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1. Introduction
1.1 Overview:
The structural design of an airplane actually begins with the flight envelope or V-n diagram, which clearly limits the maximum load factors that the airplane can withstand at any particular flight velocity. However in normal practice the airplane might experience loads that are much higher than the design loads. Some of the factors that lead to the structural overload of an airplane are high gust velocities, sudden movements of the controls, fatigue load in some cases, bird strikes or lightning strikes. So to add some inherent ability to withstand these rare but large loads, a safety factor of 1.5 is provided during the structural design.
The two major members that need to be considered for the structural design of an airplane are wings and the fuselage. As far as the wing design is concerned, the most significant load is the bending load. So the primary load carrying member in the wing structure is the spar (the front and rear spars) whose cross section is an ‘I’ section. Apart from the spars to take the bending loads, suitable stringers need to take the shear loads acting on the wings.
Unlike the wing, which is subjected to mainly unsymmetrical load, the fuselage is much simpler for structural analysis due to its symmetrical crossing and symmetrical loading. The main load in the case of fuselage is the shear load because the load acting on the wing is transferred to the fuselage skin in the form of shear only. The structural design of both wing and fuselage begin with shear force and bending moment diagrams for the respective members. The maximum bending stress produced in each of them is checked to be less than the yield stress of the material chosen for the respective member.
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1.2 Outline:
The Structural design involves:
Determination of loads acting on aircraft:
V-n diagram for the design study
Gust and maneuverability envelopes
Schrenk’s Curve
Critical loading performance and final V-n graph calculation Determination of loads acting on individual structures
Structural design study – Theory approach
Load estimation of wings
Load estimation of fuselage.
Material Selection for structural members
Detailed structural layouts
Design of some components of wings, fuselage
1.3 Parameters forwarded from ADP – 1
Take off Gross Weight, Maximum Velocity,
Cruise Velocity,
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Table 1-1: Mass ratio Split up
Components Mass Fraction
Crew 0.00053146 Landing Gear 0.042516824 Payload 0.190144687 Fixed Equipment 0.002262033 Fuselage mass 0.085033649 Horizontal Stablizer 0.012597578 Vertical Stabilizer 0.006298789 Wing Structure 0.125975776 Fuel 0.4810995 Power plant 0.053539705 Total 1 Cruise Altitude = 12 km
The airfoil used her is NACA 653 - 418 Density at cruise altitude,
Cruise C_L @ Cruise altitude,
@ 16 ˚ aoa
@ 14 ˚ since tail angle is 15.56 ˚ @ -14 ˚ aoa
4 ct = 5.797 m ( ) ( )
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2. V-n Diagram
2.1 Maneuvering Envelope:
In accelerated flight, the lift becomes much more compared to the weight of the aircraft. This implies a net force contributing to the acceleration. This force causes stresses on the aircraft structure. The ratio of the lift experienced to the weight at any instant is defined as the Load Factor (n).
Using the above formula, we infer that load factor has a quadratic variation with velocity. However, this is true only up to a certain velocity.
This velocity is determined by simultaneously imposing limiting conditions aerodynamically ((CL)max) as well as structurally (nmax). This velocity is called the Corner Velocity, and is determined using the following formula,
√
In this section, we estimate the aerodynamic limits on load factor, and attempt to draw the variation of load factor with velocity, commonly known as the V-n Diagram. The V‐n diagram is drawn for Sea level Standard conditions.
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Figure 2-1: Typical V-n diagram for a private airliner.
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V-n diagram is used primarily in the determination of combination of flight conditions and load factors to which the airplane structure must be designed. V-n diagram precisely gives the structural (maximum load factor) aV-nd aerodynamic (maximum CL) boundaries for a particular flight condition.
2.2 Construction of V-n diagram
2.2.1 Curve OA:
Maximum Load Factor, ( )
( )
Hence along the curve OA,
Using the above equation we get
Table 2-1: Velocity vs. positive load factor (n)
Velocity V (m/s) Load Factor (n)
0 0 20 0.08492371 40 0.339694842 60 0.764313394 80 1.358779368 100 2.123092762 120 3.057253578 140 4.161261814 160 5.435117472 180 6.87882055 200 8.49237105 220 10.27576897 240 12.22901431
8 At A,
2.2.2 Curve AC:
AC is a line limiting the maximum amount of load that can be withstood by the weakest structure of the aircraft
√* + √[ ( ) ]
VC= 408.32 m/s
nC=nA
2.2.3 Along CD:
The velocity at point D is given by V
D=1.5VC= 416.66 m/s n
D= 0.75nA= 4.80864 A straight line is used to join the points C and D This V
D is the dive velocity or the maximum permissible EAS in which the aircraft is at the verge of structural damage due to high dynamic pressure.
2.2.4 Along DE:
9 n= 0
For Bombers the load factor can vary from -3 to +6.5 Hence the negative load factor of aircraft is limited to -2
2.2.5 Along EF
The point F corresponds to the velocity
VC = VF = 408.32 m/s
2.2.6 Curve OG:
nF= -2 (for a typical bomber aircraft)
Hence along the curve OG,
Hence we get,
Table 2-2: Velocity vs. negative Load factor (n)
Velocity V (m/s) Load Factor (n)
0 0 20 -0.034356779 40 -0.137427115 60 -0.309211009 80 -0.549708461 100 -0.85891947 120 -1.236844037 140 -1.683482161 160 -2.198833843 180 -2.782899083 200 -3.43567788 220 -4.157170234
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2.2.7 Along GF:
Also n
G=nF Finally join GF by using a straight line
2.3 Nomenclature of curves:
• PHAA – Positive High Angle of Attack • PSL – Positive Structural Limit
• PLAA – Positive Low Angle of Attack • HSL –High Speed Limit
• NHAA – Negative High Angle of Attack • NSL – Negative Structural Limit
• NLAA – Negative Low Angle of Attack • LSL – Low Speed Limit
Figure 2-3: Four basic flight conditions showing how location of maximum stresses in wing depends on angle of attack
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2.4 Low Speed Limit:
Stall velocity is the maximum speed at which the aircraft can maintain level flight. This implies the intersection of this line at cruise n=1 with OA curve corresponds to stall velocity Vs.
