section 6: truss elements, local & global coordinates

Section 6: TRUSS ELEMENTS, LOCAL & GLOBAL COORDINATES

Introduction

The principles for the direct stiffness method are now in place. In this section of notes we will derive the stiffness matrix, both local and global, for a truss element using the direct stiffness method. Here a local coordinate system will be utilized initially anddirect stiffness method. Here a local coordinate system will be utilized initially and the element stiffness matrix will be transformed into a global coordinate system that is convenient for the overall structure.

Inclined or skewed supports will be discussed.pp


The truss element shown below is assumed to have a constant cross sectional area (A), a modulus of elasticity (E), and an initial length (L). The nodal degrees of freedom are the axial displacements directed along the length of the truss element. In addition the followingaxial displacements directed along the length of the truss element. In addition the following assumptions are made:

1. The truss element cannot sustain shear forces

2 A ff t f t di l t i d

xd2ˆ

d̂

2. Any effect of transverse displacements are ignored

3. Hooke’s law applies

Deformed shape

x2xd1

f̂f̂Node(a truss hinge)

x

Element

xf2xf1


With truss elements we again assume just as we did with spring and axial force elements that

ˆˆ

This is an approximation function. We are assuming through the use of this function that displacements are distributed in an approximately linear fashion along the element

xaau ˆˆ 21 +=

that displacements are distributed in an approximately linear fashion along the element.

( )xu ˆˆ

x̂As noted earlier the total number of unknown coefficients must equal the degrees of freedom associated with the element.

i h i l f l d i l l l di ffi iWith axial force elements and spring elements local coordinate axes were sufficient to formulate most problems. This does not work for truss elements where careful attention must be paid to local and global coordinate orientations.


Transformation of Vectors

In nearly all finite element analyses it is a necessity to introduce both local (to the element) and global (to the component) coordinate axes.

P

Y

Gl b l

XZ

Global axes


Looking at forces, the transformation from local to global coordinates is as follows:

fGlobal Local

fY

−=

y

x

Y

X

ff

ff

θθθθ

cossinsincos

Wh li d t t l t

fX

θ

ff ˆ00i θθ

GlobalLocal

When applied to a truss element:

fff

fff

x

y

x

X

Y

X

ˆˆˆ

sincos0000cossin00sincos

2

1

1

2

1

1

−

−

=

θθθθθθ

f2Y

{ } [ ] { }ff

ff yY

ˆ*T

cossin00 22

=

θθ

f1Y f2X

ˆ

yf 2̂xf 2̂

2 { } [ ] { }ff

f1Xθ

xf1̂yf1

1Note: Local x-axis is always along the element and θ is always measured counterclockwise from global x-axis to local x-axis.


From Statics we learned that a truss component is a two force member, i.e., this is an element that will not sustain shear forces: f2Y

f1Y f2X

02̂ =yfxf 2̂

2

f1Xθ

xf1̂01̂ =yf

1

2

The forces at either end must be directed along the truss component, and

ˆ00i ff θθ

Global Local

1

−

−

=

ˆ0

i00sincos0000cossin00sincos

2

1

2

1

1

x

x

X

Y

X

f

f

ffff

θθθθ

θθθθ

0cossin002Yf θθ


This leads to the following system of equations

000ˆcos 11 ++−=X ff θ

0ˆcos00

000ˆsin

000cos

22

11

11

−++=

+++=

++

xX

xY

xX

ff

ff

ff

θ

θ

θ

and the solution leads to the following matrix formulation

0ˆsin00 22 +++= xY ff θ

g

Y

X

x ff

f 1

1

1 00sincosˆ θθ

=

Y

Xx

fff

2

22sincos00ˆ θθ


A similar relationship holds for nodal displacements between local and global coordinate systems.

d2Y

d̂d1Y d2X

xd2ˆyd2

ˆd1̂

2

d1X

dd

dd xX

ˆˆ

00cossin00sincos 11

−

θθθθ

θxd1

yd1 Global Local

1

ddd

ddd

y

x

y

Y

X

Y

ˆˆ

cossin00sincos0000cossin

2

2

1

2

2

1

−=

θθθθ

θθ

{ } [ ] { }dd ˆ*T=


Inverting this last matrix expression yields:

ˆ

−

=

X

Y

X

x

y

x

ddd

ddd

2

1

1

2

1

1

sincos0000cossin00sincos

ˆˆ

θθθθθθ

or

−

Y

X

y

x

dd 2

2

2

2

cossin00ˆ θθ

It becomes obvious that:

{ } [ ] { }dd Tˆ =

Similarly

[ ] [ ] 1*TT −=

S a y

{ } [ ] { }ff Tˆ =


The element stiffness matrix can now be formulated in terms of the global coordinate system as follows.

