Indeterminate Structures The indeterminate structure rig (Figure 1) consists of three pin-jointed members made of the same type of steel. The top member is at 45 to the vertical. The middle member is horizontal and the bottom member is at an angle of 60 to the vertical; the bottom member is 1m long between pins. The outside diameter of the steel tubes is 25.4 mm and the wall thickness is 1.2 mm. Young's Modulus of the steel tubes of the rig is 210GPa. B 45 45 A 90 30 60 D Figure 1: Test rig used (left) and schematic drawing of the frame of the rig (right) F Theoretical analysis: using the analytical method. calculate the horizontal and vertical displacements of point A and the internal forces in each member of the structure if a load of 6 kN is applied vertically, as shown in Figure 1. DE 45 30 ->6KN GEN B, C, D are pin joint so there are two Reaction forces are acting at every pin joint. at point B 1By 160 45 Bx BA I apply lamis theroem By Sints = Bx Sin 225 BA = Sin 90 at point 'A' BA +6 30 AD Fx =0 GKN BACOSHC + AC +AD COS 30 = 6KN sty=0 BA Sinus = 6+AD Singo At point 'c' Fx =0 A Cz CA+Cx=0 Fy=0 Cy =0 - - - 270 at point 'D' AD Dx Apply LAMis theorem Dx Singo = Dy AD 8in 270 Sin 30 Take and equation BA COSYS + AC + AD CO330 = 6KN BA Sinus = 6+AD singo" apply BA > 6+AD Sin30 Sinys" BA value in 2nd equation 6+ ADSingo, COSHS + AC + ADCOS 30 = 6KN Sin45 (6+ADsin30) fan45+ AC + ADCOS 30" =6KN. 6 tanys+ AOsin30 tan45 + AC+ADCOS 30 = 6 KN 6 + AD + AC + 1 AD = /6KN at Point D + AC = 0 AD [+] A 1.366 AD+AC = of Sty = Dx + ADC0530 = 0 EFy by + AD Sin 30 = 0 these equation By using forces in each bar. we can calculate the Average strain in micro-strain (=10^-6 strain) Strain gauge 3 & 4 (middle member) 2.5 Load (kN) Strain gauge 1 & 2 (top member) 123456 42.5 84 125 164 -4 208 -5 JS 6 251 -6 -1-5 - 3 Strain gauge 5 & 6 (bottom member) -41 -81.5 -119-5 -158 -2015 -243-5