Oct 1, 2013 ... Assignment 1 is due. Assignment 2 is given out. Static & Fatigue Failure. Various.
Ch – 5 Shigley. Ch – 6 Shigley. 30th. Sept. 4. FMEA. Various.
FME461 Engineering Design II Dr.Hussein Jama
[email protected] Office 414 Lecture: Mon 8am -10am Tutorial Tue 3pm - 5pm
10/1/2013
1
Semester outline Date
Week
Topics
9th Sept
1
House keeping issues Introduction to mechanical design Assignment 1 is given out
16th Sept 23rd Sept
2
Ethics & safety
Various
3
Assignment 1 is due Assignment 2 is given out Static & Fatigue Failure
Various Ch – 5 Shigley Ch – 6 Shigley
30th Sept 7th Oct
4
FMEA
Various
5
Continuous Assessment Test 1 (15%) Assignment 2 is due Assignment 3 is given out
14th Oct
6
Shafts and shaft components
Ch – 7 Shigley
21st Oct
7
Welding and permanent joints
Ch – 9 Shigley
28th Oct
8
Mechanical springs
Ch – 10 Shigley
4th Nov 11th Nov
9 10
Clutches & brakes Belts and chains
Ch – 16 Shigley Ch – 17 Shigley
18th Nov
11
Statistical consideration
Ch – 20 Shigley
25th Nov
12
Continuous Assessment Test 2 (15%)
13
Presentation of assignment 2 Assignment 2 is due
10/1/2013
2nd Dec
Reference Reading Ch 1 - Norton, Shigley
2
Discussion
Shigley Chapter 5 - Static failure criteria
Ductile materials Brittle materials
Shigley – Chapter 6 Fatigue failure criteria
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3
Static & Fatigue Failure Static load – a stationary load that is gradually applied having an unchanging magnitude and direction Failure – A part is permanently distorted and will not function properly A part has been separated into two or more pieces. 10/1/2013
4
Static failure theories
Maximum shear Stress theory Distortion energy theory Ductile Coulomb-Mohr theory
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Definitions Material Strength Sy = Yield strength in tension, Syt = Syc Sys = Yield strength in shear Su = Ultimate strength in tension, Sut Suc = Ultimate strength in compression Sus = Ultimate strength in shear = .67 Su
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6
Ductile and brittle materials A ductile material deforms significantly before fracturing. Ductility is measured by % elongation at the fracture point. Materials with 5% or more elongation are considered ductile.
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Brittle material yields very little before fracturing, the yield strength is approximately the same as the ultimate strength in tension. The ultimate strength in compression is much larger than the ultimate strength in tension. 7
Failure theories – Ductile materials • Maximum shear stress theory (Tresca 1886)
( maspecimen of the same material when that specimen x )component > ( )obtained from a tension test at the yield point
Failure
= Sy
To avoid failure =
Sy 2
(
)
max component
= Sy max
=
Sy 2n
> (Sy)t Distortion contributes to failure much more than change in volume.
h
h
t
(total strain energy) – (strain energy due to hydrostatic stress) = strain energy due to angular distortion > strain energy obtained from a tension test at the yield point → failure 10/1/2013
10
Plane stress problems
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11
Stress components
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12
Failure theories – ductile materials The area under the curve in the elastic region is called the Elastic Strain Energy. 3D case
U=½ ε
UT = ½
ε +½
1 1
ε +½
2 2
ε
3 3
Stress-strain relationship
ε1 = ε2 = ε3 = 10/1/2013
UT =
1
2E
(
2 1
+
2 2
+
2 3
1
v
E 2
E 1
v
E 3
E 1
v
E
) - 2v (
2
1
E
2
+
v v
3
E 3
E 2
v 1
E
3
+
2
3)
13
Failure theories – Ductile materials Distortion strain energy = total strain energy – hydrostatic strain energy 1 UT = ( 12 + 22 + 32) - 2v 2E Substitute
Uh =
1
1
(
= 2
2
3=
2 h
+
Simplify and substitute
1
2E
3 Uh =
h
+
=
2 h
2E
(1 – 2v) =
(
1
2
+
1
3
) - 2v (
h
h
+
h
h+
+
=3
h
+ 1
+
2
3
)
(1)
h
2
(
Ud = UT – Uh
2
+
2
3
+
h
h
h
)
into the above equation
2 (1 – 2v) ) 3
6E
Subtract the hydrostatic strain energy from the total energy to obtain the distortion energy 10/1/2013 Ud = UT – Uh =
1+v
6E
(
1–
2 ) + ( 2
1–
2 ) + ( 3
2–
(2) 2 ) 3
14
Failure theories – Ductile materials Strain energy from a tension test at the yield point 1=
Sy and
2
=
U d = UT – Uh =
3
=0
1+v
6E
Substitute in equation (2)
(
2 ) + ( 2
1–
Utest = (Sy)
2
1–
2 ) + ( 3
2–
2 ) 3
1+v
3E
To avoid failure, Ud < Utest
(
1–
2 ) + ( 2
2 ) + ( 3
1–
2
10/1/2013
2–
2 ½ ) 3
Sut Mohr’s circles for compression and tension tests.
