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STRUCTURAL IDEALISATION
BY DR. MAHDI DAMGHANI
1
SUGGESTED READINGS
Chapters 19
of
Aircraft Structural Analysis
2
LEARNING OBJECTIVES
This Lecture (Lecture 6):
◾ Familiarisation with the functions of various structural components in an aircraft;
◾ Familiarisation with simplifying complex wing structure into an idealised structure for stress
analysis purposes;
Next lecture (Lecture 7):
◾ Impact of idealisation on bending of beam cross section;
◾ Impact of idealisation on shear flow and its distribution within the beam cross section;
◾ Impact of idealisation on torsion of beam cross section;
3
NOTE TO THE STUDENTS
◾ You are required to read the following paper before attending this lecture;
◾ Note that this document is uploaded on blackboard;
Odeh Dababneh, Altan Kayran, (2014) "Design, analysis and optimization of thin walled semi-
monocoque wing structures using different structural idealization in the preliminary design
phase", International Journal of Structural Integrity, Vol. 5 Issue: 3, pp.214-226,
https://doi.org/10.1108/IJSI-12-2013-0050
4
OVERVIEW
5
LIFT LOADS
◾ Lift is generated by producing a higher pressure below the wing than above it.
◾ Higher speed airflow above wing than below (streamlines closer together).
Lift
6
• Streamlines around an aerofoil.
(Above)
• Pressure Distribution around an
aerofoil. (Below)
• Resultant lift force (red arrow) acts through
centre of pressure (cop), normal to stream
• cop varies with α(angle of attack)
• Pitching Moment caused by unequal pressure
distribution around aerofoil.
WING STRUCTURE
◾ The Wing
◾ Wingbox
◾ Primary wing structure
◾ Leading edge
◾ Fixed Leading Edge
◾ Slats
◾ Droop Nose
◾ Pylon (Engine) Attachments
◾ Trailing Edge
◾ Fixed Trailing Edge
◾ Spoilers
◾ Ailerons
◾ Flaps
◾ Wing Landing Gear Attachments
◾ Wingtip and Fairings
7
Digital Mock-Up of the A350
Wing
Digital Mock-Up of a Wing-Box
(Upper Cover removed)
WATCH
8
WING-BOX
◾ Carries the main
structural loads from
the wing;
◾ Aerodynamic, inertial,
movables, fuel;
◾ Closed-cell structure
allowing resistance to
shear, torsion and
tension loads;
◾ Location of fuel tanks;
◾ Supports the landing
gear and engines, if
they are wing mounted.
9
Wing in
Plan View
A380 Wingbox
in Production
A380 Wingbox in
Production
WING LOADS
◾ Wing Finite Element Model provides
Shear , Moment and Torsion
10
WING COMPLEXITY (SPARS)
◾ Provide mounting for
WLG Fittings and Leading
and Trailing edge fittings.
11
Track Can
Cut-Outs
Rib Post
Spars
Exploded View of a generic Wing
Digital Mock-Up of Wingbox
(Upper Cover removed)
◾ Span-wise members that
carry shear loads;
◾ Fuel Tank Boundary;
WING COMPLEXITY (SPARS)
◾ For larger aircraft, the spars are usually made
up from multiple sections;
◾ Sections are normally joined together with joint
plates and straps.
12
Web Joint
Plate
Boom
Straps
Spar
Spar
Digital Mock-Up of Wingbox (Upper Cover removed)
LOAD CARRYING MECHANISM OF SPARS
13
WING COMPLEXITY (RIBS)
◾ Castellated edge allows the stringers to pass through rib feet to attached to the skin.
◾ Manholes allow access within the wingbox and movement of fuel.
◾ Stiffeners and crack stoppers are machined or bolted on to increase the strength of the overall
structure.
14
Castellation
Stringer
Manhole
Stiffener
Crack stop
/
per
View inside the Wing Box onto Rib looking
outboard
Computer Rendering of Typical Large Rib
LOAD CARRYING MECHANISM OF RIBS
15
WING COMPLEXITY (SKINS AND STRINGERS)
◾ The skin may be
assembled from multiple
panels. Joints between
skin panels are made at
with
stringer locations and
Butt
reinforced
Straps.
◾ Stringers prevent skin
buckling in compression
16
Stringer
Butt
Strap
Digital Mock-Up of Lower Wing Skins with
Stringers
Skin
Panel
Skin
Panel
Stringer being installed on an A340 Wing
and aid with bending
strength in tension.
◾ The Tension (Lower Cover) – Fatigue
& Damage Tolerance
◾ Compression (Top Cover) - Strength
◾ Stringers are riveted onto the skin or integrally
machined/formed onto the panel.
◾ Access holes on the bottom skin allow entry into the
wing-box for inspection of the internal structure and
cleaning of the fuel tanks.
LOAD CARRYING MECHANISM OF SKINS
17
ROOT JOINT
◾ Where the wing attaches
to the centre wing box in
the fuselage;
◾ The cruciform and tri-form
fittings are used to attach
the upper and lower
covers respectively to the
centre wing box. Patent
application number:
20110089292;
◾ Upper and lower joint
fittings are used to attach
the spars.
18
WING TIP
◾ Rigid structure fixed to end of the wing-box;
◾ The structure is built in a similar way to the wing-box, with spars, ribs, stringers and skin
forming the structure;
◾ Contains the navigation and strobe lights.
19
A320 Wing tip installed on end of Wing
at Broughton
Illustration of Wing Tip Structure
Navigation
Lights
Wing Tip
Fence
Wing
