The Direct Stiffness Method: Assembly and Solution IFEM Ch 3 - PDF document

Department of Engineering Mechanics PhD. TRUONG Tich Thien Introduction to FEM The Direct Stiffness Method: Assembly and Solution IFEM Ch 3 – Slide 1

Department of Engineering Mechanics PhD. TRUONG Tich Thien Introduction to FEM The Direct Stiffness Method (DSM) Steps (recalled for convenience)  Disconnection  Breakdown Localization Member (Element) Formation (Chapter 2)   Globalization Merge  Assembly & Application of BCs Solution Solution  (Chapter 3) Recovery of Derived Quantities post-processing processing conceptual steps steps steps IFEM Ch 3 – Slide 2

Department of Engineering Mechanics PhD. TRUONG Tich Thien Introduction to FEM Globalization Step: Displacement Transformation u yj u yj ¯ u x j ¯ y ¯ u x j x ¯ j y u yi u xi ¯ ϕ u yi ¯ x u xi i Node displacements transform as u xi = u xi c + u yi s , ¯ u yi = − u xi s + u yi c γ ¯ u x j = u x j c + u yj s , u yj = − u x j s + u yj c γ ¯ ¯ in which c = cos ϕ s = sin ϕ IFEM Ch 3 – Slide 3

Department of Engineering Mechanics PhD. TRUONG Tich Thien Introduction to FEM Globalization Step: Displacement Transformation (cont'd) In matrix form u xi c s 0 0 u xi ¯       u yi ¯ − s c 0 0 u yi  =       0 0 u x j ¯ c s u x j      Note: u yj ¯ 0 0 − s c u yj global on RHS, local on LHS u ( e ) = T ( e ) u ( e ) or ¯ IFEM Ch 3 – Slide 4

Department of Engineering Mechanics PhD. TRUONG Tich Thien Introduction to FEM Globalization Step: Force Transformation f yj ¯ f yj ¯ f x j f x j j y f yi ¯ f yi ¯ ϕ f xi x f xi i Node forces transform as ¯ f xi   f xi c − s 0 0     Note: ¯ f yi f yi s c 0 0   global on LHS,  =       ¯ f x j 0 0 c − s f x j    local on RHS   ¯ f yj 0 0 s c f yj ( e ) f ( e ) = ( T ( e ) ) T ¯ f IFEM Ch 3 – Slide 5

Department of Engineering Mechanics PhD. TRUONG Tich Thien Introduction to FEM Globalization: Congruential Transformation of Element Stiffness Matrices ( e ) ( e ) ¯ ¯ u ( e ) ¯ f K = f ( e ) = ( T ( e ) ) T f ( e ) u ( e ) = T ( e ) u ( e ) ¯ ¯ Exercise 3.1 ) T ¯ ( e ) T K ( e ) = ( T ( e ) ) ( e K c 2 − c 2 sc − sc   K ( e ) = E ( e ) A ( e ) s 2 − s 2 sc − sc   − c 2 c 2 L ( e ) − sc sc   − s 2 s 2 − sc sc IFEM Ch 3 – Slide 6

Department of Engineering Mechanics PhD. TRUONG Tich Thien Introduction to FEM The Example Truss - FEM Model f y 3 , u y 3 Recall from Chapter 2 f x 3 , u x 3 3 √ L ( 3 ) = 10 2 √ y E ( 3 ) A ( 3 ) = 200 2 L ( 2 ) = 10 ( 3 ) ( 2 ) E ( 2 ) A ( 2 ) = 50 x f x 1 , u x 1 f x 2 , u x 2 1 2 ( 1 ) L ( 1 ) = 10 E ( 1 ) A ( 1 ) = 100 f y 2 , u y 2 f y 1 , u y 1 IFEM Ch 3 – Slide 7

Department of Engineering Mechanics PhD. TRUONG Tich Thien Introduction to FEM Globalized Element Stiffness Equations for Example Truss f ( 1 ) u ( 1 )     1 0 − 1 0   x 1 x 1 f ( 1 ) u ( 1 ) 0 0 0 0 y 1 y 1      = 10       f ( 1 ) u ( 1 ) − 1 0 1 0      x 2 x 2 f ( 1 ) u ( 1 ) 0 0 0 0 y 2 y 2 f ( 2 ) u ( 2 )     0 0 0 0   x 2 x 2 f ( 2 ) u ( 2 ) 0 1 0 − 1 y 2 y 2      = 5       f ( 2 ) u ( 2 ) 0 0 0 0      x 3 x 3 f ( 2 ) u ( 2 ) 0 − 1 0 1 y 3 y 3 f ( 3 ) u ( 3 )     0 . 5 0 . 5 − 0 . 5 − 0 . 5 x 1   x 1 f ( 3 ) u ( 3 ) 0 . 5 0 . 5 − 0 . 5 − 0 . 5 y 1 y 1      = 20       f ( 3 ) u ( 3 ) − 0 . 5 − 0 . 5 0 . 5 0 . 5      x 3 x 3 f ( 3 ) u ( 3 ) − 0 . 5 − 0 . 5 0 . 5 0 . 5 y 3 y 3 IFEM Ch 3 – Slide 8

