Wing Layout

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

𝐖𝐛 = Where: Wb = Intensity (lb⁄ft) Wb = Gross weight (3,531.57 lb) b = Wing span (31.49 ft)

𝐖𝐨 𝐛

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

3𝜋 2 4 3𝜋

(15.75 ft)

x = 6.68 ft Reaction of support B ↑ ΣFv = 0 F − RB = 0 2,649.54 − 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=x−

a 2

d = 6.68 ft − 2.04 ft d = 4.64 ft

Moment at R B : M = (F)(d) M = (2,649.54 lb)(4.64 ft) 𝐌 = 𝟏𝟐, 𝟐𝟗𝟑. 𝟖𝟕 𝐥𝐛 𝐟𝐭 = 𝟏𝟐, 𝟐𝟗𝟑. 𝟖𝟕 𝐥𝐛 𝐟𝐭 (12in) 𝐌 = 𝟏𝟒𝟕, 𝟓𝟐𝟔. 𝟒𝟒 𝐥𝐛 𝐢𝐧

Cross Section of FRONT SPAR UNIT: inches

Thickness = 1/4 in

Section

𝐘 in

𝐀 in2

𝐀𝐘 in3

𝐘𝟐 in2

𝐀𝐘 𝟐 in4

𝐈𝐜𝐱 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

𝐈𝐜𝐱 =

𝐛𝐡𝟑 (𝐟𝐨𝐫 𝐑𝐞𝐜𝐭𝐚𝐧𝐠𝐮𝐥𝐚𝐫) 𝟏𝟐

ΣA = 3.52 in2 ΣAy = 17.43 in3 ΣAy 2 = 113.35 in4 ΣIcx = 15.51 in4

Considering the Centroid at X-Axis C= ̅ Y= ̅ Y=

ΣAy ΣA

17.43 in3 3.52 in2

̅ = 𝟒. 𝟗𝟓 𝐢𝐧 𝐘

̅)2 I = ΣIxx − (ΣA)(Y ΣIxx = ΣIcx + ΣAy 2 ΣIxx = 15.51 in4 + 113.35 in4 ΣIxx = 128.86 in4 I = 128.86 in4 – (3.52 in2 )(4.95 in)2 𝐈 = 𝟒𝟐. 𝟔𝟏 𝐢𝐧𝟒

Solving for the Bending Stress SB =

Mc I

SB =

(147,526.44 lb in)(4.95 in) 42.61 in4

𝐒𝐁 = 𝟏𝟕, 𝟏𝟑𝟖. 𝟏𝟑

𝐥𝐛 𝐢𝐧𝟐

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 𝐅𝐒 = 1.75

Cross Section of REAR SPAR UNIT: inches

Thickness = 1/4 in

Section

𝐘 in

𝐀 in2

𝐀𝐘 in3

𝐘𝟐 in2

𝐀𝐘 𝟐 in4

𝐈𝐜𝐱 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

𝐈𝐜𝐱 =

𝐛𝐡𝟑 (𝐟𝐨𝐫 𝐑𝐞𝐜𝐭𝐚𝐧𝐠𝐮𝐥𝐚𝐫) 𝟏𝟐

ΣA = 3.10 in2 ΣAy = 12.70 in3 ΣAy 2 = 70.44 in4 ΣIcx = 9.08 in4

Considering the Centroid at X-Axis C= ̅ Y= ̅ Y=

ΣAy ΣA

12.70 in3 3.10 in2

̅ = 𝟒. 𝟏𝟎 𝐢𝐧 𝐘

̅)2 I = ΣIxx − (ΣA)(Y ΣIxx = ΣIcx + ΣAy 2 ΣIxx = 9.08 in4 + 70.44 in4 ΣIxx = 79.52 in4 I = 79.52 in4 – (3.10 in2 )(4.10 in)2 𝐈 = 𝟐𝟕. 𝟒𝟏 𝐢𝐧𝟒

Solving for the Bending Stress SB =

Mc I

SB =

(147,526.44 lb in)(4.10 in) 27.41 in4

𝐒𝐁 = 𝟐𝟐, 𝟎𝟔𝟕. 𝟎𝟕

𝐥𝐛 𝐢𝐧𝟐

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

𝐅𝐒 = 1.36

WING RIB DESIGN Rib No. 1 – BUTT RIB Located 2.04 ft. from the aircraft’s 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, 𝐂𝐦 𝐜 𝟒

