Wing Layout

(1)

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

𝐖

_𝐛

=

𝐖

𝐨

𝐛

Where:

W

b

= Intensity (lb ft

⁄ )

W

b

= Gross weight (3,531.57 lb)

b = Wing span (31.49 ft)

(2)

W

_b

=

3,531.57 lb

31.49 ft

W

_b

= 112.15

lb

ft

Considering the Right Hand

F = (W

_b

) (

b₂

)

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

(3)

Get the Centroid

x =

4

3𝜋

(

b

2

) , for quarter ellipse

x =

4 3𝜋

(15.75 ft)

x = 6.68 ft

Reaction of support B

↑ ΣF

_v

= 0

F − R

B

= 0

2,649.54 − R

B

= 0

R

B

= 2,649.54 lb

(4)

Bending Stress

S

b

=

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

(5)

Cross Section of FRONT SPAR

UNIT: inches

Thickness = 1/4 in

Section

𝐘

𝐀

𝐀𝐘

𝐘

𝟐

𝐀𝐘

𝟐

𝐈𝐜𝐱

in

2

in

3

in

2

in

4

in

4 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 =

𝐛𝐡

𝟑

𝟏𝟐 (𝐟𝐨𝐫 𝐑𝐞𝐜𝐭𝐚𝐧𝐠𝐮𝐥𝐚𝐫)

ΣA = 3.52 in

2

ΣAy = 17.43 in

3

ΣAy

2

= 113.35 in

4

ΣIcx = 15.51 in

4

Considering the Centroid at X-Axis

C = Y̅ = ΣAy ΣA Y ̅ = 17.43 in3 3.52 in2 𝐘 ̅ = 𝟒. 𝟗𝟓 𝐢𝐧

(6)

I = ΣIxx − (ΣA)(Y

̅)

2

ΣIxx = ΣIcx + ΣAy

2

ΣIxx = 15.51 in

4

+ 113.35 in

4

ΣIxx = 128.86 in

4

I = 128.86 in

4

– (3.52 in

2

)(4.95 in)

2

𝐈 = 𝟒𝟐. 𝟔𝟏 𝐢𝐧

𝟒

Solving for the Bending Stress

S

_B

=

Mc

I

S

B

=

(147,526.44 lb in)(4.95 in)

42.61 in

4

𝐒

_𝐁

= 𝟏𝟕, 𝟏𝟑𝟖. 𝟏𝟑

𝐥𝐛

𝐢𝐧

𝟐

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

(7)

Cross Section of REAR SPAR

UNIT: inches

Thickness = 1/4 in

Section

𝐘

𝐀

𝐀𝐘

𝐘

𝟐

𝐀𝐘

𝟐

𝐈𝐜𝐱

in

2

in

3

in

2

in

4

in

4 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 =

𝐛𝐡

𝟑

𝟏𝟐 (𝐟𝐨𝐫 𝐑𝐞𝐜𝐭𝐚𝐧𝐠𝐮𝐥𝐚𝐫)

ΣA = 3.10 in

2

ΣAy = 12.70 in

3

ΣAy

2

= 70.44 in

4

ΣIcx = 9.08 in

4

Considering the Centroid at X-Axis

C = Y̅ = ΣAy ΣA Y ̅ = 12.70 in3 3.10 in2 𝐘̅ = 𝟒. 𝟏𝟎 𝐢𝐧

(8)

I = ΣIxx − (ΣA)(Y

̅)

2

ΣIxx = ΣIcx + ΣAy

2

ΣIxx = 9.08 in

4

_{+ 70.44 in}

4

ΣIxx = 79.52 in

4

I = 79.52 in

4

_{– (3.10 in}

2

_{)(4.10 in)}

2

𝐈 = 𝟐𝟕. 𝟒𝟏 𝐢𝐧

𝟒

Solving for the Bending Stress

S

_B

=

Mc

I

S

_B

=

(147,526.44 lb in)(4.10 in)

