6 Multi connected semi monocoque sections

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Aerospace Structure –Exercises 6 – 1 – Exercise #6 Multi-connected semi-monocoque sections Aerospace Structure –Exercises 6 – 2 – Exercise n. 1 Evaluate the position of the shear centre in the section Data: A=200 mm 2 a=300 mm b=350 mm Solution: The stringer scheme is symmetric w.r.t. a vertical and a horizontal axes. Centroid and principal axes of inertia are immediately determined Panel scheme is only symmetric w.r.t. to the horizontal axes. For such a reason the Y coordinate of the shear centre can be immediately indicated: aYCT 2 1= =150 mm The evaluation of the coordinate X of the shear cen tre involves the application a force T y at an unknown position and the solution of the section. A n additional equation to find the new unknown will be provided by invoking a zero rotation of the section. The procedure based on the reduction to an open path will avoid the need of solving a syste m with M+1 unknown. Only the calculation of circular flow and shear centre position will be coupled. 2 2 2 3 26 Aa aAJ x =    ⋅= mm 4 Aa SSS xxx 2 1 321 === mm 3 Aa SSS xxx 2 1 654 −=== mm 3 Panel 6 is cut and the open path flows are evaluated: y x x y TaJ STq ⋅− =⋅− = 3 11/ 1 y x xx y TaJ SSTq 3 221/ 2 − =+⋅− = y x xxx y TaJ SSSTq ⋅− =++⋅− = 1321/ 3 y x xxxx y TaJ SSSSTq ⋅− =+++⋅− = 3 24321/ 4 Aerospace Structure –Exercises 6 – 3 – y x x y TaJ STq ⋅− =⋅= 3 16/ 5 06 ≡/q Verification of the equivalence to the resultant of the internal forces: ( ) 0/ 1 / 2 / 4 / 5 =−−+⋅= qqqqbRx yy TqaR =⋅− = / 3 Two solving equation will be prepared. The first on e is the equivalence with the moment of the internal forces, whereas the section equal the sect ion twisting to zero. Such second condition is fulfilled only if the force is applied to the shear centre. 1. Equivalence to the moment of…