Wing Layout
May 9, 2017 | Author: Brixvon Picardo Cruz | Category: N/A
Short Description
Wing layout of the designed aircraft (Zurcx777)...
Description
Activity 3: Wing Layout
๏ To determine the appropriate wing attachment for you aircraft ๏ To determine the right position/location, material, cross-section and no.
of spar appropriate for your design ๏ To determine the right position/location, material, no. and type of wing rib appropriate for your design
๐๐ = Where: Wb = Intensity (lbโft) Wb = Gross weight (3,531.57 lb) b = Wing span (31.49 ft)
๐๐จ ๐
Wb =
3,531.57 lb 31.49 ft
Wb = 112.15
lb ft Considering the Right Hand
b
F = (Wb ) ( ) 2
F = (112.15
lb ft
) (15.75 ft)
F = 1,766.36 lb Multiply by 1.5 to get ultimate load F = (1,766.36 lb) (1.5) F = 2,649.54 lb
Get the Centroid x= x=
4
b
( ) , for quarter ellipse
3๐ 2 4 3๐
(15.75 ft)
x = 6.68 ft Reaction of support B โ ฮฃFv = 0 F โ RB = 0 2,649.54 โ R B = 0 R B = 2,649.54 lb
Bending Stress Sb =
MC I
Where: M = Moment (lb-in) C = Centroid of cross-section of the beam I = Moment of Inertia Moment Computation, M = (F)(d) Where: d=xโ
a 2
d = 6.68 ft โ 2.04 ft d = 4.64 ft
Moment at R B : M = (F)(d) M = (2,649.54 lb)(4.64 ft) ๐ = ๐๐, ๐๐๐. ๐๐ ๐ฅ๐ ๐๐ญ = ๐๐, ๐๐๐. ๐๐ ๐ฅ๐ ๐๐ญ (12in) ๐ = ๐๐๐, ๐๐๐. ๐๐ ๐ฅ๐ ๐ข๐ง
Cross Section of FRONT SPAR UNIT: inches
Thickness = 1/4 in
Section
๐ in
๐ in2
๐๐ in3
๐๐ in2
๐๐ ๐ in4
๐๐๐ฑ in4
1
9.438
0.69
6.51222
89.07584
61.46233
0.00359
2
4.7808
2.27
10.85242
22.85605
51.88323
15.50765
3
0.1248
0.56
0.069888
0.015575
0.008722
0.00292
๐๐๐ฑ =
๐๐ก๐ (๐๐จ๐ซ ๐๐๐๐ญ๐๐ง๐ ๐ฎ๐ฅ๐๐ซ) ๐๐
ฮฃA = 3.52 in2 ฮฃAy = 17.43 in3 ฮฃAy 2 = 113.35 in4 ฮฃIcx = 15.51 in4
Considering the Centroid at X-Axis C= ฬ
Y= ฬ
Y=
ฮฃAy ฮฃA
17.43 in3 3.52 in2
ฬ
= ๐. ๐๐ ๐ข๐ง ๐
ฬ
)2 I = ฮฃIxx โ (ฮฃA)(Y ฮฃIxx = ฮฃIcx + ฮฃAy 2 ฮฃIxx = 15.51 in4 + 113.35 in4 ฮฃIxx = 128.86 in4 I = 128.86 in4 โ (3.52 in2 )(4.95 in)2 ๐ = ๐๐. ๐๐ ๐ข๐ง๐
Solving for the Bending Stress SB =
Mc I
SB =
(147,526.44 lb in)(4.95 in) 42.61 in4
๐๐ = ๐๐, ๐๐๐. ๐๐
๐ฅ๐ ๐ข๐ง๐
Solving for the Factor of Safety FS =
