Download Aircraft Structures by David J. Peery PDF

By David J. Peery

Still correct a long time after its 1950 e-book, this mythical reference textual content on plane tension research is taken into account the simplest booklet at the topic. It emphasizes uncomplicated structural idea, which continues to be unchanged with the advance of latest fabrics and building tools, and the appliance of the trouble-free rules of mechanics to the research of plane structures.
Suitable for undergraduate scholars, this quantity covers equilibrium of forces, area buildings, inertia forces and cargo components, shear and bending stresses, and beams with unsymmetrical pass sections. extra issues contain spanwise air-load distribution, exterior a lot at the plane, joints and fittings, deflections of buildings, and certain equipment of research. issues regarding an information of aerodynamics seem in ultimate chapters, permitting scholars to review the prerequisite aerodynamics themes in concurrent courses.

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14. a. Find the inertia forces on the airplane. b. Find the inertia form on a 400-lb gun turret in the tail, which is 500 in. aft of the center of gravity. Neglect the moment of inertia of the turret about ita own center of gravity. c. If the nose wheel is 40 in. from the ground when the main wheels touch the ground, h d the angular velocity of the airplane and the vertical velocity of the nose wheel when the nose wheel reaches the ground, assuming no appreciable change in the moment arms. The airplane center of gravity has a vertical 69 INERTIA FORCES AND LOAD FACTORS doc i t y of 12 ft/m at the moment of impact, and the ground reactions are assumed constant until the vertical velocity reaches eero, at which time the vertical ground reaction becomes 60,OOO lb and the horizontal ground reaction becomes 20,OOO lb.

Solution. a. First COnsideFing the entire airplane 88 a free body, EF, = T COB 100 - 30,000 0 T = 30,500 Ib EF, = R - l0,Ooo 30,500Sin 100 = 0 CM, = 20 X 15,300 - 30,Mw)e R = 15,300 Ib 0 e=10in. b. considering the aft d o n of the fudage 88 B free body 88 &own m Fig. 5, it ie acted upon by an inertia force of Ma = 7x 3g = 3,000Ib The tension on section BB is found as follows: CP. = 30,000 - 3,000 Ti = 0 TI= 27,ooO Ib Sice there is no vertical acceleration,there is no vertical inertia force. Seetion BB has 8 shear force Vl of 6,300 lb, which is e q d to the sum of the weight and the vertical component of the cable force.

The wing has two drag trueses in horisontd planes 8 in. apart, which are assumed to supply enough torsional rigidity that the two spars have equal bending ddections. 41 SPACE STRUCTURE# Sbluticnr. The bending de9ections of the spars are proportional to their I d and inversely proportional to their moments of inertia. ni the rear spar must carry 40 per cent and the front spar 60 per cent of the total load in order to have the bending deflections of the beams equal. If the 4,000-lb reaultant force acted at a point 40 per cent of the distance between spars from the front spar, the former ribs would distribute the loads to the spars in the correct proportion, and there would be no torsional moment producing load in the d may be applied withdrsg-truse members.

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