s truc t aero section idealization

7/31/2019 s Truc t Aero Section Idealization

1/46

Aerospace & Mechanical engineering

University of Lige

Aerospace & Mechanical Engineering

Aircraft structures

Section idealization

Ludovic Noels

LTAS-Milieux Continus et Thermomcanique

Chemin des Chevreuils 1, B4000 Lige

[email protected]


2/46


3/46

2011-2012 Aircraft structures: Section Idealization 3

General expression for unsymmetrical beams

Stress

With

Curvature

In the principal axesIyz = 0

Euler-Bernoulli equation in the principal axis

forx in [0L]

BCs

Similar equations foruy

Pure bending: linear elasticity summary

x

z f(x) TzMxxuz =0

duz/dx =0 M>0

L

y

z

Mxx


4/46


General relationships

Two problems considered

Thick symmetrical section

Shear stresses are small compared to bending stresses ifh/L 0

L

h

L

L

h t


5/46


Shearing of symmetrical thick-section beams

Stress

With

Accurate only ifh > b

Energetically consistent averaged shear strain

with

Shear center on symmetry axes

Timoshenko equations

&

On [0L]:

Beam shearing: linear elasticity summary

h

z

y

z

b(z)

A*

h

x

z

Tz

x

Tz+ xTz x

max

x

z

x

y

y


6/462011-2012 Aircraft structures: Section Idealization 6

Shearing of open thin-walled section beams

Shear flow

In the principal axes

Shear centerS

On symmetry axes

At walls intersection

Determined by momentum balance

Shear loads correspond to

Shear loads passing through the shear center &

Torque


x

z

y

Tz

Tz

Ty

Ty

y

z

S

Tz

TyCq

s

y

t

t

h

b

z

C

t

S



Shearing of closed thin-walled section beams

Shear flow

Open part (for anticlockwise ofq, s)

Constant twist part

The completely around integrals are related to the

closed part of the section, but if there are open parts,

their contributions have been taken in qo(s)


y

z

T

Tz

Ty

C

q

s

pds

dAh

y

z

T

Tz

Ty

C

qs

p



Shearing of closed thin-walled section beams

Warping around twist centerR

With

ux(0)=0 for symmetrical section if origin on

the symmetry axis

Shear centerS

Compute q for shear passing thought S

Use

With point S=T


y

z

T

Tz

Ty

C

qs

pds

dAh

y

z

S

Tz

C

q s

p ds



Torsion of symmetrical thick-section beams

Circular section

Rectangular section

Ifh >> b

&

Beam torsion: linear elasticity summary

z

y

C

Mx

r

0.229

0.246

2

0.281

0.282

4

1/30.2310.208

1/30.1960.141

1.51h/b

z

y

C

max Mx

b

h



Torsion of open thin-walled section beams

Approximated solution for twist rate Thin curved section

Rectangles

Warping ofs-axis

Beam torsion: linear elasticity summary

y

z

l2

t2

l1t1

l3

t3

t

z

y

C

Mx

t

ns

R

pR

us


11/46



Virtual displacement (see annex of beams for demonstrations)

In linear elasticity the general formula of virtual displacement reads

(1) is the stress distribution corresponding to a (unit) load P(1)

P

is the energetically conjugated displacement to P in the direction ofP(1) that

corresponds to the strain distribution

Example: bending of a semi cantilever beam

In the principal axes

Example: shearing of a semi-cantilever beam

Deflection of open and closed section beams

x

z Tz

uz =0

duz/dx =0 M>0

L



Example 2-spar wing (one cell)

Stringers to stiffen thin skins

Angle section form spar flanges

Design stages Conceptual

Define the plane configuration

Span, airfoil profile, weights,

Analyses should be fast and simple

Formula, statistics,

Preliminary design

Starting point: conceptual design Define more variables

Number of stringers, stringer area,

Analyses should remain fast and simple

Use beam idealization

See today

FE model of thin structures

See next lectures

Detailed design

All details should be considered (rivets, ) Most accurate analyses (3D, non-linear, FE)