Suc
3
Stress state
1
Sut
Tension test Compression test
Failure envelope The component is safe if the state of stress falls inside the failure envelope. 10/1/2013 25 1 > 3 and 2 = 0
Failure theories – brittle materials Modified Coulomb-Mohr theory 3 or
2
3 or
Sut Safe
Sut Safe
I
Sut
Sut
1
Suc
1
Safe
-Sut
III
Suc
Cast iron data
II -Sut
Safe
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2
Suc
Three design zones 26
Failure theories – brittle materials Zone I
3
1> 0,
=
1
2 > 0 and
1
>
Sut 2
I
Sut
Design equation
n
1
II -Sut
Zone II
III 1
>0,
1
=
2
Sut
n
< 0 and
2
< Sut Suc
Design equation
Zone III 1
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>0,
2 < 0 and
2
> Sut
1 (
1 1 1 – )– 2 = n Sut Suc Suc Design equation
27
Summary – Brittle materials
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Example
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Solution
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Solution continued
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Static failure summary - Ductile
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Summary – Brittle materials
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Failure theories - Fatigue It is recognised that a metal subjected to a repetitive or fluctuating stress will fail at a stress much lower than that required to cause failure on a single application of load. Failures occurring under conditions of dynamic loading are called fatigue failures. Fatigue failure is characterized by three stages
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Crack Initiation
Crack Propagation
Final Fracture 34
Jack hammer component, shows no yielding before fracture. Crack initiation site
Fracture zone 10/1/2013
Propagation zone, striation
35
Example VW crank shaft – fatigue failure due to cyclic bending and torsional stresses
Propagation zone, striations
Crack initiation site 10/1/2013
Fracture area 36
928 Porsche timing pulley
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Crack started at the fillet
37
Fracture surface of a failed bolt. The fracture surface exhibited beach marks, which is characteristic of a fatigue failure.
25mm diameter steel pins from agricultural equipment. Material; AISI/SAE 4140 low allow carbon steel
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bicycle crank spider arm
This long term fatigue crack in a high quality component took a considerable time to nucleate from a machining mark between the spider arms on this highly stressed surface. However once initiated propagation was rapid and accelerating as shown in the increased spacing of the 'beach marks' on the surface caused by the advancing fatigue crack. 10/1/2013
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Crank shaft
Gear tooth failure
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40
Hawaii, Aloha Flight 243, a Boeing 737, an upper part of the plane's cabin area rips off in mid-flight. Metal fatigue was the cause of the failure.
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Fracture surface characteristics Mode of fracture
Typical surface characteristics
Ductile
Cup and Cone Dimples Dull Surface Inclusion at the bottom of the dimple
Brittle Intergranular
Shiny Grain Boundary cracking
Brittle Transgranular
Shiny Cleavage fractures Flat
Fatigue
Beachmarks Striations (SEM) Initiation sites Propagation zone Final fracture zone
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Fatigue failure – type of fluctuating stresses
Alternating stress a= min a=
m=
2
=0 Mean stress max /
2 max m
= 10/1/2013
min
max
+ 2
min
43
Fatigue Failure, S-N Curve
Test specimen geometry for R.R. Moore rotating beam machine. The surface is polished in the axial direction. A constant bending load is applied.