27.41 in

4

𝐒

_𝐁

= 𝟐𝟐, 𝟎𝟔𝟕. 𝟎𝟕

𝐥𝐛

𝐢𝐧

𝟐

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

(9)

WING RIB DESIGN

Rib No. 1 – BUTT RIB Located 2.04 ft. from the aircraft’s center

Given:

A₁= 0.4343 ft2 _t WFS= 0.0208 ft or 0.25 in l₁= 2.61 ft t_WRS = 0.0208 ft or 0.25 in A₂= 1.87 ft2 t₁= 0.0104 ft or 0.125 in l₂= 5.99 ft t₂ = 0.0104 ft or 0.125 in t₃ = 0.0104 ft or 0.125 in A₃= 0.7292 ft2 l₃= 4.73 ft

Moment at Quarter Chord,

𝐂_𝐦_𝐜

𝟒

+↷M_A= 0

(10)

P_RS= 2,649.54 lb (0.50 ft) 2.25 ft 𝐏_𝐑𝐒= 𝟓𝟖𝟖. 𝟕𝟗 𝐥𝐛 +↑ ΣF_V = 0 588.79 lb − 2,649.54 lb + P_FS= 0 P_FS= 2,649.54 lb − 588.79 lb 𝐏_𝐅𝐒= 𝟐, 𝟎𝟔𝟎. 𝟐𝟏 𝐥𝐛

Shear Flow at Front Spar (FS)

qW_FS = P_FS h_FS q_W_FS =2,060.21 lb 0.7967 ft 𝐪_𝐖_𝐅𝐒 = 𝟐, 𝟓𝟖𝟓. 𝟗𝟑 𝐥𝐛 𝐟𝐭

Shear Flow at Rear Spar (RS)

q_W_RS = PRS h_RS q_W_RS =588.79 lb 0.6730 ft 𝐪_𝐖_𝐑𝐒 = 𝟖𝟕𝟒. 𝟖𝟕 𝐥𝐛 𝐟𝐭 Shear Flow at 1: 2θ₁G = 1 A₁[q1 l1 t₁+ (q1− q2+ qWFS) hFS t_WFS ] 2θ₁G = 1 0.4343 ft2[(q1) ( 2.61 ft 0.0104 ft) + (q1− q2+ 2,585.93 lb ft) ( 0.7967 ft 0.0208 ft) ] 2θ₁G = 1 0.4343 ft2(250.96 q1+ 38.3029 q1− 38.3029 q2+ 99,048.58 lb ft) 𝟐𝛉_𝟏𝐆 = 𝟔𝟔𝟔. 𝟎𝟒 𝐪_𝟏− 𝟖𝟖. 𝟏𝟗𝟒𝟔 𝐪_𝟐+ 𝟐𝟐𝟖, 𝟎𝟔𝟒. 𝟖𝟗 𝐥𝐛 𝐟𝐭 𝐄𝐪𝐮𝐚𝐭𝐢𝐨𝐧 𝐍𝐨. 𝟏

(11)

