Ace 402 lec 7 unsymmetrical

Fuselage structure analysis
Given fuselage structure and determine
the ultimate bending strength.
 Given loads and determine the maximum
stresses applied to the fuselage structure.
 Given loads and determine the shear flow
distribution.
ACE 402- Dr Mohamed Elfarran

Shear Lags
In general shear lag results from the effect of sheet panels shear stress
that cause some stringers to resist less axial loads than those calculated
using the beam theory.
The shear lag effect is significant for:
1. Cutouts.
2. Abrupt change of area.
3. Large abrupt change in external loads.

Shear Lags
Assume three stringers as shown in figure
Cutout

Shear Lags
Fuselage with cutout
• At section 1-1 stringers 5, 6, and 7 has zero end
loads.
• As we moved toward section 3-3, the stringers’
loads increased until full effectiveness out of the
triangle that has height 3b.
• The shear lag effect is represented by the shear lag
effectiveness factor .
• can be measured from the curve or calculated as
Ksl = a/L.
Measured from figure or Ksl = a/L

Shear Lags
Shear lag due to Abrupt change of area
= you can measure Aeff from figure
+
2
• The area at the middle section is assumed to be
• The abrupt area change is calculated in terms of the
shear lag effectiveness factor .
=
+
2

Fuselage ultimate bending moment - Example
Consider the fuselage shown in figure. It is a portion of a 2024–T3 aluminum alloy
tapered fuselage of length 90 inch. The fuselage is made up of a 0.05 inch thickness
skin, 0.064 inch thickness floor plate and six stringers of the types shown.
The frame spacing is 18 inch. A cutout at one side of the fuselage is made. This cutout is
bounded by the Frames at Stations (18) and (36) and Stringers
numbered (1) and (5) (elements 2, 3 and 4 are removed).
Skin riveting are single lines countersunk of diameter 5/32" spaced at 1.0" and edge
distance 0.4”, while floor riveting are
brazier of the same diameter and spacing.
Calculate the ultimate bending moment of the fuselage cross-section at Station (0),
taking into consideration the shear lag and inter-rivet buckling effects. Calculate also
the Stringer forces at this station.

• Material: aluminum alloy 2024-T3
• Fuselage length ( ): 90 inch
• Skin thickness ( ): 0.05 inch
• Floor thickness ( ) : 0.06 inch
• Frame spacing (FS): 18 inch
• Cutout (elements 2, 3 and 4 are
removed).
• Skin riveting: single lines
countersunk =5/32“ spaced at
1.0" and edge distance 0.4”,
• floor riveting: brazier of the same
diameter and spacing.

=
−
=
− =
−
= − − − −

=
−
=
−
=
−
= − − − −
Solution strategy
=
−
• The ultimate bending moment calculations depend on the
stress distribution.
• The stress distribution depends on the effective area.
• The effective area calculations depends on the linear and
true stresses.
• So its an iterative problem.
• At least two trials should be considered.
• The fuselage is unsymmetric

Solution:
1. Determine the Inter-rivet buckling effect
Inter-Revite Buckling For upper skin elements 2,4,10,12
upper skin p 1 c 1
countersunk p/t 20 p/teq 40 Fir 21956
Floor Element 13
brazier p 1 c 3
p/t 15.625 p/teq 18.0422 Fir 36000
Trial 1

Solution:
2. Calculate the initial position of both stringers and
skins
Trial 1
No. type A Rcg Θ y' Z'
1 S2, R1 0.18 31.413 1.571 0 47.4125
3 S2, R1 0.18 31.413 2.618 -27.2 31.7063
5 S2, R2 0.18 31.413 3.142 -27.1 -0.064
6 Skin 5_7, R2 2.18 24.95 3.927 -17.64 -17.643
7 S2, R2 0.18 27.15 4.712 0 -27.15
8 Skin 7_9, R2 2.18 24.95 5.498 17.643 -17.643
9 S1, R2 0.14 31.5 0 27.163 -0.057
11 S3, R2 0.08 31.563 0.524 27.334 31.7813
13 Skin 5_9, R2 3.48 16 0 0 0
sum
Stringers and skins without buckled skins

Solution:
3. Calculate the stringer stresses, assume linear distribution
Trial 1 Stringers and skins without buckled skins
No. type
1 S2, R1 -40616
3 S2, R1 -27161
5 S2, R2 54.8257
6 Skin 5_7, R2 15113.5
7 S2, R2 23258.4
8 Skin 7_9, R2 15113.5
9 S1, R2 48.8291
11 S3, R2 -27225
13 Skin 5_9 0
sum
St. type Failing stress [psi]
S1 -37685
S2 -40616
S3 -34455