+↷ MA = 0 0 = − PRS (2.25 ft) + 2,649.54 lb (0.50 ft)

PRS =

2,649.54 lb (0.50 ft) 2.25 ft

𝐏𝐑𝐒 = 𝟓𝟖𝟖. 𝟕𝟗 𝐥𝐛 +↑ ΣFV = 0 588.79 lb − 2,649.54 lb + PFS = 0 PFS = 2,649.54 lb − 588.79 lb 𝐏𝐅𝐒 = 𝟐, 𝟎𝟔𝟎. 𝟐𝟏 𝐥𝐛

Shear Flow at Front Spar (FS) q WFS =

PFS hFS

q WFS =

2,060.21 lb 0.7967 ft 𝐪𝐖𝐅𝐒 = 𝟐, 𝟓𝟖𝟓. 𝟗𝟑

𝐥𝐛 𝐟𝐭

Shear Flow at Rear Spar (RS) q WRS =

PRS hRS

q WRS =

588.79 lb 0.6730 ft 𝐪𝐖𝐑𝐒 = 𝟖𝟕𝟒. 𝟖𝟕

𝐥𝐛 𝐟𝐭

Shear Flow at 1: 2θ1 G =

1 l1 hFS [q1 + (q1 − q2 + q WFS ) ] A1 t1 t W FS

1 2.61 ft lb 0.7967 ft [(q1 ) ( ) + (q1 − q2 + 2,585.93 ) ( )] 2 0.4343 ft 0.0104 ft ft 0.0208 ft 1 lb (250.96 q1 + 38.3029 q1 − 38.3029 q2 + 99,048.58 ) 2θ1 G = 0.4343 ft 2 ft 𝐥𝐛 𝟐𝛉𝟏 𝐆 = 𝟔𝟔𝟔. 𝟎𝟒 𝐪𝟏 − 𝟖𝟖. 𝟏𝟗𝟒𝟔 𝐪𝟐 + 𝟐𝟐𝟖, 𝟎𝟔𝟒. 𝟖𝟗 𝐄𝐪𝐮𝐚𝐭𝐢𝐨𝐧 𝐍𝐨. 𝟏 𝐟𝐭 2θ1 G =

Shear Flow at 2: 2θ2 G =

1 l2 hFS hRS [q2 + (q2 − q WFS − q1 ) ( ) + (q2 + q WRS − q3 ) ( )] A2 t2 t W FS t W RS

1 5.99 ft lb 0.7967 ft [(q2 ) ( ) + (q2 − 2,585.93 − q1 ) ( ) 2 1.87 ft 0.0104 ft ft 0.0208 ft lb 0.6730 ft )] + (q2 + 874.87 − q3 ) ( ft 0.0208 ft 1 lb (575.96 q2 + 38.3029 q2 − 99,048.58 − 38.3029 q1 + 32.36 q2 2θ2 G = 2 1.87 ft ft lb + 28,307.09 − 32.36 q3 ) ft 𝐥𝐛 𝟐𝛉𝟐 𝐆 = −𝟐𝟎. 𝟒𝟖 𝐪𝟏 + 𝟑𝟒𝟓. 𝟕𝟗 𝐪𝟐 − 𝟏𝟕. 𝟑𝟎 𝐪𝟑 − 𝟑𝟕, 𝟖𝟐𝟗. 𝟔𝟕 𝐄𝐪𝐮𝐚𝐭𝐢𝐨𝐧 𝐍𝐨. 𝟐 𝐟𝐭 2θ2 G =

Shear Flow at 3: 2θ3 G =

1 l3 hRS [q3 + (q3 − q2 − q WRS ) ] A3 t3 t W RS

1 4.73 ft lb 0.6730 ft ) [(q ( ) (q )( )] + − q − 874.87 3 3 2 0.7292 ft 2 0.0104 ft ft 0.0208 ft 1 lb (454.81 q3 + 32.36 q3 − 32.36 q2 − 28,307.09 ) 2θ3 G = 2 0.7292 ft ft 𝐥𝐛 𝟐𝛉𝟑 𝐆 = 𝟔𝟔𝟖. 𝟎𝟗 𝐪𝟑 − 𝟒𝟒. 𝟑𝟖 𝐪𝟐 − 𝟑𝟖, 𝟖𝟏𝟗. 𝟑𝟖 𝐄𝐪𝐮𝐚𝐭𝐢𝐨𝐧 𝐍𝐨 𝟑 𝐟𝐭 2θ3 G =

