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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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SUBJECT: File No -A400M/TWEOM1E3/D/57-45-046
A400M Fixed Leading Edge
AAR Snowplough Bracket Author : Date : Page
T Arden June 09 6.3.1-2
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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SUBJECT: File No -A400M/TWEOM1E3/D/57-45-046
A400M Fixed Leading Edge
AAR Snowplough Bracket Author : Date : Page
T Arden June 09 6.3.1-3
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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SUBJECT: File No -A400M/TWEOM1E3/D/57-45-046
A400M Fixed Leading Edge
AAR Snowplough Bracket Author : Date : Page
T Arden June 09 6.3.2-1
6.3.2 Portal Frame Analysis (INBD Side Deflection)
Idealisation Ref. Section 3.6
FWD
Up
Outboard
Forward
FWD
Up
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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 6.3.2-2
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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SUBJECT: File No -A400M/TWEOM1E3/D/57-45-046
A400M Fixed Leading Edge
AAR Snowplough Bracket Author : Date : Page
T Arden June 09 6.3.2-3
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
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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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SUBJECT: File No -A400M/TWEOM1E3/D/57-45-046
A400M Fixed Leading Edge
AAR Snowplough Bracket Author : Date : Page
T Arden June 09 6.3.2-5
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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File No -A400M/TWEOM1E3/D/57-45-046
SUBJECT:
A400M Fixed Leading Edge Author : Date : Page
AAR Snowplough Bracket T Arden June 09 6.3.2-6
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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File No -A400M/TWEOM1E3/D/57-45-046
SUBJECT:
A400M Fixed Leading Edge Author : Date : Page
AAR Snowplough Bracket T Arden June 09 6.3.2-7
Ref. [45]Table 4 Cases 5a & 5f
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File No -A400M/TWEOM1E3/D/57-45-046
SUBJECT:
A400M Fixed Leading Edge Author : Date : Page
AAR Snowplough Bracket T Arden June 09 6.3.2-8
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
=
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File No -A400M/TWEOM1E3/D/57-45-046
SUBJECT:
A400M Fixed Leading Edge Author : Date : Page
AAR Snowplough Bracket T Arden June 09 6.3.2-10
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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File No -A400M/TWEOM1E3/D/57-45-046
SUBJECT:
A400M Fixed Leading Edge Author : Date : Page
AAR Snowplough Bracket T Arden June 09 6.3.2-11
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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File No -A400M/TWEOM1E3/D/57-45-046
SUBJECT:
A400M Fixed Leading Edge Author : Date : Page
AAR Snowplough Bracket T Arden June 09 6.3.2-12
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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File No -A400M/TWEOM1E3/D/57-45-046
SUBJECT:
A400M Fixed Leading Edge Author : Date : Page
AAR Snowplough Bracket T Arden June 09 6.3.2-13
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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SUBJECT: File No -A400M/TWEOM1E3/D/57-45-046
A400M Fixed Leading Edge
AAR Snowplough Bracket Author : Date : Page
T Arden June 09 6.3.2-14
Portal Frame Analysis
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
Distance from edge to load a = 19.84 mm
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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 6.3.2-15
Ref. [47] Table 4 Cases 5a & 5f
SUBJECT: File No -A400M/TWEOM1E3/D/57-45-046
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A400M Fixed Leading Edge
AAR Snowplough Bracket Author : Date : Page
T Arden June 09 6.3.2-16
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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SUBJECT: File No -A400M/TWEOM1E3/D/57-45-046
A400M Fixed Leading Edge
AAR Snowplough Bracket Author : Date : Page
T Arden June 09 6.3.2-17
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=
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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 6.3.2-18
Calculation of HB, VB and MB from equilibrium equations:
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
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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 6.3.2-19
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
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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 6.3.2-20
Portal Frame Analysis
