machine design
DESCRIPTION
Notes on Machine designTRANSCRIPT
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ME311 Machine Design
W Dornfeld17Oct2013 Fairfield University
School of Engineering
Lecture 6: Fluctuating Fatigue and the Goodman Diagram;
Impact
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Fluctuating Fatigue
Last lecture we discussed loading that alternately went from tension to compression with the extremes equal and opposite.
Now we look at the more general case where there could be a mean value.
This is called Fluctuating Fatigue, and is characterized by both a mean and an alternating component.
Hamrock Section 7.3
mean
altrange
Time
StressSteadyorAveragemean =+
=2
minmax σσσ
AmplitudeStressalt =−
=2
minmax σσσ
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Fluctuating or
Fully Reversing?meanσaltσ
80
40
40
-40
40
0
2
minmax σσσ
+=mean
2
minmax σσσ
−=alt
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Effect of Mean Stress
This axis is the
fraction that σalt
is of the Endurance Strength, Se
This axis is the fraction that σmean is of the Ultimate Strength, Sut
As the Mean stress increases, the material breaks at a lower Alternating stress amplitude.
Circles represent failures
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Fluctuating Fatigue Diagrams
Several ways are available to characterize the Fluctuating Fatigue behavior. Two common ones are:• Goodman line and• Gerber line.
Both approximate the material behavior. We will use the Goodman line because it is simpler and conservative.
The Modified Goodman Diagram is the red line and is the Goodman line truncated by the Yield line.
Hamrock Section 7.10
Modified Goodman Diagram
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Drawing the Modified Goodman Diagram
The diagram is based on material properties Sut, Sy, and Se.
As with the S-N curve, Se should be derated to reflect your part:Se = kf ks kr kt km Se’
(My Part) (Test Specimen)
For torsional (shear) loading, use Ssy = 0.577 Sy,Sus = 0.67 Sut, and Se for Torsion
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Drawing the Modified Goodman Diagram
Then plot your alternating and mean stress.
If your point is below the Mod Goodman line, the part should have unlimited life.
σσσσmean
σσσσalt
Note: This is very different from the “complete” Modified Goodman Diagram that Hamrock details on P. 237 – 240.
We will not use that version – it is pretty confusing.
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Factors of Safety
The Factor of Safety depends on how the stresses behave. They might:
1. Grow proportionately2. Only grow in mean3. Only grow in alternating
How they behave depends on the actual hardware and loading involved.
σσσσmean
σσσσalt
Proportional
Mean Only
Alt. Only
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If σalt and σmean Increase Proportionately
To make the Goodman Line w/F.O.S. go through
(σM,σA) :
Se
Se/n
Sut/n Sut
Load
Line
σσσσmean
σσσσalt
σσσσA
σσσσM
mean
M
Aalt σ
σ
σσ =
)1
(:
)1()1(
ut
meanealt
ut
meanealt
ut
meaneemean
ut
ealt
SnSnofSafetyofFactoraFor
SSor
SSS
S
S
σσ
σσ
σσσ
−=
−=+−=+−=Equations of the Goodman Line:
ut
M
e
A
ut
M
e
A
ut
MeA
SSn
SnS
SnS
σσ
σσ
σσ
+=
−=
−=
1
1
)1
(
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If σalt and σmean Increase Proportionately
Equation to find alternating stress when operating point hits the FOS line or Goodman line.
Se
Se/n
Sut/n Sut
σme
an
σalt
σA
σM
Set n = 1 to get this
level
+
=
ut
e
A
M
ealt
S
Sn
SLim
σ
σσ
1
1
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If Only σalt Increases
Equation to find alternating stress when operating point hits the Goodman line.
