6.3 Snowplough

7/27/2019 6.3 Snowplough

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SUBJECT: File No -A400M/TWEOM1E3/D/57-45-046

A400M Fixed Leading Edge

AAR Snowplough Bracket Author : Date : Page

T Arden June 09 6.3.1-1

6.3 Snowplough Bracket

6.3.1 Introduction

The snow plough bracket is attached to the AAR bracket as shown below

Structure Analysed

Only the connection angles and fasteners are analysed in this report

Loading Also see page 5.7-1

The applied loads to the bracket are determined using forced displacements from the FEM

and pressure applied to the FLE lower panel

The method of analysis is given in Ref [40]

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Nodal Displacements

The following Pages present the nodal displacements given in Ref [41]

The nodes for the analysis are shown pages 6.3.1-2, -3 & -4

Summary

The values in the tables below have been extracted from those values on p. 6.3.2-3

Spanwise Average Displacement Chordwise Average Displacement

Node x disp Node y disp

mm mm

28267 -2.163 bolt 1 28268 0.125

28268 -2.155 28280 0.099

28271 -2.097 28267 0.101

28272 -2.040 28271 0.076

28273 -1.970 See page 6.15 0.100

28274 -1.892 bolt 2 28280 0.099

28280 -2.092 28281 0.050

28281 -2.026 28271 0.076

28282 -1.963 28272 0.023

28283 -1.896 average 0.062

average -2.03 bolt 3 28281 0.050

28282 -0.00228272 0.023

28273 -0.031

average 0.010

bolt 4 28282 -0.002

28283 -0.023

28284 -0.007

28273 -0.031

28274 -0.055

28275 -0.039

average -0.026

x

y

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Node Geometry And Displacements

The following table has been extracted using values in Ref [41] - Attachment 4.3, "rlg_nodes_disp_ult.txt"

Nodes Coord. (X,Y,Z)(1.3) Nodes Displ. (X,Y,Z)(1.3)

28260 14706 -280 -275 28260 -2.2871 0.0887 -2.1599

28261 14675 -280 -275 28261 -2.3477 0.1266 -2.0560

28262 14717 -255 -277 28262 -2.2383 0.1273 -2.1229

28263 14645 -264 -277 28263 -2.3887 0.2006 -1.8791

28264 14699 -257 -277 28264 -2.2803 0.1512 -2.0570

28265 14677 -259 -277 28265 -2.3232 0.1741 -2.0058

28266 14653 -252 -278 28266 -2.3604 0.2171 -1.8736

28267 14749 -256 -277 28267 -2.1631 0.1012 -2.2094

28268 14749 -246 -278 28268 -2.1553 0.1246 -2.159928269 15049 -254 -274 28269 -1.4990 0.0115 -3.5904

28270 15049 -245 -274 28270 -1.5088 0.0367 -3.5098

28271 14779 -255 -276 28271 -2.0967 0.0757 -2.3704

28272 14809 -255 -276 28272 -2.0400 0.0231 -2.5721

28273 14839 -255 -276 28273 -1.9697 -0.0314 -2.7735

28274 14869 -255 -275 28274 -1.8916 -0.0545 -2.9537

28275 14899 -255 -275 28275 -1.8213 -0.0387 -3.0909

28276 14929 -255 -275 28276 -1.7637 -0.0216 -3.1975

28277 14959 -255 -274 28277 -1.7022 -0.0235 -3.3226

28278 14989 -255 -274 28278 -1.6338 -0.0278 -3.4464

28279 15019 -254 -274 28279 -1.5635 -0.0166 -3.5327

28280 14779 -246 -277 28280 -2.0918 0.0986 -2.3243

28281 14809 -246 -277 28281 -2.0264 0.0495 -2.4823

28282 14839 -246 -277 28282 -1.9629 -0.0018 -2.6404

28283 14869 -245 -276 28283 -1.8965 -0.0233 -2.7987

28284 14899 -245 -276 28284 -1.8330 -0.0066 -2.9393

28285 14929 -245 -276 28285 -1.7705 0.0095 -3.0592

28286 14959 -245 -275 28286 -1.7051 0.0069 -3.1865

28287 14989 -245 -275 28287 -1.6387 0.0020 -3.3104

28288 15019 -245 -275 28288 -1.5732 0.0121 -3.4129

28289 15073 -268 -272 28289 -1.4541 0.0047 -3.6249

28290 15073 -251 -273 28290 -1.4580 0.0429 -3.5518

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6.3.2 Portal Frame Analysis (INBD Side Deflection)

Idealisation Ref. Section 3.6

FWD

Up

Outboard

Forward

FWD

Up

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Width of Frame

Pessimistically, assume width of frame = 162.7 mm Ref. 6.3.2-1

Height and Angle of Frame

The frame is idealised on a plane through the hinge point on a line angled through the attachment bolts

The height of the frame is pessimistically assumed as the vertical offset to the first attachment bolt

The sketch below shows the maximum offset (OB side) - the angle is the same both inbd and outbd

84.6 28.6

64.8 22.3

19.8 6.3

q = 72.35 deg

q = -17.65 deg

19.8

6.3

y

zq



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Idealised Frame

d1 = 11 mm

Distance to 1st attachment bolt = 64.8 mm

72.35 deg

d2 = 57.0 mm

162.7mm

64.8m

A B

C D

y'

z'

Inbd Outbd

z

x

q

d1

d2

d

1

sin

8.642 dd =

q



7/106

File No -A400M/TWEOM1E3/D/57-45-046

SUBJECT:

A400M Fixed Leading Edge Author : Date : Page

AAR Snowplough Bracket T Arden June 09 6.3.2-4

Section Constants

Hinge

The hinge does not have the properties of an angle since there is no moment connection at the hinge line

For the section constants use the flat strip that is attached between the connection angles

Ref page 3.3-4

Hinge is a '-12'

thkns, E = 0.109 in average of .115 & .103

thkns, E = 2.77 mm

A = 3.375 inA = 85.73 mm

Hinge leg = A/2 = 42.86 mm

I = b*d^3 / 12 Z = b*d^2 / 6

b = 42.86 mm

t = 2.77 mm

Izz = 75.8 mm^4

Iyy = 18168 mm^4

Zzz = 54.8 mm^3

Zyy = 848 mm^3

A = 119 mm^2

Section Constants in the Y'Z' plane

17.65 deg

Transformation equations can be found in [47], p.800

Note angle is negative in the clockwise direction.

72.35 deg Iy'y' = Izz(sin q) + Iyy(cos q) - Ixy*sin2q

Zy'y' = (2*Izz(sin q)/b) + (2*Iyy(cos q)/b)

q -17.65 deg -0.31 rad

Iy'y' = 16505 mm^4

Zy'y' = 770 mm^3

Product of Inertia value is zero since the reference axes lie on

the lines of symmetry of the cross section.

y y

z

z

yy

z

z

z'

z'

y'

y'

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Iz'z' = Izz(sin2q) + Iyy(cos

2q) - Ixy*sin2q

Zz'z' = (2*Iyy(sin2q)/b) + (2*Izz(cos

2q)/b)

q -17.65 deg -0.31 rad

Iz'z' = 1739 mm^4

Zz'z' = 11923 mm^3

Connection Angles Both angles are similar, ref pages 3.3-5 & 6

t = 2.0 mm Ref. [46]

Width = 22.0 mm

I = b*d^3 / 12 Z = b*d^2 / 6

Ixx = 14.7 mm^4

Iyy = 1775 mm^4Zxx = 14.7 mm^3

Zyy = 161 mm^3

A = 44 mm^2

Product of Inertia value is zero since the reference axes lie on the

lines of symmetry of the cross section.

xx

y

y

22

2



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SUBJECT:



Portal Frame Analysis

Side Load Ref. [45]Table 4 Cases 5f

Geometry parameters Ref pages 6.3.2-3 to -5

L1 = 64.80 mm Ref. p. 6.3.2-3 I1 = 14.7 mm^4 Ref. p. 6.3.2-5

L2 = 64.80 mm Ref. p. 6.3.2-3 I2 = 14.7 mm^4 Ref. p. 6.3.2-5

L3 = 162.7 mm Ref. p. 3.3-1 I3 = 1739.0 mm^4 Ref. p. 6.3.2-5

E1 = 69000 MPa Ref. p. 4.7-1

E2 = 69000 MPa Ref. p. 4.7-2

E3 = 72000 MPa Ref. p. 4.8-1

Loading

Wx = 275.04 N Load to give deflection of 2.03 mm

See page 6.3.2-11

Distance from edge to load a = 64.80 mm

Initially a unit load is applied, then the load required to give the applied deflection of 2.03 mm - Ref page

