Fa03_FEM test 1 Solution

Fa03_FEM test 1 Solution - K 5 10 3 1- 10 3 4- 10 3 1- 10 3...

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Unformatted text preview: K 5 10 3 1- 10 3 4- 10 3 1- 10 3 1.4 10 4 2- 10 3 1.1- 10 4 4- 10 3 2- 10 3 9 10 3 3- 10 3 1.1- 10 4 3- 10 3 1.4 10 4 = n3 n4 b. Displacement vector: c. Load vector: n1 n1 n2 x1 x2 x3 x4 2 pts n2 F R1 F2 F3 R4 2 pts n3 n3 n4 n4 Solution to Exam 1, Finite Element Modeling and Analysis Problem 1: Given: k1 1000 = k4 4000 = F2 20000- = k2 2000 = k5 5000 = F3 30000 = k3 3000 = k6 6000 = a. Determination of [K] by adding the contribution of each of the springs: K k1 k1- k1- k1 k2 k2- k2- k2 + k3 k3- k3- k3 + k4 k4- k4- k4 + k5 k6 + k5 k6 + ( )- k5 k6 + ( )- k5 k6 + ( ) + = n1 n2 n3 n4 n1 12 points. (2 points per each) n2 Fsum = 3 pts Fsum F F 1 + F 2 + F 3 + = h. Verify Static Equilibrium 5 pts F 1.148- 10 4 2- 10 4 3 10 4 1.475 10 3 = F K = stack 0 red , , ( 29 = g. Calculate the reaction forces 5 pts red 0.984- 3.115 = 3 pts Kred 1.4 10 4 2- 10 3 2- 10 3 9 10 3 = 3 pts x4 = red Kred 1- Fred = Fred F2 F3 = x1 = Kred submatrix K 1 , 2 , 1 , 2 , ( ) = f. Calculate the unknown displacements e. Reduced Stiffness Matrix d. Boundary Conditions: K1 augment submatrix K1 0 , 3 , 2 , 3 , ( ) submatrix K1 0 , 3 , , 1 , ( ) , ( ) = 1x 1 2x 2 Switching the rows 1x 1 K1 stack submatrix K1 2 , 3 , , 3 , ( ) submatrix K1 0 , 1 , , 3 , ( ) , ( ) = K1 6 6 6- 6- 6 8 6 4 6- 6 6 6- 6- 4 6- 8 = 2x 2 Defining some zero vectors that will be inserted into the matrix....
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This note was uploaded on 07/09/2011 for the course EML 4930 taught by Professor Schonning during the Spring '06 term at UNF.

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Fa03_FEM test 1 Solution - K 5 10 3 1- 10 3 4- 10 3 1- 10 3...

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