Introduction – Equations of motion G. Dimitriadis 08

Aeroelasticity
Lecture 8:
Supersonic Aeroelasticity
G. Dimitriadis
Introduction to Aeroelasticity


Introduction
A  All the material presented up to now
concerned incompressible flow.
A  In this lecture we will present a simple
treatment of 2D supersonic flow
aeroelasticity.
A  The discussion will concern the same
pitch-plunge airfoil treated in the 2D
incompressible case.
Introduction to Aeroelasticity


Pitch-Plunge airfoil
A  Flat plate airfoil
with pitch and
plunge degrees
of freedom.
A  Equations of
motion:
# m S & ) h

, # K h
%
(* - + %
$ S Iα ' +α

. $ 0

Introduction to Aeroelasticity

0 & ) h , ) −l(t) ,
( * - = *m (t)Kα ' +α . + xf .


Supersonic flow
A  In order to complete the model, we need to derive
expressions for the lift and moment around the flexural
axis, in the presence of a supersonic free stream.
A  The supersonic free stream is defined by:
– 
– 
– 
– 
– 

Airspeed U∞,
Pressure P∞,
Temperature T∞,
Density ρ∞,
Speed of sound a∞,

A  Furthermore, the air is described by the ratio of
specific heats at constant pressure and constant
volume, i.e.
cp
γ=
= 1.4
cv

Introduction to Aeroelasticity


Steady Potential equation
A  For supersonic flow, the steady potential
equation, in terms of perturbation
potential, is given by
2
2
∂
φ
∂
φ
2
+
1−
M
( ∞ ) ∂x 2 ∂y 2 = 0

∂φ
∂φ
= u,
=v
∂x
∂y

A  Where
A  And u, and v are small local velocity
perturbations from the free stream

Introduction to Aeroelasticity


Linearized Small Disturbance Equation
A  For unsteady flows, the potential equation includes
unsteady terms.
A  The Linerized Small Disturbance Equation is given
by:
2
2
2
2

(1 − M ∞2 )

M∞ ∂ φ
1 ∂φ
∂φ ∂φ
+
−
2
−
2
2
2 =0
2
a∞ ∂x∂t a∞ ∂t
∂x ∂y

A  Where, again, the potential represents a small

perturbation.
A  This equation is difficult to solve. As a first
approximation, a 1D method known as piston
theory can be used.
Introduction to Aeroelasticity


Piston theory
A  The aerodynamics of the moving wing are
calculated with the piston theory assumption:

–  Flow disturbances spread in a direction normal to the
wing’s surface. The wing’s movement is equivalent to
the movement of a piston in a column of air.
–  All disturbances are isentropic

A  Under this assumption, the pressure on the
surface of the wing is given by 2γ
% γ − 1 w(x,t) ( γ −1
p( x,t ) = p∞ '1 +
*
2
a∞ )
&

A  Where w(x,t) is the downwash velocity
Introduction to Aeroelasticity


Downwash

A  The downwash velocity of the wing is
given by
%'− U α ( t ) + h
( t ) + x − x α
( t )
∞
f
w ( x,t ) = &
'( U ∞α ( t ) + h
( t ) + x − x f α
( t )

(

(
(

) )
)














A  As usual, the pressure difference is
calculated from Δp = pl − pu
A  But the 2γ/γ-1 exponent makes this

calculation difficult.
Introduction to Aeroelasticity


Binomial series
A  The binomial series is a special case of a
Taylor series.
A  For |x|<1:

(1 + x )
A  where

a

" a% n
= ∑$ ' x
n =0 # n&
∞

n
" a%
a − k + 1 a( a − 1) ( a − n + 1)
=
$ '=∏
n!
k
# n& k =1

Introduction to Aeroelasticity

Binomial expansion
A  Assume that the downwash velocity is much smaller
than the speed of sound, we can use a binomial series
on the pressure equation:
2γ
γ −1

A 

# γ −1 w(x, t) &
p ( x, t ) = p∞ %1+
(
2
a∞ '
$
2
3
#
&
#
&
#
&
γ
γ
+1
γ
γ
+1

w
w
w
(
)
(
)
2
3
≈ p∞ %1+ γ λ +
% ( λ +
% ( λ ((
%
a∞
4 $ a∞ '
12 $ a∞ ' '
$
M∞
λ
=
Where
is a correction factor
2
M∞ − 1

A  Retaining only the linear# term leads
to
&
w
p ( x, t ) ≈ p∞ %1+ γ λ (

a∞ '
$

Introduction to Aeroelasticity


Pressure difference
A  Then the pressure difference is given by
)
)
U ∞α + h
+ x − x f α
,
U ∞α + h
+ x − x f α
,
Δp = pl − pu ≈ p∞ +1 + γ
λ. − p∞ +1− γ
λ.
a∞
a∞
*
*
U ∞α + h
+ x − x f α

A  So that Δp ≈ 2 p∞γ
λ
a∞

(

)

(

(

)

A  Then the total lift acting on the airfoil is
given by:
2 p∞γλ c
U ∞α + h
+ x − x f α
dx
l = ∫ Δpdx ≈
∫
0
a∞ 0
c

Introduction to Aeroelasticity

(

(

))

)