6.3.1-5

z'

x

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SUBJECT:



Ref. [45]Table 4 Cases 5a & 5f



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SUBJECT:



Constants

These constants are used in the formulas to calculate the reaction moment, horizontaland vertical end reactions, horizontal and vertical deflections and angular rotation at A:

CHH 0.184 mm/N

CVH 0.344 mm/N

CHV 0.344 mm/N

CMH 0.004 /N

CHM 0.004 /N

CVV 1.703 mm/N

CMV 0.011 /N

CVM 0.011 /N

CMM 0.000129 /Nmm

Loading Terms

LFH -12.30 mm

LFV -92.6 mm

LFM -0.57

CHH

L13

3 E1 I1

L13

L1 L2 3

3 E2 I2

L12

L3

E3 I3=

CVH

L2 L3

2 E2 I22 L1 L2

L1 L32

2 E3 I3=

CHV CVH=

CMH

L12

2 E1 I1

L2

2 E2 I22 L1 L2

L1 L3

E3 I3=

CHM CMH=

CVV

L2 L32

E2 I2

L33

3 E3 I3=

CMV

L2 L3

E2 I2

L32

2 E3 I3=

CVM CMV=

CMM

L1

E1 I1

L2

E2 I2

L3

E3 I3=

LFM W CMH a CMMa

2

2 E1 I1

=

LFV W CVH a CVM =

LFH W CHH a CHMa3

6 E1 I1

=



12/106


13/106


SUBJECT:



Calculation of HB, VB and MB from equilibrium equations:

It follows that

HB = 138 N

VB = 55 N

MB = -4471 Nmm

HA = 138 N HB = 138 N

VA = -55 N VB = 55 N

MA = -4471 Nmm MB = -4471 Nmm

Fx HA HB W 0

Fy VA VB 0

MB MB VA L3 HA L1 L2 W a L2 L1 MA 0

HB W HA=

VB V=

MB VA L3 HA L1 L2 W a L2 L1 MA=



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SUBJECT:



Portal Frame deflection

Vertical leg balance

Mint Ref page 6.3.2-6

L1 = 64.8 mm

I1 = 14.7 mm^4

E1 = 69000 MPa

64.8

4471 HA Ref Previous page

138 MA

= -4471 Nmm

MB = -4471 Nmm

55

Mint = HA*L1-MA Mint = 4441 Nmm MD = 4441 Nmm

MC = 4441 Nmm

BM Diagram

4441 4441

4441

4441

-4471 -4471

Frame Deflection

Due to direct load d = 12.30 mm =P*L^3/(3*E*I)

Due to Moment d = 9.19 mm =M*L^2/(2*E*I)

d tot = 3.11 mm Difference

FEM Deflection Ref page 6.3.1-5

The average displacement from the FEM d = 2.03 mm Absolute value shown

Unit load = 1000 N

d unit load = 7.38 mm

Then, the applied load = 275.0 N

CD

BA

CD

BA



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SUBJECT:



Frame Vertical Deflection

Pessimistically assume the frame horizontal member CD is an encastre beam

a Ref. [47] - Table 8.8, Case 1d

162.7 (Ref. p. 3.3-1)

Bolt 1 Bolt 2 Bolt 3 Bolt 4

Unit Load Deflection

Bolt P L3 I3 E3 a b Max of d max

Posn a & b

N mm mm4 MPa mm mm mm mm

1 1000 162.7 75.8 72000 19.84 142.86 142.86 0.70

2 1000 162.7 75.8 72000 61.84 100.86 100.86 3.61

3 1000 162.7 75.8 72000 103.74 58.96 103.74 3.46

4 1000 162.7 75.8 72000 145.74 16.96 145.74 0.53

Ref. p. 3.3-1 p. 6.3.2-4 p. 4.7-1

Applied Loads

The following table gives the applied loads relative to FEM displacements shown page 6.3.1-5

Unit Calculated Applied AppliedLoad d d Load

N mm mm N

1000 0.7 0.100 143.5

1000 3.61 0.062 17.1

1000 3.46 0.010 2.8

1000 0.53 -0.026 -49.4

Sum 114.0

b

C D

Inbd

C D

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SUBJECT:



Loads resolved into the Portal Frame Plane

Py'

a = 17.65

Pz'

Pappl a Py' Pz'

N deg N N

143.5 17.65 136.70 43.50

17.1 17.65 16.30 5.18

2.8 17.65 2.71 0.86

-49.4 17.65 -47.08 -14.98

Ref. p. 6.3.2-4

arm

A & B

C & D



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Vertical Load Ref. [45] Table 4 Case 5a

Geometry parameters Ref pages 6.3.2-3 & -4

L1 = 64.80 mm Ref. p. 6.3.2-3 I1 = 14.7 mm4

Ref. p. 6.3.2-5

L2 = 64.80 mm Ref. p. 6.3.2-3 I2 = 14.7 mm4

Ref. p. 6.3.2-5

L3 = 162.7 mm Ref. p. 3.3-1 I3 = 1739 mm4 Ref. p. 6.3.2-5

E1 = 69000 MPa Ref. p. 4.7-1

E2 = 69000 MPa Ref. p. 4.7-2

E3 = 72000 MPa Ref. p. 4.8-1

LoadingWx = 43.5 N Ref page 6.3.2-13

x Bolt 1 Bolt 2 Bolt 3 Bolt 4

Initial Initial Bolt Pitch X

Posn Posn mm mm

25.16 45 1 0 19.84

25.16 45 2 42 61.84

25.16 45 3 83.9 103.74

25.16 45 4 125.9 145.74


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Ref. [47] Table 4 Cases 5a & 5f




19/106




Constants

These constants are used in the formulas to calculate the reaction moment, horizontal

and vertical end reactions, horizontal and vertical deflections and angular rotation at A:

CHH 0.184 mm/N

CVH 0.344 mm/N

CHV 0.344 mm/N

CMH 0.004 /N

CHM 0.004 /N

CVV 1.703 mm/N

CMV 0.011 /N

CVM 0.011 /N

CMM 0.000129 /Nmm

Loading Terms

LFH 13.09 mm

LFV 65.0 mm

LFM 0.40

CHH

L13

3 E1 I1

L13

L1 L2 3

3 E2 I2

L12

L3

E3 I3=

CVH

L2 L3

2 E2 I22 L1 L2

L1 L32

2 E3 I3=

CHV CVH=

CMH

L12

2 E1 I1

L2

2 E2 I22 L1 L2

L1 L3

E3 I3=

CHM CMH=

CVV

L2 L32

E2 I2

L33

3 E3 I3=

CMV

L2 L3

E2 I2

L32

2 E3 I3=

CVM CMV=

CMM

L1

E1 I1

L2

E2 I2

L3

E3 I3=

LFH WL2

2 E2

I2

2 L1 L2 L3 a L1

2 E3

I3

L3 a 2

=

LFV W CVV a CVMa

3

6 E3 I3

=

LFM WL2

E2 I2L3 a

1

2 E3 I3L3 a

2

=



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Boundary Conditions

Horizontal deflection

Vertical deflection

Angular rotation

Because the above equal zero, the following three equations are solved simultaneously

Chh Chv Chm 0.184 0.344 0.004

C = Cvh Cvv Cvm = 0.344 1.703 0.011

Cmh Cmv Cmm 0.004 0.011 0.000129

Ha

X = Va

Ma

LFhx 13.09

L = LFvx = 65.0

LFmx 0.40

Now calculate the solution as: X = Cinv*L

where Cinv is the inverse of matrix C, calculated below using Excel function MINVERSE