Solution:
4. Skin effectiveness
0.8335 0.041677 0.22168
1.5242 0.076211 0.25621
0 0 0.18
0 0 2.17656
0 0 0.18
0 0 2.17656
0 0 0.135
1.5188 0.075941 0.15594
0 0 3.47604
No.
1
3
5
6
7
8
9
11
13
sum

Solution:
5. True strain and true stresses
No.
1
3
5
6
7
8
9
11
13
sum
x 10^3
-5.9 -40616.13 1 1
-3.945 -37431.953 0.2232 1.378
0.00796 85.216 1 1.554
2.195 23491.117 1 1.554
3.3786 36150.788 1 1.554
2.195 23491.117 1 1.554
0.007093 75.8958 1 1.55
-3.95485 -33998.574 1 1.248
0 0 1 1

Solution:
6. Shear lag effect
No.
1
3
5
6
7
8
9
11
13
sum
1 1
0.2232 1.378
1 1.554
1 1.554
1 1.554
1 1.554
1 1.55
1 1.248
1 1
a = 18 inch
b = 23.3
3b = 69.9

Solution:
7. First moments of area with respect to the initial axes
No.
1
3
5
6
7
8
9
11
13
sum
* Y' * Z'
0.22168 0 10.5103
0.07881 -2.1439 2.49868
0.27978 -7.582 -0.0179
3.38306 -59.686 -59.686
0.27978 0 -7.596
3.38306 59.6856 -59.686
0.20983 5.69964 -0.012
0.19474 5.32289 6.18894
3.47604 0 0
11.5068 1.29663 -107.8

Solution:
8. Calculate the buckling stresses
of skins under compression
Skin: 2,4,10,12
Trial 1 Buckled skin
the critical stress of buckled skin
upper skin r/t 639.5
5223.4

Solution:
9. First moments of area, linear stresses, and effectiveness coeffecient
with respect to the initial axes
No. b' A Θ y' Z'
2 32.331 1.6166 2.0944 -15.2789 42.4638 -36376.7 0.14359
4 32.748 1.637 3.1416 -30.5577 16 -13706.4 0.38109
10 32.751 1.637 0.523599 30.5577 16 -13706.4 0.38109
12 32.334 1.6167 1.57079 15.2789 42.4638 -36376.7 0.14359
5223.4

Solution:
10. Effective area and first moments of area for buckled skin
* y' * Z'
0.23213 -3.5466 9.85699
0.624 -19.068 9.98405
0.62405 19.0697 9.98487
0.23215 3.54693 9.85781
1.71233 0.00187 39.6837
13.2191 1.29849 -68.116
No.
2
4
10
12
sum
11. Centroid position
Y_bar 0.09823
Z_bar -5.1528

Solution:
12. Second moment of areas
Trial 1
Y Z AY^2 AZ^2 AYZ
-0.0982 52.565 0.002139 612.52 -1.1446
-27.302 36.859 58.7438 107.067 -79.306
-27.199 5.0888 206.9676 7.24511 -38.723
-17.741 -12.4897 1064.764 527.732 749.607
-0.0982 -21.997 0.002699 135.381 0.60454
17.5443 -12.489 1041.312 527.732 -741.31
27.0646 5.0958 153.70 5.44879 28.9393
27.2357 36.934 144.452 265.644 195.89
-0.0982 5.1528 0.03354 92.2942 -1.7594
2669.978 2281.06 112.801
Without Buckled skin
No.
1
3
5
6
7
8
9
11
13
sum

Solution:
13. Second moment of areas
Y Z AY^2
-15.377 47.6166 54.887
-30.656 21.1528 586.43
30.4595 21.1528 578.987
15.1806 47.6166 53.4985
1273.804
3943.782
AZ^2 AYZ
526.311 -169.965
279.205 -404.64
279.228 402.081
526.355 167.807
1611.1 -4.7175
3892.16 108.083
No.
2
4
10
12
∑
∑tot
Second Moment of area
Iyy 3892.16 Iyz 108.083 K1 7E-06
Izz 3943.78 I= 15338156 K2 0.00026

Solution:
14. Determine the position w.r.t. the Neutral axis position
Trial 1 unbuckled and buckled skin
=
+
= −
= − −
= −
52.5483
37.5932
5.83203
-11.999
-21.987
-12.966
4.35245
36.1741
5.15358
No.
2 48.0199
4 21.985
10 20.31
12 47.183
No.
1
3
5
6
7
8
9
11
13
From the first trial, we found the floor is under
buckling, so we will move the floor to the buckled skin
in the second trial ACE 402- Dr Mohamed Elfarran

