Question

is 4. Using virtuat work determine the vertical displacement at joint E assume roller and F is a pin A = 2in? E = 29000 ksi A B c 1.5 TI E ok 2. H 2

Answer 1

B Given : † cs Area of members = 2in2 1.5 ft E - 29000 ksi D F → E 6K R=3K K RF = 3K 2 ft 2 ft .. = 0 => solution : step 1) – consider actual structure & find reactions. Rox(2+2) - 6*2 = 0 .: Rp = 3k (1) Also, theV =0 => +R$ -6 + Rp=0 3-6+RE RF = 3K (A) step 2) – Find the forces in all members: FAD a) At joint D: EDE 7418V-0 => FD=3K (c) Do- 1 3K EH=0 -> FDE = 0 53.13 step 2) b) At joint A: 2 tan O = => 0 = 1.5 +1 EV =0 FAD - The COS (53-13)=0 FAE 3 (0553-13 Fie caso FE SA - FAB AE FAD - -3K . - 5K (F)

2 с FBC 7° t> FCF < =0 Fre sino - FAB = 0 5 sin 53.13 - FAB = 0 => FAB = 4K (c) c) At joint c: only two members at connected '=0 ) FBc = 0 -> +1&v=o >> FF=0 d) At joint B : El=0 For sino FAB - FF sin 53.13 = 0 4 - FBF sin 53.13 = 0 FBF = 5K (C) > +1 EV=0 FeFt Forroso : -FBE + FBF COS 53-13 = 0 FEF 4 K (F) B FAB FBE FBF COSE *FBF e) At joint F ΣH =o Feesino F FEF 3K . FBE = 5 x cos 53.13 FBE - 3K (T) step 3)- Now, consider virtual structure by removing external 6k load of apply 1k at E. A B с D F E 1K Rp=0.5K Rp=0.5K

+ - 1x2=O Reactions : CEMEE -0 -> Rp * (2+2) - RD = 0.5k (1) +5V =0 -> RD - 1+RF = 0 RF = 0.5k (4) step 4) Find forces in all members due to unit virtual load 2 FD D. FDE &) At D: > + EV-0 FAD - JK (C) > *- EH =0-> FDE = 0 0.5K A FAB b) At A : +18V = 0 FAD - FAE FAE = 1/cos 53.13 = 1.67K (T) Cos 53-13 = 0 Faesino FAD 1 FAE t> EH=0 =0 FAE sin 53.13 – FAR -0 1.67 sin 53.13 - FAB - FAB = 4.33 (0) c) At joint c: FBC = Fcp = 0 two forces meeting at c Because only

d) At joint B : o> 0 FAB FBE Forsino FAB - For sin 53.13 = 0 1:33 - FBF sin 53.13 = 0 FF = 1.66K (C) FRE Forcos >+ <V = 30 FEF sin FBFSO 3K 0 - FBE + FOF Cos 53-13 = 0 FOF - FBE = 1.66 x cas 53.13 => F e = 1K (T) sikte Forening FEF=1.32 (T) step 4) Let n → Force system due to applying 1k load (T) Tension + -ve N → Force system clue to external load Members NEI L NAL AB ( n 4 1.33 2 10.64 Вс. 0 O 2 CF 0 0 1.5 0 FE *4 -1.32 2 -10-56 E ED 0 0 2 DA 1.5 AE 4.5 -20.75 -1.67 2.5 BE - 3 -1 1.5 -4.5 BE 1.66 25 20-75 EN; 1:1 = 0.4 Displacement at joint E => A Is A = Ê Hinil k for inch conversion X12 2x2gooo AE There is no displacement of E