21.71051 0 -710.502

Cinv = 0 1.178627 -95.8813

-710.502 -95.8813 38799.6

X is calculated below using Excel function MMULT

Ha 0 N

So that X = Va = 38 N

Ma 6 Nmm

Notes Build C matrix Build L matrix

Inverse C matrix Multiply inverse C matrix by L matrix

block out matrix size (9x9), block out 3 cells vertically,

=minvers(C matrix) in top left corner =mmult in top cell

select the range containing 'C' select the ranges containing 'Cinv' & 'L'

press cntl+shift+enter press cntrl+shift+enter

CHH HA CHV VA CHM MA LFH

CVH HA CVV VA CVM MA LFV

CMH HA CMV VA CMM MA LFM

HA 0 m=

VA 0 m=

A 0 deg=



21/106






It follows that

HB = 0 N

VB = 5 N

MB = -8 Nmm

HA = -0.34 N HB = 0.34 NVA = 38.20 N VB = 5.29 N

MA = 6.44 Nmm MB = -8.37 Nmm

HB HA=

VB W VA=

MB VA L3 HA L1 L2 W L3 a MA=

Fx HA HB 0

Fy VA VB W 0

MB MB VA L3 HA L1 L2 W L3 a MA 0

DCInbd



22/106





Internal Moment

P = 43.5 N

L1 = 64.8 mm L2 = 64.8 mm

Mint = HA*L1+MA = -15.8 Nmm Mint = HB*L2+MB= 13.8 Nmm

MC = -15.8 Nmm MD = 13.8 Nmm

BM Under Vertical Load

43.519.84 142.86

-0.34 -15.8 13.8 0.34

38.20 5.29

162.7

Check

BM at Load position MP = 742.2 Nmm 742.2 Nmm

C D



23/106








L1 = 64.80 mm Ref. p. 6.3.2-3 I1 = 14.7 mm4

Ref. p. 6.3.2-5

L2 = 64.80 mm Ref. p. 6.3.2-3 I2 = 14.7 mm4

Ref. p. 6.3.2-5

L3 = 162.7 mm Ref. p. 3.3-1 I3 = 1739.0 mm4

Ref. p. 6.3.2-5

E1 = 72000 MPa Ref. p. 4.7-1

E2 = 72000 MPa Ref. p. 4.7-2

E3 = 72000 MPa Ref. p. 4.8-1

LoadingWx = 5.2 N ref page 6.3.2-13



Posn Posn mm mm

25.16 45 1 0 19.84

25.16 45 2 42 61.84

25.16 45 3 83.9 103.74

25.16 45 4 125.9 145.74




24/106





Roark 6th Table 4 Cases 5a & 5f



25/106





Constants


CHH 0.177 mm/N

CVH 0.330 mm/N

CHV 0.330 mm/N

CMH 0.004 /N

CHM 0.004 /N

CVV 1.632 mm/N

CMV 0.010 /N

CVM 0.010 /N

CMM 0.000124 /Nmm

Loading Terms

LFH 1.05 mm

LFV 5.2 mm

LFM 0.03

CHH

L13

3 E1 I1

L13

L1 L2 3

3 E2 I2

L12

L3

E3 I3=

CVH

L2 L3

2 E2 I22 L1 L2

L1 L32

2 E3 I3=

CHV CVH=

CMH

L12

2 E1 I1

L2

2 E2 I22 L1 L2

L1 L3

E3 I3=

CHM CMH=

CVV

L2 L32

E2 I2

L33

3 E3 I3=

CMV

L2 L3

E2 I2

L32

2 E3 I3=

CVM CMV=

CMM

L1

E1 I1

L2

E2 I2

L3

E3 I3=

LFH WL2

2 E2 I22 L1 L2 L3 a L

1

2 E3 I3L3 a

2

=

LFV W CVV a CVMa

3

6 E3 I3

=

LFM WL2

E2 I2L3 a

1

2 E3 I3L3 a

2

=



26/106





Boundary Conditions


Vertical deflection

Angular rotation


Chh Chv Chm 0.177 0.330 0.004

C = Cvh Cvv Cvm = 0.330 1.632 0.010

Cmh Cmv Cmm 0.004 0.010 0.000124

Ha

X = Va

Ma

LFhx 1.05

L = LFvx = 5.2

LFmx 0.03



22.625908 -2.02E-15 -740.7772

Cinv = -2.03E-15 1.2296915 -100.0354

-740.7772 -100.0354 40472


Ha 0 N

So that X = Va = 3 NMa 2 Nmm










HA 0 m=

VA 0 m=

A 0 deg=



27/106


28/106





Internal Moment

P = 5.2 N

L1 = 64.8 mm L2 = 64.8 mm


MC = -4.1 Nmm MD = 4.0 Nmm


5.2

61.84 100.86

-0.09 -4.13 3.96 0.09

3.22 1.97

162.7

Check


C D



29/106

6.3.2-26




L1 = 64.80 mm Ref. p. 6.3.2-3 I1 = 14.7 mm4

Ref. p. 6.3.2-5

L2 = 64.80 mm Ref. p. 6.3.2-3 I2 = 14.7 mm Ref. p. 6.3.2-5

L3 = 162.7 mm Ref. p. 3.3-1 I3 = 1739 mm4

Ref. p. 6.3.2-5

E1 = 69000 MPa Ref. p. 4.7-1

E2 = 69000 MPa Ref. p. 4.7-2

E3 = 72000 MPa Ref. p. 4.8-1

Loading

Wx = 0.9 N ref page 6.3.2-13



Posn Posn mm mm

25.16 45 1 0 19.84

25.16 45 2 42 61.84

25.16 45 3 83.9 103.74

25.16 45 4 125.9 145.74


SUBJECT:A400M Fixed Leading EdgeAAR Snowplough Bracket

Author :T Arden

Date :June 09

Page



30/106

6.3.2-27


SUBJECT:A400M Fixed Leading EdgeAAR Snowplough Bracket Author :

T ArdenDate :

June 09

Page

File No - A400M/TWEOM1E3/D/57-45-046



31/106

6.3.2-28

Constants



CHH 0.184 mm/N

CVH 0.344 mm/N

CHV 0.344 mm/N

CMH 0.004 /N

CHM 0.004 /N

CVV 1.703 mm/N

CMV 0.011 /N

CVM 0.011 /N

CMM 0.000129 /Nmm

Loading Terms

LFH 0.11 mm

LFV 0.5 mm

LFM 0.00

CHH

L13

3 E1 I1

L13

L1 L2 3

3 E2 I2

L12

L3

E3 I3=

CVH

L2 L3

2 E2 I22 L1 L2

L1 L32

2 E3 I3=

CHV CVH=

CMH

L12

2 E1 I1

L2

2 E2 I22 L1 L2

L1 L3

E3 I3=

CHM CM=

CVV L2 L3

2

E2 I2L3

3

3 E3 I3=

CMV

L2 L3

E2 I2

L32

2 E3 I3=

CVM CM=

CMM

L1

E1 I1

L2

E2 I2

L3

E3 I3=


T ArdenDate :

June 09

Page

LFH WL2

2 E2 I22 L1 L2 L3 a

L1

2 E3 I3L3 a

2

=

LFV W CVV a CVMa

3

6 E3 I3

=

LFM WL2

E2 I2L3 a

1

2 E3 I3L3 a

2

=

File No-A400M/TWEOM1E3/D/57-45-046



32/106

6.3.2-29

Boundary Conditions


Vertical deflection

Angular rotation


Chh Chv Chm 0.184 0.344 0.004

C = Cvh Cvv Cvm = 0.344 1.703 0.011

Cmh Cmv Cmm 0.004 0.011 0.000129

Ha

X = Va

Ma

LFhx 0.11

L = LFvx = 0.5

LFmx 0.00



21.71051 0 -710.502

Cinv = 0 1.1786275 -95.8813

-710.5022 -95.88134 38799.57


Ha 0 N


Ma 0 Nmm







CHH HA CHV VA CHM MA LF



HA 0 m=

VA 0 m=

A 0 deg=


T ArdenDate :

June 09

Page




33/106

6.3.2-30


It follows that

HB = 0 N

VB = 1 N

MB = 0 Nmm

HA = -0.01 N HB = 0.01 N

VA = 0.31 N VB = 0.55 N

MA = 0.33 Nmm MB = -0.30 Nmm

HB H=

VB W V=


Fx HA HB 0

Fy VA VB W 0



Author :T Arden

Date :June 09

Page

InbdDC




34/106

6.3.2-31

Internal Moment

P = 0.9 N

L1 = 64.8 mm L2 = 64.8 mm


MC = -0.6 Nmm MD = 0.6 Nmm


0.9

103.74 58.96

-0.01 -0.62 0.65 0.01

0.31 0.55

162.7

Check



Author :T Arden

Date :June 09

Page

C D




35/106

6.3.2-32


Vertical Load Ref [45] Table 4 Case 5a


L1 = 64.80 mm Ref. p. 6.3.2-3 I1 = 14.7 mm4

Ref. p. 6.3.2-5

L2 = 64.80 mm Ref. p. 6.3.2-3 I2 = 14.7 mm4

Ref. p. 6.3.2-5

L3 = 162.7 mm Ref. p. 3.3-1 I3 = 1739.0 mm4 Ref. p. 6.3.2-5

E1 = 69000 MPa Ref. p. 4.7-1

E2 = 69000 MPa Ref. p. 4.7-2

E3 = 72000 MPa Ref. p. 4.8-1

LoadingWx = -14.98 N ref page 6.3.2-13



Posn Posn mm mm

25.16 45 1 0 19.84

25.16 45 2 42 61.84

25.16 45 3 83.9 103.74

25.16 45 4 125.9 145.74


SUBJECT:A400M Fixed Leading Edge

AAR Snowplough BracketAuthor :