Typical testing apparatus, pure bending Motor
Load
Rotating beam machine – applies fully reverse bending stress 10/1/2013
44
Fatigue Failure, S-N Curve
Finite life
Infinite life
S′e
Se′ = endurance limit of the specimen 10/1/2013
45
Relationship Between Endurance Limit and Ultimate Strength Steel
Se′ =
0.5Sut
Sut ≤ 200 ksi (1400 MPa)
100 ksi
Sut > 200 ksi
700 MPa Sut > 1400 MPa
Cast iron
Se′ =
0.4Sut
Sut < 60 ksi (400 MPa)
24 ksi
Sut ≥ 60 ksi
Cast iron
160 MPa Sut < 400 MPa 10/1/2013
46
Relationship Between Endurance Limit and Ultimate Strength Aluminium
Se′ =
0.4Sut
Sut < 48 ksi (330 MPa)
19 ksi
Sut ≥ 48 ksi
130 MPa Sut ≥ 330 MPa
Copper alloys Copper alloys
Se′ =
0.4Sut
Sut < 40 ksi (280 MPa)
14 ksi
Sut ≥ 40 ksi
100 MPa Sut ≥ 280 MPa For N = 5x108 cycle 10/1/2013
47
For materials exhibiting a knee in the S-N curve at 106 cycles S ′ = endurance limit of the specimen (infinite life > 106) e
Se = endurance limit of the actual component (infinite life > 106) S 103
Se 106
N
For materials that do not exhibit a knee in the S-N curve, the infinite life taken at 5x108 cycles
Sf′ = fatigue strength of the specimen (infinite life > 5x108) Sf = fatigue strength of the actual component (infinite life > 5x108) S 10/1/2013
103
Sf 5x108
N
48
Correction factor’s for specimen’s endurance limit Se = Cload Csize Csurf Ctemp Crel (S′e) Sf = Cload Csize Csurf Ctemp Crel (Sf′) • Load factor, Cload
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(page 326, Norton’s 3rd ed.)
Pure bending
Cload = 1
Pure axial
Cload = 0.7
Pure torsion
Cload = 1 if von Mises stress is used, use 0.577 if von Mises stress is NOT used.
Combined loading
Cload = 1
49
Correction factor’s for specimen’s endurance limit • Size factor, Csize
(p. 327, Norton’s 3rd ed.)
Larger parts fail at lower stresses than smaller parts. This is mainly due to the higher probability of flaws being present in larger components.
For rotating solid round cross section
d ≤ 0.3 in. (8 mm)
Csize = 1
0.3 in. < d ≤ 10 in.
Csize = .869(d)-0.097
8 mm < d ≤ 250 mm
Csize = 1.189(d)-0.097
If the component is larger than 10 in., use Csize = .6 10/1/2013
50
Correction factor’s for specimen’s endurance limit For non rotating components, use the 95% area approach to calculate the equivalent diameter. Then use this equivalent diameter in the previous equations to calculate the size factor. A95 = (π/4)[d2 – (.95d)2] = .0766 d2
d95 = .95d
d
Solid or hollow non-rotating parts
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dequiv = .37d
dequiv = (
A95 0.0766
)1/2
Rectangular parts
51
dequiv = .808 (bh)1/2
Correction factor’s for specimen’s endurance limit I beams and C channels
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52
Correction factor’s for specimen’s endurance limit • surface factor, Csurf
(p. 328-9, Norton’s 3rd ed.)
The rotating beam test specimen has a polished surface. Most components do not have a polished surface. Scratches and imperfections on the surface act like a stress raisers and reduce the fatigue life of a part. Use either the graph or the equation with the table shown below.
Csurf = A (Sut)b
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Correction factor’s for specimen’s endurance limit • Temperature factor, Ctemp
(p.331, Norton’s 3rd ed.)
High temperatures reduce the fatigue life of a component. For accurate results, use an environmental chamber and obtain the endurance limit experimentally at the desired temperature. For operating temperature below 450 oC (840 oF) the temperature factor should be taken as one.
Ctemp = 1
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for T ≤ 450 oC (840 oF)
54
Correction factor’s for specimen’s endurance limit • Reliability factor, Crel (p. 331, Norton’s 3rd ed.) The reliability correction factor accounts for the scatter and uncertainty of material properties (endurance limit).
