Solution of Equilibrium Equations in Static Analysis: LDLT Solution ...

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L LDL T Solution Recall the Gauss multiplication factors – they enter into L i ‐1 , i = 1, 2, 3: −1 1 Step 1: Step 2: r3 = r3 + 8/7 r2; r4 = r4 + (‐5/14) r2; r2 = r2 + 4/5 r1; r3 = r3 + (‐1/5) r1; r4 = r4; Step 3: r4 = r4 + 4/3 r3; ⎡ 1 ⎤ ⎡1 ⎤ ⎡1 ⎤ ⎢ 4/5 1 ⎥ ⎢ ⎢ ⎥ 1 0 1 ⎥ ⎢ − = ⎢ ⎥ 1 0 1 ⎥ − L ⎢ −1/5 0 1 ⎥ 2 = L ⎢ 0 8/7 1 ⎥ 3 = ⎢ ⎥ ⎢ 0 0 1 ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎣ 0 0 0 1 ⎦ ⎣ 0 − 5/ 14 0 1 ⎦ ⎣ 0 0 4/ 3 1 ⎦ (i.e. the i th column of L i ‐1 contains the multipliers of the i th step) 15-Jun-07 L ⎡ 1 ⎤ ⎢ 4/5 1 ⎥ ⎢ − ⎥ 1/5 − 8/7 1 ⎢ ⎥ ⎣ 0 5/14 −4/3 1⎦ = L1L2L3 = ⎢ ⎥ Method of Finite Elements I 14
LDL T Solution Recall the pivots in the Gauss elimination – they enter into S (i.e. reduced K): S ⎡5 −4 1 0 ⎤ ⎢ 14 / 5 −16 / 5 1 ⎥ = ⎢ ⎥ ⎢ 15/7 − 20/7 ⎥ ⎢ ⎥ ⎣ 5/6 ⎦ First row of K Second row of K after step 1 Third row of K after step 2 Fourth row of K after step 3 For the matrix D: d ij =δδ ij s ij ⎡5 ⎤ ⎢ 14 / 5 ⎥ D = ⎢ ⎥ ⎢ 15 / 7 ⎥ ⎢ ⎥ ⎣ 5/6⎦ V is the right‐hand ha side after the reduction of K to upper tia triangular form V = [ 0 1 8/7 7/6] T 15-Jun-07 Method of Finite Elements I 15
Page 1 and 2: Solution of Equilibrium Equations i
Page 3 and 4: Review ‐ Matrices • Positive‐
Page 5 and 6: Gauss elimination In a Gauss elimin
Page 7 and 8: Physical Interpretation A physical
Page 9 and 10: Physical Interpretation 15-Jun-07 M
Page 11 and 12: LDL T Solution The successive matri
Page 13: LDL T Solution K = LS Now = % where
Page 17 and 18: LDL T Solution Algorithm: • Colum
Page 19 and 20: LDL T Solution Practical issues (2)
Page 21 and 22: LDL T Solution Example: K First, ge
Page 23 and 24: LDL T Solution j= 5: g 15 = k 15 =
Page 25 and 26: LDL T Solution V = V − l U r = m
Page 27 and 28: Error Estimate •Now we use t=3 di
Page 29 and 30: Error Estimate Bathe, page 749: Sum

Solution of Equilibrium Equations in Static Analysis: LDLT Solution ...

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