Question

For truss shown below a vertical load of 25 KN and Horizontal Load of 30 KN applied at Node 3 ( Use FEM Nodal displacement, Direct stiffness method)

1). Calculate clearly the member length and distance between members A = 5 x 10^-4 m^2 and E = 200 GPa

2). Determine the member and global stiffness matrix and show the calculation fot Sinθ and Cosθ clearly

3). determine the displacement and member forces

7 600 30 160 0

All Load and dimensions are in meter and KN

Answer 1

A D8 Da Du -4 A = 5 x 10 m2 25 KN Im 60' D3 B 30 KN E = 200 G Pa Dz © 'DE 60' D5 D :) from Geometry we get Member, AB = 7 m ce for equilateral treiangle Member Be=7 m 7 Membere, BD = tan60° 4.04 m and. 8.08 m CD distance will be Sincere 7 sin 60° 2 The stiffness matreix of treuss [K] EA L Coso Caso sino -cost - liso sino coso sind sino arso sino asino Custo -caso sino coso Coso sino Coso sino - sino casosino sino

Membere o Da 1 D3 30 www 7 m De 1 TAB DI 2 در 4 [K ] FA 0.107 0.061 - 01/07 -0.06111 0.061 0.035 -0.061 -0.035 2 -0.107 -0.061 0.107 0.061 3 -0.061 -0.035 0cool 0.0354 Member Da 0 120° om d = 4:04 m DE ho 5 6 3 4 0061 - 0.107 -0.06 0.107 5 Ki] - EA -0.107 0.185 0.107 -0.18516 -0.061 0.107 0.061 -0.1073 0.107 -0.185 -0.107 0.1851 4

Member (3 60' D8 Da d 7m D4 f 8 3 4 [Kg] = EA 0035 0.061 -0.035 0.0617 0.06 0.107 -0.061 -0.1078 -0.035 -0.061 0.035 0.061 | 3 -0.061 -0.107 0.061 0110714 Global stiffness matrix [kg] = [ki] + [k₂] + [kg] After applying boundary condition Dg = 0 (zero) NOW, Do D7 D5 Di = D2 = 3 4 EA 0.203 0.105 ka 3 0.105 0.327 4 matrix atter applying This is the Global stiffness matrix boundary Condition.

NOW 3 4 0.105 80 EA 0.203 0.327 Du 25 01105 0.203 01105 30 =) D3 EA 01327 0.105 25 Du 129.79 com n EA 34.77 Du D2 - 129.79 EA E = 200x10% kN/m = 5x10-4 Da 34.77 EA FA = 200 x10 x 5 x 104 100000 KN 129.79 Ds = 0.00129 m 100000 1.29 mm Da 34.77 100000 3,47x10-4 m - 0.34 mm Displacemeort are - D2 = 1129 mm DU 0.347 mm

calculation fore me mbese force Member L-7 m { EA 0.8660 -015 D2 (-0.866 615 fi 7 Da Du EA (-0.866 -0.5 0.866 0.6]. 0 EA 124.79 34.77 18.529 CC KN Membere L= 4.04 m o 120° DS Ds F2 = EA L [ coso -sino asso Cosso sino] D3 Du -0.866 -0.5 [0:5 0.866] 4104 EL 1291779 34.77 ll 8.60 KN

Member 3 L 7 m 60' 0.5 -0,866 0.866 EA 015 F2 X [ o 7 0 SA 129,79 34.77 13:57 fa KN Conclusion o- Displacement:- D2 = 1129 mm Du 01 347 mm Membere forces FBC 18.529 KN FBD (Tensile) (compressive) KN (Tensile) 1) -8.60 KN FAB 13.57 KN