Aerospace Structures: Chapter 2 (Internal Loads)

8/22/2019 Aerospace Structures: Chapter 2 (Internal Loads)

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CHAPTER 2

INTERNAL LOADS IN AEROSPACE STRUCTURES

2.1 Force and Moment Distributions

Slender body under axial force

Slender body under torqueSlender body under lateral loads

2.2 Inertia Loads

Load factorExamples


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2.1 Force and Moment Distributions

Aircraft with high aspect ratio wings and rockets can be modeled as a slender structures

subjected to external loads in the form of axial forces, lateral forces and moments. These

external loads in turn induce internal forces and moments. In the following sections, we

will look at the distributions of forces and moments along slender structures. We will

consider only statically determinate cases.

Recall:

1) For statically determinate structures, force and moment distributions can bedetermined considering only equilibrium equations.

2)For statically indeterminate structures, it is necessary to consider thedeformation under applied load to determine force and moment distributions.


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2.1.1 Resultant Forces and Moments

Consider a slender body with the x-axis placed along the longest dimension. Now letsintroduce an imaginary cut normal to the x axis and consider the stress components acting

over a cross-section located at coordinatex as shown below.

Then three resultant forces and three resultant moments acting over the cross-section aredefined as follows :

dAxF xxV)( : axial force in thex direction

dAxV xyy W)( : (transverse) shear force in they direction

dAxV xzz W)( : (transverse) shear force in thezdirection

dAzxM xxy V)( : moment around they axis

dAyxM xxz V)( : moment around thezaxis

dAzyxT xyxz )()( WW : torque or moment around thex axis

Wxz

Wxyy

z

dA

y

z

x

y

z

xVz( ) Vy ( )

T(x)

y ( )

xz( )

z

y

x

Vxx

This figure shows positive forces and

moments on the positivex-surface.

F(x)

positivex-surface


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( ) 1000F x lb

( ) 1000F x lb

xx

1000 1000

xx

1000 1000


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5000zV lb

( ) 5000zV x lb xx

5000 5000

z

xx

5000 5000

z


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5000y lb in

5000y lb in

xx

5000 5000

z

xx

5000 5000

z


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2.1.2 Slender Body under Axial Force

A rocket or a helicopter blade can be modeled as a slender body under axial force.

f x( ) : applied force per unit length, e.g. gravity

A x( ) : cross-sectional area

To look at equilibrium, lets create a free body by introducing imaginary cut(s).

Introduce a cut atx and consider the free body on the right hand side of the cut.

(Axial forces) = 0 for the free body.

P lb

x=0

f lb/in

x=L

x

P

x

[

[[ d

P lb

0x x

f

[[ df )(

F x

( )F x ( )F x

x

0

0

x

[

d[


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( ) ( ) 0

L

x

F x P f d

[

[

[ [

(1)

( ) ( )

L

x

F x P f d

[

[

[ [

(2)

Example :

Consider a rocket on a launch pad modeled as a slender body under its own weight.

Mg: payload weight, m(x): mass per length, g: gravity

Introducing a cut atx,

( ) ( ) ( )

L L

x x

F x f d Mg mg d Mg

[ [

[ [

[ [ [

(1)

For constant m,

( )

( ) ( )

L

xF x mg Mg

F x mg L x Mg

[

o (2)

x=0

f lb/in

x=L

x

payload g


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2.1.3 Slender Body under Torque

Consider a high aspect ratio wing subject to aerodynamic moment and, possibly, wing tip

moment due to a wing tip fuel tank or an engine. Thex-axis is along the wingspan.

( )Tf x : applied torsional moment per unit length

T : torsional moment applied atx = L

Using double arrows to indicate torque,

To look at equilibrium, lets create a free body by introducing imaginary cut atx andconsider the right hand side of the cut.

(torques acting over the free body) = 0

( )Tf x

T

x=0 x

x

x=0

x=L

( ) :Tlb in

f xin

inlbT :

( )Tf x

Tx


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For the free body, lets look at the span between [ [ [and d .

( ) ( ) 0

( ) ( )

L

Tx

L

T

x

T x T f d

T x T f d

[

[

[

[

[ [

[ [

(1)

Example: A straight wing fixed at the root (x = 0) is subjected to a torque produced by

the aileron deflection. The aileron extends fromx = L/2 to wing tip (x = L). Torque ( )Tf x

per unit aileron span is assumed constantfo. Determine torque T(x).