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 = 1739.0 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
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
Distance from edge to load a = 61.84 mm
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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 6.3.2-21
Roark 6th Table 4 Cases 5a & 5f
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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 6.3.2-22
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 0.177 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
=
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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 6.3.2-23
Boundary Conditions
Horizontal deflection
Vertical deflection
Angular rotation
Because the above equal zero, the following three equations are solved simultaneously
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
Now calculate the solution as: X = Cinv*L
where Cinv is the inverse of matrix C, calculated below using Excel function MINVERSE
22.625908 -2.02E-15 -740.7772
Cinv = -2.03E-15 1.2296915 -100.0354
-740.7772 -100.0354 40472
X is calculated below using Excel function MMULT
Ha 0 N
So that X = Va = 3 NMa 2 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=
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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 6.3.2-25
Internal Moment
P = 5.2 N
L1 = 64.8 mm L2 = 64.8 mm
Mint = HA*L1+MA = -4.1 Nmm Mint = HB*L2+MB= 4.0 Nmm
MC = -4.1 Nmm MD = 4.0 Nmm
BM Under Vertical Load
5.2
61.84 100.86
-0.09 -4.13 3.96 0.09
3.22 1.97
162.7
Check
BM at Load position MP = 194.7 Nmm 194.7 Nmm
C D
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6.3.2-26
Portal Frame Analysis
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 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
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
Distance from edge to load a = 103.74 mm
SUBJECT:A400M Fixed Leading EdgeAAR Snowplough Bracket
Author :T Arden
Date :June 09
Page
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6.3.2-27
Roark 6th Table 4 Cases 5a & 5f
SUBJECT:A400M Fixed Leading EdgeAAR Snowplough Bracket Author :
T ArdenDate :
June 09
Page
File No - A400M/TWEOM1E3/D/57-45-046
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6.3.2-28
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 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=
SUBJECT:A400M Fixed Leading EdgeAAR Snowplough Bracket Author :
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
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6.3.2-29
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 0.11
L = LFvx = 0.5
LFmx 0.00
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 1.1786275 -95.8813
-710.5022 -95.88134 38799.57
X is calculated below using Excel function MMULT
Ha 0 N
So that X = Va = 0 N
Ma 0 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 LF
CVH HA CVV VA CVM MA LFV
CMH HA CMV VA CMM MA LFM
HA 0 m=
VA 0 m=
A 0 deg=
SUBJECT:A400M Fixed Leading EdgeAAR Snowplough Bracket Author :
T ArdenDate :
June 09
Page
File No - A400M/TWEOM1E3/D/57-45-046
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6.3.2-30
Calculation of HB, VB and MB from equilibrium equations:
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=
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
SUBJECT:A400M Fixed Leading EdgeAAR Snowplough Bracket
Author :T Arden
Date :June 09
Page
InbdDC
File No - A400M/TWEOM1E3/D/57-45-046
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6.3.2-31
Internal Moment
P = 0.9 N
L1 = 64.8 mm L2 = 64.8 mm
Mint = HA*L1+MA = -0.6 Nmm Mint = HB*L2+MB= 0.6 Nmm
MC = -0.6 Nmm MD = 0.6 Nmm
BM Under Vertical Load
0.9
103.74 58.96
-0.01 -0.62 0.65 0.01
0.31 0.55
162.7
Check
BM at Load position MP = 31.8 Nmm 31.8 Nmm
SUBJECT:A400M Fixed Leading EdgeAAR Snowplough Bracket
Author :T Arden
Date :June 09
Page
C D
File No - A400M/TWEOM1E3/D/57-45-046
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6.3.2-32
Portal Frame Analysis
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 = 1739.0 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
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
Distance from edge to load a = 145.74 mm
SUBJECT:A400M Fixed Leading Edge
AAR Snowplough BracketAuthor :
T ArdenDate :
June 09
Page
File No - A400M/TWEOM1E3/D/57-45-046
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6.3.2-33
Roark 6th Table 4 Cases 5a & 5f
SUBJECT:
A400M Fixed Leading EdgeAAR Snowplough Bracket Author :
T ArdenDate :
June 09
Page
File No - A400M/TWEOM1E3/D/57-45-046
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6.3.2-34
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 -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=
SUBJECT:A400M Fixed Leading Edge
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
