By David J. Peery
Suitable for undergraduate scholars, this quantity covers equilibrium of forces, area constructions, inertia forces and cargo components, shear and bending stresses, and beams with unsymmetrical move sections. extra subject matters comprise spanwise air-load distribution, exterior rather a lot at the aircraft, joints and fittings, deflections of buildings, and designated equipment of research. issues concerning an information of aerodynamics look in ultimate chapters, permitting scholars to check the prerequisite aerodynamics subject matters in concurrent courses.
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16. Assume rigid bulkhesds in planes BCDE 35 &PAC% STRUCTURBS and B'C'LYE'. Assume alao that the vertical load of 1O , OO lb ie equivalent to vertical loads of 500 lb on each side truss plus a torsional moment of 10,OOO in-lb, 88 ahown in Fig. 16. Fro. 16. no. 17. Sd& The aide trueses are first d y a e d 89 coplanar trusses CSrrJring loads of 500 Ib. The forces obtained are shown in Fig. 17. Thew! 3 Member ER BE' BB' BC' CC' I Forcea resulting from Mlolbperade Forces resulting from 10,ooO in-lb, torsion Total forcerr -632 +a8 +266 -286 +319 0 0 0 Co' +638 Do' -532 0 Eo' -319 0 +3€9 -286 +319 0 -319 d +319 -266 +a7 -m -819 This structure is in reality statically indeterminate, and the t o r s i o d analysis gives only approximate values of the forces in the members.
Assume that fbe airplane Shown in Fig. 3 weighs 20,OOO Ib. and that the braking force F is 8,000 Ib. a. Find the wheel resctions RI and Rz. b. 7 ft/m). 3. - sblution. a. EF, = 8,000 Ma - Ma = 0 8,OOO lb ZMa, = 120Ra - 8,000 X 50 - 20,000 X 20- 0 Ri = 6,670 lb ZFV = 6,670 20,OOO & = 13,330 Ib b. From Eqs. 2, - 0 From Eq. 9, c! i! 88 W ft/W 2C4OOO lP-d=2a8 o - ( i ~ . ss)~ rrP835ft When 8 is measured aa positive to the left, it is neceeaesy to consider u aa negativq since it is an acceleration to the right.
The vertical load of 20 Ibfin. is distributed to the spars in inverse proportion to the distance of the center of pressure from the spars. , and that on the rear spar 4 Ib/h. If the front spar is considered as a free body, as shown in Fig. 21(u), the vertical forces at A and G may be obtained. SPACE STRUCTURES 39 AIRCRAFT STRUCTURES 40 ZMA 5;-16 X 180 X 90+ loocT. p 0 0, = 2,590 Ib Gu = 4,320 Ib 23’. 16 X 180 - 2,590 - A. = 0 A, = 290 Ib Form A, cannot be found at this point in the analysis, since the drag-truss m e m k exert forces on the front spar which are not shown in FEg.