(1) 02

Lxd d

0 0 0 0

2

2 2 2

1( ) ( )

2

L L LL

LT

L L L

T x f d f d f d f f L

[ [ [[

[

[ [ [

[ [ [ [ [

T

( )Tf d[ [T(x)

d[

[[ d[0,

0x

[

x

( )Tf d[ [T(x)

d[

[[ d[0,

0x

[

x

2

L


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(2)2

Lx Ld d

0 0 0 0( ) ( ) ( )

L L LL

T x

x x x

T x f d f d f d f f L x

[ [ [[

[

[ [ [

[ [ [ [ [

( )Tf d[ [

T(x)

d[

[[ d[0,

0x

[

x

x

L

0

T

f L

0.5

0.5 1.0


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1.4.4 Slender Body under Lateral Loads

High aspect ratio aircraft wing, tails and fuselage can be modeled as a beam.

Example: A wing of a VTOL aircraft subjected to an engine thrust at the wingtip can be

modeled as a cantilever beam under a tip force as shown below.

To determine shear force and moment distributions due to the tip force, lets introduce an

imaginary cut located at x to create a free body as shown below.

Consider the free body on the right hand side of the cut.

(the sum of vertical forces) = 0

( ) 0 ( )z zV x P V x P o (1)

(the sum of moments) = 0

( ) ( ) 0

( ) ( )

y

y

M x P L x

x P L x

o (2)

0x x L

Pz

x

Vz( )

y ( )

V xz( )

x

xy ( )

P

L


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Example: A high aspect ratio wing with a podded engine can be modeled as a cantilever

beam under a downward force as shown below.

1) 0 x ad d

Consider an imaginary cut located atx as shown below.

Consider the free body on the right hand side of the cut.

(the sum of vertical forces) = 0, ( ) 0 ( )z E z EV x W V x W o

(the sum of moments) = 0, ( ) ( ) 0 ( ) ( )y E y Ex W a x M x W a x o

2) a x Ld d

0x x L

z

EW

x a

xVz( )

y ( )

Vz( )

x

xy ( )

L

EW

a

x

Vz( )

y ( )

V xz( )

x

y ( )

x L

EW

x a

EW : engine weight

( ) 0

( ) 0

z

y

V x

M x


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Example:

A wing subjected to a lift, its own weight or fuel weight can be modeled as a cantilever

beam under a distributed load ( )zp x as shown below.

Introduce a cut atx to isolate a free body.

In addition, consider a section between and +d[ [ [ of the free body,

oL

x

zz dpxVz

[

[

[[ 0)()(0)in(Force (1)

L

x

zz dpxV

[

[

[[)()( (2)

oL

x

zy dpxxMx

[

[

[[[ 0)()()(0)about(Moment (3)

( ) ( ) ( )

L

y z

x

x x p d

[

[

[ [ [

(4)

x

x=0

p x lbinz

( ),

x=L

x

V xz( )

xy ( )

V xz( )

x

d[

[

xy ( )

( )zp [


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Example: A lift over the wingspan may be interpolated as a polynomial function as

follows:2( )zp x A Bx Cx "

where coefficients ,A B and C are determined via matching the actual lift distribution

with the quadratic function at three points along the wing span.

Now lets look at shear force and moment corresponding to the individual terms.

(1) ( ) ( )z zp x A p A[ o

( ) ( )

( ) ( )

L L LL

z z x

x x x

z

V x p d Ad A d A

V x A L x

[ [ [[

[

[ [ [

[ [ [ [ [

2

2

( ) ( ) ( ) ( )

( )2

1( ) ( )

2

L L

y z

x x

LL

x x

y

x x p d x Ad

A x d A x

M x A L x

[ [

[ [

[[

[ [

[ [ [ [ [

[[ [ [

(2) ( ) ( )z z

p x Bx p B[ [ o

2

2 2

( ) ( )2

( ) ( )2

LL L L

z z

x x x x

z

V x p d B d B d B

BV x L x

[[ [ [

[ [ [ [

[[ [ [ [ [ [

3 22

3 2 3

( ) ( ) ( ) ( )

( )3 2

( ) (2 3 )6

L L

y z

x x

LL

x x

y

x x p d x B d

B x d B x

BM x L L x x

[ [

[ [

[[

[ [

[ [ [ [ [ [

[ [[ [ [


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(3) 2 2( ) ( )z zp x Cx p C[ [ o

32 2

3 3

( ) ( ) 3

( ) ( )3

LL L L

z z

x x x x

z

V x p d C d C d C

CV x L x

[[ [ [

[ [ [ [

[[ [ [ [ [ [

2

4 33 2

( ) ( ) ( ) ( )

( )4 3

( )

L L

y z

x x

LL

x x

y

x x p d x C d

C x d B x

M x

[ [

[ [

[[

[ [

[ [ [ [ [ [

[ [[ [ [

"

Note:

( ) (1 )zx

p x a A BxL

,a

A a BL

zp

a

xL


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2.2 Inertia Loads

Load Factor

For an aircraft in flight, load factorn is defined as

W

Ln

whereL = lift, W= vehicle weight.

