2. 2
1. Key points of L9 & L10
2. Momentum Theory and Blade Element Theory
2.1 Momentum theory
2.2 Blade element theory
2.3 Blade shape for ideal rotor without wake
rotation
2.4 Strip Theory or BEM Theory
(A combination of momentum theory
and blade element theory)
2
L11 and L12
3. 3
3
3. Power coefficient Cp including the
effects of wake rotation and profile
drag
4. Advanced aerodynamics topics
(# 3.13, pp. 141-145)
5. Chapter 4: Mechanics and Dynamics
5.1 An overview of basic mechanics
5. 5
1. Range of validity of momentum theory
a = axial induction factor
for a ≤ 0.4, momentum theory is considered
valid. 2nd Edition (3.12, 3.14), pp. 94-95
For a ≥ 0.4, momentum theory is considered
not valid
P.87
),
14
.
2
.
3
(
..........
..........
)
1
(
4
P.87
),
17
.
2
.
3
..(
..........
).........
1
(
4
2
a
a
C
a
a
C
p
t
-
=
-
=
L10)
L9
empirical,
.(
..........
2
)
1
(
4 +
+
-
= a
Ct
6. 6
α
c/4
Pitching moment = M
Drag force = D
Chord
Air flow
Drag and lift forces and pitching moment on stationary airfoil:
α : angle of attack ; c: chord (p. 97)
Airfoil Basics
7. 7
L = Lift force over the reference area c.l
Cl = lift coefficient
l
C
l
U )
.
c
(
2
1 2
÷
ø
ö
ç
è
æ
= r
( )
length
unit
per
force
Dynamic
length
unit
per
force
Lift
2
1
1
.
2
1
.
2
1
2
2
2
=
÷
ø
ö
ç
è
æ
=
÷
ø
ö
ç
è
æ
÷
ø
ö
ç
è
æ
=
÷
ø
ö
ç
è
æ
=
c
U
l
L
l
l
c
U
l
L
l
c
U
L
r
r
r
8. 8
Drag force = D
Drag coefficient:
Pitching moment = M
Pitching moment coefficient:
)
area)(
Ref.
(
2
1 2
d
C
U ÷
ø
ö
ç
è
æ
= r ….(3.43) p. 104
)
(c)(C
U m
area)
Ref.
(
2
1 2
÷
ø
ö
ç
è
æ
= r
….(3.44) p. 104
length
unit
per
force
Dynamic
length
unit
per
force
Drag
2
1 2
=
=
c
U
l
D
Cd
r
moment
Dynamic
moment
Pitching
)
.
.(
2
1 2
=
=
c
l
c
U
M
Cm
r
9. 9
μ = coefficient of viscosity
= kinematic viscosity
r
m
u =
÷
ø
ö
ç
è
æ
=
÷
÷
ø
ö
ç
ç
è
æ
»
ï
þ
ï
ý
ü
ï
î
ï
í
ì
u
m
r Uc
f
Uc
f
c
c
c
m
d
l
u
m
r Uc
Uc
=
=
= number
Reynolds
Re ….(3.41)
p. 103
(* we take chord length c as reference length)
10. 10
Stream tube model of flow behind rotating wind blade. Picture of stream
tube with wake rotation.
Momentum Theory with Wake Rotation (#3.3, pp. 96-101)
11. 11
An important statement in the text (p. 97)
“Note that when wake rotation is included in
the analysis, the induced velocity at the rotor
consists of not only the axial component,
Ua, But also a component in the rotor plane
rΩa'.” (a' = angular induction factor)
Momentum Theory with Wake Rotation
12. 12
rΩ
Plane of Rotation
U
aU
Wind Direction
U(1-a)
For future reference:
)
'
1
(
)
1
(
'
tan
a
r
a
U
aU
r
a
+
W
-
=
W
=
f
Flow velocity diagram at an annulus in a HAWT rotor disk
f
a'r Ω
f f
16. 16
( )
( ) 106
p.
),
1
.
5
.
3
(
..........
..........
dr
r
)
a
1
(
a
4
U
91
p.
),
5
.
3
.
3
.....(
..........
dr
r
2
U
2
1
)
a
1
(
a
4
dT
2
2
p
-
r
=
p
r
-
=
106
p.
),
2
.
5
.
3
.(
..........
).........
(
)
1
(
'
4
91
p.
),
10
.
3
.
3
...(
).........
