L5/1

LECTURE 5/11; PRINCIPLE OF VIRTUAL WORK

5.1 IMPROVED RECIPE

5.2 PRINCIPLE OF VIRTUAL WORK

5.3 ELEMENT CONTRIBUTION

E,

x,X

gL

L5/2

LEARNING OUTCOMES

Student knows the basic building blocks –element approximation and virtual work

expression– of an element contribution,

knows the role of principle of virtual work in derivation of element contributions of the

finite element method,

is able to derive the element contributions of bar and beam elements in the material

coordinate system, when the virtual work expression per unit length and the element

approximation are given.

L5/3

EXAMPLE L5/P1. Consider the beam truss of the figure. Determine the displacements

and rotations of nodes 2 and 4. Assume that the beams are rigid in the axial directions.

Cross-sections and lengths are the same and Young’s modulus E is constant.

Answer 3

27

900z

Yzz

f LEI

& 3

411

1800z

Yzz

f LEI

Z

1

1

fz

X 2

3

2

3

4

L

z x

x z

x z

L5/4

The Math3300 solution is given by

In :

BEAM EE, GG, ρρ , AA, Irr, Iyy, Izz, 0,1,0 , 0,0,0 1,2

BEAM EE, GG, ρρ , AA, Irr, Iyy, Izz, 0,1,0 , 0,0, ff 1 2,4

BEAM EE, GG, ρρ , AA, Irr, Iyy, Izz, 0,1,0 , 0,0,0 4,3

In :

0,0, 0,0,0 0,0,00,0,0 0,0,0 0, 1 , 0, 0, 0,0,0 0,0,0, 0,0 0,0,0 0, 2 , 0

Out :

0,0, 0,0,0 0,0,0

0,0,0 0,0,0 0, , 0, 0, 0,0,0 0,0,0

, 0,0 0,0,0 0, , 0

Parameters of the problem can be functions of x. Then the element contribution may not

be obtained from the exact solution and a there is need for a more general manner of

derivation!

L5/5

5.1 IMPROVED RECIPE

Derive the element contributions eW from virtual work expression of the model and

polynomial approximations to the unknown functions in terms of nodal displacements

and rotations of the material coordinate system.

Express the nodal displacements and rotations of the material coordinate system in

terms of those in the structural coordinate system (symbols ai ).

Sum the element contributions over the elements and their loading modes to end up

with the virtual work expression ( )e eme E e E mW W W of structure.

Restructure to get the form T ( )W a Ka F Use the principle of virtual work 0W a , basic lemma of variational calculus for

n a , and solve the dofs from the system equations 0 Ka F .

new step !

L5/6

ELEMENT APPROXIMATION

Element approximation is a polynomial interpolation of the nodal values. Roughly

speaking, approximation depends on the nodal values used, number of nodes associated

with the element, the geometry of element, and dimension of the problem. T

2

2 1 1

21

2 1

x

x

x xx x u

uux x

x x

T2

12 12 2

2 2

(1 ) (1 2 )

(1 )

(3 2 )

( 1)

z

y

z

y

u

hw

u

h

& 1

2 1

x xx x

shape functions

1xu

u 2xu

2 1 1zu

1y w

2y

2zu

2 1

L5/7

5.2 PRINCIPLE OF VIRTUAL WORK

Principle of virtual work int ext 0W W W u U is just another representation

of the balance laws of continuum mechanics. It is important due to its wide applicability

and physical meanings of the terms.

int int intc( : )VV V

W w dV dV w d

ext ext ext( )V VV V

W w dV f u dV w d

ext ext ext( )A AA A

W w dA t u dA w d

The details of the final expressions vary case by case, but the principle itself does not!

V u

t dA

f dV

P

P’

0u

A

L5/8

EQUIVALENT REPRESENTATIONS In connection with the finite element method, one may think that approximating a function

f in means actually solving the problem 0u f in by using an approximation of

type Tu N a. In the approximation, the elements of N are given (shape functions).

Strong form Find 0 ( )u C s.t 0u f in

Weak form Find 0 ( )u C s.t. ( ) 0W u u f d

0 ( )u C

Discrete form Find na such that T ( ) 0W a Ka F n a

The strong and the weak forms are just equivalent representations of the same problem. To

prove the equivalence, one needs the basic lemma of variational calculus!

L5/9

The discrete form follows by substituting the approximation Tu N a and its variation Tu a N to the weak form (as the shape functions are given T Tu N a a N )

( )W u u f d

T T T T( ) ( ) 0W f d d fd

a N N a a NN a N

T T( ) 0W a Ka F a R

The basic lemma of variational calculus in the form T 0v u n v & nu

0u (here v a and u a ), gives the linear system of equations 0 R Ka F

when applied to the last form above.