Department of Engineering Mechanics PhD. TRUONG Tich Thien Introduction to FEM Assembly Rules 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 augment the element stiffness equations as shown for the example truss in the next slide. IFEM Ch 3 – Slide 9

Department of Engineering Mechanics PhD. TRUONG Tich Thien Introduction to FEM Expanded Element Stiffness Equations of Example Truss f ( 1 ) u ( 1 )      10 0 − 10 0 0 0  x 1 x 1 f ( 1 ) u ( 1 ) 0 0 0 0 0 0     y 1 y 1       f ( 1 ) u ( 1 )   − 10 0 10 0 0 0      x 2  x 2 =     f ( 1 )   u ( 1 )   0 0 0 0 0 0     y 2 y 2        f ( 1 )  0 0 0 0 0 0  u ( 1 )        x 3 x 3 0 0 0 0 0 0 f ( 1 ) u ( 1 ) y 3 y 3 f ( 2 ) u ( 2 )       0 0 0 0 0 0 x 1 x 1 f ( 2 ) u ( 2 ) 0 0 0 0 0 0  y 1   y 1        f ( 2 ) u ( 2 )   0 0 0 0 0 0     x 2   x 2 =     f ( 2 )   u ( 2 ) 0 0 0 5 0 − 5       y 2 y 2        f ( 2 )  0 0 0 0 0 0  u ( 2 )        x 3 x 3 0 0 0 − 5 0 5 f ( 2 ) u ( 2 ) y 3 y 3 f ( 3 ) u ( 3 )       10 10 0 0 − 10 − 10 x 1 x 1 f ( 3 ) u ( 3 ) 10 10 0 0 − 10 − 10  y 1   y 1        f ( 3 ) u ( 3 )   0 0 0 0 0 0     x 2   x 2 =     f ( 3 )   u ( 3 ) 0 0 0 0 0 0       y 2 y 2        f ( 3 )  − 10 − 10 0 0 10 10  u ( 3 )        x 3 x 3 − 10 − 10 0 0 10 10 f ( 3 ) u ( 3 ) y 3 y 3 IFEM Ch 3 – Slide 10

Department of Engineering Mechanics PhD. TRUONG Tich Thien Introduction to FEM Reconnecting Members by Enforcing Compatibility Rule f ( 1 ) u x 1       10 0 − 10 0 0 0 x 1 To apply compatibility, drop f ( 1 ) u y 1 0 0 0 0 0 0  y 1    the member index from the       f ( 1 )   u x 2 − 10 0 10 0 0 0     x 2   nodal displacements =     f ( 1 )   0 0 0 0 0 0    u y 2    y 2        f ( 1 )  0 0 0 0 0 0     u x 3     x 3 0 0 0 0 0 0 f ( 1 ) u y 3 y 3 f ( 1 ) K ( 1 ) u = f ( 2 ) u x 1      0 0 0 0 0 0  x 1 f ( 2 ) u y 1 0 0 0 0 0 0     y 1       f ( 2 )   0 0 0 0 0 0 u x 2       x 2 = f ( 2 ) K ( 2 )     = f ( 2 )   u 0 0 0 5 0 − 5    u y 2    y 2       0 0 0 0 0 0  f ( 2 )    u x 3       x 3 0 0 0 − 5 0 5 f ( 2 ) u y 3 y 3 f ( 3 ) K ( 3 ) f ( 3 ) u = u x 1      10 10 0 0 − 10 − 10  x 1 f ( 3 ) u y 1 10 10 0 0 − 10 − 10  y 1          f ( 3 )   0 0 0 0 0 0 u x 2     x 2   =       f ( 3 ) 0 0 0 0 0 0    u y 2    y 2         − 10 − 10 0 0 10 10   f ( 3 ) u x 3       x 3 − 10 − 10 0 0 10 10 f ( 3 ) u y 3 y 3 IFEM Ch 3 – Slide 11

Department of Engineering Mechanics PhD. TRUONG Tich Thien Introduction to FEM Next, Apply Equilibrium Rule f 3 3 3 f ( 3 ) − f ( 3 ) 3 − f ( 2 ) f ( 2 ) 3 3 3 (3) (2) Be careful with + directions of internal forces! Applying this to all joints (see Notes): f = f + f + f (2) (3) (1) IFEM Ch 3 – Slide 12

Department of Engineering Mechanics PhD. TRUONG Tich Thien Introduction to FEM Forming the Master Stiffness Equations through Equilibrium Rule f = f + f + f = (K + K + K ) u = K u (2) (2) (3) (1) (3) (1)  f x 1   20 10 − 10 0 − 10 − 10   u x 1  f y 1 10 10 0 0 − 10 − 10 u y 1             f x 2 − 10 0 10 0 0 0 u x 2       =       f y 2 0 0 0 5 0 − 5 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 13

Department of Engineering Mechanics PhD. TRUONG Tich Thien Introduction to FEM Applying Support and Loading Boundary Conditions to Example Truss 1 3 2 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 14