2
(
2
1
)
1
(
'
4
3
2
dr
Ur
a
a
rdr
r
U
a
a
dQ
W
-
=
W
-
=
p
r
p
r
2.1 Momentum Theory (L9 + L10)
2nd Edition (3.58-59) p. 118
dL = 4b[1-b*cos(φ)](1/2)ρUΩr2(2πrdr)
= 4b[1-a]ρUr3(πΩdr) . . . . . . . . . . Peters
19. 19
dL = dFL
Ω dFN
dD= dFD
dFT
U(1-a)
Ω r ( 1+ a') Ur
Ф
α
Plane of Rotation
θ (class) = θP (text)
= angle between
the chord line and
the plane of rotation
α = angle of attack
= angle between the
chord line and the relative wind Ur
Ф = θ + α = θp + α = relative wind angle
= angle between the plane
of rotation and relative wind
velocity Ur
Chord line
Blade Geometry for HAWT Aerodynamic Analysis
A typical blade element of length dr and width c
20. 20
dL = dFL
Ω
FN = Ftransit
D
FT
U(1-a)
a'Ω r
Ur
Ф
α
Plane of Rotation
Chord line
Ω r
FTorque
U(1-a) Ω
21. 21
dL = dFL
Ω dFN
dD= dFD
dFT
U(1-a)
Ur
Ф
α
Plane of Rotation
Ur = relative wind velocity
dFL= incremental lift force
dFD = incremental drag force
dFN = Incremental drag force
normal to the plane of
rotation (this contributes
to thrust)
dFT = Incremental drag force
tangential to the
plane of rotation (this
contributes to torque)
Chord line
Ω r ( 1+ a')
Blade Geometry for HAWT Aerodynamic Analysis
23. 23
Let c(Δr) = dA
B = Number of blades
91
p.
.(3.3.7),
..........
..........
ratio.....
speed
tip
U
R
91
p.
3.8),
.......(3.
ratio.....
speed
local
R
r
.
U
r
p.109
(3.5.6),
........
..........
)
'
a
1
(
)
a
1
(
)
'
a
1
(
r
)
a
1
(
U
tan
cos
)
'
a
1
(
r
sin
)
a
1
(
U
U
r
r
r
=
W
=
l
=
l
=
W
=
l
l
+
-
=
+
W
-
=
f
f
+
W
=
f
-
=
2nd Edition
(3.63) p. 120
(3.26-27) p. 98
24. 24
Incremental lift force: (3.55-56) p. 120
( )
( ) 109
p.
),
9
.
5
.
3
.(
..........
..........
.
.
2
1
force
drag
l
Incrementa
109
p.
),
8
.
5
.
3
.(
..........
..........
.
.
2
1
2
2
dr
c
U
C
dF
dr
c
U
C
dF
r
d
D
r
l
L
÷
ø
ö
ç
è
æ
=
÷
ø
ö
ç
è
æ
=
r
r
25. 25
Force FT in the plane of blade rotation
generates useful torque: Q
109
p.
),
10
.
5
.
3
..(
..........
..........
sin
cos
:
thrust
to
s
contribute
dF
component
The
109
p.
),
11
.
5
.
3
..(
..........
..........
cos
sin
N
f
f
f
f
D
L
N
D
L
T
dF
dF
dF
dF
dF
dF
+
=
-
=
(FN exerts a thrust load on the rotor)
2nd Edition (3.67-68) p. 120
26. 26
We consider a rotor with B blades
The elemental normal force (thrust) on the
section at a distance r from the center:
Similarly, the elemental torque due to the
tangential force dFT operating at a distance r
from the center :
109
p.
),
12
.
5
.
3
....(
..........
)
sin
cos
(
2
1
. 2
dr
c
C
C
U
B
dF D
L
r
N f
f
r +
÷
ø
ö
ç
è
æ
=
110
.
p
),
13
.
5
.
3
.(
..........
..........
..........
.
. T
dF
r
B
dQ =
2nd Edition (3.69-70) p. 120-121
28. 28
Summary of Blade Element theory
We considered an annular rotor section
dA = 2 π r dr. Then we obtained one equation for
normal force (thrust) dFN: (3.69, 3.71) pp. 120-121
and one equation for dFT that gives the torque dQ:
109
p.
),
12
.
5
.
3
....(
..........
dr
c
)
sin
C
cos
C
(
U
2
1
.
B
dF d
l
2
r
N f
+
f
÷
ø
ö
ç
è
æ
r
=
( ) 110
p.
),
14
.