Allowable Bending Stress Computed Bending Stress Where:
Type of Material: Aluminum Alloy 6061-T6 Ultimate Bending Stress: 30000 psi
FS =
30,000 psi 17,138.13 psi ๐
๐ = 1.75
Cross Section of REAR SPAR UNIT: inches
Thickness = 1/4 in
Section
๐ in
๐ in2
๐๐ in3
๐๐ in2
๐๐ ๐ in4
๐๐๐ฑ in4
1
7.95
0.625
4.96875
63.2025
39.50156
0.00326
2
4.04
1.895
7.6558
16.3216
30.92943
9.07332
3
0.1248
0.575
0.07176
0.015575
0.008956
0.00299
๐๐๐ฑ =
๐๐ก๐ (๐๐จ๐ซ ๐๐๐๐ญ๐๐ง๐ ๐ฎ๐ฅ๐๐ซ) ๐๐
ฮฃA = 3.10 in2 ฮฃAy = 12.70 in3 ฮฃAy 2 = 70.44 in4 ฮฃIcx = 9.08 in4
Considering the Centroid at X-Axis C= ฬ
Y= ฬ
Y=
ฮฃAy ฮฃA
12.70 in3 3.10 in2
ฬ
= ๐. ๐๐ ๐ข๐ง ๐
ฬ
)2 I = ฮฃIxx โ (ฮฃA)(Y ฮฃIxx = ฮฃIcx + ฮฃAy 2 ฮฃIxx = 9.08 in4 + 70.44 in4 ฮฃIxx = 79.52 in4 I = 79.52 in4 โ (3.10 in2 )(4.10 in)2 ๐ = ๐๐. ๐๐ ๐ข๐ง๐
Solving for the Bending Stress SB =
Mc I
SB =
(147,526.44 lb in)(4.10 in) 27.41 in4
๐๐ = ๐๐, ๐๐๐. ๐๐
๐ฅ๐ ๐ข๐ง๐
Solving for the Factor of Safety FS =
Allowable Bending Stress Computed Bending Stress Where:
Type of Material: Aluminum Alloy 6061-T6 Ultimate Bending Stress: 30000 psi
FS =
30,000 psi 22,067.07 psi
๐
๐ = 1.36
WING RIB DESIGN Rib No. 1 โ BUTT RIB Located 2.04 ft. from the aircraftโs center Given: A1 = 0.4343 ft 2
t W FS = 0.0208 ft or 0.25 in
l1 = 2.61 ft
t W RS = 0.0208 ft or 0.25 in
A2 = 1.87 ft 2
t1 = 0.0104 ft or 0.125 in
l2 = 5.99 ft
t 2 = 0.0104 ft or 0.125 in t 3 = 0.0104 ft or 0.125 in
A3 = 0.7292 ft 2 l3 = 4.73 ft
Moment at Quarter Chord, ๐๐ฆ ๐ ๐
+โท MA = 0 0 = โ PRS (2.25 ft) + 2,649.54 lb (0.50 ft)
PRS =
2,649.54 lb (0.50 ft) 2.25 ft
๐๐๐ = ๐๐๐. ๐๐ ๐ฅ๐ +โ ฮฃFV = 0 588.79 lb โ 2,649.54 lb + PFS = 0 PFS = 2,649.54 lb โ 588.79 lb ๐๐
๐ = ๐, ๐๐๐. ๐๐ ๐ฅ๐
Shear Flow at Front Spar (FS) q WFS =
PFS hFS
q WFS =
2,060.21 lb 0.7967 ft ๐ช๐๐
๐ = ๐, ๐๐๐. ๐๐
๐ฅ๐ ๐๐ญ
Shear Flow at Rear Spar (RS) q WRS =
PRS hRS
q WRS =
588.79 lb 0.6730 ft ๐ช๐๐๐ = ๐๐๐. ๐๐
๐ฅ๐ ๐๐ญ
Shear Flow at 1: 2ฮธ1 G =
1 l1 hFS [q1 + (q1 โ q2 + q WFS ) ] A1 t1 t W FS
1 2.61 ft lb 0.7967 ft [(q1 ) ( ) + (q1 โ q2 + 2,585.93 ) ( )] 2 0.4343 ft 0.0104 ft ft 0.0208 ft 1 lb (250.96 q1 + 38.3029 q1 โ 38.3029 q2 + 99,048.58 ) 2ฮธ1 G = 0.4343 ft 2 ft ๐ฅ๐ ๐๐๐ ๐ = ๐๐๐. ๐๐ ๐ช๐ โ ๐๐. ๐๐๐๐ ๐ช๐ + ๐๐๐, ๐๐๐. ๐๐ ๐๐ช๐ฎ๐๐ญ๐ข๐จ๐ง ๐๐จ. ๐ ๐๐ญ 2ฮธ1 G =