Structural idealization


14/46


Principle of idealization

Booms

Stringers, spar flanges,

Have small sections compared to airfoil

Direct stress due to wing bending is

almost constant in each of these

They are replaced by concentrated areacalled booms

Booms

Have their centroid on the skin

Are carrying most direct stress due

to beam bending

Skin

Skin is essentially carrying shear stress

It can be assumed

That skin is carrying only shear stress

If direct stress carrying capacity of skin is

reported to booms by appropriate

modification of their area

Wing section idealization


15/46


Panel idealization

Skin panel

Thickness tD, width b

Carrying direct stress linearly distributed

Replaced by

Skin without thickness

2 booms of areaA1 andA2

Booms area depending on loading

Moment around boom 2

Total axial loading


tD

b

y

z

xxx1 xx

2

b

xx1

xx2

A1 A2

y

z

x


16/46


Example

2-cell box wing section

Simply symmetrical

Angle section of 300 mm2

Idealization of this section

to resist to bending moment?

Bending moment alongy-axis

6 direct-stress carrying

booms

Shear-stress-only carrying

skin panels


hl=0.4m

h

r=0.2m

la = 0.6 mlb = 0.6 m

tr= 2 mmtl = 3 mm tm = 2.5 mm

ta

= 2 mm

tb = 1.5 mm

hl=0.4m

hr

=0.2m

la = 0.6 mlb = 0.6 m


ta = 2 mm tb = 1.5 mm

A1 A2 A3

A6

A5

A4

y

z


17/46


Booms area Bending moment

Alongy-axis

Stress proportional toz

stress distribution is

linear on each section edge

Contributions Flange(s) area

Reported skin parts

Use formula for linear distribution


hl=0.4m

hr

=0.2m

la = 0.6 mlb = 0.6 m


ta = 2 mm tb = 1.5 mm

A1 A2 A3

A6A5

A4

y

z


18/46


Consequence on bending

Idealization depends on the loading case

Booms area are dependent on the loading case

Direct stress due to bending is carried by booms only

For bending the axial load is equal to zero

But direct stress depends on the distancez from neutral axis

It can be concluded that for open or closed sections, the position of the

neutral axis, and thus the second moments of area

Refer to the direct stress carrying area only

Depend on the loading case only

Section idealization consequences


19/46


20/46


Centroid

Of idealized section


z8= 0.038 m

A1

= 640 mm2

y

z

C

y

z

O

A2 = 600 mm2

A3 = 600 mm2

A4 = 600 mm2

A5 = 620 mm2

A6 = 640 mm2

A7 = 640 mm2

A8 = 850 mm2

A9 = 640 mm2

z7

= 0.144 m

z6

= 0.336 m

z5 = 0.565 m

z4

= 0.768 m

z3

= 0.960 m

z2

= 1.14 m

z1

= 1.2 m

My


21/46


Second moment of area

Of idealized section


z8= 0.038 m

A1

= 640 mm2

y

z

C

y

z

O

A2 = 600 mm2

A3 = 600 mm2

A4 = 600 mm2

A5 = 620 mm2

A6 = 640 mm2

A7 = 640 mm2

A8 = 850 mm2

A9 = 640 mm2

z7

= 0.144 m

z6

= 0.336 m

z5 = 0.565 m

z4

= 0.768 m

z3

= 0.960 m

z2

= 1.14 m

z1

= 1.2 m

My

S i id li i


22/46


Stress distribution

Stress assumed constant in each boom

As we are in the principal axes


z8= 0.038 m

A1

= 640 mm2

y

z

C

y

z

O

A2 = 600 mm2

A3 = 600 mm2

A4 = 600 mm2

A5 = 620 mm2

A6 = 640 mm2

A7 = 640 mm2

A8 = 850 mm2

A9 = 640 mm2

z7

= 0.144 m

z6

= 0.336 m

z5 = 0.565 m

z4

= 0.768 m

z3

= 0.960 m

z2

= 1.14 m

z1

= 1.2 m

My

S ti id li ti


23/46


Consequence on open-thin-walled section shearing

Classical formula

Results from integration of balance

equation

With

So consequences are

Terms & should account for the direct

stress-carrying parts only (which is not the case of shear-carrying-only skin

panels)

Expression of the shear flux should be modified to account for discontinuities

encountered between booms and shear-carrying-only skin panels


q + sq s

s

x

x

s

qq

q + xq x

s

s + ss s

xx + xxx x

xx

S ti id li ti


24/46


Consequence on open-thin-walled section shearing (2)

