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Fa03_FEM test 1 Solution

# Fa03_FEM test 1 Solution - Solution to Exam 1 Finite...

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K 5 10 3 × 1 - 10 3 × 4 - 10 3 × 0 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 × 0 1.1 - 10 4 × 3 - 10 3 × 1.4 10 4 × = n3 n4 b. Displacement vector: c. Load vector: n1 n1 n2 δ 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 - 0 0 k1 - k1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 k2 k2 - 0 0 k2 - k2 0 0 0 0 0 + 0 0 0 0 0 0 0 0 0 0 k3 k3 - 0 0 k3 - k3 + k4 0 k4 - 0 0 0 0 0 k4 - 0 k4 0 0 0 0 0 + 0 0 0 0 0 k5 k6 + 0 k5 k6 + ( ) - 0 0 0 0 0 k5 k6 + ( ) - 0 k5 k6 + ( ) + = n1 n2 n3 n4 n1 12 points. (2 points per each) n2

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Fsum 0 = 3 pts Fsum F 0 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 δ = δ stack 0 δ red , 0 , ( 29 = g. Calculate the reaction forces 5 pts δ red 0.984 - 3.115 = 3 pts Kred 1.4 10 4 × 2 - 10 3 × 2 - 10 3 × 9 10 3 × = 3 pts x4 0 = δ red Kred 1 - Fred = Fred F2 F3 = x1 0 = 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 , 0 , 1 , ( ) , ( ) = 1x 1 θ 2x 2 θ Switching the rows 1x 1 θ K1 stack submatrix K1 2 , 3 , 0 , 3 , ( ) submatrix K1 0 , 1 , 0 , 3 , ( ) , ( ) = K1 6 6 6 - 6 - 6 8 6 4 6 - 6 6 6 - 6 - 4 6 - 8 = 2x 2 θ Defining some zero vectors that will be inserted into the matrix. z_1_4 0 0 0 0 ( ) = z_12_1 0 0 0 0 0 0 0 0 0 0 0 0 ( ) T = z_2_4 stack z_1_4 z_1_4 , ( ) = z_12_3 augment z_12_1 z_12_1 , z_12_1 , ( ) = z_3_4 stack z_1_4 z_1_4 , z_1_4 , ( ) = z_6_4 stack z_3_4 z_3_4 , ( ) = z_12_6 augment z_12_3 z_12_3 , ( ) = Problem 2: Element 1 (beam): Use 2nd template, replace A with n2 and B with n3 4 pts (2 for using the right templete, 2 for matching the right labels) Given: EA 6 = EI 4 = L 2 = 2x 2 θ 1x 1 θ 2x 2 θ K1 2 EI L 3 6 3 - L 6 - 3 L 3 - L 2 L 2 3 - L L 2 6 - 3 - L 6 3 L 3 L L 2 3 L 2 L 2

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