section 4: truss elements, local & global coordinates info on web... · 2013-03-20 · section...

Section 4: 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 principle of minimum potential energy will be utilized to re-derive the stiffness matrices.


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

d

2. Any effect of transverse displacements are ignored

3. Hooke’s law applies

Deformed shape

x2xd1

ffNode(a truss hinge)

x

Element

xf2xf1


Assume

This is an approximation function. We are assuming through the use of this function h di l di ib d i i l li f hi l h l

xaau ˆˆ 21 +=

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

( )xu ˆˆ

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


To establish the unknown coefficients consider the boundary conditions of the truss element

D f d hDeformed shape

Thus

( ) 110ˆ adu x == ( ) LaddLu xx 212ˆˆˆ +==

Thus

−=

Ldda xx 12

2

ˆˆ

and in a matrix format

− dxxdd 1ˆˆˆˆˆˆˆ

−=+=

+=

x

xxx

xxx

d

dLx

LxdNdNx

Ldddu

2

12211

121 ˆ

1ˆˆˆˆˆ


Here

xNˆ

1−=xNˆ

=

are shape functions. Consider the following guidelines as they relate to one dimensional structural elements when selecting a displacement function.

LN 11 =

LN2 =

g p

1. Common approximation functions are usually polynomials where the function can be expressed in terms of shape functions.

2 The approximation functions must be continuous through the element2. The approximation functions must be continuous through the element.

3. The approximation functions must provide inter-element continuity for all degrees of freedom at each node of every element.

4. The approximation functions must allow for rigid body motion, because rigid body motion can occur in a structure.

With

xaau ˆˆ 21 +=


The strain displacement relationship can now be cast as

d ˆ( )

dNNd

xdudx

x1

ˆˆ

=ε

d

dNN

xd

x

x

1

2

21

ˆ11

ˆˆ

=

=

dB

dLL

x

x

1

2

ˆ

ˆ

=

−=

Ldd

dB

xx

x

12

2

ˆˆ

ˆ

−=

=

which is constant along the length of the truss element.

L


( ) ( )ˆˆ

The stress-strain relationship becomes

( ) ( )

=

=

d

dBE

xEx

x1

ˆ

ˆˆˆ εσ

−=

LddE

d

xx

x

12

2

ˆˆ

From elementary mechanics

−

=

=

ddAE

AT

xx 12ˆˆ

σ

L

AE


The nodal forces become

TfTf xx =−= 21ˆˆ

thus

( )xxx ddL

AEfT 121ˆˆˆ −=−=

ff xx 21

or

( )xxx ddL

AEfT

L

122ˆˆˆ −==

or

( )( )

xxx

AE

ddL

AEf 211ˆˆˆ −=

In a matrix format

df ˆˆ

( )xxx ddL

AEf 122ˆˆˆ −=

−

−=

x

x

x

x

d

dL

AEf

f

2

1

2

1

ˆ1111

ˆ


Consider the situation where

xx dd 21ˆˆ =

The case where both nodes move the same amount in the same direction. This is definition of rigid body motion. Here

xx 21

121

21

ˆˆˆˆ

ˆˆ

xL

ddd

xaau

xxx

−+=

+=

221 ˆ

ˆˆˆ xL

ddd

L

xxx

−+=

1

ˆ

ˆ0ˆ

d

xL

d x

+=

1

1

ad x

==


With

2211ˆˆˆ dNdNu +=

( ) 121

1211

2211

aNNaNaNdNdNu xx

+=+=+

this infers that

1ˆ au =

Thus for rigid body motion

( ) 121 aNN +=

g y

211 NN +=


In addition

( )L

ddx xxˆˆ

ˆ 12 −=ε ( )

−=

LddEx xxˆˆ

ˆ 12σ

−=

LddAET xxˆˆ12

Ldd xx

0

ˆˆ22 −

=

−=

LddE xxˆˆ

22

−=

LddAE xx

0

ˆˆ22

S i d h i l f i h l ill ll b f i id b d i

L=

=

LE 0

=

LAE 0

Stress, strain and the axial force in the truss element will all be zero for rigid body motion.