Shear Flow at 2: 2θ₂G = 1 A₂[q2 l₂ t₂+ (q2− qWFS− q1) ( h_FS t_WFS) + (q2+ qWRS− q3) ( h_RS t_WRS) ] 2θ₂G = 1 1.87 ft2[(q2) ( 5.99 ft 0.0104 ft) + (q2− 2,585.93 lb ft− q1) ( 0.7967 ft 0.0208 ft) + (q₂+ 874.87 lb ft− q3) ( 0.6730 ft 0.0208 ft) ] 2θ₂G = 1 1.87 ft2(575.96 q2+ 38.3029 q2− 99,048.58 lb ft− 38.3029 q1+ 32.36 q2 + 28,307.09 lb ft − 32.36 q3) 𝟐𝛉_𝟐𝐆 = −𝟐𝟎. 𝟒𝟖 𝐪_𝟏+ 𝟑𝟒𝟓. 𝟕𝟗 𝐪_𝟐− 𝟏𝟕. 𝟑𝟎 𝐪_𝟑− 𝟑𝟕, 𝟖𝟐𝟗. 𝟔𝟕 𝐥𝐛 𝐟𝐭 𝐄𝐪𝐮𝐚𝐭𝐢𝐨𝐧 𝐍𝐨. 𝟐 Shear Flow at 3: 2θ₃G = 1 A₃[q3 l₃ t₃+ (q3− q2− qWRS) h_RS t_WRS ] 2θ₃G = 1 0.7292 ft2[(q3) ( 4.73 ft 0.0104 ft) + (q3− q2− 874.87 lb ft) ( 0.6730 ft 0.0208 ft) ] 2θ₃G = 1 0.7292 ft2(454.81 q3+ 32.36 q3− 32.36 q2− 28,307.09 lb ft) 𝟐𝛉_𝟑𝐆 = 𝟔𝟔𝟖. 𝟎𝟗 𝐪_𝟑− 𝟒𝟒. 𝟑𝟖 𝐪_𝟐− 𝟑𝟖, 𝟖𝟏𝟗. 𝟑𝟖 𝐥𝐛 𝐟𝐭 𝐄𝐪𝐮𝐚𝐭𝐢𝐨𝐧 𝐍𝐨 𝟑

Assuming that the Twist Angles are equal in the cells:

𝜽_𝟏= 𝜽_𝟐= 𝜽_𝟑

Equate

𝐄𝐪𝐮𝐚𝐭𝐢𝐨𝐧 𝐍𝐨. 𝟏

and

𝐄𝐪𝐮𝐚𝐭𝐢𝐨𝐧 𝐍𝐨 𝟑

:

666.04 q₁− 88.1946 q₂+ 228,064.89 lb ft =668.09 q3− 44.38 q2− 38,819.38 lb ft −44.38 q₂+ 88.1946 q₂= 666.04 q₁− 668.09 q₃+ 38,824.45 lb ft+ 228,064.89 lb ft 43.81 q₂= 666.04 q₁− 668.09 q₃+ 266889.34lb ft q₂=666.04 q1− 668.09 q3+ 266,889.34 lb ft 43.81 𝐪_𝟐 = 𝟏𝟓. 𝟐𝟎 𝐪_𝟏− 𝟏𝟓. 𝟐𝟓 𝐪_𝟑+ 𝟔, 𝟎𝟗𝟏. 𝟗𝟕𝐥𝐛 𝐟𝐭 𝐄𝐪𝐮𝐚𝐭𝐢𝐨𝐧 𝐍𝐨 . 𝟒

(12)

Substitute 𝐄𝐪𝐮𝐚𝐭𝐢𝐨𝐧 𝐍𝐨 . 𝟒 to 𝐄𝐪𝐮𝐚𝐭𝐢𝐨𝐧 𝐍𝐨. 𝟏

666.04 q₁− 88.1946 q₂+ 228,064.89 lb ft  𝐄𝐪𝐮𝐚𝐭𝐢𝐨𝐧 𝐍𝐨. 𝟏 666.04 q₁ = 88.1946 (15.20 q₁− 15.25 q₃+ 6,091.97lb ft) − 228,064.89 lb ft 666.04 q₁ = 1,340.56 q₁− 1,344.97 q₃+ 537,278.86 lb ft− 228,064.89 lb ft 674.52 q₁ = 1,349.38 q₃− 309,213.97lb ft q₁=1,349.38 q3− 309,213.97 lb ft 674.52 𝐪𝟏= 𝟐. 𝟎𝟎 𝐪𝟑− 𝟒𝟓𝟖. 𝟒𝟐 𝐥𝐛 𝐟𝐭 𝐄𝐪𝐮𝐚𝐭𝐢𝐨𝐧 𝐍𝐨. 𝟓

