5 Open and closed semi monoque sections

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Aerospace Structure –Exercise 5 – 1 – Exercise #5 Open and closed semi- monocoque sections Aerospace Structure –Exercise 5 – 2 – Exercise n. 1 Evaluate the equivalent shear flow in the following section: Data: A=200 mm 2 ; a=180 mm ; b=300 mm ; Ty=2000 N Solution: - Centroid and principal inertia axes are immediate ly found thanks to section symmetry 2 2 44 Aa aAJ x =⋅= mm 4 Aa SS xx 2 1 41 == mm 3 Aa SS xx 2 1 32 − == mm 3 - The shear flows are evaluated by applying N-1 tim es the shear flow equation: yy x x y TaAa Aa TJ STq ⋅− =⋅     ⋅− =⋅− = 2 11 2 1 2 1 1 021 2 =+⋅− = x xx y J SSTq y x xxx y TaJ SSSTq ⋅=++⋅− = 2 1321 3 Verification is carried out as far as equilibrium in the x and y direction is concerned: 02 =⋅= qbRx ( ) y yy y Ta T a TaqqaR =      +⋅=−⋅= 22 13 Numerical values are substituted: Aerospace Structure –Exercise 5 – 3 – 6480000 =xJ mm 4 ; 18000 41 == xx SS mm 3 ; 18000 32 − == xx SS mm 3 556 . 51 −=q N/mm; 556 . 53 =q N/mm; 02=q N/mm Exercise n. 2 Evaluate the equivalent shear flows in the section and the horizontal distance of the shear centre from the stringer that is indicated in the figure: Data: A=180 mm 2 ; a=200 mm ; Ty=3500 N Solution: - Centroid and principal inertia axes are immediately found thanks to section symmetry 24 aAJ x ⋅= mm 4 Aa SS xx == 41 mm 3 Aa SS xx − == 32 mm 3 - Shear flows are evaluated: yy x x y TaAa Aa TJ STq ⋅− =⋅⋅− =⋅− = 4 1 4 1 2 1 1 021 2 =+⋅− = x xx y J SSTq y x xxx y TaJ SSSTq ⋅=++⋅− = 4 1321 3 Verification is performed: Aerospace Structure –Exercise 5 – 4 – 02 321 =⋅+⋅+⋅= qaqaqaRx ( ) y yy y Ta T a TaqqaR =      +⋅=−⋅= 4444 13 In order to evaluate the shear centre, the force can be considered apply at a unknown distance d, and the equivalence between the moment of the in ternal force and of the…