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Exercise 1: V-n diagramsThe 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-n diagram. The particular 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 Vs1Vs1= Design maneuvering speedVAVs1VA need not exceed VCThe design maneuvering speed for maximum gust intensity VBThe design maneuvering speed for maximum gust intensity VB need not be greater than the cruise speed VCHowever, it may not also be less than the speed determined from the intersection of the CN max posline and the gust line marked VB.Design cruising speed VCThe design cruising speed VC (in keas) must be selected by the designer, but must satisfy the following relationshipVC VB + 43 keasDesign diving speed VDVD 1.25 VCNegative 1-g stall speed Vs1 negVs1 neg= Design limit load factors, nlimposand nlimnegnlimpos 2.1 + nlimpos may not be less than 2.5nlimposneed not be greater than 3.8 at W=WFDGWnlimneg for speeds VCnlimnegvaries linearly between VC and VDGust load factor linesFor VB gust lines: Ude= 66 fps for altitudes from sea-level to 23000 feetUde= 84.67 0.000933h for altitudes from 20000 feet to 50000 feetFor VC gust lines: Ude= 50 fps for altitudes from sea-level to 23000 feetUde= 66.67 0.000833h for altitudes from 20000 feet to 50000 feetFor VD gust lines: Ude= 25 fps for altitudes from sea-level to 23000 feetUde= 33.34 0.000417h for altitudes from 20000 feet to 50000 feet

Calculations1-g stall speed, Vs1= For take-off,CNmax= 1.1xCLmax takeoff = 1.1x1.81 = 1.991Vs1= = 16.954 m/s (considered Vs1 neg)For landing,CNmax= 1.1xCLmax landing = 1.1x1.61 = 1.771Vs1= = 17.976m/s (considered Vs1 pos)nlimpos = 2.1 + 24000/11516 = 3.2698VA takeoff= Vs1VA landing =Vs1 = 16.954 = 30.657 m/sDesign cruising speed Vc = 117.222 m/sDesign diving speed VD = 1.25 x Vc = 146.5275m/s

= =17.96Kg for subsonic airplanes = 0.88/5.3+ = 0.6794

Gust line velocities for different values of flight speedFor VB :Ude= 84.67 0.000933h = 84.67 0.000933x26240 = 60.17758 m/sFor VC :Ude= 66.67 0.000833h = 66.67 0.000833x26240 = 44.842m/sFor VB :Ude= 33.34 0.000417h = 33.34 10.942 = 22.397 m/sSolving for , = 1 +For VB, = 1 + = 1 + 0.040 = 1.040, 0.96For VC, = 1 + 0.037 = 1.037, 0.963For VD, = 1 + 0.023 = 1.023, 0.9768The required graphs are shown below

Result: The required velocities are calculated and the related graphs have been plotted.

Exercise 2: Schrenks CurveThe lift acting on a wing varies across the wings 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 both distributions, another lift distribution curve is made which serves an intermediate between the previous two distributions. This curve is known as the Schrenks curve. The Schrenks 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 LoOnce Lo is found, we use the equation of an ellipse

and obtain an expression for L2L2 = LoThe next step is to consider a linear/trapezoidal distributionAssuming 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, Hence, the expression for L1 is attainedL1 = L10 - Calculations:Wmax = 10516 kgnmax = 1.040L = nW = 1.040*10516* 9.81 = 107288.4383Croot = 2.4147mCtip = 0.9649mLtip = 0.5*1.225*(4222)*1.81*0.9659 = 190696.0669 N/mLroot =0.5*1.225*(4222)*1.81*2.4147 =476730.2959 N/m b = 8.29

L0= 2.4999Lb/2

For the elliptic distribution, =

Lo = 16486.51025 N/mHence, the lift distribution obtained is L1 = 3977.448

For the linear distribution,=

53644.2192 = L1b/2 = 7395.588N/mL10 = 18488.23044 N/mHence, the lift distribution attained is L2 = 18488.23 2676.150y.

The lift at various points across the semi-wingspan for all distributions are given belowFraction of semi wingspanLift as per the elliptical distribution (N)Lift as per the linear distribution (N)Mean value (N)

016486.50918488.2304417484.36972

115999.52815812.0804415905.80422

214440.38213135.9304413788.15622

311376.474710459.7804410918.12757

44322.4487783.630446053.03762

4.144907395.5969863697.798493

The required graphical representation of data:

Result: Hence, the distribution of lift across the wing of our plane has been determined through the use of Schrenks method.