Maximum bending moment at the root 1768011725Nm On considering the safety

Maximum bending moment at the root 1768011725nm on

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Maximum bending moment at the root =1768011.725Nm On considering the safety factor (1.5) and the load factor (2.5) into account, we have Maximum design bending moment = 1.5*n*Mmax =6630043.969 Nm
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Area required for the stringers and spars are calculated from the bending formula Y=0.3808m M=6630043.969 Nm σb = 476*10^6 Nm^-2 From this we can get the area required Hence A=0.03658 m^2 the area required to support the maximum bending moment coming on the wing is 0.03658 m^2 Total area required is divided between spar and stringers as follows; Total area required is summation of 60 % of this area is allotted for spars and 40 % for stringers. Area of the spar =0.02195 m^2 Area of the stringers =0.01
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463m 2
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Front spar: 60% of the spar area is taken for the front spar. A=0.35C- 0.15C B=0.65C-0.35C Area of the top flange = ( a/a+b)*A fs =8.78*10^-3 Area of the bottom flange = (b/a+b)*A fs =4.39*10^-3 Rear spar: 40% of the spar area is taken for the front spar. Area of the top flange = ( c/c+d)* A rs =6.585*10^-3 m 2 Area of the bottom flange = (d/c+d)* A rs =2.195*10^-3 m 2 Ssitable angle section are selected with respect to area for front spar and rear spar.
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The properties of the angle section selected are tabulated. Component Area m 2 Width m Thickness m Radius of Curvature M Front Spar Top Flange 94 17.98 2.81 2.25 Bottom Flange 47 12.75 12.75 1.59 Rear Spar Top Flange 70.5 15.60 2.44 1.95 Bottom Flange 23.5 8.99 1.41 1.21 T W Stringer Selection: For a double riveted joint we have K c = 10.2 R
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We choose aluminium Al – 2024 material whose yield stress is 340 MPa, E = 70 MPa and n = 0.3375 Let us assume the skin thickness be 2.5 mm On substituting these values in the above formula we get Then the stringer spacing b =10, 89 Cm No of stringers at top surface, Top=((perimeter)/2)/10.5 = 40 stringers No of stringers at bottom surface, Bottom=((perimeter)/2)/14.0 = 30 stringers Area of one stringers, Astr = 4.4857 cm^2 the stringers are usually Z SECTIONS The specification of the Z –section chosn for the stringersare given below; B T R A
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The various dimensions are as follows: A=2.765 cm B=7.374 cm T=0.376 cm T1=0.346 cm R=0.356 cm Area = 4.4857 cm^2 Redrived data from tabulation is given by Ixy =425647.8612 cm^4 Iyy =15930092.35 cm^4 I xx =640587.3284 cm^4 Xcg=22.9420cm Ycg=18.3754cm ΣA=827.19cm^2 ΣAx=184415.3561cm^2 ΣAy=15200.2041cm^2 ΣAxy=3814409.798 cm^4 ΣIGX==43176.9810 cm^4 ΣIGY=12916,521 cm^4 For the airfoil selected, cruise angle of attack =5 0. M x =M cos
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= 7103065.13*cos 5 0 =7076035.823Nm M y =M sin =713065.13*sin 5 0 =619072.9172Nm Substituting M x , M z , I xx ,I zz and I xz in the formula, we get, b =11.0199*y -0.2563*x For the top most stringers, (15th) X=227.5596-222.9420 =54.6176cm Y=53.1250-18.3757 = 34.7493 cm b=368.9353MPa, this is less than the allowable stress, Marigen of safety = ( b allowable/ b max )-1=0.2902 =29.02%
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FUSELAGE DESIGN: The fuselage is the main structure or body of the fixed-wing aircraft. It provides space for cargo, controls, accessories, passengers, and other equipment. In single-engine aircraft, the fuselage houses the power plant. In multiengine aircraft, the engines may be either in the fuselage, attached to the fuselage, or suspended from the wing structure. There are two general types of fuselage construction: truss and monocoque. Truss Type A truss is a rigid framework made up of members, such as beams, struts, and bars to resist deformation by applied loads.
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  • Winter '12
  • Aerodynamics, Fixed-wing aircraft, Airfoil, Wing design, aircraft wing

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