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Exercise 1: V-n diagrams The V-n diagrams are used primarily in the determination of combinations of flight conditions and load factors to which the aircraft structure must be designed. This depicts the aircraft limit load as a function of airspeed. In practice, the load factors due to maneuver and gust are indicated by a diagram called “velocity load factor” or the “V“V-n diagram”. The particular V-n V-n diagram to which an airplane must be designed depends on the certification basis selected by the manufacturer or customer. V-n diagram for FAR 23, FAR 25 and military aircrafts are some of them. 1-g stall speed V s1
Vs1=
Design maneuvering speed
VA Vs1
VA need not exceed VC The design maneuvering speed for maximum gust intensity V B
The design maneuvering speed for maximum gust intensity VB need not be greater than the cruise speed VC However, it may not also be less than the speed determined from the intersection o f the C N max posline and the gust line marked VB. Design cruising speed V C
The design cruising speed VC (in keas) must be selected by the designer, but must satisfy the following relationship
VC VB + 43 keas Design diving speed V D
VD 1.25 VC Negative 1-g stall speed V s1 neg
Vs1 neg=
2
Design limit load factors, n limposand nlimneg
nlimpos 2.1 +
nlimpos may not be less than 2.5 nlimposneed not be greater than 3.8 at W=WFDGW
nlimneg for speeds
VC
nlimnegvaries linearly between VC and VD Gust load factor lines
For VB gust lines: Ude= 66 fps for altitudes from sea-level to 23000 feet Ude= 84.67 – 0.000933h for altitudes from 20000 feet to 50000 feet For VC gust lines: Ude= 50 fps for altitudes from sea-level to 23000 feet Ude= 66.67 – 0.000833h for altitudes from 20000 feet to 50000 feet For VD gust lines: Ude= 25 fps for altitudes from sea-level to 23000 feet Ude= 33.34 – 0.000417h for altitudes from 20000 feet to 50000 feet
Calculations
1-g stall speed, Vs1=
For take-off, C Nmax= 1.1xCLmax takeoff = 1.1x1.81 = 1.991
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Vs1=
= 16.954 m/s (considered Vs1 neg) For landing, C Nmax= 1.1xCLmax landing = 1.1x1.61 = 1.771 Vs1=
= 17.976m/s (considered Vs1 pos) nlimpos = 2.1 + 24000/11516 = 3.2698 VA takeoff = Vs1
√
VA landing =Vs1
= 16.954
= 30.657 m/s Design cruising speed Vc = 117.222 m/s Design diving speed VD = 1.25 x Vc = 146.5275m/s
=
=17.96 K g for subsonic airplanes = 0.88
= 0.6794
/5.3+
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Gust line velocities for different values of flight speed For VB :Ude= 84.67 – 0.000933h = 84.67 – 0.000933x26240 = 60.17758 m/s For VC :Ude= 66.67 – 0.000833h = 66.67 – 0.000833x26240 = 44.842m/s For VB :Ude= 33.34 – 0.000417h = 33.34 – 10.942 = 22.397 m/s Solving for
For VB,
,
= 1 +
= 1 +
= 1 + 0.040 = 1.040, 0.96 For VC,
For VD,
= 1 + 0.037 = 1.037, 0.963
= 1 + 0.023 = 1.023, 0.9768
The required graphs are shown below
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V-n gust diagram 1.04
C'
B'
1.03
D' 1.02 ) n ( 1.01 r o t c 1 a f d a 0.99 o L
0.98
E' G'
0.97
F'
0.96 0
50
100
150
200
250
300
350
Speed (V)
Result: The required velocities are calculated and the related graphs have been plotted.
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Exercise 2: Schrenk’s Curve The lift acting on a wing varies across the wing’s span. In order to determine the distribution of lift across the wingspan, we first assume the distribution to be elliptical. Once the elliptical distribution has been obtained, we go on to make a second assumption that the distribution is linear/trapezoidal. Using the data obtained from bo th distributions, another lift distribution curve is made which serves an intermediate between the previous two distributions. This curve is known as the Schrenk’s curve. The Schrenk’s curve attained defines the lift distribution across the span of the wing of our plane. Initially, we assume the distribution to be elliptic. Assuming the lift per unit span to be Lo, the total lift acting across the semi-wing span can be given by
∫ =
→
=
We use this relation to solve Lo Once Lo is found, we use the equation of an ellipse
()
and obtain an expression for L2 L2 = Lo
The next step is to consider a linear/trapezoidal distribution Assuming the lift per unit span to be L10 and L1b/2, the total lift acting across the semi-wing span can be given by
∫ =
→
Now,
=
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Hence, the expression for L1 is attained L1 = L10 -
*⁄+
Calculations:
Wmax = 10516 kg nmax = 1.040 L = nW = 1.040*10516* 9.81 = 107288.4383 Croot = 2.4147m Ctip = 0.9649m 2
Ltip = 0.5*1.225*(422 )*1.81*0.9659 = 190696.0669 N/m 2
Lroot =0.5*1.225*(422 )*1.81*2.4147 =476730.2959 N/m b = 8.29
L0= 2.4999L b/2
For the elliptic distribution,
=
Lo = 16486.51025 N/m
Hence, the lift distribution obtained is L1 = 3977.448
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For the linear distribution,
=
53644.2192 =
L1b/2 = 7395.588N/m L10 = 18488.23044 N/m Hence, the lift distribution attained is L2 = 18488.23 – 2676.150y.
The lift at various points across the semi-wingspan for all distributions are given below Fraction of semi wingspan
Lift as per the elliptical distribution (N)
Lift as per the linear distribution (N)
Mean value (N)
0 1
16486.509 15999.528
18488.23044 15812.08044
17484.36972 15905.80422
2
14440.382
13135.93044
13788.15622
3
11376.4747
10459.78044
10918.12757
4 4.1449
4322.448 0
7783.63044 7395.596986
6053.03762 3697.798493
The required graphical representation of data:
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Result: Hence, the distribution of lift across the wing of our plane has been determined through the use of Schrenk’s method.
