REPORT #

4

Load per Unit Span for Front Spar and Rear Spar at Different Conditions

January 28, 11 PAGE 1

CHECKED BY DATE CHECKED

Condition I. POSITIVE HIGH ANGLE OF ATTACK

1.) Center of Pressure, CP ;

CP=C p'× c

−¿¿

where : C p' is center of pressure from report # 3

c−¿=¿ ¿

wing geometric chord

c−¿=5.375 ft ¿

Solution;

CP=C p'× c

−¿¿

CP=0 .245×5.375 ft

CP=1.3 ft

2.) Load per Unit Span, P

P=Cnqc–

q=113.45 lbf t 2

from report # 3 Condition I

Cn=2.64from report # 3 Condition I

Solution;

P=Cnqc–

P= (2.64 )(113.45 lbf t 2 ) (5.375 ft )

P=1609.9 lbft

*Convert to Nm

, where 1Nm

is equal to

14.5939lbft

P=23494.78 Nm

3.) Load per Unit Span on the Rear Spar, PRS

PRS=P (E−x )D

where D=0.7c–

−0.2c–

¿0.7 (5.375 ft )−0.2¿)

¿0.7c–

−0.2c–

¿2.6875 ft

where E=0.2c–

+ D2

¿2.41875 ft

where x=0.2c–

¿1.075 ft

Solution;

PRS=P (E−x )D

PRS=1609.9

lbft

(2.41875 ft−1.075 ft )

2.6875 ft

PRS=804.95lbft

*Convert to Nm

, where 1Nm

is equal to

14.5939lbft

PRS=11747.39Nm

4.) Moment due to Torque Load, PT

PT=±M T

D

where M T is moment torqe due to load P

M T=P (E−CP )

M T=1609.9lbft

(2.41875 ft−1.3 ft )

M T=1773.9lb

f t 2

Solution;

PT=±M T

D

PT=1773.9

lb

f t 2

2.6875 ft

PT=660.06lbft

*Convert to Nm

, where 1Nm

is equal to

14.5939lbft

PT=9632 .86Nm

5.) Load per Unit Span on the Front Spar, PFS

PFS=P−PRS

Solution;

PFS=P−PRS

PFS=[1609.9 lbft ]−[804.95 lbft ]PFS=804.95

lbft

*Convert to Nm

, where 1Nm

is equal to

14.5939lbft

PFS=11747.4Nm

6.) Load per Unit Spar with Torque on the Rear Spar, PRST

PRST=PT (E−x )D

Solution;

PRST=PT (E−x )D

PRST=660.06

lbft

(2.41875 ft−1.075 ft )

2.6875 ft

PRST=330.0304lbft

*Convert to Nm

, where 1Nm

is equal to 14.5939

lbft

PRST=4816.43Nm

7.) Load per Unit Span with Torque on the Front Spar, PFST

PFST=PT−PRST

Solution;

PFST=PT−PRST

PFST=[660.06 lb / ft ]−[330.0304 lb / ft ]

PFST=330.0303lbft

*Convert to Nm

, where 1Nm

is equal to

14.5939lbft

PFST=4816.43Nm

8.) Total load per Unit Span on the Spar, PTOTAL

PTOTAL=PT+(PRST+PFST )

Solution;

PTOTAL=9632.86Nm

+[{4816.43 Nm }+{4816.43 Nm }]PTOTAL=19265.7

Nm

Condition II. NEGATIVE HIGH ANGLE OF ATTACK

1.) Center of Pressure, CP ;

CP=C p'× c

−¿¿

where : C p' is center of pressure from report # 3

c−¿=¿ ¿

wing geometric chord

c−¿=5.375 ft ¿

Solution;

CP=C p'× c

−¿¿

CP=0 .3×5.375 ft

CP=1.6 ft

2.) Load per Unit Span, P

P=Cnqc–

q=113.45 lbf t 2

from report # 3 Condition I

Cn=2.64from report # 3 Condition I

Solution;

P=Cnqc–

P= (2.64 )(113.45 lbf t 2 ) (5.375 ft )