5
.
3
..(
..........
dr
r
c
cos
C
sin
C
U
2
1
.
B
dQ d
l
2
r f
-
f
÷
ø
ö
ç
è
æ
r
=
30. 30
Ideal Rotor (p.121)
1. Axial induction factor a = 1/3 for each
annular stream tube as in a Betz rotor
2. No wake rotation (angular induction
factor a' =0)
3. no losses from a finite number of blades
4. No drag, Cd = 0
31. 31
2nd Edition (3.72) 121, (3.74) p. 122, 124
Recall from momentum theory
( )
111
p.
),
3
.
6
.
3
.......(
..........
..........
..........
sin
3
2
sin
)
1
(
111
p.
),
2
.
6
.
3
.(
..........
..........
)
0
cos
(
2
1
B.
109
p.
),
12
.
5
.
3
....(
..........
)
sin
cos
(
2
1
.
:
theory
element
blade
From
9
8
110
p.
),
1
.
6
.
3
....(
..........
..........
3
1
1
3
1
4
106
p.
),
1
.
5
.
3
........(
..........
..........
2
)
1
(
4
2
1
2
2
2
2
2
f
f
f
r
f
f
r
p
r
p
r
p
r
U
a
U
U
dr
c
C
U
dr
c
C
C
U
B
dF
dr
r
U
dr
r
U
dr
r
a
a
U
dT
r
L
r
D
L
r
N
=
-
=
+
=
+
÷
ø
ö
ç
è
æ
=
=
÷
÷
ø
ö
ç
ç
è
æ
÷
ø
ö
ç
è
æ
-
=
-
=
34. 34
Cl Bc
4p r
= tanfsinf............(3.6.4),p. 111
=
2
3
.
1
lr
æ
è
ç
ö
ø
÷sinf.
c =
8p rsinf
3BCl lr
.................( 3.6.8),p. 111
Alternative Method: μ = λr
c = [4πr/BCl]*[(3b-1)/b] =
[16πr/BCl]*[b2/(1 + λr
2)]
Equation to find optimum chord. (3.75, 3.79) p. 122
tan(ϕ) = 2/(3λr)
35. 35
Example (p.122) and table 3.2 (p. 123)
λ = 7 , R = 5 m, Cl = 1
8
.
0
R
r
consider
on,
illustrati
for
and,
7
Take
3
B
,
7
at
C
C
o
o
min
l
d
=
=
a
=
=
a
÷
÷
ø
ö
ç
ç
è
æ
36. 36
r/R c θ
0.8 0.236 6.8 o -0.2 o 7º
f a = f - = 7o
( )
o
1
1
1
8
.
6
8
.
0
x
7
x
3
2
tan
111
p.
),
7
.
6
.
3
..(
..........
8
.
0
x
7
x
3
2
tan
R
r
3
2
tan
=
÷
ø
ö
ç
è
æ
=
÷
ø
ö
ç
è
æ
=
÷
÷
ø
ö
ç
ç
è
æ
l
=
f
-
-
-
2nd Edition (3.76), p. 122
Book method to get inflow angle.
θ
49. 49
To combine the momentum theory with the
blade element theory, we have to express
Ur = V in terms of U and Ω r.
50. 50
dL = dFL
Ω
FN = thrust
D
FT
U(1-a)
a'Ω r
Ur
Ф
α
Plane of Rotation
Chord line
Ω r
FTorque
U(1-a) Ω
51. 51
General Blade-Element/Momentum Theory
dL = 2ρ(2πrdr)(U-wcosΦ)w = (B/2)ρ(cdr)V2Cl
Cl = s*sin(ϕ - θ) = s*[sinϕcosθ – cosϕsinθ)]
s = slope of lift curve, 2π for flat plate, 5.73 for 0012
8πr(1-b*cosϕ)b = Bcs(1+μ2-b2)[sinϕcosθ – cosϕsinθ)]
For large λr, ϕ is small and we can approximate.
We also define σ’= sλr/4 = [Bc/(2πr)]sλr/4 = A.
52. 52
BEM with Small Angles
For small angles, cosΘ=1, sinΘ=Θ, cosϕ=1, sinϕ=(1-b)/λr
This results in the BEM balance of: b(1-b) =A[1-b-λrθ]
Which gives a quadratic: b2 – b(1+A) +A(1 – λrθ)
b = {(1+A) – [(1-A)2 +4λrθ]1/2}/2
This b can be used either in momentum or blade element to
get the lift which resolves into thrust and torque.
