The Direct Stiffness Method Part II IFEM Ch 3 Slide 1 - PDF document

Introduction to FEM The Direct Stiffness Method Part II IFEM Ch 3 – Slide 1

Introduction to FE The Direct Stiffness Method (DSM) Steps (repeated here for convenience)  Disconnection  Localization Breakdown Member (Element) Formation   Globalization Merge  Assembly & Application of BCs Solution Solution  Recovery of Derived Quantities post-processing processing conceptual steps steps steps IFEM Ch 3 – Slide 2

Introduction to FEM Rules That Govern Assembly 1. Compatibility : The joint displacements of all members meeting at a joint must be the same 2. Equilibrium : The sum of forces exerted by all members that meet at a joint must balance the external force applied to that joint. To apply these rules in assembly by hand , it is convenient to expand or augment the element stiffness equations as shown for the example truss on the next slide. IFEM Ch 3 – Slide 3

Introduction to FEM Expanded Element Stiffness Equations of Example Truss (1) (1) f x 1 u x 1      10 0 − 10 0 0 0  (1) (1) f y 1 u y 1 0 0 0 0 0 0           (1) (1)   f x 2 − 10 0 10 0 0 0 u x 2       =       (1) (1) 0 0 0 0 0 0  f y 2   u y 2           (1)  0 0 0 0 0 0   (1) f x 3 u x 3       0 0 0 0 0 0 (1) (1) f y 3 u y 3 (2) (2) f x 1 u x 1      0 0 0 0 0 0  (2) (2) f y 1 u y 1 0 0 0 0 0 0           (2) (2)   f x 2 0 0 0 0 0 0 u x 2       =       (2) (2)  f y 2  0 0 0 5 0 − 5  u y 2           (2)  0 0 0 0 0 0   (2) f x 3 u x 3       0 0 0 − 5 0 5 (2) (2) f y 3 u y 3 (3) (3) f x 1 u x 1      10 10 0 0 − 10 − 10  (3) (3) f y 1 u y 1 10 10 0 0 − 10 − 10           (3)   (3) f x 2 0 0 0 0 0 0 u x 2       =     (3)   (3)  f y 2  0 0 0 0 0 0  u y 2          (3)   − 10 − 10 0 0 10 10  (3)  f x 3 u x 3       − 10 − 10 0 0 10 10 (3) (3) f y 3 u y 3 IFEM Ch 3 – Slide 4

Introduction to FEM Reconnecting Members by Enforcing Compatibility Rule (1) f x 1 u x 1      10 0 − 10 0 0 0  To apply compatibility, drop (1) f y 1 u y 1 0 0 0 0 0 0       the member index from the     (1)   f x 2 − 10 0 10 0 0 0 u x 2       =     nodal displacements (1)   0 0 0 0 0 0  f y 2   u y 2          0 0 0 0 0 0  (1)    f x 3 u x 3       0 0 0 0 0 0 (1) f y 3 u y 3 (1) (1) u (2) f = K f x 1 u x 1       0 0 0 0 0 0 (2) f y 1 u y 1 0 0 0 0 0 0           (2)   f x 2 u x 2 0 0 0 0 0 0         =     (2) (2) (2) 0 0 0 5 0 − 5  f y 2   u y 2  f   = K u         0 0 0 0 0 0   (2) f x 3   u x 3     0 0 0 − 5 0 5 (2) f y 3 u y 3 (3) (3) (3) f x 1 u x 1       f K u 10 10 0 0 − 10 − 10 = (3) f y 1 u y 1 10 10 0 0 − 10 − 10           (3)   f x 2 u x 2 0 0 0 0 0 0         =   (3)   0 0 0 0 0 0  f y 2   u y 2           (3)  − 10 − 10 0 0 10 10   f x 3   u x 3     − 10 − 10 0 0 10 10 (3) f y 3 u y 3 IFEM Ch 3 – Slide 5

Introduction to FEM Next, Apply Equilibrium Rule 3 f 3 3 (3) f 3 (2) f 3 (3) − f 3 (2) (3) − f 3 (2) Be careful with + directions of internal forces! Applying this to all joints (see Notes): f = f + f + f (2) (1) (3) IFEM Ch 3 – Slide 6

Introduction to FEM Forming the Master Stiffness Equations through Equilibrium Rule f = f + f + f = (K + K + K ) u = K u (2) (3) (1) (2) (3) (1)       f x 1 20 10 − 10 0 − 10 − 10 u x 1 10 10 0 0 − 10 − 10 f y 1 u y 1             f x 2 − 10 0 10 0 0 0 u x 2       =       0 0 0 5 0 − 5 f y 2 u y 2             f x 3 − 10 − 10 0 0 10 10 u x 3       f y 3 − 10 − 10 0 − 5 10 15 u y 3 IFEM Ch 3 – Slide 7