Shear Flow at 2: 2ฮธ2 G =
1 l2 hFS hRS [q2 + (q2 โ q WFS โ q1 ) ( ) + (q2 + q WRS โ q3 ) ( )] A2 t2 t W FS t W RS
1 5.99 ft lb 0.7967 ft [(q2 ) ( ) + (q2 โ 2,585.93 โ q1 ) ( ) 2 1.87 ft 0.0104 ft ft 0.0208 ft lb 0.6730 ft )] + (q2 + 874.87 โ q3 ) ( ft 0.0208 ft 1 lb (575.96 q2 + 38.3029 q2 โ 99,048.58 โ 38.3029 q1 + 32.36 q2 2ฮธ2 G = 2 1.87 ft ft lb + 28,307.09 โ 32.36 q3 ) ft ๐ฅ๐ ๐๐๐ ๐ = โ๐๐. ๐๐ ๐ช๐ + ๐๐๐. ๐๐ ๐ช๐ โ ๐๐. ๐๐ ๐ช๐ โ ๐๐, ๐๐๐. ๐๐ ๐๐ช๐ฎ๐๐ญ๐ข๐จ๐ง ๐๐จ. ๐ ๐๐ญ 2ฮธ2 G =
Shear Flow at 3: 2ฮธ3 G =
1 l3 hRS [q3 + (q3 โ q2 โ q WRS ) ] A3 t3 t W RS
1 4.73 ft lb 0.6730 ft ) [(q ( ) (q )( )] + โ q โ 874.87 3 3 2 0.7292 ft 2 0.0104 ft ft 0.0208 ft 1 lb (454.81 q3 + 32.36 q3 โ 32.36 q2 โ 28,307.09 ) 2ฮธ3 G = 2 0.7292 ft ft ๐ฅ๐ ๐๐๐ ๐ = ๐๐๐. ๐๐ ๐ช๐ โ ๐๐. ๐๐ ๐ช๐ โ ๐๐, ๐๐๐. ๐๐ ๐๐ช๐ฎ๐๐ญ๐ข๐จ๐ง ๐๐จ ๐ ๐๐ญ 2ฮธ3 G =
Assuming that the Twist Angles are equal in the cells: ๐ฝ๐ = ๐ฝ๐ = ๐ฝ๐
Equate ๐๐ช๐ฎ๐๐ญ๐ข๐จ๐ง ๐๐จ. ๐ and๐๐ช๐ฎ๐๐ญ๐ข๐จ๐ง ๐๐จ ๐: lb lb = 668.09 q3 โ 44.38 q2 โ 38,819.38 ft ft lb lb โ44.38 q2 + 88.1946 q2 = 666.04 q1 โ 668.09 q3 + 38,824.45 + 228,064.89 ft ft lb 43.81 q2 = 666.04 q1 โ 668.09 q3 + 266889.34 ft lb 666.04 q1 โ 668.09 q3 + 266,889.34 ft q2 = 43.81 ๐ฅ๐ ๐ช๐ = ๐๐. ๐๐ ๐ช๐ โ ๐๐. ๐๐ ๐ช๐ + ๐, ๐๐๐. ๐๐ ๐๐ช๐ฎ๐๐ญ๐ข๐จ๐ง ๐๐จ . ๐ ๐๐ญ 666.04 q1 โ 88.1946 q2 + 228,064.89
Substitute ๐๐ช๐ฎ๐๐ญ๐ข๐จ๐ง ๐๐จ . ๐ to ๐๐ช๐ฎ๐๐ญ๐ข๐จ๐ง ๐๐จ. ๐ 666.04 q1 โ 88.1946 q2 + 228,064.89
lb ft
๏ ๐๐ช๐ฎ๐๐ญ๐ข๐จ๐ง ๐๐จ. ๐
lb lb 666.04 q1 = 88.1946 (15.20 q1 โ 15.25 q3 + 6,091.97 ) โ 228,064.89 ft ft lb lb 666.04 q1 = 1,340.56 q1 โ 1,344.97 q3 + 537,278.86 โ 228,064.89 ft ft lb 674.52 q1 = 1,349.38 q3 โ 309,213.97 ft lb 1,349.38 q3 โ 309,213.97 ft q1 = 674.52 ๐ฅ๐ ๐ช๐ = ๐. ๐๐ ๐ช๐ โ ๐๐๐. ๐๐ ๐๐ช๐ฎ๐๐ญ๐ข๐จ๐ง ๐๐จ. ๐ ๐๐ญ
Substitute ๐๐ช๐ฎ๐๐ญ๐ข๐จ๐ง ๐๐จ . ๐ to๐๐ช๐ฎ๐๐ญ๐ข๐จ๐ง ๐๐จ ๐: 668.09 q3 โ 44.38 q2 โ 38,819.38
lb ft
๏ ๐๐ช๐ฎ๐๐ญ๐ข๐จ๐ง ๐๐จ ๐
