caed lec5 truss sept2014

Truss Element

Outline

Introduction to 2-D FE Analysis

Truss Bar problem

2-D Cantilever problem

Frame element

Objectives Understand what truss element is and its degree of

freedom

Understandhow to develop the transformation matrix

Know how to formulate the stiffness matrix of truss element

CAE (FEM) Syllabus Structure

1-D Problem

1-D element

-spring

-bar

2-D Problem

1-D element

-Truss

-Beam

-Frame

2-D element

-Linear

-Quadratic

Two-dimensional analysis allows us to

determine detailed information concerning

deformation, stress and strain, within a

complex shaped two-dimensional elastic body.

Why 2-D Analysis?

Introduction

In 1-D problems, we have equations in 1-D axis

Force

In x-Dir.

x

Force

In y-Dir.

y

Force1

In the bar longitudinal-Dir.

Force2

Introduction

E1 E2 100 N

Cantilever

(Variable or constant area)

force

Metal plate with cut

In all 1-D cases, we applied one equilibrium principles

at two node for each element.

And we have obtained 2 equations for each

Element.

Introduction

For a 1-D , the 2 equilibrium equations

are written in local element coordinates as:

e2

e1

e2

e1

F

F

u

u

kk

kk

ee

ee

}F{}u{]K[ eee

Element e

1e 2e

u1e

u2e

F1e F2

e

ke

As a reminder:

e2

e2

e1

e1

e2

e1

F)uu(k

F)uu(k

e

e

And transferred to Matrix, as:

IntroductionStep from

1-D to 2-D

From local coordinates to Global Coordinates

Global coordinates

Fixed in space

x`Local coordinates

Directed according

to the element direction

Y

X

Look, now for each node we will write 2 equilibrium equations

Hence, For each element, they will be 4 equations

y`

Introduction

Original before

deformation

After

deformation

In 1-D analysis, assuming small displacements and elastic material behavior we have:

a) A strain-displacement eqn.,

and,

b) A stress-strain eqn.

(1D Hooke’s Law)

dx

du

L

LL

0

0

E

Question: What is the 2-D or 3-D form of these equations?

2-D Stress Analysis

Three cases will be considered under

the 2-D simulation :

•Truss Problem

•Beam Problem

•Frame Problem

2-D FE Analysis

2-D Truss Bar Problem

Definition: Truss An engineering structure consisting of straight members

connected at their ends by means of bolts, rivets, pins or welding

Tension and compression capability only (no bending).

Degree of Freedom

Truss has tension and compression capability

No bending

Therefore, in 2-D problem, truss element has 2 dof at each node

Displacements in x and y directions

U1y

U2

x

U2y

q

U1x

ke

node 1

node 2

Typical plane 2-D trusses

A truss structure consists only of two-force members:

direct tension and compression.

Truss Bar Problem

What is the size of the element stiffness matrix for a 2-Dtruss element?

U1y

U2x

U2y

F1x

F1y

F2x

F2y

q

U1x

ke

node 1

node 2

Truss Bar Problem

The element stiffness matrix is now a 4 x 4 instead of a 2 x 2 in 1-D analysis, because each node can now move in the x- or y- direction. In this case each node has 2 DOFs.

U1y

U2x

U2y

F1x

F1y

F2x

F2y

q

U1x

ke

node 1

node 2

The four nodal equilibrium

equations for the element can be

written in matrix form as:

Where k11, …, k44 are sixteen

unknown stiffness coefficients

to be determined.

Truss Bar Problem

y2

x2

y1

x1

F

F

F

F

y2

x2

y1

x1

U

U

U

U

44434241

34333231

24232221

14131211

kkkk

kkkk

kkkk

kkkk

Xx

y

θ4(= 0)

Xθ1

X

θ2 (= 90)X

θ3

X

Y

Since there are many cases, I need to generalize

Truss Bar Problem

x

y

There are two frames of references for the truss problems:

• Global coordinate system (X,Y)

• Local coordinate system (x, y)

1. Choose the global (X,Y) system to represent the

location of each joint (or, node).

2. Track the orientation of each member (or, element)

by angle, θ.

3. Write the transformation of the displacement from

local to global

Truss Bar Problem

UiX

UiY

X

Yi

j

θ

Uiy

UiX

θ

UiX = uix cos θ - uiy sin θ

UiY = uix sin θ + uiy cos θ

UjX = ujx cos θ - ujy sin θ

UjY = ujx sin θ + ujy cos θ

UiX = uix cos θ - uiy sin θ

UiY = uix sin θ + uiy cos θ

UjX = ujx cos θ - ujy sin θ

UjY = ujx sin θ + ujy cos θ

In Matrix form: {U}=[T]{u}

Displacements of nodes

Truss Bar Problem

UjX

UjY

UiX

UiY

X

Yi

j

θ

Local displacements in Global CS

jy

jx

iy

ix

jyjx

jyjx

iyix

iyix

u

u

u

u

uu

uu

uu

uu

qq

qq

qq

qq

qq

qq

qq

qq

cossin00

sincos00

00cossin

00sincos

cossin00

sincos00

00cossin

00sincos

jX

jX

iY

iX

jX

jX

iY

iX

U

U

U

U

form;matrix in written be can equationslinear These

U

U

U

U

=> {U}=[T]{u}

[T] is called the transformation matrix

Is it only the displacement can be analyzed at the nodes??

