Question

20 kN 5 Determine the vertical displacement of joint B. Assume the members are pin connected at their end points. Take A = 200 mm and E= 200 GPa for each member. Use the method of virtual work. 12 kN #m B 4 m

Answer 1

Step 1:

Apply wait Load at B n B C & Considel Joinc 4 fis < cal FCO fco=o Fcs=0 Considee Joint B foto {fnão F BC = fśA foc Efj=0 Foo •Fo'o =-w Golf 60 Dal knd Comp)

Efxro FCB + Fio Coso- o - FD=26.83>> Gs (26.565) Considel Joint D Foo=1 Efnão FOCO o tan (1) Fóa Gosotrok co so to For 2-26-565 Fóa --POE {fyro o fod to be sino - fóa Sino=0 F20 + F'Sexo Sin (26.565) = 0 > F'DE = -1 2 x Sin06-565) =YPE =- DE =-11118 Comp TEDA= 1:118) tension Stepa) Now Considel Given Loading i Considel Joint C logo C 12 0:26.565 Feel Fico {fg=0 > 12 + fcd Sind=0 => fas: -12 Sin(26.565) FD-26.837 Comp fosz24 (tens)

() Considel Joint B 12o ->F3=24 {faro FBA |fbA=FBc =24 / (tension) WEBD =) Fy=0 F&D = -20) Comp) (ii) Considee Joint D {fg=0 FOA r FBBS-20 FAA Coso + FoE Coso - Falcoso =0 Foca-26-832 -> foa +FoE = -26.807 (1) -FDE =) { fy=0 FDA sino at FDB & FOL sino - FDE 8ino=0 -> fra -20 Sin (26:56.1) -26-832 FOE co oroa-foe = 71.553 () Solve (1) { (2) FDA = 22:3605) (tension) Foe=-49.19 (Corp)

Length menbel (ive to load criotual) (FxF'x2) AB 4m 24 kn o BC 4m 24 KN O hiz - 414721. -26-837 o BD 2. -20 +40 DA 4.721m 22-3605 1-188 118.02 AE 4-721 -49.14 -1*118 259.62 417-6436) A= {FxFxL | foom vistual work) ({fxFx 2) = - Get -)*(417-64 417-6486x103) AE 2oo x 10 % 200x109 A-0.0104 m Con A-10:44mm 42s

Hope this helps

Please feel free to comment on any further queries

All the best :)