P=1609.9 lbft

*Convert to Nm

, where 1Nm

is equal to

14.5939lbft

P=23494.78 Nm

3.) Load per Unit Span on the Rear Spar, PRS

PRS=P (E−x )D

where D=0.7c–

−0.2c–

¿0.7 (5.375 ft )−0.2¿)

¿0.7c–

−0.2c–

¿2.6875 ft

where E=0.2c–

+ D2

¿2.41875 ft

where x=0.2c–

¿1.075 ft

Solution;

PRS=P (E−x )D

PRS=1609.9

lbft

(2.41875 ft−1.075 ft )

2.6875 ft

PRS=804.95lbft

*Convert to Nm

, where 1Nm

is equal to

14.5939lbft

PRS=11747.39Nm

4.) Moment due to Torque Load, PT

PT=±M T

D

where M T is moment torqe due to load P

M T=P (E−CP )

M T=1609.9lbft

(2.41875 ft−1.6 ft )

M T=1297.98lb

f t 2

Solution;

PT=±M T

D

PT=1773.9

lb

f t 2

2.6875 ft

PT=482.97lbft

*Convert to Nm

, where 1Nm

is equal to

14.5939lbft

PT=7048.435Nm

5.) Load per Unit Span on the Front Spar, PFS

PFS=P−PRS

Solution;

PFS=P−PRS

PFS=[1609.9 lbft ]−[804.95 lbft ]PFS=804.95

lbft

*Convert to Nm

, where 1Nm

is equal to

14.5939lbft

PFS=11747.4Nm

6.) Load per Unit Spar with Torque on the Rear Spar, PRST

PRST=PT (E−x )D

Solution;

PRST=PT (E−x )D

PRST=482.97

lbft

(2.41875 ft−1.075 ft )

2.6875 ft

PRST=241.4856lbft

*Convert to Nm

, where 1Nm

is equal to 14.5939

lbft

PRST=3524.217Nm

7.) Load per Unit Span with Torque on the Front Spar, PFST

PFST=PT−PRST

Solution;

PFST=PT−PRST

PFST=[482.97 lb / ft ]−[241.4856 lb / ft ]

PFST=241.4856lbft

*Convert to Nm

, where 1Nm

is equal to

14.5939lbft

PFST=3524.217Nm

8.) Total load per Unit Span on the Spar, PTOTAL

PTOTAL=PT+(PRST+PFST )

Solution;

PTOTAL=7048.435Nm

+[{3524.217 Nm }+{3524.217 Nm }]PTOTAL=14096.8696

Nm

Condition III. POSITIVE LOW ANGLE OF ATTACK

1.) Center of Pressure, CP ;

CP=C p'× c

−¿¿

where : C p' is center of pressure from report # 3

c−¿=¿ ¿

wing geometric chord

c−¿=5.375 ft ¿

Solution;

CP=C p'× c

−¿¿

CP=0 .3×5.375 ft

CP=1.6 ft

2.) Load per Unit Span, P

P=Cnqc–

q=381.87 lbf t 2

from report # 3 Condition I

Cn=−2.187from report # 3 Condition I

Solution;

P=Cnqc–

P= (−2.187 )(381.87 lbf t 2 ) (5.375 ft )

P=−4488.96 lbft

*Convert to Nm

, where 1Nm

is equal to

14.5939lbft

P=−65511.5 Nm

3.) Load per Unit Span on the Rear Spar, PRS

PRS=P (E−x )D

where D=0.7c–

−0.2c–

¿0.7 (5.375 ft )−0.2¿)

¿0.7c–

−0.2c–

¿2.6875 ft

where E=0.2c–

+ D2

¿2.41875 ft

where x=0.2c–

¿1.075 ft

Solution;

PRS=P (E−x )D

PRS=−4488.96 lb

ft(2.41875 ft−1.075 ft )

2.6875 ft

PRS=−2244.48 lbft

*Convert to Nm

, where 1Nm

is equal to

14.5939lbft

PRS=−32755.75 Nm

4.) Moment due to Torque Load, PT

PT=±M T

D

where M T is moment torqe due to load P

M T=P (E−CP )