1) For an aircraft in level flight,L = Wand n = 1.

2) For an aircraft accelerating in vertical direction,

Newtons 2nd

law

WLMa

where M: vehicle mass, a : acceleration

(1 )W a

L Ma W a W Wg g

o

or L nW

with 1a

ng

L

a

W


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Inertia Force and Inertia moment

Inertia Force:

Consider a massMunder applied forceF.

a: acceleration

Newtons second law: a F (1)

Equation (1) can be rewritten as ( ) 0F Ma (2)

which corresponds to the static condition shown below:

So, it is observed that, with the inertia force ( )a ,

(forces) 0 : force equilibrium======================================================Note: Consider a mass accelerating vertically as shown in the sketch

From the sketch on the right hand side,

0 (1 ) 0a

F W Ma F W F nW

g

o o

where 1a

ng

: load factor

M

F

a

a F

F

W Mg a

F

W Mg

a


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Inertia Moment:

T: Applied moment

I: Mass moment of inertia

T : Angular acceleration

Newtons second law for a rotating mass:

I TT (3)

Equation (3) can be rewritten as

( ) 0T IT (4)

which corresponds to a static condition shown as follows:

So, it is observed that, with the inertia moment( )IT ,

(moments) 0 : moment equilibrium

T T

I

IT

T


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Example:

Consider a single-stage rocket in vertical flight at a high altitude. The rocket is subjectedto thrust T = 7 MgwhereMis the total mass. Total length of the rocket isL, and mass per

unit length is constant om . Determine axial force ( )F x . Assume no aerodynamic loads.

a : acceleration

7 6 6a T Mg Mg Mg Mg a g o (1)

0

0

( ) ( )( ) 0

x

T F x m d g a

[

[

[

(2)

0 0

0 0

0 0

0

0

( ) ( )( ) ( ) ( )

7 7

7 7 7 7

x x

x

F x T m d g a T m g a d

T m g d T m gx

MMg m gx Mg gx

L

[ [

[ [

[

[

[ [

[

(3)

or

( ) 7 ( 1) ( 1)x x

F x Mg TL L

(4)

0x

T

x

T ( )F x

[ d[

x/L

F/T

1.0

1.0


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Example:

The rocket shown in the figure experiences a wind gust during its vertical ascent. The

gust results in the load )2

3(0L

xppz per length. The pitching is prevented by

vectoring thrust Tas shown. Assume m, mass per length, is constant.

Note: 1) 0 , ,p T m andL are given quantities.

2) The c.g. of the rocket is at2

Lx

.

(a)Express angle T in terms ofT, 0p and lengthL.(b)Determine shear force zV and moment y along the length of the rocket. Express

them in terms of 0p .

(c)Plot 0( )zV p L vs. Lx and 20( )y p L vs. Lx .

)

2

3(0 L

x

ppz

03p

0p

T T

z 0x

L


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Solution:

(a)

Force equilibrium: 0zF ,

0 0sin 0

L L

zamdx p dx T T

For constant m, 00

23 sin 0

L xmaL p dx T

LT

2

0

0

13 sin 0

L

maL p x x T L

T

02 sin 0maL Lp T To (1)

Moment equilibrium: 0 0xM

0 00

L L

zxp dx xamdx

T

)2

3(0L

xppz

T

ma

z

a: horizontal acceleration


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23

2

00 0

23 0

L L

p x x dx xamdxL

2 3 2

0

0 0

3 2 10

2 3 2

L L

p x x ma xL

2 2

0

50

6 2

map L Lo

0

5

3a p

m? (2)

Placing equation (2) into equation (1),

0 0

0

52 sin 0

31sin 0

3

p L Lp T

p L T

T

T

0

1sin

3T p LTo (3)

(b) Introduce a cutx to create a free body.