Tip
WINGTIP (WINGLET & WING FENCE)
◾ A winglet or wing fence can be added
to the end of the wingtip to reduce the
induced drag effect of the wing;
◾ A winglet generates more load than a
wing fence but design of the wing and
wingtip considers these extra forces;
◾ “Sharklet” is Airbus’s trade name for
the winglets being added to the A320
family.
20
Wing Tip with blended Winglet installed
Illustration of vortices
created at the Wing Tip
Blended
Winglet
SLATS
◾ Extend out from the leading edge to increase lift and
allow the wing to be flown at a higher angle of attack
(i.e. slower speeds);
◾ Extended by a rack and pinion arrangement.
Protection exists to avoid inadvertent asymmetric
deployment of slats;
◾ Supported by slat tracks which run along a set of
rollers carrying the vertical and side loads;
◾ Slat Cans house the slat tracks when retracted and
act as a fuel boundary;
21
Cross-Sectional diagram of Slat arrangement
Aircraft Wing with Slats deployed
Slats
FLAPS
◾ Extend to increase the
effective wing area and
camber;
◾ This increases wing lift
and also increases drag to
enable a steeper descent
when landing without the
increase in airspeed.
22
Diagram of Flap in un-deployed and deployed states
Flaps
View of wing with flaps and spoilers fully deployed
AILERONS
◾ Controls the roll rate of
the aircraft, but may
also be used for Load
Alleviation Function in
conjunction with the
spoilers;
◾ Larger aircraft may have
more than one aileron
on each wing;
◾ Attached onto trailing
edge ribs aft of the rear
spar;
◾ Mass weights are
usually added forward of
the hinge line to reduce
flutter.
23
Aileron
Static discharger
View on underside of wing, looking up
Exploded view of Aileron composite structure
WING STRUCTURE CATEGORISATION
◾ Category A;
Structure identified as Principal Structural Elements (PSE).
These are the elements that contribute significantly to carrying
flight, ground or pressurisation loads and whose failure could
result in catastrophic failure of the aircraft.(Ref. ACJ 25.571(a) 2.2;
AC 25.571-1C 6d.)
These structures must be replaced or repaired on the discovery of
any damage unless specific rational is provided.
24
◾ Category B;
Structures whose failure or detachment could indirectly
compromise continued safe flight or landing by an adverse effect
on a CategoryA structure.
These structures must be replaced or repaired on the discovery of
any damage unless specific rational is provided.
◾ Category C;
Structures whose failure or detachment will not compromise
continued safe flight or landing but where the potentially large size
of released elements needs to be considered.
As these structures are not identified as either category A or B,
any failure or departure from the aircraft must be demonstrated as
not preventing continued safe flight and landing and the probability
of occurrence is acceptably low (Ref. ACJ 25C- 571 (a)2.1.1e.).
No detachment of structure is allowed (Ref. NPA25C-290).
These structures must be replaced or repaired on discovery of
element failure at an appropriate time.
◾ Category D;
Structures whose failure or detachment has no airworthiness
consequence but only has an economic impact.
These structures may have to be replaced or repaired on
discovery of failure if they form part of the aircraft external envelop.
WATCH
25
WHAT IS THE POINT?
◾ So far, we have been dealing with simple
structural components such as plates and
beams;
◾ Real life structures are complex and in order
to analyse them simplification must be
made;
◾ The behaviour of the simplified/idealised
structure must be as close as that of the real
structure;
◾ Stresses/strains obtained from the idealised
structure are representative of the real
complex structure.
26
Actual cross-section
of a thin-walled beam
Sheet-stringer
idealisation of the
same section
IDEALISATION
27
Stringers and spar flanges
have small cross-sectional
dimensions compared to
the complete section
R
R
e
e
p
p
l
l
a
a
c
c
e
e
w
w
i
i
t
t
h
hb
b
o
o
o
o
m
m
s
s
(concentration of area) at
the midline of skin
Carrying direct
stresses only
Carrying direct and
shear stresses
We will further assume all
direct stresses are taken
by stringers and spar
flanges. Skin takes all the
shear
The variation of stress
(due to bending) over the
cross section is small
PANEL IDEALISATION
◾ We would like to idealise the
panel into the following;
◾ Direct stress carrying booms
◾ Shear stress carrying skins
◾ All direct stresses are given
to booms;
◾ Note that the distribution of
stress has disappeared in
idealised structure though;
◾ As long as we can get the
extremes of stress, it is fine;
◾ What should be the area of
booms?
28
By putting the thickness as zero, i.e.
𝐴 = 0, the direct stress carrying
ability of skin vanishes (𝜎 = 𝑃/𝐴)
◾ For idealisation;
PANEL IDEALISATION
29
M1  M2
Taking moment about
the orange line
tDb
3
b
2
1