Assuming that the Twist Angles are equal in the cells: 𝜽𝟏 = 𝜽𝟐 = 𝜽𝟑

Equate 𝐄𝐪𝐮𝐚𝐭𝐢𝐨𝐧 𝐍𝐨. 𝟏 and𝐄𝐪𝐮𝐚𝐭𝐢𝐨𝐧 𝐍𝐨 𝟑: lb lb = 668.09 q3 − 44.38 q2 − 38,819.38 ft ft lb lb −44.38 q2 + 88.1946 q2 = 666.04 q1 − 668.09 q3 + 38,824.45 + 228,064.89 ft ft lb 43.81 q2 = 666.04 q1 − 668.09 q3 + 266889.34 ft lb 666.04 q1 − 668.09 q3 + 266,889.34 ft q2 = 43.81 𝐥𝐛 𝐪𝟐 = 𝟏𝟓. 𝟐𝟎 𝐪𝟏 − 𝟏𝟓. 𝟐𝟓 𝐪𝟑 + 𝟔, 𝟎𝟗𝟏. 𝟗𝟕 𝐄𝐪𝐮𝐚𝐭𝐢𝐨𝐧 𝐍𝐨 . 𝟒 𝐟𝐭 666.04 q1 − 88.1946 q2 + 228,064.89

Substitute 𝐄𝐪𝐮𝐚𝐭𝐢𝐨𝐧 𝐍𝐨 . 𝟒 to 𝐄𝐪𝐮𝐚𝐭𝐢𝐨𝐧 𝐍𝐨. 𝟏 666.04 q1 − 88.1946 q2 + 228,064.89

lb ft

 𝐄𝐪𝐮𝐚𝐭𝐢𝐨𝐧 𝐍𝐨. 𝟏

lb lb 666.04 q1 = 88.1946 (15.20 q1 − 15.25 q3 + 6,091.97 ) − 228,064.89 ft ft lb lb 666.04 q1 = 1,340.56 q1 − 1,344.97 q3 + 537,278.86 − 228,064.89 ft ft lb 674.52 q1 = 1,349.38 q3 − 309,213.97 ft lb 1,349.38 q3 − 309,213.97 ft q1 = 674.52 𝐥𝐛 𝐪𝟏 = 𝟐. 𝟎𝟎 𝐪𝟑 − 𝟒𝟓𝟖. 𝟒𝟐 𝐄𝐪𝐮𝐚𝐭𝐢𝐨𝐧 𝐍𝐨. 𝟓 𝐟𝐭

Substitute 𝐄𝐪𝐮𝐚𝐭𝐢𝐨𝐧 𝐍𝐨 . 𝟒 to𝐄𝐪𝐮𝐚𝐭𝐢𝐨𝐧 𝐍𝐨 𝟑: 668.09 q3 − 44.38 q2 − 38,819.38

lb ft

 𝐄𝐪𝐮𝐚𝐭𝐢𝐨𝐧 𝐍𝐨 𝟑

lb lb 668.09 q3 = 44.38 (15.20 q1 − 15.25 q3 + 6,091.97 ) + 38,819.38 ft ft lb lb 668.09 q3 = 674.58 q1 − 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 𝐪𝟑 = 𝟎. 𝟓𝟎𝟏𝟔 𝐪𝟏 + 𝟐𝟐𝟗. 𝟖𝟗

𝐥𝐛 𝐟𝐭

𝐄𝐪𝐮𝐚𝐭𝐢𝐨𝐧 𝐍𝐨. 𝟔

Substitute 𝐄𝐪𝐮𝐚𝐭𝐢𝐨𝐧 𝐍𝐨. 𝟓 to𝐄𝐪𝐮𝐚𝐭𝐢𝐨𝐧 𝐍𝐨. 𝟔: q3 = 0.5016 q1 + 229.89

lb ft

 𝐄𝐪𝐮𝐚𝐭𝐢𝐨𝐧 𝐍𝐨. 𝟔

lb lb ) + 229.89 ft ft lb lb q3 = 1.0032 q3 − 229.94 + 229.89 ft ft lb −0.01q3 = −0.05 ft lb 0.05 ft q3 = 0.0032 q3 = 0.5016 (2.00 q3 − 458.42