df ˆ0101ˆ

In local coordinates Element nodal forces and displacements in local coordinates

−

−

=

x

y

x

x

y

x

ddd

LEA

fff

2

1

1

2

1

1

ˆˆ

010100000101

ˆˆ { } [ ] { }df ˆk̂ˆ =→

y

x

y

x

df 2

2

2

2ˆ0000ˆ Element stiffness matrix

in local coordinates

With

{ } [ ] { }dd Tˆ = { } [ ] { }ff Tˆ =


then { } [ ] { }[ ] { } [ ] [ ] { }df

df

=

=

Tk̂T

ˆk̂ˆ

[ ] { } [ ] [ ] { }

{ } [ ] [ ] [ ] { }

{ } [ ] { }dkf

df

df

=

=

− Tk̂T

TkT1

Thus the element stiffness matrix in terms of global coordinates is

{ } [ ] { }dkf =

[ ] [ ] [ ][ ]ˆ1

and in a matrix format the element stiffness matrix is

[ ] [ ] [ ][ ]TT 1 kk −=

θθθθθθ 22 sincoscossincoscos

[ ] [ ] [ ][ ]

−−−−

−−

== −

θθθθθθθθθθθθθθθθθθ

22

221

sincoscossincoscossinsincossinsincos

sincoscossincoscos

TˆTL

EAkk

Here again θ is the angle between the local and global x-coordinate axes, measured counterclockwise.

−− θθθθθθ 22 sinsincossinsincos


− 0101In order to formulate the component stiffness

In Class Example

Node number[ ]

−=

000001010000

1 LEAk

matrix from the individual element stiffness matrices consider the following simple truss

23

2

0000

−45°

−− 4

242

42

42

P (kN)Y

global

1

23

3

90°

[ ]

−−−−−−

=

42

42

42

42

42

42

42

42

42

42

42

42

2 LEAk

(kN)X

global

Elementnumber

1 3

[ ]

−10100000

EAk

L (m) 4444

number [ ]

−

=

101000003 L

EAk


Begin creating component stiffness matrix

from element stiffness matrices using element #1from element stiffness matrices using element #1

−

XX

dd

EAff 11

00000101

1,31,1

−=

Y

X

Y

Y

X

Y

ddd

LEA

fff

3

3

1

3

3

1

000001010000

3,1 3,3

Global force-displacement relationship still has the form

−

Y

X

Y

X

dd

FF

1

1

1

1

00000101

and for the moment we focus on

=

Y

X

Y

X

ddd

LEA

FFF

2

2

2

2

0101

{ } [ ] { }dF K=

and for the moment we focus on the contribution of the element nodal forces (f) to the global nodal forces (F).

−

Y

X

Y

X

dd

FF

3

3

3

3

00000101


df 2222

−−−−

−−

=

X

Y

X

X

Y

X

ddd

LEA

fff

3

2

2

42

42

42

42

42

42

42

42

42

42

42

42

3

2

2 2,32,2

−−

Y

X

Y

X

dff

3

3

42

42

42

42

4444

3

33,2 3,3

−

Y

X

Y

X

dd

FF

1

1

1

1

00000101

−−−−

=

Y

X

Y

Y

X

Y

ddd

LEA

FFF

2

2

1

222242

42

42

42

42

42

42

42

2

2

1

101

−−−+−−

Y

X

Y

X

dd

FF

3

3

42

42

42

42

42

42

42

42

3

3

00101


df 0000

−=

X

Y

X

X

Y

X

ddd

LEA

fff

2

1

1

2

1

1

00001010

00002,11,1

−

Y

X

Y

X

dff

2

2

2

2

1010 2,12,2

−−

Y

X

Y

X

dd

FF

1

1

1

1

001010010001

−+−−−+−−

−−=

Y

X

Y

X

ddd

LEA

FFF

2

2

222242

42

42

42

42

42

42

42

2

2

101110

00

−−−+−−

Y

X

Y

X

dd

FF

3

3

42

42

42

42

4444

3

3

00101


− dF 010001

−−−

−

=

X

Y

X

X

Y

x

ddd

EAFFF

2

1

1

42

42

42

42

2

1

1

00001010010001

−+−−−+−−

=

X

Y

X

Y

dddL

FFF

3

2

222242

42

42

42

42

42

42

42

3

2

00101

110

−− YY dF 344443 00

vector of nodal displacementsglobal coordinate system

vector of nodal forcesl b l di t t

Component stiffness matrix

global coordinate systemglobal coordinate system { } [ ]{ }dKF =


• How to deal with the unknown support reactions ?• These are nodes where the displacements are known, i.e., zero in the perfectly rigid

support case. Supports can have proscribed displacements.