L5/10

EXAMPLE. Approximate ( ) 2min{ / ,1 / }f x x L x L in [0, ]x L . Use sinusoidal shape

functions and T{1,2 }( ) a sin( / )jj nu x j x L N a . Illustrate the behavior of ( )u x in

cases {1,5,100}n , when 1L .

Answer In the figures, ( )f x ~red and ( )u x ~blue.

0 0.2 0.4 0.6 0.8 1 0 0.2 0.4 0.6 0.8 1 0 0.2 0.4 0.6 0.8 1

n=1 n=5 n=100

L5/11

Here ( ) sin( / )jN x j x L and therefore (Mathematica may prove to be useful)

302 [1 cos( )]

( )L

i iLF N fdx U i

i

&

0/ 2

Lij i j ijK N N dx L

The solution to the parameters is in the symbolic form 1a K F . As K is diagonal

34a / [1 cos( )]

( )j j jjF K U j

j

and

{1,2 } 34( ) [1 cos( )]sin( / )

( )ju x U j j x Lj

A method based on a weak (integral) form is quite reliable in producing a close fit. The

same approach works also when applied to finding an approximation to the exact solution

of a differential equation (even when the exact solution is not known)!

L5/12

5.3 ELEMENT CONTRIBUTION

Derivation of the virtual work expression T ( )eW a Ka F can be performed on

various ways:

(a) Start from the boundary value problem, find the exact solution, and use the constitutive

equation to derive the relationship between the nodal displacements, internal forces,

and external forces.

(b) Start from the virtual work expression of the model and use a polynomial

approximation of displacement and rotation.

(c) Start from the generic virtual work expression, apply kinematical and kinetic

assumptions of the model, and use a polynomial approximation of displacement and

rotation.

L5/13

ELEMENT CONTRIBUTION; BAR (x-axis)

Virtual work: int, ,x xw u EAu & ext

xw uf (per unit domain)

Approximation: T

1

2

1( ) ( ) x

xx

uu x u x

u

, where

1x xh

Cross-sectional area A, Young’s modulus E, and force per unit length xf (acting on the x-

axis) may depend on position. Virtual work expression depends on the model only but the

approximation can be chosen in various ways!

x EA

h

2xu

z

1xuxf

L5/14

ELEMENT CONTRIBUTION; BAR (x-axis)

T1 1

2 2

1 1 1( )

1 1 12x x x

x x

u u f hEAWu uh

&

TX

x Y

Z

uu u

u

i

&

1XY

hZ

i

Above, xf and EA are assumed constants and the elements of matrix i (1 1, 2 1, 3 1 )

are the components of the unit vector i

in the structural coordinate system ( xua i). The

algorithm of Math3300 is based on element contributions in the variational form!

x EA

h

2xu

z

1xuxf

L5/15

First, element approximation Tu N a and its variation T Tu N a a N are

substituted to the virtual work expression to get

, ,e ee

x x xW u EAu d uf d

T T T

, ,e ee

x x xW EA d f d

a N N a a N

T T, ,( )e e

ex x xW EA d f d

a N N a N

If the approximation inside an element is taken to be linear, shape functions and the

parameters of element e with 1 2[ , ]e x x are given by

2

1

1 x xx xh

N

& ,

111x h

N

& 1

2

x

x

uu

a

& 1

2

x

x

uu

a

L5/16

Assuming that Young’s modulus E, cross-sectional area A, and the distributed force xf

are constants in the element, integration over the element domain gives (the

expressions of the shape functions need to be substituted first)

2 2

1 1

T T1 1 2

2 2 1

1 11 1 1( )1 1

x xx xexx xx x

u u x xW EA dx f dx

u u x xh h h

T1 1

2 2

1 1 1( )

1 1 12x xe x

x x

u u f hEAWu uh

A derivation along these lines is valid also when Young’s modulus E, cross-sectional area

A, and the distributed force xf are not constants!

L5/17

EXAMPLE. Consider the bar model and a piecewise linear approximation. Determine the

equivalent nodal forces F of the element contribution T ( )eW a Ka F when

(a) xf is constant,

(b) xf is piecewise linear Txf N f , where the nodal values are T

1 2{ }x xf ff ,

(c) ( 1/ 2)x xf F , where is the Dirac-delta and xF the value of the force.