55. 55
Now, we express the blade element theory as
follows (after some algebra):
)
....(
2
2
1
tan
1
1
cos
sin
'
)
'
1
(
)
.......(
2
2
1
tan
1
sin
cos
'
)
1
(
2
2
2
2
2
2
D
dr
r
r
r
C
C
C
a
dQ
C
dr
r
U
C
C
C
a
dT
l
d
l
l
d
l
p
r
f
f
f
s
p
r
f
f
f
s
W
÷
÷
ø
ö
ç
ç
è
æ
-
+
=
÷
÷
ø
ö
ç
ç
è
æ
+
-
=
Combining Eq. (D) and 2nd eq. p. 124 and Eq. (B), we get
)
.......(
..........
..........
tan
1
1
cos
'
1
'
4
F
C
C
C
a
a
l
d
l
÷
÷
ø
ö
ç
ç
è
æ
-
¢
=
+ f
f
s
Combining Eq. (c), Eq. (3.5.1) we get
)
.(
..........
..........
tan
1
sin
cos
.
1
4
2
E
C
C
C
a
a
l
d
l ÷
÷
ø
ö
ç
ç
è
æ
+
¢
=
+
f
f
f
s
56. 56
It is conventional (“accepted practice” according
to the text) to neglect drag force (Cd = 0) in the
BEM theory: 2nd Edition (3.88-89) p. 125
With Cd = 0 in Eqs. (E) and (F), we get
( )
[ ]
( )
[ ] 115
.
p
),
10
.
7
.
3
..(
..........
1
C
'
cos
4
1
'
a
cos
C
'
'
a
1
'
a
4
115
.
p
),
9
.
7
.
3
....(
cos
C
'
sin
4
1
1
a
sin
cos
C
'
a
1
a
4
l
l
l
2
2
l
-
s
f
=
Þ
f
s
=
+
f
s
f
+
=
Þ
f
f
s
=
-
57. 57
Power Coefficient CP including the effects of
wake rotation and profile drag (pp. 116-119)
We begin with the basic equation:
(3.92) p. 127
ò
ò W
=
=
W
=
R
r
R
r h
h
dQ
dP
P
dQ
dP
power
Total
p.117
),
13
.
7
.
3
..(
..........
..........
where rh is the rotor radius at the hub
58. 58
Power Coefficient Cp:
2nd Edition (3.94) p. 127
?
dQ
117
p.
,
)
15
.
7
.
3
.....(
2
1 3
2
=
W
=
=
=
ò
U
R
dQ
P
P
P
P
C
R
r
dynamic
wind
p
h
p
r
59. 59
We already have an expression for dQ from
the blade element theory: (3.95) p. 127
( ) p.114
.),
2
.
7
.
3
....(
cos
sin
sin
)
1
(
' 2
2
2
2
dr
r
C
C
a
U
dQ d
l f
f
f
r
p
s -
-
=
Also, we wish to replace dr by dλr by using our earlier
definition of local speed ratio λr :
r
r
d
R
r
U
r
l
l
l
l
×
=
=
W
=
R
dr
p.91
),
8
.
3
.
3
....(
..........
..........
60. 60
We make use of these two equations for dQ
and λr, and express CP (after some algebra)
as follows:
117
p.
),
16
.
7
.
3
....(
cot
1
sin
1
)
1
(
'
2 2
2
2 r
r
l
d
l
p d
C
C
a
C
C
h
l
l
f
f
s
l
l
l
ú
û
ù
ê
ë
é
÷
ø
ö
ç
è
æ
-
÷
÷
ø
ö
ç
ç
è
æ
-
= ò
Where λh, is the local speed ratio at r = rh.
To further simplify the above integral equation for
CP, we revisit two earlier-derived equation for
a/(1-a) and a/a' from the blade element theory:
61. 61
That is from p. 125,
p.117
),
18
.
7
.
3
.......(
..........
..........
'
tan
p.115
),
8
.
7
.
3
........(
..........
..........
tan
a'
a
and
p.115
),
7
.
7
.
3
.....(
).........
1
(
'
cos
sin
a
4
p.115
),
5
.
7
.
3
.(
..........
sin
4
1
cos
'
1
2
2
r
r
l
l
a
a
a
C
C
a
a
l
f
f
l
s
f
f
f
f
s
=
Þ
=
-
=
Þ
×
=
-
62. 62
Substituting these two equations into integral for
Cp: see bottom of page 127
117
p.
