Title: CH18 LEC 30 Slide 1
1Chapter 18
Shafts and Axles
Dr. A. Aziz Bazoune King Fahd University of
Petroleum Minerals Mechanical Engineering
Department
2Chapter Outline
18-1 Introduction .92218-2 Geometric
Constraints .92718-3 Strength Constraints
.93318-4 Strength Constraints Additional
Methods .940 18-5 Shaft Materials
.94418-6 Hollow Shafts .94418-7 Critical
Speeds (Omitted) .945 18-8 Shaft Design
.950
3LECTURE 30
18-1 Introduction .92218-2 Geometric
Constraints .92718-3 Strength Constraints
.933
4Fatigue Analysis of shafts
Neglecting axial loads because they are
comparatively very small at critical locations
where bending and torsion dominate. Remember the
fluctuating stresses due to bending and torsion
are given by
Mm Midrange bending moment, sm Midrange
bending stress Ma alternating bending
moment, sa alternating bending stress Tm
Midrange torque, tm Midrange shear stress Ta
alternating torque, tm Midrange shear stress
Kf fatigue stress concentration factor for
bending Kfs fatigue stress concentration factor
for torsion
CH-18 LEC 30 Slide 4
5Fatigue Analysis of shafts
For solid shaft with round cross section,
appropriate geometry terms can be introduced for
C, I and J resulting in
Mm Midrange bending moment, sm Midrange
bending stress Ma alternating bending
moment, sa alternating bending stress Tm
Midrange torque, tm Midrange shear stress Ta
alternating torque, tm Midrange shear stress
Kf fatigue stress concentration factor for
bending Kfs fatigue stress concentration factor
for torsion
CH-18 LEC 30 Slide 5
6Fatigue Analysis of shafts
Combining these stresses in accordance with the
DE failure theory the von-Mises stress for
rotating round, solid shaft, neglecting axial
loads are given by
(18-12)
where A and B are defined by the radicals in Eq.
(8-12) as
The Gerber fatigue failure criterion
CH-18 LEC 30 Slide 6
7Fatigue Analysis of shafts
The critical shaft diameter is given by
(18-13)
or, solving for 1/n, the factor of safety is
given by
(18-14)
CH-18 LEC 30 Slide 7
8Fatigue Analysis of shafts
where
(18-15)
CH-18 LEC 30 Slide 8
9Fatigue Analysis of shafts
Particular Case
For a rotating shaft with constant bending and
torsion, the bending stress is completely
reversed and the torsion is steady. Previous
Equations can be simplified by setting Mm 0 and
Ta 0, which simply drops out some of the terms.
Critical Shaft Diameter
(18-16)
Safety Factor
(18-17)
CH-18 LEC 30 Slide 9
10Fatigue Analysis of shafts
(18-18)
CH-18 LEC 30 Slide 10
11Shaft Diameter Equation for the DE-Elliptic
Criterion
Remember
(18-12)
where A and B are defined by
The Elliptic fatigue-failure criterion is defined
by
CH-18 LEC 30 Slide 11
CH-18 LEC 30 Slide 11
12Shaft Diameter Equation for the DE-Elliptic
Criterion
Substituting for A and B gives expressions for d,
1/n and r
Critical Shaft Diameter
(18-19)
Safety Factor
(18-20)
CH-18 LEC 30 Slide 12
13Shaft Diameter Equation for the DE-Elliptic
Criterion
CH-18 LEC 30 Slide 13
14Shaft Diameter Equation for the DE-Elliptic
Criterion
Particular Case
For a rotating shaft with constant bending and
torsion, the bending stress is completely
reversed and the torsion is steady. Previous
Equations can be simplified by setting Mm 0 and
Ta 0, which simply drops out some of the terms.
Critical Shaft Diameter
(18-21)
Safety Factor
(18-22)
CH-18 LEC 30 Slide 14
15Shaft Diameter Equation for the DE-Elliptic
Criterion
- At a shoulder Figs. A-15-8 and A-15-9 provide
information about Kt and Kts. - For a hole in a solid shaft, Figs. A-15-10 and
A-15-11 provide about Kt and Kts . - For a hole in a solid shaft, use Table A-16
- For grooves use Figs. A-15-14 and A-15-15
CH-18 LEC 30 Slide 15
16Shaft Diameter Equation for the DE-Elliptic
Criterion
- The value of slope at which the load line
intersects the junction of the failure curves is
designated rcrit. - It tells whether the threat is from fatigue or
first cycle yielding - If r gt rcrit, the threat is from fatigue
- If r lt rcrit, the threat is from first cycle
yielding.
CH-18 LEC 30 Slide 16
17Shaft Diameter Equation for the DE-Elliptic
Criterion
- For the Gerber-Langer intersection the strength
components Sa and Sm are given in Table 7-10 as
(18-23)
CH-18 LEC 30 Slide 17
18Shaft Diameter Equation for the DE-Elliptic
Criterion
- For the DE-Elliptic-Langer intersection the
strength components Sa and Sm are given by
(18-24)
CH-18 LEC 30 Slide 18
19Note that in an analysis situation in which the
diameter is known and the factor of safety is
desired, as an alternative to using the
specialized equations above, it is always still
valid to calculate the alternating and mid-range
stresses using the following Eqs. and
substitute them into the one of the equations for
the failure criteria , Eqs. (7-48) to (7-51) and
solve directly for n.
CH-18 LEC 30 Slide 19
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22- In a design situation, however, having the
equations pre-solved for diameter is quite
helpful. - It is always necessary to consider the
possibility of static failure in the first load
cycle. - The Soderberg criteria inherently guards against
yielding, as can be seen by noting that its
failure curve is conservatively within the yield
(Langer) line on Fig. 727, p. 348. - The ASME Elliptic also takes yielding into
account, but is not entirely conservative
throughout its range. This is evident by noting
that it crosses the yield line in Fig. 727. - The Gerber and modified Goodman criteria do not
guard against yielding, requiring a separate
check for yielding. A von Mises maximum stress is
calculated for this purpose.
23- To check for yielding, this von Mises maximum
stress is compared to the yield strength, as
usual - For a quick, conservative check, an estimate for
smax can be obtained by simply adding sa and sm
. (sa sm ) will always be greater than or equal
to smax, and will therefore be conservative.
CH-18 LEC 30 Slide 23
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25Example
Solution
7-20
7-20
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26Figure 7-20
Figure 7-21
CH-18 LEC 30 Slide 26
27CH-18 LEC 30 Slide 27
2818-22
Solderberg
18-14
18-24
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