Question

imple 6 – Truss 6000 lb 12 000 lb Given: Truss structure as shown. Find: Forces in each member: AB, BC, AD, EF, BD, BE, CE, CF - 6 ft- - 6 ft- Member specifications: JL AD, DE, EF L2x2xy-doubled BD, CE, BE L2 x 2 x 18-single- AB, BC, CF C3 x 4.1 - doubled -

Answer 1

Thuss 600016 12.000 lb. Question - messing tono = 6st I can For stability Elmo z + St=0 Extra Ety = 0 3 - Fynt fyn = 6000+12.000 Expat hy = 18,000 - 0 utapon EMAC0 = 18 = 6000x 6 + 12000 x 12 fy = [6000 112000x2]x6 to3 2000 + 8000 Why 10000144 from en = 800035/45

At point A Efno FARCOso + FAD= 0 - Efy=0 An Sino+fy = 0 - 3 from Eqr ♡ & onda - FAD x sino & fay=5a FAD tomo=fy Coso » PAD - fyA800086 tono tono = 8 to Ita = 6000 Ib. 1 In egn ♡ - - 6000 x 62782 FAR= - 6000 Coso P fasa - 10000 Ib. Ans At point Efy = 0 FAD = FRE [FRE = 6000 lb. Ay SA

At point ß 1600056. FAB & FBE Efx= 0 =) Fancoso - Foct Beloso - . Efy=0 Fansino + FRESINO + Frog + 6000=0 put the voluesin tar - 10000 x 6 = fine & FBEXG =) 10 2 6 FBE + 10 Foc - - 60,000 put the values in - - 10000XS + FREX 3 +0+6000 50 10 FBE = 80,000-60000 KESE = 2500 lblogas Put the value fe in egr 6 68 2500 + 10 fac = -60,000 Fue = -60,oop - 15,000 Frue = -75,000 16.4

At point 8 Ef y=0 fyrr sino =0 Ron de fer -1000o XIOR [Ffc = -12,500 IblA coso Efy=0 Frosinot fer to ref = +12,500 X 8 Teef = 7,500 lb. As At pointe 112000 fic Ro {fy=0 & fre & Persino +2000 =0 fel = -127000 + 12,500 X8 = 12,000 +10000 FFEC= -2,000 Ib. I

10,000 lb ņ forces in member i AB Fan - 10,000. BC F – 7,500 I b. - AD FAD = 6,000 lb. - Ef a fer = 7,500 lb - BD > F= 0 lb. I BE - FBE = 6,000 - CE - FCE = -2,000 sh. - CF = For +12,5001b - 6,000 Ib. AX Answer indicate direction of here negative sign force is opposite.

CHAPTER THREE Stress and Deformation Analysis 10 6000 lb 12 000 lb Member specifications: AD, DE, EF L2 X 2 X 1/8 - doubled — BD, CE, BE L2 X 2 X 1/8-single- AB, BC, CF C3 X 4.1 - doubled — JL 02.13 (Problem 13)