Wind Energy Systems MEMS 5705
Spring 2017
Lecture 11, Feb. 22
Lecture 12, Feb. 27
1
L11 and L12
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
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
3
1. Key points of L9 & L10 (Feb. 15 and 20)
4
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
Ct = 4a (1 - a)......... .......... ..(3.2.17),P.87
Cp = 4a(1 - a)2 .......... .......... (3.2.14),P.87
For a ≥ 0.4, momentum theory is considered
not valid
Ct = 4 (1- a) + 2.......... .(empirical, L9 + L10)
5
Airfoil Basics
Pitching moment = M
Air flow
α
Drag force = D
Chord
c/4
Drag and lift forces and pitching moment on stationary airfoil:
α : angle of attack ; c: chord (p. 97)
6
L = Lift force over the reference area c.l
æ1
2ö
= ç rU ÷(c.l ) Cl
è2
ø
Cl = lift coefficient
=
L
æ1
2ö
ç rU ÷(c.l )
è2
ø
Ll
Ll
=
=
æ1
öæ 1 ö æ 1
2
2 ö
r
U
c
.
l
r
U
c÷
ç
÷ç ÷ ç
è2
øè l ø è 2
ø
Lift force per unit length
=
Dynamic force per unit length
7
Drag force = D
æ1
2ö
= ç rU ÷(Ref. area)(Cd )
è2
ø
Drag coefficient:
Cd =
D
l
1
rU 2c
2
=
….(3.43) p. 104
Drag force per unit length
Dynamic force per unit length
Pitching moment = M
æ1
ö
= ç rU 2 ÷(Ref. area)(c)(Cm )
è2
ø
Pitching moment coefficient:
Cm =
M
1
rU 2 .(c.l )c
2
=
Pitching moment
Dynamic moment
….(3.44) p. 104
8
μ = coefficient of viscosity
m
= kinematic viscosity
u=
r
rUc Uc
Re = Reynolds number =
=
m
u
….(3.41)
p. 103
ì cl ü
ï ï æ rUc ö æ Uc ö
÷÷ = f ç
÷
í cd ý » f çç
ïc ï è m ø è u ø
î mþ
(* we take chord length c as reference length)
9
Momentum Theory with Wake Rotation (#3.3, pp. 96-101)
Stream tube model of flow behind rotating wind blade. Picture of stream
tube with wake rotation.
10
Momentum Theory with Wake Rotation
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)
11
a'r Ω
Wind Direction
f
f aU
For future reference:
a' rW
tan f =
aU
U (1 - a )
=
Wr (1 + a' )
U
f
U(1-a)
rΩ
Plane of Rotation
Flow velocity diagram at an annulus in a HAWT rotor disk
12
tan f =
a' r W
U (1 - a )
=
aU
Wr (1 + a' )
2
a(1 - a )
æ Wr ö
=
ç
÷
U
a' (1 + a' )
è
ø
Wr
= lr = local speed ratio
2nd Edtion (3.26-27) p. 98
U
Wr r
r
=
× = l ........ .......... .......... ...(3.3.8) , p. 91
U R
R
WR
l=
= tip speed ratio..... .......... .......(2. 5.4), p. 60
U
2nd Edition (2.76) p. 63
blade tip speed
=
wind speed
2nd Edition (3.25) p. 98
lr =
2
a(1 - a )
.......... ...... .......... .......... ..(3.3.6), p. 91
a' (1 + a' )
We will further verify this geometrical derivation analytically.