M T=−4488.96 lbft

(2.41875 ft−1.6 ft )

M T=−3619.2 lbf t 2

Solution;

PT=±M T

D

PT=−3619.2 lb

f t 2

2.6875 ft

PT=−1346.7 lbft

*Convert to Nm

, where 1Nm

is equal to

14.5939lbft

PT=−19653.45 Nm

5.) Load per Unit Span on the Front Spar, PFS

PFS=P−PRS

Solution;

PFS=P−PRS

PFS=[−4488.96 lbft ]−[−2244.48 lbft ]PFS=−2244.48 lb

ft

*Convert to Nm

, where 1Nm

is equal to

14.5939lbft

PFS=−32755.75 Nm

6.) Load per Unit Spar with Torque on the Rear Spar, PRST

PRST=PT (E−x )D

Solution;

PRST=PT (E−x )D

PRST=−1346.7 lb

ft(2.41875 ft−1.075 ft )

2.6875 ft

PRST=−673.34 lbft

*Convert to Nm

, where 1Nm

is equal to 14.5939

lbft

PRST=−9826.7 Nm

7.) Load per Unit Span with Torque on the Front Spar, PFST

PFST=PT−PRST

Solution;

PFST=PT−PRST

PFST=[−19653.45 Nm ]−[−9826.7 Nm ]PFST=−9826.7 N

m

8.) Total load per Unit Span on the Spar, PTOTAL

PTOTAL=PT+(PRST+PFST )

Solution;

PTOTAL=−19653.45 Nm

+[{−9826.7 Nm }+{−9826.7 Nm }]

PTOTAL=−39306.9 Nm

Condition IV. NEGATIVE LOW ANGLE OF ATTACK

1.) Center of Pressure, CP ;

CP=C p'× c

−¿¿

where : C p' is center of pressure from report # 3

c−¿=¿ ¿

wing geometric chord

c−¿=5.375 ft ¿

Solution;

CP=C p'× c

−¿¿

CP=0 .245×5.375 ft

CP=1.32 ft

2.) Load per Unit Span, P

P=Cnqc–

q=381.87 lbf t 2

from report # 3 Condition I

Cn=.73121from report # 3 Condition I

Solution;

P=Cnqc–

P= (.73121 )(381.87 lbf t2 ) (5.375 ft )

P=1500.86 lbft

*Convert to Nm

, where 1Nm

is equal to

14.5939lbft

P=21903. 4 Nm

3.) Load per Unit Span on the Rear Spar, PRS

PRS=P (E−x )D

where D=0.7c–

−0.2c–

¿0.7 (5.375 ft )−0.2¿)

¿0.7c–

−0.2c–

¿2.6875 ft

where E=0.2c–

+ D2

¿2.41875 ft

where x=0.2c–

¿1.075 ft

Solution;

PRS=P (E−x )D

PRS=1500.86

lbft

(2.41875 ft−1.075 ft )

2.6875 ft

PRS=750.43lbft

*Convert to Nm

, where 1Nm

is equal to

14.5939lbft

PRS=10951.7Nm

4.) Moment due to Torque Load, PT

PT=±M T

D

where M T is moment torqe due to load P

M T=P (E−CP )

M T=1500.86lbft

(2.41875 ft−1.32 ft )

M T=1653.75lb

f t 2

Solution;

PT=±M T

D

PT=1653.75

lb

f t 2

2.6875 ft

PT=615.35lbft

*Convert to Nm

, where 1Nm

is equal to

14.5939lbft

PT=8980.38Nm

5.) Load per Unit Span on the Front Spar, PFS

PFS=P−PRS

Solution;

PFS=P−PRS

PFS=[1500.86 lbft ]−[750.43 lbft ]PFS=750.43

lbft

*Convert to Nm

, where 1Nm

is equal to

14.5939lbft

PFS=10951.7Nm

6.) Load per Unit Spar with Torque on the Rear Spar, PRST

PRST=PT (E−x )D

Solution;

PRST=PT (E−x )D

PRST=615.35

lbft

(2.41875 ft−1.075 ft )