Vs = 68.630 m/s
Figure 2-4: Rough V-n Diagram
From the V-n diagram, it is observed that the stall curve corresponds to maximum value of CLmax and any point beyond this curve for a particular velocity is not achievable in flight as it enters the stall region there. The upper horizontal line corresponds to limit load factor as well as ultimate load factor. It
-10
-5
0
5
10
0
100
200
300
400
500
Load
F
act
or
Velocity (m/s)
Rough V-n Diagram
PHAA
NHAA
PSL
HSL
NSL
PLAA
NLAA
LSL
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shows that there is outright structural failure when the aircraft is flown beyond this value of load factor.
n=-2 gives the negative limit load factor and negative ultimate load factor.
From the figure, it is clear that for a particular velocity, it is not possible to fly at a value of CL higher than the CLmax corresponding to that velocity. If we wish to increase the lift of the airplane to that value of CLmax, then we should increase the flying speed of the airplane.
Figure 2-5: Maneuvering Envelope
-3 -2 -1 0 1 2 3 4 5 6 7 0 50 100 150 200 250 300 350 400 450 Load Fac to r Velocity
Maneuvering Envelope
LSL PIAA NIAA HS L13
Figure 2-6: Maneuvering Envelope
Figure 2-7: Maneuvering envelope with coordinates
Hence for the strategic bomber aircraft we get, Safety Factor = 1.5 -3 -2 -1 0 1 2 3 4 5 6 7 0 50 100 150 200 250 300 350 400 450 Load Fac to r Velocity
Maneuvering Envelope
A H C E D B G F -3 -2 -1 0 1 2 3 4 5 6 7 0 50 100 150 200 250 300 350 400 450 Load Fac to r VelocityManeuvering Envelope with coordinates
68.63, -0.4046 68.630 2,1 408.32, 6.41152 173.77, 6.41152 159.5944,-2 416.66 , 0 416.66, 4.80864 408.32,-2
14 Caution Speed = 325 m/s
Corner Velocity = 173.77 m/s Stall speed = 59.669 m/s
Safety load factor limit i.e., indications given to pilot n = -2/ 1.5 = -1.3333
n = 6.41152/ 1.5 = 4.2743 Dive Velocity = 416.66 m/s
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3. Gust Envelope
3.1 Description:
Gust is a sudden, brief increase in the speed of the wind. Generally, winds are least gusty over large water surfaces and most gusty over rough land and near high buildings. With respect to aircraft turbulence, a sharp change in wind speed relative to the aircraft; a sudden increase in airspeed due to fluctuations in the airflow, resulting in increased structural stresses upon the aircraft.
Sharp-edged gust (u) is a wind gust that results in an instantaneous change in direction or speed.
Derived gust velocity (U or Umax) is the maximum velocity of a sharp-edged gust that would produce a given acceleration on a particular airplane flown in level flight at the design cruising speed of the aircraft and at a given air density. As a result a 25% increase is seen in lift for a longitudinally disturbing gust. The effect of turbulence gust is to produce a short time change in the effective angle of attack. These changes produce a variation in lift and thereby load factor For velocities up to Vmax, cruise, a gust velocity of 15 m/s at sea level is assumed. For Vdiv, a gust velocity of 10 m/s is assumed.
Effective gust velocity: The vertical component of the velocity of a
sharp-edged gust that would produce a given acceleration on a particular airplane flown in level flight at the design cruising speed of the aircraft and at a given air density.
Reference Gust Velocity (Uref ) —at sea level 15m/s.
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Figure 3-1: Variation in Aerodynamic limits due to gust
3.2 Construction
The increase in the load factor due to the gust can be calculated by For curve above V-axis:
Where
K Gust Alleviation Factor
U max Maximum derived Gust Velocity a Lift Curve Slope for wing
For curve below V-axis:
-6 -4 -2 0 2 4 6 8 10 12 0 50 100 150 200 250 Load fac to r Velocity
Variation in aerodynamic limits
Normal Stall curve Gust stall curve Normal neg stall curve
Gust neg stall curve Flaps Retracted
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Gust Alleviation Factor:
Gust Alleviation Factor (K):
Lateral Mass Ratio (µ):
̂ Where
g Acceleration due to Gravity ĉ Mean Aerodynamic Chord
̂ ( ) ct Chord at tip cr Chord at root cr = 11.593 m ct = 5.797 m √ a= 0.1213507 /degree
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a= 6.9528829 /radian
for a’ =0.15/ degree where a’ is lift curve slope for the chosen airfoil NACA 65-(3) 418
a’ lift curve slope for airfoil
Sweep angle at leading Edge of Wing
( )
Table 3-1: Equivalent air speed and corresponding Derived Gust Velocity
For Velocity at points Equivalent air speed
V (m/s)
Derived Gust Velocity
Umax (m/s)
B,G 173.77 15
C,F 408.32 10
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By using the equations and for various speeds of Umax we get the following gust lines
Figure 3-2: Gust Lines
-2 -1 0 1 2 3 4 0 50 100 150 200 250 300 350 400 450 Load fac to r Velocity
Gust Lines
U=15m/s U=10m/s U=5m/s U=0m/s U=-5m/s U=-10./s U=-15m/s
Level
Design speed 277.77 m/s
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Figure 3-3: Overlapped maneuvering envelope and gust lines.
The load factors at the various points can be found using the formula using the corresponding values of Umax
n B’ = 1.4195966 n G’= 0.846720 nC’ = 2.5617017 nF’ = -0.5617017 n D’ = 1.796799 n E’ = 0.203200 -3 -2 -1 0 1 2 3 4 5 6 7 0 50 100 150 200 250 300 350 400 450 Load fac to r Velocity
Overlaped Maneuver envelope and gust lines
Gust stall curve Gust neg stall curve HSL
PLAA NLAA PIAA
NIAA U=15m/s U=10m/s
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The positive load factor along the curve OB’ is given by the equation
Hence along the curve OA,
But also
Equating the above two equations we get an intersecting point B where velocity is
VB = 73.1379 m/s
Since the velocities and load factors at C, D, E and F are known and straight lines are used to join these points in sequence
3.2.1 Line FG:
It is found that negative gust line of U= -15 m/s intersects the positive high angle of attack condition at G.
