Question

Chapter 4, Problem 4/042 Multistep Determine the force in member BE of the loaded truss. Enter a positive number if in tension, negative if in compression. 15' 19 15 B 20° 200 F E 20 1.0L 2.OL 20 45° 45° A D

Answer 1

Determine Reaction forces at A and Di load at f=1.06 M Load at E=2.0L/B с E ILIM 12L HA a b a .. Vo = VA B EFx=0 HA=0 Efy=0 VA + VD = 3L EMA = o... (static condt) -La-2L(a+b) + Vo(zatb)=0 = La + 21a + 2Lb D (2atb) L(3a+26) from above calculations; (2atb) a = 1.631 ft 6 - 0.879 3 — Substitute in above V = 6.52L equation = 1.6L 4.07 IVO = 1.6L]() VA = 3L-1-6L = 1.4L TVA- 1.4L) () Consider section M-M above : FBC (Assume all forces are compressive) 1.6 L +(FBE) sine = 2L :: (Fire) sin(42.02°) - 0.4L FBE = 0.609L -0-6094 . Efy=0 FBE E FEF 24 - Therefore - D 1.6L negative sign for compression Therefore, force acting in member BE is 0.6094 (compressive in nature)

Solving Part I: Determine a, b, ando. CELO526 B 191 с 20° 70 30 200 tan 65 = h 15 n. o - E h -th = 32.16" (ED sin 45°) Esin 20° AM $50 D :: CE cos 2*+ Ensin 45º = h K 15/ 15 K Also; b a kapcosas (ED cosáse) -(CE sin20) = 15° solving equations 1 and 2 simultaneously, we get 0.939 (16)+ 0.707(ED)= 32.16 0 0.707 (ED) - 0.342(CE) - 15' - CE = 13.395" - BF (symmetrical) ED = 27.696" - AF (symmetrical) .. Therefore; from above bigwee; la = ED COC 45° = 19.58 b =(199- 26Esin 20° = 19-9.1627 -9.8373" i.[6=9.837" Cews 20° tan 0 = 12-587 bt Bfsin20° 14.418 CELO520 o - tant (0.87) (Besin 20°)+b i-o = 41.02 B 0 E