Question

Q3: Determine the vertical and horizontal displacement of joint A for the truss shown in Fig. (3), each bar is made of steel and has the cross-sectional area of 400mm?. Take E = 200 GPa. Use the method of virtual work.

E D 00. 2 m СІ Ao cocoa B 1.5 m 1.5 m 20 KN 40 kN Fig. (3)

Answer 1

Given loading (N) E D Hot al |४ 2m A TL B and he Tuc Since there is only one member at support D, which is horizmtal (ED) , there exists horizontal reaction Only > Ho = NED VD = 0 40 20 * * 1.5m 1.5 m < Fy = 0 Ve = 60 KN Efx=0 He = HD EMC=0 4013) + 2011.5) – HD (2) = 0 Ho = 75 KN 1>) He = 75 KN IK) At Joint D Efm = 0 - HD NED = 75 KM At Joint B NBE Fr=0 => NAB = NBC MBC NBE = 20 KN Efy=0 NAB 20 At Joint A 2. 5 2 7 MARE E fy=o => NAE NAR 1.5 HO sind cosa = 3/5 sinx= 4/5 40 MAE = 40x5 to y NAE = 50 KN Efn = 0 MAE COS X + NAB = 0 MAB = NBC 01 DX Ž = -30 KN

Joint c Fy=0 → NEC sind + 60=0 -60 NEC NEC($) NEL 30 x 75 = -75 KM Too 60 Vertical Displacement (ni) Apply a withal unit Wad (I kn) at A acting down D E 1.5 KM (ni), = 1.5 KN DE & a 1.5 KN o (MI) BE A = 0 KN B Тk I KN =1 • (11) sina AE (MDAE 5 KN = 1.25 KN H =0 cosx + (N. DAB ("DAE 5 x 7 / X IN AB = -0.75 KN (M) BC (MIDAB = -0.75 KM - 0 - 1x 5 = -1.25 KN H (MI) E sin x + 1 (Milee Horizontal Displacemeat (ne] Apply a virtual unit load (1km) at A acting towarde left £ D (n.) DE = 0 KM A B (N2) BE = 0 KM EKN I KN

(12) A sinx=0 AE (2) A 2 OKN 'AE (N2) AB (12) BC = I KN (12) CE = 0 KN Member Length() NIKN) n, (kN) n2 (KN) N.NL n,NL D₂NL AB 1.5 m -30 -0.75 33.75 -45 1.5 m -30 Be -0.75 33.15 -45 50 1.25 156.25 6 2.5 m O AE BE 2 m 20 O 0 0 0 CE 2.5 m -75 -1.25 0 234.375 DE 1.5 m 75 0 166.75 Arial rigidity (AE) (400x106m2) x (200 x10 kN/m2) = 80000 KN -> constant for all members Ση, Νι = 626.875 kN)? m ENNL 7.836x103 kNm 626.875 80000 AE En NL 90 (km)? m -90 -3 1.125 X 10 En NL KN m - AE 80 000

By principle of virtual work (IKN) * Avim) = 5 n NL KN-m AE OVA -3 7.386 x 10 m. 7.386 mm ( down) (1 km) x DHA m) E MNL KN-m AE THA = -1.125 x 103 m (opposite to the direction 11 -10125 mm of applied load) => AHA 1:125 mm (right)