The general relationship from the previous page holds for an truss element oriented along the x axis. Thus

[ ]

−

−=

1111ˆ

LAEk

is the local stiffness matrix. We note that the local stiffness matrix is symmetric, i.e.,

and square. The global stiffness matrix and global force matrix are assembled using nodal

jiij kk ˆˆ =

force equilibrium equations, force deformation equations and compatibility equations. Examples of assembling these equations will be given.

[ ] [ ] ( )∑N

ekK ( )∑N

efF[ ] [ ] ( )∑=

=e

ekK1

( )∑=

=e

efF1


One can quickly populate the global stiffness matrix for a truss structure using the methodology developed for the spring element. Consider the ith truss element:

bith Element

ith element stiffness propertiestransfers into the global stiffness

ba baa

ii Element

matrix as follows

abaa

kkbkkaba

00LLL

MM

kka

ba

bbba kkb0

LLL

MM

LLL abaa

kkb

kka

00 MM

LLL bbba kkbith element’s stiffness matrix with node numbers a and b


I l lIn class example


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.

Y

Gl b l

XZ

Global axes


In general for forces, the transformation is as follows:

FGlobal Local

Fy

−=

v

u

Y

X

ff

FF

θθθθ

cossinsincos

which appears as follows when appied to a truss element:

Fx

θ

which appears as follows when appied to a truss element:

ff

ff

y

x

y

x

ˆˆ

00cossin00sincos

1

1

1

1

−

θθθθf2y

fff

ff

y

x

y

y

x

y

ˆˆ

cossin00sincos00

2

2

1

2

2

−=

θθθθ

f1y f2x

ˆ

yf 2xf 2

[ ] ff ˆ*T=f1x

θxf1

yf1


For a true truss element, an element that will not sustain shear forces at the end, thenf2y

f1y f2x

02 =yfxf 2

f1xθ

xf101 =yf

and

ˆ00i ff θθ

Global Local

−

−

=

ˆ0

i00sincos0000cossin00sincos

2

1

2

1

1

x

x

x

y

x

f

f

ffff

θθθθ

θθθθ

0cossin002 yf θθ


This leads to the following system of equations

000ˆcos ++−= ff θ

0ˆcos00

000ˆsin

000cos

22

11

11

−++=

+++=

++=

xx

xy

xx

ff

ff

ff

θ

θ

θ

solution leads to (show for homework)

0ˆsin00 22 +++= xy ff θ

( )

=

x

y

x

x

x

fff

f

f

2

1

1

2

1

sincos0000sincos

ˆ

ˆ

θθθθ

y

xf

f2

2


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

d2y

dd1y d2x

xd2ˆyd2

ˆd1 Global

d1x

dd

dd

xx

ˆˆ

00cossin00sincos 11

−

θθθθ

θxd1

yd1 Global Local

ddd

ddd

y

x

y

y

x

y

ˆˆ

cossin00sincos0000cossin

2

2

1

2

2

1

−=

θθθθ

θθ

[ ] dd ˆ*T= (9)


Inverting this last matrix expression yields (show for homework):

−

=

x

y

x

x

y

x

ddd

ddd

2

1

1

2

1

1

sincos0000cossin00sincos

ˆˆˆ

θθθθθθ

or

−

yydd 22 cossin00ˆ θθ

where it is obvious that:

[ ] dd Tˆ =

Similarly

[ ] [ ] 1*TT −=

S a y

(9) [ ] ff Tˆ =


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

df ˆ0101ˆIn 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 ˆkˆ =→

y

x

y

x

df 2

2

2

2ˆ0000ˆ Element stiffness matrix

in local coordinates (extended)

