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Shafts and Shaft Components 377
Note that we could have used Eq. (7–7) directly.
Check yielding.
S y S y 57 000
n y = > = = 2.64
σ max σ + σ m 12 910 + 8659
a
Also check this diameter at the end of the keyway, just to the right of point I,
and at the groove at point K. From moment diagram, estimate M at end of
keyway to be M 3750 lbf-in.
Assume the radius at the bottom of the keyway will be the standard
r d 0.02, r 0.02 d 0.02 (1.625) 0.0325 in.
K t = 2.14 (Table 7–1), q 0.65 (Fig. 6–20)
K f = 1 + 0.65(2.14 − 1) = 1.74
K ts = 3.0 (Table 7–1), q s = 0.71 (Fig. 6–21)
K fs = 1 + 0.71(3 − 1) = 2.42
32K f M a 32(1.74)(3750)
σ = = = 15 490 psi
a
πd 3 π(1.625) 3
√
√ 3(16)(2.42)(3240)
K fs T m
σ = 3(16) = = 16 120 psi
m
πd 3 π(1.625) 3
1 σ a σ m 15 490 16 120
= + = + = 0.854
n f S e S ut 25 100 68 000
n f = 1.17
The keyway turns out to be more critical than the shoulder. We can either
increase the diameter or use a higher strength material. Unless the deflection
analysis shows a need for larger diameters, let us choose to increase the
strength. We started with a very low strength and can afford to increase it
some to avoid larger sizes. Try 1050 CD with S ut = 100 kpsi.
Recalculate factors affected by S ut , i.e., k a → S e ; q → K f → σ a
k a = 2.7(100) −0.265 = 0.797, S e = 0.797(0.835)(0.5)(100) = 33.3 kpsi
q = 0.72, K f = 1 + 0.72(2.14 − 1) = 1.82
32(1.82)(3750)
σ = = 16 200 psi
a
π(1.625) 3
1 16 200 16 120
= + = 0.648
n f 33 300 100 000
n f = 1.54
Since the Goodman criterion is conservative, we will accept this as close enough
to the requested 1.5.
Check at the groove at K, since K t for flat-bottomed grooves are often very
high. From the torque diagram, note that no torque is present at the groove.
From the moment diagram, M a = 2398 lbf in, M m = T a = T m = 0. To quickly
check if this location is potentially critical, just use K f = K t = 5.0 as an
estimate, from Table 7–1.
32(5)(2398)
32K f M a
σ a = 3 = 3 = 28 460 psi
πd π(1.625)
S e 33 300
n f = = = 1.17
σ a 28 460