2.6875 ft

PRST=307.7lbft

*Convert to Nm

, where 1Nm

is equal to 14.5939

lbft

PRST=4490.19Nm

7.) Load per Unit Span with Torque on the Front Spar, PFST

PFST=PT−PRST

Solution;

PFST=PT−PRST

PFST=[8980.38 Nm ]−[4490.19 Nm ]PFST=4490.19

Nm

8.) Total load per Unit Span on the Spar, PTOTAL

PTOTAL=PT+(PRST+PFST )

Solution;

PTOTAL=8980.38Nm

+(4490.19 Nm +4490.19 Nm )

PTOTAL=−39306.9 Nm

Condition V. INVERTED FLIGHT

1.) Center of Pressure, CP ;

CP=C p'× c

−¿¿

where : C p' is center of pressure from report # 3

c−¿=¿ ¿

wing geometric chord

c−¿=5.375 ft ¿

Solution;

CP=C p'× c

−¿¿

CP=0 .3×5.375 ft

CP=1.6125 ft

2.) Load per Unit Span, P

P=Cnqc–

q=113.45 lbf t 2

from report # 3 Condition I

Cn=0.5484from report # 3 Condition I

Solution;

P=Cnqc–

P= (.5484 )(113.45 lbf t 2 ) (5.375 ft )

P=334.4 lbft

*Convert to Nm

, where 1Nm

is equal to

14.5939lbft

P=4880.5 Nm

3.) Load per Unit Span on the Rear Spar, PRS

PRS=P (E−x )D

where D=0.7c–

−0.2c–

¿0.7 (5.375 ft )−0.2¿)

¿0.7c–

−0.2c–

¿2.6875 ft

where E=0.2c–

+ D2

¿2.41875 ft

where x=0.2c–

¿1.075 ft

Solution;

PRS=P (E−x )D

PRS=334.4

lbft

(2.41875 ft−1.075 ft )

2.6875 ft

PRS=167.2lbft

*Convert to Nm

, where 1Nm

is equal to

14.5939lbft

PRS=2440.25Nm

4.) Moment due to Torque Load, PT

PT=±M T

D

where M T is moment torqe due to load P

M T=P (E−CP )

M T=334.4lbft

(2.41875 ft−1.61 ft )

M T=269.6lb

f t2

Solution;

PT=±M T

D

PT=269.6

lb

f t 2

2.6875 ft

PT=100.326lbft

*Convert to Nm

, where 1Nm

is equal to

14.5939lbft

PT=1464.15Nm

5.) Load per Unit Span on the Front Spar, PFS

PFS=P−PRS

Solution;

PFS=P−PRS

PFS=[334.4 lbft ]−[167.2 lbft ]PFS=167.2

lbft

*Convert to Nm

, where 1Nm

is equal to

14.5939lbft

PFS=2440.25Nm

6.) Load per Unit Spar with Torque on the Rear Spar, PRST

PRST=PT (E−x )D

Solution;

PRST=PT (E−x )D

PRST=100.326

lbft

(2.41875 ft−1.075 ft )

2.6875 ft

PRST=50.16lbft

*Convert to Nm

, where 1Nm

is equal to 14.5939

lbft

PRST=732.08Nm

7.) Load per Unit Span with Torque on the Front Spar, PFST

PFST=PT−PRST

Solution;

PFST=PT−PRST

PFST=[732.08 Nm ]−[732.08 Nm ]PFST=732.076

Nm

8.) Total load per Unit Span on the Spar, PTOTAL

PTOTAL=PT+(PRST+PFST )

Solution;

PTOTAL=732.08Nm

+(732.08 Nm +732.076 Nm )

PTOTAL=2928.3Nm

Condition VI. GLIDING AT THE LIMIT DIVING SPEED

1.) Center of Pressure, CP ;

CP=C p'× c

−¿¿

where : C p' is center of pressure from report # 3

c−¿=¿ ¿

wing geometric chord

c−¿=5.375 ft ¿

Solution;