Equilibrium of a boom of an idealized section

Lecture on beam shearing

Direct stress reads

With &

Eventually

(no sum on i)


Tzy

z

x

x

Ty

x

qi

qi+1qi

qi+1

xx+xxx

xx

y

z

x

qi

qi+1

S ti id li ti


25/46


Consequence on open-thin-walled section shearing (3)

Shear flow


Tzy

z

x

x

Ty

S ti id li ti


26/46


Example

Idealized U shape

Booms of 300 mm2- area each

Booms are carrying all the direct stress

Skin panels are carrying all the shear flow

Shear load passes through the shear center

Shear flow?


S

Tz = 4.8 kN

y

h

=0.4m

b =0.2 m

z

y

z

C

A1A2

A3 A4

S ti id li ti


27/46


Shear flow Simple symmetry principal axes

Only booms are carrying direct stress


Shear flow


S

Tz = 4.8 kN

y

h=0.4m

b = 0.2 m

z

y

z

C

A1A2

A3 A4

q



28/46


Comparison with uniform U section We are actually capturing the average value in each branch


S

Tz

y

h

b

z

y

z

C

A1A2

A3 A4

q

q

s

S

Tz

y

t

t

t

h

b

z

y

z

C

q

q

O



29/46


Consequence on closed-thin-walled section shearing

Classical formula

With

And

for anticlockwise q and s

So consequences are the same as for open section




30/46


Example

Idealized wing section

Simply symmetrical

Booms are carrying all the direct stress

Skin panels are carrying all the shear flow

Shear load passes through booms 3 & 6

Shear flow?


A1 = 200mm2

y

z

C y

z

O

A2 = 250 mm2

A5 =A4

bm = 0.24 m

Tz = 10 kN

A3 = 400 mm2

A4 = 100 mm2

A6 =A3A7 =A2

A8 =A1

br= 0.24 mbl = 0.12 m

hl = 0.1 m

hm

=

0.

2m

hr= 0.06 m



31/46


Open part of shear flow

Symmetrical section

Shear center & centroid on Cy axis

Ixy = 0 (we are in the principal axes)

Only booms are carrying direct stress



A1 = 200mm2

y

z

C y

z

O

A2 = 250 mm2

A5 =A4

bm = 0.24 m

Tz = 10 kN

A3 = 400 mm2

A4 = 100 mm2

A6 =A3A7 =A2

A8 =A1

br= 0.24 mbl = 0.12 m

hl = 0.1 m

hm

=

0.

2m

hr= 0.06 m



32/46


Open part of shear flow (2)

Choose (arbitrarily) the origin

between boom 2 and 3


A1 = 200 mm2

y

z

C y

z

O

A2 = 250 mm2

A5 =A4

bm = 0.24 m

Tz = 10 kN

A3 = 400 mm2

A4 = 100 mm2

A6 =A3A7 =A2

A8 =A1

br= 0.24 mbl = 0.12 m

hl = 0.1 m

hm

=

0.

2m

hr= 0.06 m

s0



33/46


Open part of shear flow (3)

Choose (arbitrarily) the origin between boom 2 and 3 (2)


y

z

C y

z

O

2

bm = 0.24 m

-qo=28.9 kN.m-1

3

br= 0.24 mbl = 0.12 m

hl = 0.1 m

hm

=

0.

2m

hr= 0.06 m

4

5

6 7

8

1

-qo=32.5 kN.m-1

qo=18 kN.m-1

qo=22.4 kN.m-1

qo=18 kN.m-1

-qo=28.9 kN.m-1



34/46


Constant part of shear flow

(anticlockwise s, q)

If origin is chosen at point O

With

&


y

z

C y

z

O

2

bm = 0.24 m

-qo=28.9 kN.m-1

3

br= 0.24 mbl = 0.12 m

hl = 0.1 m

hm

=

0.