b2
M1 
2tD
2

2

1 
2 
Taking moment about
the orange line
M2  B1

1b
Actual thickness of
skin
Direct stress carrying
thickness of skin
NOTES
30
WING IDEALISATION AS DONE IN INDUSTRY (GLOBAL FEM)
31
View from the top
◾ Skin is modelled as shell elements
◾ Stringers as bar elements
◾ Spar caps as bar elements
View from the bottom
GFEM OF A SECTION THROUGH THE WING
32
Upper skin
Lower skin
Stringer
Spar cap
Rib
EXAMPLE
◾ Part of a wing section is in the form of the
two-cell box shown in figure, in which the
vertical spars are connected to the wing
skin through angle sections, all having a
cross-sectional area of 300mm2. Idealise
the section into an arrangement of direct
stress-carrying booms and shear-stress-
only-carrying panels suitable for resisting
bending moments in a vertical plane.
Position the booms at the spar/skin
junctions.
33
SOLUTION
34
The idealised structure with booms modelled as
concentration of area located at the mid-plane of skins
What are the cross section areas?
From symmetry we know 𝐵1 = 𝐵6, 𝐵2 = 𝐵5, 𝐵3 = 𝐵4
Note that 𝜎6/𝜎1 = −1. When 1 is in tension 6 will be in
equal compression and vice versa due to bending
SOLUTION
35
The idealised structure with booms modelled as
concentration of area located at the mid-plane of skins
What are the cross section areas?
From symmetry we know 𝐵1 = 𝐵6, 𝐵2 = 𝐵5, 𝐵3 = 𝐵4
𝑦1/𝑦2 𝑦3/𝑦2
𝑦5/𝑦2
SOLUTION
36
The idealised structure with booms modelled as
concentration of area located at the mid-plane of skins
What are the cross section areas?
From symmetry we know 𝐵1 = 𝐵6, 𝐵2 = 𝐵5, 𝐵3 = 𝐵4
FEA OF ORIGINAL SECTION
37
Load applied
at shear centre
7 ribs equally
spaced
Von Mises
stresses
SECOND MOMENT OF AREA FOR IDEALISED SECTIONS
◾ We have n booms with areas B1, B2, … Bn the second moment of areas are;
◾ The next two examples demonstrate this.
38
EXAMPLE
◾ Construct an idealised cross-section by evaluating the boom areas at points A, B, C, D, E and
F. Use the method based on the equilibrium of bending stresses. Moreover, calculate the
second moment of area of the idealised section.
39
SOLUTION
40
The area of the booms for F=A, E=B and D=C based
on symmetry.
SOLUTION
41
BA
BF
BB
BE
BC
BD
EXAMPLE
42
◾ Construct the idealised section for the following airfoil. It can be assumed that stringers are
spaced at 50 mm interval and 45o in the straight and curved section, respectively. It can be
further assumed that the only loading is Mx. Obtain centroidal location and second moment of
areas for the idealised section. Calculate direct stress in the booms under bending moment of
500,000 N.mm.All dimensions are in millimetres.
y
x
SOLUTION (BOOM CROSS SECTION AREAS)
43
x
y
7
11 9 8
6
2 4 5
1
12
2
21 66.67mm
B6  B7 
6 6
21
1001.6
501.6
3 4 5 8 9 10
6 6
B  B  B  B  B  B 
501.6
2 1
501.6
2 1 80mm2
3
10
SOLUTION (BOOM CROSS SECTION AREAS)
◾ The vertical distance between boom 1 and 2
(blue arrow in opposite figure) is as below;
◾ The length of chord from boom 1 to boom 2
is a quarter of circumference of the semi
circle;
b12  0.25
R 12.5
◾ Therefore, we have;
44
x
y
7
11 9 8
6
2 4 5
1
12
3
10
2
2 1
y  y  50 50
2
14.65mm
B2  B11 
50
50
6
6
501.6







