12
Computational mechanics
The algebraic difficulties of analytic mechanical solutions are rather over-
whelming and become insurmountable when solving real-world problems. Com-
putational mechanics is based on the discretization of the geometric continuum
and describing its physical behavior in terms of generalized coordinates. Its
focus is on computing numerical solutions to practical problems of engineering
mechanics.
12.1 Three-dimensional elasticity
One of the fundamental concepts necessary to understanding continuum me-
chanical systems is a generic treatment of elasticity described in detail in the
classical reference of the subject [17]. When an elastic continuum undergoes
a one-dimensional deformation, like in the case of the bar discussed in Section
11.3, Young’s modulus was adequate to describe the changes.
For a general three-dimensional elastic continuum we need another coeffi-
cient, introduced by Poisson, to capture the three-dimensional elastic behav-
ior. Poisson’s ratio measures the contraction of the cross-section while an
object such as a beam is stretched. The ratio ν is defined as the ratio of the
relative contraction and the relative elongation:
ν =
dr
r
/
dl
l
.
Here a beam with circular cross-section and radius r is assumed. Poisson’s
ratio is in the range of zero to 1/2 and expresses the compressibility of the
material. The two constants are also often related as
μ =
E
2(1 + ν)
,
and
λ =
(1 + ν)(1 2ν)
.
177
178 Applied calculus of variations for engineers
Here μ and λ are the so-called Lam´e constants.
In a three-dimensional elastic body, the elasticity relations could vary sig-
nificantly. Let us consider isotropic materials, whose elastic behavior is inde-
pendent of the material orientation. In this case Young’s modulus is replaced
by an elasticity matrix whose terms are only dependent on the Lam´e constants
as follows
D =
λ +2μλ λ000
λλ+2μλ000
λλλ+2μ 000
000μ 00
0000μ 0
00000μ
.
Viewing an infinitesimal cube of the three-dimensional body, there are six
stress components on the element,
σ
=
σ
x
σ
y
σ
z
τ
yz
τ
xz
τ
xy
.
The first three are normal and the second three are shear stresses. There are
also six strain components
=
x
y
z
γ
yz
γ
xz
γ
xy
.
The first three are extensional strains and the last three are rotational strains.
The stress-strain relationship is described by the generalized Hooke’s law as
σ
= D.
This will be the fundamental component of the computational techniques for
elastic bodies. Let us further designate the location of an interior point of the
elastic body with
r
(x, y, z)=xi + yj + zk =
x
y
z
,
Computational mechanics 179
and the displacements of the point with
u
(x, y, z)=ui + vj + wk =
u
v
w
.
Then the following strain relations hold:
x
=
∂u
∂x
,
y
=
∂v
∂y
,
and
z
=
∂w
∂z
.
These extensional strains manifest the change of rate of the displacement of
an interior point of the elastic continuum with respect to the coordinate di-
rections.
The rotational strains are computed as
γ
yz
=
∂v
∂z
+
∂w
∂y
,
γ
xz
=
∂u
∂z
+
∂w
∂x
,
and
γ
xy
=
∂u
∂y
+
∂v
∂x
.
These terms define the rate of change of the angle between two lines crossing
at the interior point that were perpendicular in the un-deformed body and
get distorted during the elastic deformation.
The strain energy contained in the three-dimensional elastic continuum is
E
s
=
1
2
V
σ
T
dV =
1
2
V
σ
x
σ
y
σ
z
τ
yz
τ
xz
τ
xy
x
y
z
γ
yz
γ
xz
γ
xy
dV.
We will also consider distributed forces acting at every point of the volume
(like the weight of the beam in Section 11.3), described by
180 Applied calculus of variations for engineers
f
= f
x
i + f
y
j + f
z
k =
f
x
f
y
f
z
.
The work of these forces is based on the displacements they caused at the
certain points and computed as
W =
V
u
T
fdV. =
V
uvw
f
x
f
y
f
z
dV.
The above two energy components constitute the total potential energy of the
volume as
E
p
= E
s
W.
In order to evaluate the dynamic behavior of the three-dimensional body,
the kinetic energy also needs to be computed. Let the velocities at every point
of the volume be described by
˙u
(x, y, z)= ˙ui vj wk =
˙u
˙v
˙w
.
With a mass density of ρ, assumed to be constant throughout the volume, the
kinetic energy of the body is
E
k
=
1
2
ρ
V
˙u
T
˙udV.
We are now in the position to write the variational statement describing
the equilibrium of the three-dimensional elastic body:
I(u
(x, y, z)) =
V
(E
k
E
p
)dV = extremum,
which is of course Hamilton’s principle.
The unknown displacement solution of the body at every (x, y, z)pointis
the subject of the computational solution discussed in the next sections.
12.2 Lagrangian formulation
The equations will be obtained by finding an approximate solution of the vari-
ational problem based on the total energy of the system as follows. For

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