method of finite elements i · method for . structural analysis • matrix method . for computing...

Institute of Structural Engineering

Method of Finite Elements I

Chapter 2

The Direct Stiffness Method




Direct Stiffness Method (DSM)• Computational method for structural analysis• Matrix method for computing the member forces

and displacements in structures• DSM implementation is the basis of most commercial

and open-source finite element software• Based on the displacement method (classical hand

method for structural analysis)• Formulated in the 1950s by Turner at Boeing and

started a revolution in structural engineering



Learning Goals of this Chapter• DSM formulation• DSM software workflow for …

• linear static analysis (1st order)• 2nd order linear static analysis• linear stability analysis

Lecture 2b



Computational Structural Analysis

Modelling is the most important step in the process of a structural analysis !

X

Y

Physical problem Continuousmathematical model

Discretecomputational model

strong form weak form



System Identification (Modelling)

Global Coordinate SystemNodesElementsBoundary conditionsLoads

X

Y

1

3 4

25

6

1 2

3

4

Element numbersand orientation

Node numbers

5 6



Deformations – Degrees of Freedom

System Deformations

System identification

Nodal Displacements

nodes, elements, loads and supportsdeformed shape

(deformational, nodal)degrees of freedom = dofs



Degrees of Freedom

7 * 2 = 14 dof

Frame Structure

8 * 3 = 24 dof

Truss Structure

ui = ( udx , udy )

uiui

ui = ( udx , udy , urz )dof per node

dof of structure



Elements: Truss

compatibility

ux = displacement in directionof local axis X

ε =𝐷𝐷𝐷𝐷𝐿𝐿

σ = 𝐸𝐸 εconst. equation

equilibrium

𝑁𝑁 = 𝐴𝐴σ=𝐴𝐴𝐸𝐸 ε =𝐴𝐴𝐸𝐸𝐿𝐿𝐷𝐷𝐷𝐷

𝐹𝐹2 = −𝐹𝐹1 = 𝑁𝑁

𝐷𝐷𝐷𝐷 = (u2 −u1)

F2F1

𝐹𝐹1 =𝐴𝐴𝐸𝐸𝐿𝐿

(u1 −u2)

𝐹𝐹2 =𝐴𝐴𝐸𝐸𝐿𝐿

(−u1 + u2)

f = k uk : (element) stiffness matrixf : (element) nodal forcesu : (element) displacement vector