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Equating the above equation with the OA curve equation we get the point G where
VG = 51.52026m/s
Figure3-4: Gust Envelope
-2 -1.5 -1 -0.5 0 0.5 1 1.5 2 2.5 3 0 50 100 150 200 250 300 350 400 450 Load fac to r Velocity (m/s)
Gust Envelope
B C D G O E F24
Figure 3-5: Gust Envelope with coordinates
-2 -1.5 -1 -0.5 0 0.5 1 1.5 2 2.5 3 0 50 100 150 200 250 300 350 400 450 Load fac to r Velocity
Gust Envelope with coordinates
416.66, 1.7968 410.9975, -1.3791 416.66, 0 416.66, 1.7968 408.32, 2.561702 73.13798, 1.419596 51.52026, 0.704426
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4. Schrenk’s Curve
4.1 Description
Lift varies along the wing span due to the variation in chord length, angle of attack and sweep along the span. Schrenk’s curve defines this lift distribution over the wing span of an aircraft, also called simply as Lift Distribution Curve. Schrenk’s Curve is given by
Where
y1 is Linear Variation of lift along semi wing span also named as L1 y2 is Elliptic Lift Distribution along the wing span also named as L2
a = 44.8285 m
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Figure 4-1: Wing geometry showing sweep angle and semi span along the root.
4.2 Linear Lift Distribution:
Lift at root Lroot = 90978.038 N/m Lift at tip Ltip = 45492.942 N/m
By representing this lift at sections of root and tip we can get the equation for the wing.
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Figure 4-2: Linear lift distribution
Equation of linear lift distribution for starboard wing
Equation of linear lift distribution for port wing we have to replace x by –x in general,
Twice the area under y1= Total lift= 2491907.5 N ≈ Take off Gross Weight
Figure 4-3: Linear Variation of lift along wing semi span
0 10 20 30 40 50 60 70 80 90 100 0 10 20 30 40 50 Li ft p e r m e te r (N /m ) Tho u san d s
Wing Semi Span (m)
Linear variation of Lift along wing Semi span
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For the Schrenk’s curve we only consider half of the linear distribution of lift and hence we derive y1/2
4.3 Elliptic Lift Distribution:
Twice the area under the curve or line will give the lift which will be required to overcome weight
Considering an elliptic lift distribution we get
Where b1 is Actual lift at root
And a is wing semi span Lift at tip
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Figure 4-4: Elliptic lift distribution
Equation of elliptic lift distribution
√
Figure 4-5: Elliptic lift distribution
√ 0 10 20 30 40 50 60 70 80 0 10 20 30 40 50 Lift p e r m e te r (N /m ) Th o u san d s
Wing Semi Span (m)
Elliptic variation of Lift along wing Semi span
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4.4 Construction of Schrenk’s Curve:
Schrenk’s Curve is given by
√
√
Substituting different values for x we can get the lift distribution for the wing semi span
Table 4-1: Lift distribution table along semi span x L1 L2 L 0 90978.04 70776.189 80877.11 1 89963.39 70758.577 80360.98 2 88948.75 70705.716 79827.23 3 87934.1 70617.525 79275.81 4 86919.45 70493.872 78706.66 5 85904.81 70334.571 78119.69 6 84890.16 70139.378 77514.77 7 83875.52 69907.993 76891.75 8 82860.87 69640.055 76250.46 9 81846.22 69335.14 75590.68 10 80831.58 68992.759 74912.17 11 79816.93 68612.349 74214.64 12 78802.29 68193.275 73497.78 13 77787.64 67734.819 72761.23 14 76772.99 67236.175 72004.58 15 75758.35 66696.443 71227.4 16 74743.7 66114.615 70429.16 17 73729.06 65489.57 69609.31 18 72714.41 64820.058 68767.23 19 71699.76 64104.686 67902.22 20 70685.12 63341.899 67013.51 21 69670.47 62529.962 66100.22 x L1 L2 L 22 68655.83 61666.935 65161.38 23 67641.18 60750.639 64195.91 24 66626.53 59778.626 63202.58 25 65611.89 58748.129 62180.01 26 64597.24 57656.013 61126.63 27 63582.6 56498.706 60040.65 28 62567.95 55272.111 58920.03 29 61553.3 53971.505 57762.4 30 60538.66 52591.398 56565.03 31 59524.01 51125.351 55324.68 32 58509.37 49565.739 54037.55 33 57494.72 47903.426 52699.07 34 56480.07 46127.307 51303.69 35 55465.43 44223.675 49844.55 36 54450.78 42175.265 48313.02 37 53436.14 39959.82 46697.98 38 52421.49 37547.784 44984.64 39 51406.84 34898.417 43152.63 40 50392.2 31952.741 41172.47 41 49377.55 28619.406 38998.48 42 48362.91 24742.227 36552.57 43 47348.26 20007.494 33677.88 44 46333.61 13543.872 29938.74 44.8285 45492.98 0 22746.49
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Figure 4-6: Schrenk’s curve with linear and elliptic lift distribution
Replacing x by –x for port wing we can get lift distribution for entire span.
Figure 4-7: Schrenk’s curve
0 10 20 30 40 50 60 70 80 90 100 0 10 20 30 40 50 Li ft p e r m e te r sp an ( N /m ) Tho u san d s
Wing span loaction (m)
Schrenk's Curve
L L1 L2 0 10 20 30 40 50 60 70 80 90 -60 -40 -20 0 20 40 60 Li ft p e r m e te r sp an ( N /m ) Tho u san d sWing span loaction (m)
Schrenk's Curve
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5. Load Estimation on wings
5.1 Description:
The solution methods which follow Euler’s beam bending theory (σ/y=M/I=E/R) use the bending moment values to determine the stresses developed at a particular section of the beam due to the combination of aerodynamic and structural loads in the transverse direction. Most engineering solution methods for structural mechanics problems (both exact and approximate methods) use the shear force and bending moment equations to determine the deflection and slope at a particular section of the beam. Therefore, these equations are to be obtained as analytical expressions in terms of span wise location. The bending moment produced here is about the longitudinal (x) axis.
5.2 Loads acting on wing:
As both the wings are symmetric, let us consider the starboard wing at first. There are three primary loads acting on a wing structure in transverse direction which can cause considerable shear forces and bending moments on it. They are as follows:
Lift force (given by Schrenk’s curve)
Self-weight of the wing
Weight of the power plant
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5.3 Shear force and bending moment diagrams due to
loads along transverse direction at cruise condition:
Lift Force given by Schrenk’s Curve:
Linear lift distribution (trapezium):
Elliptic lift distribution (quarter ellipse) √ √
Figure 5-1: Lift distribution (linear)
0 5000 10000 15000 20000 25000 30000 35000 40000 45000 50000 0 5 10 15 20 25 30 35 40 45 50 Li ft p e r u n it le n gth (N /m )
Span wise location (m)
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Figure 5-2: Lift distribution (Elliptic)
Self-Weight (y3): Self-weight of the wing,
Assuming parabolic weight distribution
( ( )) Where b span 0 5000 10000 15000 20000 25000 30000 35000 40000 0 10 20 30 40 50 Li ft p e r u n it le n gth (N /m )
Span wise location (m)
Elliptic lift distribution (y2/2)
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When we integrate from x=0 (root location) to x=b (tip location) we get the net weight of port wing.