 2 
2  2   75.44mm2
2 1
12.5 2

SOLUTION (BOOM CROSS SECTION AREAS)
◾ The length of chord from boom 1 to boom 12 is
half of the circumference of semi circle as
below;
b112  0.5
R  25
◾ Therefore, we have;
45
B1  B12 
50
2 

  70.87mm2
50 
2 
6
6
25 2





2 
2 1
12.5 2

x
y
7
11 9 8
6
2 4 5
1
12
3
10
SOLUTION (SECOND MOMENT OF AREAS)
◾ Centroidal location can be found by taking the
moment of concentrated areas about 6-7 and 11-7:
◾ Second moment of areas can be readily calculated
as below:
46
x
y
7
11 9 8
4 5 6
1
12
2 3
10
x 
B6 0  B5 50  B4 100  B3 150  B2 200 B1 235.35
B6  B5  B4  B3  B2  B1
x 123.11mm
y  50mm
I  266.67502
380502
 75.44502
 70.8735.352
 2087671mm4
xx
Iyy  266.67123.11 38073.11  75.4476.89 70.87112.24  7264173mm
2 2 2 2 4
Ixy  0
SOLUTION (DIRECT STRESSES)
◾ Normal stress can be calculated as;
◾ At booms 2-6, we have;
◾ At booms 7-11, we have;
◾ At booms 1, we have;
◾ At boom 12, we have;
47
yy xy z
xx
I
M  0, I  0  
Mxx y
208767.1
z
 
500,00050
 11.9MPa
 
500,00050  11.9MPa
208767.1
z
x
y
7
11 9 8
4 5 6
1
12
2 3
10
208767.1
z
 
500,00035.35
 8.46MPa
 
500,00035.35 8.46MPa
208767.1
z
SOLUTION (FEA-DISPLACEMENTS)
◾ In the FEM, an upward 1,000N force is applied at the tip at
the location of the shear centre.
◾ By plotting the displacement
contours in the vertical
direction, i.e. U2, it can be
confirmed that at each section
the vertical displacements are
almost equal suggesting that
the load is applied at the
shear centre with no twist of
the section.
48
SOLUTION (FEA-DIRECT STRESSES)
◾ At a section 500mm from the tip, the
direct stresses are plotted in the
opposite figure.
◾ The direct stress on the upper skin is
-12.13MPa whereas for the bottom
skin it reads as +12.13MPa.
◾ This value is 1.2% more than hand
calculation for the idealised section.
49
SOLUTION (FEA-SHEAR FLOW)
◾ Let’s look at shear flow distribution, i.e.
SF3, in the section.
◾ The shear flow is as the result of shear
force only as the force was applied at the
shear centre meaning no twist of the
section, hence no shear stresses due to
twist.
◾ Pay attention to the location of zero shear
flow and linear distribution of shear flow
in flanges and quadratic in the webs.
50
Quadratic shear
flow distribution
Linear shear
flow distribution
Point with zero
shear flow
TUTORIAL 1
51
◾ Idealise the box section into an arrangement of direct stress-carrying booms positioned at the
four corners and panels which are assumed to carry only shear stresses. Find the centroid
location of the idealised section and then calculate second moment of area for the idealised
section about x and y axis.
SOLUTION
52
1 2
3
4
500 mm
300
mm
6 6
2
B  508 328
50010
2 1
3008
2 1 B  3556mm2
 B
2 3
6 6
1
B  6010  4010 
50010
21
30010
21 B  4000mm2
 B
1 4
SOLUTION
◾ To obtain the second moment of area, it is
essential to find the neutral axis location;
◾ Taking moment about bottom skin, line 43;
4 4
yBi  Bi yi 
i1 i1
◾ Taking moment about spar 14;
4 4
xBi  Bi xi 
i1 i1
53
1 2
3
4
500 mm
300
mm
24000  3556
y 
4000300  3556300
150mm
23556500
24000  3556
 235.3mm
x 
SOLUTION
54
1 2
3
4
500 mm
300
mm
x
y
234.75
150
4
2 2
xx
I  24000150  23556150  340,020,000 mm
yy
 941,238,752.08 mm4
I  24000235.32
 23556500  235.32
 0
Ixy  235.31504000 235.31504000
 500 235.31503556500 235.31503556 No need to calculate
as it is singly
symmetric

