By Jacob Fish, Ted Belytschko
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Those notes are from a path taught through Michael FUaseta within the Spring of 1996.
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Extra resources for A First Course in Finite Elements [With CDROM]
The scatter operations on the elements then give the following (I and J give the global node numbers of the element): Element 1, I ¼ 1, J ¼ 4: 2 ð1Þ 3 k 0 0 Àkð1Þ ! 6 1 À1 0 0 0 7 ~ ð1Þ ¼ 6 0 7: Kð1Þ ¼ kð1Þ )K 4 0 À1 1 0 0 0 5 Àkð1Þ 0 0 kð1Þ Element 2, I ¼ 4, J ¼ 2: 2 ! 1 À1 ~ ð2Þ )K À1 1 Kð2Þ ¼ kð2Þ 0 60 ¼6 40 0 3 0 0 ð2Þ 7 0 Àk 7 : 0 0 5 ð2Þ 0 k 0 k ð2Þ 0 Àkð2Þ Element 3, I ¼ 1, J ¼ 3: 2 ! 2: mechanical, electrical and hydraulic systems with an identical network structure. TWO-DIMENSIONAL TRUSS SYSTEMS Element 4, I ¼ 4, J ¼ 3: 2 !
P3 p4 ! ¼ ! e3 ; e4 "1 "E ¼ u d " u2 ! ¼ " p1 " p2 ! ¼ "e1 "e2 ! ¼ 0 10=kð1Þ ! Partitioning above after two rows and columns gives kð3Þ þ kð4Þ þ kð5Þ Àkð4Þ ! ! 10 Àkð5Þ 0 Àkð4Þ d ¼ À F 0 kð1Þ þ kð2Þ þ kð4Þ kð1Þ Àkð2Þ Letting ke ¼ 1 for e ¼ 1 to 5 and solving above gives ! ! 1, consist of bar elements positioned at arbitrary angles in space joined by pin-like joints that cannot transmit moments. In order to analyze such general truss 2 Recommended for Structural Mechanics Track. 12 A two-dimensional truss element in the local coordinate system x0e 1 , y1 .
17 is: k¼ 5Etab ; ða þ bÞl where E is the Young’s modulus and t is the width of the bar (Hint: subdivide the bar with a square hole into 3 elements). 2. 18. Nodes A and B are fixed. A force equal to 10 N acts in the positive x-direction at node C. Coordinates of joints are given in meters. Young’s modulus is E ¼ 1011 Pa and the cross-sectional area for all bars are A ¼ 2 Á 10À2 m2 . a. b. c. d. Number the elements and nodes. Assemble the global stiffness and force matrix. Partition the system and solve for the nodal displacements.
A First Course in Finite Elements [With CDROM] by Jacob Fish, Ted Belytschko