Answer Happy

100 KN 75 KN A B с FI, ul F2, u2 F3, u3 - WWW k1 1 2 3 k2 Fig. 21 Fig. Q1 shows bar assembly consisting of two dissimilar cylindrical rods joined together at B. The assembly is fastened to a rigid support at end A and it is subjected to 100 kN and 75 kN axial loads at B and C, respectively. All bar elements and nodes should be numbered as shown in the equivalent spring system schematic above. Table QI below shows the necessary information required for the material properties and the assembly geometry Table 1 STEEL (1) ALUMINIUM (2) Area - 60mm Area - 120mm E200GPa = Length = 80mm Length = 40mm E = 70GP The force-displacement equation for each element may take the form: ke -kel 6)= has ΕΑ ke EA =
Using the direct stiffness formulation approach, determine (a) The stiffness matrix for each element, Element 1: () = 10" (-15 151603], Element 2: (63= 109 (2. zi1 Cuz {} " 2.11 : = -2.1 (b) 1.5 c) (c) The global stiffness matrix for the bar assembly -1.5 0 = 108-1.5 3.6 -2.12 -2.1 2.1 (uz) The displacement of nodes 2 and 3. uy = -11.7 x 10-4 m uy = -15.2 x 10 m The reaction at the left support (at A) F; = 1.75 x 105 N, The direction of Fl is to the right The stresses in the aluminium and steel bars Os = 2.92 x 10° Pa (Compression) a = 6.25 x 10" Pa (Compression) (d) (e)