Post on 22-Apr-2017
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.