Substitute 𝐄𝐪𝐮𝐚𝐭𝐢𝐨𝐧 𝐍𝐨 . 𝟒 to𝐄𝐪𝐮𝐚𝐭𝐢𝐨𝐧 𝐍𝐨 𝟑:

668.09 q₃− 44.38 q₂− 38,819.38 lb ft  𝐄𝐪𝐮𝐚𝐭𝐢𝐨𝐧 𝐍𝐨 𝟑 668.09 q₃= 44.38 (15.20 q₁− 15.25 q₃+ 6,091.97lb ft) + 38,819.38 lb ft 668.09 q₃= 674.58 q₁− 676.80 q₃+ 270,361.63lb ft+ 38,819.38 lb ft 668.09 q₃+ 676.80 q₃ = 674.58 q₁+ 270,361.63lb ft+ 38,819.38 lb ft 1,344.89 q₃ = 674.58 q₁+ 309,181.01 lb ft q3= 674.58 q₁+ 309,181.01 lb_ft 1,344.89 𝐪_𝟑= 𝟎. 𝟓𝟎𝟏𝟔 𝐪_𝟏+ 𝟐𝟐𝟗. 𝟖𝟗 𝐥𝐛 𝐟𝐭 𝐄𝐪𝐮𝐚𝐭𝐢𝐨𝐧 𝐍𝐨. 𝟔

(13)

Substitute

𝐄𝐪𝐮𝐚𝐭𝐢𝐨𝐧 𝐍𝐨. 𝟓

to𝐄𝐪𝐮𝐚𝐭𝐢𝐨𝐧 𝐍𝐨. 𝟔:

q₃= 0.5016 q₁+ 229.89 lb ft  𝐄𝐪𝐮𝐚𝐭𝐢𝐨𝐧 𝐍𝐨. 𝟔 q₃= 0.5016 (2.00 q₃− 458.42 lb ft) + 229.89 lb ft q3= 1.0032 q3− 229.94 lb ft+ 229.89 lb ft −0.01q₃= −0.05lb ft q₃= 0.05 lb ft 0.0032 𝐪_𝟑= 𝟏𝟓. 𝟔𝟑 𝐥𝐛

𝐟𝐭 , the shear flow direction is same as the assumption

From

:

q₁= 2.00 q₃− 458.42 lb ft q₁= 2.00 (15.63 lb ft) − 458.42 lb ft 𝐪_𝟏= 𝟒𝟐𝟕. 𝟏𝟔𝐥𝐛

𝐟𝐭 , the shear flow is opposite from the assumption

From

𝐄𝐪𝐮𝐚𝐭𝐢𝐨𝐧 𝐍𝐨 . 𝟒

:

q₂= 15.20 q₁− 15.25 q₃+ 6,091.97lb ft q₂= 15.20 (427.16lb ft) − 15.25(15.63 lb ft) + 6,091.97 lb ft 𝐪_𝟐= 𝟏𝟐, 𝟑𝟒𝟔. 𝟒𝟒𝐥𝐛

(14)

Torsional Shearing Stress at Cell 1: τ1= q₁ t₁ τ1= 427.16lb_ft 0.0104 ft 𝛕_𝟏= 𝟒𝟏, 𝟎𝟕𝟑. 𝟎𝟖 𝐥𝐛 𝐟𝐭𝟐

Torsional Shearing Stress at Cell 2:

τ₂ = q2 t₂ τ₂ =12,346.44 lb ft 0.0104 ft 𝛕_𝟐= 𝟏, 𝟏𝟖𝟕, 𝟏𝟓𝟕. 𝟔𝟗 𝐥𝐛 𝐟𝐭𝟐

Torsional Shearing Stress at Cell 3:

