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hw3-2008sol

# hw3-2008sol - Chapter 17 Solution 9 We have E = 210 GPa A =...

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Chapter 17, Solution 9 We have E GPa A m = = × 210 5 10 4 2 Table P17.9 Data for the truss of Fig.P17.9 Element Length c s c cs s o o θ 2 2 1 5 5313 0 6 08 036 0 48 0 64 2 4 90 0 1 0 0 1 . . . . . . ( a ) Apply Eq.(17.14): u v u v 1 1 2 2 [ ] . . . . . . . . . . . . . . . . k AE u v u v 1 1 1 2 2 5 036 0 48 036 0 48 0 48 0 64 0 48 0 64 036 0 48 036 0 48 0 48 0 64 0 48 0 64 = u v u v 1 1 3 3 [ ] k AE u v u v 2 1 1 3 3 4 0 0 0 0 0 1 0 1 0 0 0 0 0 1 0 1 = ( b ) Global Stiffness Matrix u v u v u v 1 1 2 2 3 3 [ ] . . . . . . . . . . . . . . . . . . . K u v u v u v = 10 0 756 1008 0 756 1008 0 0 1008 3969 1008 1344 0 2 625 0 756 1008 0 756 1008 0 0 1008 1344 1008 1344 0 0 0 0 0 0 0 0 0 2 625 0 0 0 2 625 7 1 1 2 2 3 3 ( c ) F v x 1 7 1 10000 10 0 756 1008 1008 3969 0 025 = . . . . . 100 10 1008 10 0 025 3696 3 7 1 × = × + ( . )( . ) . v or v m 1 0 00887 = . F v x 1 7 7 1 0 756 10 0 025 100887 10 = × + × ( . )( . ) or F kN x 1 99 6 = − .

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