𝐪𝟑 = 𝟏𝟓. 𝟔𝟑

𝐥𝐛 𝐟𝐭

, the shear flow direction is same as the assumption

From𝐄𝐪𝐮𝐚𝐭𝐢𝐨𝐧 𝐍𝐨. 𝟓: q1 = 2.00 q3 − 458.42 q1 = 2.00 (15.63 𝐪𝟏 = 𝟒𝟐𝟕. 𝟏𝟔

𝐥𝐛 𝐟𝐭

lb ft

lb lb ) − 458.42 ft ft , the shear flow is opposite from the assumption

From𝐄𝐪𝐮𝐚𝐭𝐢𝐨𝐧 𝐍𝐨 . 𝟒: q2 = 15.20 q1 − 15.25 q3 + 6,091.97

lb ft

lb lb lb q2 = 15.20 (427.16 ) − 15.25(15.63 ) + 6,091.97 ft ft ft 𝐪𝟐 = 𝟏𝟐, 𝟑𝟒𝟔. 𝟒𝟒

𝐥𝐛 𝐟𝐭

, the shear flow direction is the same as the assumption

Torsional Shearing Stress at Cell 1: q1 τ1 = t1 lb 427.16 ft τ1 = 0.0104 ft 𝛕𝟏 = 𝟒𝟏, 𝟎𝟕𝟑. 𝟎𝟖

𝐥𝐛 𝐟𝐭 𝟐

Torsional Shearing Stress at Cell 2: q2 t2

τ2 =

lb ft τ2 = 0.0104 ft 12,346.44

𝛕𝟐 = 𝟏, 𝟏𝟖𝟕, 𝟏𝟓𝟕. 𝟔𝟗

𝐥𝐛 𝐟𝐭 𝟐

Torsional Shearing Stress at Cell 3: q3 t3

τ3 =

lb ft τ3 = 0.0104 ft 15.63

𝛕𝟑 = 𝟏, 𝟓𝟎𝟐. 𝟖𝟖

𝐥𝐛 𝐟𝐭 𝟐

Torsional Shearing Stress at Front Spar (FS): τFS = τFS

qFS

t W FS

lb ft = 0.0208 ft 2,585.93

𝛕𝐅𝐒 = 𝟏𝟐𝟒, 𝟑𝟐𝟑. 𝟓𝟔

𝐥𝐛 𝐟𝐭 𝟐

Torsional Shearing Stress at Rear Spar (RS): τRS = τRS

qRS

t W RS

lb ft = 0.0208 ft 874.87

𝛕𝐑𝐒 = 𝟒𝟐, 𝟎𝟔𝟏. 𝟎𝟔

𝐥𝐛 𝐟𝐭 𝟐

Factor of Safety at Cell 1: FS1 =

τALLOWABLE τCOMPUTED

FS1 =

4,320,000 psf 41,073.08 psf

𝐅𝐒𝟏 = 𝟏𝟎𝟓. 𝟏𝟖

Factor of Safety at Cell 2: FS2 =

τALLOWABLE τCOMPUTED

FS2 =

4,320,000 psf 1,187,157.69 psf

𝐅𝐒𝟐 = 𝟑. 𝟔𝟒

Factor of Safety at Cell 3: FS3 =

τALLOWABLE τCOMPUTED

FS3 =

4,320,000 psf 1,502.88 psf

𝐅𝐒𝟑 = 𝟐, 𝟖𝟕𝟒. 𝟒𝟖

Factor of Safety at Front Spar (FS): τALLOWABLE τCOMPUTED 4,320,000 psf = 124,323.56 psf

FSFS = FSFS

𝐅𝐒𝐅𝐒 = 𝟑𝟒. 𝟕𝟓

Factor of Safety at Rear Spar (RS): τALLOWABLE τCOMPUTED 4,320,000 psf = 42,061.06 psf

