Question

The frame supports a distributed load and a tension force of P = 120 lb, as shown in the figure below. 75 lb/ft 7 7 7 7 7 7 e DB -2 ft -2 ft Part A Determine the resultant internal loadings acting on the cross sections at point D of the frame. Express your answers, separated by commas, to three significant figures. VALO vecmo e ? ND= Vp=, Mp= lb, lb, lb-ft Submit Previous Answers Request Answer X Incorrect; Try Again; 5 attempts remaining

Answer 1

Drawing the FBO of member AG ! 300 lb 6 Foc k 3ft Beau 2 at 214 - 1309 120 16 EMA =0 445 Foc (3) - *300*5 - 120 Co$ 30 (1) = 0 4 FBC = 928.4 lb EFxzo: Ax - 315*(928-1) + 120 Sin 30 =0 Ax = 496.86 lb EFy=0; Ay + 4/5(928.1) - 300 - 120 C08 30 =0 Ay = - 338.55 lb Internal forces at point oi- E FxPo' 6.86 11 No = - 496-861b 18+ 338.55.b E Fy = 0: Vo = - 338.55 lb

EMDEO Mo = - 338.55*1 MO -338.5 :- Noz, VDE, Mg = (-496.86, - 338.55,- 338-55) Ab, lb, lb-ft Internal loadings at boint E- ( 15x3) lb ME VE 1.5ft ae - 341 3 ft EFX =O Rolb NE = Ro Sin 30 > NE = 60 lb E Fy=0; VE - 75x3 - 120 COA30 = 0 VE = 328.92 lb EME=0 - ME - 75x3x1.5 - RoxCOX 30 x 3 = 0 ME = -649.27 lb-ft NĘ =, VE =, Me= (60, 388-92,-649-27) lb, db, db-ft tu