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Structural idealisation.pptx

  • 3. LEARNING OBJECTIVES This Lecture (Lecture 6): ◾ Familiarisation with the functions of various structural components in an aircraft; ◾ Familiarisation with simplifying complex wing structure into an idealised structure for stress analysis purposes; Next lecture (Lecture 7): ◾ Impact of idealisation on bending of beam cross section; ◾ Impact of idealisation on shear flow and its distribution within the beam cross section; ◾ Impact of idealisation on torsion of beam cross section; 3
  • 4. NOTE TO THE STUDENTS ◾ You are required to read the following paper before attending this lecture; ◾ Note that this document is uploaded on blackboard; Odeh Dababneh, Altan Kayran, (2014) "Design, analysis and optimization of thin walled semi- monocoque wing structures using different structural idealization in the preliminary design phase", International Journal of Structural Integrity, Vol. 5 Issue: 3, pp.214-226, https://doi.org/10.1108/IJSI-12-2013-0050 4
  • 6. LIFT LOADS ◾ Lift is generated by producing a higher pressure below the wing than above it. ◾ Higher speed airflow above wing than below (streamlines closer together). Lift 6 • Streamlines around an aerofoil. (Above) • Pressure Distribution around an aerofoil. (Below) • Resultant lift force (red arrow) acts through centre of pressure (cop), normal to stream • cop varies with α(angle of attack) • Pitching Moment caused by unequal pressure distribution around aerofoil.
  • 7. WING STRUCTURE ◾ The Wing ◾ Wingbox ◾ Primary wing structure ◾ Leading edge ◾ Fixed Leading Edge ◾ Slats ◾ Droop Nose ◾ Pylon (Engine) Attachments ◾ Trailing Edge ◾ Fixed Trailing Edge ◾ Spoilers ◾ Ailerons ◾ Flaps ◾ Wing Landing Gear Attachments ◾ Wingtip and Fairings 7 Digital Mock-Up of the A350 Wing Digital Mock-Up of a Wing-Box (Upper Cover removed)
  • 9. WING-BOX ◾ Carries the main structural loads from the wing; ◾ Aerodynamic, inertial, movables, fuel; ◾ Closed-cell structure allowing resistance to shear, torsion and tension loads; ◾ Location of fuel tanks; ◾ Supports the landing gear and engines, if they are wing mounted. 9 Wing in Plan View A380 Wingbox in Production A380 Wingbox in Production
  • 10. WING LOADS ◾ Wing Finite Element Model provides Shear , Moment and Torsion 10
  • 11. WING COMPLEXITY (SPARS) ◾ Provide mounting for WLG Fittings and Leading and Trailing edge fittings. 11 Track Can Cut-Outs Rib Post Spars Exploded View of a generic Wing Digital Mock-Up of Wingbox (Upper Cover removed) ◾ Span-wise members that carry shear loads; ◾ Fuel Tank Boundary;
  • 12. WING COMPLEXITY (SPARS) ◾ For larger aircraft, the spars are usually made up from multiple sections; ◾ Sections are normally joined together with joint plates and straps. 12 Web Joint Plate Boom Straps Spar Spar Digital Mock-Up of Wingbox (Upper Cover removed)
  • 13. LOAD CARRYING MECHANISM OF SPARS 13
  • 14. WING COMPLEXITY (RIBS) ◾ Castellated edge allows the stringers to pass through rib feet to attached to the skin. ◾ Manholes allow access within the wingbox and movement of fuel. ◾ Stiffeners and crack stoppers are machined or bolted on to increase the strength of the overall structure. 14 Castellation Stringer Manhole Stiffener Crack stop / per View inside the Wing Box onto Rib looking outboard Computer Rendering of Typical Large Rib
  • 16. WING COMPLEXITY (SKINS AND STRINGERS) ◾ The skin may be assembled from multiple panels. Joints between skin panels are made at with stringer locations and Butt reinforced Straps. ◾ Stringers prevent skin buckling in compression 16 Stringer Butt Strap Digital Mock-Up of Lower Wing Skins with Stringers Skin Panel Skin Panel Stringer being installed on an A340 Wing and aid with bending strength in tension. ◾ The Tension (Lower Cover) – Fatigue & Damage Tolerance ◾ Compression (Top Cover) - Strength ◾ Stringers are riveted onto the skin or integrally machined/formed onto the panel. ◾ Access holes on the bottom skin allow entry into the wing-box for inspection of the internal structure and cleaning of the fuel tanks.