lb lb 668.09 q3 = 44.38 (15.20 q1 โ 15.25 q3 + 6,091.97 ) + 38,819.38 ft ft lb lb 668.09 q3 = 674.58 q1 โ 676.80 q3 + 270,361.63 + 38,819.38 ft ft lb lb 668.09 q3 + 676.80 q3 = 674.58 q1 + 270,361.63 + 38,819.38 ft ft lb 1,344.89 q3 = 674.58 q1 + 309,181.01 ft lb 674.58 q1 + 309,181.01 ft q3 = 1,344.89 ๐ช๐ = ๐. ๐๐๐๐ ๐ช๐ + ๐๐๐. ๐๐
๐ฅ๐ ๐๐ญ
๐๐ช๐ฎ๐๐ญ๐ข๐จ๐ง ๐๐จ. ๐
Substitute ๐๐ช๐ฎ๐๐ญ๐ข๐จ๐ง ๐๐จ. ๐ to๐๐ช๐ฎ๐๐ญ๐ข๐จ๐ง ๐๐จ. ๐: q3 = 0.5016 q1 + 229.89
lb ft
๏ ๐๐ช๐ฎ๐๐ญ๐ข๐จ๐ง ๐๐จ. ๐
lb lb ) + 229.89 ft ft lb lb q3 = 1.0032 q3 โ 229.94 + 229.89 ft ft lb โ0.01q3 = โ0.05 ft lb 0.05 ft q3 = 0.0032 q3 = 0.5016 (2.00 q3 โ 458.42
๐ช๐ = ๐๐. ๐๐
๐ฅ๐ ๐๐ญ
, the shear flow direction is same as the assumption
From๐๐ช๐ฎ๐๐ญ๐ข๐จ๐ง ๐๐จ. ๐: q1 = 2.00 q3 โ 458.42 q1 = 2.00 (15.63 ๐ช๐ = ๐๐๐. ๐๐
๐ฅ๐ ๐๐ญ
lb ft
lb lb ) โ 458.42 ft ft , the shear flow is opposite from the assumption
From๐๐ช๐ฎ๐๐ญ๐ข๐จ๐ง ๐๐จ . ๐: q2 = 15.20 q1 โ 15.25 q3 + 6,091.97
lb ft
lb lb lb q2 = 15.20 (427.16 ) โ 15.25(15.63 ) + 6,091.97 ft ft ft ๐ช๐ = ๐๐, ๐๐๐. ๐๐
๐ฅ๐ ๐๐ญ
, the shear flow direction is the same as the assumption
Torsional Shearing Stress at Cell 1: q1 ฯ1 = t1 lb 427.16 ft ฯ1 = 0.0104 ft ๐๐ = ๐๐, ๐๐๐. ๐๐
๐ฅ๐ ๐๐ญ ๐
Torsional Shearing Stress at Cell 2: q2 t2
ฯ2 =
lb ft ฯ2 = 0.0104 ft 12,346.44
๐๐ = ๐, ๐๐๐, ๐๐๐. ๐๐
๐ฅ๐ ๐๐ญ ๐
Torsional Shearing Stress at Cell 3: q3 t3
ฯ3 =
lb ft ฯ3 = 0.0104 ft 15.63
๐๐ = ๐, ๐๐๐. ๐๐
๐ฅ๐ ๐๐ญ ๐
Torsional Shearing Stress at Front Spar (FS): ฯFS = ฯFS
qFS
t W FS
lb ft = 0.0208 ft 2,585.93
๐๐
๐ = ๐๐๐, ๐๐๐. ๐๐
๐ฅ๐ ๐๐ญ ๐
Torsional Shearing Stress at Rear Spar (RS): ฯRS = ฯRS
qRS
t W RS
lb ft = 0.0208 ft 874.87
๐๐๐ = ๐๐, ๐๐๐. ๐๐
๐ฅ๐ ๐๐ญ ๐
Factor of Safety at Cell 1: FS1 =
ฯALLOWABLE ฯCOMPUTED
FS1 =
4,320,000 psf 41,073.08 psf
๐
๐๐ = ๐๐๐. ๐๐
Factor of Safety at Cell 2: FS2 =
ฯALLOWABLE ฯCOMPUTED
FS2 =
4,320,000 psf 1,187,157.69 psf
๐
๐๐ = ๐. ๐๐
Factor of Safety at Cell 3: FS3 =
ฯALLOWABLE ฯCOMPUTED
FS3 =
4,320,000 psf 1,502.88 psf
๐
๐๐ = ๐, ๐๐๐. ๐๐
Factor of Safety at Front Spar (FS): ฯALLOWABLE ฯCOMPUTED 4,320,000 psf = 124,323.56 psf
FSFS = FSFS
๐
๐๐
๐ = ๐๐. ๐๐
Factor of Safety at Rear Spar (RS): ฯALLOWABLE ฯCOMPUTED 4,320,000 psf = 42,061.06 psf
FSRS = FSRS
๐
๐๐๐ = ๐๐๐. ๐๐