FSRS = FSRS

𝐅𝐒𝐑𝐒 = 𝟏𝟎𝟐. 𝟕𝟏

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, 𝐂𝐦 𝐜 𝟒

+↷ MA = 0 0 = − PRS (1.05 ft) + 2,649.54 lb (0.2340 ft) PRS =

2,649.54 lb (0.2340 ft) 1.05 ft

𝐏𝐑𝐒 = 𝟓𝟗𝟎. 𝟒𝟕 𝐥𝐛

+↑ ΣFV = 0 590.47 lb − 2,649.54 lb + PFS = 0 PFS = 2,649.54 lb − 590.47 lb 𝐏𝐅𝐒 = 𝟐, 𝟎𝟓𝟗. 𝟎𝟕 𝐥𝐛

Shear Flow at Front Spar (FS) q WFS =

PFS hFS

q WFS =

2,059.07 lb 0.3729 ft 𝐪𝐖𝐅𝐒 = 𝟓, 𝟓𝟐𝟏. 𝟕𝟖

𝐥𝐛 𝐟𝐭

Shear Flow at Rear Spar (RS) q WRS =

PRS hRS

q WRS =

590.47 lb 0.3150 ft 𝐪𝐖𝐑𝐒 = 𝟏, 𝟖𝟕𝟒. 𝟓𝟏

𝐥𝐛 𝐟𝐭

Shear Flow at 1: 2θ1 G =

1 l1 hFS [q1 + (q1 − q2 + q WFS ) ] A1 t1 t W FS

1 1.2196 ft lb 0.3729 ft [(q ( ) (q )( )] ) + − q + 5,521.78 1 1 2 0.0951 ft 2 0.0104 ft ft 0.0208 ft 1 lb (117.27 ) 2θ1 G = q + 17.93 q − 17.93 q + 98,993.83 1 1 2 0.0951 ft 2 ft 𝐥𝐛 𝟐𝛉𝟏 𝐆 = 𝟏, 𝟒𝟐𝟏. 𝟔𝟔 𝐪𝟏 − 𝟏𝟖𝟖. 𝟓𝟒 𝐪𝟐 + 𝟏, 𝟎𝟒𝟎, 𝟗𝟒𝟒. 𝟓𝟖 𝐄𝐪𝐮𝐚𝐭𝐢𝐨𝐧 𝐍𝐨. 𝟏 𝐟𝐭 2θ1 G =

Shear Flow at 2: 2θ2 G =

1 l2 hFS hRS [q2 + (q2 − q WFS − q1 ) ( ) + (q2 + q WRS − q3 ) ( )] A2 t2 t W FS t W RS

1 2.8014 ft lb 0.3729 ft [(q2 ) ( ) + (q2 − 5,521.78 − q1 ) ( ) 2 0.4089 ft 0.0104 ft ft 0.0208 ft lb 0.3150 ft )] + (q2 + 1,874.51 − q3 ) ( ft 0.0208 ft 1 lb (269.37 q2 + 17.93 q2 − 98,993.83 − 17.93 q1 + 15.14 q2 2θ2 G = 2 0.4089 ft ft lb + 28,388.01 − 15.14 q3 ) ft 𝐥𝐛 𝟐𝛉𝟐 𝐆 = −𝟒𝟑. 𝟖𝟓 𝐪𝟏 + 𝟕𝟑𝟗. 𝟔𝟒 𝐪𝟐 − 𝟑𝟕. 𝟎𝟑 𝐪𝟑 − 𝟕𝟎, 𝟔𝟎𝟓. 𝟖𝟐 𝐄𝐪𝐮𝐚𝐭𝐢𝐨𝐧 𝐍𝐨. 𝟐 𝐟𝐭 2θ2 G =

Shear Flow at 3: 2θ3 G =

1 l3 hRS [q3 + (q3 − q2 − q WRS ) ] A3 t3 t W RS

1 2.2158 ft lb 0.3150 ft [(q3 ) ( ) + (q3 − q2 − 1,874.51 ) ( )] 2 0.1597 ft 0.0104 ft ft 0.0208 ft 1 lb (213.06 q3 + 15.14 q3 − 15.14 q2 − 28,388.01 ) 2θ3 G = 2 0.1597 ft ft 𝐥𝐛 𝟐𝛉𝟑 𝐆 = 𝟏, 𝟒𝟐𝟖. 𝟗𝟑 𝐪𝟑 − 𝟗𝟒. 𝟖𝟎 𝐪𝟐 − 𝟏𝟕𝟕, 𝟕𝟓𝟖. 𝟑𝟔 𝐄𝐪𝐮𝐚𝐭𝐢𝐨𝐧 𝐍𝐨 𝟑 𝐟𝐭 2θ3 G =