=
File No - A400M/TWEOM1E3/D/57-45-046
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6.3.2-35
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 -0.53
L = LFvx = -2.6LFmx -0.02
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 -710.5022
Cinv = 0 1.1786275 -95.88134
-710.5022 -95.88134 38799.57
X is calculated below using Excel function MMULT
Ha 0 N
So that X = Va = -2 N
Ma -3 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=
SUBJECT:
A400M Fixed Leading EdgeAAR Snowplough Bracket
Author :T Arden
Date :June 09
Page
File No - A400M/TWEOM1E3/D/57-45-046
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6.3.2-36
Calculation of HB, VB and MB from equilibrium equations:
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=
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
SUBJECT:
A400M Fixed Leading EdgeAAR Snowplough Bracket
Author :
T ArdenDate :
June 09
Page
InbdD
C
File No - A400M/TWEOM1E3/D/57-45-046
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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
BM Under Vertical Load
-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
SUBJECT:A400M Fixed Leading Edge
AAR Snowplough BracketAuthor :
T ArdenDate :
June 09
Page
C D
File No - A400M/TWEOM1E3/D/57-45-046
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6.3.2-38
Portal Frame Analysis
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 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
SUBJECT:A400M Fixed Leading EdgeAAR Snowplough Bracket
Author :T Arden
Date :June 09
Page
File No - A400M/TWEOM1E3/D/57-45-046
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6.3.2-39
Applied load in Portal Frame plane
a =
Pz'
Distr a
Distr Load
Resolving Load into Protal frame plane
Running Load, wa = 2.69 N/mm
Resolved Running Load, wa = 2.69 N/mm = wa/cos a
SUBJECT:A400M Fixed Leading EdgeAAR Snowplough Bracket
Author :T Arden
Date :June 09
Page
A & B
C & D
File No - A400M/TWEOM1E3/D/57-45-046
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6.3.2-40
Roark 6th Table 4 Cases 5a & 5f
SUBJECT:A400M Fixed Leading EdgeAAR Snowplough Bracket Author :
T ArdenDate :
June 09
Page
File No - A400M/TWEOM1E3/D/57-45-046
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6.3.2-41
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 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=
SUBJECT:A400M Fixed Leading EdgeAAR Snowplough Bracket
Author :T Arden
Date :June 09
Page
File No - A400M/TWEOM1E3/D/57-45-046
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6.3.2-42
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 74.79
L = LFvx = 372.4
LFmx 2.29
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 -710.5022
Cinv = 0 1.1786275 -95.88134
-710.5022 -95.88134 38799.57
X is calculated below using Excel function MMULT
Ha -5 N
So that X = Va = 219 N
Ma 116 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=
SUBJECT:A400M Fixed Leading EdgeAAR Snowplough Bracket
Author :T Arden
Date :June 09
Page
File No - A400M/TWEOM1E3/D/57-45-046
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6.3.2-43
Calculation of HB, VB and MB from equilibrium equations:
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=
SUBJECT:A400M Fixed Leading EdgeAAR Snowplough Bracket
Author :T Arden
Date :June 09
Page
InbdD
C
File No - A400M/TWEOM1E3/D/57-45-046
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6.3.2-44
Internal Moment
P = 438.2 N
L1 = 64.8 mm L2 = 64.8 mm
Mint = HA*L1+MA = -232.3 Nmm Mint = HB*L2+MB= 232.3 Nmm
MC = -232.3 Nmm MD = 232.3 Nmm
BM Under Vertical Load
-5.38 -232.25 232.25 5.38
219.11 219.11
162.7
C D
SUBJECT:A400M Fixed Leading EdgeAAR Snowplough Bracket
Author :T Arden
Date :June 09
Page
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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
SUBJECT:A400M Fixed Leading EdgeAAR Snowplough Bracket
Author :T Arden
Date :June 09
Page
A B
C D
z'
Inbd Outbd
xHB
VBVA
HA
MAMB
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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
SUBJECT:A400M Fixed Leading EdgeAAR Snowplough Bracket
Author :T Arden
Date :June 09
Page
Ry Ry
Py'
arm
A & B
C & D
y'
x
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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:
A400M Fixed Leading EdgeAAR Snowplough Bracket
Author : Date :Page
22
2
y y
x
x
22
2
Snowplough deflects
Outboard for thisanalysis
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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
SUBJECT:A400M Fixed Leading EdgeAAR Snowplough Bracket Author :
T ArdenDate :
June 09
Page
y'
z'
Mx
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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:
A400M Fixed Leading EdgeAAR Snowplough Bracket
Author : Date :Page
File No - A400M/TWEOM1E3/D/57-45-046
t