1 dof per nodeDX

𝐿𝐿,𝐸𝐸,𝐴𝐴N

F1 F2

ux

X/Y = local coordinate system

DX = displacement of truss end1 1

2 2

EA EAF uL LF uEA EA

L L

− =

−



Elements: Beam3 dof per node

DX

DY

RZ

𝐿𝐿,𝐸𝐸,𝐴𝐴uy

uy

ux

k u

ux = displacement in directionof local axis X

uy = displacement in directionof local axis Y

3 2 3 2

2

0 0 0 0

12 6 12 6 0 0

6 4 0 0

EA EAL L

EI EI EI EIL L L LEI EI

LL

−

−

2

3 2 3 2

2

6 2

0 0 0 0

12 6 12 6 0 0

6 2 0

EI EILL

EA EAL L

EI EI EI EIL L L LEI EI

LL

−

−

− − −

1

2

3

4

5

26

6 4 0

u

u

u

u

uEI EI

uLL

−



The Beam Stiffness Matrix

B

A´

B´

υA

υB

MAB

VAB

VBA

MBA

φB

φA

A

3 2 3 2

2 2

3 2 3 2

2 2

12 6 12 6

6 4 6 2

12 6 12 6

6 2 6 4

AB

AB

BA

BA

EI EI EI EIL L L L

V EI EI EI EIM L L L LV EI EI EI EI

L L L LMEI EI EI EIL L L L

− − = − − −

−

A

A

B

B

υϕυϕ

υA υBφA φB

4 DOFs φA, φB, υA, υΒ



υA

3 2 3 2

2 2

3 2 3 2

2 2

12 6 12 6

6 4 6 2

12 6 12 6

6 2 6 4

AB

AB

BA

BA

EI EI EI EIL L L L

V EI EI EI EIM L LL LV EI EI EI EI

L L L LMEI EI EI EI

L LL L

− − − − = − − − −

A

A

B

B

υϕυϕ

υB

* υΑ

* υB




* φA

φA

3 2 3 2

2 2

3 2 3 2

2 2

12 6 12 6

6 4 6 2

12 6 12 6

6 2 6 4

AB

AB

BA

BA

EI EI EI EIL L L L



− − = − − −

−

A

A

B

B

υϕυϕ

* φB BφBA

4EI L

26EL L2EI L

26EL L

φB




3 2 3 2

2 2

3 2 3 2

2 2

12 6 12 6

6 4 6 2

12 6 12 6

6 2 6 4

AB

AB

BA

BA

EI EI EI EIL L L L



− − = − − −

−

A

A

B

B

υϕυϕ

direct

: Moment/Shear in that occurs for a rotation/displacement / acting on node : Moment/Shear in that occurs for a rotation/displacement / acting on node

ik k k

ki i i

t i kt k i

ϕ δ

ϕ δ

Coupling Stiffness Terms

coupling

: Moment/Shear in that is required for inducing in a rotation/displacement / 1 : Moment/Shear in that is required for inducing in a rotation/displacement / 1

ik i i

ki k k

s i is k k

ϕ δ =

ϕ δ =

Direct Stiffness Terms




Elements: Global Orientation

local

global

uloc = R uglob

uglob = u = RT uloc

𝜃𝜃

kglob = k = RT kloc R

cos sin 0 0 0 0sin cos 0 0 0 00 0 1 0 0 00 0 0 cos sin 00 0 0 sin cos 00 0 0 0 0 1

R

θ θθ θ

θ θθ θ

−

= −



Beam Stiffness Matrixe.g. k24 =

reactionin global direction Yat start node S

due to a

unit displacementin global direction Xat end node E

UXE=1

FYS

S

EFXS =

FYS =

MS =

FXS =

FYS =

ME =

UXS UYS RS UXE UYE RE

k14 k15 k16

k24 k25 k26

k34 k35 k36

k44 k45 k46

k55 k56

k66

k11 k12 k13

k22 k23

k33

symm.

Element stiffness matrixin global orientation

[ ] [ ][ ] [ ]

iS iSSS SE

iE iEES EE

f uf u

=

i i

i i

k kk k

f = k u



Nodal Equilibrum

3 4

2

5

6

f4r4: Vector of all forces acting at node 4

r4 = - k6ES u3 + contribution of element 6 due tostart node displacement u3

- k6EE u4 + contribution of element 6 due toend node displacement u4

- k5EE u4 + contribution of element 5 due toend node displacement u4

- k5ES u2 + contribution of element 5 due tostart node displacement u2

f4 external load

Equilibrum at node 4: r4 = - k5SE u2 -k6ES u3 - k5EE u4 - k6EE u4 + f4 = 0



Global System of Equations

r1 = -

u1

r2 = -

r3 = -

r4 = -

u2 u3 u4

k5ES k6ES k5EE+k6EE

1

3 4

25

6

1 2

34

k1EE+k3SS+k4SS

k3SE k4SE

k3ES k2EE+k3EE+k5SS

k5SE

k4ES k4EE+k6SS

k6SE

+ f1 = 0

+ f2 = 0

+ f3 = 0

+ f4 = 0

- K U + F = 0 F = K U



K = global stiffness matrix = Assembly of all ke

F = K U

Global System of Equations

= equilibrium at every node of the structure

F = global load vector = Assembly of all fe

U = global displacement vector = unknown



Solving the Equation System

K U = F

U = K-1 F

What are the nodal displacements fora given structure (= stiffness matrix K )due to a given load (= load vector F ) ?

