Question

③ using virtual work the vertical displacement at Joint E assume Des a roller and F is apin. A = wina E- 29000 ksi A B C D wts ANL 6K at- { А, Е

Answer 1

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Let, reaction art D and F be by and o (Ey, fx), respectively. Efy =0 Dy + Fy - 4=0 fr Ex T 2' YR o tan" (1.s) G Emp=0 - 36.870 Cufy) - (4x2)=0 fy= 2K EFXO fx=0 Dy= - 2K Using method of joink. EDA at joint D, E fx=0 DH FDE ↑ Dy

Fre=0 Efy=0 FDA + Dy=0 2 FDA-- Dy - -26 Co Zero force members, Fac- fef=o] FBF FEF I fuc Take joint f. E fy=0 fy + FBF sino =0 FBF sin 36.87 =0 [FBF = – 3.33 K (C) T fy 2 + EFX - - FEF - FBA caso =0 - FEF -(-3.33 Cas 36.87 =0 [fe FEF - 2.67K (T) Take joint E. Efy=0 FAE FBE FEF FACsino & FBE – 4 = 0 -4=07 FDE UK EFx=0 را - FAC Coso & FEF = 0 FAE COS 36.87 +2.67 = 0 FRE - 3. 33k (T)

FAB + (3.33 ces 36.87) = 0 member FBC Foc=0 from (3.33) sin 86.87 + fee - 4 = FBE & - 2* (T)) Take joint A Efx=0 FAB FDA FAE FAG + FAECOSO = 0 FAB=- 2.66k (c) 1K vertical. for determining vertical displacemento at E!. replace all load with External unit Load С GEMD=0 n F (Fyx4) - (182) = 0 D 21 21 T Dy IR - O.SK Dy - O.SK zero perce FAD At joint D EFX -o ↑ Dy [fe=o

Efy=0 FAD+ Dy' =0 FAD - - Dy – – O.SK (c) At joint F FBF E fy=0 fy't Forsino =0 For Fffy FBF = – 0.5 Sin 36.87 = – 0.833K (c) EFX -o -FEF – F BFCOSO = 0 -FER-(-0.833 Cas 36.87)=0 = Fee = 0.6 67k T FAE -> FBE At joint E. Feod -FDE - FAECOS @ +FER=0 FER IR - FARCOS 36.87 +0.667=0 FAE = 0.833K Efy=0 FAC Sino + FBE – 1 = 0 (0.833 Sin 36.87) + FBE - 1 = 0 Еве O.SK (T)

FAB + FAE Coso =0 Fast Co. 833 x cos 36.87)=0 1 joint a FAB E Fx=0 Сре FABE O. 667 K tc ) we have, 2 ina 29 ooo k 81 E = let, ferce in member der te External load unit LA LI 11 member Length den to NNL Force External load, N L unit AE leadin los - 2 - O.S AB - 2.66 -0.667 3.548 AF DE 2 o o AE 2.5 3.33 0.833 6.93/AG 1.5/AE BE 1.5 2 ois ВС 2 0 CF EE 2 2.67 0.667 3.su8lne 6.93/AG BE 2.5 - 3.33 - 0.833

at E be Vertical deflection AE = ENL = 1 AE [ 1. St 3.548 + 6.93 t 1.5 +3.548 +6.93 23.956 23.956 AE (**) [ 29000 x122) =) 4613 x 10-4 ft 2) 4.9564 x 10-3 in