RIB Located at the tip of the Wing Given: A1 = 0.0951 ft 2
t W FS = 0.0208 ft or 0.25 in
l1 = 1.2196 ft
t W RS = 0.0208 ft or 0.25 in
A2 = 0.4089 ft 2
t1 = 0.0104 ft or 0.125 in
l2 = 2.8014 ft
t 2 = 0.0104 ft or 0.125 in t 3 = 0.0104 ft or 0.125 in
A3 = 0.1597ft 2 l3 = 2.2158ft
Moment at Quarter Chord, ๐๐ฆ ๐ ๐
+โท MA = 0 0 = โ PRS (1.05 ft) + 2,649.54 lb (0.2340 ft) PRS =
2,649.54 lb (0.2340 ft) 1.05 ft
๐๐๐ = ๐๐๐. ๐๐ ๐ฅ๐
+โ ฮฃFV = 0 590.47 lb โ 2,649.54 lb + PFS = 0 PFS = 2,649.54 lb โ 590.47 lb ๐๐
๐ = ๐, ๐๐๐. ๐๐ ๐ฅ๐
Shear Flow at Front Spar (FS) q WFS =
PFS hFS
q WFS =
2,059.07 lb 0.3729 ft ๐ช๐๐
๐ = ๐, ๐๐๐. ๐๐
๐ฅ๐ ๐๐ญ
Shear Flow at Rear Spar (RS) q WRS =
PRS hRS
q WRS =
590.47 lb 0.3150 ft ๐ช๐๐๐ = ๐, ๐๐๐. ๐๐
๐ฅ๐ ๐๐ญ
Shear Flow at 1: 2ฮธ1 G =
1 l1 hFS [q1 + (q1 โ q2 + q WFS ) ] A1 t1 t W FS
1 1.2196 ft lb 0.3729 ft [(q ( ) (q )( )] ) + โ q + 5,521.78 1 1 2 0.0951 ft 2 0.0104 ft ft 0.0208 ft 1 lb (117.27 ) 2ฮธ1 G = q + 17.93 q โ 17.93 q + 98,993.83 1 1 2 0.0951 ft 2 ft ๐ฅ๐ ๐๐๐ ๐ = ๐, ๐๐๐. ๐๐ ๐ช๐ โ ๐๐๐. ๐๐ ๐ช๐ + ๐, ๐๐๐, ๐๐๐. ๐๐ ๐๐ช๐ฎ๐๐ญ๐ข๐จ๐ง ๐๐จ. ๐ ๐๐ญ 2ฮธ1 G =
Shear Flow at 2: 2ฮธ2 G =
1 l2 hFS hRS [q2 + (q2 โ q WFS โ q1 ) ( ) + (q2 + q WRS โ q3 ) ( )] A2 t2 t W FS t W RS
1 2.8014 ft lb 0.3729 ft [(q2 ) ( ) + (q2 โ 5,521.78 โ q1 ) ( ) 2 0.4089 ft 0.0104 ft ft 0.0208 ft lb 0.3150 ft )] + (q2 + 1,874.51 โ q3 ) ( ft 0.0208 ft 1 lb (269.37 q2 + 17.93 q2 โ 98,993.83 โ 17.93 q1 + 15.14 q2 2ฮธ2 G = 2 0.4089 ft ft lb + 28,388.01 โ 15.14 q3 ) ft ๐ฅ๐ ๐๐๐ ๐ = โ๐๐. ๐๐ ๐ช๐ + ๐๐๐. ๐๐ ๐ช๐ โ ๐๐. ๐๐ ๐ช๐ โ ๐๐, ๐๐๐. ๐๐ ๐๐ช๐ฎ๐๐ญ๐ข๐จ๐ง ๐๐จ. ๐ ๐๐ญ 2ฮธ2 G =
Shear Flow at 3: 2ฮธ3 G =
1 l3 hRS [q3 + (q3 โ q2 โ q WRS ) ] A3 t3 t W RS
1 2.2158 ft lb 0.3150 ft [(q3 ) ( ) + (q3 โ q2 โ 1,874.51 ) ( )] 2 0.1597 ft 0.0104 ft ft 0.0208 ft 1 lb (213.06 q3 + 15.14 q3 โ 15.14 q2 โ 28,388.01 ) 2ฮธ3 G = 2 0.1597 ft ft ๐ฅ๐ ๐๐๐ ๐ = ๐, ๐๐๐. ๐๐ ๐ช๐ โ ๐๐. ๐๐ ๐ช๐ โ ๐๐๐, ๐๐๐. ๐๐ ๐๐ช๐ฎ๐๐ญ๐ข๐จ๐ง ๐๐จ ๐ ๐๐ญ 2ฮธ3 G =
Assuming that the Twist Angles are equal in the cells: ๐ฝ๐ = ๐ฝ๐ = ๐ฝ๐