∫ ∫ ( )
( )
Substituting various values of x in the above equation we get the self-weight of the wing.
Figure 5-3: Self weight of wing
Power plant weight:
Power plant is assumed to be a point load,
-25000 -20000 -15000 -10000 -5000 0 0 5 10 15 20 25 30 35 40 45 50 WE ig h t o f e m p ty wi n g (N /m )
Span wise location (m)
Self Weight
36 Acting at x= 8 m and x= 14 m from the root. Fuel weight:
This design has fuel in the wing so we have to consider the weight of the fuel in the wing.
Again by using general formula for straight line y=mx + c we get,
Figure 5-4: Fuel Distribution
-20000 -18000 -16000 -14000 -12000 -10000 -8000 -6000 -4000 -2000 0 0 5 10 15 20 25 30 35 40 45 50 Fu e l w e ig h t (N /m )
Span wise location (m)
Fuel distribution
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Figure 5-5: Overall Load distribution Table 5-1: Loads simplified as point loads
-80000 -60000 -40000 -20000 0 20000 40000 60000 0 5 10 15 20 25 30 35 40 45 50 Load ac ting o n wi n g (N /m )
Span wise location (m)
Load distribution
Curve / component Area enclosed / structural weight (N)
Centroid (from wing root)
y1/2 1529447.31 19.923 m
y2/2 1245953.75 3.510534 m
Wing 313917 16.8107 m
Fuel 365752.803 16.4606 m
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Figure 5-6: Reaction force and Bending moment calculations
Now we know VA and MA, using this we can find out shear force and Bending moment.
5.3.1 Shear Force:
39 √ ( ) ( ) ∫ ∫
By using the corresponding values of x in appropriate equations we get the plot of shear force
Note: Shear force is a discrete function along y axis so in order to make it continuous we introduce straight lines.
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Figure 5-7: Shear force diagram - discrete
Figure 5-8: Shear force diagram- continuous
-3000 -2500 -2000 -1500 -1000 -500 0 500 1000 -44.8285 -34.8285 -24.8285 -14.8285 -4.8285 5.1715 15.1715 25.1715 35.1715 Sh e ar Fo rc e (N ) Th o u san d s Location in wing (m)
Shear Force
-3000 -2500 -2000 -1500 -1000 -500 0 500 1000 -44.8285 -34.8285 -24.8285 -14.8285 -4.8285 5.1715 15.1715 25.1715 35.1715 Sh e ar Fo rc e (N ) Th o u san d s Location in wing (m)Shear Force (Actual)
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5.3.2 Bending moment:
∫ ∫ ( ( √ ) ( )) ( ) ∫ ∫By substituting the values of x for the above equations of bending moments obtained we can get a continuous bending moment curve for the port wing. Note: if we replace the x by -x in each term we get the distribution of starboard wing
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Figure 5-9: Bending moment diagram
5.4 Shear force and bending moment diagrams due to
loads along chordwise direction at cruise condition:
Aerodynamic center- This is a point on the chord of an airfoil section where the bending moment due to the components of resultant aerodynamic force (Lift and Drag) is constant irrespective of the angle of attack. Hence the forces are transferred to this point for obtaining constant Ma.c
Shear center- This is a point on the airfoil section where if a force acts, it produces only bending and no twisting. Hence the force is transferred to this point and the torque is found.
Cruise CL=0.204908 @ V= 250 m/s Cruise CD= 0.0055
Angle of attack= -0.811439˚ (obtained from the lift curve slope) Angle of attack @ zero lift= -3o
0 10 20 30 40 50 60 70 -50 -40 -30 -20 -10 0 10 20 30 40 50 B e n d in g M o m e n t (N m ) M ill ion s Location in wing (m)
Bending Moment
43 Wing lift curve slope (a)= 0.1213507 /degree
Co-efficient of moment about aerodynamic centre= -0.0543 Location of aerodynamic centre:
Location of shear centre:
Lift and drag are the components of resultant aerodynamic force acting normal to and along the direction of relative wind respectively. As a result, components of them act in the chordwise direction also which produce a bending moment about the normal (z) axis.
Figure 5-10: Normal and chord wise coefficients
Co-efficient of force along the normal direction,
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Chordwise force at root
Chordwise force at tip
By using y = mx +c again we get the equation as
The above equation gives the profile of load acting chordwise, by integrating this above equation we get a component of Shear force and again by integrating the same we get the component of Bending Moment
∫
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Figure 5-11: Load along chordwise direction
To find fixing moment and the reaction force,
5.4.1 Shear Force:
0 200 400 600 800 1000 1200 1400 0 5 10 15 20 25 30 35 Load alon g c h o rd wi se d ir e ction ( N ) Spanwise location (m)46
Figure 5-12: Shear force
5.4.2 Bending Moment:
Figure 5-13: Bending moment
-35 -30 -25 -20 -15 -10 -5 0 0 5 10 15 20 25 30 35 Sh e ar Fo rc e (N ) Th o u san d s Spanwise location (m)
Shear Force
0 100 200 300 400 500 600 0 5 10 15 20 25 30 35 B e n d in g m o m e n t (N m ) Th o u san d s Spanwise location (m)Bending Moment
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Torque due to normal forces and constant pitching moment at cruise condition:
Figure 5-14: Moment about aerodynamic center
The lift and drag forces produce a moment on the surface of cross-section of the wing, otherwise called a torque, about the shear center. Moment about the aerodynamic center gets transferred to the shear center. The powerplant also produces a torque about the shear center on the chord under which it is located.
Figure 5-15:Torque due to normal force and moment
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5.5 Torque at cruise condition:
5.5.1 Torque due to normal force:
Where c chord
the equation for chord can also be represented in terms of x by taking c= mx +k, Therefore torque ∫ ∫
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Figure 5-16: Torque due to normal force
5.5.2 Torque due to chord wise force:
5.5.3 Torque due to moment:
∫ 0 100 200 300 400 500 600 700 800 900 1000 0 5 10 15 20 25 30 35 40 To rq u e (N m ) Th o u san d s Spanwise location (m)
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Figure 5-17: Torque due to moment
5.5.4 Torque due to powerplant:
The powerplant is situated under a chord (8 m and 14 m from the wing root; chord length 10.7504 m and 10.1184m ) from 0.1c to 0.5c at 10.7504m and from 0.1 c to 0.5249c an Uniformly Distributed Load of 15513.488 N/m is assumed to be present for this 4.3 m since the powerplant weight is 66708N. The centroid of the applied UDL is at 0.3c for first case and at 0.31245c at second location.