  • 17. LOAD CARRYING MECHANISM OF SKINS 17
  • 18. ROOT JOINT ◾ Where the wing attaches to the centre wing box in the fuselage; ◾ The cruciform and tri-form fittings are used to attach the upper and lower covers respectively to the centre wing box. Patent application number: 20110089292; ◾ Upper and lower joint fittings are used to attach the spars. 18
  • 19. WING TIP ◾ Rigid structure fixed to end of the wing-box; ◾ The structure is built in a similar way to the wing-box, with spars, ribs, stringers and skin forming the structure; ◾ Contains the navigation and strobe lights. 19 A320 Wing tip installed on end of Wing at Broughton Illustration of Wing Tip Structure Navigation Lights Wing Tip Fence Wing Tip
  • 20. WINGTIP (WINGLET & WING FENCE) ◾ A winglet or wing fence can be added to the end of the wingtip to reduce the induced drag effect of the wing; ◾ A winglet generates more load than a wing fence but design of the wing and wingtip considers these extra forces; ◾ “Sharklet” is Airbus’s trade name for the winglets being added to the A320 family. 20 Wing Tip with blended Winglet installed Illustration of vortices created at the Wing Tip Blended Winglet
  • 21. SLATS ◾ Extend out from the leading edge to increase lift and allow the wing to be flown at a higher angle of attack (i.e. slower speeds); ◾ Extended by a rack and pinion arrangement. Protection exists to avoid inadvertent asymmetric deployment of slats; ◾ Supported by slat tracks which run along a set of rollers carrying the vertical and side loads; ◾ Slat Cans house the slat tracks when retracted and act as a fuel boundary; 21 Cross-Sectional diagram of Slat arrangement Aircraft Wing with Slats deployed Slats
  • 22. FLAPS ◾ Extend to increase the effective wing area and camber; ◾ This increases wing lift and also increases drag to enable a steeper descent when landing without the increase in airspeed. 22 Diagram of Flap in un-deployed and deployed states Flaps View of wing with flaps and spoilers fully deployed
  • 23. AILERONS ◾ Controls the roll rate of the aircraft, but may also be used for Load Alleviation Function in conjunction with the spoilers; ◾ Larger aircraft may have more than one aileron on each wing; ◾ Attached onto trailing edge ribs aft of the rear spar; ◾ Mass weights are usually added forward of the hinge line to reduce flutter. 23 Aileron Static discharger View on underside of wing, looking up Exploded view of Aileron composite structure
  • 24. WING STRUCTURE CATEGORISATION ◾ Category A; Structure identified as Principal Structural Elements (PSE). These are the elements that contribute significantly to carrying flight, ground or pressurisation loads and whose failure could result in catastrophic failure of the aircraft.(Ref. ACJ 25.571(a) 2.2; AC 25.571-1C 6d.) These structures must be replaced or repaired on the discovery of any damage unless specific rational is provided. 24 ◾ Category B; Structures whose failure or detachment could indirectly compromise continued safe flight or landing by an adverse effect on a CategoryA structure. These structures must be replaced or repaired on the discovery of any damage unless specific rational is provided. ◾ Category C; Structures whose failure or detachment will not compromise continued safe flight or landing but where the potentially large size of released elements needs to be considered. As these structures are not identified as either category A or B, any failure or departure from the aircraft must be demonstrated as not preventing continued safe flight and landing and the probability of occurrence is acceptably low (Ref. ACJ 25C- 571 (a)2.1.1e.). No detachment of structure is allowed (Ref. NPA25C-290). These structures must be replaced or repaired on discovery of element failure at an appropriate time. ◾ Category D; Structures whose failure or detachment has no airworthiness consequence but only has an economic impact. These structures may have to be replaced or repaired on discovery of failure if they form part of the aircraft external envelop.
