Oct 15, 2003, 08:28 PM Registered User Singapore Joined Aug 2002 759 Posts Fuselage Bending moment What is the maximum bending moment on a fuselage? I supppose it would be a pullout after a vertical dive. So the lift force pulls the wing up at the Mean Aerodynamic Chord or the Neutral point or wherever....and the CG pulls the plane down. The distance between these 2 points multiplied with the maximum lift force on the wing is the max bending moment? How do I find the mean point where the lift force acts? Thanks Shane
Oct 16, 2003, 01:14 AM
Registered User
Singapore
Joined Aug 2002
759 Posts
Quote:
 The aerodynamic forces and moments of the wing and tail are applied to the aerodynamic centers which are at 25% of the respective mean aerodynamic chords
I understood that this is an approximation only. The Allegro Lite plan has a marked "Neutral Point", 3.75" from the LE, and the CG is supposed to be 0.3" to 0.6" from this "Neutral Point". Could this be the ACTUAL Mean Aerodynamic Center? If it is then the max bending moment would be 0.6" x 150lb ( max lift force )= 90 inch lbs.

Shane
 Oct 16, 2003, 05:15 AM Registered User Punta Gorda, FL Joined Apr 2002 4,952 Posts The neutral point is the aerodynamic center of the whole aircraft. The trouble is that at model sizes and speeds the neutral point is not always fixed but may shift some with changes in the pitch atitude of the model. This is caused by shifts in the aerodynamic centers of the flying surfaces. Those shifts are caused by shifts in laminar seperation bubbles and boundary layer thickness along the airfoil chord with changes in angle of attack. The flow beyond the boundary layer behaves as though the shape of the airfoil were changing with angle of attack and the boundary layer were not. The maximum coefficient of lift occurs at a particular angle of attack and the exact aerodynamic center of the wing can be found at that angle of attack. The lift force of the wing acts through that aerodynamic center. When the plane is in dynamic equilibrium the sum of all the moments is zero. And the sum of all the force vectors is zero. Knowing every force and moment will allow you to find the bending moment on the fuselage. During a high speed pull out, the inertial forces come into play and the mass distribution along the fuselage determines the inertial force distribution. Those inertial forces have to be added to the aerodynamic forces and moments to determine the bending load on the fuselage. If the plane weighed 1.5 pounds and the lift force was 150 pounds then the plane would experience roughly 100 G's of inertial forces. If the tail weighed 0.8 ounces, then during this maneuver, the inertia of the tail would contribute five pounds to the bending load on the tail boom. In addition to that, the aerodynamic down load by the horizontal tail would contribute an additional bending load on the tail boom. The mass of the tail boom itself would also contribute to its distributed inertial bending load. At 57 meters per second, how much down load by the stab would be needed to balance the nose down pitching moment of the wing?
Oct 16, 2003, 06:52 AM
Registered User
East Anglia, UK
Joined Sep 2002
29,714 Posts
Actually I'd say the worst forces are put on a fuselage in a high speed snap roll. You may need to dive to get the high speed, admittedly..
Speed + control surface deflection will do the worst. And a heavy engine on a long nose will load that up as well in pitch and yaw excursions.

Of course heaviest loads are in crash situations.