WithWith

[ ] dd Tˆ = [ ] ff Tˆ =


then

[ ] df = ˆkˆ [ ] [ ] [ ][ ]

[ ] [ ][ ] [ ]

df

df

f

=

=− TkT

TkT1

where the element stiffness matrix in terms of global coordinates is

[ ] dkf =

in a matrix format:

[ ] [ ] [ ][ ]TˆT 1 kk −=

[ ] [ ] [ ][ ]

−−−−

−−

== −

θθθθθθθθθθθθθθθθθθ

22

22

22

1

sincoscossincoscossinsincossinsincos

sincoscossincoscos

TˆTL

EAkk

−− θθθθθθθθθθθθ

22 sinsincossinsincossincoscossincoscosL


− 0101

In order to load up the component stiffness matrix consider the following simple truss

Node number

2

[ ]

−=

000001010000

1 LEAk

23

2 0000−45

[ ]

−−−−

42

42

42

42

42

42

42

42

EAkP [kN]

11 3

90

L m

[ ]

−−−−

=

42

42

42

42

42

42

42

42

44442 L

k

Elementnumber [ ]

−=

00001010

0000

3 LEAk

L m

[ ]

− 101000003 L


Create component stiffness matrix

from element stiffness matrices

−

xx

dd

EAff 11

00000101

1,31,1

−=

x

y

x

y

ddd

LEA

fff

3

3

1

3

3

1

000001010000

3,1 3,3 yyf 33 0000

−

y

x

y

x

dd

FF

1

1

1

1

00000101

=

y

x

y

x

ddd

LEA

FFF

2

2

2

2

0101

−

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


−−

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

42

42

42

42

3

3

00101

vector of nodal displacementsglobal coordinate system

vector of nodal forcesglobal coordinate system [ ] dKF =

Component stiffness matrix


• How to deal with the problem of supports (restraints) ?h d h h di l k i h f l i id• These are nodes where the displacements are known, zero in the perfectly rigid

support case.

unknown known

−−

y

x

EAFFF

22221

1

000

00001010010001support

Reactions

known supportDisplacements

−+−−−+−−

−−=

x

y

x

dL

EAFF

342

42

42

42

42

42

42

42

4444

2

2

00

101110

00

0

−−

− ydP 34

242

42

4200

k d lknown applied nodal loads unknown nodal Displacements


With the stiffness matrix partitioned as follows

f1 0

=

x

y

x

KKEAfff

12112

1

1

000

−x

y

dd

KKL

P

f

3

3

22212 00

where ydP 3

01100001

−

0001

−

−

=42

4200

0110

11K

−=42

42

00

12K

+−

421

4210

−

42

42


−−2201

−+ 221

−

=

42

4200

4401

21KJSK =

−

=

42

42

4422

44

d

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

[ ]

+

−

=

−

−

01

0

122

1

223

3

L

PK

dd

y

x

−

−

−

−+=

1

01

42

42

42

42

PL

PEAL

−−=

221AE


Next determine the nodal forces:

F 0010001

−−−

−

PLEAFFF

x

y

x

000

00001010010001

42

42

42

42

2

1

1

−=

−+−−−+−−

=

dd

AEPL

LEA

F

x

y

x

2211

0101

1100 3

222242

42

42

42

42

42

42

42

4444

2

2

−−=

−−

−

P

dP y

0

22100 342

42

42

42

−

=PP

0

− P0


In general, with [ ] dd Tˆ =

−

=

y

x

y

x

ddd

dd

1

1

1

1

i0000cossin00sincos

ˆˆˆ

θθθθθθ

−

y

x

y

x

dd

dd

2

2

2

2

cossin00sincos00

ˆ θθθθ

000011d x

The for element #1 in the truss

PL

ˆ

−−−

=

2211

0

100001000010

ˆˆˆ

3

3

1

AEPL

ddd

y

x

y ( )