T

0

2( ) (3 )

z

xp x p

L

T

0

5

3ma p

0

2( ) (3 )zp p

L

[[

0

5

3p

[

( )zV x

( )y x

d[


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Consider the free body shown in the right sketch,

Force equilibrium:

05( ) ( ) 03

Lz z z

xF V x p p d[ [

2

0 0 0

2 2

0

2 5 1 5( ) 3 3

3 3

4 1

3

LL

zx

x

V x p p d pL L

p L x L xL

[[ [ [ [

-

-

2

0

1 4( )

3 3z

x xV x p L

L L

-

Moment equilibrium:

05

( ) ( ) 03

L L

y y zx x xM M x p x d p x d[ [ [ [ [

0 0

2

0 0

2 3

0

2 2 3 3

0

2 5

( ) 3 3

2 5 4 4 2 23

3 3 3

4 2 2

3 3 3

4 2 2

3 3 3

L L

y x x

L L

x x

L

x

M x p x d p x dL

xp x d p x d

L L L

xp x

L L

xp x L x L x L x

L L

[

[ [ [ [

[[ [ [ [ [

[ [ [

-

-

2 32

0

1 2 1( )

3 3 3y

x x xM x p L

L L L-


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Example:Consider a cargo plane in flight as shown in the sketch. The nose is located at a bodystation (BS) of 0 in. The loaded plane weighs 150,000 lb, and its c.g. is at BS 250 in. The

fuselage is 600 in long and together with the payload weighs a constant 150 lb/in. The tailweighs 2,000 lb and has a c.g. at BS 560in.

(a) Determine the c.g. location of the wing including the engines and the fuel in the wing.(b) The aircraft is at a trimmed (i.e. no pitching acceleration) maneuver with a load factor

of n = 3. The resultant aerodynamic forces WL and TL on the wing and tail are

respectively at BS 200 in and 550 in. Determine WL and TL .

(c) Determine the shear and bending moment distribution on the fuselage.

-----------------------------------------------------------------------------------------------------

(a)

Determination of c.g. location for (wing + engine + fuel in the wing)

600

0

150,000 150 2,000 58,000

wef wing engine fuel AC fuselage tail

wef

W W W W W W W

W dx lb

o (1)

WLTL

ACW

B.S. 0 200 250 550 600

x

z

250B.S.@000,150 lbWAC xWwef B.S.@2,000

@ 560B.S.

tailW lb

x dx

{

150 lb/in

tailW

560


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0c.g.)aboutMoment( 600

0

(250 ) (560 250) ( 250)150 0wef tail W x W x dx (2)

where x : c.g. location for wefW

600

0

58,000(250 ) 2,000(560 250) ( 250)150 0x x dx (3)

inx 724.161)000,500,4000,620(000,56

1250 (4)

(b)

lbWAC 000,150

3

n

gg

WM ACAC

000,150

Maneuver load factor 3n (5)

Total AC L nW (6)

3 150, 000 450, 000W TL L lbo u (7)

Taking moment about the c.g. of the aircraft in flight,

TWTWy LLLLM 6030050 o (8)

By solving equations (1) and (2),

385,714.3 , 64,285.7W TL lb L lb (9)

WL TL

ACW

aMAC


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(c) To determine shear force and moment distribution, the whole domain is divided into

five regions as shown in the sketch.

I) Region (1) )724.1610( dd x

xxVdVF z

x

zz 450)(04500

o [

2

0

225)(

0)(450)(

xxM

dxMM

y

x

yAy

o

[[

200B.S.@

3.714,385 lbLW

724.161B.S.@3 wefW 3 @ 560B.S.tailW

x

inlb /)150(3

550B.S.@

7.285,64 lbLT

region (1) (2) (3)

(4)

(5)

yM

[

[dx

zV

A


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II) Region (2) )200724.161( dd x

000,174450)(0450)000,58(3 0o

xxVdVFz

x

zz [

000,140,28000,174225)(

0)(450)724.161)(000,58(3)(

2

0

o

xxxM

dxxMM

y

x

yAy [[

III) Region (3) )550200( dd x

3.714,211450)(

0450)000,2(37.285,64

600

o

xxV

dVF

z

xzz

[

600

2

( ) 64,285.7(550 ) 3(2,000)(560 ) 450( ) 0

( ) 225 211,714.3 49,002,865

y A y

x

y

M M x x x d

M x x x

[ [

o

3 @ 560B.S.tailW

64,285.7

@ 550

B.S.

TL lb

yM

zV

[

A

x

[

[d x

724.161B.S.@3 wefW

yM

zV

A

[d


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IV) Region (4) )560550( dd x

000,276450)(

0450)000,2(3

600

o

xxV

dVF

z

xzz [

000,360,84000,276225)(

0)(450)560)(000,2(3)(

2

600

o

xxxM

dxxMM

y

x

yAy [[

V) Region (5) )600560( dd x

000,270450)(

0450600

o

xxV

dVF

z

x

zz [

000,000,81000,270225)(

0)(450)(

2

600

o

xxxM

dxMM

y

x

yAy [[

3 @ 560B.S.tailW

yM

zV

[

A

x

yM

zV

[

A

x

[d

[d


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Documents

Aerospace Structures: Chapter 2 (Internal Loads)