Solution: Trial 2 Unbuckled skin
N0. type A y' Z' dN.A σlinear Weff Atot εlinear x 10^3 σtrue Ksl Keff Aeff Aeff * Y' Aeff * Z'
1 S2, R1 0.18 -0.098 52.57 52.54828 -40616 0.834 0.22168 -5.900 -40616.133 1.000 1.000 0.222 -0.022 11.653
3 S2, R1 0.18 -27.3 36.86 37.593197 -29057 1.378 0.24888 -4.221 -38449.086 0.223 1.323 0.074 -2.007 2.709
5 S2, R2 0.18 -27.2 5.089 5.832030 -4507.7 2.314 5.924 0.51089 -0.655 -7006.434 1.000 1.554 0.794 -21.598 4.041
6 Skin 5_7 2.18 -17.74 -12.5 -11.99899 9274.4 0 2.17656 1.347 14415.254 1.000 1.554 3.383 -60.018 -42.253
7 S2, R2 0.18 -0.098 -22 -21.98654 16994 0 0.18 2.469 26414.008 1.000 1.554 0.280 -0.027 -6.154
8 Skin 7_9 2.18 17.544 -12.5 -12.9656558 10021.5 0 2.17656 1.456 15576.568 1.000 1.554 3.383 59.353 -42.253
9 S1, R2 0.14 27.0655.096 4.3524539 -3364.1 2.679 6.858 0.51399 -0.489 -5228.914 1.000 1.554 0.799 21.622 4.071
11 S3, R1 0.08 27.23636.93 36.1740663 -27960 1.459 0.15297 -4.062 -34130.444 1.000 1.221 0.187 5.086 6.897
sum 9.121 2.389 -61.291
No. b' A y' Z' σlinear Keff Aeff Aeff * y' Aeff * Z'
2 32.4048 1.62 -15.38 47.62 -37115.968 0.141 0.228 -3.506 10.858
4 30.5073 1.53 -30.66 21.15 -16992.590 0.307 0.469 -14.374 9.918
10 30.1018 1.51 30.46 21.15 -15698.479 0.333 0.501 15.254 10.593
12 32.3639 1.62 15.181 47.62 -36468.912 0.143 0.232 3.518 11.036
13 47.9220 3.07 -0.1 5.153 -3983.339 0.038 0.118 -0.012 0.607
sum 1.547 0.880 43.012
∑tot 10.668 3.269 -18.278
Buckled skin
Y_bar 0.31
Z_bar -1.71
Centroid position

Buckled skin
N0. Y Z AY^2 AZ^2 AYZ
1 -0.4047 54.278698 0.0363026 653.101 -4.8692
3 -27.609 38.572448 56.025724 109.358 -78.274
5 -27.505 6.8021976 600.747005 36.7424 -148.57
6 -18.047 -10.776327 1101.86619 392.872 657.945
7 -0.4047 -20.284115 0.04581709 115.113 2.29655
8 17.2378 -10.776327 1005.25264 392.872 -628.44
9 26.7581 6.8091976 572.016809 37.0415 145.562
11 26.9293 38.647448 135.411698 278.901 194.336
sum 3471.40219 2016 139.988
Y Z AY^2 AZ^2 AYZ
-15.6836 49.3300 56.0868 554.8732 -176.4116
-30.9624 22.8662 449.5085 245.1635 -331.9685
30.1531 22.8662 455.3246 261.8459 345.2896
14.8742 49.3300 51.2776 564.0049 170.0613
-0.4064 6.8662 0.0195 5.5546 -0.3288
1012.2170 1631.4422 6.6419
4483.6192 3647.4412 146.6302
No.
2
4
10
12
13
sum
∑tot
Iy 3647.441221 Iz 4483.62 Iyz 146.63
K1 8.97796E-06 K2 0.00027 K3 0.00022
Second moments of area

Buckled skin
No.
2
4
10
12
13
sum
∑tot
dN.A σlinear F Mz My
54.2629 -40616 -9003.7 -3643.5787 488707.937
39.4543 -29532 -2170.6 -59928.138 83726.3252
7.69759 -5761.7 -4575.3 -125843.4 31122.0515
-10.181 7620.3 25779.9 465254.99 277812.602
-20.26 15164.8 4242.75 1716.9404 86060.3507
-11.334 8483.57 28700.4 -494733.16 309284.915
5.93094 -4439.3 -3546.6 94901.249 24149.7147
37.7466 -28254 -5275.7 142070.85 203892.639
34151.1 19795.754 1504756.54
dN.A σlinear F Mz My
49.8163 -37287.7783 -8502.3 -133346.83 419420.144
22.8540 -17106.3449 -8020.9 -248347.72 183408.372
22.8540 -17106.3449 -8566.7 258313.29 195888.585
49.3036 -36904.0689 -8553.3 127223.91 421935.602
6.8625 -5136.6452 -605.2 -245.9808 4155.39799
-34249 3596.6625 1224808.1
-97.425 23392.416 279948.435
N0.
1
3
5
6
7
8
9
11
sum
= ∗ = ∗ , = ∗ y
Myult 2725086
=
−

Ace 402 lec 7 unsymmetrical

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