section 5 truss elements.ppt - cleveland state universityin nearly all finite element analyses it is...

Section 5: TRUSS ELEMENTS, LOCAL & GLOBAL COORDINATES

Washkewicz College of Engineering

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 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.

1



X

Y

Z

Global axes

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

2



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

When applied to a truss element:

fX

fY

y

x

Y

X

ff

ff

cossinsincos

ff

ffff

ffff

y

x

y

x

Y

X

Y

X

ˆ*T

ˆˆˆˆ

cossin00sincos0000cossin00sincos

2

2

1

1

2

2

1

1

f1X

f1Y f2X

f2Y

xf1̂

yf1̂

yf 2̂xf 2̂

Global Local

GlobalLocal

1

2

Note: Local x-axis is always along the element and is always measured counterclockwise from global x-axis to local x-axis. 3



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:

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

0

ˆ0

ˆ


2

1

2

2

1

1

x

x

Y

X

Y

X

f

f

ffff

f1X

f1Y f2X

f2Y

xf1̂

01̂ yf

02̂ yfxf 2̂

Global Local

1

2

4



This leads to the following system of equations

and the solution leads to the following matrix formulation

0ˆsin00

0ˆcos00

000ˆsin

000ˆcos

22

22

11

11

xY

xX

xY

xX

ff

ff

ff

ff

Y

X

Y

X

x

x

ffff

f

f

2

2

1

1

2

1

sincos0000sincos

ˆ

ˆ

5



d1X

d1Y d2X

d2Y

dd

dddd

dddd

y

x

y

x

Y

X

Y

X

ˆ*T

ˆˆˆˆ


2

2

1

1

2

2

1

1

xd2ˆyd2

ˆ

xd1̂yd1̂

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

Global Local

1

2

6



Inverting this last matrix expression yields:

or

It becomes obvious that:

Similarly

Y

X

Y

X

y

x

y

x

dddd

dddd

2

2

1

1

2

2

1

1

cossin00sincos00

00cossin00sincos

ˆˆˆˆ

dd Tˆ

1*TT

ff Tˆ 7



y

x

y

x

y

x

y

x

dddd

LEA

ffff

2

2

1

1

2

2

1

1

ˆˆˆˆ

0000010100000101

ˆˆˆˆ

df ˆk̂ˆ

In local coordinates Element nodal forces and displacements in local coordinates

Element stiffness matrix in local coordinates

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

With

dd Tˆ ff Tˆ

8



then

Thus the element stiffness matrix in terms of global coordinates is

and in a matrix format the element stiffness matrix is

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

dkf

df

df

df

Tk̂T

Tk̂T

ˆk̂ˆ

1

TˆT 1 kk

22

22

22

22

1

sinsincossinsincossincoscossincoscos

sinsincossinsincossincoscossincoscos

TˆTL

EAkk

9



Elementnumber

1

23

Node number

1

2

3

0000010100000101

1 LEAk

101000001010

0000

3 LEAk

°

°

L (m)

42

42

42

42

42

42

42

42

42

42

42

42

42

42

42

42

2 LEAk

P (kN)

In order to formulate the component stiffness matrix from the individual element stiffness matrices consider the following simple truss

X

In Class Example

Yglobal

10



Global force-displacement relationship still has the form

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

Begin creating component stiffness matrix from element stiffness matrices using element #1