Torque produced about shear center
-8 -7 -6 -5 -4 -3 -2 -1 0 0 5 10 15 20 25 30 35 40 To rq u e (N m ) M ill io n s Spanwise location (m)
Torque due to Moment
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Figure 5-18: Torque due to powerplant
Then the different torque components are brought together in a same graph to make a comparison
Figure 5-19: Torque comparison
0 10 20 30 40 50 60 70 0 5 10 15 20 25 30 35 40 To rq u e (N m ) Th o u sa n d s Spanwise location (m)
Torque due to Powerplant
-8000 -7000 -6000 -5000 -4000 -3000 -2000 -1000 0 1000 2000 0 10 20 30 40 To rq u e (N m ) Th o u san d s Spanwise location (m)
Torque comparison
Torque due to Normal Forces Torque due to moment Torque due to powerplant
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The net torque will be sum of all the above torques i.e. torque due to normal force, chordwise force, powerplant and aerodynamic moment
Figure 5-20: Net torque
-7 -6 -5 -4 -3 -2 -1 0 0 5 10 15 20 25 30 35 40 To rq u e (N m ) M ill io n s Spanwise location (m)
Torque
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5.6 Load at Critical flight condition:
Optimum Wing structural design consists of determining that stiffness distribution which is proportional to the local load distribution. The aerodynamic forces of lift and drag are resolved into components normal and parallel to the wing chord. The distribution of shear force, bending moment and torque over the aircraft wing are considered for wing structural analysis.
Identification of critical points from the maneuvering and gust envelopes: 1. Maneuvering envelope
Table 5-1: Coordinates of V-n diagram
Point Load factor E.A.S. (m/s)
A 6.41152 173.77 C 6.41152 408.32 D 4.80864 416.66 E 0 416.66 F -2 408.16 G -2 159.5944 2. Gust envelope
Table 5-2: Coordinates of gust envelope
Point Load factor E.A.S. (m/s)
B’ 1.41959 173.137975 C’ 2.5617 408.16 D’ 1.7968 416.66 E’ 0.2032 416.66 F’ -1.3255 408.16 G’ 0.5822 51.52026
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Corner points are representative of critical flight load conditions – a summary is given below.
Table 5-3: Coordinates of critical conditions
Critical flight condition Point (‘n’, E.A.S.)
‘n’ max point C’ (2.5617, 408.16)
Positive H.A.A. A (6.41152, 173.77)
Positive L.A.A D (4.80864, 416.66)
Negative H.A.A G (-2, 408.16)
Negative L.A.A E’ (0, 416.66)
Shear force and bending moment diagrams of a wing due to normal forces at critical flight condition:
In the preliminary stage of structural analysis, the critical flight loading condition of positive high angle of attack (represented by point A in v-n diagram) will be investigated.
It is seen that lift has increased by 6.41152 times.
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Figure 5-21: Linear Variation of lift along wing semi span
Figure 5-22: Elliptic variation of lift along wing semi span
0 100 200 300 400 500 600 700 0 10 20 30 40 50 Li ft p e r m e te r (N /m ) Th o u san d s
Wing Semi Span (m)
Linear variation of Lift along wing Semi span
(critical condition)
L1 0 50 100 150 200 250 300 350 400 450 500 0 10 20 30 40 50 Li ft p e r m e te r (N /m ) Th o u san d sWing Semi Span (m)
Elliptic variation of Lift along wing Semi span
(Critical Condition)
56
The aim is to find the shear forces and bending moments due to normal forces in critical flight condition. There are three primary loads acting on a wing structure in transverse direction which can cause considerable shear forces and bending moments on it. They are as follows:
Lift force (given by Schrenk’s curve)
Self-weight of the wing
Weight of the power plant
Weight of the fuel in the wing
Now, the proportionality constant influences the lift force alone and other factors remain unaffected.
Table 5-4: Loads simplified as point loads at critical flight condition
Curve / component Area enclosed / structural weight (N)
Centroid (from wing root)
y1/2 1529447.31×6.41152 19.923 m
y2/2 1245953.75 ×6.41152 3.510534 m
Wing 313917 16.8107 m
Fuel 365752.803 16.4606 m
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Figure 5-23: Critical schrenk’s curve
Figure 5-24: load distribution at critical condition
0 100 200 300 400 500 600 -60 -40 -20 0 20 40 60 Lift p e r m e te r sp an ( N /m ) Th o u san d s
Wing span loaction (m)
Schrenk's Curve (Critical Condition)
58
Figure 5-25: load distribution at critical condition
Now we know VA and MA, using this we can find out shear force and Bending moment,
5.7 Shear force and bending moment diagrams due to
loads along transverse direction at critical condition:
∫ -0.1 -0.05 0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0 5 10 15 20 25 30 35 40 45 50 Load ( N ) M ill io n s Location in wing (m)
59 ( ( √ ( ))) ( ) ∫ ∫
By using the corresponding values of x in appropriate equations we get the plot of shear force
60
Figure 5-26: Transverse Shear force diagram at critical condition
Figure 5-27: Transverse Shear force diagram at critical condition
5.7.1 Bending moment:
∫ ∫ -20000 -18000 -16000 -14000 -12000 -10000 -8000 -6000 -4000 -2000 0 2000 -44.8285 -34.8285 -24.8285 -14.8285 -4.8285 5.1715 15.1715 25.1715 35.1715 Sh e ar Fo rc e (N ) Tho u sa n d s Location in wing (m)Shear Force (Critical condition)
-20000 -18000 -16000 -14000 -12000 -10000 -8000 -6000 -4000 -2000 0 2000 -44.8285 -34.8285 -24.8285 -14.8285 -4.8285 5.1715 15.1715 25.1715 35.1715 Sh e ar Fo rc e (N ) Tho u sa n d s Location in wing (m)
61 ( ( ( √ ) ( ))) ( ) ∫ ∫
By substituting the values of x for the above equations of bending moments obtained we can get a continuous bending moment curve for the port wing.