),
11
.
7
.
3
.....(
d
cot
C
C
1
)
a
1
(
'
a
8
d
cot
C
C
1
)
a
1
(
'
a
8
d
cot
C
C
1
)
a
1
(
tan
a
8
d
cot
C
C
1
sin
)
a
1
(
cos
sin
a
4
2
d
cot
C
C
1
sin
1
)
a
1
(
C
'
2
C
r
l
d
3
r
2
r
2
r
l
d
r
2
r
2
r
l
d
2
r
2
r
l
d
2
2
r
2
r
l
d
2
l
2
p
h
h
h
h
h
l
ú
û
ù
ê
ë
é
f
÷
÷
ø
ö
ç
ç
è
æ
-
-
l
l
=
l
l
ú
û
ù
ê
ë
é
f
÷
÷
ø
ö
ç
ç
è
æ
-
-
l
l
=
l
l
ú
û
ù
ê
ë
é
f
÷
÷
ø
ö
ç
ç
è
æ
-
-
f
l
=
l
l
ú
û
ù
ê
ë
é
f
÷
÷
ø
ö
ç
ç
è
æ
-
f
-
f
f
l
=
l
l
ú
û
ù
ê
ë
é
f
÷
÷
ø
ö
ç
ç
è
æ
-
f
-
s
l
=
ò
ò
ò
ò
ò
l
l
l
l
l
l
l
l
l
l
63. 63
It is instructive to set Cd = 0 in previous eq.
92
p.
),
14
.
3
.
3
.......(
..........
d
)
a
1
(
'
a
8
C r
3
r
2
p
h
l
l
-
l
= ò
l
l
This is Cp based on momentum theory including
wake rotation.
64. 64
Project #2 part b
Due along with part a
March 6
Repeat table 3.2 and the figures on
slides 40 – 42 including wake rotation and
using the Peters’ method. Include both θT
and θP.
65. 65
4. Advanced Aerodynamics Topics
In later lectures, we treat the following
four advanced topics (this treatment goes
well beyond the text):
1.Dynamic wake or inflow
2.Dynamic stall
3.Tip-loss factor
4.Rotational sampling
66. 66
5. Chapter 6: Wind Turbine
Materials and Components
Please review basics of statics, mechanics
of materials and dynamics; this is typically
covered in the undergraduate engineering
program.
67. 67
In this lecture we address a very
important topic: Fatigue
For many components of WT, fatigue is the
design driver.
69. 69
Overview (p. 257)
“It is well known that many materials which can
withstand a load when applied once, will not survive
if that load is applied and then removed ('cycled') a
number of times. This increasing inability to withstand
loads applied multiple times is called fatigue damage.
The underlying causes of fatigue damage are
complex, but they can be most simply conceived of
as deriving from the growth of tiny cracks. With each
cycle, the cracks grow a little, until the material fails.
This simple view is also consistent with another
observation about fatigue: the lower the magnitude of
the load cycle, the greater the number of cycles that
the material can withstand.”
70. 70
“Like all rotating machines, windmills are
generators of fatigue. Every revolution of its
components (i.e., rotor, transmission,
generator, yaw column, etc.) produces a
load cycle, known as fatigue cycle. Each of
these cycles causes a finite amount of
damage, resulting in a reduction in the
component fatigue life.”
72. 72
Typical pattern of design drivers for the primary structural and mechanical
components in a HAVVT. (ASME publication, ibid)
73. 73
Let
nL= the total number of blade revolutions
over the turbine’s life time
Since the number of many of the fatigue-
producing stress cycles is proportional to nL,
it is instructive to evaluate nL in a typical
scenario:
74. 74
Consider a large turbine with an RPM of
approximately 30 to 70 operating 4000
hours per year. This turbine would
experience from 108 to 109 cycles over a 20-
year lifetime. How?
75. 75
Total number of cycles
over a lifetime = nL
nL = 60 k nrotor(Hop)(Y)…………. ………..(6.1), p. 258
k = no. of cyclic events/revolution
(atleast equal to 1; see text p. 157)
nrotor = rotational speed (rpm)
Hop = operating hours/year
Y = years of operation
76. 76
Ω = 30 to 70 rpm ≈ 50 rpm
Hop= 4000 hours/year
Y = 20 years
nL = 60(1)(50)(4000)(20)
= (240)106) cycles
77. 77
S ≡ stress
N ≡ number of cycles to failure
This is evaluated by conducting a rotating
beam test
S-N Curve (p. 259)
83. 83
Rotating Beam Test
1) Loaded with known weight W
2) Cycled until fracture
3) The value of max. stress (σa) at fracture
and at a specific number of cycles N is
recorded. This gives fatigue strength at
that N.