Y

X

Y

X

Y

X

Y

X

dddd

LEA

ffff

3

3

1

1

3

3

1

1

0000010100000101

Y

X

Y

X

Y

X

Y

X

Y

X

Y

X

dddddd

LEA

FFFFFF

3

3

2

2

1

1

3

3

2

2

1

1

00000101

00000101

1,31,1

3,1 3,3

dF K

11



Y

X

Y

X

Y

X

Y

X

Y

X

Y

X

dddddd

LEA

FFFFFF

3

3

2

2

1

1

42

42

42

42

42

42

42

42

42

42

42

42

42

42

42

42

3

3

2

2

1

1

00101

00000101

Y

X

Y

X

Y

X

Y

X

dddd

LEA

ffff

3

3

2

2

42

42

42

42

42

42

42

42

42

42

42

42

42

42

42

42

3

3

2

2 2,32,2

3,2 3,3

12



Y

X

Y

X

Y

X

Y

X

Y

X

Y

X

dddddd

LEA

FFFFFF

3

3

2

2

1

1

42

42

42

42

42

42

42

42

42

42

42

42

42

42

42

42

3

3

2

2

1

1

00101

11000

001010010001

Y

X

Y

X

Y

X

Y

X

dddd

LEA

ffff

2

2

1

1

2

2

1

1

101000001010

00002,11,1

2,1 2,2

13



Y

X

Y

X

Y

X

Y

X

Y

X

Y

x

dddddd

LEA

FFFFFF

3

3

2

2

1

1

42

42

42

42

42

42

42

42

42

42

42

42

42

42

42

42

3

3

2

2

1

1

00101

11000

001010010001

Component stiffness matrix

vector of nodal displacementsglobal coordinate system

vector of nodal forcesglobal coordinate system dKF

14



• 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 31 1 1 1

1 31 1 1 1

2 2 2 22 34 4 4 42 2 2 2

2 2 2 22 34 4 4 42 2 2 2

1 2 2 2 2 23 3 3 4 4 4 4

1 2 2 2 2 23 3 3 4 4 4 4

1 0 0 0 1 00 1 0 1 0 0

0 0

0 1 10 1 0 1

0 0

X X X X

Y Y Y Y

X X X X

Y Y Y Y

X X X

Y Y Y

F R f fF R f f

F R f f EALF R f f

F f fF P f f

3

3

0000

X

Y

dd

known applied nodal loads unknown nodal displacements

unknown support reactions known support displacements

15



With the stiffness matrix partitioned as follows

where

Y

X

Y

X

Y

X

dd

KK

KK

LEA

P

RRRR

3

3

2221

1211

2

2

1

1

0000

0

421

4210

42

4200

01100001

11K

42

42

42

42

0001

12K

16



42

4200

42

4201

21K JSK

42

42

42

421

22

221

1

01

0

1

42

42

42

42

122

3

3

AEPL

PEAL

PK

dd

Y

X

and

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

17



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

or simply

P

PP

P

dd

AEPL

LEA

P

RRRR

Y

X

Y

X

Y

X

0

0

2211

0000

00101

11000

001010010001

0

3

3

42

42

42

42

42

42

42

42

42

42

42

42

42

42

42

42

2

2

1

1

PP

P

RRRR

Y

X

Y

X

0

2

2

1

1

18



1 1

11

33

33

ˆ 01 0 0 0ˆ 00 1 0 0ˆ 10 0 1 0

0 0 0 1 1 2 2ˆ

x X

Yy

Xx

Yy

d ddd PL

dAEddd

The local nodal displacements for element #1 are

AEPL

AEPL

d

d

y

x

221ˆ

ˆ

3

3

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

dd Tˆ

Y

X

Y

X

y

x

y

x

dddd

dddd

2

2

1

1

2

2

1

1

cossin00sincos00

00cossin00sincos

ˆˆˆˆ

19



EAP

AEPL

AEPL

LL

dB

22

22111

ˆ

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

The axial strain in element #1 is

AP

E

22

1 1

1 3

2 2

x xf fA

P

1 11 3

0y yf f

20




ˆ

01 10

0

B d

L L

The axial strain in element #3 zero

And element #3 is a zero force member.

0000

0000

0100100000010010

ˆˆˆˆ

2

2

1

1

2

2

1

1

Y

X

Y

X

y

x

y

x

dddd

AEPL

dddd

21



The axial strain in element #2 is


1221112211

1221112211

2211

00

ˆˆˆˆ

42

42

42

42

22

42

42

42

42

3

3

2

2

42

42

42

42

42

42

42

42

42

42

42

42

42

42

42

42

3

3

2

2

Y

X

Y

X

y

x

y

x

dddd

AEPL

dddd

AEP

dd

AEPL

LL

dB

X

X

1011

ˆ

3

2

22



The axial stress in element #2 is

Finally, the nodal forces associated with element #2 are

AP

E

PAff xx

23

22

0

23

22

yy ff

23



In class example(s)

24

section 5 truss elements.ppt - cleveland state universityin nearly all finite element analyses it is...

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