Page 325 - Shigley's Mechanical Engineering Design

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300 Mechanical Engineering Design
The next step is to estimate the bending stress at point B. The bending moment
is

225F 225(6.8)
M B = R 1 x = 250 = 250 = 695.5N · m
550 550
3
3
3
3
Just to the left of B the section modulus is I/c = πd /32 = π32 /32 = 3.217 (10 ) mm .
The reversing bending stress is, assuming infinite life,
695.5
M B −6 6
σ rev = K f = 1.55 (10) = 335.1(10 ) Pa = 335.1MPa
I/c 3.217
This stress is greater than S e and less than S y . This means we have both finite life and
no yielding on the first cycle.
For finite life, we will need to use Eq. (6–16). The ultimate strength, S ut = 690
MPa = 100 kpsi. From Fig. 6–18, f = 0.844. From Eq. (6–14)

( fS ut ) 2 [0.844(690)] 2
a = = = 1437 MPa
S e 236
and from Eq. (6–15)

1 fS ut 1 0.844(690)
b =− log =− log =−0.1308
3 S e 3 236
From Eq. (6–16),
−1/0.1308

1/b

335.1
3
Answer N = σ rev = = 68(10 ) cycles
a 1437

6–11 Characterizing Fluctuating Stresses
Fluctuating stresses in machinery often take the form of a sinusoidal pattern because
of the nature of some rotating machinery. However, other patterns, some quite irreg-
ular, do occur. It has been found that in periodic patterns exhibiting a single maxi-
mum and a single minimum of force, the shape of the wave is not important, but the
peaks on both the high side (maximum) and the low side (minimum) are important.
Thus F max and F min in a cycle of force can be used to characterize the force pattern.
It is also true that ranging above and below some baseline can be equally effective
in characterizing the force pattern. If the largest force is F max and the smallest force
is F min , then a steady component and an alternating component can be constructed
as follows:

F max + F min
F max − F min
F m = F a =
2
2
where F m is the midrange steady component of force, and F a is the amplitude of the
alternating component of force.