K-1left multiply

K-1 K U = K-1 F

Inversion possible only if K is non-singular(i.e. the structure is sufficiently supported = stable)



Beam Element Results

2. Element end forcesCalculate element end forces = f = k u

4. Element deformations along axis

1. Element nodal displacementsDisassemble u (indiv. element displacement vectors) from resulting global displacements U

3. Element stress and strain along axisCalculate moment/shear from end forces (equilibrium equation)Calculate curvature/axial strain from moments/axial force

Calculate displacements from strain (direct integration)

Institute of Structural Engineering Page 22


PP/2

PL/8 P/2

PL/8

l/2 l/2

P/2

PL/8P/2

PL/8

DSM Gleichung

Pfx1 fy2

1 2P/2

PL/8

P/2

PL/8

ux1

uy1

ux2

uy2

φ2φ1

intu + PK = ffx1

fy2

Special Considerations

Treatment of Loads acting within Elements

In this case, we need to transfer the loads within the element to equivalent nodal loads, and accordingly modify the DSM equation:



PP/2

PL/8 P/2

PL/8

l/2 l/2

P/2

PL/8P/2

PL/8

P/2

PL/8

P/2

PL/8

ux1

uy2

ux2

uy2

φ2φ1

= int+ Pf Ku

fx1fy2Lokale Elementkräfte

0

2

80

2

8

P

Pl

P

Pl

= −

intP






ql/2

qL2/12 ql/2l/2 l/2

ql/2

ql/2

ql/2 ql/2

qL2/12

ux1

uy2

ux2

uy2

φ2φ1fx1

fy2Lokale Elementkräfte

2

2

0

2

120

2

12

ql

ql

ql

ql

= −

intP

q

qL2/12

qL2/12 qL2/12

qL2/12

= int+ Pf Ku






- Discretization- Individualization/Localization- Member Formulation

- Global Coordinates- Assembly- Boundary Conditions- Solution for Displacements- Element Moment/Shears

The DSM-Steps



Illustrative Example: Truss System

Step # 0: Idealized System with Loads and Supports

1 2

3

fy3=1

fx3=2

Loads

Supports

The DSM-Steps


Method of Finite Elements I@C. Felippa, Uni Colorado

Step #1: Define a global coordinate SystemEnumerate the Nodes, Elements and Degrees of Freedom

1 2

3

(3)(2)

(1)ux2, fx2

uy2, fy2

ux3, fx3

uy3, fy3

ux1, fx1

uy1, fy1

DOFs on the global system

x

y3

3 3

10 2

200 2

L

E A

=

=

2

2 2

1050

LE A

==

1

1 1

10100

LE A==


The DSM-Steps

https://www.colorado.edu/engineering/Aerospace/CAS/courses.d/IFEM.d/



Step #2: Decomposition and definition of Local Systems

1 2

3

(3) (2)

(1)

In this step we ignore external loads and supports and wedefine the local coordinate systems for every element

( ),i ix y

3x3y

2x

2y

1x1y


The DSM-Steps




Step #3: Definition of DOFs on the Local System

i j(e)

In this step, we define the DOFs for each element in their local system (according to each element type, e.g. beam or truss)

exey

, yyi iu f

, xxi iu f, yyj ju f

, xxj ju f


The DSM-Steps




Step #4: Define the Local Stifness Matrix

For a Truss Element:

@C. Felippa, Uni Colorado

1 0 1 00 0 0 01 0 1 0

0 0 0 0

xixi

yiyi

xjxj

yjyj

ufuf EAuf Luf

− = −

1 0 1 00 0 0 01 0 1 0

0 0 0 0

eEAL

− = −

KTruss Stiffness Matrix:

xiu yiu xju yju

xiu yiu xju yju

xiu

yiu

xju

yju


The DSM-Steps




Step #5: Global Coordinates – Rotation & Translation

@C. Felippa, Uni Coloradoi

j

exey

, yyi iu f

, xxi iu f

, yyj ju f , xxj ju f

xiu

yiu

xju

yju

( ) ( ), ,e ex y x y→

ϕ

cos sin 0 0sin cos 0 00 0 cos sin0 0 sin cos

xi xi

yi yi

xj xj

yj yj

u uu uu uu u

ϕ ϕϕ ϕ

ϕ ϕϕ ϕ

− =

−

& Te e e e e e= =

⇓

u R u u R u

Global

Local

Example: The truss element

In this step, we move from the local to the global coordinate system using the Rotation Matrix

The DSM-Steps




Step #5: Global Coordinates – Rotation & Translation

0 90

1 2

1 0 0 0 0 1 0 00 1 0 0 1 0 0 0

, 0 0 1 0 0 0 0 10 0 0 1 0 0 1 0

oϕ ϕ= =

− = =

−

R R

45

3

2 2 2 2 0 0

2 2 2 2 0 0

0 0 2 2 2 2

0 0 2 2 2 2

oϕ=

−

= −

R

For the elements of our simple example:


