The Direct Stiffness Method Part I

Introduction to FEM

The Direct Stiffness Method Part I

IFEM Ch 2 – Slide 1

Introduction to FEM

The Direct Stiffness Method (DSM) Importance: DSM is used by all major commercial FEM codes A democratic method, works the same no matter what the element:

Bar (truss member) element, 2 nodes, 4 DOFs

Quadratic thin-plate element, 6 nodes, 12 DOFs

Tricubic brick element, 64 nodes, 192 DOFs

Obvious decision: use the truss to teach the DSM

IFEM Ch 2 – Slide 2

Introduction to FEM

Model Based Simulation (a simplification of diagrams of Chapter 1)

IDEALIZATION

Physical system

DISCRETIZATION

Mathematical model

FEM

SOLUTION

Discrete model

Discrete solution

Solution error Discretization + solution error Modeling + discretization + solution error VERIFICATION & VALIDATION

IFEM Ch 2 – Slide 3

Introduction to FEM

Idealization Process (a) Physical System

member support joint

IDEALIZATION

; ; ;

; ; ;

(b) Idealized Sytem: FEM-Discretized Mathematical Model

IFEM Ch 2 – Slide 4

Introduction to FEM

DSM: Breakdown Steps

;; ;;

;; ;;

FEM model:

Remove loads & supports:

Disassemble:

battens

Localize: Generic element:

IFEM Ch 2 – Slide 5

longerons diagonals longerons

Introduction to FEM

DSM: Assembly & Solution Steps Form elements:

battens

longerons diagonals longerons

Globalize:

Merge:

; ;

; ;

Apply loads and supports:

; ;

; ;

Solve for joint displacements:

IFEM Ch 2 – Slide 6

Introduction to FEM

The Direct Stiffness Method (DSM) Steps Starting with: Idealization Breakdown (Chapter 2)

Assembly & Solution (Chapter 3)

  Disconnection 

  

Localization Member (Element) Formation

Globalization Merge Application of BCs Solution Recovery of Derived Quantities

IFEM Ch 2 – Slide 7

Introduction to FEM

A Physical Plane Truss Typical of those used for building roofs and short span bridges. member support joint

Too complicated to do by hand. We will use a simpler one to illustrate DSM steps

IFEM Ch 2 – Slide 8

Introduction to FEM

The Example Truss: Physical and Pin-Jointed Idealization 3

Idealization as Pin-Jointed Truss and FEM Discretization

IFEM Ch 2 – Slide 9

2

; ;

; ;

1

Introduction to FEM

The Example Truss - FEM Model: Nodes, Elements and DOFs fy3 , uy3 3(10,10) E (3) = 100, A(3) =  2 2, (3)  (3) L = 10 2, ρ = 3/20

fx1 , ux1 1(0,0) fy1 , uy1

45o E (2) = 50, A(2) = 1, (2) L = 10, ρ(2) = 1/5

(3)

y 45 o

x

E (1) = (1)

fx3 , ux3

(2)

(1) (1)

50, A = 2, L = 10, ρ(1) = 1/5

fx2 , ux2 2(10,0) fy2 , uy2

IFEM Ch 2 – Slide 10

Introduction to FEM

The Example Truss - FEM Model BCs: Applied Loads and Supports (Saved for Last) f y3 = 1 3

(3)

x

;; ;; 1

(2)

(1)

;; ;;

y

f x3 = 2

IFEM Ch 2 – Slide 11

2

Introduction to FEM

Master (Global) Stiffness Equations     f=    Linear structure:    f x1 K x1x1  f y1   K y1x1     f  K  x2  =  x2x1  f  K  y2   y2x1  f  K x3 x3x1 f y3 K y3x1

Nodal forces

 u x1  u y1    u  x2  u= u   y2  u  x3 u y3 

 f x1 f y1   f x2   f y2   f x3  f y3 K x1y1 K y1y1 K x2y1 K y2y1 K x3y1 K y3y1

K x1x2 K y1x2 K x2x2 K y2x2 K x3x2 K y3x2

K x1y2 K y1y2 K x2y2 K y2y2 K x3y2 K y3y2

K x1x3 K y1x3 K x2x3 K y2x3 K x3x3 K y3x3

Master stiffness matrix or

f = Ku

IFEM Ch 2 – Slide 12

  K x1y3 u x1   K y1y3    u y1    K x2y3    u x2    K y2y3    u y2  K x3y3   u x3  K y3y3 u y3

Nodal displacements

Introduction to FEM

Member (Element) Stiffness Equations _

¯f = K u¯  f¯   K¯ xi xi xi K¯ yi xi  f¯yi     ¯ = ¯ fx j K x j xi f¯y j K¯ y j xi

K¯ xi yi K¯ yi yi K¯ x j yi K¯ y j yi

K¯ xi x j K¯ yi x j K¯ x j x j K¯ y j x j

IFEM Ch 2 – Slide 13

 K¯ xi y j  u¯ xi  K¯ yi y j  u¯ yi     K¯ x j y j  u¯ x j u¯ y j K¯ y j y j

Introduction to FEM

First Two Breakdown Steps: Disconnection and Localization 3

3

Remove loads and supports, _ and disconnect pins y_ (3) x (3)