13
2nd Edition (3.32-33) pp. 98-99
dP
dC p =
.......... .......... .......... ........( 3.3.13), p. 92
1
rAU 3
2
l
8
C p = 2 ò a' (1 - a )l3r dlr .......... .......... ..(3.3.14), p. 92
l
0
14
2. Momentum Theory and Blade
Element Theory (# 3.6, pp. 117-121)
15
2.1 Momentum Theory (L9 + L10)
2nd Edition (3.58-59) p. 118
1 2
dT = 4a(1 - a) rU (2p r dr ).......... .....( 3.3.5),p. 91
2
= rU2 4a(1 - a)(p r dr ).......... .......... (3.5.1),p. 106
1
dQ = 4a' (1 - a ) r U Wr 2 (2p rdr )......... ...(3.3.10), p. 91
2
= 4a' (1 - a )rUr 3 (p W dr )......... .......... .(3.5.2), p. 106
dL = 4b[1-b*cos(φ)](1/2)ρUΩr2(2πrdr)
= 4b[1-a]ρUr3(πΩdr) . . . . . . . . . . Peters
16
2.2 Blade element theory
17
18
Blade Geometry for HAWT Aerodynamic Analysis
A typical blade element of length dr and width c
Plane of Rotation
Chord line
Ф
dFT
Ω r ( 1+ a')
Ur
dL = dFL
α
θ (class) = θP (text)
U(1-a)
= 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
Ω
dFN
dD= dFD
19
Plane of Rotation
Chord line
Ф
U(1-a)
a'Ω r
Ωr
α
FT
FTorque
Ω
dL = dFL
Ur
FN = Ftransit
Ω
U(1-a)
D
20
Blade Geometry for HAWT Aerodynamic Analysis
Plane of Rotation
Ф
Chord line
dFT
Ur
dL = dFL
α
Ω r ( 1+ a')
Ω
U(1-a)
dD= dFD
dFN
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)
21
1 2 ü nd
CL = L rUr A ï 2 Ed. p. 104
2
Old
ï
1 2 ï
Cd = D rUr A ý Text. p. 97
2
ï
1 2 ï
Cm = M rUr A.c
ïþ
2
22
2nd Edition
(3.63) p. 120
(3.26-27) p. 98
Let c(Δr) = dA
B = Number of blades
Ur = U(1 - a) sin f
= W r (1 + a' ) cos f
U(1 - a)
(1 - a)
tan f =
=
.......... ........(3.5.6), p.109
W r (1 + a' ) (1 + a' )l r
Wr
r
lr =
= l. = local speed ratio..... .......(3. 3.8), p. 91
U
R
WR
l=
= tip speed ratio..... .......... .......... .(3.3.7), p. 91
U
23
Incremental lift force: (3.55-56) p. 120
2
æ1
ö
dFL = Cl ç rU r ÷ .(c.dr ).......... .......... .(3.5.8), p. 109
è2
ø
Incrementa l drag force
2
æ1
ö
dFD = Cd ç rU r ÷ .(c.dr ).......... .......... .(3.5.9), p. 109
è2
ø
24
Force FT in the plane of blade rotation
generates useful torque: Q
dFT = dFL sin f - dFD cos f.......... .......... ..(3.5.11),p. 109
The component dFNcontributes to thrust :
dFN = dFL cos f + dFD sin f.......... .......... ..(3.5.10), p. 109
(FN exerts a thrust load on the rotor)
2nd Edition (3.67-68) p. 120
25
We consider a rotor with B blades
The elemental normal force (thrust) on the
section at a distance r from the center:
æ1
ö
dFN = B.ç r Ur2 ÷(CL cos f + CD sinf ) c dr .......... ....( 3.5.12),p. 109
è2
ø
Similarly, the elemental torque due to the
tangential force dFT operating at a distance r
from the center :
dQ = B. r . dFT .......... .......... .......... .(3.5.13),p.110
2nd Edition (3.69-70) p. 120-121
26
2nd Edition (3.71) p. 121
dQ = elemental torque
æ1
2ö
= B.ç r U r ÷(Cl sin f - Cd cos f )c r dr.......... .(3.5.14), p.110
è2
ø
DP = WDQ = elemental power
27
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
æ1 2ö
dFN = B.ç r Ur ÷(Cl cos f + Cd sin f) c dr.......... ....(3.5.12),p. 109
è2
ø
and one equation for dFT that gives the torque dQ:
æ1 2ö
dQ = B.ç r Ur ÷(Cl sin f - Cd cos f)c r dr.......... ..(3.5.14),p. 110
è2
ø
28
2.3 Blade shape for ideal rotor
without wake rotation (pp. 121-123)
29
Ideal Rotor (p.121)
1.
2.
3.
4.
