mech1641 - w2008 lecture 04 - fatigue iii
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MECH1641 – Machine Design II
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Lecture 4: Fatigue III
Designing for Fatigue spreadsheet example
Fluctuating Stresses
Mean Stress & Stress Amplitude
Maximum & Minimum Stress
Constant Life Diagrams
Homework & Reading Assignment
MECH1641 – Machine Design II
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Objectives
After this lecture and associated homework, you will:
Use a spreadsheet to design a hollow axle for cyclic bending loads.
Characterize fluctuating stresses in terms of mean stress and stress amplitude.
Characterize fluctuating stresses in terms of maximum and minimum stress.
Apply the relationship between the two methods of characterizing fluctuating stresses.
Interpret and construct Constant Life diagrams from information obtained from the S-N curve.
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Example Problem 1Create a spreadsheet to design a trailer axle made of 4340 steel, heat-treated to 160 ksi. The total trailer weight W is 4200 lb, divided evenly between identical front and rear axles. The wheel-to-bearing distance x is 6 in, and the minimum required safety factor against fatigue failure is 2. The axle surface will be machined to an outside diameter of 2.5 in, and then coated to protect it from corrosion in service, which will be at room temperatures. Reliability should be 95%.
½ W
x
di do
x48"
First, calculate the axle’s endurance limit from the material ultimate strength and appropriate modifying factors. Than calculate the bending stress from the applied load and axle dimensions. Then calculate a safety factor n by dividing the endurance limit by the bending stress.Set the outside diameter to 2.5 in, and use trail and error to find the ID that results in a safety factor of 2.Modify the spreadsheet to calculate the mass of the axle, using the density of steel and the calculated volume of the axle.Determine the mass of a suitable axle with an OD of 3 in, and compare it to the mass of the 2.5 in. axle. (Check MatWeb.com for the density.) Which axle is preferable?
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Fluctuating Stresses
Machine components seldom encounter true completely reversing stresses; fluctuatingstresses are far more common.
Fluctuating stresses are induced by a combination of static and reversing cyclic loads.
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Characterizing Fluctuating Stresses
Fluctuating stresses can be characterized, or specified, in two ways:
Maximum and minimum stresses σmax and σmin.
Mean stress and stress amplitude σm and σa.
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Special Cases of Fluctuating Stresses
For fully reversingloads:
σm=0
σmax = -σmin
σa=σmax
For zero-to-max-to-zero loads:
σmin=0
σmax = 2σa
σa=σm
amax
m
min
amax
m
min
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Fluctuating Stress Parameter Relationships
2
2
max minm
max mina
σ σσ
σ σσ
+=
−=
Note that if any two of these stress parameters are known, the other two are readily calculated.
Stress amplitude σa is always positive.
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Example Problem 2What is mean stress σm and stress amplitude σa of an application that cycles between -20 and +60 ksi? Sketch the stress vs. time profile.
( ) ( )
( ) ( )
: , : 60 ; 2060 20
2 220
60 202 2
40
m a max min
max minm
m
max mina
a
Find Given ksi ksiksi ksi
ksi
ksi ksi
ksi
σ σ σ σ
σ σσ
σ
σ σσ
σ
= = −
+ −+= =
∴ =
− −−= =
∴ =
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Example Problem 3Find the minimum stress σmin and stress amplitude σaof an application that has a maximum stress of 975 MPs and a mean stress of 350 MPa. Sketch the stress vs. time profile.
( )
( ) ( )
: , : 975 ; 350
22
2 350 975
275
975 2752 2
625
min a max m
max minm min m max
min
min
max mina
a
Find Given MPa MPa
MPa MPa
MPa
MPa MPa
MPa
σ σ σ σσ σσ σ σ σ
σ
σ
σ σσ
σ
= =
+= → = −
= −
∴ = −
− −−= =
∴ =
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Fluctuating Stresses with Infinite LifeThe existence of a static tensile stress (the mean stress σm) reduces the amplitude of alternating stress σa that can be applied.
As the mean stress σm approaches the material ultimate strength Su, the allowable stress amplitude σa decreases towards zero.
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Fluctuating Stresses with Infinite LifeCase (a) shows a fully reversing cyclic stress at the level of the endurance limit, i.e., infinite life.
For infinite life with some tensile load superimposed, the stress amplitude must decrease.
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Fluctuating Stresses with Infinite LifeFrom a, to b, to c and so on, the superimposed tensile stress σmincreases; to maintain infinite life, the stress amplitude σa must decrease.
