Question

Determine the forces in members DE, EI, FI, and HI of the arched truss. The forces are positive if in tension, negative if in compression. 

Answer 1

Draw the free body diagram of the structure 135 KN 30 kN 30,KN 20 kN 20 kN 16 Dimensions in meters G. Calculate the vertical reaction at point G. take the moment about point A G. x2820 x25+30x 21+135 x14+30x 7+20x3 G. 117.5kN The horizontal reaction at point G is, G. 0 Calculate the length of the member HH 16 HH HH 5.818m

Cut the section along the members EF, FI, and HI as showing in the following figure 20 kN F. FI 5,818 HI 16 Calculate the force in the member HI. Take the moment about point F. 5.818 4 7.06 7.06 b-ssis x4)+(FHlx 706 x 5.818)=(117.5 Hl 706 3)-(0x16) FH = 53.47 KN (7) Calculate the force in the member EF. Apply vertical equilibrium equations 5.818 7.06 -20 + G ' = 0 5.818 7.06 Hlx 6.4. FEF X--+ 53.47 × 20+117.5=0 6.4 FEF 181.201 FEr 181.201kN(C)

Calculate the force in the member FI. Apply horizontal equilibrium equations 4 4 6.4 7.06 7 4 4 181.201× _ + 53.47x-+ F, + 0 = 0 6.4 7.06 FR 82.96 F 82.96kN(T) Cut the section along the members ED, DL and JI as showing in the following figure 4-3 30 KN 20 kN 5 16 Calculate the force in the member DE. Take the moment about point I DE 7.61 DE 7.61 ×761 x4HF FDE163.94 FDE 163.94kN (C) FDE X--X4_ FDE ×ー×5 |+(117.5×11)-(30x4)+(20x8)-(0x16) DE 7.61 x5j+(1175x11)-(30x4)+(20x8)+(0x16)

Calculate the force in the member DI. Apply vertical equilibrium equations #xx 20-30 + G. 8.54 FD1×ー-163.94x- FDI3.07 Calculate the force in the member J. Apply horizontal equilibrium -20-30+ 117.5=0 8.54 7.61 equations 854 DE x 761 + Gx + F, = 0 8.54 7.61 F, = 151.87 kN(7) Consider joint I. IE IF LJ 5.818 IH Calculate the force in the member IE. Apply vertical equilibrium equations 5.818 7.06 8.54 5.818 -3.07 × _ +Fe ×-= 53.47× 8.54 6.4 7.06 Fe = 60,08 FIE 60.08kN(T)