The DSM-Steps



Step #6: Transform the local stiffness matrix to global coordinates


1

e e e

e e e

Te e

e e e e e e e e

Te e e e e

Te e e e

−

==

=

= → =

→ = ⇒

=

u R uf R f

R R

K u f K R u R f

R K R u f

K R K R


The DSM-Steps




For our example:


1 2

3

(3) (2)

(1)

ux2, fx2

uy2, fy2

ux3, fx3

uy3, fy3

ux1, fx1

uy1, fy1

x

y

3

3 3

10 2

200 2

L

E A

=

=2

2 2

1050

LE A

==

1

1 1

10100

LE A==



The DSM-Steps



Method of Finite Elements I@C.

Fel

ippa

, Uni

Col

orad

o

Example:

1 =T1 1 1K R K R

2 2 2 2= TK R K R

3 3 3 3= TK R K R



The DSM-Steps




DisassemblyIndividualization/LocalizationMember Formulation

Global CoordinatesAssemblyApplication of BCsSolutionResponse Quantities

Back to the Summary – DSM Steps


What have we done so far?

Next Steps:




Step #7: Assembly - Bring together the individual element stiffness matrices Keinto the System's stiffness Matrix Ksys

In our example we have 6 DOFs

1 =K

2 =K

3 =K

ux1 uy1 ux2 uy2

ux2 uy2 ux3 uy3

ux1 uy1 ux3 uy3

Stiffness Matrix remains Symmetrical!(Maxwell-Betti Prinzip)

U = [ux1 uy1 ux2 uy2 ux3 uy3]T

1 2 3 4 5 6

u1 u2 u3 u4

u3 u4 u5 u6

u1 u2 u5 u6

u1 u2u3u4

u3 u4u5u6

u1 u2u5u6


The DSM-Steps



1 1 1 111 21 31 411 1 1 121 22 32 42

1 1 1 1 131 32 33 431 1 1 141 42 43 44

k k k kk k k kk k k kk k k k

=

K

ux2 uy2 ux3 uy3

ux1 uy1 ux3 uy3

2 2 2 233 43 53 632 2 2 243 44 54 64

2 2 2 2 253 54 55 652 2 2 263 64 65 66


=

K

3 3 3 311 21 51 613 3 3 321 22 52 62

3 3 3 3 351 52 55 653 3 3 361 62 65 66


=

K


1 2 3 4 5 6ux1 uy1 ux2 uy2




The DSM-Steps

Stiffness Matrix remains Symmetrical!(Maxwell-Betti Prinzip)



1 3 1 3 1 1 3 311 11 21 21 31 41 51 611 3 1 3 1 1 3 321 21 22 22 32 42 52 62

1 1 1 2 1 2 2 231 32 33 33 43 43 53 631 1 1 2 1 2 2 241 42 43 43 44 44 54 643 3 2 2 2 3 2 351 52 53 54 55 55 65 653 3 2 2 2 361 62 63 64 65 65

sys

k k k k k k k kk k k k k k k k

k k k k k k k kk k k k k k k kk k k k k k k kk k k k k k

+ ++ +

+ +=

+ ++ ++

K

2 366 66k k

+

ux1 uy1 ux2 uy2 ux3 uy3

1 2 3 4 5 6


Stiffness Matrix remains Symmetrical! (Maxwell-Betti Prinzip)




The DSM-Steps



10 10 0 10 10 0 10 100 10 0 10 0 0 10 10

10 0 10 0 0 0 00 0 0 0 5 0 510 10 0 0 0 10 0 1010 10 0 5 0 10 5 10

sys

+ + − − − + + − − − +

= + −

− − + + − − − + +

K

ux1 uy1 ux2 uy2 ux3 uy3

where sys sys=K u f




The DSM-Steps

Stiffness Matrix remains Symmetrical! (Maxwell-Betti Prinzip)



1

1

2

2

21

x

y

xsys

y

RRR

fR

=

Step #8: Assemnle the nodal loads Fi into the System-Load Vektor Fsys

1 2

3

fy3=1

fx3=2

ux3

uy3

ux2

uy2ux1

uy1

Rx2

Ry2Ry1

Rx1

ux1

uy1

ux2

uy2

ux3

uy3


The DSM-Steps



Step #9: Impose the Boundar Conditions by eliminating the corresponding Rows and Columns of the fixed (restrained) DOFs from the system‘s Stiffness Matrix, and the system load vector.

ff fs f f

sys sys sys fs ss s s

= ⇒ =

K K u fK u f

K K u f

Why can we do this?