62
Figure 5-28: Transverse bending moment diagram at critical condition
5.8 Shear force and bending moment diagrams due to
loads along chordwise direction at critical condition:
Critical CL=2.71925 @ V= 250 m/s Critical CD= 0.0084
Angle of attack= 16˚ (obtained from the lift curve slope) Wing lift curve slope (a) = 0.1213507 /degree
Co-efficient of moment about aerodynamic centre= -0.025 Location of aerodynamic centre:
Location of shear centre:
-100 0 100 200 300 400 500 600 -50 -40 -30 -20 -10 0 10 20 30 40 50 B e n d in g M o m e n t (N m ) M ill io n s Location in wing (m)
63
Figure 5-29: Determination of chordwise force components at critical condition
Co-efficient of force along the normal direction,
Chordwise force at root
Chordwise force at tip
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By using y = mx +c again we get the equation as
The above equation gives the profile of load acting chordwise, by integrating this above equation we get a component of Shear force and again by integrating the same we get the component of Bending Moment
∫
∬
Figure 5-30: Load along chord wise direction at critical condition
0 50000 100000 150000 200000 250000 300000 350000 400000 0 5 10 15 20 25 30 35 Load alon g c h o rd wi se d ir e ction ( N ) Spanwise location (m)
Load along Chordwise direction (critical
condition)
65 To find fixing moment and the reaction force,
5.8.1 Shear Force:
Figure 5-31: Chordwise Shear force diagram at critical condition
5.8.2 Bending Moment:
-12000 -10000 -8000 -6000 -4000 -2000 0 0 5 10 15 20 25 30 35 Sh e ar Fo rc e (N ) Th o u san d s Spanwise location (m)Shear Force
66
Figure 5-32: Chordwise Bending moment diagram at critical condition
Torque due to normal forces and constant pitching moment at cruise condition:
Figure 5-33: Determination of various components of torque
0 20000 40000 60000 80000 100000 120000 140000 160000 180000 0 5 10 15 20 25 30 35 B e n d in g m o m e n t (N m ) Th o u san d s Spanwise location (m)
Bending Moment
67
Figure 5-34: Determination of various components causing torque
5.9 Torque at critical flight condition:
5.9.1 Torque due to normal force:
Where c chord
the equation for chord can also be represented in terms of x by taking c= mx +k,
Therefore torque
∫
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Figure 5-35: Torque due to normal force at critical condition
5.9.2 Torque due to chord wise force:
5.9.3 Torque due to moment:
∫ 0 2000 4000 6000 8000 10000 12000 0 5 10 15 20 25 30 35 40 To rq u e (N m ) Th o u san d s Spanwise location (m)
Torque due to normal forces
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Figure 5-36: Torque due to moment at critical condition
5.9.4 Torque due to powerplant:
The powerplant is situated under a chord (8 m and 14 m from the wing root; chord length 10.7504 m and 10.1184m ) from 0.1c to 0.5c at 10.7504m and from 0.1 c to 0.5249c an Uniformly Distributed Load of 15513.488 N/m is assumed to be present for this 4.3 m since the powerplant weight is 66708N. The centroid of the applied UDL is at 0.3c for first case and at 0.31245c at second location.
Torque produced about shear center
Hence tis is weight this will remain same as that of the cruise condition.
-2 -1.8 -1.6 -1.4 -1.2 -1 -0.8 -0.6 -0.4 -0.2 0 0 5 10 15 20 25 30 35 40 To rq u e ( N m ) M ill io n s Spanwise location (m)
Torque due to Moment
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Figure 5-37: Torque due to powerplant at critical condition unchanged
Then the different torque components are brought together in a same graph to make a comparison
The net torque will be sum of all the above torques i.e. torque due to normal force, chordwise force, powerplant and aerodynamic moment
0 10 20 30 40 50 60 70 0 5 10 15 20 25 30 35 40 To rq u e (N m ) Th o u sa n d s Spanwise location (m)
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Figure 5-38: Torque comparison at critical condition
Figure 5-39: Net torque at critical condition
-4000 -2000 0 2000 4000 6000 8000 10000 12000 0 10 20 30 40 To rq u e ( N m ) Th o u san d s Spanwise location (m)
Torque comparison
Torque due to Normal Forces Torque due to moment Torque due to powerplant
0 1 2 3 4 5 6 7 8 9 10 0 5 10 15 20 25 30 35 40 To rq u e (N m ) M ill io n s Spanwise location (m)
Torque
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5.10 Interim Summary:
DUE TO NORMAL FORCES:
Table 5-5: Determination of maximum values of normal force
Cruise
condition
+ve high AOA
condition
At
Max. Shear force (N)
-2461484.863-17480623.2
Wing rootMax. Bending
moment (Nm)
65115382.95
486567918.1
Wing rootDUE TO CHORDWISE FORCES:
Table 5-6: Determination of maximum values of chordwise force
Cruise
condition
+ve high AOA
condition
At
Max. Shear force (N)
-31590.839-10102394.13
Wing rootMax. Bending
moment (Nm)
51270.9081
163971959.1 Wing root73
6. Material Selection:
6.1 Description:
Aircraft structures are basically unidirectional. This means that one dimension, the length, is much larger than the others - width or height. For example, the span of the wing and tail spars is much longer than their width and depth; the ribs have a much larger chord length than height and/or width; a whole wing has a span that is larger than its chords or thickness; and the fuselage is much longer than it is wide or high. Even a propeller has a diameter much larger than its blade width and thickness, etc.... For this simple reason, a designer chooses to use unidirectional material when designing for an efficient strength to weight structure.
Unidirectional materials are basically composed of thin, relatively flexible, long fibers which are very strong in tension (like a thread, a rope, a stranded steel wire cable, etc.)
An aircraft structure is also very close to a symmetrical structure. That means the up and down loads are almost equal to each other. The tail loads may be down or up depending on the pilot raising or dipping the nose of the aircraft by pulling or pushing the pitch control; the rudder may be deflected to the right as well as to the left (side loads on the fuselage). The gusts hitting the wing may be positive or negative, giving the up or down loads which the occupant experiences by being pushed down in the seat ... or hanging in the belt.
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structural material that can withstand both tension and compression. Unidirectional fibers may be excellent in tension, but due to their small cross section, they have very little inertia (we will explain inertia another time) and cannot take much compression. They will escape the load by bucking away. As in the illustration, you cannot load a string, or wire, or chain in compression. In order to make thin fibers strong in compression, they are "glued together" with some kind of an "embedding". In this way we can take advantage of their tension strength and are no longer penalized by their individual compression weakness because, as a whole, they become compression resistant as they help each other to not buckle away. The embedding is usually a lighter, softer "resin" holding the fibers together and enabling them to take the required compression loads. This is a very good structural material.
