (2.65)(5.29)2/32 = 7.28 in3, and the bending stress in the spokes, from Eq. (7.38), would be = 4000(48 5.2) = 61 psi 7.28(8)(48) (7.45)
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The combined stress for the spoke is then = 2850 + 61 = 2910 psi. 7.3.3 Thin Disk If a thin disk has a large radius in comparison to its thickness, the stress can be assumed constant across the thickness. Defining the stress function F as F = rz r the stress equation for a thin disk of variable thickness z(r) is r2 and t = 1 dF + 2 r 2 z dr (7.48) dF 2 r 3z r dz d 2F dF F + (3 + ) r F = 0 2 + r dr dr g z dr dr (7.47) (7.46)
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TABLE 7.5 Rim Stress for Example 7
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Downloaded from Digital Engineering Library @ McGraw-Hill (www.digitalengineeringlibrary.com) Copyright 2004 The McGraw-Hill Companies. All rights reserved. Any use is subject to the Terms of Use as given at the website.
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MACHINE ELEMENTS THAT ABSORB AND STORE ENERGY
The boundary conditions for the center or inner radius and the outer radius are both F = 0. If there is a central bore, a finite value for ri is used in the formulation, and for an infinitesimal hole, the limit is taken later. At the center of the solid disk, r = t. The inertia for a disk is given by J=
ro ri
2 r 3z dr g
(7.49)
Equations (7.46) through (7.48) can be solved explicitly for the uniform and hyperbolic profiles [7.4]. For the general case, using a difference approximation for dz/dr and F leads to a tractable banded set of linear simultaneous equations. Divide the radius into equal intervals of length r. Determine the z derivative for each point using z = j zj + 1 zj 1 2 r (7.50)
except at the two boundaries, where, for the same accuracy, z = j 2zj + 3 9zj + 2 + 18zj + 1 11zj 6 r 11zj 18zj 1 + 9zj 2 2zj 3 6 r (7.51)
at the center or inner radius and z = j (7.52)
at the outer radius. The stress equation then becomes the set of equations Aj Fj 1 + Bj Fj + Cj Fj + 1 = Dj where Aj = rj r rj 1 rj z j + r 2 2 zj rj r
(7.53)
(7.54) (7.55) (7.56) (7.57)
Bj = 2 Cj = rj r
+ rj
z j 1 zj
rj 1 rj z j + r 2 2 zj r 3zj 2 j g
Dj = (3 + )
At the two boundary points the equation is identically satisfied, leaving (ro ri )/ r 1 equations. For the inner equation, Aj = 0, and for the outer equation, Cj = 0. The simultaneous equations are solved for Fj , and the stresses are found from Eqs. (7.46) and (7.48). See the user s guide for your computer for a convenient method of solving linear simultaneous equations. 7.3.4 Disk of Constant Thickness The solutions for a disk of uniform thickness are simplified if a constant is defined as 0 =
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