Lift force
A  So that the lift force becomes
2 p∞γλc *
*c
, 
,



l = ∫ Δpdx ≈
. U ∞α + h + + − x f - α /
0
2
a∞ +
c

A  Noting that the the speed of sound is:
p∞
a∞ = γ
ρ∞

A  We finally obtain

2 ρ∞U ∞ λc '
'c
) 
)


l=
+ U ∞α + h + ( − x f * α ,
*
2
M∞ (
Introduction to Aeroelasticity


Moment around flexural axis
A  The moment around the flexural axis is
given by

2 p γλ
mxf = − ∫ Δp(x − x f )dx ≈ − ∞
0
a∞
c

c

∫ (U α + h + ( x − x )α ) (x − x
0

∞

f

f

)dx

A  Which leads to:
&
&
#c
& 1 2
2 ρ∞U∞ λ c # # c
2
mxf = −
%U∞ % − x f (α + % − x f ( h + ( c − 3cx f + 3x f ) α (
'
$2

' 3
M∞ $ $ 2
'

Introduction to Aeroelasticity


Lift and moment
A  Remembering from lecture 1 that
m 2
#c
%
S = m − x f , Iα =
c − 3cx f + 3x 2f
$2
&
3

(

)

A  We can simplify the lift and moment
expressions such that:
2 ρ∞U∞ λ c "
S %

l=
$U∞α + h + α '
m &

M∞ #
2 ρ∞U∞ λ c " S
S  Iα %
mxf = −
$U∞ α + h + α '
m
M∞ # m
m &

Introduction to Aeroelasticity


Equations of motion
A  Substituting the lift and moment
expressions into the aeroelastic equations
of motion gives:
! m S
#
# S Iα
"

$ ')
&(
&)
%*

h
α

+) ! K h

,+#
)- #" 0

'
!
S $

)
− #U∞α + h + α &
$
0 ' h + 2 ρ∞U∞ λ c )
"
m %
&(
(
,=
&
M∞ ) ! S
Kα % * α S  Iα $
− U
α + h + α &
) #" ∞ m
m
m %
*

A  i.e. the complete supersonic aeroelastic
model
Introduction to Aeroelasticity

+
)
)
,
)
)
-


In matrix form
A  In matrix form the equations of motion can
be written as:
! m S
#
# S Iα
"

$ ')
&(
&)
%*

h
α

+) 2 ρ U λ c ! 1
S/m
∞ ∞
#
,+

M ∞ #" S / m Iα / m
)-

! 0 U
$
∞
& ' 0 +
2 ρ∞U∞ λ c #
+
,
#
S &=(
M ∞ # 0 U∞
* 0 &
m %
"

$ ')
&(
&)
%*

h
α

+) ! K h
,+#
)- #" 0

0 $' h +

&(
,
&
Kα % * α -

A  These are quasi-steady, small disturbance
equations. They are valid for M∞>1.2.

Introduction to Aeroelasticity


Solution
A  The equations of motion are 2nd order linear
ODEs and can be solved as usual.
A  At each value of the Mach number and
airspeed, the eigenvalues, χi, i=1,…,4, can be
evaluated.
A  From the eigenvalues, natural frequencies,
ωi, and damping ratios, ζi, can be calculated.
A  The natural frequency and damping ratio
variation with airspeed can be plotted for
each Mach number.
Introduction to Aeroelasticity


Example 1
Eigensolution for a free
stream Mach number of
1.5 . The natural
frequencies and

damping are plotted for
all airspeeds between 0
and 700m/s.
There is a flutter point at
approximately 607m/s.
Is this a physically
possible flutter speed?

Introduction to Aeroelasticity


Unmatched flutter speeds
A  The flutter speed calculated in this example may or may
not be physical, it depends on the system’s flight
condition.
A  Consider the case where the wing is flying at sea level and
the atmospheric pressure is 1bar:
–  ρ∞=1.225kg/m3
–  p∞=101325Pa

A  Then the speed of sound is 340m/s. Therefore, the flutter
speed of 607m/s corresponds to a Mach number of 1.8.
A  But the Mach number used for the simulation is 1.5. This
case is an example of an unmatched flutter speed. The
system can flutter but not at an attainable Mach number.
A  As the flutter Mach number is higher than the simulation
Mach number, this is a safe flight condition.
Introduction to Aeroelasticity

Example 2
The free stream Mach
number is still 1.5 and
the flight condition is
the same. However,
the spring stiffnesses
have been decreased
so that the flutter
speed is now 511m/s.

Introduction to Aeroelasticity


Matched flutter speed
A  The speed of sound is still 340m/s.
However, the new flutter speed is 511m/s.
A  This flutter speed occurs at a Mach
number of 1.5, the same as the simulation
Mach number.
A  This is an example of a matched flutter
speed: flutter occurs at the simulation
Mach number.
A  Clearly the flight condition is not safe.
Introduction to Aeroelasticity


Example 3
The free stream Mach
number is still 1.5 and
the flight condition is

the usual. The spring
stiffnesses have been
further decreased so
that the flutter speed
is now 420m/s.
This corresponds to a
Mach number of 1.2.
The flight condition is
very unsafe.

Introduction to Aeroelasticity