T ArdenDate :

June 09

Page




36/106

6.3.2-33


SUBJECT:

A400M Fixed Leading EdgeAAR Snowplough Bracket Author :

T ArdenDate :

June 09

Page




37/106

6.3.2-34

Constants



CHH 0.184 mm/N

CVH 0.344 mm/N

CHV 0.344 mm/N

CMH 0.004 /N

CHM 0.004 /N

CVV 1.703 mm/N

CMV 0.011 /N

CVM 0.011 /N

CMM 0.000129 /Nmm

Loading Terms

LFH -0.53 mm

LFV -2.6 mm

LFM -0.02

CHH

L13

3 E1 I1

L13

L1 L2 3

3 E2 I2

L12

L3

E3 I3=

CVH

L2 L3

2 E2 I22 L1 L2

L1 L32

2 E3 I3=

CHV CVH=

CMH

L12

2 E1 I1

L2

2 E2 I22 L1 L2

L1 L3

E3 I3=

CHM CMH=

CVV

L2 L32

E2 I2

L33

3 E3 I3

=

CMV

L2 L3

E2 I2

L32

2 E3 I3=

CVM CMV=

CMM

L1

E1 I1

L2

E2 I2

L3

E3 I3=


AAR Snowplough BracketAuthor :T Arden

Date :June 09

Page

LFH WL2

2 E2 I22 L1 L2 L3 a

L1

2 E3 I3L3 a

2

=

LFV W CVV a CVMa

3

6 E3 I3

=

LFM WL2

E2 I2L3 a

1

2 E3 I3L3 a

2

=




38/106

6.3.2-35

Boundary Conditions


Vertical deflection

Angular rotation


Chh Chv Chm 0.184 0.344 0.004

C = Cvh Cvv Cvm = 0.344 1.703 0.011

Cmh Cmv Cmm 0.004 0.011 0.000129

Ha

X = Va

Ma

LFhx -0.53

L = LFvx = -2.6LFmx -0.02



21.71051 0 -710.5022

Cinv = 0 1.1786275 -95.88134

-710.5022 -95.88134 38799.57


Ha 0 N

So that X = Va = -2 N

Ma -3 Nmm










HA 0 m=

VA 0 m=

A 0 deg=

SUBJECT:

A400M Fixed Leading EdgeAAR Snowplough Bracket

Author :T Arden

Date :June 09

Page




39/106

6.3.2-36


It follows that

HB = 0 N

VB = -13 N

MB = 2 Nmm

HA = 0.10 N HB = -0.10 N

VA = -1.56 N VB = -13.42 N

MA = -2.53 Nmm MB = 1.92 Nmm

HB HA=

VB W VA=


Fx HA HB 0

Fy VA VB W 0


SUBJECT:


Author :

T ArdenDate :

June 09

Page

InbdD

C




40/106

6.3.2-37

Internal Moment

P = -15.0 N

L1 = 64.8 mm L2 = 64.8 mm

Mint = HA*L1+MA = 4.1 Nmm Mint = HB*L2+MB= -4.8 Nmm

MC = 4.1 Nmm MD = -4.8 Nmm


-15.0145.74 16.96

0.10 4.14 -4.75 -0.10

-1.56 -13.42

162.7

Check

BM at Load position MP = -222.9 Nmm -222.9 Nmm



T ArdenDate :

June 09

Page

C D




41/106

6.3.2-38


Vertical Load Ref [45] Table 4 Case 5a


L1 = 64.80 mm Ref. p. 6.3.2-3 I1 = 14.7 mm^4 Ref. p. 6.3.2-5

L2 = 64.80 mm Ref. p. 6.3.2-3 I2 = 14.7 mm^4 Ref. p. 6.3.2-5

L3 = 162.7 mm Ref. p. 3.3-1 I3 = 1739.0 mm^4 Ref. p. 6.3.2-5

E1 = 69000 MPa Ref. p. 4.7-1E2 = 69000 MPa Ref. p. 4.7-2

E3 = 72000 MPa Ref. p. 4.8-1

Loading Applied Aero Loading:-

Reference:- File:-

Pressure for all 176 cases-new case files-LE Fix for Sonaca.xls

From Report:-

M57RP0404691_v2 Target Wingbox & FLE Air Pressures.doc

For point 661, the nearest location to the centre of the access panel 4 held by the snowplough,for case CMR#27500F-TLL2-L, the Largest Pressure value from either Left or Right Hand Wing

Max Pressure = 89196 Pa

Max Pressure = 0.089 MPa

Total load on Snowplough side = 438.22 N Ref AAR_Pod_Snowplough_Calcs_v5.xmcd

Frame width = 162.7 mm

Running Load, wa = 2.69 N/mm


Author :T Arden

Date :June 09

Page


160429985.xls.ms_office 6.3.2 Frame Vert Distr Load Printed 7/27/2013


42/106

6.3.2-39

Applied load in Portal Frame plane

a =

Pz'

Distr a

Distr Load

Resolving Load into Protal frame plane


Resolved Running Load, wa = 2.69 N/mm = wa/cos a


Author :T Arden

Date :June 09

Page

A & B

C & D




43/106

6.3.2-40



T ArdenDate :

June 09

Page




44/106

6.3.2-41

Constants



CHH 0.184 mm/N

CVH 0.344 mm/N

CHV

0.344 mm/N

CMH 0.004 /N

CHM 0.004 /N

CVV 1.703 mm/N

CMV 0.011 /N

CVM 0.011 /N

CMM 0.000129 /Nmm

Loading Terms Note Wa = Wb, then Wa-Wb = 0

LFH 74.79 mm

LFV 372.4 mm^2

LFM 2.29 mm

CHH

L13

3 E1 I1

L13

L1 L2 3

3 E2 I2

L12

L3

E3 I3=

CVH

L2 L3

2 E2 I22 L1 L2

L1 L32

2 E3 I3=

CHV

CVH

=

CMH

L12

2 E1 I1

L2

2 E2 I22 L1 L2

L1 L3

E3 I3=

CHM CMH=

CVV

L2 L32

E2 I2

L33

3 E3 I3=

CMVL2 L3E2 I2

L32

2 E3 I3=

CVM CM=

CMML1

E1 I1

L2

E2 I2

L3

E3 I3=


Author :T Arden

Date :June 09

Page




45/106

6.3.2-42

Boundary Conditions


Vertical deflection

Angular rotation


Chh Chv Chm 0.184 0.344 0.004

C = Cvh Cvv Cvm = 0.344 1.703 0.011

Cmh Cmv Cmm 0.004 0.011 0.000129

Ha

X = Va

Ma

LFhx 74.79

L = LFvx = 372.4

LFmx 2.29



21.71051 0 -710.5022

Cinv = 0 1.1786275 -95.88134

-710.5022 -95.88134 38799.57


Ha -5 N


Ma 116 Nmm










HA 0 m=

VA 0 m=

A 0 deg=


Author :T Arden

Date :June 09

Page




46/106

6.3.2-43


It follows that Note Wa = Wb

HB = 5 N

VB = 219 N

MB = -116 Nmm

HA = -5 N HB = 5 N

VA = 219 N VB = 219 N

MA = 116 Nmm MB = -116 Nmm

HB H=


Author :T Arden

Date :June 09

Page

InbdD

C




47/106

6.3.2-44

Internal Moment

P = 438.2 N

L1 = 64.8 mm L2 = 64.8 mm


MC = -232.3 Nmm MD = 232.3 Nmm


-5.38 -232.25 232.25 5.38

219.11 219.11

162.7

C D


Author :T Arden

Date :June 09

Page




48/106

6.3.3-1

6.3.3 Portal Frame Summary

Reactions in the XZ' plane

Constraints in the XZ' plane

At A & B

Load HA VA MA HB VB MB Ref

Posn Page

N N Nmm N N Nmm

Side Load 137.5 -48 -3935.1 137.5 48.0 -3935.1 6.3.2-10 & -11

Bolt 1 -0.5 38 7.6 0.5 5.3 -9.8 6.3.2-18

Bolt 2 -0.1 3 2.2 0.1 2.0 -2.4 6.3.2-24

Bolt 3 -0.02 0 0.39 0.02 0.55 -0.35 6.3.2-30

Bolt 4 0.1 -2 -3.0 -0.1 -13.4 2.3 6.3.2-36

Distr Load -7.2 219 136.5 7.2 219.1 -136.5 6.3.2-43Sum 130 211 -3791 145 261 -4082

