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FITS AND TOLERANCES 27.12
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FIGURE 27.3 A press-fitted assembly. Inner member has hole of radius a. Contact surface has radius b. Outer member has outside radius c.
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Solution. The worst condition would occur when the hole is minimum and the shaft is maximum. From Table 27.7, we find C = 0 and C = +4.310 for the lower limit of the hole and upper limit of the shaft, respectively. Using Eq. (27.5), we find L = CD1/3 = +4.310(1.5)1/3 = 0.0049 in 1000
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Therefore, the maximum shaft has a diameter di = 1.5 + 0.0049 = 1.5049 in. Similarly, the minimum hole is do = 1.5000 in. The radial interference is = 0.5(0.0049) = 0.00245 in. For use in Eq. (27.10), we observe that b = 0.75 in and c = 1.25 in based on the nominal dimensions. Using E = 30 Mpsi, we find the contact pressure to be p= = E 2 (c b2 ) bc2 30(10)6(0.00245) [(1.25)2 (0.75)2] 0.75(1.25)2
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= 62.7 kpsi Using Eq. (27.11) to get the stresses in the outer member gives ot = p c2 + b2 (1.25)2 + (0.75)2 = 133.2 kpsi 2 2 = 62.7 c b (1.25)2 (0.75)2
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or = p = 62.7 kpsi
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FITS AND TOLERANCES 27.13
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For the inner member, the worst stress is given by it = and the result is ir = it = 62.7 kpsi pb2 = p = 62.7 kpsi b2 a2
27.5 ABSOLUTE TOLERANCES
When an aggregate of several parts is assembled, the gap, grip, or interference is related to dimensions and tolerances of the individual parts. Consider an array of parallel vectors as depicted in Fig. 27.4, the x s directed to the right and the y s directed to the left. They may be treated as scalars and represented algebraically. Let ti be the bilateral tolerance on xi and tj be the bilateral tolerance on yj, all being positive numbers. The gap remaining short of closure is called w and may be viewed as the slack variable permitting summation to zero. Thus, (x1 + x3 + ) (y2 + y4 + ) w = 0 or w = xi yj (27.13)
The largest gap w exists when the right-tending vectors are the largest possible and the left-tending vectors are the smallest possible. Expressing Eq. (19.13) in terms of the greatest deviations from the means gives wmax = (xi + ti) (yj tj) = xi y j +
(27.14)
See Ref. [27.4].
FIGURE 27.4 An array of parallel vectors.
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FITS AND TOLERANCES 27.14
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Similarly, for the smallest gap, wmin = (xi ti) (yj + tj) = xi yj
(27.15)
The mean of w is w= 1 1 (z + zmin ) = [( xi yj + t) + ( xi yj t)] 2 max 2 (27.16)
w = xi yj The bilateral tolerance of w is 1 1 tw = (wmax wmin ) = [( xi yj + t) ( xi yj t)] 2 2 tw =
(27.17)
Equation (27.15) gives rise to expressions such as the stacking of tolerances in describing the conditions at the gap.All the bilateral tolerances of the constituent x s and y s add to the tolerance of the gap. If the gap is an interference, then w is a righttending vector (negative). For all instances to be interference fits, both wmax and wmin have to be negative. Example 3. In the pin-washer-sleeve snap-ring assembly depicted in Fig. 27.5, identify the mean gap w, gap tolerance tw, maximum gap wmax, and minimum gap wmin if x1 = 1.385 0.005, y2 = 0.125 0.001, y3 = 1.000 0.002, and y4 = 0.250 0.001 in.
FIGURE 27.5 (a) A pin-washer snap-ring assembly and associated gap; (b) parallel vectors describing gap.
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