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Extra resources for Aircraft Structures 3E
For horizontal equilibrium of the element cos e = U U~~ x + T~ x . ( - S x ) ( U - (4 Now substituting the values nX = 160N/mm2, gy = -120N/mm 200 N/mm2 we have 2 and u = c1= T~~= f113N/mm2 Replacing cot B in Eq. (ii) by 1/ tan 8 from Eq. (i) yields a quadratic equation in u 2 - U(Ux - U y ) + UxUy - Try 2 =0 The numerical solutions of Eq. (iii) corresponding to the given values of ox, are the principal stresses at the point, namely aI= 200 N/mm2 (given), aJr= - 160 N/mm 2 (iii) and 74,; 16 Basic elasticity Fig.
These equations are sufficient, when supplemented by appropriate boundary conditions, to obtain unique solutions for the six stress, six strain and three displacement functions. It is found, however, that exact solutions are obtainable only for some simple problems. g. finite differences) or by the Rayleigh-Ritz method based on energy principles (Chapter 5). Two approaches are possible in the solution of elasticity problems. We may solve initially either for the three unknown displacements or for the six unknown stresses.
Substituting these values of u and Y in Eq. s. of this equation is a constant which means that F1( x )and F2(y ) must be constants, otherwise a variation of either x or y would destroy the equality. Denoting Fl ( x )by C and F2(y ) by D gives C+D=-- Pb2 8IG (viii) and so that and Py3 uPy3 h ( Y )=a-=+DY+H Therefore from Eqs (vii) u = - - - -Px2y 2EI uPy3 Py3 - Dy 6EI 6IG + + +H uPxy2 Px3 v = -+-+Cx+F 2EI 6EI The constants C , D, F and H are now determined from Eq. (viii) and the displacement boundary conditions imposed by the support.
Aircraft Structures 3E by Megson