Analogy of the Approximate Methods
So far we have seen that calculus of variations and the finite element method both refer to the integral equations. The finite difference method on the other hand deals primarily with the differential equations. At any stage of the problem however, these
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So far we have seen that calculus of variations and the finite element method both refer to the integral equations. The fmite difference method on the other hand deals primarily with the differential equations. At any stage of the problem however, these three methods follow very analogous paths. In order to illustrate the similarities between them, let us consider an elementary problem such as a simple prismatic beam subject to a distributed transverse load p(x). The integral and the differential equations of this problem are, respectively,
~
=
fi
(EIy2 - 2 py) dx
(5.1)
and ElyIV = p.
(5.2)
Substituting for y in Eq. (5.1) its equivalence in the form of central differences, the functional (the total potential energy of the system) becomes n
1T
= EI I
2 i=l
(Yi-1-2Yi + Yi+1)2 b2
where the expression for the derivative is the order of h 2 .
H. Kardestuncer, Discrete Mechanics, A Unified Approach © Springer-Verlag Wien 1975
43
The stationary value of
i)1f
i)Yi
1f
,then, indicates that
= 0 = EI
~
(
Yi -2 - 4 Yi-l + 6 Yi - 4 Yi+l + Yi+2 - Pi)' (5.3)
which is precisely the fmite difference approximation of the differential equation given by Eq. (5.2). One would certainly anticIpate this since Eq. (5.2) is the corresponding HulerLagrange equation ofEq. (5.1). Because
where F = EI y2 _ PY , 2
yields Eq. (5.2). It can also be shown that the same difference equation [Eq. 5.2] is encountered (implicitly) in the ftnite element method. Take, for instance, the localized displacement function in the form of a cubic polynomial
(5.4)
u =
After writing the total potential energy of the system 1f
=
t J e:*
D e:
with the stress and strain expressions
d x -
fp u
dx
44 the minimum potential energy principle yields the following stiffness matrix equation for a line element. 12 -6
vLZ kif
Pe
!k i ,i+1 I
= ------~--------
I I
k i +1,i
k i +1,i+1
4 L
I
I I
I I
-12
6
V LZ
I -6 I L2"
L
I I
LZ
2
---------~----------symm.
I I I I I I I
12
V
-6
4 L
The assembly of the two adjacent elements at any nodal point i indicates that
which for the elements of equal length of h becomes
Substituting the central differences for the slopes
Yi+1 - Yi-l 2h
45
and carrying the multiplication, one will obtain
which possesses the same difference molecule as that of Eq. (5.3). The problem posed by Eq. (5.1) is solved by the fmite element method using six line elements. The complete stiffness matrix after the introduction of the boundary conditions i.e., u1 = u7 = 0, takes the following form.
4
-6 24
2 0 8
0 -12 -6 24
6 2 0 8
symm.
0 -12 -6 24
6 2 0 8
0 -12 -6 24
6 2 0 8
0 -12 -6 24
The solution of this set yields u u
2 3
= u = u u
6 5 4
0.04428 • 0.076032 • = 0.08748 •
which represent the displacements at the corresponding points.
6 2 0 8
0 6 2
4
-ph 2 /12 ph 0 ph 0 ph 0 ph 0 ph 0 ph 2 /12
46 These, of course, are in agreement with those obtained before. The discrepancies, however, are expected to increase for large systems since the finite element method requires the so
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