Se
σσσσa max
Sutσσσσmean
σσσσalt
σσσσA
σσσσM
)1(
)1(
max
ut
Mea
ut
meanealt
SS
SS
σσ
σσ
−=
−=
Equation of the
Goodman Line:
A
an
σ
σ max=
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0
10
20
30
40
50
60
70
0 10 20 30 40 50 60 70
Mean Stress (ksi)
Alte
rna
tin
g S
tre
ss (
ksi)
Fatigue Exercise
Yield 40 ksi
UltimateTensile 65 ksi
Endurance 30 ksi
Min Max Mean Alt
0 36 ksi
-27 37 ksi
14 32 ksi
On a Goodman Diagram, predict
fatigue for these loadings:
Given a bar of steel
with these properties
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Fatigue Exercise
Yield 40 ksi
UltimateTensile 65 ksi
Endurance 30 ksi
Min Max Mean Alt
0 36 18 18 ksi
-27 37 5 32 ksi
-18 46 14 32 ksi
0
5
10
15
20
25
30
35
40
45
0 5 10 15 20 25 30 35 40 45 50 55 60 65 70
Mean Stress (ksi)
Alt
ern
ati
ng
Str
ess (
ksi)
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Fatigue Diagram for Finite Life
From EngRasp ETBX
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Another Type of Fatigue Diagram
σalt = 50 ksiσmean = 70 ksi
Note Finite Life curves
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Stress Ratios for This Diagram
σalt = 50 ksiσmean = 70 ksi
mean
altAratioAmplitude
RratioStress
σ
σ
σ
σ
=
=
max
min
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More Cantilever Beam
Ignore Stress
Concentration
Details
12 Gauge (0.1094” thick)
0.75 in. wide
4 in. longHigh Strength Steel, with
Sut = 245 ksi
Machined finish
Room TemperatureSe = kf ks kr S’e = (0.63)(1)(0.87)(100) = 54.8 ksi
CASE 2: Tip is flexed between 0.075 in and 0.225 in. What is life for 95% survival?
By proportioning, the force now fluctuates between 8.631 lb and 3 x 8.631 = 25.893 lb.
Stresses go from +23.1 ksi to +69.3 ksi.
σalt
0
σmean0.075
0.150
0.225Tip
Deflection(in)
0
23.1
46.2
69.3
Stress
(ksi)
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Cantilever Beam, contd.
σalt
0
σmean0.075
0.150
0.225Tip
Deflection(in)
0
23.1
46.2
69.3
Stress
(ksi)
ksi
ksi
alt
mean
1.232
1.233.69
2
2.462
1.233.69
2
minmax
minmax
=−
=−
=
=+
=+
=
σσσ
σσσ
0
20
40
60
0 20 40 60 80 100 120 140 160 180 200 220 240 260
Mean Stress (ksi)
Alt
ern
ati
ng
Str
es
s (k
si)
Sut
Se
σσσσM
σσσσA
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Cantilever Beam FOS
ksi
ksi
alt
mean
1.23
2.46
=
=
σ
σ
0
20
40
60
0 20 40 60 80 100 120 140 160 180 200 220 240 260
Mean Stress (ksi)
Alt
ern
ati
ng
Str
es
s (k
si)
Sut
Se
σσσσM
σσσσA
What is the Factor of Safety?
A. If both alternating and mean stresses increase proportionately:
B. If only alternating stress increases:
64.1611.0
1611.0189.0422.0
245
2.46
8.54
1.231===+=+=+= n
SSn ut
M
e
A σσ
92.11.23
4.44
4.44)189.01(8.54)245
2.461(8.54)1(
max
max
===
=−=−=−=
A
a
ut
Mea
n
ksiS
S
σ
σ
σσ
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Impact
W
W
h
δk
Static displacement of Weight on Spring of stiffness, k
k
Wst =δ
Impact factor is
ststatic
m
h
W
PI
δδ
δ 211maxmax
++===
Effect is dependent on Spring Stiffness. A soft spring means large static deflection, which means smaller Impact factor.
Impact significantly increases forces!
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Impact For this situation, what is:1. The static deflection2. The Impact factor3. The max deflection4. The max force
10Lb
10”
δ100 Lb/in
10Lb
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Impact For the case of a weight sliding horizontally with a velocity, V, and hitting the spring
δ
k
W
W
Vgk
WV2
max =δ
where g is the gravitational constant, 386 in/s2 or 9.8 m/s2.
Recognizing that W/k = δst
g
Vst
2
max
δδ =
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Impact
For this setup, what velocity gives the same max force as the falling weight just did?
δ
100Lb/in
10Lb
V
10Lb
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Impact
Read Hamrock’s Example 7.10 of a diver landing on a diving board.
Note that the spring here is a beam, whose stiffness is calculated as Force/Deflection.
Also note that he deflects the end of the board 4 inches, and sees a max force of 2844 Lb!