CHAPTER-4

LOAD ESTIMATION OF WINGS

4.1 INTRODUCTION

Wing design has several goals related to the wing performance and lift

distribution. One would like to have a distribution of CL(y) that is relatively flat so

that the airfoil sections in one area are not "working too hard" while others are at

low CL. The induced drag depends solely on the lift distribution, so one would like

to achieve a nearly elliptical distribution of section lift. On the other hand

structural weight is affected by the lift distribution also so that the ideal shape

depends on the relative importance of induced drag and wing weight.

4.2 TO FIND THE LIFT DISTRIBUTION OF THE WING

Span = 10.2 m

WTO = 17789.7679 N

For the rectangular wing for the utility aircraft,

Root chord (CROOT) = 1.457 m

Tip chord (CTIP) = 1.457 m

VCRUISE = 62.222 m/s

ρsealevel = 1.2256 kg/m3

CL = 1.6

Using the lift formula,

LTIP = 5530.777967 N/m

LROOT = 5530.777967 N/m

a = Span of wing/2 = 10.2/2

a = 5.1 m

The elliptical lift distribution of the wing is given by

x2/a2 + y2/b2 = 1

y2 = b2(1 – x2/a2)

where, a = semispan; b = Lift at root

Area under the curve = πab/4 = WTO/2

From the above relation,

b = 4 WTO /2πa = 4 * 17789.7679/2π * 5.1

b = 2220.65 N

4.2.1 ELLIPTICAL LOAD DISTRIBUTION

Varying the ‘x’ value and find the value of ‘y’ for elliptical load distribution

y2 = b2 (1 – x2/a2)

where a = 5.1 m

b = 2220.65 N

x y x y

0 2221.777 2.55 1924.115

0.051 2221.666 2.805 1855.55

0.255 2218.998 3.06 1777.422

0.51 2210.64 3.315 1688.404

0.765 2196.64 3.57 1586.666

1.02 2176.888 3.825 1469.567

1.275 2151.226 4.08 1333.066

1.53 2119.44 4.335 1170.394

1.785 2081.249 4.59 968.4501

2.04 2036.292 4.845 693.7496

2.295 1984.11 5.1 0

We obtain the following graph using the above table

0 0.3570.7141.0711.4281.7852.1422.4992.8563.213 3.57 3.9274.2844.6414.9980

500

1000

1500

2000

2500

Elliptic lift distribution

GRAPH ‘4.1’

4.2.1 TRAPEZOIDAL LOAD DISTRIBUTION

Now to find the trapezoidal lift distribution, we plot the Lift at root, Lift at tip

against the semi span and obtain the following graph.

x y

0 0

0 5530.778

5.1 5530.778

5.1 0

0 1 2 3 4 5 60

1000

2000

3000

4000

5000

6000

Trapezoidal lift distribution

GRAPH ‘4.2’

4.2.3 ACTUAL LOAD DISTRIBUTIONNow adding both the graphs, we obtain the following load distribution over the semi span.

0 0.255 0.51 0.765 1.02 1.275 1.53 1.785 2.04 2.295 2.55 2.805 3.06 3.315 3.57 3.825 4.08 4.335 4.59 4.845 5.10

500

1000

1500

2000

2500

Load estimation over wings

GRAPH ‘4.3’

4.3 TO FIND THE VALUE OF k:

Structural weight of our Wings,

WWing = 0.25*WTO = 0.25*17789.7679

WWing = 4447.4419 kg.

Weight of each wing, WPORT = WSTARBOARD = WWing/2 = 2223.721 N

Assume the load distribution to be parabolic. Then, y = k(x – b/2)2

- WPORT= k ∫(x – b/2)2 dx where b = total span

- WPORT = kb3/12

k = -12 WPORT/ b3 = -12*4447.4419/10.23

k = -25.14

4.4 TO FIND THE LOAD DISTRIBUTION DUE TO SELF

WEIGHT OF THE WING

Sub ‘k’ values in y = k(x – b/2)2, and plotting the load values (y) vs semi span, we

obtain the following values

0 1 2 3 4 5 6

-700

-600

-500

-400

-300

-200

-100

0

Load distribution due to self weight of the wing

GRAPH ‘4.4’

4.5 TO FIND LIFT LOAD INTENSITY:

Lift on each point =√ {WPORT2 (1 – x2/b2)}

Where, x = point on span

b = semi span

X Lift Load intensity

0 2221.7769

0.5 2211.073606

1 2178.648246

1.5 2123.505984

2 2043.808929

2.5 1936.527823

3 1796.72838

3.5 1615.993047

4 1378.311278

4.5 1045.542071

5 437.8153236

5.1 0

On choosing the interval, the shear increment, shear force, Bending moment increment and bending moment are found.

X Load intensity

Interval

Shear increment

Shear force

Bending increment

Bending moment

0 2221.7769 0.5 8851.8585 19100.67846

0.5 2211.073606 0.5 1108.21262

67743.6458

84148.87609

5 14951.80236

1 2178.648246 0.5 1097.43046

36646.2154

13597.46532

3 11354.33704

1.5 2123.505984 0.5 1075.53855

75570.6768

63054.22306

8 8300.113974

2 2043.808929 0.5 1041.82872

84528.8481

32524.88124

7 5775.232727

2.5 1936.527823 0.5 995.084188

13533.7639

42015.65301

7 3759.57971

3 1796.72838 0.5 933.3140508

2600.44989

1533.553458 2226.026252

3.5 1615.993047 0.5 853.180356

91747.2695

31086.92985

6 1139.096396

4 1378.311278 0.5 748.576081

3998.69345

2686.490746

2 452.6056502

4.5 1045.542071 0.5 605.963337

1392.73011

5347.855891

6 104.7497585

5 437.8153236 0.5 370.839348

521.890766

2103.655220

2 1.094538309

5.1 0 0.1 21.89076618

-2.224E-12

1.094538309

-1.83298E-12

4.6 SHEAR FORCE DIAGRAM:

0 1 2 3 4 5 6

-2000

0

2000

4000

6000

8000

10000

SHEAR FORCE DIAGRAM

GRAPH ‘4.5’

4.7 BENDING MOMENT DIAGRAM:

0 1 2 3 4 5 6

-5000

0

5000

10000

15000

20000

25000

BENDING MOMENT DIAGRAM

GRAPH ‘4.6’

4.8 LOCATION OF ENGINE: Engine selection

Number of engines = 1

Type of engine = TCM Tsio-360 Piston-propeller engine

Engine Specification = 210 hp (157 kW) at 2,800 rpm for take-off

There is always a controversy whether to have engines with tractor or pusher

configuration. Here, the tractor configuration is chosen.

Advantages Disadvantages

The heavy engine at the front which helps to move the centre of gravity forward and therefore allows a smaller tail for stability considerations.

The propeller slipstream disturbs the quality of the airflow over the fuselage and the wing root.

The propeller is working in an undisturbed free stream.

The increased velocity and flow turbulence over the fuselage due to the propeller slipstream increase the local skin friction on the fuselage.

There is more effective flow of cooling air for the engine

Noisy, vortex laden prop wash

RESULT:

Thus the load estimation on wings has been done successfully.