2m

hr= 0.06 m

4

5

6 7

8

1

-qo=32.5 kN.m-1

qo=18 kN.m-1

qo=22.4 kN.m-1

qo=18 kN.m-1

-qo=28.9 kN.m-1



35/46


Constant part of shear flow (2)


y

z

C y

z

O

2

bm = 0.24 m

-qo=28.9 kN.m-1

3

br= 0.24 mbl = 0.12 m

hl = 0.1 m

hm

=

0.2m

hr= 0.06 m

4

5

6 7

8

1

-qo=32.5 kN.m-1

qo=18 kN.m-1

qo=22.4 kN.m-1

qo=18 kN.m-1

-qo=28.9 kN.m-1



36/46


Total shear flow


y

z

C y

z

O

2

bm = 0.24 m

-qo=28.9 kN.m-1

3

br= 0.24 mbl = 0.12 m

hl = 0.1 m

hm

=

0.

2m

hr= 0.06 m

4

5

6 7

8

1

-qo=32.5 kN.m-1

qo=18 kN.m-1

qo=22.4 kN.m-1

qo=18 kN.m-1

-qo=28.9 kN.m-1

y

z

C y

z

O

2-q=34.2 kN.m-1

3

4

5

6 7

8

1

-q=37.8 kN.m-1

q=12.7 kN.m-1

q=17.1 kN.m-1

q=12.7 kN.m-1

-q=34.2 kN.m-1

-q=5.3 kN.m-1

-q=5.3 kN.

m-1



37/46


Consequence on torsion

If no axial constraint

Torsion analysis does not involve axial stress

So torsion is unaffected by the structural idealization


Exercise: Structural idealization


38/46



Box section

Arrangement of

Direct stress carrying booms positioned at the four corners and

Panels which are assumed to carry only shear stresses

Constant shear modulus

Shear centre?10 mm

10 mm

8 mm

10 mm

300 mm

500 mm

Angles50 x 40 x 8 mm

Angles

60 x 50 x 10 mm

References


39/46


References

Lecture notes

Aircraft Structures for engineering students, T. H. G. Megson, Butterworth-

Heinemann, An imprint of Elsevier Science, 2003, ISBN 0 340 70588 4

Other references

Books

Mcanique des matriaux, C. Massonet & S. Cescotto, De boek Universit, 1994,ISBN 2-8041-2021-X



40/46



As shear center lies on Oy by symmetry we considerTZ Section is required to resist bending moments in a vertical plane

Direct stress at any point is directly proportional to the distance from the

horizontal axis of symmetry, i.e. axis y

The distribution of direct stress in all the panels will be linear so that we can

use the relation below

In addition to contributions from adjacent panels, booms areas include the

existing spar flanges

b

xx1

xx2

A1 A2

y

z

x

21

34

300 mm

500 mm

yS

Tz

yT

z

O



41/46



Booms area

By symmetry

A3 =A2 = 3540 mm2

A4 =A1 = 4000 mm2

10 mm

10 mm

8 mm

10 mm

300 mm

500 mm

Angles50 x 40 x 8 mm

Angles

60 x 50 x 10 mm

21

34

300 mm

500 mm

yS

Tz

yT

z

O



42/46



Shear flow

Booms area

A3 =A2 = 3540 mm2

A4 =A1 = 4000 mm2

By symmetryIyz = 0

As only booms resist direct stress

21

34

300 mm

500 mm

yS

Tz

yT

z

O



43/46


e c se S uc u a dea a o

Open shear flow

21

34

300 mm

500 mm

yS

Tz

yT

z

O>0

s



44/46


Constant shear flow

Load through the shear center

no torsion

With

and

21

34

300 mm

500 mm

yS

Tz

yT

z

O

>0

s

-1.77 x 10-3 Tz 1.57 x 10-3 Tz



45/46


Total shear flow

21

34

300 mm

500 mm

yS

Tz

yT

z

O>0

s

-1.77 x 10-3 Tz 1.57 x 10-3 Tz

21

34

300 mm

500 mm

yS

Tz

yT

z

O>0

-0.034 x 10-3 Tz

-0.034 x 10-3 Tz

1.536 x 10-3 Tz-1.804 x 10-3 Tz



46/46

Shear center

Moment around O

Due to shear flow

Should be balanced by the external loads

21

34

300 mm

500 mm

yS

Tz

yT

z

O>0

-0.034 x 10-3 Tz

-0.034 x 10-3 Tz

1.536 x 10-3 Tz-1.804 x 10-3 Tz

s truc t aero section idealization

Documents