−−

−=

AEPL

AEd

d

y

x

221ˆ3

3

y


( ) dBx ˆ=ε

The strain in element #1 is

( )

( )PLAEPL

LL

dBx

22111

−

−=

ε

( )

PAE

PLLL

22

221

−=

−−

EA=

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

( ) ( )P

xEx

22−=

= εσ ( ) ( )P

AxxF

22−=

= σ

A−=


Potential Energy Approach in Deriving Bar Element Equations

The principle of minimum potential energy is now used to derive the bar element equations presented earlier. Recall that

Ω+Uπ

For the internal strain energy component consider the following figure:

Ω+= Upπ

P


The differential of strain energy is given by the expression:

( ) ( ) ( )( ) ( ) ( )dVd

dxzydU

xx

xx

εσεσ

=∆∆∆=

For the entire bar:

dVdUV xx

x

εσε

∫ ∫

= ~

( ) dVdEV xx

V

x

εεε

∫ ∫

∫ ∫

=

~~0

0

( ) dVEV

xx

εε

∫

=

2

~

0

2

0

( )( ) dVdVExV xV

xx εσεε∫∫ =

=

21

2

0


The potential energy term is associated with externally applied loads and is given by the expression

where the first term represents contributions from any body forces, the second term

∑∫∫=

−−−=ΩM

iixixS xV b dfdSuTdVuX

1

ˆˆˆˆˆˆ

p y y ,represents any contribution from surface tractions, and the third term represents the contribution from concentrated nodal forces.

The total potential energy for a bar element of length L and constant cross sectional area p gy gA becomes:

p U Ω+=π

L

M

iixixS xV bxV x

A

dfdSuTdVuXdV1

ˆˆˆˆˆˆ

ˆˆˆˆˆˆ21

−−−=

∫∫∫

∑∫∫∫=

εσ

xxxxS xV bxx dfdfdSuTdVuXxdA2211

0

ˆˆˆˆˆˆˆˆˆ2

−−−−= ∫∫∫ εσ


With( )

APxσ −=

( )[ ] ( )xD

xEA

εε

==

and

[ ] [ ]ED =

then formally in matrix notation

[ ] D εσ = [ ] xx D εσ =

dPdPdSTudVXuxdAxxS x

T

V bT

L

xT

xpˆˆˆˆˆˆˆ

2 210

−−−−= ∫∫∫ εσπ

PddSTudVXuxdA T

S xT

V bT

L

xT

xˆˆˆˆˆˆ

2 0

0

−−−= ∫∫∫ εσ


which leads to

[ ] [ ] [ ] ( ) ddBDBdA LTTTT ˆˆˆ ∫ [ ] [ ] [ ] ( )

[ ] [ ] dSTNdudVXNdu

xddBDBdA

S xTTT

V bTTT

xTTT

xp

ˆˆˆˆˆˆ

ˆ2 0

=−

=−

=

==

∫∫

∫ εσπ

PdT

SV

ˆ−

∫∫

or

[ ] [ ] [ ] [ ] dVXNdxddBDBdAV b

TTL

TTT

pˆˆˆˆˆ

2−= ∫∫π

[ ] PddSTNdT

S xTT

Vp

ˆˆˆ

2 0

−− ∫

∫∫


Integration yields

[ ] [ ] [ ] A LTTT ˆˆ ∫ [ ] [ ] [ ]

[ ] [ ] PdSTNdVXNd

xddBDBdA

S xT

V bTT

TTT

p

ˆˆˆ

ˆ2 0

−−−

=

∫∫

∫π

[ ] [ ] [ ] fddBDBdAL TTTT

SV

ˆˆˆˆ2

−=

∫∫

where

[ ] [ ] PdSTNdVXNf TT ++= ∫∫ ˆˆˆ

now one can definitely see that

[ ] [ ] PdSTNdVXNfS xV b ++= ∫∫

( )( ) ( )d

dd

p

xxpp

ˆ, 21

π

ππ

=

=


The minimization of potential energy requires

0ˆ0ˆ21

=

∂

∂=

∂

∂

x

p

x

p

ddππ

With

[ ] [ ] [ ] ˆˆ* TTTdBDBdU [ ] [ ] [ ]