The forces can be analyzed as well

Truss Bar Problem

i

jFjX

FjY

FiX

FiY

X

Y

θ FiX = fix cos θ - fiy sin θ

FiY = fix sin θ + fiy cos θ

FjX = fjx cos θ - fjy sin θ

FjY = fjx sin θ + fjy cos θ

In Matrix form: {F}=[T]{f}

Forces at nodes

Truss Bar Problem

Similar to displacement

jy

jx

iy

ix

jyjx

jyjx

iyix

iyix

f

f

f

f

ff

ff

uu

uu

qq

qq

qq

qq

qq

qq

qq

qq

cossin00

sincos00

00cossin

00sincos

cossin00

sincos00

00cossin

00sincos

jX

jX

iY

iX

jX

jX

iY

iX

F

F

F

F

form;matrix in written be can equationslinear These

F

F

U

U

=> {F}=[T]{f}

[T] is called the transformation matrix

y

x

(x2,y2)

(x1,y1)1

2

θ

φle

x2-x1

y2-y1

Direction cosines?

212

2

12 yyxxle q sincos 12

el

yym

qcosl

el

xxl 12cos

q

qsinmthen

Truss Bar Problem

Now, the Transformation matrix becomes

lm

ml

lm

ml

T

00

00

00

00

cossin00

sincos00

00cossin

00sincos

qq

qq

qq

qq

Local force-stiffness-displacement matrices

For 1-d problem:

jx

ix

jx

ix

u

u

kk

kk

f

f

For 2-d problem:

jy

jx

iy

ix

jy

jx

iy

ix

u

u

u

u

kk

kk

f

f

f

f

0000

00

0000

00

jy

jx

iy

ix

jy

jx

iy

ix

u

u

u

u

k

f

f

f

f

0000

0101

0000

0101k is defined as equivalent

stiffness

L

EAk

We want to relate the forces to the deformation by using the spring eqn.

What is the element (spring) equation?

f = k . u

If we have multi forces, multi displacements in the same element,

it should be written in matrix format

{f}= [k] {u}Remember: this is

in local coordinates

But, we have{f}=[T]¯¹{F} & {u}=[T]¯¹{U}

Truss Bar Problem

Compensate for {f} and {u}, getting:

[T]¯¹{F} = [k] [T]¯¹{U}

Multiply both sides by [T], getting:

{F} = [T][k] [T]¯¹{U}

where, [T][k] [T]¯¹= [K] , element stiffness matrix

This will result in the general 2-D element relation, as

Truss Bar Problem

{F} = [K] {U}

FiX

FiY

FjX

FjY

22

22

22

22

mlmmlm

lmllml

mlmmlm

lmllml

k

UiX

UiY

UjX

UjY

=

l

AEk ee

e

Force matrix

Displacement

matrix

Element stiffness matrix

Truss Bar Problem

After getting the forces and displacements, Further information:

• the Reaction forces

From the equilibrium equation applied on the element,

R = ku – f

Generally, for set of elements with 2-D,

{R} = [K]{U} – {F}

• the Stresses in the element

From the definition of the stress

σe = fe /Ae = ke u / Ae = ke (ui – uj)e / Ae

Truss Bar Problem

Reactions and Stresses

Reaction forces

FUKRG

y

x

y

x

y

x

y

x

y

x

y

x

y

x

y

x

y

x

y

x

Y

X

Y

X

Y

X

Y

X

Y

X

F

F

F

F

F

F

F

F

F

F

U

U

U

U

U

U

U

U

U

U

R

R

R

R

R

R

R

R

R

R

5

5

4

4

3

3

2

2

1

1

5

5

4

4

3

3

2

2

1

1

5

5

4

4

3

3

2

2

1

1

KG

Stress in each element

𝜎 =𝑖𝑛𝑡𝑒𝑟𝑛𝑎𝑙 𝑓𝑜𝑟𝑐𝑒

𝑎𝑟𝑒𝑎=𝑓

𝐴

𝜎 =𝑓

𝐴=𝑘(𝑢𝑖𝑥 − 𝑢𝑗𝑥)

𝐴=

𝐸𝐴𝐿 (𝑢𝑖𝑥 − 𝑢𝑗𝑥)

𝐴

𝜎 =𝐸(𝑢𝑖𝑥 − 𝑢𝑗𝑥)

𝐿

𝑈 = 𝑇 𝑢𝑢 = 𝑇 −1 𝑈𝑢𝑖𝑥𝑢𝑖𝑦𝑢𝑗𝑥𝑢𝑗𝑦

=

cos 𝜃 sin 𝜃 0−sin 𝜃 cos 𝜃 0

00

00

cos 𝜃−sin 𝜃

00

sin 𝜃cos 𝜃

𝑈𝑖𝑥

𝑈𝑖𝑦

𝑈𝑗𝑥𝑈𝑗𝑦

Any questions? Transformation matrix

Directional cosines

Element stiffness matrix (for Truss)

UjX = ujx cos θ - ujy sin θ

UjY = ujx sin θ + ujy cos θ

UiX = uix cos θ - uiy sin θ

UiY = uix sin θ + uiy cos θ

UjX = l ujx – m ujy

UjY = m ujx + l ujy

UiX = l uix - m uiy

UiY = m uix + l uiy

The 4 displacement equations

can be re-written as

In Matrix form: {U}=[T]{u}

or {u}=[T]¯¹{U}

In Matrix form, still same:

{U}=[T]{u}

or {u}=[T]¯¹{U}

Truss Bar Problem

FjX = fjx cos θ - fjy sin θ

FjY = fjx sin θ + fjy cos θ

FiX = fix cos θ - fiy sin θ

FiY = fix sin θ + fiy cos θ

FjX = l fjx – m fjy

FjY = m fjx + l fjy

FiX = l fix - m fiy

FiY = m fix + l fiy

The 4 force equations

are re-written as

In Matrix form: {F}=[T]{f}

or {f}=[T]¯¹{F}

In Matrix form, still same:

{F}=[T]{f}

or {f}=[T]¯¹{F}

Truss Bar Problem