Axial induction factor a = 1/3 for each
annular stream tube as in a Betz rotor
No wake rotation (angular induction
factor a' =0)
no losses from a finite number of blades
No drag, Cd = 0
30
2nd Edition (3.72) 121, (3.74) p. 122, 124
Recall from momentum theory
dT =
1
rU 2 4a(1 - a )(2p r dr ).......... .......... ........( 3.5.1), p. 106
2
æ 1 æ 1 öö
= rU 2 4çç ç1 - ÷ ÷÷ p r dr .......... .......... ....( 3.6.1), p. 110
è 3 è 3 øø
8
= rU 2 p r dr
9
From blade element theory :
æ1
2ö
dFN = B.ç r U r ÷(CL cos f + CD sin f ) c dr .......... ....( 3.5.12),p. 109
è2
ø
1
2
= B. r U r (CL cos f + 0) c dr .......... .......... .(3.6.2),p. 111
2
U (1 - a ) 2 U
Ur =
=
.......... .......... .......... .......( 3.6.3), p. 111
sin f
3 sin f
31
2nd Edition (3.63) p. 120, (3.76) p. 122
U(1 - a)
(1 - a)
tan f =
=
.......... .......... ....(3.5.6), p. 109
W r (1 + a' ) (1 + a' )l r
1
(1 - )
2
3
tan f =
=
.......... .......... .......... ....( 3.6.5), p. 111
(1 + 0)l r 3l r
32
2
æ 2 ö
U÷
ç
1
dFN = B. rç 3 ÷ (Cl cos f) c dr
2 ç sin f ÷
ç
÷
è
ø
(dFN ) from b.e. theory
= (dT) from momentum theory
æ8ö
= rU ç ÷ p r dr
è9ø
2nd Edition (3.77) p. 122
ClBc
Þ
= tan f sin f.......... ....( 3.6.4), p. 111
4p r
2
33
Equation to find optimum chord. (3.75, 3.79) p. 122
Cl Bc
= tan f sin f ............( 3.6.4),p. 111
4p r
æ2 1 ö
= ç . ÷ sin f .
tan(ϕ) = 2/(3λr)
è 3 lr ø
8p r sin f
c=
.................( 3.6.8),p. 111
3B Cl lr
Alternative Method: μ = λr
c = [4πr/BCl]*[(3b-1)/b] =
[16πr/BCl]*[b2/(1 + λr2)]
34
Example (p.122) and table 3.2 (p. 123)
λ = 7 , R = 5 m, Cl = 1
æ Cd ö
çç ÷÷ at a = 7o , B = 3
è Cl ømin
Take a = 7 and, for illustration, consider
r
= 0 .8
R
o
35
Book method to get inflow angle.
r/R
c
0.8
0.236
f
6.8 o
θ
a =f
-0.2 o
-θ
=7
o
7º
æ 2 ö
2
ö
-1æ
÷÷ = tan ç
f = tan çç
÷.......... ..(3.6.7),p. 111
è 3 x 7 x 0 .8 ø
è 3l(r R )ø
2
ö
-1æ
o
= tan ç
÷ = 6 .8
è 3 x 7 x 0 .8 ø
-1
2nd Edition (3.76), p. 122
36
Now try Peters’ New Method for angle.
λ = 7 ; λr = 7(0.8) = 5.6
Φ = π/3 –(1/3)*arccos[(1-5.62)/(1+5.62)]
= 60º - (1/3)(180º- 20.25º) = 6.8º
α = 7º = 6.8º - Θ;
Θ = -0.2º
37
Book method for chord.
8p r sin f
c=
3B c l lr
8p (0.8 R ) sin 6.8o
=
3(3) (1)(7 ´ 0.8)
8p (0.8 ´ 5) sin 6.8o
=
= 0.236
3(3) (1)(7 ´ 0.8)
c = 0.236
m
38
PETERS’ METHOD FOR CHORD
b = 1/[1+2*cos(6.8º)] = .335
c = 16π(.8)(5)(.335)2/[(3)(1)(1+5.62)]
c = 0.2323 m
39
1.4
Note: Twist angle θt = θp – θp, o
where θp, o = pitch at tip, r = R.
at r/R = 0.8, θt = θp – θp, o = -0.2º – (-1.6) º
= 1.4º
40
[Figure 3.25. p. 123]
41
Figure 3.26, p. 123
, p. 113
42
3.5 page 623
(not clear to us,
hence not done)
43
We define θP,0 as pitch at tip.
