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.

Answer 1

method of Joint FAE At Joint - A8 o = 53.139 A FAB Ostan' in ) -53:13 40 kn 0 = 53.13 + +1€fy=0 FAE Sin (53:13) = 40 FALE 40 Sin (53613) - FAE = 50 kN; it, &fa=0 M for sofAE COS (53-13) • 50 80.6 =-30 kr fAB = -30 kn At Joint B² FBE 4 FAB FBC 20KM for = 20 KM foc = FAB fece = -30 kn

At Joint Elo FED 112 10- 53.13 o A FAE FEC COSO = 0.6 sino - 0.8 for I &fa=0 FAE Coso = fee coso + FED 50 x 0.6 - Oro fec +FED 0.6 fec+fen = 30 KM TO +7 & fy=0 fre Sine +foe +fEc sino = 0 50 x0.8 +90 + fec (0.8) = 0 40+20 +0.8fEc=0 fec -60 0.8 fec = -75 kn - @ Ris for you ② in FED 30-0.6ffc - 30-0.6675) - 75 kr fED=75 kn

remove all external loads kål values & apply IkH load at 'A' in verti - cal direction a ter stenlos 2,5 m am 3 0 53.13 COSO= 0.6 A let 1.5m sino=0:8 115 I kn At Joint A KAE k 20 +1 Efya po *KAB co m IKN - = kae sino KAE I = 1:25 = 018 KAE = 1,25kN It Efx=0 T KAB Coso KAE KAB - 3 - (0.6) (1.25) w *- 0-257 KAB

Joint Blo KBE 4 KAB KBC KBE = OK KBC = KAB KBC = -0.75 KM I J - At Joint Els KOE - A e = pefy = O 0 *KCE OKN KAE Sino+IBE+KCE Sino=0 KAE og og SKOLE 1-25 X018 to + KCE (0-8)=0 KCE =-1195 kn 0.8 KCE al. 25 KN Is fa=0 T KDE + KCE COSO = KAE Cose koe 1125 x 0.6 - (-1,25 x0,6 0.75 t 0.75 = 1.5 kry KOE 1.5 km 112 =

ki remore external loods horizontal lood at 'A' -> apply const lood (hen) in " Horizontal" and apply 12 Pech 0-53-13 coso=0.6 sino=0.8 9.59 2 m 0 et I ku a 4 1.5m B с Joint i KA KAB 1 IKN 4 + A & fy- = 0 ران =0 K'AE Sino KAE o ky t e fa=0 11 K'ng = 4 KN AKBE Joint 18!! KAB ¢ KiBC B KBE - Okal KBC Ik ky

At Joint 'Els L2 داو AKDE o 4 Efy=0 t *KCE NC K AE Kae sino+k'ce sino +k8e = 0 KBE k (O) sine f K'CE Sino + 0 = 0 klee = 0 km; cos Is é fa=0 k'se cose = kép cose + KDE O COSO = 0 COSO + kbe k'pE = 0 Kn 7 mm? x 200 X 200 Gp a 2 - 400 mm AE IT १ 400 mm x 200 kN/m2 mm 80000 kn y T 8x 104 kn AE

(u) load (P) Member k' K Ct). Pkl Length (m) P.k'l AE Cky 2 .Cm) AE (KM (kal AE cm2 - 30 -4. -5.625x16 8 X10 1.5 AB -0.75 1 4.9187 x104 8x104 105 -5.625x104 Вс -30 4.9187x10 4 -0.75 BE q 90 1) CE 9.5 -75 -1.95 en 298 X10-4 o DE 15 75 115 0 21.09 x1014 o = A E 2. 5 50 1.95 19.53x10-4

=) (SA) E Pkl AE - 104(4.2187 +42187+0+29.3 +2109 +19.53) = 104x (18.3574) m = 7.8357 mm (Salv 7.836 m Verhoceed T (San & Pk'l АК m - (-5.625-5.625)8104 = -11.95.410-4 - 1.125 mm (SA) - 1.195 mm - Io Horizontal