barcode generator vb.net source code FLYWHEELS 7.17 in Software

Generator European Article Number 13 in Software FLYWHEELS 7.17

f4 = =

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

(7.42) (7.43)

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)

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)

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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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=

(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)

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 =