Assuming that the Twist Angles are equal in the cells: 𝜽𝟏 = 𝜽𝟐 = 𝜽𝟑

Equate 𝐄𝐪𝐮𝐚𝐭𝐢𝐨𝐧 𝐍𝐨. 𝟏 and𝐄𝐪𝐮𝐚𝐭𝐢𝐨𝐧 𝐍𝐨 𝟑: lb lb = 1,428.93 q3 − 94.80 q2 − 177,758.36 ft ft lb lb −94.80 q2 + 188.54 q2 = 1,421.66 q1 − 1,428.93 q3 − 177,758.36 + 1,040,944.58 ft ft lb 93.74 q2 = 1,421.66 q1 − 1,428.93 q3 + 863,186.22 ft lb 1,421.66 q1 − 1,428.93 q3 + 863,186.22 ft q2 = 93.74 𝐥𝐛 𝐪𝟐 = 𝟏𝟓. 𝟏𝟕 𝐪𝟏 − 𝟏𝟓. 𝟐𝟒 𝐪𝟑 + 𝟗, 𝟐𝟎𝟖. 𝟑𝟎 𝐄𝐪𝐮𝐚𝐭𝐢𝐨𝐧 𝐍𝐨 . 𝟒 𝐟𝐭 1,421.66 q1 − 188.54 q2 + 1,040,944.58

Substitute 𝐄𝐪𝐮𝐚𝐭𝐢𝐨𝐧 𝐍𝐨 . 𝟒 to 𝐄𝐪𝐮𝐚𝐭𝐢𝐨𝐧 𝐍𝐨. 𝟏 1,421.66 q1 − 188.54 q2 + 1,040,944.58

lb ft

 𝐄𝐪𝐮𝐚𝐭𝐢𝐨𝐧 𝐍𝐨. 𝟏

lb lb ) − 1,040,944.58 ft ft lb lb 1,421.66 q1 = 2,860.15 q1 − 2,873.35 q3 + 1,736,132.88 − 1,040,944.58 ft ft lb 1,438.49 q1 = 2,873.35 q3 − 695,188.30 ft lb 2,873.35 q3 − 695,188.30 ft q1 = 1,438.49 1,421.66 q1 = 188.54 (15.17 q1 − 15.24 q3 + 9,208.30

𝐪𝟏 = 𝟐. 𝟎𝟎 𝐪𝟑 − 𝟒𝟖𝟑. 𝟐𝟖

𝐥𝐛 𝐟𝐭

𝐄𝐪𝐮𝐚𝐭𝐢𝐨𝐧 𝐍𝐨. 𝟓

Substitute 𝐄𝐪𝐮𝐚𝐭𝐢𝐨𝐧 𝐍𝐨 . 𝟒 to𝐄𝐪𝐮𝐚𝐭𝐢𝐨𝐧 𝐍𝐨 𝟑: 1,428.93 q3 − 94.80 q2 − 177,758.36

lb ft

 𝐄𝐪𝐮𝐚𝐭𝐢𝐨𝐧 𝐍𝐨 𝟑

lb ) + 177,758.36 ft lb 1,428.93 q3 = 1,438.12 q1 − 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 − 15.24 q3 + 9,208.30

𝐪𝟑 = 𝟎. 𝟓𝟎𝟔𝟖 𝐪𝟏 + 𝟑𝟕𝟎. 𝟐𝟕

𝐥𝐛 𝐟𝐭

𝐄𝐪𝐮𝐚𝐭𝐢𝐨𝐧 𝐍𝐨. 𝟔

lb ft lb ft lb ft

Substitute 𝐄𝐪𝐮𝐚𝐭𝐢𝐨𝐧 𝐍𝐨. 𝟓 to𝐄𝐪𝐮𝐚𝐭𝐢𝐨𝐧 𝐍𝐨. 𝟔: q3 = 0.5068 q1 + 370.27

lb ft

 𝐄𝐪𝐮𝐚𝐭𝐢𝐨𝐧 𝐍𝐨. 𝟔

lb lb ) + 370.27 ft ft lb lb q3 = 1.0136 q3 − 244.93 + 370.27 ft ft lb 0.0136 q3 = −125.34 ft lb −125.34 ft q3 = 0.0136 q3 = 0.5068 (2.00 q3 − 483.28