τ₃ = q3 t₃ τ₃ = 15.63 lb ft 0.0104 ft 𝛕_𝟑= 𝟏, 𝟓𝟎𝟐. 𝟖𝟖 𝐥𝐛 𝐟𝐭𝟐

Torsional Shearing Stress at Front Spar (FS):

τ_FS = qFS t_WFS τ_FS = 2,585.93 lb ft 0.0208 ft 𝛕_𝐅𝐒= 𝟏𝟐𝟒, 𝟑𝟐𝟑. 𝟓𝟔 𝐥𝐛 𝐟𝐭𝟐

Torsional Shearing Stress at Rear Spar (RS):

τ_RS = qRS t_WRS τ_RS =874.87 lb ft 0.0208 ft 𝛕_𝐑𝐒= 𝟒𝟐, 𝟎𝟔𝟏. 𝟎𝟔 𝐥𝐛 𝐟𝐭𝟐

(15)

Factor of Safety at Cell 1:

FS₁ = τALLOWABLE τ_COMPUTED FS₁ = 4,320,000 psf 41,073.08 psf 𝐅𝐒_𝟏=𝟏𝟎𝟓. 𝟏𝟖

Factor of Safety at Cell 2:

FS₂= τALLOWABLE

τCOMPUTED FS₂= 4,320,000 psf

1,187,157.69 psf 𝐅𝐒_𝟐=𝟑. 𝟔𝟒

Factor of Safety at Cell 3:

FS₃= τALLOWABLE

τ_COMPUTED FS₃= 4,320,000 psf

1,502.88 psf 𝐅𝐒_𝟑=𝟐, 𝟖𝟕𝟒. 𝟒𝟖

Factor of Safety at Front Spar (FS):

FS_FS= τALLOWABLE

τ_COMPUTED FS_FS= 4,320,000 psf

124,323.56 psf 𝐅𝐒_𝐅𝐒= 𝟑𝟒. 𝟕𝟓

Factor of Safety at Rear Spar (RS):

FS_RS= τALLOWABLE

τ_COMPUTED FS_RS= 4,320,000 psf

42,061.06 psf 𝐅𝐒_𝐑𝐒=𝟏𝟎𝟐. 𝟕𝟏

(16)

RIB Located at the tip of the Wing

Given:

A₁= 0.0951 ft2 _t WFS= 0.0208 ft or 0.25 in l₁= 1.2196 ft t_WRS= 0.0208 ft or 0.25 in A₂= 0.4089 ft2 t₁= 0.0104 ft or 0.125 in l₂= 2.8014 ft t₂= 0.0104 ft or 0.125 in t₃= 0.0104 ft or 0.125 in A₃= 0.1597ft2 l₃= 2.2158ft

Moment at Quarter Chord,

𝐂_𝐦_𝐜

𝟒 +↷M_A = 0 0 = − P_RS (1.05 ft) + 2,649.54 lb (0.2340 ft) PRS= 2,649.54 lb (0.2340 ft) 1.05 ft 𝐏_𝐑𝐒= 𝟓𝟗𝟎. 𝟒𝟕 𝐥𝐛

(17)

+↑ ΣFV = 0

590.47 lb − 2,649.54 lb + PFS= 0 P_FS= 2,649.54 lb − 590.47 lb

𝐏𝐅𝐒= 𝟐, 𝟎𝟓𝟗. 𝟎𝟕 𝐥𝐛

Shear Flow at Front Spar (FS)

q_W_FS = PFS h_FS q_W_FS =2,059.07 lb 0.3729 ft 𝐪_𝐖_𝐅𝐒 = 𝟓, 𝟓𝟐𝟏. 𝟕𝟖 𝐥𝐛 𝐟𝐭

Shear Flow at Rear Spar (RS)