  • 26. WHAT IS THE POINT? ◾ So far, we have been dealing with simple structural components such as plates and beams; ◾ Real life structures are complex and in order to analyse them simplification must be made; ◾ The behaviour of the simplified/idealised structure must be as close as that of the real structure; ◾ Stresses/strains obtained from the idealised structure are representative of the real complex structure. 26 Actual cross-section of a thin-walled beam Sheet-stringer idealisation of the same section
  • 27. IDEALISATION 27 Stringers and spar flanges have small cross-sectional dimensions compared to the complete section R R e e p p l l a a c c e e w w i i t t h hb b o o o o m m s s (concentration of area) at the midline of skin Carrying direct stresses only Carrying direct and shear stresses We will further assume all direct stresses are taken by stringers and spar flanges. Skin takes all the shear The variation of stress (due to bending) over the cross section is small
  • 28. PANEL IDEALISATION ◾ We would like to idealise the panel into the following; ◾ Direct stress carrying booms ◾ Shear stress carrying skins ◾ All direct stresses are given to booms; ◾ Note that the distribution of stress has disappeared in idealised structure though; ◾ As long as we can get the extremes of stress, it is fine; ◾ What should be the area of booms? 28 By putting the thickness as zero, i.e. 𝐴 = 0, the direct stress carrying ability of skin vanishes (𝜎 = 𝑃/𝐴)
  • 29. ◾ For idealisation; PANEL IDEALISATION 29 M1  M2 Taking moment about the orange line tDb 3 b 2 1  b2 M1  2tD 2  2  1  2  Taking moment about the orange line M2  B1  1b Actual thickness of skin Direct stress carrying thickness of skin
  • 31. WING IDEALISATION AS DONE IN INDUSTRY (GLOBAL FEM) 31 View from the top ◾ Skin is modelled as shell elements ◾ Stringers as bar elements ◾ Spar caps as bar elements View from the bottom
  • 32. GFEM OF A SECTION THROUGH THE WING 32 Upper skin Lower skin Stringer Spar cap Rib
  • 33. EXAMPLE ◾ Part of a wing section is in the form of the two-cell box shown in figure, in which the vertical spars are connected to the wing skin through angle sections, all having a cross-sectional area of 300mm2. Idealise the section into an arrangement of direct stress-carrying booms and shear-stress- only-carrying panels suitable for resisting bending moments in a vertical plane. Position the booms at the spar/skin junctions. 33
  • 34. SOLUTION 34 The idealised structure with booms modelled as concentration of area located at the mid-plane of skins What are the cross section areas? From symmetry we know 𝐵1 = 𝐵6, 𝐵2 = 𝐵5, 𝐵3 = 𝐵4 Note that 𝜎6/𝜎1 = −1. When 1 is in tension 6 will be in equal compression and vice versa due to bending
  • 35. SOLUTION 35 The idealised structure with booms modelled as concentration of area located at the mid-plane of skins What are the cross section areas? From symmetry we know 𝐵1 = 𝐵6, 𝐵2 = 𝐵5, 𝐵3 = 𝐵4 𝑦1/𝑦2 𝑦3/𝑦2 𝑦5/𝑦2
  • 36. SOLUTION 36 The idealised structure with booms modelled as concentration of area located at the mid-plane of skins What are the cross section areas? From symmetry we know 𝐵1 = 𝐵6, 𝐵2 = 𝐵5, 𝐵3 = 𝐵4
  • 37. FEA OF ORIGINAL SECTION 37 Load applied at shear centre 7 ribs equally spaced Von Mises stresses
  • 38. SECOND MOMENT OF AREA FOR IDEALISED SECTIONS ◾ We have n booms with areas B1, B2, … Bn the second moment of areas are; ◾ The next two examples demonstrate this. 38
  • 39. EXAMPLE ◾ Construct an idealised cross-section by evaluating the boom areas at points A, B, C, D, E and F. Use the method based on the equilibrium of bending stresses. Moreover, calculate the second moment of area of the idealised section. 39
  • 40. SOLUTION 40 The area of the booms for F=A, E=B and D=C based on symmetry.
  • 42. EXAMPLE 42 ◾ Construct the idealised section for the following airfoil. It can be assumed that stringers are spaced at 50 mm interval and 45o in the straight and curved section, respectively. It can be further assumed that the only loading is Mx. Obtain centroidal location and second moment of areas for the idealised section. Calculate direct stress in the booms under bending moment of 500,000 N.mm.All dimensions are in millimetres. y x