CP=C p'× c

−¿¿

CP=0 .3×5.375 ft

CP=1.6125 ft

2.) Load per Unit Span, P

P=Cnqc–

q=255.4 lbf t 2

from report # 3 Condition I

Cn=−0.5958from report # 3 Condition I

Solution;

P=Cnqc–

P= (−0.5958 )(255.4 lbf t 2 ) (5.375 ft )

P=−817.9 lbft

*Convert to Nm

, where 1Nm

is equal to

14.5939lbft

P=−11936.5 Nm

3.) Load per Unit Span on the Rear Spar, PRS

PRS=P (E−x )D

where D=0.7c–

−0.2c–

¿0.7 (5.375 ft )−0.2¿)

¿0.7c–

−0.2c–

¿2.6875 ft

where E=0.2c–

+ D2

¿2.41875 ft

where x=0.2c–

¿1.075 ft

Solution;

PRS=P (E−x )D

PRS=−817.9 lb

ft(2.41875 ft−1.075 ft )

2.6875 ft

PRS=−408.95 lbft

*Convert to Nm

, where 1Nm

is equal to

14.5939lbft

PRS=−5968.2 Nm

4.) Moment due to Torque Load, PT

PT=±M T

D

where M T is moment torqe due to load P

M T=P (E−CP )

M T=−817.9 lbft

(2.41875 ft−1.61 ft )

M T=−659.44 lbf t2

Solution;

PT=±M T

D

PT=−659.44 lb

f t 2

2.6875 ft

PT=−245.4 lbft

*Convert to Nm

, where 1Nm

is equal to

14.5939lbft

PT=−3580.94 Nm

5.) Load per Unit Span on the Front Spar, PFS

PFS=P−PRS

Solution;

PFS=P−PRS

PFS=[−817.9 lbft ]−[−408.95 lbft ]PFS=−408.95 lb

ft

*Convert to Nm

, where 1Nm

is equal to

14.5939lbft

PFS=−5968.228 Nm

6.) Load per Unit Spar with Torque on the Rear Spar, PRST

PRST=PT (E−x )D

Solution;

PRST=PT (E−x )D

PRST=−245.4 lb

ft(2.41875 ft−1.075 ft )

2.6875 ft

PRST=−122.7 lbft

*Convert to Nm

, where 1Nm

is equal to 14.5939

lbft

PRST=−1790.5 Nm

7.) Load per Unit Span with Torque on the Front Spar, PFST

PFST=PT−PRST

Solution;

PFST=PT−PRST

PFST=[−3580.9 4 Nm ]−[−1790.5 Nm ]PFST=−1790.47 N

m

8.) Total load per Unit Span on the Spar, PTOTAL

PTOTAL=PT+(PRST+PFST )

Solution;

PTOTAL=−3580.94 Nm

+(−1790.5 Nm+−1790.47 Nm )

PTOTAL=2928.3Nm

Condition VII. POSITIVE GUST WITH FLAPS DOWN

1.) Center of Pressure, CP ;

CP=C p'× c

−¿¿

where : C p' is center of pressure from report # 3

c−¿=¿ ¿

wing geometric chord

c−¿=5.375 ft ¿

Solution;

CP=C p'× c

−¿¿

CP=0 .245×5.375 ft

CP=1.31 ft

2.) Load per Unit Span, P

P=Cnqc–

q=673.95 lbf t 2

from report # 3 Condition I

Cn=−0.99131from report # 3 Condition I

Solution;

P=Cnqc–

P= (−0.991 )(673.95 lbf t2 ) (5.375 ft )

P=−3591.0032 lbft

*Convert to Nm

, where 1Nm

is equal to

14.5939lbft

P=−52406.7 Nm

3.) Load per Unit Span on the Rear Spar, PRS

PRS=P (E−x )D

where D=0.7c–

−0.2c–

¿0.7 (5.375 ft )−0.2¿)

¿0.7c–

−0.2c–

¿2.6875 ft

where E=0.2c–

+ D2

¿2.41875 ft

where x=0.2c–

¿1.075 ft

Solution;

PRS=P (E−x )D

PRS=−3591 lb

ft(2.41875 ft−1.075 ft )