qW_RS = P_RS h_RS q_W_RS =590.47 lb 0.3150 ft 𝐪_𝐖_𝐑𝐒 = 𝟏, 𝟖𝟕𝟒. 𝟓𝟏 𝐥𝐛 𝐟𝐭 Shear Flow at 1: 2θ₁G = 1 A₁[q1 l₁ t₁+ (q1− q2+ qWFS) h_FS t_WFS ] 2θ₁G = 1 0.0951 ft2[(q1) ( 1.2196 ft 0.0104 ft) + (q1− q2+ 5,521.78 lb ft) ( 0.3729 ft 0.0208 ft) ] 2θ₁G = 1 0.0951 ft2(117.27 q1+ 17.93 q1− 17.93 q2+ 98,993.83 lb ft) 𝟐𝛉_𝟏𝐆 = 𝟏, 𝟒𝟐𝟏. 𝟔𝟔 𝐪_𝟏− 𝟏𝟖𝟖. 𝟓𝟒 𝐪_𝟐+ 𝟏, 𝟎𝟒𝟎, 𝟗𝟒𝟒. 𝟓𝟖 𝐥𝐛 𝐟𝐭 𝐄𝐪𝐮𝐚𝐭𝐢𝐨𝐧 𝐍𝐨. 𝟏

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Shear Flow at 2: 2θ2G = 1 A₂[q2 l₂ t₂+ (q2− qWFS− q1) ( h_FS t_WFS) + (q2+ qWRS− q3) ( h_RS t_WRS) ] 2θ₂G = 1 0.4089 ft2[(q2) ( 2.8014 ft 0.0104 ft) + (q2− 5,521.78 lb ft− q1) ( 0.3729 ft 0.0208 ft) + (q₂+ 1,874.51 lb ft− q3) ( 0.3150 ft 0.0208 ft) ] 2θ2G = 1 0.4089 ft2(269.37 q2+ 17.93 q2− 98,993.83 lb ft− 17.93 q1+ 15.14 q2 + 28,388.01 lb ft − 15.14 q3) 𝟐𝛉_𝟐𝐆 = −𝟒𝟑. 𝟖𝟓 𝐪_𝟏+ 𝟕𝟑𝟗. 𝟔𝟒 𝐪_𝟐− 𝟑𝟕. 𝟎𝟑 𝐪_𝟑− 𝟕𝟎, 𝟔𝟎𝟓. 𝟖𝟐 𝐥𝐛 𝐟𝐭 𝐄𝐪𝐮𝐚𝐭𝐢𝐨𝐧 𝐍𝐨. 𝟐 Shear Flow at 3: 2θ3G = 1 A₃[q3 l₃ t₃+ (q3− q2− qWRS) h_RS t_WRS ] 2θ₃G = 1 0.1597 ft2[(q3) ( 2.2158 ft 0.0104 ft) + (q3− q2− 1,874.51 lb ft) ( 0.3150 ft 0.0208 ft) ] 2θ₃G = 1 0.1597 ft2(213.06 q3+ 15.14 q3− 15.14 q2− 28,388.01 lb ft) 𝟐𝛉_𝟑𝐆 = 𝟏, 𝟒𝟐𝟖. 𝟗𝟑 𝐪_𝟑− 𝟗𝟒. 𝟖𝟎 𝐪_𝟐− 𝟏𝟕𝟕, 𝟕𝟓𝟖. 𝟑𝟔 𝐥𝐛 𝐟𝐭 𝐄𝐪𝐮𝐚𝐭𝐢𝐨𝐧 𝐍𝐨 𝟑

Assuming that the Twist Angles are equal in the cells:

𝜽𝟏= 𝜽𝟐= 𝜽𝟑

Equate

𝐄𝐪𝐮𝐚𝐭𝐢𝐨𝐧 𝐍𝐨. 𝟏

and

:

