AAE556 lecture 05 MDOF response

AAE 556
Aeroelasticity
Lecture 5 –
1) Compressibility;
2) Multi-DOF systems
Reading: Sections 2-13 to 2-15

5-1
Purdue Aeroelasticity


Homework for Monday?
i

Prob. 2.1

–
–
i

Uncambered (symmetrical sections) MAC = 0
Lift acts at aero center (AC) a distance e ahead to the shear center

Problem 2.3 – wait to hand in next Friday

5-2
Purdue Aeroelasticity


Aeroelasticity matters
Reflections on the feedback process

5-3
Purdue Aeroelasticity


Topic 1 - Flow compressibility (Mach number) has an effect on divergence because the liftcurve slope depends on Mach number

Approximate the effect of Mach number by adding the Prandtl-Glauert correction factor for sub-sonic flow

KT
qD =
SeC Lα
qDo

C Lα =

KT
=
SeC Lα0

C Lα0
1− M

2

KT 1 − M 2
qD =
= qDo 1 − M 2
SeC Lα0
Plots as a curve vs. M

5-4

Purdue Aeroelasticity


But wait! – there’s more!
Mach number depends on altitude and airspeed so two expressions must be
reconciled

qatmosphere = qa

Physics

1
1 2 2
2
qa = ρV = ρ a M
2
2

M=V/a

Speed of sound, “a," depends on temperature and temperature depends on altitude

KT 1 − M 2
qD =
= q Do 1 − M 2
SeC Lα
0

5-5
Purdue Aeroelasticity


The divergence equation which contains Mach number
must be consistent with the “physics” equation

qDo 1 − M D

2

1 2
2
2
= ρa M D = q1M D
2
1.
2.
3.

1 2
q1 = ρa
2

Choose an altitude
Find the speed of sound
Square both sides of the above equation and solve for
MD

2

(

)

 qDo 
2
4

 1 − M D = M D
 q1 
5-6
Purdue Aeroelasticity


Determining MD requires solving a quadratic equation

2

 qDo 
 qDo 
2
 M D − 
 = 0
+ 
 q1 
 q1 

qatmosphere = qa
1

1 2 2
2
qa = ρV = ρa M
2
2

qDo 1 − M 2

250

dynamic pressure
(lb/sq. ft)

MD

4

2

200

sea level q
divergence q

150 20,000 f t.
100

40,000 f t.

50

0
0.00

0.25

0.50

0.75

Mach number
Purdue Aeroelasticity

1.00
5-7


If we want to increase the divergence Mach number we must increase
stiffness (and weight) to move the math line upward

dynamic pressure
(lb/sq. ft)

400
350
300

sea level q

divergence q

250
200

20,000 f t.

150
100

40,000 ft.

50
0
0.00

0.25

0.50

0.75

1.00

Mach number
5-8
Purdue Aeroelasticity


Summary
i

Lift curve slope is one strong factor that determines divergence dynamic
pressure

– depends on Mach number
i

Critical Mach number solution for divergence dynamic pressure must be added
to the solution process

5-9
Purdue Aeroelasticity


Topic 2 – Multi-degree-of-freedom (MDOF) systems

i

Develop process for analyzing MDOF systems

i

Define theoretical stability conditions for MDOF systems

i

Reading - Multi-degree-of-freedom systems – Section 2.14

5-10
Purdue Aeroelasticity

Here is a 2 DOF, segmented, aeroelastic finite wing model - two discrete aerodynamic surfaces with flexible
connections used to represent a finite span wing (page 57)

A

2KT

3KT

Torsional springs

fuselage

panel 2

panel 1
e

b/2

V

shear
centers

A

b/2

aero
centers

αο + θ2
αο + θ1

wing root

view A-A

wing tip

Torsional degrees of freedom
5-11
Purdue Aeroelasticity


Introduce “strip theory” aerodynamic modeling to represent twist dependent airloads

i

Strip theory assumes that lift depends only on local angle of attack of the strip of aero
surface

–

why is this an assumption?

L1 = qSCLα (α o + θ1 )

q twist angles are measured from
a common reference

L2 = qSCLα (α o + θ2 )
5-12
Purdue Aeroelasticity


The two twist angles are unknowns - we have to construct two free body diagrams to
develop equations to find them
Structural restoring torques depend on the difference between elastic twist angles

Wing root

Wing tip

Internal shear forces are present, but not drawn

Double arrow vectors are torques

5-13
Purdue Aeroelasticity


This is the eventual lift re-distribution equation due to aeroelasticity – let’s
see how we find it

 L1 
W 2 − q 
 =



 L2  2(4 − q )  2 

Observation - Outer wing panel carries more of the total load than the inner panel as q increases

5-14
Purdue Aeroelasticity


Torsional static equilibrium is
a special case of dynamic equilibrium



M
=
I
θ
∑ 1 1 1 = 0 = L1e − 3KTθ1 + 2 KT (θ 2 − θ1 )