Answer (a) 112

xf h

F (b) 1

2

2 11 26

x

x

fhf

F (c)

112

xF

F

L5/18

The equivalent nodal forces are obtained by using Tu N a in the virtual work

expression of the external forces

ext

e xW uf d

& Tu a N

ext T

e xW f d

a N

& 1

N

& 1x x

h

2

1

2ext T

1

1xxx

x xW f dx

x xh

a

The given expressions of the distributed forces correspond to equivalent nodal forces

(a) 2

1

2ext T T T

1

1112

x xxx

x x f hW f dxx xh

a F a a

L5/19

ext T T 112

xf hW

a F a

112

xf h

F

(b) 2

1

T1 12 2ext T T

2 21 1

2 11 11 26

x x xx x x

f fx x x x hW dxf fx x x xh h

a a

1ext T T

2

2 11 26

x

x

fhWf

a F a

1

2

2 1

1 26x

x

fhf

F

(c) 2

1

2ext T T1 2

1

11 ( )12 2

x xxx

x x Fx xW F x dxx xh

a a

ext T T 112

xFW

a F a

1

12xF

F

L5/20

EXAMPLE L5/P2. The bar of the figure (EA is constant) is loaded by a distributed force.

Determine the axial displacement ( )u x by the finite element method, when 0u at 0x ,

, 0x x LEAu n F at x L and 2( / 2) sin( / (2 ))xf EA U x L . Use three elements of the

same length and a piecewise linear approximation.

Answer 1 0xu & 212xu U & 3

32xu U & 4xu U (exact sin( )

2xu UL

)

L EA x

xf

L/3

1 2 3

1 2 3 4 L/3 L/3

L5/21

In Math3300 distributed force xf , Young’s modulus E , and the cross-sectional area A

may be functions of x . Here 2( / 2 ) sin( / 2 )xf UEA L x L , and the solution is given

by

Out :

BAR EE, ρρ , AA ,AAEE Sin 2

4 1,2

BAR EE, ρρ , AA ,AAEE Sin 2

4 2,3

BAR EE, ρρ , AA ,AAEE Sin 2

4 3,4

0,0,0 0,0,0 0,0,0

3 , 0,0 2 , 0,0 0,0,0

23 , 0,0

√32 , 0,0 0,0,0

, 0,0 , 0,0 0,0,0

Here, the numerical solution to the nodal values is exact no matter the force (explanation

of superconvergence phenomenon is not a part of this course)!

L5/22

EXAMPLE L5/P3. The cross sectional area of a bar is given by 0/ 1 / (2 )A A x L .

Assuming that the approximation of displacement u is (piecewise) linear, Young’s

modulus E and density of the material are constants and distributed loading xf is due to

the gravity, determine the displacement at the free end of the bar. Use two elements of

equal size.

Answer 24( )

9g Lu L

E

, 229( )

70g Lu L

E

... 23 log 4( )

4g Lu L

E

E,

x,X

gL

one element exact

L5/23

In hand calculations, the starting point is the element approximation. Approximation, Tu N a, cross-sectional area (in terms of its nodal values), and weight per unit length

are here

1

2

1( ) x

x

uu x h x x

uh

1,

2

1 1 1 xx

x

uu

uh

1

2

1( )A

A x h x xAh

& 1

2x

Agf gA h x xAh

Virtual works of internal and external forces per unit length of a bar are given by

T T

1 1 1 1int2 2

2 2 2 2

1 1 11 1

1 1 1x x x x

x x x x

u u u uEA EAwu u u uh h

L5/24

T

1 11int3

2 22

1 11 1

x x

x x

u uAEw h x xu uAh

T

1 1ext2

2 2

1x

x

u Ah xw h x x g

u Axh

Element contribution of a typical element is obtained as integral over the domain

occupied by the element

T1 1int int 1 2

0 2 2

1 1( )1 12

h x x

x x

u uE A AW w dxu uh

T

1 1ext ext0 2 2

2 11 26

h x

x

u AghW w dxu A

L5/25

T1 1 1int ext 1 2

2 2 2

1 1 2 1( )[ ]1 1 1 22 6

x x

x x

u u AE A A ghW W Wu u Ah

Element contributions of the two elements

T1 0 0

1 1

0 01 1 117( )a a1 1 104 48

EA gLAWL

( ,2Lh 1 0,A A

02

34AA )

T1 12 0 0

2 2

a a1 1 85( )a a1 1 74 48

EA gLAWL

( ,2Lh

01

3 ,4AA

02 2

AA )

Virtual work of the structure is the sum over elements 1 2eW W W W

T1 10 0 0 0

1 1 12 2

a a5 5 87a a a 10 ( )a a5 5 74 48 4 48

EA A gL A gLAWL L

L5/26

T1 10 0

2 2

a a12 5 18( )

a a5 5 74 48EA gLAW

L

Principle of virtual work and the basic lemma of variational calculus give

T1 10 0

2 2

a a12 5 18( ) 0

a a5 5 74 48EA gLAW

L

1

2

aa

10 0

2

a12 5 180

a5 5 74 48EA gLA

L

12 2 2

1

2

a 12 5 18 5 5 18 25 / 84a 5 5 7 5 12 7 29 / 7012 12 35

gL gL gLE E E

L5/27

In Math3300 solution, the given quantities may depend on the material coordinate of

the bar (the coordinate system is associated with a structural part which may consist of

one or several elements)