1,421.66 q₁− 188.54 q₂+ 1,040,944.58 lb ft =1,428.93 q3− 94.80 q2− 177,758.36 lb ft −94.80 q₂+ 188.54 q₂ = 1,421.66 q₁− 1,428.93 q₃− 177,758.36 lb ft+ 1,040,944.58 lb ft 93.74 q₂= 1,421.66 q₁− 1,428.93 q₃+ 863,186.22lb ft q₂=1,421.66 q1− 1,428.93 q3+ 863,186.22 lb ft 93.74 𝐪_𝟐 = 𝟏𝟓. 𝟏𝟕 𝐪_𝟏− 𝟏𝟓. 𝟐𝟒 𝐪_𝟑+ 𝟗, 𝟐𝟎𝟖. 𝟑𝟎 𝐥𝐛 𝐟𝐭 𝐄𝐪𝐮𝐚𝐭𝐢𝐨𝐧 𝐍𝐨 . 𝟒

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Substitute 𝐄𝐪𝐮𝐚𝐭𝐢𝐨𝐧 𝐍𝐨 . 𝟒 to 𝐄𝐪𝐮𝐚𝐭𝐢𝐨𝐧 𝐍𝐨. 𝟏

1,421.66 q₁− 188.54 q₂+ 1,040,944.58 lb ft  𝐄𝐪𝐮𝐚𝐭𝐢𝐨𝐧 𝐍𝐨. 𝟏 1,421.66 q₁= 188.54 (15.17 q₁− 15.24 q₃+ 9,208.30 lb ft) − 1,040,944.58 lb ft 1,421.66 q₁= 2,860.15 q₁− 2,873.35 q₃+ 1,736,132.88 lb ft− 1,040,944.58 lb ft 1,438.49 q₁= 2,873.35 q₃− 695,188.30lb ft q₁=2,873.35 q3− 695,188.30 lb ft 1,438.49 𝐪_𝟏= 𝟐. 𝟎𝟎 𝐪_𝟑− 𝟒𝟖𝟑. 𝟐𝟖 𝐥𝐛 𝐟𝐭 𝐄𝐪𝐮𝐚𝐭𝐢𝐨𝐧 𝐍𝐨. 𝟓

Substitute

to

:

1,428.93 q₃− 94.80 q₂− 177,758.36 lb ft  𝐄𝐪𝐮𝐚𝐭𝐢𝐨𝐧 𝐍𝐨 𝟑 1,428.93 q₃ = 94.80 (15.17 q₁− 15.24 q₃+ 9,208.30 lb ft) + 177,758.36 lb ft 1,428.93 q₃ = 1,438.12 q₁− 1,444.75 q₃+ 872,946.84lb ft+ 177,758.36 lb ft 1,444.75 q₃+ 1,428.93 q₃ = 1,438.12 q₁+ 872,946.84lb ft+ 177,758.36 lb ft 2,837.68 q₃ = 1,438.12 q₁+ 1,050,705.20 lb ft q₃=1,438.12 q1+ 1,050,705.20 lb ft 2,837.68 𝐪_𝟑= 𝟎. 𝟓𝟎𝟔𝟖 𝐪_𝟏+ 𝟑𝟕𝟎. 𝟐𝟕 𝐥𝐛 𝐟𝐭 𝐄𝐪𝐮𝐚𝐭𝐢𝐨𝐧 𝐍𝐨. 𝟔

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Substitute

to𝐄𝐪𝐮𝐚𝐭𝐢𝐨𝐧 𝐍𝐨. 𝟔:

q₃= 0.5068 q₁+ 370.27 lb ft  𝐄𝐪𝐮𝐚𝐭𝐢𝐨𝐧 𝐍𝐨. 𝟔 q₃= 0.5068 (2.00 q₃− 483.28 lb ft) + 370.27 lb ft q3= 1.0136 q3− 244.93 lb ft+ 370.27 lb ft 0.0136 q₃= −125.34lb ft q₃=−125.34 lb ft 0.0136 𝐪_𝟑= 𝟗, 𝟐𝟏𝟔. 𝟏𝟖 𝐥𝐛