84. 84
4 ) a number of specimens of a given
material are tested for different values of
W
5) Prepare a plot of σa vs. N
88. 88
Fatigue Strength = σfs
σfs ≡ purely alternating stress σa
that produces fracture in a
given number of cycles N
89. 89
Fatigue life
≡ The number of cycles N that
produces failure at a given
stress level σfs
90. 90
N
σa
σe
Typically in machine design 100x106 or 10x106
10x109 for wind turbines
σe ≡ value of σa where the σa vs. N curve is
horizontal . If not horizontal, σa value at
N = 1010 cycles
100. 100
The text takes σfs ≈ σe and SF or safety
factor ≈ 1. The reasons are:
1) For wind turbines N ≥10x109 cycles for
continuous operation. For such a high
value of N, σfs → σe
2) For wind turbines SF varies from 1 to 1.1
( SF is partially accounted for by taking
σfs ≈ σe )
101. 101
Consider Eq. (6.6) on p. 261:
1
)
44
.
2
.
4
.........(
..........
..........
1
=
+
Þ
-
=
u
m
e
a
u
m
a
e
s
s
s
s
s
s
s
s
This agrees with conventional Goodman design rule for
σe = σfs and SF =1
106. 106
Fatigue analysis (Goodman’s rule)
addresses σa and σm . But in
actuality combined loading gives 6
stresses (3-D) or 3 principal
stresses.
(Note: stresses are always 3-D)
108. 108
von Mises effective stress σe :
( ) ( ) ( )
( )
( ) ( ) ( )2
1
3
2
3
2
2
2
1
2
zx
2
yz
2
xy
2
x
z
2
z
y
2
y
x
2
e
6
2
s
-
s
+
s
-
s
+
s
-
s
=
t
+
t
+
t
+
s
-
s
+
s
-
s
+
s
-
s
=
s
110. 110
( ) ( ) ( )
( )
( ) ( ) ( )
( )
2
2
2
2
2
2
2
2
2
2
2
2
2
2
6
2
,
6
2
zxm
yzm
xym
xm
zm
zm
ym
ym
xm
em
zxa
yza
xya
xa
za
za
ya
ya
xa
ea
and
t
t
t
s
s
s
s
s
s
s
t
t
t
s
s
s
s
s
s
s
+
+
+
-
+
-
+
-
=
+
+
+
-
+
-
+
-
=
112. 112
( ) ( ) ( )
( ) ( ) ( )2
1
3
2
3
2
2
2
1
2
2
1
3
2
3
2
2
2
1
2
2
2
m
m
m
m
m
m
em
a
a
a
a
a
a
ea
and
s
s
s
s
s
s
s
s
s
s
s
s
s
s
-
+
-
+
-
=
-
+
-
+
-
=
115. 115
Cumulative Damage and Miner’s Rule:
A component may experience multiple load cycles of
different amplitudes, as illustrated in Fig. 6.6, p. 262.
0 0.5 1 1.5 2 2.5 3
-3
-2
-1
0
1
2
3
Time
Stress,
arbitrary
units
119. 119
Equation (6.8), p.262, is based on the
assumptions:
1. fatigue damage at any stress level is
directly proportional to the number of
cycles at that level;
2. the ordering of stress level is of no
consequence.
120. 120
Summary of Miner’s rule (hypothesis) for
cumulative damage; that is, eq. (6.8),
p.262:
1.A widely used hypothesis (not derivable)
2.Simple
3.generally, accuracy is adequate
4.No better alternatives as a design tool.
122. 122
0 0.5 1 1.5 2 2.5 3
-3
-2
-1
0
1
2
3
Time
Stress,
arbitrary
units
σ2
σ3
σ1
8667
.
0
15
4
10
4
20
4
15
4
10
4
failure
to
cycles
of
no.
applied
cycles
of
no.
20
4
3
2
1
=
+
+
=
=
=
=
=
D
and s
s
s
Text, p. 262
123. 123
Let us answer the following question about
cumulative damage for the material
depicted by the uppermost curve in the
next figure σm = 0.
a) If σa = 500 MPa is applied for 8000
cycles, for how many additional cycles can
σa = 400 MPa be applied?