Note that cases e and f have yielding that occurs at the very first load cycle.
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Fluctuating Stresses with Infinite Life
Strangely, if the superimposed stress σm is compressive, it doesn’t require a decrease in amplitude to maintain equivalent life.
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Constant Life Diagram
The horizontal axis is mean stress σm.
The vertical axis is stress amplitude σa.
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Constant Life Diagram Horizontal Axis Since the values along the horizontal axis have a stress amplitude of zero, it represents simple static loading, i.e. no cycling.
Sy
0
-Sy
Sy
0
-Sy
Sy
0
-Sy
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Constant Life Diagram Vertical Axis Since the values along the vertical axis have a mean stress of zero, it is represents fully-reversed cyclic loading.
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Constant Life Diagram
The horizontal axis is mean stress σm.
The vertical axis is stress amplitude σa.
Sy
0
-Sy
0
Sy
-Sy
Sy
0
-Sy
Sy
0
-Sy
Sy
0
-Sy
Sy
0
-Sy
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Constant Life Diagram The lines of constant life show the combinations of σmin and σmax that allow for equal life. The 106 line can be considered infinite life, corresponding to the endurance limit.
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Constant Life Diagram All points on the red 106 life line represent combinations of mean stress and stress amplitude that will result in a life of 106 cycles.
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Constant Life Diagram All points below the red 106 life line represent combinations of mean stress and stress amplitude that will result in a life greater than 106 cycles, essentially infinite.
Infinite Life for combinations of σm & σa in this region.
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Constant Life Diagram
Combinations of σm & σain this region will fail between 103 and 104
cycles.
Similarly, all points between the the red 103 and 104
lines line represent combinations of mean stress and stress amplitude that will result in a life between these limits.
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Constant Life Diagram The values at points C, D, E, F on the vertical axis come from the S-N curve. They are fatigue strengths at various values for N.
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Constant Life Diagram All points along the line A-A’’ represent fluctuating load profiles with a tensile peak of σmax=Sy.
Sy
0
-Sy
Sy
0
-Sy
Sy
0
-Sy
Sy
0
-Sy
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Constant Life Diagram Similarly, all points along the line A’-A’’ represent fluctuating load profiles with a compressive peak of σmin=-Sy.
Sy
0
-Sy
Sy
0
-Sy
Sy
0
-Sy
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Constant Life Diagram All points along lines parallel to A-A’’ represent fluctuating load profiles with a fixed value for σmax.
σmax =Sy
σmax =Sn
σmax = ½Sn
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Constant Life Diagram
σmax =Sy
σmax =Sn
σmax = ½Sn
This type of Constant Life diagram shows only mean stress and stress amplitude. It could be reconstructed to show maximum and minimum stresses.
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Fatigue Strength Diagram for Alloy SteelsApplicable to AISI 4340, 4130, 2330, 8630.
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Example Problem 4If a test specimen made from normalized 4130 steel with an ultimate strength of 100 ksi is subject to a maximum bending stress of 90 ksi and a minimum stress of 0, what is the expected fatigue life? Sketch the stress vs. time profile.
4: 90 90% 0 10max u minFind N Given ksi S N cyclesσ σ= = = → =
0
90 ksi
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Example Problem 5For an normalized 4130 test specimen with Su=100 ksi, what is the maximum stress that can be applied for 106 cycles if the loading is of the zero-to-max-to-zero type? Sketch the stress vs. time profile.
610 : 0
70% 70max min
max u
Find for N cycles Given
S ksi
σ σ
σ
= =
= =
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Homework and Reading Assignment
For next class, read Juvinall Chapter 8 Sections 8.9, then answer the questions below.1. Determine the mean stress and stress amplitude of an
application that cycles between -350 and 780 MPa? Sketch the stress vs. time profile.
2. Determine the minimum stress and stress amplitude of an application with a mean stress of 925 MPa and a maximum stress of 1190 MPa. Sketch the stress vs. time profile.
3. For 4340 test specimen with Su=150 ksi, what is the predicted life for a fully reversing stress of 90 ksi? (3×104 cycles)
4. What is the maximum stress that will provide the same life if the load profile is changed to a zero-to-max-to-zero type? (86 ksi.)
5. What stress amplitude will provide the same life if the load profile is fluctuating with a maximum stress of 60 ksi? (27 ksi)
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Closing Notes
Quiz 2 (5%) is on Friday January 25.
Test 1 (20%) is on Friday February 22.
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