WOOD
Historically, wood has been used as the first unidirectional structural raw material. They have to be tall and straight and their wood must be strong and light. The dark bands (late wood) contain many fibers, whereas the light bands (early wood) contain much more "resin". Thus the wider the dark bands, the stronger and heavier the wood. If the dark bands are very narrow and the light bands quite wide, the wood is light but not very strong. To get the most efficient strength to weight ratio for wood we need
a definite numbers of bands per inch.
Some of our aircraft structures are two-dimensional (length and width are large with respect to thickness). Plywood is often used for such structures. Several thin boards (foils) are glued together so that the fibers of the various layers cross
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them at 30 and 45 degrees as well). Plywood makes excellent "shear webs" if the designer knows how to use plywood efficiently. (We will learn the basis of stress analysis sometime later.)
Today good aircraft wood is very hard to come by. Instead of using one good board for our spars, we have to use laminations because large pieces of wood are practically unavailable, and we no longer can trust the wood quality. From an availability point of view, we simply need a substitute for what nature has supplied us with until now.
ALUMINUM ALLOYS
So, since wood may not be as available as it was before, we look at another material which is strong, light and easily available at a reasonable price (there's no point in discussing Titanium - it's simply too expensive). Aluminum alloys are certainly one answer. We will discuss the properties of those alloys which are used in light plane construction in more detail later. For the time being we will look at aluminum as a construction material.
Extruded Aluminum Alloys: Due to the manufacturing process for aluminum
we get a unidirectional material quite a bit stronger in the lengthwise direction than across. And even better, it is not only strong in tension but also in compression. Comparing extrusions to wood, the tension and compression characteristics are practically the same for aluminum alloys so that the linear stress analysis applies. Wood, on the other hand, has a tensile strength about twice as great as its compression strength; accordingly, special stress analysis methods must be used and a good understanding of wood under stress is essential if stress concentrations are to be avoided!
Aluminum alloys, in thin sheets (.016 to .125 of an inch) provide an excellent two dimensional material used extensively as shear webs - with or without
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stiffeners - and also as tension/compression members when suitably formed (bent).
It is worthwhile to remember that aluminum is an artificial metal. There is no aluminum ore in nature. Aluminum is manufactured by applying electric power to bauxite (aluminum oxide) to obtain the metal, which is then mixed with various strength-giving additives. (In a later article, we will see which additives are used, and why and how we can increase aluminum's strength by cold work hardening or by tempering.) All the commonly used aluminum alloys are available from the shelf of dealers. When requested with the purchase, you can obtain a "mill test report" that guarantees the chemical and physical properties as tested to accepted specifications.
As a rule of thumb, aluminum is three times heavier, but also three times stronger than wood. Steel is again three times heavier and stronger than aluminum.
STEEL
The next material to be considered for aircraft structure will thus be steel, which has the same weight-to-strength ratio of wood or aluminum.
Apart from mild steel which is used for brackets needing little strength, we are mainly using a chrome-molybdenum alloy called AISI 413ON or 4140. The common raw materials available are tubes and sheet metal. Steel, due to its high density, is not used as shear webs like aluminum sheets or plywood. Where we would need, say.100" plywood, a .032 inch aluminum sheet would be required, but only a .010 steel sheet would be required, which is just too thin to handle with any hope of a nice finish. That is why a steel fuselage uses tubes also as diagonals to carry the shear in compression or tension and the whole structure is then covered with fabric (light weight) to give it the required
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aerodynamic shape or desired look. It must be noted that this method involves two techniques: steel work and fabric covering.
We will be discussing tubes and welded steel structures in more detail later and go now to "artificial wood" or composite structures.
COMPOSITE MATERIALS
The designer of composite aircraft simply uses fibers in the desired direction exactly where and in the amount required. The fibers are embedded in resin to hold them in place and provide the required support against buckling. Instead of plywood or sheet metal which allows single curvature only, the composite designer uses cloth where the fibers are laid in two directions .(the woven thread and weft) also embedded in resin. This has the advantage of freedom of shape in double curvature as required by optimum aerodynamic shapes and for very appealing look (importance of esthetics).
Today's fibers (glass, nylon, Kevlar, carbon, whiskers or single crystal fibers of various chemical compositions) are very strong, thus the structure becomes very light. The drawback is very little stiffness. The structure needs stiffening which is achieved either by the usual discreet stiffeners, -or more elegantly with a sandwich structure: two layers of thin uni- or bi-directional fibers are held apart by a lightweight core (foam or "honeycomb"). This allows the designer to achieve the required inertia or stiffness.
From an engineering standpoint, this method is very attractive and supported by many authorities because it allows new developments which are required in case of war. But this method also has its drawbacks for homebuilding: A mold is needed, and very strict quality control is a must for the right amount of fibers and resin and for good adhesion between both to prevent too "dry" or "wet" a structure. Also the curing of the resin is quite sensitive to temperature,
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humidity and pressure. Finally, the resins are active chemicals which will not only produce the well-known allergies but also the chemicals that attack our body (especially the eyes and lungs) and they have the unfortunate property of being cumulatively damaging and the result (in particular deterioration of the eye) shows up only years after initial contact.
Another disadvantage of the resins is their limited shelf life, i.e., if the resin is not used within the specified time lapse after manufacturing, the results may be unsatisfactory and unsafe.
HEAVY AIRCRAFT RAW MATERIALS
The focus of our article is our Table which gives typical values for a variety of raw materials.
Column 1 lists the standard materials which are easily available at a reasonable cost. Some of the materials that fall along the borderline between practical and impractical are:
Magnesium: An expensive material. Castings are the only readily available
forms. Special precaution must be taken when machining magnesium because this metal burns when hot.
Titanium: A very expensive material. Very tough and difficult to machine.
Carbon Fibers: Still very expensive materials.
Kevlar Fibers: Very expensive and also critical to work with because it is hard
to "soak" in the resin. When this technique is mastered, the resulting structure is very strong, but it also lacks in stiffness.
Columns 2 through 6:
Columns 2 through 6 list the relevant material properties in metric units. Column 2, the density (d), is the weight divided by the volume.
