notes_34_ex_pg_192_03

notes_34_ex_pg_192_03 - Matrix Analysis Example Hughes...

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Matrix Analysis Example Hughes figure 5.12 page 191 ff ORIGIN := 1 F YB C 6 b b F XB 3 A B 2 4 5 c 1 a input data 1 2 4 1 2 c 1 2 3 4 3 4 a 3 n_elements := 3 n_nodes := 3 ie := 1 .. n_elements f, δ element; F, structure; m = element n_free := 2 number of degrees of freedom per node n_dof := n_nodes n_free n_dof = 6 total number of degrees of freedom in structure input for the class and text problem: nod_el := 2 nodes per element in := 1 .. nod_el 1 2 0 0 1 1 elem := 1 3 nodal map of elements XY := 1 0 location of nodes A := 1 E := 1 2 3 0 1 1 1 __________________________________________________________________________________________________________________ element stiffness matrix: geometry 0 1 0 0 X ie 2 , X ie 1 , ( 2 Y ie 2 , Y ie 1 , ( 2 + ) ) 1 X ie , in := XY elem ie , in , 1 Y ie , in := XY elem ie , in , 2 X = 0 0 Y = 0 1 L ie := L = 1 1 0 0 1 1.414 notes_34_ex_pg_192_2003.mcd 1
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Y ie , 2 Y ie , 1 π 0 angle ie := if X ie , 2 X ie , 1 > 0 , atan X ie , 2 X ie , 1 , 2 gets angle - π /2 < angle < π /2 angle = 90 don't need angle now but will deg later for T 45 0 angle ie := if ( X ie , 2 X ie , 1 < 0 , angle ie + π, angle ie ) gets angle in appropriate quadrant angle = 90 deg 135 element stiffness, element coordinates 1 0 1 0
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This note was uploaded on 02/24/2012 for the course MECHANICAL 2.082 taught by Professor Davidburke during the Spring '03 term at MIT.

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notes_34_ex_pg_192_03 - Matrix Analysis Example Hughes...

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