44
45
46
2.4 Strip Theory or BEM Theory
(Blade Element + Momentum Theory)
(pp. 124-127)
47
Momentum Theory :
2nd Edition pp. 124, 125
1 2
dT = 4a(1 - a) r U 2 p r dr.......... .......... .....( 3.5.1), p. 106
2
1
dT = 4a' (1 + a' ) r W 2 r 2 2 p r dr.......... .........( 3.3.4), p. 90
2
1
dQ = 4a' (1 - a) r U W r 2 2 p r dr.......... .........( 3.5.2), p. 106
2
Blade element theory :
1
dFN = dT = (Cl cos f + Cd sin f ) r U r2Bc dr ...(3.5.12), p. 109
2
1
dQ = (Cl sin f - Cd cos f ) r U r2Bc r dr .......... .(3.5.14), p. 110
2
48
To combine the momentum theory with the
blade element theory, we have to express
Ur = V in terms of U and Ω r.
49
Plane of Rotation
Chord line
Ф
U(1-a)
a'Ω r
Ωr
α
FT
FTorque
Ω
dL = dFL
Ur
FN = thrust
Ω
U(1-a)
D
50
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.
51
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.
52
Ф
Ur
W r(1 + a' )
U(1-a)
U (1 - a )
.......... .......... ....( 3.5.7), p. 109
sin f
W r (1 + a' )
Ur =
cos f
Ur =
2nd Edition (3.64) p. 120
53
Thus,
U (1 - a ) W r (1 + a' )
Ur =
=
.......... ........( A)
sin f
cos f
U (1 - a )
(1 - a ) 1
and also : tan f =
=
× .......... ....( B )
W r (1 + a' ) (1 + a' ) lr
Define local solidity ratio s ' :
Bc
s'=
.......... .......... ......( 3.7.3), p.114
2p r
2nd Edition (3.82) p. 125
54
Now, we express the blade element theory as
follows (after some algebra):
ö1
s ' Cl cos f æ Cd
2
ç
÷
dT = (1 - a )
1
+
tan
f
r
U
2 p r dr .......(C )
2
ç
÷
sin f è Cl
ø2
2
dQ = (1 + a' )
s ' Cl sin f æ Cd 1 ö 1
2 2
ç
÷
1
r
W
r r 2 p r dr ....( D )
2
ç
÷
cos f è Cl tan f ø 2
Combining Eq. (c), Eq. (3.5.1) we get
ö
4a
cos f æ Cd
= s ¢Cl . 2 çç1 +
tan f ÷÷.......... .......... .(E )
1+ a
sin f è Cl
ø
Combining Eq. (D) and 2nd eq. p. 124 and Eq. (B), we get
4a'
s ¢Cl æ Cd 1 ö
çç1 ÷÷.......... .......... .......( F )
=
1 + a' cos f è Cl tan f ø
55
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
4a
cos f
= s' Cl
1- a
sin2 f
Þ a= 1
....( 3.7.9), p.115
2
1 + 4 sin f (s' Cl cos f)
[
]
4a'
s' Cl
=
1 + a' cos f
Þ a' = 1
.......... ..(3.7.10 ), p.115
[4 cos f (s' Cl )- 1]
56
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
dP = WdQ.......... .......... ..(3.7.13), p.117
R
R
rh
rh
Total power P = ò dP = ò W dQ
where rh is the rotor radius at the hub
57
Power Coefficient Cp:
nd
2 Edition (3.94) p. 127
R
ò W dQ
P
P
r
Cp =
=
= h
.....( 3.7.15), p. 117
Pwind Pdynamic 1 r p R 2U 3
2
dQ = ?