M
=
I
θ
∑ 2 2 2 = 0 = L2e − 2 KT (θ 2 − θ1 )
Arrange these two simultaneous equations in matrix form

 5 − 2 θ1 
1 0 θ1 
1
KT 
  − qSeC Lα 
  = qSeC Lα α o  


− 2 2  θ 2 
0 1 θ 2 
1
5-15
Purdue Aeroelasticity


Summary
i

The equilibrium equations are written in terms of unknown displacements and known applied
loads due to initial angles of attack. These lead to matrix equations.

i

Matrix equation order, sign convention and ordering of unknown displacements (torsion angles) is
important

 5 − 2 θ1 
− 1 0  θ1 
1
KT 

  + qSeC Lα 
  = qSeC Lα α o  


− 2 2  θ 2 
 0 − 1 θ 2 
1
5-16
Purdue Aeroelasticity


Problem solution outline

 5 − 2 θ1 
− 1 0  θ1 
1
KT 
  + qSeCLα 
  = qSeC Lα α o  


− 2 2  θ 2 
 0 − 1 θ 2 
1
Combine structural and aero stiffness matrices on the left hand side

θ1 
1
[ KT ]θ  = qSCLα α o 1


 2

 KT 

The aeroelastic stiffness matrix is

Invert matrix and solve for θ1 and θ2

{θ i } = qSeCLα α o [ KT ] {1}
−1

5-17
Purdue Aeroelasticity


The solution for the θ’s requires inverting the aeroelastic stiffness matrix

 K11
 KT  = 
 K 21

 K11

 K 21

K12 

K 22 

−1

K12 
1  K 22
 = 
∆  − K 21
K 22 

− K12 

K11 

∆ = K11 K 22 − K12 K 21
5-18
Purdue Aeroelasticity


The aeroelastic stiffness matrix determinant is a
function of q

i

∆ = K11 K 22 − K12 K 21

The determinant is

∆ = q − 7q + 6
2

where

q=

qSeCLα

∆ = ( 1− q ) ( 6 − q )

KT

When dynamic pressure increases, the determinant ∆ tends to zero – what happens to the
system then?
5-19
Purdue Aeroelasticity


Plot the aeroelastic stiffness determinant D against
dynamic pressure (parameter)

6
4

determinant

2

STABLE

0

UNSTABLE

-2
-4
-6
-8
0

2

4

6

8

Dynamic pressure parameter

The determinant of the stiffness matrix is always positive until the air is turned on

5-20
Purdue Aeroelasticity


Solve for the twist angles created by an input angle of attack
αo

θ1  qα o
 =
∆
θ 2 

4 − q 


7 − q 

q=

∆ = ( 1− q ) ( 6 − q )

5-21
Purdue Aeroelasticity

qSeCLα
KT


Twist deformation vs. dynamic pressure parameter

A

2KT
panel 2

panel 1
e

b/2

V

A

b/2

αο + θ2
αο + θ1

shear
centers

view A-A

divergence

aero
centers

al angle
panel twist/initi
panel twist,
θi/αo of attack

3KT

outboard
panel
10
8

outboard

panel

unstable
Unstable
q region
region
divergence

6
4
2
0
-2

inboard
panel

-4
-6

inboard
panel
outboard
Outboard
panel
panel (2)

-8

-10

0

1

2

3

4

5

6

7

dynamic pressure parameter, q

determinant ∆ is zero
5-22
Purdue Aeroelasticity


Panel lift computation on each segment gives:


q
(
4

−
q
)



1+
 L1 
∆
=
qSC
α
 

Lα o 
q
7
−
q
) 
 L2 
1 + (

∆

L flex
Lrigid

L1 + L2
L1 + L2

=
=
q ( 2 S ) CLα α o qStotal CLα α o

Note that

Stotal = 2 S
5-23
Purdue Aeroelasticity


More algebra - Flexible system lift

L flex
Lrigid

q
= 1 + ( 5.5 − q )
∆

L flex

 q

= qStotal CLα α o 1 + ( 5.5 − q ) ÷
 ∆


Set the wing lift equal to half the airplane weight

L flex

 q
 Weight
= qStotal CLα α o 1 + ( 5.5 − q ) ÷ =
2
 ∆

5-24
Purdue Aeroelasticity


Lift re-distribution due to aeroelasticity (originally presented on slide 13)

W∆
αo =
2qStotal CLα ( ∆ + q ( 5.5 − q ) )

1 + q (4 − q ) 

 L1 
  = qSCLα α o  q 7 − q ∆ 
) 
 L2 
1 + (

∆

 L1 
W 2 − q 

 =


 L2  2(4 − q )  2 

Observation - Outer wing panel carries more of the total load than the inner panel as q increases

5-25
Purdue Aeroelasticity