Question

B. Solve the following problems longhand (use calculator/computer, but no FEA software).

For the plane truss shown below, determine the displacements and reactions. Let E=160 GPa, A=2×10-4 m2 for elements 1 and 3, and A=4×10-4 m2 for element 2.Note that the roller boundary condition means that the corresponding node stays on the surface without any separation or penetration. (Using FEA)

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Answer 1

Let F1  ,F2, F3 be the reaction forces in member 1, 2 and 3 and R1, R2 and R3 be the reactions at joints 1, 2, 3 respectively.

L1, L2, L3 be the lengths of member 1, 2, 3 respectively.

Given E for members = 160 GPa, cross section area A1,A2 = 2* 10-4 m2 , A= 4* 10-4 m2

L3 = 2 metre

Applied Force = 1200 KN

Consider Joint 3 Free body diagram of Jomt 3 1200 KN by vertical equilibrium & consider 1200 KN = F2 [R3=07 joint 30 - 451 body diagram of soint 2 Te By Lani's Theorem F2 - = 1 S Sim 120 sin 135° sim (105) on soling Fi= 1075.9KN R2 = 878.5KN

Deflection in number ④ ₂ = 1200810 x L2 4x10 4 x 1608 109 by Right A rule in A123 we will get L2 = 2m & L, = 2.83 m) solving a=.0375m on for member A1 = EXLI Aix a t 1075.9x10 3 x 2.83 2x10-4 x 160x109 42 00951m. ③ in downward y' Therefore deflection of Joint is equal to ₂ = 0.0375m Deflection of Joint 2 x Hong X axis =-.0951 cos 60° - 47.57mm Along y aris = - .0951 sim 60 = -82.3 mm