5.1 Introduction 249Figure 5.1 Planar truss: (a) nodal and member numbers; (b) nodal degrees of freedom.zero, as they correspond to the fixed nodes)
( 4 x 4 +x 6 +x 7 )u 1 +√
3 (x 6 −x 7 )u 2 − 4 x 4 u 3 −x 7 u 7 +√
3 x 7 u 8 = 0 (E 1 )
√
3 (x 6 −x 7 )u 1 + 3 (x 6 +x 7 )u 2 +√
3 x 7 u 7 − 3 x 7 u 8 = −4 Rl
E(E 2 )
− 4 x 4 u 1 +( 4 x 4 + 4 x 5 +x 8 +x 9 )u 3 +√
3 (x 8 −x 9 )u 4 − 4 x 5 u 5−x 8 u 7 −√
3 x 8 u 8 −x 9 u 9 +√
3 x 9 u 10 = 0 (E 3 )
√
3 (x 8 −x 9 )u 3 + 3 (x 8 +x 9 )u 4 −√
3 x 8 u 7− 3 x 8 u 8 +√
3 x 9 u 9 − 3 x 9 u 10 = 0 (E 4 )− 4 x 5 u 3 +( 4 x 5 +x 10 +x 11 )u 5 +√
3 (x 10 −x 11 )u 6−x 10 u 9 −√
3 x 10 u 10 =4 Ql
E(E 5 )
√
3 (x 10 −x 11 )u 5 + 3 (x 10 +x 11 )u 6 −√
3 x 10 u 9 − 3 x 10 u 10 = 0 (E 6 )−x 7 u 1 +√
3 x 7 u 2 −x 8 u 3 −√
3 x 8 u 4 +( 4 x 1 + 4 x 2+x 7 +x 8 )u 7 −√
3 (x 7 −x 8 )u 8 − 4 x 2 u 9 = 0 (E 7 )
√
3 x 7 u 1 − 3 x 7 u 2 −√
3 x 8 u 3 − 3 x 8 u 4 −√
3 (x 7 −x 8 )u 7
+ 3 (x 7 +x 8 )u 8 = 0 (E 8 )−x 9 u 3 +√
3 x 9 u 4 −x 10 u 5 −√
3 x 10 u 6 − 4 x 2 u 7+( 4 x 2 + 4 x 3 +x 9 +x 10 )u 9 −√
3 (x 9 −x 10 )u 10 = 0 (E 9 )
√
3 x 9 u 3 − 3 x 9 u 4 −√
3 x 10 u 5 − 3 x 10 u 6 −√
3 (x 9 −x 10 )u 9+ 3 (x 9 +x 10 )u 10 = −4 Sl
E(E 10 )
the vector of loads for thejth member. The formulation of the equilibrium equations for the complete truss
follows fairly standard procedure [5.1].