_

x (2) (3)

y

(2) _

y (2)

_ (1)

y 2

; ;

; ;

1

1

_

x (1)

x (1)

These steps are conceptual (not actually programmed as part of the DSM)

IFEM Ch 2 – Slide 14

2

Introduction to FEM

The 2-Node Truss (Bar) Element

Equivalent spring stiffness k s = EA / L _ _ fyi, u yi _ _ fxi , uxi

i

_ y

_ _ fyj, u yj _ _ fxj , uxj

_ x

j

i

j

−F

F L

IFEM Ch 2 – Slide 15

d

Introduction to FEM

Truss (Bar) Element Formulation by Mechanics of Materials (MoM) F = ks d =

EA d, L

 1  f¯  xi EA  0  f¯yi  =   ¯  −1 fx j L 0 f¯y j from which

 1 EA  0 K=  −1 L 0

F = f¯x j = − f¯xi ,

d = u¯ x j − u¯ xi

Exercise 2.3 0 −1 0   u¯ xi  0 0 0   u¯ yi  Element stiffness   equations in local 0 1 0 u¯ x j coordinates 0 0 0 u¯ y j 0 0 0 0

−1 0 1 0

0 0  0 0

IFEM Ch 2 – Slide 16

Element stiffness matrix in local coordinates

Introduction to FEM

Where We Are So Far in the DSM we are done with this ...

  Disconnection Breakdown (Chapter 2)  Localization

Member (Element) Formation

we finish Chapter 2 with

Assembly & Solution (Chapter 3)

  

Globalization Merge Application of BCs Solution Recovery of Derived Quantities

IFEM Ch 2 – Slide 17

Introduction to FEM

Globalization: Displacement Transformation _

uyj

uyj

_

uxj _

_

y y

_

uyi

u yi

x _

uxj

ϕ

uxi

x

j

uxi

i Node displacements transform as

in which

u¯ xi = u xi c + u yi s,

u¯ yi = −u xi s + u yi cγ

u¯ x j = u x j c + u y j s,

u¯ y j = −u x j s + u y j cγ

c = cos ϕ

s = sin ϕ

IFEM Ch 2 – Slide 18

Introduction to FEM

Displacement Transformation (cont'd) In matrix form _

uxi _ u yi _ uxj _ uyj

or

c −s = 0 0

_e

u

s c 0 0

=

0 0 c −s

0 0 s c

Te u e

IFEM Ch 2 – Slide 19

uxi uyi uxj uyj Note: global on RHS, local on LHS

Introduction to FEM

Globalization: Force Transformation _

fyj

fyj

_

fxj

y

_

fyi

fyi

_

ϕ

fxi

x

i

fxi

Node forces transform as f xi c −s 0 s c 0 f yi = 0 0 c fx j 0 0 s fyj or

f

fxj

j

e

_

0 0 −s c

_ e T e

= (T ) f

IFEM Ch 2 – Slide 20

f_xi f_yi f_x j fyj

Note: global on LHS, local on RHS

Introduction to FEM

Globalization: Congruential Transformation of Element Stiffness Matrices

¯

¯

e

K ue = f

¯

e

¯

f e = ( T e) T f e

u e = T e ue

¯

e

K e = ( T e )T K T e

e e A E e K = Le

 c2  sc  2 −c −sc

sc s2 −sc −s 2

−c2 −sc c2 sc

IFEM Ch 2 – Slide 21

−sc  −s 2   sc s2

Exercise 2.8

Introduction to FEM

The Example Truss - FEM Model (Recalled for Convenience) fy3 , uy3 3(10,10) E (3) = 100, A(3) =  2 2, (3)  (3) L = 10 2, ρ = 3/20

fx1 , ux1 1(0,0) fy1 , uy1

45o E (2) = 50, A(2) = 1, (2) L = 10, ρ(2) = 1/5

(3)

y 45 o

x

E (1) = (1)

fx3 , ux3

(2) fx2 , ux2

(1) (1)

50, A = 2, L = 10, ρ(1) = 1/5

2(10,0) fy2 , uy2

IFEM Ch 2 – Slide 22

Insert the geometric & physical properties of this model into the globalized member stiffness equations

Introduction to FEM

We Obtain the Globalized Element Stiffness Equations of the Example Truss  f (1)   u (1)    1 0 −1 0 x1 x1 (1) (1)   f y1  u y1 0 0 0 0   = 10      (1)   f (1)  −1 0 1 0 u x2  x2 (1) 0 0 0 0 f y2 u (1) y2  u (2)   f (2)    0 0 0 0 x2 x2 (2) (2)  u y2  In the next class  f y2  0 1 0 −1        f (2)  = 5  0 0 0 0   u (2)  we will put these to good use x3 x3 (2) (2) 0 −1 0 1 f y3 u y3  (3)   f (3)   0.5 0.5 −0.5 −0.5  u x1 x1 (3)    u (3)  f y1 0.5 0.5 −0.5 −0.5   y1   = 20      (3)   f (3)  −0.5 −0.5 0.5 0.5 u x3 x3 (3) −0.5 −0.5 0.5 0.5 f y3 u (3) y3

IFEM Ch 2 – Slide 23