  • 43. SOLUTION (BOOM CROSS SECTION AREAS) 43 x y 7 11 9 8 6 2 4 5 1 12 2 21 66.67mm B6  B7  6 6 21 1001.6 501.6 3 4 5 8 9 10 6 6 B  B  B  B  B  B  501.6 2 1 501.6 2 1 80mm2 3 10
  • 44. SOLUTION (BOOM CROSS SECTION AREAS) ◾ The vertical distance between boom 1 and 2 (blue arrow in opposite figure) is as below; ◾ The length of chord from boom 1 to boom 2 is a quarter of circumference of the semi circle; b12  0.25 R 12.5 ◾ Therefore, we have; 44 x y 7 11 9 8 6 2 4 5 1 12 3 10 2 2 1 y  y  50 50 2 14.65mm B2  B11  50 50 6 6 501.6         2  2  2   75.44mm2 2 1 12.5 2 
  • 45. SOLUTION (BOOM CROSS SECTION AREAS) ◾ The length of chord from boom 1 to boom 12 is half of the circumference of semi circle as below; b112  0.5 R  25 ◾ Therefore, we have; 45 B1  B12  50 2     70.87mm2 50  2  6 6 25 2      2  2 1 12.5 2  x y 7 11 9 8 6 2 4 5 1 12 3 10
  • 46. SOLUTION (SECOND MOMENT OF AREAS) ◾ Centroidal location can be found by taking the moment of concentrated areas about 6-7 and 11-7: ◾ Second moment of areas can be readily calculated as below: 46 x y 7 11 9 8 4 5 6 1 12 2 3 10 x  B6 0  B5 50  B4 100  B3 150  B2 200 B1 235.35 B6  B5  B4  B3  B2  B1 x 123.11mm y  50mm I  266.67502 380502  75.44502  70.8735.352  2087671mm4 xx Iyy  266.67123.11 38073.11  75.4476.89 70.87112.24  7264173mm 2 2 2 2 4 Ixy  0
  • 47. SOLUTION (DIRECT STRESSES) ◾ Normal stress can be calculated as; ◾ At booms 2-6, we have; ◾ At booms 7-11, we have; ◾ At booms 1, we have; ◾ At boom 12, we have; 47 yy xy z xx I M  0, I  0   Mxx y 208767.1 z   500,00050  11.9MPa   500,00050  11.9MPa 208767.1 z x y 7 11 9 8 4 5 6 1 12 2 3 10 208767.1 z   500,00035.35  8.46MPa   500,00035.35 8.46MPa 208767.1 z
  • 48. SOLUTION (FEA-DISPLACEMENTS) ◾ In the FEM, an upward 1,000N force is applied at the tip at the location of the shear centre. ◾ By plotting the displacement contours in the vertical direction, i.e. U2, it can be confirmed that at each section the vertical displacements are almost equal suggesting that the load is applied at the shear centre with no twist of the section. 48
  • 49. SOLUTION (FEA-DIRECT STRESSES) ◾ At a section 500mm from the tip, the direct stresses are plotted in the opposite figure. ◾ The direct stress on the upper skin is -12.13MPa whereas for the bottom skin it reads as +12.13MPa. ◾ This value is 1.2% more than hand calculation for the idealised section. 49
  • 50. SOLUTION (FEA-SHEAR FLOW) ◾ Let’s look at shear flow distribution, i.e. SF3, in the section. ◾ The shear flow is as the result of shear force only as the force was applied at the shear centre meaning no twist of the section, hence no shear stresses due to twist. ◾ Pay attention to the location of zero shear flow and linear distribution of shear flow in flanges and quadratic in the webs. 50 Quadratic shear flow distribution Linear shear flow distribution Point with zero shear flow
  • 51. TUTORIAL 1 51 ◾ Idealise the box section into an arrangement of direct stress-carrying booms positioned at the four corners and panels which are assumed to carry only shear stresses. Find the centroid location of the idealised section and then calculate second moment of area for the idealised section about x and y axis.
  • 52. SOLUTION 52 1 2 3 4 500 mm 300 mm 6 6 2 B  508 328 50010 2 1 3008 2 1 B  3556mm2  B 2 3 6 6 1 B  6010  4010  50010 21 30010 21 B  4000mm2  B 1 4
  • 53. SOLUTION ◾ To obtain the second moment of area, it is essential to find the neutral axis location; ◾ Taking moment about bottom skin, line 43; 4 4 yBi  Bi yi  i1 i1 ◾ Taking moment about spar 14; 4 4 xBi  Bi xi  i1 i1 53 1 2 3 4 500 mm 300 mm 24000  3556 y  4000300  3556300 150mm 23556500 24000  3556  235.3mm x 
  • 54. SOLUTION 54 1 2 3 4 500 mm 300 mm x y 234.75 150 4 2 2 xx I  24000150  23556150  340,020,000 mm yy  941,238,752.08 mm4 I  24000235.32  23556500  235.32  0 Ixy  235.31504000 235.31504000  500 235.31503556500 235.31503556 No need to calculate as it is singly symmetric