Denote with s the fixed DOFs and with f the fre DOFs, then theglbal equilibrium equations can be rewritten as:.

But for restrained DOFs we have: s =u 0

0

ff fs f f ff f f

sf ss s sf f s

== ⇒ =

K K u f K u fK K f K u f

This means we can solve the equation only for the free DOFs


The DSM-Steps



1

1

2

2

21

x

y

xsys

y

RRRR

=

f

1

1

2

2

3

3

x

y

xsys

y

x

y

uuuuuu

=

u

10 10 0 10 10 0 10 100 10 0 10 0 0 10 10

10 0 10 0 0 0 00 0 0 0 5 0 510 10 0 0 0 10 0 1010 10 0 5 0 10 5 10

sys

+ + − − − + + − − − +

= + −

− − + + − − − + +

K

ux1 = uy1 = ux2 = uy2 = 0

ffKfsK

sfKssK



The DSM-Steps



1

1

2

2

21

x

y

xsys

y

RRRR

=

f

1

1

2

2

3

3

x

y

xsys

y

x

y

uuuuuu

=

u

10 10 0 10 10 0 10 100 10 0 10 0 0 10 10

10 0 10 0 0 0 00 0 0 0 5 0 510 10 0 0 0 10 0 1010 10 0 5 0 10 5 10

sys

+ + − − − + + − − − +

= + −

− − + + − − − + +

K

ux1 = uy1 = ux2 = uy2 = 0



The DSM-Steps



21f

=

f

Step #10: Solution: Solve for the unknow displacements uff :

3

3

xf

y

uu

=

u 10 1010 15ff

= ⇒

K 1ff f f f ff f

−= ⇒ =K u f u K f

1 1für a b d bc d c aad bc

− − = ⇒ = −−

A AReminder: Inverse of a 2x2 Matrix:

Hence 1

3

3

15 10 2110 10 1150 100

0.40.2

f ff f

xf

y

uu

− − = = −−

⇒ = =

u K f

u

0000

0.40.2

sys

⇒ =

u


The DSM-Steps



Step #11: Back-Calculate the Individual Element forces(a) Reorganize the system (global) displacements usys into the respective elemental vectors ue

1 2 1

1 2 11 2 3

2 3 3

2 3 3

0 0 00 0 0

, , ,0 0.4 0.40 0.2 0.2

x x x

y y y

x x x

y y y

u u uu u uu u uu u u

= = = = = =

u u u


The DSM-Steps




The DSM-Steps

Step #11: Back Calculate the Element Forces(b) Rotation from global back to local coordinates

e e e=u R u

1 1 1 2 2 2

3 3 3

0 0 1 0 0 0 00 1 0 0 0 0 0

, 0 0 0 0 1 0.4 0.20 0 0 1 0 0.2 0.4

2 2 2 2 0 0 0 02 2 2 2 0 0 0 0

0.4 0.420 0 2 2 2 20.2 0.140 0 2 2 2 2

− = = = = =

− − − = = = − −

u R u u R u

u R u



Step #11: Back Calculate the Element Forces(b) Calculate the element forces

1 1 1

2 2 2

3 3 3

01 0 1 0 0 10 0 0 0 0 0

51 0 1 0 0.2 1

0 0 0 0 0.4 0

1 0 1 0 0 8.40 0 0 0 0 0

201 0 1 0 0.42 8.4

0 0 0 0 0.14 0

= =

− − = = = −

− − −

= = = −

−

F K u

F K u

F K u

eF


The DSM-Steps



Summary: Linear Static Analysis (1st order)

Workflow of computer program

1. System identification: Elements, nodes, support and loads2. Build element stiffness matrices and load vectors3. Assemble global stiffness matrix and load vector4. Solve global system of equations (=> displacements)5. Calculate element results

Exact solution for displacements and stresses

method of finite elements i · method for . structural analysis • matrix method . for computing...

Documents