Internal Moments At C & D

Load MC MD MP Ref

Posn Page

Nmm Nmm Nmm

Side Load 3904 3904 3904 6.3.2-11

Load MC MD MP

Posn

Nmm Nmm Nmm

Bolt 1 -19 16 739 6.3.2-19

Bolt 2 -5 4 194 6.3.2-25

Bolt 3 -1 1 32 6.3.2-31

Bolt 4 5 -6 -222 6.3.2-37

Sum -19 16 743

Load MC MD MP Ref

Posn Page

Nmm Nmm Nmm

Distr Load -273 273 273 6.3.2-44


Author :T Arden

Date :June 09

Page

A B

C D

z'

Inbd Outbd

xHB

VBVA

HA

MAMB


160429985.xls.ms_office 6.3.3 Summary Printed 7/27/2013


49/106

6.3.3-2

Reactions in the Z'Y' plane

The moments below cause a torsion on the connection angles

Py'

Fixing Moments Moments due to Py'

Member CD in the XY' Plane

Below, A=C & B = D

Bolt Py' L3 a Rc Mc RD MD

Posn

N mm mm N Nmm N Nmm

1 137 162.7 19.84 131 2091 6 290

2 16 162.7 61.84 11 387 5 237

3 3 162.7 103.74 1 37 2 65

4 -47 162.7 145.74 -1 -75 -46 -641

Sum 109 142 2441 -33 -48

Maximum Total Ry = 142 N Max Abs of Rc & RD

Maximum Torsion, Mz = 2441 Nmm

McMd


Author :T Arden

Date :June 09

Page

Ry Ry

Py'

arm

A & B

C & D

y'

x


160429985.xls.ms_office 6.3.3 Summary Printed 7/27/2013


50/106

6.3.4-1

6.3.4 Connection Angles Both angles are similar, ref pages 3.6-1 & -2

t = 2.0 mm

Width = 22 mm

Hole dia = 4.83 mm

Net Width = 17.17 mm

Bending about yy

About yy - bending about the edge of the AAR bracket flange

Use the Gross Section

I = b*d^3 / 12 Z = b*d^2 / 6

Iyy = 14.7 mm4

Gross

Ixx = 1775 mm4

Gross

Zyy = 14.7 mm3

Gross

Zxx = 161.3 mm3

Gross

A = 44.0 mm2

Gross

MB = Myy = 4082 Nmm From Portal Frame reactions, page 6.3.3-1

Zyy = 14.7 mm3

Bending sy= 278 MPa =M/Z

End Load VA =Pel = 261 N From Portal Frame reactions, page 6.3.3-1

End Load sel= 5.94 MPa =P/A

Total sy + sel = 284 MPa

SUBJECT:


Author : Date :Page

22

2

y y

x

x

22

2

Snowplough deflects

Outboard for thisanalysis


160429985.xls.ms_office 6.3.4 Stress Levels Printed 7/27/2013


51/106

6.3.4-2

Bending about xx

For this section, through the fastner hole, use NET Sections

I = b*d^3 / 12 Z = b*d^2 / 6

Iyy = 11.4 mm^4 Net

Ixx = 844 mm^4 Net

Zyy = 11.4 mm^3 Net

Zxx = 98.3 mm^3 Net

A = 34.3 mm^2 Net

Maximum Total Ry = 142 N Ref Rc page 6.3.3-2

Py'

57.00

Ry'

Mx = 8066 Nmm = Py'*arm

Zxx = 98.3 mm^3 Previous Page

sx= 82 MPa =M/Z

End Load VA =Pel = 261 N Previous Page

End Load sel= 7.61 MPa =P/A

sy + sel = 284 MPa Previous page

s total = 366 MPa Combination is Pressimistic

Ftu = 425 MPa Ref Section 4 RF = 1.16


T ArdenDate :

June 09

Page

y'

z'

Mx




52/106

6.3.4-3

Shear

Torsional t = 2.0 mm

Width = 17.17 mm

Maximum Torsion, Mz = 2441 Nmm Ref Mc page 6.3.3-2

t tors= 114 MPa

Direct Ry' = 142 N

Rx = 145 N = max H, page

Resultant direct shear = 203 N

Area = w*t = 34.34 mm^2

t direct= 5.9 MPa

t tot= 120 MPa

Ftu = 425 MPa Previous Page

Fsu = 245 MPa =Ftu/Sqrt(3) RF = >2

Combined Shear & Bending

Maximum Principle Stress

s princ = 402 MPa

Ftu = 425 MPa RF = 1.06

Ref Roark 6th Table 20 Case 18

SUBJECT:


Author : Date :Page


t

sss

ss

= 2xy

yx

2

yx

2,142



53/106

6.3.5-1

6.3.5 Fastener Analysis

Loading

Maximum Loading on one leg

Shear Loads in the ZY Plane

Mx = 8066 Nmm Ref Page 6.3.4-2

Hy = 142 N Ref Page 6.3.3-2

V max = 261 N Max of VA & VB - ref page

Pitch Per fastener

20.78

Ry max = 530 N Mx/pitch + Hy

Rz max = 131 N Vmax/2

R res = sqrt (Ry 2+Rz^2)

R res = 546

Ref AP6-020-00 Table 12 - 2024-T351 3B Protruding Head Bolts

3/16" in t = 2.0 mm P allow = 7520 N RF = >2

Bolt Shear


Author : Date :Page

Hy

V

Mx

y'

z'

A B

C D

z'

Inbd Outbd

xHB

VBVA

HA

MAMB

Hy

V

Mx


160429985.xls.ms_office 6.3.5 Fasteners Printed 7/27/2013


54/106

6.3.5-2

Tension Loads in the XZ Plane

My = 4082 Nmm Ref Page 6.3.4-1

Hx = 145 N Max (HA ,HB) Ref Page 6.3.3-2

Per fastener

Pitch Rx max = 342 N My/pitch + Hx

20.78

R tens= 342 N

Ref AP6-020-00 Table 13

Minimum 3/16" allowable, P allow = 7110 N RF = >2

Bolt Tension

Shear & Tension Interaction

RF = PBI


T ArdenDate :

June 09

Page

Hx

My


160429985.xls.ms_office 6.3.5 Fasteners Printed 7/27/2013


55/106

6.3.6-1

6.3.6 Portal Frame Analysis (OTBD Side Deflection)

Idealisation

Ref. Section 3.6


Author : Date :Page

y

z




56/106

6.3.6-2

Width of Frame

Assume width of frame = 162.7 mm Ref. 6.3.6-1

Height and Angle of Frame

The frame is idealised on a plane through the hinge point on a line angled through the attachment bolts

The height of the frame is pessimistically assumed as the vertical offset to the first attachment bolt

The sketch below shows the maximum offset (OB side) - the angle is the same both inbd and outbd

84.6 28.6

64.8 22.3

19.8 6.3

q = 72.35 deg

q= 17.65 deg

Idealised Frame

Offset to 1st attachment bolt = 64.8 mm

72.35 deg


Author : Date :Page

19.8

6.3

y

zq

162.7mm

64.8mm

A B

C D

y'

z'

Inbd Outbd

z

x




57/106

6.3.6-3

Section Constants

Hinge

The hinge does not have the properties of an angle since there is no moment connection at the hinge line

For the section constants use the flat strip the is attached between the connection angles

