mae 342 lecture 10

11
Spacecraft Structures Space System Design, MAE 342, Princeton University Robert Stengel Copyright 2008 by Robert Stengel. All rights reserved. For educational use only. http://www.princeton.edu/~stengel/MAE342.html Discrete (lumped-mass) structures Distributed structures Buckling Structural dynamics Finite-element analysis Spacecraft Mounting for Launch Spacecraft protected from atmospheric heating and loads by fairing Fairing jettisoned when atmospheric effects become negligible Spacecraft attached to rocket by adapter, which transfers loads between the two Spacecraft (usually) separated from rocket at completion of thrusting Clamps and springs for attachment and separation Fairing Constraints for Various Launch Vehicles Static envelope Dynamic envelope accounts for launch vibrations, with sufficient margin for error Various appendages stowed for launch Large variation in spacecraft inertial properties when appendages are deployed STEREO Spacecraft Primary Structure Configuration Spacecraft structure typically consists of Beams Flat and cylindrical panels Cylinders and boxes Primary structure is the “rigid” skeleton of the spacecraft Secondary structure may bridge the primary structure to hold components Solar TErrestrial RElations Observatory

Upload: avinash-k-saurav

Post on 20-Jul-2016

230 views

Category:

Documents


0 download

DESCRIPTION

lecture

TRANSCRIPT

Spacecraft StructuresSpace System Design, MAE 342, Princeton University

Robert Stengel

Copyright 2008 by Robert Stengel. All rights reserved. For educational use only.http://www.princeton.edu/~stengel/MAE342.html

• Discrete (lumped-mass)structures

• Distributed structures

• Buckling

• Structural dynamics

• Finite-element analysis

Spacecraft Mounting

for Launch

• Spacecraft protected fromatmospheric heating andloads by fairing

• Fairing jettisoned whenatmospheric effects becomenegligible

• Spacecraft attached torocket by adapter, whichtransfers loads between thetwo

• Spacecraft (usually)separated from rocket atcompletion of thrusting

• Clamps and springs forattachment and separation

Fairing Constraints for

Various Launch Vehicles

• Static envelope

• Dynamic envelope accountsfor launch vibrations, withsufficient margin for error

• Various appendages stowedfor launch

• Large variation inspacecraft inertialproperties whenappendages are deployed

STEREO Spacecraft Primary

Structure Configuration

• Spacecraft structuretypically consists of

– Beams

– Flat and cylindrical panels

– Cylinders and boxes

• Primary structure is the“rigid” skeleton of thespacecraft

• Secondary structure maybridge the primary structureto hold components

Solar TErrestrial RElations Observatory

UARS Primary and

Secondary Structure

• Instrument Module provides

– Support for 10 scientificinstruments

– Maintains instrumentalignment boresights

– Interfaces to launch vehicle(SSV)

• Secondary Structuresupports

– 6 equipment benches

– 1 optical bench

– Instrument mounting links

– Solar array truss

– Several instruments havekinematic mounts

Structural Material Properties

• Stress, !: Force per unit area

• Strain, ": Elongation per unit length

!

" = E #

• Proportionality factor, E: Modulus of elasticity, or Young!s modulus

• Strain deformation is reversible below the elastic limit

• Elastic limit = yield strength

• Proportional limit ill-defined for many materials

• Ultimate stress: Material breaks

• Poisson!s ratio, #:

!

" =#lateral

#axial,

typically 0.1 to 0.35

• Thickening under compression

• Thinning under tension

Uniform Stress Conditions

• Average axial stress, !

!

" = P A = Load Cross Sectional Area

• Average axial strain, "

!

" = #L L

• Effective spring constant, ks

!

" = P A = E# = E$L

L

P =AE

L$L = k

s$L

Springs and Dampers

!

fx = "ks#x = "ks x " xo( ) ; k = springconstant

!

fx = "kd#˙ x = "kd#v = "kd v " vo( ) ; k = dampingconstant

• Force due to linear spring

• Force due to linear damper

Mass, Spring, and Damper

!

"˙ ̇ x = fx m = #kd"˙ x # ks"x + forcing function( ) m

!

"˙ ̇ x +kd

m"˙ x +

ks

m"x =

forcing function

m

!

"˙ ̇ x + 2#$n"˙ x +$

n

2

"x =$n

2

"u

!

"n = natural frequency, rad /s

# = damping ratio

$x = displacement

$u = disturbance or control

• Newton!s second law leads to asecond-order dynamic system

Response to Initial Condition

• Lightly dampedsystem has adecaying, oscillatorytransient response

• Forcing by step orimpulse produces asimilar transientresponse

!

"n

= 6.28 rad /sec

# = 0.05

Oscillations

!

"x = Asin #t( )

!

"˙ x = A# cos #t( )

= A# sin #t + $ 2( )

!

"˙ ̇ x = #A$ 2sin $t( )

= A$ 2sin $t + %( )

• Phase angle of velocity (wrt displacement) is $/2 rad (or 90°)

• Phase angle of acceleration is $ rad (or 180°)

• As oscillation frequency, %, varies

– Velocity amplitude is proportional to %

– Acceleration amplitude is proportional to %2

Response to

Oscillatory Input

!