sss
ss
= 2xy
yx
2
yx
2,142
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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
SUBJECT:A400M Fixed Leading EdgeAAR Snowplough Bracket
Author : Date :Page
Hy
V
Mx
y'
z'
A B
C D
z'
Inbd Outbd
xHB
VBVA
HA
MAMB
Hy
V
Mx
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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
SUBJECT:A400M Fixed Leading EdgeAAR Snowplough Bracket Author :
T ArdenDate :
June 09
Page
Hx
My
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6.3.6-1
6.3.6 Portal Frame Analysis (OTBD Side Deflection)
Idealisation
Ref. Section 3.6
SUBJECT:A400M Fixed Leading EdgeAAR Snowplough Bracket
Author : Date :Page
y
z
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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
SUBJECT:A400M Fixed Leading EdgeAAR Snowplough Bracket
Author : Date :Page
19.8
6.3
y
zq
162.7mm
64.8mm
A B
C D
y'
z'
Inbd Outbd
z
x
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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'
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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
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June 09
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6.3.6-5
Portal Frame Analysis
Ref Roarks 6th Ed Table 4
Side Load Ref Roarks 6th Ed Table 4 Case 5f
Geometry parameters Ref pages 6.3.6-2 & -3
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
Distance from edge to load a = 64.80 mm
SUBJECT:A400M Fixed Leading EdgeButtstraps
Author :T Arden
Date :June 09
Page
z'
x
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6.3.6-6
Roark 6th Table 4 Cases 5a & 5f
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6.3.6-7
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 -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=
SUBJECT:A400M Fixed Leading EdgeButtstraps
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
=
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6.3.6-8
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 -8.38
L = LFvx = -63.1
LFmx -0.39
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 -710.5022
Cinv = 0 1.1786275 -95.88134
-710.5022 -95.88134 38799.57
X is calculated below using Excel function MMULT
Ha 94 N
So that X = Va = -37 N
Ma -3047 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=
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Date :June 09
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6.3.6-9
Calculation of HB, VB and MB from equilibrium equations:
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=
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June 09
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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
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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
SUBJECT:A400M Fixed Leading EdgeAAR Snowplough Bracket Author :
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b
C D
Inbd
C D
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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
SUBJECT:A400M Fixed Leading EdgeAAR Snowplough Bracket Author :
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arm
A & B
C & D
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6.3.6-13
Portal Frame Analysis
Ref Roarks 6th Ed Table 4
Vertical Load Ref Roarks 6th Ed Table 4 Case 5a
Geometry parameters Ref pages 6.3.6-2 & -3
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
LoadingWx = 43.5 N ref page 6.3.6-12
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
Distance from edge to load a = 19.84 mm
SUBJECT:A400M Fixed Leading EdgeAAR Snowplough Bracket
Author :T Arden
Date :June 09
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6.3.6-14
Roark 6th Table 4 Cases 5a & 5f
SUBJECT:A400M Fixed Leading EdgeAAR Snowplough Bracket
Author :T Arden
Date :June 09
Page
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6.3.6-15
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 0.177 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=
SUBJECT:A400M Fixed Leading EdgeAAR Snowplough Bracket
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
=
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6.3.6-16
Boundary Conditions
Horizontal deflection
Vertical deflection
Angular rotation
Because the above equal zero, the following three equations are solved simultaneously
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
Now calculate the solution as: X = Cinv*L
where Cinv is the inverse of matrix C, calculated below using Excel function MINVERSE
22.627273 -6.74E-16 -740.8067
Cinv = -2.03E-15 1.2297002 -100.0361
-740.8067 -100.0361 40472.7
X is calculated below using Excel function MMULT
Ha 0 N
So that X = Va = 38 N
Ma 7 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=