Ref page 3.3-5

Hinge is a '-12'

thkns, E = 0.109 in average of .115 & .103

thkns, E = 2.77 mm

A = 3.375 in

A = 85.73 mm

Hinge leg = A/2 = 42.86 mm

I = b*d^3 / 12 Z = b*d^2 / 6

b = 42.86 mm

t = 2.77 mm

Ixx = 75.8 mm^4

Iyy = 18168 mm^4

Zxx = 54.8 mm^3

Zyy = 848 mm^3A = 119 mm^2

Section Constants in the Y'Z' plane Ref. page 6.15

Iyy = t*b3*

(sin2q/12 + b*t

3*(cosq)/12

17.65 Zyy = t*b2*

(sinq/6 + b*t2*(cosq)/6

b = 42.86 mm

t = 2.77 mm

72.35 For y'y' q = 72.35 deg

I y'y' = 16521 mm4

Z y'y' = 824 mm3

For z'z', q = 17.65 deg

I z'z' = 1742 mm4

Z z'z' = 309 mm3

A = 119 mm2

SUBJECT:A400M Fixed Leading EdgeButtstraps

Author :T Arden

Date :June 09

Page

y y

z

z

y y

z

z

z'

z'

y'

y'




58/106

6.3.6-4

Connection Angles Both angles are similar, ref pages 3.3-6 & -7

t = 2.0 mm

Width = 22.0 mm

I = b*d^3 / 12 Z = b*d^2 / 6

Ixx = 14.7 mm4

Iyy = 1775 mm4

Zxx = 14.7 mm3

Zyy = 161 mm3

A = 44 mm2

x x

y

y

22

2

SUBJECT:A400M Fixed Leading EdgeButtstraps Author :

T ArdenDate :

June 09

Page




59/106

6.3.6-5


Ref Roarks 6th Ed Table 4

Side Load Ref Roarks 6th Ed Table 4 Case 5f


L1 = 64.80 mm Ref. p. 6.3.2-3 I1 = 14.7 mm4

Ref. p. 6.3.2-5

L2 = 64.80 mm Ref. p. 6.3.2-3 I2 = 14.7 mm

4

Ref. p. 6.3.2-5L3 = 162.7 mm Ref. p. 3.3-1 I3 = 1739.0 mm

4Ref. p. 6.3.2-5

E1 = 69000 MPa Ref. p. 4.7-1

E2 = 69000 MPa Ref. p. 4.7-2

E3 = 72000 MPa Ref. p. 4.8-1

Loading

Initially a unit load is applied, then the load required to give the applied deflection of 2.03 mm - Ref page 6.4

Wx = 187.47 N Load to give deflection of 2.03 mm

See page 6.3.6-10



Author :T Arden

Date :June 09

Page

z'

x




60/106

6.3.6-6



T ArdenDate :

June 09

Page




61/106

6.3.6-7

Constants


CHH 0.184 mm/N

CVH 0.344 mm/N

CHV 0.344 mm/N

CMH 0.004 /N

CHM 0.004 /N

CVV 1.703 mm/N

CMV 0.011 /N

CVM 0.011 /N

CMM 0.000129 /Nmm

Loading Terms

LFH -8.38 mm

LFV -63.1 mm

LFM -0.39

CHHL1

3

3 E1 I1

L13

L1 L2 3

3 E2 I2

L12

L3

E3 I3=

CVH

L2 L3

2 E2 I22 L1 L2

L1 L32

2 E3 I3=

CHV CVH=

CMH

L12

2 E1 I1

L2

2 E2 I22 L1 L2

L1 L3

E3 I3=

CHM CMH=

CVV

L2 L32

E2 I2

L33

3 E3 I3=

CMV

L2 L3

E2 I2

L32

2 E3 I3=

CVM CM=

CMML1

E1 I1

L2

E2 I2

L3

E3 I3=


Author :T Arden

Date :June 09

Page

LFM W CMH a CMMa

2

2 E1 I1

=

LFV W CVH a CVM =

LFH W CHH a CHMa

3

6 E1 I1

=




62/106

6.3.6-8

Boundary Conditions


Vertical deflection

Angular rotation


Chh Chv Chm 0.184 0.344 0.004

C = Cvh Cvv Cvm = 0.344 1.703 0.011

Cmh Cmv Cmm 0.004 0.011 0.000129

Ha

X = Va

Ma

LFhx -8.38

L = LFvx = -63.1

LFmx -0.39



21.71051 0 -710.5022

Cinv = 0 1.1786275 -95.88134

-710.5022 -95.88134 38799.57


Ha 94 N


Ma -3047 Nmm










HA 0 m=

VA 0 m=

A 0 deg=


Author :T Arden

Date :June 09

Page




63/106

6.3.6-9


It follows that

HB = 94 N

VB = 37 N

MB = -3047 Nmm

HA = 94 N HB = 94 N

VA = -37 N VB = 37 N

MA = -3047 Nmm MB = -3047 Nmm

Fx

HA HB W 0

Fy VA VB 0

MB MB VA L3 HA L1 L2 W a L2 L1 MA 0

HB W HA=

VB VA=

MB VA L3 HA L1 L2 W a L2 L1 MA=


T ArdenDate :

June 09

Page




64/106

6.3.6-10

Portal Frame deflection

Vertical leg balance

Mint Ref page 6.3.6-2

L1 = 64.8 mm

I1 = 14.7 mm^4

E1 = 69000 MPa

64.8

3047 HA Ref Previous page

94 MA = -3047 Nmm

MB = -3047 Nmm

37

Mint = HA*L1-MA Mint = 3027 Nmm MD = 3027 Nmm

MC = 3027 Nmm

BM Diagram

3027 3027

3027

3027

-3047 -3047

Frame Deflection

Due to direct load = 8.38 mm =P*L^3/(3*E*I)

Due to Moment = 6.27 mm =M*L^2/(2*E*I)

tot= 2.12 mm Difference

FEM Deflection Ref page 6.3-5

The average displacement form the FEM = 2.03 mm Absolute value shown

Unit load = 1000 N

unit load = 10.82 mm

Then, the applied load = 187.5 N

CD

BA

CD

BA


T ArdenDate :

June 09

Page




65/106



66/106

6.3.6-11

Frame Vertical Deflection

Pessimistically assume the frame horizontal member CD is an encastre beam

a

162.7

Bolt 1 Bolt 2 Bolt 3 Bolt 4

Unit Load Deflection

Bolt P L3 I3 E3 a b Max of max

Posn a & b

N mm mm MPa mm mm mm mm1 1000 162.7 75.8 72000 19.84 142.86 142.86 0.70

2 1000 162.7 75.8 72000 61.84 100.86 100.86 3.61

3 1000 162.7 75.8 72000 103.74 58.96 103.74 3.46

4 1000 162.7 75.8 72000 145.74 16.96 145.74 0.53

Applied Loads

The following table gives the applied loads relative to FEM displacements shown page 6.3-5

Unit Calculated Applied Applied

Load Load

N mm mm N

1000 0.7 0.100 143.5

1000 3.61 0.062 17.1

1000 3.46 0.010 2.8

1000 0.53 -0.026 -49.4

Sum 114.0

For the loads applied in the plane of the portal frame - see next page


T ArdenDate :

June 09

Page

b

C D

Inbd

C D




67/106

6.3.6-12

Loads resolved into the Portal Frame Plane

Py'

a = 17.65

Pz'

Pappl a Py' Pz'

N deg N N

143.5 17.65 136.70 43.50

17.1 17.65 16.30 5.18

2.8 17.65 2.71 0.86

-49.4 17.65 -47.08 -14.98

Ref. p. 6.3.2-4


T ArdenDate :

