Question

Section LU LeerLUL Sample Problem 600 lb 600 lb 600 TL What are the forces in AB, AC, BC, DE (using method of joints) and DF, EF, EG (using method of section). 300 lb R 2 ft 4 in. 300 lb 6 ft A 121.5? C EL G Н. 8f- St- St-test- 11200 (1) Method of joints MA = -8660014 16(600) – 600 (24)_3263 00 000 lb 0001 Free body: Truss: EF, = 0: H Because of the symmetry of the truss and loading, = 0 30022 A=H, = total load shoe 1 1111 A = H, = 1200 167 FAC + FAB Cos 36,8 4 6 10 Sample Problem FX=

Answer 1

ANS- 600 lb 600 lb Coolb 2 ft.yin Oo V T 30016 30db Glt! E G Kate Start att EFx=0 =[Mx = 0] IMA = 0 » - 600 x 8 - 600 x 16 - 600 < 24 - 300 x 32 + y^3) = 0 600X8 + 600 X16+ 60004 + 300 x32 =Hy = 2 32 Hy = 1200 lb a хғу во → Ay + Hy + 300 + 600 + 600+600 +300 = 0 Ay = 2400 - 1200 Ay = 1200 lb il

At joint a 900.06 1300.16 > FAB La Fans : AAAAC {fy =0 = 900 + Far sino = 0 А 8 с tam (O) = = o = temi" 87181.19.0 0 - 36.869° = - 900 sin ( 36.869) : -1500 16 Toi FAB = 1500 lb (compressive )) EFx = 0 → FAR COS O = Fac → FAC = 1500 cos (36.869") FAL= 1200] (Tensile)

At point 'c Fy = 0 T: FCB=0 | At point B:- Golb A BCE 36.369 VO2 GMA Haze 844 FBA = Iscolb - FBE E Fa = 0 => 1500 60g ( 36-869) – F&D Cosa, - FSE6030, 50 ton az = 8 Q2 = tan! ( 5 ) = 36.869 JABXD → 400 – COS(16.298) Fan - COS(36-369) FgE =0 and 2ft.4in = B 8ft X (2.33 kt) = 400 - 0.96 Fgp - 0.8 Fee = 0 = 400 – 0.96 Føp → tom 0.8 FSE - y & = 400 - 0.96 FBD FRE 0.8 2i = tem" (2333) y = 16.258°) Please Fron tero

1 FRE = 500- 102 Føp Efy =0 -> → - 600 + Fem sin (36-869) + Free sin ( 36-869) – Fan sin(16.258) = 0 FBA (0.599) + FBE (0.599) - 6.2799) Fiso = 600 5) 1500 (0.599) + (500-12 Fen) 0.999 – (0.2748) F603 600 3 -0.7188 Fon - 0.2799 Fan = 600 - 900 – 300 –> (0.9987) FBD = 600 FBD - Goo fast FBD = 600 (c)] D 1 At point 'D':- T 600lb 16.253 171216-258 FOR FOB = 60016 FRE 20 h 600 Sin (16.258) + For Sin (16.298) +-=0 - 600 + 0 → FOE + Fpf (0.2399) = 432.022

= FDF (0.2799) = 432-022 - FDE 432.022 - FDE 0:2799 > 0 FDE = 1543:48 - 3.57 Foe ΣF, το + For. cos (16-258) = 600 (0) (16.253) FDF =600 db Then from equation 0 600 = 1543.48 -3.57 FOE → 3.57 fpe : 1543.49 - 600 FDE 7 264.31 lb (c) 1600.16 16.2009 30016 6 FEF- I 36.39 H IME = 0 -) - 600 x8 – 300716 + 1200X16 + F_Sin (16-255 )*8+ FEG G For cos (16•250)*6 = 0 Ker seu > 1200 16 KA2X Poko - 8 Fe = 4800 + 4300 - 19200

= - 9600 FDE = -1200 db : For = 1200 16 EMA = 0 =) - 300 8 + 1200 x 9 - Fax 6 =0 → Fra x6 = - 2400 + 9600 FEG 3 7200 Fra = 1200 16 (1) EFQ = 0 =) FDe Cos (16-258) + FEE COS(36.869) + FEG = 0 ->-1200 cos (16.258) + Fee COS (36.869) + 1200 - Fre = 152.01 - 1200 cos (36-869) FEF = -60.16 To Fee = 60.4b c)