Equate ๐๐ช๐ฎ๐๐ญ๐ข๐จ๐ง ๐๐จ. ๐ and๐๐ช๐ฎ๐๐ญ๐ข๐จ๐ง ๐๐จ ๐: lb lb = 1,428.93 q3 โ 94.80 q2 โ 177,758.36 ft ft lb lb โ94.80 q2 + 188.54 q2 = 1,421.66 q1 โ 1,428.93 q3 โ 177,758.36 + 1,040,944.58 ft ft lb 93.74 q2 = 1,421.66 q1 โ 1,428.93 q3 + 863,186.22 ft lb 1,421.66 q1 โ 1,428.93 q3 + 863,186.22 ft q2 = 93.74 ๐ฅ๐ ๐ช๐ = ๐๐. ๐๐ ๐ช๐ โ ๐๐. ๐๐ ๐ช๐ + ๐, ๐๐๐. ๐๐ ๐๐ช๐ฎ๐๐ญ๐ข๐จ๐ง ๐๐จ . ๐ ๐๐ญ 1,421.66 q1 โ 188.54 q2 + 1,040,944.58
Substitute ๐๐ช๐ฎ๐๐ญ๐ข๐จ๐ง ๐๐จ . ๐ to ๐๐ช๐ฎ๐๐ญ๐ข๐จ๐ง ๐๐จ. ๐ 1,421.66 q1 โ 188.54 q2 + 1,040,944.58
lb ft
๏ ๐๐ช๐ฎ๐๐ญ๐ข๐จ๐ง ๐๐จ. ๐
lb lb ) โ 1,040,944.58 ft ft lb lb 1,421.66 q1 = 2,860.15 q1 โ 2,873.35 q3 + 1,736,132.88 โ 1,040,944.58 ft ft lb 1,438.49 q1 = 2,873.35 q3 โ 695,188.30 ft lb 2,873.35 q3 โ 695,188.30 ft q1 = 1,438.49 1,421.66 q1 = 188.54 (15.17 q1 โ 15.24 q3 + 9,208.30
๐ช๐ = ๐. ๐๐ ๐ช๐ โ ๐๐๐. ๐๐
๐ฅ๐ ๐๐ญ
๐๐ช๐ฎ๐๐ญ๐ข๐จ๐ง ๐๐จ. ๐
Substitute ๐๐ช๐ฎ๐๐ญ๐ข๐จ๐ง ๐๐จ . ๐ to๐๐ช๐ฎ๐๐ญ๐ข๐จ๐ง ๐๐จ ๐: 1,428.93 q3 โ 94.80 q2 โ 177,758.36
lb ft
๏ ๐๐ช๐ฎ๐๐ญ๐ข๐จ๐ง ๐๐จ ๐
lb ) + 177,758.36 ft lb 1,428.93 q3 = 1,438.12 q1 โ 1,444.75 q3 + 872,946.84 + 177,758.36 ft lb 1,444.75 q3 + 1,428.93 q3 = 1,438.12 q1 + 872,946.84 + 177,758.36 ft lb 2,837.68 q3 = 1,438.12 q1 + 1,050,705.20 ft lb 1,438.12 q1 + 1,050,705.20 ft q3 = 2,837.68 1,428.93 q3 = 94.80 (15.17 q1 โ 15.24 q3 + 9,208.30
๐ช๐ = ๐. ๐๐๐๐ ๐ช๐ + ๐๐๐. ๐๐
๐ฅ๐ ๐๐ญ
๐๐ช๐ฎ๐๐ญ๐ข๐จ๐ง ๐๐จ. ๐
lb ft lb ft lb ft
Substitute ๐๐ช๐ฎ๐๐ญ๐ข๐จ๐ง ๐๐จ. ๐ to๐๐ช๐ฎ๐๐ญ๐ข๐จ๐ง ๐๐จ. ๐: q3 = 0.5068 q1 + 370.27
lb ft
๏ ๐๐ช๐ฎ๐๐ญ๐ข๐จ๐ง ๐๐จ. ๐
lb lb ) + 370.27 ft ft lb lb q3 = 1.0136 q3 โ 244.93 + 370.27 ft ft lb 0.0136 q3 = โ125.34 ft lb โ125.34 ft q3 = 0.0136 q3 = 0.5068 (2.00 q3 โ 483.28
๐ช๐ = ๐, ๐๐๐. ๐๐
๐ฅ๐
, the shear flow is opposite from the assumption
๐๐ญ
From๐๐ช๐ฎ๐๐ญ๐ข๐จ๐ง ๐๐จ. ๐: q1 = 2.00 q3 โ 483.28 q1 = 2.00 (9,216.18 ๐ช๐ = ๐๐, ๐๐๐. ๐๐
lb ft
lb lb ) โ 483.28 ft ft
๐ฅ๐ ๐๐ญ
, the shear flow direction is the same as the assumption
From๐๐ช๐ฎ๐๐ญ๐ข๐จ๐ง ๐๐จ . ๐: q2 = 15.17 q1 โ 15.24 q3 + 9,208.30 q2 = 15.17(17,949.08 ๐ช๐ = ๐๐๐, ๐๐๐. ๐๐
๐ฅ๐ ๐๐ญ
lb ft
lb lb lb ) โ 15.24 (9,216.18 ) + 9,208.30 ft ft ft , the shear flow direction is the same as the assumption
