Question

Find Delta By and Delta Cx using virtual work method. Bar areas ; Top Chords = 4; Bottom Chords = 2.5; BD = 3.

Problem 2: Find ABy and ACx. Bar areas: top chords = 4; bottom chords = 2.5; BD = 3. 40 kB 10ft 40 k 10 f T 20ft 20ft —

Answer 1

цок р .28.28! 4 Q Le 2212 O=Tau id=tant to yo) d = 26-56° Be Tum ( 0) – 26560 P = Tun'() =45° EMAO ř 20lt zo' Coq Sne)(40) + righe coso) lid) O=Tan () 0-63-43° 740 x 20 = Hox20 [Rezo) At joont.. Sfxzo ffeo=o] fo Sfyzo fisc=0 ALD Efx=0 FAD to JB=0 TAD co с о [fъв е чор) соттрг ATB FAB I to Efyão FAB Sinus = 40 THAB = 56.56 K tension to foB 40 Despencemet Se Sole So we need to Calculate forces in BD and AB cs die drithe unit dowds. CamScanner

Virtual work method for by) Apply an unit load at B in y-direction EMA=o BK Rc Coso) (10) +Resino) (40) = 120 o AK 2054 20' Re= 0.4969k At joint ac PKBC Ifyo KBCS2D2856 = 0.4969 506343 Re=0.4969 KBC= 0.993k (C) 0 =63.417° x = 26-56 Koc BRAD {fp=0 KALI KREE on? KOL + KBC CO12657 - 0-4963 C056343 Kpc -0.667K CT) At jand Sko Koe Cms 26-5% ерего KAD CO326.56 = 0.667 KAO = 0.7457K ) {fyzo KAD SN 26:56 2 RBD KBD = 0:333 K (T) At jant 'B CBC Sfaxzo Kec C0326.5% = KAR LOJUS RAB= 1-256K (Ⓡ) Scanned with CamScanner

Sertran P | K | Lidl AE UZ PKYAE (is) -0.2018 56.56 1.256 [28.2842| 4829000 -0.0183 40 H0.333 LOX1Z3 X 220VO -0.226. lo SPKL Defleurs - AE ABy = To By = -0.226 in wad est & in & Apply an anut x-directa SMAZO Checolo) (60) +Rejsinc)(40) – ICLO) Re = 0.24846 ا وب 6=63-43 Styco Ay+ Re SD 63.43 =0 Ay ca0-222k -Sfxão Axt Rcc0283.43 = 1 CS Scanned, witbay e 0.888 K CamScanner

At joints Efx=0 0.888 KAO 10-222 KAB cos 45 + kao Cos26:56 = 0.888 Chyzo KAB foru5+ KAD $1.26:56 = 0.222 V=45 B=2656 Еве - -осове Ево = 1 мектр т ato Styco KBD = KAD S1026-52 Kep = 0.66614] () Membe Tk I L AE 1 PKLAE en -0.627 | 28.82x12 uxarosso 1-0.1032 0.666 | 20/12: 37629020 -0.0367 -0.1404 Ces) Deflethan uta 1. Ac = -0.1404 in Scanned with CamScanner