Method of Finite Elements I - ETH Zürich - Homepage · Method of Finite Elements I Direct Stiffness Method (DSM) • Computational method for structural analysis • Matrix method

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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



Goals of this Chapter

• DSM formulation

• DSM software workflow for …

• linear static analysis (1st order)

• 2nd order linear static analysis

• linear stability analysis



Computational Structural Analysis

Modelling is the most important step in

the process of a structural analysis !

X

Y

Physical problem Continuous

mathematical modelDiscrete

computational model

strong form weak form



System Identification (Modelling)

Global Coordinate System

Nodes

Elements

Boundary conditions

Loads

X

Y

1

3 4

2

5

6

1 2

3

4

Element numbers

and orientation

Node numbers

5 6



Deformations

System Deformations

System identification

Nodal Displacements

nodes, elements, loads and supports

deformed 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 direction

of local axis X

e =𝐷𝑋

𝐿

s = 𝐸 econst. equation

equilibrum

𝑁 = ʃ 𝐸 s = 𝐸𝐹 s =𝐸𝐹

𝐿𝐷𝑋

𝑃2 = −𝑃1 = 𝑁

𝐷𝑋 = (u2−u1)

P2P1

𝑃1 =𝐸𝐹

𝐿(u1−u2)

𝑃2 =𝐸𝐹

𝐿(−u1 + u2)

p = k u

p : (element) stiffness matrix

k : (element) nodal forces

u : (element) displacement vector

1 dof per node

DX

𝐿, 𝐸, 𝐹N

P1 P2

ux

X/Y = local coordinate system

DX = displacement of truss end



Elements: Beam3 dof per node

DX

DY

RZ

𝐿, 𝐸, 𝐹

uy

uy

ux

k u

ux = displacement in direction

of local axis X

uy = displacement in direction

of local axis Y



Elements: Global Orientation

local

global

uglob = u = R uloc

𝑅 𝜃 =

cos 𝜃 − sin 𝜃 0 0 0sin 𝜃 cos 𝜃 0 0 00 0 1 0 00 0 cos 𝜃 − sin 𝜃 00 0 sin 𝜃 cos 𝜃 00 0 0 0 1

𝜃

kglob = k = RT kloc R



Beam Stiffness Matrixe.g. k24 =

reaction

in global direction Y

at start node S

due to a

unit displacement

in global direction X

at end node E

UXE=1

FYS

S

EFXS =

FYS =

MZS =

FXS =

FYS =

MZE =

UXS UYS UZS UXE UYE UZE

k14 k15 k16

k24 k25 k26

k34 k35 k36

k44 k45 k46

k55 k56

k66

k11 k12 k13

k22 k23

k33

symm.

Element stiffness matrix

in global orientation

iE

iS

EEES

SESS

iE

iS

u

u

p

p

ii

ii

kk

kk

p = k u



Nodal Equilibrum

3 4

2

5

6

f4

r4: 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 tostart 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

2

5

6

1 2

3

4

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 for

a 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 forces

Calculate element end forces = p = k u

4. Element deformations along axis

1. Element nodal displacements

Disassemble u from resulting global displacements U

3. Element stress and strain along axis

Calculate moment/shear from end forces (equilibrium equation)

Calculate curvature/axial strain from moments/axial force

Calculate displacements from strain (direct integration)



Lateral Load

1. Adjust global load vector

f = local load vector => add to global load vector F

2. Adjust element stresses

M due to u

M due to f M diagrame.g. bending moment M:



Linear Static Analysis (1st order)

Workflow of computer program

1. System identification: Elements, nodes, support and loads

2. Build element stiffness matrices and load vectors

3. Assemble global stiffness matrix and load vector

4. Solve global system of equations (=> displacements)

5. Calculate element results

Exact solution for displacements and stresses



2nd Order Effectsor the influence of the axial normal force

Normal forces change the stiffness of the structure !



Geometrical Stiffness Matrix

kG = geometrical stiffness matrix of a truss element

p = ( k + kG ) u

Very small element rotation

=> Member end forces (=nodal forces p )

perpendicular to axis due to initial N

Truss

NOTE:

It’s only a

approximation



Beams: Geometrical Stiffness

kG = geometrical stiffness matrix of a beam element

kG =



Linear Static Analysis (2nd order)

Global system of equations

( K + KG ) U = F U = ( K + KG )-1 F

Inversion possible only if K + KG is non-singular, i.e.

- the structure is sufficiently supported (= stable)

- initial normal forces are not too big

What are the 2nd order nodal displacements for

a given structure due to a given load ?



Linear Static Analysis (2nd order)


1. Perform 1st order analysis

2. Calculate resulting axial forces in elements (=Ne)

3. Build element geometrical stiffness matrices due to Ne

4. Add geometrical stiffness to global stiffness matrix

5. Solve global system of equations (=> displacements)

6. Calculate element results

NOTE: Only approximate solution !



Stability Analysis

How much can a given load be increased until a

given structure becomes unstable ?

(K + λmax KG0) U = F

Nmax = λmax N0

KG = f(Nmax)KG(Nmax) = λmax KG(N0) = λmax KG0

2nd order analysis No additional load possible

(K + λmax KG0) ΔU = ΔF = 0

linear algebra

(A - λ B) x = 0 Eigenvalue problem



Stability Analysis

Eigenvalue problem

(A - λ B) x = 0

λ = eigenvalue

x = eigenvector

(K - λ KG0) x = 0

λ = critical load factor

x = buckling mode

e.g. Buckling of a column

λ N0

λ F

x

Solution



Stability Analysis


1. Perform 1st order analysis

2. Calculate resulting axial forces in elements (=N0)

3. Build element geometrical stiffness matrices due to N0

4. Add geometrical stiffness to global stiffness matrix

5. Solve eigenvalue problem

NOTE: Only approximate solution !

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Method of Finite Elements I - ETH Zürich - Homepage · Method of Finite Elements I Direct Stiffness Method (DSM) • Computational method for structural analysis • Matrix method