A" 2sin "t + #( )[ ] + 2$"

nA" sin "t + # 2( )[ ] +"

n

2

Asin "t( )[ ] ="n

2

Bsin "t( )[ ]

• ... however, A must be a complex number for this to work

• A better way: Compute Laplace transform to find transferfunctions

!

L "x(t)[ ] = "x(s) = "x(t)e#stdt0

$

% , s =& + j', ( j = i = #1)

!

L "˙ x (t)[ ] = s"x(s)

L "˙ ̇ x (t)[ ] = s2"x(s)

• Neglecting initial conditions

Transfer Function

• or

!

L "˙ ̇ x + 2#$n"˙ x +$

n

2

"x( ) = L $n

2

"u( )

!

s2 + 2"#

ns+#

n

2( )$x(s) =#n

2$u(s)

• Transfer function from input to displacement

!

"x(s)

"u(s)=

#n

2

s2 + 2$#

ns+#

n

2( )

Transfer Functions of Displacement,

Velocity, and Acceleration

!

"x(s)

"u(s)=

#n

2

s2 + 2$#

ns +#

n

2( )"˙ x (s)

"u(s)=

#n

2s

s2 + 2$#

ns +#

n

2( )"˙ ̇ x (s)

"u(s)=

#n

2s

2

s2 + 2$#

ns +#

n

2( )

• Input tovelocity:multiply by s

• Input toacceleration:multiply by s2

• Transfer functionfrom input todisplacement

From Transfer Function to

Frequency Response

• Displacement frequency response (s = j%)

!

"x( j#)

"u( j#)=

#n

2

j#( )2

+ 2$#n j#( ) +#n

2!

"x(s)

"u(s)=

#n

2

s2 + 2$#

ns+#

n

2( )

• Displacement transfer function

Real and imaginary components

Frequency Response

• %n: natural frequency of the system

• %: frequency of a sinusoidal input to the system

!

"x( j#)

"u( j#)=

#n

2

j#( )2

+ 2$#n j#( ) +#n

2

=#n

2

#n

2 %# 2( ) + 2$#n j#( )&

#n

2

c #( ) + jd #( )

=#n

2

c #( ) + jd #( )

'

( )

*

+ , c #( ) % jd #( )c #( ) % jd #( )

'

( )

*

+ , =

#n

2c #( ) % jd #( )[ ]

c2 #( ) + d2 #( )

& a(#) + jb(#) & A(#)e j- (# )

• Frequency response is a complex function

– Real and imaginary components, or

– Amplitude and phase angle

Frequency

Response of the

2nd-Order System

• Bode plot– 20 log(Amplitude Ratio) [dB] vs. log !

– Phase angle (deg) vs. log !

• Natural frequency characterized by

– Peak (resonance) in amplituderesponse

– Sharp drop in phase angle

• Acceleration frequency responsehas the same peak

• Convenient to plot response onlogarithmic scale

!

ln A(")e j# (" )[ ] = lnA(") + j#(")

Acceleration

Response of the

2nd-Order System

• Important points:– Low-frequency acceleration

response is attenuated

– Sinusoidal inputs at naturalfrequency resonate, I.e.,they are amplified

– Component naturalfrequencies should be highenough to minimizelikelihood of resonantresponse

Spacecraft Stiffness* Requirements

for Primary Structure

* Natural frequency

Example 8.1(Fundamentals of Space Systems, 2005)

• Atlas IIAS launch vehicle

• Spacecraft structure meetsprimary stiffnessrequirements

• What are axial stiffnessrequirements for Units Aand B?

– Support deck naturalfrequency = 50 Hz

Octave Rule: Componentnatural frequency ! 2 xnatural frequency ofsupporting structure

• Unit A: 2 x 15 Hz = 30 Hz, supported by primary structure

• Unit B: 2 x 50 Hz = 100 Hz, supported by secondary structure

Factors and Margins of Safety

• Factor of Safety– Typical values: 1.25 to 1.4

!

Allowable load (yield stress)

Expected limit load (stress) "Design factor of safety#1

• Margin of Safety– “the amount of margin that exists above the

material allowables for the applied loadingcondition (with the factor of safety included)”,Skullney, Ch. 8, Pisacane, 2005

!

Load (stress) that causes yield or failure

Expected service load

Worst-Case Axial Stress

on a Simple Beam

• Axial stress due to bending

!

" = My I

!

" =M h /2( )

I

• Maximum stress

• Worst-case axial stress due to bending and axial force

!

"wc

= ±P

A

#

$ %

&

' ( max

±M h /2( )

I

Example 8.3(Fundamentals of Space Systems, 2005)

• Spacecraft weight = 500 lb

• Atlas IIAS launch vehicle

• Factor of safety = 1.25

• Maximum stress onspacecraft adapter?

Atlas IIAS Limit Loads (g)

Example 8.3, con’t.

• Worst-case axial load at BECO (5±0.5 g)

!