June 09

Page

arm

A & B

C & D




68/106

6.3.6-13



Vertical Load Ref Roarks 6th Ed Table 4 Case 5a


L1 = 64.80 mm Ref. p. 6.3.2-3 I1 = 14.7 mm^4 Ref. p. 6.3.2-5

L2 = 64.80 mm Ref. p. 6.3.2-3 I2 = 14.7 mm^4 Ref. p. 6.3.2-5

L3 = 162.7 mm Ref. p. 3.3-1 I3 = 1742.5 mm^4 Ref. p. 6.3.2-5

E1 = 72000 MPa Ref. p. 4.7-1

E2 = 72000 MPa Ref. p. 4.7-2

E3 = 72000 MPa Ref. p. 4.8-1




Posn Posn mm mm

25.16 45 1 0 19.84

25.16 45 2 42 61.84

25.16 45 3 83.9 103.74

25.16 45 4 125.9 145.74



Author :T Arden

Date :June 09

Page




69/106

6.3.6-14



Author :T Arden

Date :June 09

Page




70/106

6.3.6-15

Constants



CHH 0.177 mm/N

CVH 0.330 mm/N

CHV

0.330 mm/N

CMH 0.004 /N

CHM 0.004 /N

CVV 1.632 mm/N

CMV 0.010 /N

CVM 0.010 /N

CMM 0.000124 /Nmm

Loading Terms

LFH 12.56 mm

LFV 62.3 mm

LFM 0.38

CHH

L13

3 E1 I1

L13

L1 L2 3

3 E2 I2

L12

L3

E3 I3=

CVH

L2 L3

2 E2 I22 L1 L2

L1 L32

2 E3 I3=

CHV

CVH

=

CMH

L12

2 E1 I1

L2

2 E2 I22 L1 L2

L1 L3

E3 I3=

CHM CMH=

CVV

L2 L32

E2 I2

L33

3 E3 I3=

CMVL2 L3E2 I2

L32

2 E3 I3=

CVM CM=

CMML1

E1 I1

L2

E2 I2

L3

E3 I3=


Author :T Arden

Date :June 09

Page

LFH W

L2

2 E2 I2 2 L1

L2

L3 a

L1

2 E3 I3 L3 a

2

=

LFV W CVV a CVMa

3

6 E3 I3

=

LFM WL2

E2 I2L3 a

1

2 E3 I3L3 a

2

=




71/106

6.3.6-16

Boundary Conditions


Vertical deflection

Angular rotation


Chh Chv Chm 0.177 0.330 0.004

C = Cvh Cvv Cvm = 0.330 1.632 0.010

Cmh Cmv Cmm 0.004 0.010 0.000124

Ha

X = Va

Ma

LFhx 12.56

L = LFvx = 62.3

LFmx 0.38



22.627273 -6.74E-16 -740.8067

Cinv = -2.03E-15 1.2297002 -100.0361

-740.8067 -100.0361 40472.7


Ha 0 N


Ma 7 Nmm










HA 0 m=

VA 0 m=

A 0 deg=


T ArdenDate :

June 09

Page




72/106

6.3.6-17


It follows that

HB = 0 N

VB = 5 N

MB = -9 Nmm

HA = -0.36 N HB = 0.36 N

VA=

38.20 N VB = 5.29 NMA = 6.69 Nmm MB = -8.71 Nmm

HB H=

VB W VA=


Fx HA HB 0

Fy VA VB W 0



T ArdenDate :

June 09

Page

DCInbd




73/106

6.3.6-18

Internal Moment

P = 43.5 N

L1 = 64.8 mm L2 = 64.8 mm


MC = -16.4 Nmm MD = 14.4 Nmm


43.5

19.84 142.86

-0.36 -16.4 14.4 0.36

38.20 5.29

162.7

Check



Author :T Arden

Date :June 09

Page

C D




74/106

6.3.6-19





L1 = 64.80 mm Ref. p. 6.3.2-3 I1 = 14.7 mm^4 Ref. p. 6.3.2-5

L2 = 64.80 mm Ref. p. 6.3.2-3 I2 = 14.7 mm^4 Ref. p. 6.3.2-5

L3 = 162.7 mm Ref. p. 3.3-1 I3 = 1739.0 mm^4 Ref. p. 6.3.2-5

E1 = 69000 MPa Ref. p. 4.7-1

E2 = 69000 MPa Ref. p. 4.7-2

E3 = 72000 MPa Ref. p. 4.8-1




Posn Posn mm mm

25.16 45 1 0 19.84

25.16 45 2 42 61.84

25.16 45 3 83.9 103.74

25.16 45 4 125.9 145.74


SUBJECT:


Author :T Arden

Date :

June 09

Page




75/106

6.3.6-20


SUBJECT:


Author :

T ArdenDate :

June 09

Page




76/106

6.3.6-21

Constants



CHH 0.184 mm/N

CVH 0.344 mm/N

CHV 0.344 mm/N

CMH 0.004 /N

CHM 0.004 /N

CVV 1.703 mm/N

CMV 0.011 /N

CVM 0.011 /N

CMM 0.000129 /Nmm

Loading Terms

LFH 1.10 mm

LFV 5.5 mm

LFM 0.03

CHH

L13

3 E1 I1

L13

L1 L2 3

3 E2 I2

L12

L3

E3 I3=

CVH

L2 L3

2 E2 I22 L1 L2

L1 L32

2 E3 I3=

CHV CVH=

CMH

L12

2 E1 I1

L2

2 E2 I22 L1 L2

L1 L3

E3 I3=

CHM CMH=

CVV

L2 L32

E2 I2

L33

3 E3 I3=

CMV

L2 L3

E2 I2

L32

2 E3 I3

=

CVM CMV=

CMML1

E1 I1

L2

E2 I2

L3

E3 I3=

SUBJECT:


Author :T Arden

Date :

June 09

Page

LFH WL2

2 E2 I22 L1 L2 L3 a

L1

2 E3 I3L3 a

2

=

LFV W CVV a CVMa

3

6 E3 I3

=

LFM WL2

E2 I2L3 a

1

2 E3 I3L3 a

2

=

e o - A400M/TWEOM1E3/D/57-45-046



77/106

6.3.6-22

Boundary Conditions


Vertical deflection

Angular rotation


Chh Chv Chm 0.184 0.344 0.004

C = Cvh Cvv Cvm = 0.344 1.703 0.011

Cmh Cmv Cmm 0.004 0.011 0.000129

Ha

X = Va

Ma

LFhx 1.10

L = LFvx = 5.5

LFmx 0.03



21.71051 0 -710.5022

Cinv = 0 1.1786275 -95.88134

-710.5022 -95.88134 38799.57


Ha 0 N


Ma 2 Nmm










HA 0 m=

VA 0 m=

A 0 deg=

SUBJECT:


T ArdenDate :

June 09

Page

e o - A400M/TWEOM1E3/D/57-45-046



78/106

6.3.6-23


It follows that

HB = 0 N

VB = 2 N

MB = -2 Nmm

HA = -0.09 N HB = 0.09 N

VA = 3.22 N VB = 1.97 N

MA = 1.86 Nmm MB = -2.02 Nmm

HB HA=

VB W VA=


Fx HA HB 0

Fy VA VB W 0



AAR Snowplough Bracket Author :T Arden

Date :June 09

Page

DCInbd




79/106

6.3.6-24

Internal Moment

P = 5.2 N

L1 = 64.8 mm L2 = 64.8 mm


MC = -4.0 Nmm MD = 3.8 Nmm


5.2

61.84 100.86

-0.09 -3.97 3.80 0.09

3.22 1.97

162.7

Check


SUBJECT:


Author :

T ArdenDate :

June 09

Page

C D




80/106


81/106

6.3.6-26


SUBJECT:


T ArdenDate :

June 09

Page




82/106

6.3.6-27

Constants



CHH 0.184 mm/N

CVH 0.344 mm/N

CHV 0.344 mm/N

CMH 0.004 /N

CHM 0.004 /N

CVV 1.703 mm/N

CMV 0.011 /N

CVM 0.011 /N

CMM 0.000129 /Nmm

Loading Terms

LFH 0.11 mm

LFV 0.5 mm

LFM 0.00

CHH

L13

3 E1 I1

L13

L1 L2 3

3 E2 I2

L12

L3

E3 I3=

CVH

L2 L3

2 E2 I22 L1 L2

L1 L32

2 E3 I3=

CHV CVH=

CMH

L12

2 E1 I1

L2

2 E2 I22 L1 L2

L1 L3

E3 I3=

CHM CMH=

CVV

L2 L32

E2 I2

L33

3 E3 I3=

CMV

L2 L3

E2 I2

L32

2 E3 I3=

CVM CMV=

CMML1

E1 I1

L2

E2 I2

L3

E3 I3=



T ArdenDate :

June 09

Page

LFH WL2

2 E2 I22 L1 L2 L3 a

L1

2 E3 I3L3 a

2

=

LFV W CVV a CVMa

3

6 E3 I3

=

LFM WL2

E2 I2L3 a

1

2 E3 I3L3 a

2

=




83/106

6.3.6-28

Boundary Conditions


Vertical deflection

Angular rotation


Chh Chv Chm 0.184 0.344 0.004C = Cvh Cvv Cvm = 0.344 1.703 0.011

Cmh Cmv Cmm 0.004 0.011 0.000129

Ha

X = Va

Ma

LFhx 0.11

L = LFvx = 0.5

LFmx 0.00



21.71051 0 -710.5022

Cinv = 0 1.1786275 -95.88134

-710.5022 -95.88134 38799.57


Ha 0 N


Ma 0 Nmm










HA

0 m=

VA 0 m=

A 0 deg=


AAR Snowplough Bracket Author :