SUBJECT:A400M Fixed Leading EdgeAAR Snowplough Bracket Author :
T ArdenDate :
June 09
Page
File No - A400M/TWEOM1E3/D/57-45-046
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6.3.6-17
Calculation of HB, VB and MB from equilibrium equations:
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=
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
SUBJECT:A400M Fixed Leading EdgeAAR Snowplough Bracket Author :
T ArdenDate :
June 09
Page
DCInbd
File No - A400M/TWEOM1E3/D/57-45-046
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6.3.6-18
Internal Moment
P = 43.5 N
L1 = 64.8 mm L2 = 64.8 mm
Mint = HA*L1+MA = -16.4 Nmm Mint = HB*L2+MB= 14.4 Nmm
MC = -16.4 Nmm MD = 14.4 Nmm
BM Under Vertical Load
43.5
19.84 142.86
-0.36 -16.4 14.4 0.36
38.20 5.29
162.7
Check
BM at Load position MP = 741.6 Nmm 741.6 Nmm
SUBJECT:A400M Fixed Leading EdgeAAR Snowplough Bracket
Author :T Arden
Date :June 09
Page
C D
File No - A400M/TWEOM1E3/D/57-45-046
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6.3.6-19
Portal Frame Analysis
Ref Roarks 6th Ed Table 4
Vertical Load Ref Roarks 6th Ed Table 4 Case 5a
Geometry parameters Ref pages 6.3.6-2 & -3
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 = 5.2 N ref page 6.3.6-12
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
Distance from edge to load a = 61.84 mm
SUBJECT:
A400M Fixed Leading EdgeAAR Snowplough Bracket
Author :T Arden
Date :
June 09
Page
File No - A400M/TWEOM1E3/D/57-45-046
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6.3.6-20
Roark 6th Table 4 Cases 5a & 5f
SUBJECT:
A400M Fixed Leading EdgeAAR Snowplough Bracket
Author :
T ArdenDate :
June 09
Page
File No - A400M/TWEOM1E3/D/57-45-046
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76/106
6.3.6-21
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 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:
A400M Fixed Leading EdgeAAR Snowplough Bracket
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
=
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6.3.6-22
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 1.10
L = LFvx = 5.5
LFmx 0.03
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 -710.5022
Cinv = 0 1.1786275 -95.88134
-710.5022 -95.88134 38799.57
X is calculated below using Excel function MMULT
Ha 0 N
So that X = Va = 3 N
Ma 2 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=
SUBJECT:
A400M Fixed Leading EdgeAAR Snowplough Bracket Author :
T ArdenDate :
June 09
Page
e o - A400M/TWEOM1E3/D/57-45-046
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6.3.6-23
Calculation of HB, VB and MB from equilibrium equations:
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=
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
SUBJECT:A400M Fixed Leading Edge
AAR Snowplough Bracket Author :T Arden
Date :June 09
Page
DCInbd
File No - A400M/TWEOM1E3/D/57-45-046
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6.3.6-24
Internal Moment
P = 5.2 N
L1 = 64.8 mm L2 = 64.8 mm
Mint = HA*L1+MA = -4.0 Nmm Mint = HB*L2+MB= 3.8 Nmm
MC = -4.0 Nmm MD = 3.8 Nmm
BM Under Vertical Load
5.2
61.84 100.86
-0.09 -3.97 3.80 0.09
3.22 1.97
162.7
Check
BM at Load position MP = 194.9 Nmm 194.9 Nmm
SUBJECT:
A400M Fixed Leading EdgeAAR Snowplough Bracket
Author :
T ArdenDate :
June 09
Page
C D
File No - A400M/TWEOM1E3/D/57-45-046
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6.3.6-26
Roark 6th Table 4 Cases 5a & 5f
SUBJECT:
A400M Fixed Leading EdgeAAR Snowplough Bracket Author :
T ArdenDate :
June 09
Page
File No - A400M/TWEOM1E3/D/57-45-046
160429985.xls.ms_office 6.3.6 Frame Vert Load 3 Printed 7/27/2013
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82/106
6.3.6-27
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 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=
SUBJECT:A400M Fixed Leading Edge
AAR Snowplough BracketAuthor :
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
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6.3.6-28
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 0.004C = 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
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 -710.5022
Cinv = 0 1.1786275 -95.88134
-710.5022 -95.88134 38799.57
X is calculated below using Excel function MMULT
Ha 0 N
So that X = Va = 0 N
Ma 0 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=
SUBJECT:A400M Fixed Leading Edge
AAR Snowplough Bracket Author :
T ArdenDate :
June 09
Page
File No - A400M/TWEOM1E3/D/57-45-046
160429985.xls.ms_office 6.3.6 Frame Vert Load 3 Printed 7/27/2013
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6.3.6-29
Calculation of HB, VB and MB from equilibrium equations:
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=
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
SUBJECT:A400M Fixed Leading Edge
AAR Snowplough Bracket Author :
T ArdenDate :
June 09
Page