Torsional Shearing Stress at Cell 1: q1 ฯ1 = t1 lb 17,949.08 ft ฯ1 = 0.0104 ft ๐๐ = ๐, ๐๐๐, ๐๐๐. ๐๐
๐ฅ๐ ๐๐ญ ๐
Torsional Shearing Stress at Cell 2: q2 t2
ฯ2 = ฯ2 =
lb ft 0.0104 ft
141,041.26
๐๐ = ๐๐, ๐๐๐, ๐๐๐. ๐๐
๐ฅ๐ ๐๐ญ ๐
Torsional Shearing Stress at Cell 3: q3 t3
ฯ3 =
lb ft ฯ3 = 0.0104 ft 9,216.18
๐ฅ๐ ๐๐ญ ๐
๐๐ = ๐๐๐, ๐๐๐. ๐๐
Torsional Shearing Stress at Front Spar (FS): ฯFS = ฯFS
qFS
t W FS
lb ft = 0.0208 ft 5,521.78
๐๐
๐ = ๐๐๐, ๐๐๐. ๐๐
๐ฅ๐ ๐๐ญ ๐
Torsional Shearing Stress at Rear Spar (RS): ฯRS = ฯRS
qRS
t W RS
lb ft = 0.0208 ft 1,874.51
๐๐๐ = ๐๐, ๐๐๐. ๐๐
๐ฅ๐ ๐๐ญ ๐
Factor of Safety at Cell 1: FS1 =
ฯALLOWABLE ฯCOMPUTED
FS1 =
4,320,000 psf 1,725,873.08 psf
๐
๐๐ = ๐. ๐๐
Factor of Safety at Cell 2: FS2 =
ฯALLOWABLE ฯCOMPUTED
FS2 =
4,320,000 psf 13,561,659.62 psf
๐
๐๐ = ๐. ๐๐๐๐
Factor of Safety at Cell 3: FS3 =
ฯALLOWABLE ฯCOMPUTED
FS3 =
4,320,000 psf 886,171.15 psf
๐
๐๐ = ๐. ๐๐
Factor of Safety at Front Spar (FS): FSFS =
ฯALLOWABLE ฯCOMPUTED
FSFS =
4,320,000 psf 265,470.19 psf
๐
๐๐
๐ = ๐๐. ๐๐
Factor of Safety at Rear Spar (RS): ฯALLOWABLE ฯCOMPUTED 4,320,000 psf = 90,120.67 psf
FSRS = FSRS
๐
๐๐๐ = ๐๐. ๐๐
There are four major types of wing carrythrough structure. The "box carrythrough" is virtually standard for high-speed transports and general-aviation aircraft. The box carrythrough simply continues the wing box through the fuselage. The fuselage itself is not subjected to any of the bending moment of the wing, which minimizes fuselage weight. However, the box carrythrough occupies a substantial amount of fuselage volume, and tends to add cross-sectional area at the worst possible place for wave drag, as discussed above. Also, the box carrythrough interferes with the longeron load-paths. The "ring-frame" approach relies upon large, heavy bulkheads to carry the bending moment through the fuselage. The wing panels are attached to fittings on the side of these fuselage bulkheads. While this approach is usually heavier from a structural viewpoint, the resulting drag reduction at high speeds has led to the use of this approach for most modern fighters.
