2010/10/17 Gijs de Rooy <gijsr...@hotmail.com>: > For example, the current FG747 wing flex is calculated on various thing, one > of which is the amount > of fuel in each wing section, but how much lift is required to contradict > this "fuel-force"? >
Gijs, The structural analysis of a wing is a very complex field of engineering but for the sake of a visual representation of the wing bending in Flight Gear, I would say that the wing can be satisfyingly modelled as a cantilever beam of uniform stiffness submitted to a uniform loading. According to Timoshenko's "Element of Strength of Materials"(1940), the deflection of such a beam at any spanwise station 'x' is d = w*x^2*(x^2+6*b^2-4*b*x)/(E*I) The wing span is b, the wing root is assumed to be at x=0 and the wing tip at x=b. The distributed load 'w' is assumed to be uniformly distributed and should be expressed in lbf/ft or lbf/in In this formula, the sign convention is w>0 means the loads are applied downward and d>0 means the wing deflects downward as well. The Young modulus E=10,000,000 psi can be guessed because the 747 wings are likely made of aluminium but the beam inertia I (in ft^4 or in^4) is quite difficult to assess without the blue prints (and even with the blue prints by the way). However, here is a small suggestion to use this formula: 1. The loading on the wing is basically the algebraic sum of the wing weight and of the aerodynamic lift. Assuming that the wing deflection under its own weight is of second order (a wing is sized by aerodynamic loads, inertia/manoeuvre/'g' loads and landing loads if the landing gears are attached to the wing), 'w' can be assumed to be proportional to the lift coefficient CL i.e. w = CL*0.5*rho*S*V^2/b where rho is the air density, S the wing area, V the air speed and b the wing span. 2. The maximum deflection at the wing tip being dmax = w*b^4/(8*E*I), if we assume that during a take-off, the wing deflection is 3ft then you can calculate E*I = CL*0.5*rho*S*V*b^3/(8*3ft) (replace V, rho and CL by the appropriate values at take-off). 3. Use the above deflection formula with the values computed in the previous steps. This is a very rough approximation because the wing inertia is not uniform across the span (a wing structure is made of spars and ribs/bulkheads/webs) and the distributed loading is not uniform either (lift forces are not uniform, weight is not uniformly distributed - think about the engines), etc. I hope this helps however. Cheers, Bertrand. ------------------------------------------------------------------------------ Download new Adobe(R) Flash(R) Builder(TM) 4 The new Adobe(R) Flex(R) 4 and Flash(R) Builder(TM) 4 (formerly Flex(R) Builder(TM)) enable the development of rich applications that run across multiple browsers and platforms. Download your free trials today! http://p.sf.net/sfu/adobe-dev2dev _______________________________________________ Flightgear-devel mailing list Flightgear-devel@lists.sourceforge.net https://lists.sourceforge.net/lists/listinfo/flightgear-devel