58
We already have an expression for dQ from
the blade element theory: (3.95) p. 127
U 2 (1 - a)2
2
(
)
dQ = s ' p r
C
sin
f
C
cos
f
r
dr ....( 3.7.2.), p.114
l
d
2
sin f
Also, we wish to replace dr by dλr by using our earlier
definition of local speed ratio λr :
Wr
r
lr =
= l .......... .......... ....( 3.3.8), p.91
U
R
R
dr = × dlr
l
59
We make use of these two equations for dQ
and λr, and express CP (after some algebra)
as follows:
l
æ 1 ö é æ Cd ö
ù 2
÷÷ ê1 - ç
Cp = 2 ò s ' Cl (1 - a) çç
cot
f
lr dlr ....(3.7.16),p. 117
÷
ú
C
l lh
l ø
û
è sinf ø ë è
2
2
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:
60
That is from p. 125,
a
1
= s ' Cl cos f ×
.......... .(3.7.5), p.115
2
1- a
4 sin f
4 a sin f
Þ
= s ' Cl (1 - a )......... .....( 3.7.7), p.115
cos f
and
2
a
lr
=
.......... .......... ........( 3.7.8), p.115
a' tan f
Þ a tan f = a' lr .......... .......... .......( 3.7.18), p.117
61
Substituting these two equations into integral for
Cp: see bottom of page 127
l
ù 2
2
1 é æ Cd ö
2
÷÷ cot fú l r dl r
Cp = 2 ò s' Cl (1 - a)
ê1 - çç
l lh
sin f ë è Cl ø
û
2
= 2
l
ù 2
4 a sin2 f (1 - a) é æ Cd ö
òl cos f sin f ê1 - ççè Cl ÷÷ø cot fú lr dlr
ë
û
h
l
l
é æ Cd ö
ù 2
8
= 2 ò a tan f (1 - a)ê1 - çç ÷÷ cot fú l r dl r
l lh
ë è Cl ø
û
l
é æ Cd ö
ù 2
8
= 2 ò a' l r (1 - a)ê1 - çç ÷÷ cot fú l r dl r
l lh
ë è Cl ø
û
l
é æ Cd ö
ù
8
3
÷÷ cot fú dl r .....( 3.7.11), p. 117
= 2 ò l r a' (1 - a)ê1 - çç
l lh
ë è Cl ø
û
62
It is instructive to set Cd = 0 in previous eq.
l
8
3
Cp = 2 ò a' (1 - a)lr dlr .......... .......( 3.3.14), p. 92
l lh
This is Cp based on momentum theory including
wake rotation.
63
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.
64
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
65
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.
66
In this lecture we address a very
important topic: Fatigue
For many components of WT, fatigue is the
design driver.
67
Fatigue: Chapter 6
(# 6.1 and 6.2, pp.257-266)
68
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.”
69
“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.”
70
See also Fig. 6.1 page 259.
(Wind Turbine Technology, David Spera (Ed.), ASME publication, 1994, p. 549)
71
Typical pattern of design drivers for the primary structural and mechanical
components in a HAVVT. (ASME publication, ibid)
72
Let
nL= the total number of blade revolutions
over the turbine’s life time
Since the number of many of the fatigueproducing stress cycles is proportional to nL,
it is instructive to evaluate nL in a typical
scenario:
73
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 20year lifetime. How?
74
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
75
Ω = 30 to 70 rpm ≈ 50 rpm
Hop= 4000 hours/year
Y = 20 years
nL = 60(1)(50)(4000)(20)
= (240)106) cycles
76
S-N Curve (p. 259)
S ≡ stress
N ≡ number of cycles to failure
This is evaluated by conducting a rotating
beam test
77
78
w/2
w/2
w/2
Specimen
L1
wL1/2
w/2
B.M
79
32M
s max, min = ±
p d3
32(WL1 2)
=±
p d3
Typically
d = 0.3 in
80
one cycle
s max
Time
s min
(Purely) alternating stress, that is, the mean stress is zero
and the stresses are fully reversing = σa
81
Rotating Shaft under constant B.M
A
Ω
A
Point A experiences alternating stresses
82
Rotating Beam Test
1)
2)
3)
Loaded with known weight W
Cycled until fracture
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.
83
4 ) a number of specimens of a given
material are tested for different values of
W
5) Prepare a plot of σa vs. N
84
(p. 259)
σe=endurance ‘strength’ (details to follow)
85
86
Fatigue life for purely alternating
stress σa
87
Fatigue Strength = σfs
σfs ≡ purely alternating stress σa
that produces fracture in a
given number of cycles N
88
Fatigue life
≡ The number of cycles N that
produces failure at a given
stress level σfs
89
σa
Typically in machine design 100x106 or 10x106
10x109 for wind turbines
σe
N
σe ≡ value of σa where the σa vs. N curve is
horizontal . If not horizontal, σa value at
N = 1010 cycles
90
σa
σfs
N= 10x109 cycles
σe
N
N
Fatigue life N at σa= σfs
91
σa (purely alternating)
σfs
N= 10x109 cycles
σe
N
N
(N>= 10x109 cycles)=continuous operation
92
Fatigue Formula or
Goodman’s Rule
(Text, p. 260)
93
Typically, stresses do not have a zero
mean.
 σmax , σmin , σalternating , and σmean
What are these?