𝐪𝟑 = 𝟗, 𝟐𝟏𝟔. 𝟏𝟖

𝐥𝐛

, the shear flow is opposite from the assumption

𝐟𝐭

From𝐄𝐪𝐮𝐚𝐭𝐢𝐨𝐧 𝐍𝐨. 𝟓: q1 = 2.00 q3 − 483.28 q1 = 2.00 (9,216.18 𝐪𝟏 = 𝟏𝟕, 𝟗𝟒𝟗. 𝟎𝟖

lb ft

lb lb ) − 483.28 ft ft

𝐥𝐛 𝐟𝐭

, the shear flow direction is the same as the assumption

From𝐄𝐪𝐮𝐚𝐭𝐢𝐨𝐧 𝐍𝐨 . 𝟒: q2 = 15.17 q1 − 15.24 q3 + 9,208.30 q2 = 15.17(17,949.08 𝐪𝟐 = 𝟏𝟒𝟏, 𝟎𝟒𝟏. 𝟐𝟔

𝐥𝐛 𝐟𝐭

lb ft

lb lb lb ) − 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 τ1 = t1 lb 17,949.08 ft τ1 = 0.0104 ft 𝛕𝟏 = 𝟏, 𝟕𝟐𝟓, 𝟖𝟕𝟑. 𝟎𝟖

𝐥𝐛 𝐟𝐭 𝟐

Torsional Shearing Stress at Cell 2: q2 t2

τ2 = τ2 =

lb ft 0.0104 ft

141,041.26

𝛕𝟐 = 𝟏𝟑, 𝟓𝟔𝟏, 𝟔𝟓𝟗. 𝟔𝟐

𝐥𝐛 𝐟𝐭 𝟐

Torsional Shearing Stress at Cell 3: q3 t3

τ3 =

lb ft τ3 = 0.0104 ft 9,216.18

𝐥𝐛 𝐟𝐭 𝟐

𝛕𝟑 = 𝟖𝟖𝟔, 𝟏𝟕𝟏. 𝟏𝟓

Torsional Shearing Stress at Front Spar (FS): τFS = τFS

qFS

t W FS

lb ft = 0.0208 ft 5,521.78

𝛕𝐅𝐒 = 𝟐𝟔𝟓, 𝟒𝟕𝟎. 𝟏𝟗

𝐥𝐛 𝐟𝐭 𝟐

Torsional Shearing Stress at Rear Spar (RS): τRS = τRS

qRS

t W RS

lb ft = 0.0208 ft 1,874.51

𝛕𝐑𝐒 = 𝟗𝟎, 𝟏𝟐𝟎. 𝟔𝟕

𝐥𝐛 𝐟𝐭 𝟐

Factor of Safety at Cell 1: FS1 =

τALLOWABLE τCOMPUTED

FS1 =

4,320,000 psf 1,725,873.08 psf

𝐅𝐒𝟏 = 𝟐. 𝟓𝟎

Factor of Safety at Cell 2: FS2 =

τALLOWABLE τCOMPUTED

FS2 =

4,320,000 psf 13,561,659.62 psf

𝐅𝐒𝟐 = 𝟎. 𝟑𝟏𝟖𝟓

Factor of Safety at Cell 3: FS3 =

τALLOWABLE τCOMPUTED

FS3 =

4,320,000 psf 886,171.15 psf

𝐅𝐒𝟑 = 𝟒. 𝟖𝟕

Factor of Safety at Front Spar (FS): FSFS =

τALLOWABLE τCOMPUTED

FSFS =

4,320,000 psf 265,470.19 psf

𝐅𝐒𝐅𝐒 = 𝟏𝟔. 𝟐𝟕

Factor of Safety at Rear Spar (RS): τALLOWABLE τCOMPUTED 4,320,000 psf = 90,120.67 psf

FSRS = FSRS

𝐅𝐒𝐑𝐒 = 𝟒𝟕. 𝟗𝟒

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, Zurcx777’s designer preferred to use “bending beam”. 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 “former rib” 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, Zurcx777’s 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 “load paths”. 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.