Many light aircraft and slower transport aircraft use an external strut to carry the bending moments. While this approach is probably the lightest of all, it obviously has a substantial drag penalty at higher speeds. Aircraft wings usually have the front spar at about 20-300Jo of the chord back from the leading edge. The rear spar is usually at about the 60-75% chord location. Additional spars may be located between the front and rear spars forming a "multispar" structure. Multispar structure is typical for large or high-speed aircraft. If the wing skin over the spars is an integral part of the wing structure, a "wing box" is formed which in most cases provides the minimum weight.
The "bending beam" carrythrough can be viewed as a compromise between these two approaches. Like the ring-frame approach, the wing panels are attached to the side of the fuselage to carry the lift forces. However, the bending moment is carried through the fuselage by one or several beams that connect the two wing panels. This approach has less of a fuselage volume increase than does the box-carrythrough approach.
Among the types of wing attachment, Zurcx777โs designer preferred to use โbending beamโ. Zurcx777 is a low wing aircraft therefore it can be attached by a slot on the undercarriage and can be connected together by a bending beam. The bending beam is utilized to avoid adding thickness to the bulkhead of the fuselage which is critical to a passenger aircraft. The moment of the wing will in turn be carried by the fuselage with the help of several beams connecting the two wings surfaces. With this, ideal space for the cabin would be utilized.
Aircraft with the landing gear in the wing will usually have the gear located aft of the wing box, with a single trailing-edge spar behind the gear to carry the flap loads. (See left figure below) Zurcx777 will use a โformer ribโ type for the wing rib. (See right figure below)
Source: P. Raymer, Daniel. "Special Considerations in Configuration Layout." Aircraft Design: A Conceptual Approach. Second ed. Washington, DC: American Institute of Aeronautics and Astronautics, 1992. 154-164. Print.
This activity discusses a number of important considerations, such as the wing attachment appropriate for the designed aircraft, the right position/location, material, crosssection and number of spar appropriate for the design and also the right position/location, material, number and type of wing rib appropriate for the design. All of these are numerically analyzed in the previous stages of the design process. During configuration layout, Zurcx777โs designer considered the impact in a qualitative sense. A poorly designed aircraft can have excessive drag and can cause lift losses or disruption. This activity may help aircraft designers during their concept layout to know what requirements to consider in order to attain aerodynamic efficiency. The development of a good structural arrangement has the primary concern of the provision of efficient โload pathsโ. These load paths are the structural elements by which opposing forces are connected. As mentioned earlier, this activity focuses on the position/location and the materials to be used for the designed aircraft it is because the size and weight of the structural members will be minimized by locating these opposing forces near to each other. Choosing for the material to be used is an important factor to be consider during this design process. The yield and ultimate strength, stiffness, density, fracture toughness, fatigue crack resistance, creep, corrosion resistance, temperature limits, producibility, repairability, cost and availability are the factors to be consider during the selection of the material. The wing loading is important in determining how rapidly the climb is established. A lightly laden wing has a further effective cruising performance for less thrust is required to maintain lift for level flight. Conversely, a heavily laden wing is more appropriate for higher speed flight because smaller wings offer less drag. A lighter loaded wing will have a superior rate of climb compared to a heavier loaded wing as less airspeed is required to generate the additional lift to increase altitude. Wing generates the lift required for flight. Spars, ribs and skins are the major structural elements of the wing. Spars carry the major wing bending loads. Ribs carry the shear loads on the wing. Ideal proportion of the weight and payload is important in design of an aircraft. It needs to be strong and stiff enough to withstand the exceptional circumstances in which it has to operate. Aerodynamic shape of the wing is important in creating the required lift for the aircraft. Ribs also help the wing to maintain its aerodynamic shape under loaded condition. The aerodynamic distributed load on the wing creates shear force, bending moment and torsional moment at wing stations. The "bending beam" carrythrough used by Zurcx777 has the wing panels attached to the side of the fuselage to carry the lift forces.
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