94
Completely Reversed Stress
σa = σalternating
σa
s max
s min
0
N
s max = s min
s average = s mean = 0 = s m
95
σ
σa = σalternating
s mean = s m
s max
σa
s min
0
N
s m = (s max + s min ) 2
s a = (s max - s min ) 2
96
σ = cyclic stress
σ = σm + σa
σe= σa (for continuous operation, N = 1010
cycles)
σfs= σa (N = Nstipulated)
σm = σaverage = σmean
97
σa
σfs
Goodman
sa
Line
s fs
sm
+
=1
su
s fs
σa
s max
sm
SF
σa
0
Test data
sa sm
1
+
=
s fs s u SF
su
SF
σu
σm
98
Goodman Design Rule:
sa sm
1
+
=
s fs s u SF
99
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 )
100
Consider Eq. (6.6) on p. 261:
sa
se =
.......... .......... .........( 4.2.44)
sm
1su
sa sm
Þ
+
=1
se su
This agrees with conventional Goodman design rule for
σe = σfs and SF =1
101
Combined Loading
and
Goodman’s Rule
102
Wind turbines experience
combined fluctuating bending,
torsion, and axial loading.
103
For example,
s x => s xa + s xm
s y => s ya + s ym
t xy => t xya + t xym
104
Combined Loading
és x t xy t xz
ê
ê t yx s y t yz
êt t s
ë zx zy z
ù
ú
ú º 6 s' s
ú
û
or equivalently,
és1 0 0
ê
ê 0 s2 0
êë 0 0 s3
ù
ú
ú º Three principal stresses : s1 ³ s 2 ³ s3
úû
105
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)
106
Calculate mean and
alternating stresses w.r.t the
von Mises effective stress: σe
and then calculate :
σea and σem
107
von Mises effective stress σe :
2s = (s x - s y ) + (s y - s z ) + (s z - s x )
2
2
e
(
2
+6 t +t +t
2
xy
2
yz
2
zx
2
)
= (s1 - s 2 ) + (s 2 - s3 ) + (s3 - s1 )
2
2
2
108
s x => s xa + s xm
s y => s ya + s ym
s z => s za + s zm
t xy =>t xya + t xym
t yz =>t yza + t yzm
t zx =>t zxa + t zxm
s ea + s em
109
2s
2
ea
= (s xa - s ya ) + (s ya - s za ) + (s za - s xa )
2
(
+6t
and ,
2s
2
em
2
xya
+t
2
2
yza
+t
2
zxa
2
)
= (s xm - s ym ) + (s ym - s zm ) + (s zm - s xm )
2
(
+6t
2
xym
+t
2
yzm
2
+t
2
zxm
2
)
110
or equivalently,
s 1 => s 1a + s 1m
s 2 => s 2 a + s 2 m
s 3 => s 3a + s 3m
s ea
s em
111
2s
2
ea
= (s 1a - s 2a ) + (s 2a - s 3 a ) + (s 3a - s 1a )
2
2
2
and
2s
2
em
= (s 1m - s 2 m ) + (s 2 m - s 3 m ) + (s 3 m - s 1m )
2
2
2
112
Goodman rule
sa sm
1
+
=
s fs s u SF
changes to
s ea s em
1
+
=
s fs s u SF
113
Cumulative Damage
and
(Palmgren)–Miner Rule
(Text: p. 262)
114
Cumulative Damage and Miner’s Rule:
A component may experience multiple load cycles of
different amplitudes, as illustrated in Fig. 6.6, p. 262.
Stress, arbitrary units
3
2
1
0
-1
-2
-3
0
0.5
1
1.5
2
2.5
3
Time
115
Idealized types of fatigue cycles
116
2nd Edtion (6.7) p. 262
117
The
M
i=1
2nd Edition (6.8) p. 262
118
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.
119
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.
120
Illustrating the application of cumulative
damage rule, Eq. (6.8), p. 262.
121
Stress, arbitrary units
3
2
1
0
-1
σ1
-2
-3
0
0.5
1
1.5
2
2.5
3
Time
σ2
4
no. of cycles applied
=
20 no. of cycles to failure
4
4
s2 =
and
s3 =
10
15
4
4
4
D=
+
+
= 0.8667
20 10 15
σ3
s1 =
Text, p. 262
122
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?
123
Aluminium
8000
n2
D=
+
=1
4
4
1.4 ´ 10
6 ´ 10
n2 = 25,700
(Overly approximate; explains the method)
124