1 0 0 0 1 0 unknown support reactions known support displacements

1 31 1 1 1

1 31 1 1 1

2 2 2 22 3

1 0 0 0 1 00 1 0 1 0 0

0 0

X X X X

Y Y Y Y

F R f fF R f f

F R f f EA

− = = + −= = + − − +

000

4 4 4 42 2 2 22 2 2 22 3

4 4 4 42 2 2 21 2 2 2 2 2

3 3 3 4 4 4 4

0 0

0 1 10 1 0 1

X X X X

Y Y Y Y

X X X

F R f f EALF R f f

F f f

= = + = − − + −= = + = = + − − + − 3

00

Xd

4 4 4 41 2 2 2 2 23 3 3 4 4 4 40 0Y Y YF P f f

= − = + − − 3Yd

k d lknown applied nodal loads unknown nodal displacements


With the stiffness matrix partitioned as follows

XR1 0

=

X

Y

X

KKEARRR

12112

1

1

000

−X

Y

dd

KKL

P

R

3

22212 00

where YdP 3

01100001

−

0001

−

−

=42

4200

0110

11K

−=42

42

00

12K

+−

421

4210

−

42

42


−−2201

−+221

and

−

=

42

4200

4401

21K JSK =

−

+=

42

42

441

22

then determine nodal displacements first by solving the following subsystem of equations:

[ ]

−

=

−

− 0

1

122

3

3

PK

dd

Y

X

−

−

−−+

=

1

01

42

42

42

42

PL

PEAL

−−=

221AEPL


At this point the unknown unknown reaction forces can be computed from

− PR X 00100011

−=

−−−

=

P

P

PLEARRR

X

Y

X

0000

00001010010001

42

42

42

42

2

1

1

−

−−=−=

−−−+−−−+−−

− P

P

dd

AEL

P

R

Y

X

Y

0221

10

00101

1100

3

3222242

42

42

42

42

42

42

42

2

or simply

PdP Y 22100 34444

PR

−=

P

P

RRR

X

Y

X

0

2

1

1

PR Y2


At this point the local displacements are extracted from the global displacements through

{ } [ ] { }dd Tˆ ={ } [ ] { }dd T

−

Y

Xx

dd

dd

1

1

1

1

00cossin00sincos

ˆˆ

θθθθ

−

=

Y

X

Y

y

x

y

dd

ddd

2

2

1

2

2

1

cossin00sincos00

coss

ˆˆ

θθθθ

θθ

1 1ˆ 01 0 0 0x Xd d =

The local nodal displacements for element #1 are

PL

ˆ11

33

3

ˆ 00 1 0 0ˆ 10 0 1 0

0 0 0 1 1 2 2ˆ

Yy

Xx

Y

dd PLdAEd

dd

= = = − = − −

( )

−−

−=

AEPL

AEd

d

y

x

221ˆ

ˆ

3

3

33 Yyd


{ }{ }dB ˆ=ε

The axial strain in element #1 is

{ }{ }

( )PLAEPL

LL

dB

22111

−

−=

ε

( )

PAE

PLLL

22

221

−=

−−

EA=

The negative sign indicates compression. Calculate the stress and the force in element #1

AP

E

22−=

= εσ1 1

1 3

2 2

x xf fA

P

σ= −=

1 11 3

0y yf f= −

=A 2 2P= −



=

===

−

=

000

000

100000010010

ˆˆˆ

1

1

1

1

Y

X

y

x

ddd

AEPL

ddd

=

=

−

00

00

01001000

ˆ2

2

2

2

Y

X

y

x

ddAE

dd

{ }{ }ˆB dε =

The axial strain in element #3 zero

{ }{ }01 10L L = −

0=

And element #3 is a zero force member.



=

−−

0ˆ

242

42

42

42

2 Xx dd

−−=−=

=

−−−

−=

2211

0

ˆˆˆ

3

3

2

42

42

42

42

42

42

42

42

42

42

42

42

3

3

2

Y

X

Y

x

y

ddd

AEPL

ddd

( )( ) ( )( )( ) ( )

( ) ( )

−−=−−+−−=−−+−−

=

1221112211

22

42

42

42

42

344443 Yyd

( )( ) ( )( )( ) ( )

−=−−+−−=−−−−

=

1221112211

42

42

42

42

The axial strain in element #2 is

{ }{ }d

dB

=

0

ˆε

AEP

dd

AEPL

LL X

X −=

−==

−=

1011

3

2


The axial stress in element #2 is

E= εσ

Fi ll th d l f i t d ith l t #2

AP

−=

Finally, the nodal forces associated with element #2 are

Aff xx −= 2

32

2

0

23

22 −= yy ff

PA

−== σ 0=


In class example(s)

section 6: truss elements, local & global coordinates

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