2.6875 ft

PRS=−1795.5 lbft

*Convert to Nm

, where 1Nm

is equal to

14.5939lbft

PRS=−26203.4 Nm

4.) Moment due to Torque Load, PT

PT=±M T

D

where M T is moment torqe due to load P

M T=P (E−CP )

M T=−3591 lbft

(2.41875 ft−1.31 ft )

M T=−3956.8 lbf t2

Solution;

PT=±M T

D

PT=−3956.8 lb

f t 2

2.6875 ft

PT=−1472.3 lbft

*Convert to Nm

, where 1Nm

is equal to

14.5939lbft

PT=−21486. 8 Nm

5.) Load per Unit Span on the Front Spar, PFS

PFS=P−PRS

Solution;

PFS=P−PRS

PFS=[−3591 lbft ]−[−1795.5 lbft ]PFS=−1795.5 lb

ft

*Convert to Nm

, where 1Nm

is equal to

14.5939lbft

PFS=−26203. 3 Nm

6.) Load per Unit Spar with Torque on the Rear Spar, PRST

PRST=PT (E−x )D

Solution;

PRST=PT (E−x )D

PRST=−1472.3 lb

ft(2.41875 ft−1.075 ft )

2.6875 ft

PRST=−736 lbft

*Convert to Nm

, where 1Nm

is equal to 14.5939

lbft

PRST=−10743. 4 Nm

7.) Load per Unit Span with Torque on the Front Spar, PFST

PFST=PT−PRST

Solution;

PFST=PT−PRST

PFST=[−21486.8 Nm ]−[−10743.4 Nm ]PFST=−10743. 4 N

m

8.) Total load per Unit Span on the Spar, PTOTAL

PTOTAL=PT+(PRST+PFST )

Solution; PTOTAL=−21486.8 Nm

+[−21486.8 Nm ]+[−10743.4 Nm ]

PTOTAL=−42973.5 NmCondition VIII. NEGATIVE GUST WITH

FLAPS DOWN

1.) Center of Pressure, CP ;

CP=C p'× c

−¿¿

where : C p' is center of pressure from report # 3

c−¿=¿ ¿

wing geometric chord

c−¿=5.375 ft ¿

Solution;

CP=C p'× c

−¿¿

CP=0 .3×5.375 ft

CP=1.6125 ft

2.) Load per Unit Span, P

P=Cnqc–

q=673.95 lbf t 2

from report # 3 Condition I

Cn=−0.71116from report # 3 Condition I

Solution;

P=Cnqc–

P= (−0.711)(673.95 lbf t 2 ) (5.375 ft )

P=−2576.16 lbft

*Convert to Nm

, where 1Nm

is equal to

14.5939lbft

P=−37596.3 Nm

3.) Load per Unit Span on the Rear Spar, PRS

PRS=P (E−x )D

where D=0.7c–

−0.2c–

¿0.7 (5.375 ft )−0.2¿)

¿0.7c–

−0.2c–

¿2.6875 ft

where E=0.2c–

+ D2

¿2.41875 ft

where x=0.2c–

¿1.075 ft

Solution;

PRS=P (E−x )D

PRS=−2576.16 lb

ft(2.41875 ft−1.075 ft )

2.6875 ft

PRS=−1288.08 lbft

*Convert to Nm

, where 1Nm

is equal to

14.5939lbft

PRS=−18798.15 Nm

4.) Moment due to Torque Load, PT

PT=±M T

D

where M T is moment torqe due to load P

M T=P (E−CP )

M T=−2576.16 lbft

(2.41875 ft−1.61 ft )

M T=−2077.03 lbf t2

Solution;

PT=±M T

D

PT=−2077.03 lb

f t 2

2.6875 ft

PT=−772.8 lbft

*Convert to Nm

, where 1Nm

is equal to

14.5939lbft

PT=−11278.9 Nm

5.) Load per Unit Span on the Front Spar, PFS

PFS=P−PRS

Solution;

