metode schrenk
DESCRIPTION
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Metoda Schrenk
External Loads
LOADSLOADS1 LOADSLOADSLOADSLOADS1
AIRFRAMEAIRFRAME2 AIRFRAMEAIRFRAMEAIRFRAMEAIRFRAME2
STRUCTURAL STRUCTURAL
RESPONSERESPONSE3STRUCTURAL STRUCTURAL
RESPONSERESPONSE
STRUCTURAL STRUCTURAL
RESPONSERESPONSE3
LOCAL LOCAL
EFFECTSEFFECTS4
LOCAL LOCAL
EFFECTSEFFECTS
LOCAL LOCAL
EFFECTSEFFECTS4
MATERIALSMATERIALS
5
MATERIALSMATERIALSMATERIALSMATERIALS
5
FAILURE FAILURE
THEORIESTHEORIES
6
FAILURE FAILURE
THEORIESTHEORIES
FAILURE FAILURE
THEORIESTHEORIES
6
FAILURE FAILURE
MODESMODES
7
FAILURE FAILURE
MODESMODES
FAILURE FAILURE
MODESMODES
7
Schrenk Method
Air load distribution for wing with aerodynamic twist is obtained in two parts. The first part called the basic lift distribution. The second part called additional lift distribution.
Lift = Basic Lift + Additional Lift
where • Basic Lift is obtained for the angle of attack at which
the entire wing has no lift.
• Additional lift is calculated if there is no twist and constant airfoil section.
Additional Lift
The method consist of simply averaging
the lift forces obtained from an elliptical lift
distribution with those obtained from a
planform lift distribution
22
14
2
1
b
y
b
Sccc planformlplanform
Basic Lift
a0lb cm2
1cc
2/S
cdym
m
2b
0
0
02b
0
0
2b
0
aR0
0w
cdym
cdym
The wing angle of attack for zero lift is found from the following equation
Schrenk: Total Lift Distributions
Exercise
Calculate Shear Force, Bending Moment Distribution at the following flight condition:
Cruise speed, Vc 100 m/s
Aircraft mass = 22,000 kg
Engine mass = 1000 kg each and located at ws3835
2000 Kg Fuel is equally distributed in each side of the inboard wing tank from ws550 to ws4245
Wing Area, S = 65 m2; mean chord, c = 2.464 m
• CLmax = 1.82; dCL/d = 4.7 /rad; Cm.ac = -0.04
g = 10 m/s2 ; air density, = 1.25 kg/m3
distance between ac-wing and ac-horizontal tail is 10 m;
centre of gravity the aircraft is located at 0.4 m behind the ac-wing.
Additional Lift at CL = 1.0
Calculation of SF, BM
Geometry CL = 1 CL = 0.53 SF BM