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Fatigue Stress Concentration Factor, Kf Experimental data shows that the actual stress concentration factor is not as high as indicated by the theoretical value, Kt. The stress concentration factor seems to be sensitive to the notch radius and the ultimate strength of the material. Notch sensitivity factor Fatigue stress Kf = 1 + (Kt – 1)q concentration factor rd (p. 340, Norton’s 3 ed.)
Steel
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Fatigue Stress Concentration Factor, Kf for Aluminum (p. 341, Norton’s 3rd ed.)
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Design process •
Determine the maximum alternating applied stress ( the size and cross sectional profile
•
Select material → Sy, Sut
•
Choose a safety factor → n
•
Determine all modifying factors and calculate the endurance limit of the component → Se
•
Determine the fatigue stress concentration factor, Kf
•
Use the design equation to calculate the size
Kf
a)
in terms of
Se a= n
•
Investigate different cross sections (profiles), optimize for size or weight
•
You may also assume a profile and size, calculate the alternating stress and determine the safety factor. Iterate until you obtain the desired 58 safety factor
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Design for finite life
Sn = a (N)b equation of the fatigue line A S
B Se 106
103
Point A
Point B 10/1/2013
N
Sn = .9Sut
Point A
Sn = .9Sut
N = 10
N = 103
Sn = Se
Sn = Sf
3
6
N = 10
Point B
N = 5x108
59
Design for finite life Sn = a (N)b
log Sn = log a + b log N Apply boundary conditions for point A and B to find the two constants “a” and “b”
log .9Sut = log a + b log 10
a=
3
log Se = log a + b log 106
b=
N
Sn = Se ( 106 ) Calculate Sn
2
Se 1 3
log
.9Sut Se
Se log ( .9S ) ut
and replace Se in the design equation Kf
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⅓
(.9Sut)
a=
Sn n
Design equation 60
The effect of mean stress on fatigue life Mean stress exist if the loading is of a repeating or fluctuating type.
a
Mean stress is not zero
Gerber curve Alternating stress
Se Goodman line
Soderberg line 10/1/2013
Sy Mean stress
Sut
m 61
The effect of mean stress on fatigue life goodman diagram a
Yield line
Alternating stress
Goodman line Safe zone
C
Sy
Sut
m
Mean stress
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62
The Effect of Mean Stress on Fatigue Life Modified Goodman Diagram a
Sy
Yield line
Goodman line Safe zone -
m
10/1/2013
- Syc
Safe zone
Sy
Sut
+
m
63
The Effect of Mean Stress on Fatigue Life Modified Goodman Diagram Fatigue,
m
≤0
Fatigue, a
a
Se a=
Se
nf
m
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Sut m
Sut
m
=
>0 = = 1
1
Infinite life
nf Finite life Yield
Syc
ny Safe zone
-
Sn
+
m
Se
Yield a+
a
+
m
Safe zone
C
a+
m
=
Sy
ny
- Syc
64
Applying Stress Concentration factor to Alternating and Mean Components of Stress •
Determine the fatigue stress concentration factor, Kf, apply directly to the alternating stress → Kf
•
If Kf
max
a
< Sy then there is no yielding at the notch, use Kfm = Kf
and multiply the mean stress by Kfm → Kfm •
m
> Sy then there is local yielding at the notch, material at the notch is strain-hardened. The effect of stress concentration is reduced. If Kf
max
Calculate the stress concentration factor for the mean stress using the following equation, Kfm =
Sy
Kf
a
m
Fatigue design equation Kf a Kfm m = +
Se
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Sut
1
nf
Infinite life 65
Combined loading All four components of stress exist, xa
alternating component of normal stress
xm
mean component of normal stress
xya
alternating component of shear stress
xym
mean component of shear stress
Calculate the alternating and mean principal stresses, 1a,
1m, 10/1/2013
2a
=(
xa /2)
(
2 / 2 ) +( xa
2 ) xya
2m
=(
xm /2)
(
2 / 2 ) +( xm
2 ) xym 66
Combined loading Calculate the alternating and mean von Mises stresses, a′ = (
2
1a
m′ = (
2
1m
+ +
2
2a -
1a
2
2m -
1m
1/2 ) 2a 1/2 ) 2m
Fatigue design equation
′a Se 10/1/2013
+
′m Sut
=
1
nf
Infinite life
67
Example 6-7
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68
Example 6-9
Shaft is made from SAE1050 steel and is cold formed
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Solution
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Solution
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