"wc

= ±P

A

#

$ %

&

' ( max

±Mc

I

!

A = 2"rt = 7.1 in2

I = "r3t = 286 in4

• Worst-case stress

!

"wc =500 # 5.5

7.1+500 # 0.5 # 42 # 9

286

$

% & '

( ) #1.25 = 897.1 psi

• Worst-case lateral load at BECO (2.5 ± 1 g) or Maximum FlightWinds (2.7 ± 0.8 g)

!

"wc =500 # 3.5

7.1+500 # 2 # 42 # 9

286

$

% & '

( ) #1.25 =1960 psi

Force and Moments on a Slender

Cantilever (Fixed-Free) Beam

• Idealization of

– Launch vehicle tied-down to alaunch pad

– Structural member of apayload

• For a point force

– Force and moment must beopposed at the base

– Shear distribution is constant

– Bending moment increases asmoment arm increases

– Torsional moment andmoment arm are fixed

Structural Stiffness

• Geometric stiffening property of a structure is portrayed by thearea moment of inertia

• For bending about a y axis (producing distortion along an x axis)

• Area moment of inertia for simple cross-sectional shapes

• Solid rectangle of height, h, andwidth, w:

• Solid circle of radius, r:

• Circular cylindrical tube with innerradius, ri: and outer radius, ro:

!

Iy = wh3/12

!

Iy = "r4 /4

!

Iy = " ro4

# ri4( ) /4

!

Ix

= x z( )z2 dzzmin

zmax

"

Bending Stiffness

• Neutral axis neither shrinks nor stretches in bending

• For small deflections, the bending radius of curvature ofthe neutral axis is

!

r =EI

M

• Deflection at a point characterized bydisplacement and angle:

Bending

Deflection

• Second derivative of z and first derivative of & areinversely proportional to the bending radius:

!

d2z

dx2

=d"

dx=My

EIy

Maximum Deflection and

Bending Moment of Beams(see Fundamentals of Space Systems for additional cases)

Fixed-Free Beam Fixed-Fixed Beam Pinned-Pinned Beam

Ymax = maximum deflection Mmax = maximum bending moment

Maximum Deflection and

Bending Moment of Plates(see Fundamentals of Space Systems for additional cases)

Circular Plate

Rectangular Plate!

m =1/"

Typical Cross-Sectional Shear Stress

Distribution for a Uniform Beam

• Shear stress due to bending moment ishighest at the neutral axis

• Maximum values for various cross sectons(see Fundamentals of Space Systems)

Buckling

• Predominant steady stressduring launch is compression

• Thin columns, plates, andshells are subject to elasticinstability in compression

• Buckling can occur below thematerial!s elastic limit

!

"cr

=C# 2

E

L /$( )2

=P

A

Critical buckling stress of a column(Euler equation)

!

C = function of end " fixity"

E = modulus of elasticity

L = column length

" = I A = radius of gyration

Pcr = critical buckling load

A = cross sectional area

Effect of “Fixity” on Critical

Loads for Beam Buckling

• Euler equation

– Slender columns

– Critical stress below theelastic limit

– Relatively thick column walls

• Local collapse due to thinwalls is called crippling

Crippling vs. Buckling

Critical Stress for Plate and Cylinder Buckling

Bending Moment and Linear Deflection

due to a Distributed Normal Force

• Deflection is found by four integrations of the deflectionequation

!

d2

dx2EIy

d2z

dx2

"

# $

%

& ' x= xs

= N 'y xs( )

N!y(xs)

!

My x( ) = Ny (x)xmin

xmax

" x # xcm( )dx

= N 'y (x)xmin

xmax

" dx x # xcm( )dxxmin

xmax

"

!

N '(x) = normal force variation with length

Bending Vibrations of a

Free-Free Uniform Beam

!

EIyd4z

dx4

x= xs

= k = "m'd2z

dt2

x= xs

!

EIy = constant

m'= mass variation with length (constant)

k = effective spring constant

• Solution by separation of variables requires that left andright sides equal a constant, k

• An infinite number of separation constants, ki, exist

• Therefore, there are an infinite number of vibrationalresponse modes

Bending Vibrations of a

Free-Free Uniform Beam

• In figure, (u = z, y = x)

• Left side determinesvibrational mode shape

• Right side describesoscillation

• Natural frequency ofeach mode proportionalto (ki)

1/2

Mode shapes of bending vibrations

!

EIyd4z

dx4

= ki = "m'd2z

dt2

Fundamental Vibrational

Frequencies of Circular Plates

f = natural frequencyof first mode, Hz

Finite-Element

Structural Model

TIMED Spacecraft • Grid of elements, each with

– mass, damping, andelastic properties

– 6 degrees of freedom

Vibrational Mode Shapes

for the X-30 Vehicle (Raney

et al, J. Aircraft, 1995)

Computational Grid Model

• Body elastic deflection distorts the shape ofscramjet inlet and exhaust ramps

• Aeroelastic-propulsive interactions

• Impact on flight dynamics

Shapes of the First Seven Modes

Next Time:Flight Path Guidance,

Navigation, and Control