PFS=P−PRS

PFS=[−2576.16 lbft ]−[−1288.08 lbft ]PFS=−1288.08 lb

ft

*Convert to Nm

, where 1Nm

is equal to

14.5939lbft

PFS=−18798.15 Nm

6.) Load per Unit Spar with Torque on the Rear Spar, PRST

PRST=PT (E−x )D

Solution;

PRST=PT (E−x )D

PRST=−772.8 lb

ft(2.41875 ft−1.075 ft )

2.6875 ft

PRST=−386.4 lbft

*Convert to Nm

, where 1Nm

is equal to 14.5939

lbft

PRST=−5639.4 Nm

7.) Load per Unit Span with Torque on the Front Spar, PFST

PFST=PT−PRST

Solution;

PFST=PT−PRST

PFST=[−11278. 9 Nm ]−[−5639. 4 Nm ]

PFST=−5639.4 Nm

8.) Total load per Unit Span on the Spar, PTOTAL

PTOTAL=PT+(PRST+PFST )

Solution;

PTOTAL=−11278.9 Nm

+[−5639.4 Nm ]+[−5639 .4 Nm ]

PTOTAL=−22557.77 NmCondition IX. DIVE WITH FLAPS DOWN

1.) Center of Pressure, CP ;

CP=C p'× c

−¿¿

where : C p' is center of pressure from report # 3

c−¿=¿ ¿

wing geometric chord

c−¿=5.375 ft ¿

Solution;

CP=C p'× c

−¿¿

CP=0 .245×5.375 ft

CP=1.31 ft

2.) Load per Unit Span, P

P=Cnqc–

q=673.95 lbf t 2

from report # 3 Condition I

Cn=0.538468from report # 3 Condition I

Solution;

P=Cnqc–

P= (0.5385 )(673.95 lbf t 2 )(5.375 ft )

P=1950.59 lbft

*Convert to Nm

, where 1Nm

is equal to

14.5939lbft

P=28466.7 3 Nm

3.) Load per Unit Span on the Rear Spar, PRS

PRS=P (E−x )D

where D=0.7c–

−0.2c–

¿0.7 (5.375 ft )−0.2¿)

¿0.7c–

−0.2c–

¿2.6875 ft

where E=0.2c–

+ D2

¿2.41875 ft

where x=0.2c–

¿1.075 ft

Solution;

PRS=P (E−x )D

PRS=1950.59

lbft

(2.41875 ft−1.075 ft )

2.6875 ft

PRS=975.3lbft

*Convert to Nm

, where 1Nm

is equal to

14.5939lbft

PRS=14233. 4Nm

4.) Moment due to Torque Load, PT

PT=±M T

D

where M T is moment torqe due to load P

M T=P (E−CP )

M T=1950.59lbft

(2.41875 ft−1.31 ft )

M T=2149.3lb

f t 2

Solution;

PT=±M T

D

PT=2149.3

lb

f t 2

2.6875 ft

PT=799.7lbft

*Convert to Nm

, where 1Nm

is equal to

14.5939lbft

PT=11671.36Nm

5.) Load per Unit Span on the Front Spar, PFS

PFS=P−PRS

Solution;

PFS=P−PRS

PFS=[1950.59 lbft ]−[975.3 lbft ]PFS=975.3

lbft

*Convert to Nm

, where 1Nm

is equal to

14.5939lbft

PFS=14233.36Nm

6.) Load per Unit Spar with Torque on the Rear Spar, PRST

PRST=PT (E−x )D

Solution;

PRST=PT (E−x )D

PRST=799.7

lbft

(2.41875 ft−1.075 ft )

2.6875 ft

PRST=399.87lbft

*Convert to Nm

, where 1Nm

is equal to 14.5939

lbft

PRST=5835.68Nm

7.) Load per Unit Span with Torque on the Front Spar, PFST

PFST=PT−PRST

Solution;

PFST=PT−PRST

PFST=[11671.36 Nm ]−[5835.68 Nm ]PFST=5835.68

Nm

8.) Total load per Unit Span on the Spar, PTOTAL

PTOTAL=PT+(PRST+PFST )

Solution;

PTOTAL=22672.36Nm

+[5835.68 Nm ]+[5835.68 Nm ]PTOTAL=23342.7

Nm