Question

1) A three-member plane truss is shown in the figure. Member (2) has a cross section of 65 mm' and is made of steel (E = 207 GPa) while members (1) and (3) have a cross section of 75 mm² and are made of Aluminum (E = 70 GPa). Determine displacement at the free end (node 3) and stresses in each member. Solve this problem by analytical method discussed in the class (and lecture notes). 12 KN (1) 30 2.50 m

Answer 1

ANSWER:

KW 60H (2) 2.50 m Á do 1.25 m DE Given that: Member 2 has a cross section (A2): 65 m2 and made up of steel E2= 207 Gpa Member O and are cross sections A1 =43: 75 mm2 and made up of Huminium E;=E Simce licos Boo do 22 m die l3: 2m , daarom I coceso sino liso CHC018 Si no 22 127 (2 gooi 0.43 0.5 10.8.1 0.75 3) 0.25 60

w Stiffnes materia for eleve matrix for element o bre no de 0-6 Now stiffness matrix KB) b/w (2-3) L-GIGI where G=[e2 es? • 5.25 03/25 .435 -0.3-0.435 435 0-757 -6-43 -157 L 6.25 -6.435 0.25 0.43 L 7.435 -0.77 0.435 6-757 :5210 1 1.3125 - 2.29 -2.23 -1.3125 2-23 397 -2.26 -397 v2 -1.03125 -26 1327 2:28 13 [ -2.25 -3.97 2.28 2.97.303 How since hode 1,344 is fined Therefore U, EU 2 = Us="4 0000] O10- 0 ooo Oto! , vi V2 V2 0 0 0 0 o $250 -5250 1 0 0 0 0 Lo -5256 o $250) know falz) berwen node 2-4) Kle) - 12 Ez (G2 thel L2 L-G2 G2] 12 V2 vu Vy ay 1031 21 0 -2 6 102 oooo 7-21 020 uy Loooo don Eff]-{ Ke][u] up = 0.15 mm) 42= 0.188 mm

Now swisse in each nyemben, 486e oseala 0.138 :44Vyn global stiffness mabic (alter negleet above) lowf 22.11 a5 2.2f ? 1 2.28 9.22] v2 2 42 = 0.115 mm U₂=61)88 mm Now shesses in each member 0.138 6 = -6.53 Mga IX. 103 uy=0 62 = -24.15 MPa 163 = =24 15-pa) oz- 70 [005 -0.37 05 087 16: -7.73 Mpa ร***