𝐟𝐭 , the shear flow is opposite from the assumption

From

:

q₁= 2.00 q₃− 483.28 lb ft q₁= 2.00 (9,216.18 lb ft) − 483.28 lb ft 𝐪_𝟏= 𝟏𝟕, 𝟗𝟒𝟗. 𝟎𝟖 𝐥𝐛

𝐟𝐭 , the shear flow direction is the same as the assumption

From

:

q₂= 15.17 q₁− 15.24 q₃+ 9,208.30 lb ft q₂= 15.17(17,949.08 lb ft) − 15.24 (9,216.18 lb ft) + 9,208.30 lb ft 𝐪_𝟐= 𝟏𝟒𝟏, 𝟎𝟒𝟏. 𝟐𝟔𝐥𝐛

𝐟𝐭 , the shear flow direction is the same as the assumption

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Torsional Shearing Stress at Cell 1: τ₁= q1 t₁ τ₁= 17,949.08 lb ft 0.0104 ft 𝛕_𝟏= 𝟏, 𝟕𝟐𝟓, 𝟖𝟕𝟑. 𝟎𝟖 𝐥𝐛 𝐟𝐭𝟐

Torsional Shearing Stress at Cell 2:

τ₂ = q2 t₂ τ₂ =141,041.26 lb ft 0.0104 ft 𝛕_𝟐= 𝟏𝟑, 𝟓𝟔𝟏, 𝟔𝟓𝟗. 𝟔𝟐 𝐥𝐛 𝐟𝐭𝟐

Torsional Shearing Stress at Cell 3:

τ₃ = q3 t₃ τ₃ =9,216.18 lb ft 0.0104 ft 𝛕_𝟑= 𝟖𝟖𝟔, 𝟏𝟕𝟏. 𝟏𝟓 𝐥𝐛 𝐟𝐭𝟐

Torsional Shearing Stress at Front Spar (FS):

τ_FS = qFS t_WFS τ_FS = 5,521.78 lb ft 0.0208 ft 𝛕_𝐅𝐒= 𝟐𝟔𝟓, 𝟒𝟕𝟎. 𝟏𝟗 𝐥𝐛 𝐟𝐭𝟐

Torsional Shearing Stress at Rear Spar (RS):

τ_RS = qRS t_WRS τ_RS =1,874.51 lb ft 0.0208 ft 𝛕_𝐑𝐒= 𝟗𝟎, 𝟏𝟐𝟎. 𝟔𝟕 𝐥𝐛 𝐟𝐭𝟐

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Factor of Safety at Cell 1:

FS₁ = τALLOWABLE τ_COMPUTED FS₁ = 4,320,000 psf 1,725,873.08 psf 𝐅𝐒_𝟏=𝟐. 𝟓𝟎

Factor of Safety at Cell 2:

FS₂= τALLOWABLE

τ_COMPUTED FS₂= 4,320,000 psf

13,561,659.62 psf 𝐅𝐒_𝟐=𝟎. 𝟑𝟏𝟖𝟓

Factor of Safety at Cell 3:

FS₃= τALLOWABLE

τ_COMPUTED FS₃= 4,320,000 psf

886,171.15 psf 𝐅𝐒_𝟑=𝟒. 𝟖𝟕

Factor of Safety at Front Spar (FS):

FS_FS= τALLOWABLE

τ_COMPUTED FS_FS= 4,320,000 psf

265,470.19 psf 𝐅𝐒_𝐅𝐒= 𝟏𝟔. 𝟐𝟕

Factor of Safety at Rear Spar (RS):

FS_RS= τALLOWABLE

τ_COMPUTED FS_RS= 4,320,000 psf

90,120.67 psf 𝐅𝐒_𝐑𝐒=𝟒𝟕. 𝟗𝟒

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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.

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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.

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This activity discusses a number of important considerations, such as the wing attachment appropriate for the designed aircraft, the right position/location, material, cross-section 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.