T ArdenDate :

June 09

Page




84/106

6.3.6-29


It follows that

HB = 0 N

VB = 1 N

MB = 0 Nmm

HA = -0.01 N HB = 0.01 N

VA = 0.31 N VB = 0.55 N

MA = 0.33 Nmm MB = -0.30 Nmm

HB HA=

VB W VA=


Fx HA HB 0

Fy VA VB W 0



AAR Snowplough Bracket Author :

T ArdenDate :

June 09

Page

InbdDC




85/106

6.3.6-30

Internal Moment

P = 0.9 N

L1 = 64.8 mm L2 = 64.8 mm


MC = -0.6 Nmm MD = 0.6 Nmm


0.9

103.74 58.96

-0.01 -0.62 0.65 0.01

0.31 0.55

162.7

Check



AAR Snowplough Bracket Author :T Arden

Date :June 09

Page

C D




86/106

6.3.6-31





L1 = 64.80 mm Ref. p. 6.3.2-3 I1 = 14.7 mm^4 Ref. p. 6.3.2-5

L2 = 64.80 mm Ref. p. 6.3.2-3 I2 = 14.7 mm^4 Ref. p. 6.3.2-5

L3 = 162.7 mm Ref. p. 3.3-1 I3 = 1739.0 mm^4 Ref. p. 6.3.2-5

E1 = 69000 MPa Ref. p. 4.7-1

E2 = 69000 MPa Ref. p. 4.7-2

E3 = 72000 MPa Ref. p. 4.8-1

LoadingWx = -14.98 N ref page 6.3.6-12



Posn Posn mm mm

25.16 45 1 0 19.84

25.16 45 2 42 61.84

25.16 45 3 83.9 103.74

25.16 45 4 125.9 145.74



Author :T Arden

Date :June 09

Page




87/106

6.3.6-32



Author :T Arden

Date :June 09

Page




88/106

6.3.6-33

Constants



CHH 0.184 mm/N

CVH 0.344 mm/N

CHV

0.344 mm/N

CMH 0.004 /N

CHM 0.004 /N

CVV 1.703 mm/N

CMV 0.011 /N

CVM 0.011 /N

CMM 0.000129 /Nmm

Loading Terms

LFH -0.53 mm

LFV -2.6 mm

LFM -0.02

CHH

L13

3 E1 I1

L13

L1 L2 3

3 E2 I2

L12

L3

E3 I3=

CVH

L2 L3

2 E2 I22 L1 L2

L1 L32

2 E3 I3=

CHV

CVH

=

CMH

L12

2 E1 I1

L2

2 E2 I22 L1 L2

L1 L3

E3 I3=

CHM CMH=

CVV

L2 L32

E2 I2

L33

3 E3 I3=

CMVL2 L3E2 I2

L32

2 E3 I3=

CVM CM=

CMML1

E1 I1

L2

E2 I2

L3

E3 I3=


Author :T Arden

Date :June 09

Page

LFH W

L2

2 E2 I2 2 L1

L2

L3 a

L1

2 E3 I3 L3 a

2

=

LFV W CVV a CVMa

3

6 E3 I3

=

LFM WL2

E2 I2L3 a

1

2 E3 I3L3 a

2

=




89/106

6.3.6-34

Boundary Conditions


Vertical deflection

Angular rotation


Chh Chv Chm 0.184 0.344 0.004

C = Cvh Cvv Cvm = 0.344 1.703 0.011

Cmh Cmv Cmm 0.004 0.011 0.000129

Ha

X = Va

Ma

LFhx -0.53

L = LFvx = -2.6

LFmx -0.02



21.71051 0 -710.5022

Cinv = 0 1.1786275 -95.88134

-710.5022 -95.88134 38799.57


Ha 0 N


Ma -3 Nmm










HA 0 m=

VA 0 m=

A 0 deg=


Author :T Arden

Date :June 09

Page




90/106

6.3.6-35


It follows that

HB = 0 N

VB = -13 N

MB = 2 Nmm

HA = 0.10 N HB = -0.10 N

VA = -1.56 N VB = -13.42 N

MA = -2.53 Nmm MB = 1.92 Nmm

HB H=

VB W VA=


Fx HA HB 0

Fy VA VB W 0



T ArdenDate :

June 09

Page

InbdD

C




91/106

6.3.6-36

Internal Moment

P = -15.0 N

L1 = 64.8 mm L2 = 64.8 mm

Mint = HA*L1+MA = 4.1 Nmm Mint = HB*L2+MB= -4.8 Nmm

MC = 4.1 Nmm MD = -4.8 Nmm


-15.0

145.74 16.96

0.10 4.14 -4.75 -0.10

-1.56 -13.42

162.7

Check

BM at Load position MP = -222.9 Nmm -222.9 Nmm


Author :T Arden

Date :June 09

Page

C D




92/106

6.3.6-37





L1 = 64.80 mm Ref. p. 6.3.2-3 I1 = 14.7 mm^4 Ref. p. 6.3.2-5

L2 = 64.80 mm Ref. p. 6.3.2-3 I2 = 14.7 mm^4 Ref. p. 6.3.2-5

L3 = 162.7 mm Ref. p. 3.3-1 I3 = 1742.5 mm^4 Ref. p. 6.3.2-5

E1 = 72000 MPa Ref. p. 4.7-1

E2 = 72000 MPa Ref. p. 4.7-2E3 = 72000 MPa Ref. p. 4.8-1

Loading Applied Aero Loading:-

Reference:- File:-

Pressure for all 176 cases-new case files-LE Fix for Sonaca.xls

From Report:-

M57RP0404691_v2 Target Wingbox & FLE Air Pressures.doc

For point 661, the nearest location to the centre of the access panel 4 held by the snowplough,

for case CMR#27500F-TLL2-L, the Largest Pressure value from either Left or Right Hand Wing

Max Pressure = 89196 Pa

Max Pressure = 0.089 MPa

Total load on Snowplough side = 438.22 N Ref AAR_Pod_Snowplough_Calcs_v5.xmcd

Frame width = 162.7 mm



Author :T Arden

Date :June 09

Page




93/106

6.3.6-38

Applied load in Portal Frame plane

a =

Pz'

Distr a

Distr Load

Resolving Load into Protal frame plane


Resolved Running Load, wa = 2.69 N/mm = wa/cos a


Author :T Arden

Date :June 09

Page

A & B

C & D




94/106

6.3.6-39



T ArdenDate :

June 09

Page




95/106

6.3.6-40

Constants



CHH 0.177 mm/N

CVH 0.330 mm/N

CHV

0.330 mm/N

CMH 0.004 /N

CHM 0.004 /N

CVV 1.632 mm/N

CMV 0.010 /N

CVM 0.010 /N

CMM 0.000124 /Nmm

Loading Terms Note Wa = Wb, then Wa-Wb = 0

LFH 71.71 mm

LFV 357.0 mm^2

LFM 2.20 mm

CHH

L13

3 E1 I1

L13

L1 L2 3

3 E2 I2

L12

L3

E3 I3=

CVH

L2 L3

2 E2 I22 L1 L2

L1 L32

2 E3 I3=

CHV

CVH

=

CMH

L12

2 E1 I1

L2

2 E2 I22 L1 L2

L1 L3

E3 I3=

CHM CMH=

CVV

L2 L32

E2 I2

L33

3 E3 I3=

CMVL2 L3E2 I2

L32

2 E3 I3=

CVM CM=

CMML1

E1 I1

L2

E2 I2

L3

E3 I3=


Author :T Arden

Date :June 09

Page




96/106

6.3.6-41

Boundary Conditions


Vertical deflection

Angular rotation


Chh Chv Chm 0.177 0.330 0.004

C = Cvh Cvv Cvm = 0.330 1.632 0.010

Cmh Cmv Cmm 0.004 0.010 0.000124

Ha

X = Va

Ma

LFhx 71.71

L = LFvx = 357.0

LFmx 2.20



22.627273 -6.74E-16 -740.8067

Cinv = -2.03E-15 1.2297002 -100.0361

-740.8067 -100.0361 40472.7


Ha -6 N


Ma 121 Nmm






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6.3 Snowplough