Introduction to FEM Applying Support and Loading Boundary Conditions to Example Truss 1 2 3 Recall: Displacement BCs: u x 1 = u y 1 = u y 2 = 0 Force BCs: f x 2 = 0 , f x 3 = 2 , f y 3 = 1 1 2 �� �� �� �� IFEM Ch 3 – Slide 8

Introduction to FEM Where Do Boundary Conditions Go? u x 1 = u y 1 = u y 2 = 0 Recall f x 2 = 0 , f x 3 = 2 , f y 3 = 1       f x 1 20 10 − 10 0 − 10 − 10 u x 1 10 10 0 0 − 10 − 10 f y 1 u y 1             f x 2 − 10 0 10 0 0 0 u x 2       =       0 0 0 5 0 − 5 f y 2 u y 2             f x 3 − 10 − 10 0 0 10 10 u x 3       − 10 − 10 0 − 5 10 15 f y 3 u y 3 IFEM Ch 3 – Slide 9

Introduction to FEM Reduced Master Stiffness Equations for Hand Computation Strike out rows and columns pertaining to known displacements:         10 0 0 u x 2 f x 2 0  =  = 0 10 10 u x 3 f x 3 2       0 10 15 u y 3 f y 3 1 Reduced or ^ ^ ^ K u = f stiffness equations Solve by Gauss elimination for unknown node displacements IFEM Ch 3 – Slide 10

Introduction to FEM Solve for Unknown Node Displacements and Complete the Displacement Vector     u x 2 0  = u x 3 0 . 4    u y 3 − 0 . 2 Expand with known displacement BCs   0 0     0 u =     0     0 . 4   − 0 . 2 IFEM Ch 3 – Slide 11

Introduction to FEM Recovery of Node Forces Including Reactions  20 10 − 10 0 − 10 − 10   0    − 2 10 10 0 0 − 10 − 10 0 − 2             − 10 0 10 0 0 0 0 0       f = Ku = =       0 0 0 5 0 − 5 0 1             − 10 − 10 0 0 10 10 0 . 4 2       − 10 − 10 0 − 5 10 15 − 0 . 2 1 1 3 2 Reaction Forces �� �� 1 �� �� 2 IFEM Ch 3 – Slide 12

Introduction to FEM Recovery of Internal Forces (Axial Forces in Truss Members) For each member (element) e = (1), (2), (3) 3 1. extract u from u e (3) F 2. transform to local (element) displacements (2) F _ e e e u = T u (1) F 1 2 d e = ¯ u e u e x j − ¯ 3. compute elongation xi direction of arrows F e = E e A e e is for +F (tension) L e d e 4. compute axial force See Example 3.1 of Notes for a detailed calculation IFEM Ch 3 – Slide 13

Introduction to FEM Computer Oriented Assembly and Solution in Actual FEM Codes (delayed until Part III of course) K stored in special sparse format (for example "skyline format") Assembly done by "freedom pointers" (Sec 3.5.1) Equations for supports are not physically deleted (Sec 3.5.2) Next slide explains this for the example truss IFEM Ch 3 – Slide 14

Introduction to FEM Computer Oriented Modification of Master Stiffness Equations (delayed until Part III of course) Recall u x 1 = u y 1 = u y 2 = 0 (freedoms 1, 2, 4) f x 2 = 0 f x 3 = 2 f y 3 = 1 , ,      20 10 − 10 0 − 10 − 10 f x 1  0 10 10 0 0 − 10 − 10 0 f y 1             − 10 0 10 0 0 0 u x 2 0       =       0 0 0 5 0 − 5 f y 2 0             − 10 − 10 0 0 10 10 u x 3 2       − 10 − 10 0 − 5 10 15 u y 3 1 zero out rows and columns 1, 2 and 4 store 1's on diagonal IFEM Ch 3 – Slide 15

Introduction to FEM Computer Oriented Modification of Master Stiffness Equations (delayed until Part III of course) 1 0 0 0 0 0 u x 1 0       0 1 0 0 0 0 u y 1 0       0 0 10 0 0 0 u x 2 0       =       0 0 0 1 0 0 u y 2 0             0 0 0 0 10 10 u x 3 2 0 0 0 0 10 15 u y 3 1 ^ ^ Modified master same u as in K u = f stiffness equations original equations IFEM Ch 3 – Slide 16