In :BAR EE, ρρ , A0 1 2 , A0 1 2 1,2

BAR EE, ρρ , A0 1 2 , A0 1 2 2,3

In ∶ 0,0,0 0,0,0 0,0,0, 0,0 1 , 0,0 0,0,0, 0,0 2 , 0,0 0,0,0

Out:

0,0,0 0,0,0 0,0,0

, 0,0 , 0,0 0,0,0

, 0,0 , 0,0 0,0,0

L5/28

ELEMENT CONTRIBUTION; BEAM (xz-plane)

Virtual work: int, ,xx zz xxw w EI w & ext

zw wf (per unit length)

Approximation:

T21

2 12 2

2 2

(1 ) (1 2 )

(1 )( )

(3 2 )

( 1)

z

y

z

y

u

hw x

u

h

, where

1x xh

Above, the x-axis is chosen to coincide with the neutral axis of the beam!

x EIzz

h 2zu

z

2y1y 1zu

zf

L5/29

ELEMENT CONTRIBUTION; BEAM (xz-plane)

T

1 12 21 1

32 2

2 22 2

12 6 12 6 66 4 6 2

( )12 6 12 6 612

6 2 6 4

z z

y yzz z

z z

y y

h hu u

h h h h hEI f hWu h h uh

hh h h h

&

TX

z Y

Z

uu u

u

k

etc.

Above, zf and zzEI are assumed constants and the elements of matrices i, j and k (

1 1, 2 1, 3 1 ) are the components of the unit vectors ,i j

and k

in the structural

coordinate system.

x EIzz

h 2zu

z

2y 1y 1zu

zf

L5/30

First, element approximation Tw N a and its variation Tw a N are substituted to

the virtual work expression to get

, ,e ee

xx zz xx zW w EI w d wf d

T T T

, ,e ee

xx zz xx zW EI d f d

a N N a a N

T T, ,( )e e

exx zz xx zW EI d f d

a N N a N

The shape function expressions 1( ) /x x h & 2 1h x x (note that

, ,/xw d dx w & dx hd and therefore , , /xw w h and 2, , /xxw w h ):

L5/31

2

2

2

2

(1 ) (1 2 )

(1 )

(3 2 )

( 1)

h

h

N

& 2

,

6 (1 )

1 (1 4 3 )6(1 )

(3 2)

xh

hh

N

& , 2

6(2 1)2(3 2)16(1 2 )(3 1)

xxh

hh

N

In the next step, the shape function expressions are substituted into the virtual work

expression. Integration over the domain occupied by the element gives the element

contribution. A derivation along these lines is valid also when the given functions are

not constants!

L5/32

EXAMPLE. The integral representation of equivalent nodal forces is e zf d

F N .

Determine the equivalent nodal forces of a beam element for (a) const.zf , (b)

1 2(1 )z z zf f f , and (c) ( / 2)z zf F x h (Dirac delta at the midpoint), when

T2 2 2 2(1 ) (1 2 ) (1 ) (3 2 ) ( 1)h h N

Answer

6

612z hhf

h

F & 1 2

21 93 29 2160 60

2 3

z zh hhf hf

h h

F &

4

48z hF

h

F

L5/33

Constant

2

210 2

2

(1 ) (1 2 ) 6(1 )

612(3 2 )

( 1)

ez

z zh hhff d f h d

hh

F N

Linear 1 1 1 2

1 20 0

21 93 2

(1 )9 2160 60

2 3

z zz z

h hhf hff hd f hd

h h

F N N

Dirac

2

2

2

21/2

(1 ) (1 2 ) 4(1 )

48(3 2 )

( 1)

ez

z zh hFf d F

hh

F N

L5/34

STRUCTURAL PARTS AND ELEMENTS

Structural part is just part of a structure idealized as a rigid body, bar, beam, plate etc.

The proper idealization depends on the aim of the analysis. Understanding the

underlying assumptions and built-in restrictions of the engineering models is quite

essential in the modeling of a structure or a machine part. “A good model is as simple

as possible but not simpler than that”.

Element is a part of a structural part. In principle, element division is chosen to reduce

the discretization error somewhat below the modeling error. The use of a finer division

than necessary is waste of resources (money or time) as the total error cannot be

reduced below the modeling error no matter the numerical effort.