InbdDC
File No - A400M/TWEOM1E3/D/57-45-046
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6.3.6-30
Internal Moment
P = 0.9 N
L1 = 64.8 mm L2 = 64.8 mm
Mint = HA*L1+MA = -0.6 Nmm Mint = HB*L2+MB= 0.6 Nmm
MC = -0.6 Nmm MD = 0.6 Nmm
BM Under Vertical Load
0.9
103.74 58.96
-0.01 -0.62 0.65 0.01
0.31 0.55
162.7
Check
BM at Load position MP = 31.8 Nmm 31.8 Nmm
SUBJECT:A400M Fixed Leading Edge
AAR Snowplough Bracket Author :T Arden
Date :June 09
Page
C D
File No - A400M/TWEOM1E3/D/57-45-046
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6.3.6-31
Portal Frame Analysis
Ref Roarks 6th Ed Table 4
Vertical Load Ref Roarks 6th Ed Table 4 Case 5a
Geometry parameters Ref pages 6.3.6-2 & -3
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
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
Distance from edge to load a = 145.74 mm
SUBJECT:A400M Fixed Leading EdgeAAR Snowplough Bracket
Author :T Arden
Date :June 09
Page
File No - A400M/TWEOM1E3/D/57-45-046
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6.3.6-32
Roark 6th Table 4 Cases 5a & 5f
SUBJECT:A400M Fixed Leading EdgeAAR Snowplough Bracket
Author :T Arden
Date :June 09
Page
File No - A400M/TWEOM1E3/D/57-45-046
160429985.xls.ms_office 6.3.6 Frame Vert Load 4 Printed 7/27/2013
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88/106
6.3.6-33
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 -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=
SUBJECT:A400M Fixed Leading EdgeAAR Snowplough Bracket
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
=
File No - A400M/TWEOM1E3/D/57-45-046
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6.3.6-34
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 -0.53
L = LFvx = -2.6
LFmx -0.02
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 -710.5022
Cinv = 0 1.1786275 -95.88134
-710.5022 -95.88134 38799.57
X is calculated below using Excel function MMULT
Ha 0 N
So that X = Va = -2 N
Ma -3 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=
SUBJECT:A400M Fixed Leading EdgeAAR Snowplough Bracket
Author :T Arden
Date :June 09
Page
File No - A400M/TWEOM1E3/D/57-45-046
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6.3.6-35
Calculation of HB, VB and MB from equilibrium equations:
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=
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
SUBJECT:A400M Fixed Leading EdgeAAR Snowplough Bracket Author :
T ArdenDate :
June 09
Page
InbdD
C
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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
BM Under Vertical Load
-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
SUBJECT:A400M Fixed Leading EdgeAAR Snowplough Bracket
Author :T Arden
Date :June 09
Page
C D
File No - A400M/TWEOM1E3/D/57-45-046
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6.3.6-37
Portal Frame Analysis
Ref Roarks 6th Ed Table 4
Vertical Load Ref Roarks 6th Ed Table 4 Case 5a
Geometry parameters Ref pages 6.3.6-2 & -3
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
Running Load, wa = 2.69 N/mm
SUBJECT:A400M Fixed Leading EdgeAAR Snowplough Bracket
Author :T Arden
Date :June 09
Page
File No - A400M/TWEOM1E3/D/57-45-046
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6.3.6-38
Applied load in Portal Frame plane
a =
Pz'
Distr a
Distr Load
Resolving Load into Protal frame plane
Running Load, wa = 2.69 N/mm
Resolved Running Load, wa = 2.69 N/mm = wa/cos a
SUBJECT:A400M Fixed Leading EdgeAAR Snowplough Bracket
Author :T Arden
Date :June 09
Page
A & B
C & D
File No - A400M/TWEOM1E3/D/57-45-046
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6.3.6-39
Roark 6th Table 4 Cases 5a & 5f
SUBJECT:A400M Fixed Leading EdgeAAR Snowplough Bracket Author :
T ArdenDate :
June 09
Page
File No - A400M/TWEOM1E3/D/57-45-046
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6.3.6-40
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 0.177 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=
SUBJECT:A400M Fixed Leading EdgeAAR Snowplough Bracket
Author :T Arden
Date :June 09
Page
File No - A400M/TWEOM1E3/D/57-45-046
160429985.xls.ms_office 6.3.6 Frame Vert Distr Load Printed 7/27/2013
7/27/2019 6.3 Snowplough
96/106
6.3.6-41
Boundary Conditions
Horizontal deflection
Vertical deflection
Angular rotation
Because the above equal zero, the following three equations are solved simultaneously
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
Now calculate the solution as: X = Cinv*L
where Cinv is the inverse of matrix C, calculated below using Excel function MINVERSE
22.627273 -6.74E-16 -740.8067
Cinv = -2.03E-15 1.2297002 -100.0361
-740.8067 -100.0361 40472.7
X is calculated below using Excel function MMULT
Ha -6 N
So that X = Va = 219 N
Ma 121 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'
pre
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