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Table 6-1: Material property table
Materials d fy fu e E/103 E/d Root2 of
N/d Root3 of E/d fu/d 1 2 3 4 5 6 7 8 9 10 Wood Spruce .45 - 3.5/11 - 1.4 220 0 70 22.0 (15) Poplar .43 - 30/12 - 1.0 220 0 70 22.0 (15) Oregon Pine .56 - 4.0/13 - 1.5 220 0 70 22.0 (15) Fiberglas s Matte 2.2 - 15 - 1.5 700 17 5.0 7 (70% Glass) Woven 2.2 - 35 - 2.0 900 20 6.0 16 Unidirectio nal 2.2 - 60 - 3.5 150 0 27 7.0 27 Alum. Alloy 5052-H34 2.7 16 24 4 7.1 260 0 30 7.0 11 8086-H34 2.7 22 31 5 7.1 260 0 30 7.0 11 6061 -T6 2.7 24 26 9 7.1 260 0 30 7.0 11 6351 -T6 2.7 25 28 9 7.1 260 0 30 7.0 11 6063-T6 2.7 17 21 9 7.1 260 0 30 7.0 11 7075-T3 2.8 25 41 1 2 7.2 260 0 30 7.0 14 Steel AISI 1026 7.8 25 38 1 5 21.0 270 0 18 3.5 5 4130 N (4140) 7.8 42 63 1 0 21.0 270 0 18 3.5 7 Lead 11.3 - - - -Magnesium Alloy 1.8 20 30 - 4.5 250 0 37 9.0 16 Titanium 4.5 50 80 - 11.0 240 0 23 5.0 18
Units for above kg/d
m3 kg/m m2 kg/m m2 % kg/m m2 km kg-m2 kg2/3m1/3 km to obtain: lbs/cu
3 KSI KSI % KSI
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Column 3, the yield stress (fy), is the stress (load per area) at which there will be a permanent deformation after unloading (the material has yielded, given way ... )
Column 4, the ultimate stress (fu), is the stress (load per area) at which it cannot carry a further load increase. It is the maximum load before failure.
Column 5, the elongation at ultimate stress (e), in percentage gives an indication of the 'Toughness" of the material.
Column 6 lists the Yongs Modular or Modulus of Elasticity (E), which is the steepness of the stress/strain diagram as shown in Figure 1.
Important Note: For wood, the tension is much greater (2 to 3 times) than the compression. Both values are given in the Table. For fiberglass, the same applies, but the yield is so dependent on the manufacturing process that we cannot even give 'Iypical values'.
Figure 6-1: Stress strain curves for different materials
Columns 7 to 10: Columns 7 to 10 are values which allow the comparison of materials from a weight standpoint (the above referenced text by Timoshenko will also show you why we use those "funny" looking values).
Column 7 gives the stiffness of a sandwich construction. The higher the value, the stiffer the construction. From the Table, we see that metals are high wood comes
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close, but fiberglass is low: which means fiberglass will be heavier for the same stiffness.
Figure 6-2: Stress strain curve
Column 8 shows the column buckling resistance for the same geometric shapes. This time, wood is better than the light alloys, coming before steel and fiberglass. (Surprisingly, the usual welded steel tube fuselage is not very weight efficient.) Column 9 gives the plate buckling stiffness, which is also a shear strength
measure. Here again, wood (plywood) is in a very good position before aluminum and fiberglass, with steel not very good.
Column 10 provides a crude way of measuring the strength to weight ratio of materials because it does not take into account the various ways the material is used in "light structures". According to this primitive way of looking,
unidirectional fibers are very good, followed by high strength (2024) aluminum and wood, then the more common aluminum alloys and finally steel.
From just this simple table, we find there is not one material that provides an overwhelming solution to all the factors that must be considered in designing a light aircraft. Each material has some advantage somewhere. The designer's choice (no preconceived idea) will make a good aircraft structure ... if the choice is good!
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7. Detailed wing design
7.1 Spar design:
Spars are members which are basically used to carry the bending and shear loads acting on the wing during flight. There are two spars, one located at 15-20% of the chord known as the front spar, the other located at 60-70% of the chord known as the rear spar. Some of the functions of the spar include: They form the boundary to the fuel tank located in the wing.
The spar flange takes up the bending loads whereas the web carries the shear loads.
The rear spar provides a means of attaching the control surfaces on the wing.
Considering these functions, the locations of the front and rear spar are fixed at 0.17c and 0.65c respectively. The NACA 65 (3) 418 airfoil is drawn to scale using any design software and the chord thickness at the front and rear spar locations are found to be 1.9708 m and 7.5354 m respectively.
7.1.1 Geometric dimensions:
The spar design for the wing root has been taken because the maximum bending moment and shear force are at the root. It is assumed that the flanges take up all the bending and the web takes all the shear effect. The maximum bending moment for high angle of attack condition is 486567918.4 Nm. the ratio in which the spars take up the bending moment is given as
83 h1 height of front spar
h2 height of rear spar
From the above two equations,
The yield tensile stress σy for 7075 Al Alloy is 455.053962 MPa. The area of the flanges is determined using the relation
where M is bending moment taken up by each spar, A is the flange area of each spar,
z is the centroid distance of the area = h/2. Using the available values,
Area of front spar,
Area of rear spar,
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Each flange of the spar is made of two angle sections. For the front spar, the length of the angle is 6t, angle height is 5t with angle thickness t. Area for each angle of front spar is found to be 0.1799507 m2 and hence value of t is found to be
Length of the front angle section:
Height of the front angle section:
For the rear spar, the length of the angle is 8t, angle height is 3.5t with vertical thickness t and horizontal thickness t/2. Area for each angle of rear spar is found to be 0.164486 m2 and hence value of t is found to be.
Length of the rear angle section:
Height of the rear angle section:
Now to determine the thickness of the web portion, the ultimate shear stress of 7075 Al Alloy is 317.1588MPa. The maximum shear force at root of the wing
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for high angle of attack condition is 17480623.2 N. The wing chord is assumed to be a simply supported beam supported at the two spars. The maximum shear force acts at the centre of pressure which can be located by using the formula,
Figure 7-1: Reaction force determination at spars
Considering force and moment equilibriums for the given simply supported configuration, the reactive shear force at the spar supports are found to be
We know that,
V shear force at the spar
86 t thickness of the web.
Thus,
FOS = 1.5
z is the centroidal distance of the area = h/2 Thus the thicknesses of the web portions are,
All dimensions are in m
It becomes necessary to check whether the shear stress due to this thickness is less than the allowable of the material.
1.00128 0.02346 0.1251 Rear spar 0.43806 0.0763 0.13414 0.8048 0.670 Front Spar