[ ] 121 ˆ

ˆ111

ˆˆ

*

xxx

dELdd

dBDBdU

−

−

=

=

[ ]

( )2221

21

2

21

ˆˆˆ2ˆ

ˆ1x

xx

ddddE

dLLL

+−=

( )22112 2 xxxx ddddL

+


and

xxxx

Tfdfdfd 2211ˆˆˆˆˆˆ +=

then

xxxx fff 2211

( ) ( )

( )

xxxxxxxxxx

p

EAL

fdfdddddLEAL

dd 22112

2212

1211

ˆˆˆˆˆˆˆ2ˆ2ˆˆ0

+−+−

∂∂

=∂

∂=

π

( ) xxx fddLEAL

1212ˆˆ2ˆ2

2−

−=

( ) ( )xxxxxxxxxx

p fdfdddddLEAL

dd 22112

2212

1222

ˆˆˆˆˆˆˆ2ˆ2ˆˆ0

+−+−

∂∂

=∂

∂=

π

( ) xxx fddLEAL

2212ˆˆ2ˆ2

2−

+−=


These last two equations. can be put into a matrix format as follows

=

−

−

−=

∂

∂00

ˆ

ˆ

ˆ

ˆ

1111

ˆ2

1

2

1

x

x

x

xp

f

f

d

dL

AEdπ

or

−

xx dAEf 11ˆ11ˆ

where

−

=

xx dLf 22ˆ11ˆ

where

[ ]

−

−=

1111ˆ

LAEk

as before.


Comparison of a Finite Element Solution to an Exact Solution

It is time to make comparisons. In the previous in-class problem we solved the problem using first one and then two elements. Here we will make comparisons of two, four and eight element models to an exact solution. The exact solution for axial di l t i bt i d f th l ti f th f ll i idisplacements is obtained from the solution of the following expression:

( ) ( )∫ dP1

with

( ) ( )∫= dxxPAE

xu

( ) ( )1021 xxxP

=

25x=


then

( ) 251 dxxxu = ∫( )

1

3

35 CAEx

AE

+=

∫

Evaluating this expression at the support yields

( ) 0L( )

1

3

35

0

CAEL

Lxu

+=

==

or

LC 5 3

( )3335x

AELC

35

1 −= ( )33

35 LxAExu −=


The following figure plots the exact expression for the displacement along the rod along with the finite element calculations:

The finite element analysis match the exact solutions at the node points. Although the nodal displacements match the exact solution, the displacement values at i di l i b i li di l f iintermediate locations are poor because we are using linear displacement functions within each element. As we increase the number of elements we converge on the exact solution.


The exact solution for stress is:

( ) dEduEExP =

===

δεσ

22

5.22

5 xxdx

Edx

EEA

==

=

=== εσ

Since stress is derived from the slope of the displacement, and since the displacement is linear within each element, then the stress is constant in each element. This can be seen in the following figure:

2

seen in the following figure:

The best approximation of the stress occurs at the midpoint of the element, not at the nodes The reason for this is that theat the nodes. The reason for this is that the derivative of displacement is better predicted between the nodes then at the nodes.


As we saw in the example problem the axial stress is not continuous across element boundaries. Recall that equilibrium is not satisfied within each element, only at the

d A h b f l i h di i i i d d hnodes. As the number of elements increase the discontinuity in stress decreases and the approximation of equilibrium improves.

Finally in the figure below the convergence of the axial stress at the fixed end is y g gdepicted. Again, as the number of elements increase the stress at the fixed end converges to the exact solution from below.

section 4: truss elements, local & global coordinates info on web... · 2013-03-20 · section...

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