11
Analytic mechanics
Analytic mechanics is a mathematical science, but it is of high importance
for engineers as it provides analytic solutions to fundamental problems of
engineering mechanics. At the same time it establishes generally applicable
procedures. Mathematical physics texts, such as [5] and [6], laid the founda-
tion for these analytic approaches addressing physical problems.
In the following sections we find analytic solutions for classical mechanical
problems of elasticity utilizing Hamilton’s principle. The most fitting applica-
tion is the excitation of an elastic system by displacing it from its equilibrium
position. In this case the system will vibrate with a frequency characteristic
to its geometry and material, while constantly exchanging kinetic and poten-
tial energy.
The case of non-conservative systems, where energy loss may occur due to
dissipation of the energy, will not be discussed. Hamilton’s principle may be
extended to non-conservative systems, but the added difficulties do not en-
hance the discussion of the variational aspects, which is our main focus.
11.1 Elastic string vibrations
We now consider the vibrations of an elastic string. Let us assume that the
equilibrium position of the string is along the x axis, and the endpoints are
located at x =0andx = L. We will stretch the string (since it is elastic) by
displacing it from its equilibrium with some
ΔL
value, resulting in a certain force F exertedonbothendpointstoholditin
place. We assume there is no damping and the string will vibrate indefinitely
if displaced, i.e., the system is conservative.
A particle of the string located at the coordinate value x at the time t has
a yet unknown displacement value of y(x, t). The boundary conditions are:
157
158 Applied calculus of variations for engineers
y(0,t)=y(L, t)=0,
in other words the string is clamped at the ends. In order to use Hamilton’s
principle, we need to compute the kinetic and potential energies.
With unit length mass of ρ, the kinetic energy is of the form
E
k
=
1
2
L
0
ρ(
∂y
∂t
)
2
dx.
The potential energy is related to the elongation (stretching) of the string.
The arc length of the elastic string is
L
0
1+(
∂y
∂x
)
2
dx,
and the elongation due to the transversal motion is
ΔL =
L
0
1+(
∂y
∂x
)
2
dx L.
Assuming that the elongation is small, i.e.,
|
∂y
∂x
| < 1,
it is reasonable to approximate
1+(
∂y
∂x
)
2
1+
1
2
(
∂y
∂x
)
2
.
The elongation by substitution becomes
ΔL
1
2
L
0
(
∂y
∂x
)
2
dx.
Hence the potential energy contained in the elongated string is
E
p
=
1
2
F ΔL =
F
2
L
0
(
∂y
∂x
)
2
dx.
We are now in a position to apply Hamilton’s principle. The variational prob-
lem becomes
I(y)=
t
1
t
0
(E
k
E
p
)dt =
1
2
t
1
t
0
L
0
(ρ(
∂y
∂t
)
2
F (
∂y
∂x
)
2
)dxdt = extremum.
The Euler-Lagrange differential equation for a function of two variables, de-
rived in Section 3.2, is applicable and results in
F
2
y
∂x
2
= ρ
2
y
∂t
2
. (11.1)
Analytic mechanics 159
This is the well-known differential equation of the elastic string, also known
as the wave equation.
The solution of the problem may be solved by separation. We seek a solu-
tion in the form of
y(x, t)=a(t)b(x),
separating it into time and geometry dependent components. Then
2
y
∂x
2
= b

(x)a(t)
and
2
y
∂t
2
= a

(t)b(x),
where
b

(x)=
2
b
∂x
2
,
a

(t)=
2
a
∂t
2
.
Substituting into Equation (11.1) yields
b

(x)
b(x)
=
1
f
2
a

(t)
a(t)
,
where for future convenience we introduced
f
2
=
F
ρ
.
The two sides of this differential equation are dependent on x and t, respec-
tively. Their equality is required at any x and t values, which implies that
the two sides are constant. Let’s denote the constant by λ and separate the
(partial) differential equation into two ordinary differential equations:
2
b
∂x
2
+ λb(x)=0,
and
2
a
∂t
2
+ f
2
λa(t)=0.
The solution of these equations may be obtained by the techniques learned in
Chapter 5 for the eigenvalue problems. The first equation has the eigensolu-
tions of the form
b
k
(x)=sin(
L
x); k =1, 2,...,
160 Applied calculus of variations for engineers
corresponding to the eigenvalues
λ
k
=
k
2
π
2
L
2
.
Applying these values we obtain the time-dependent solution from the second
equation by means of classical calculus in the form of
a
k
(t)=c
k
cos(
kπf
L
t)+d
k
sin(
kπf
L
t),
with c
k
,d
k
arbitrary coefficients. Considering that at t = 0 the string is in a
static equilibrium position
a
(t =0)=0
we obtain d
k
= 0 and the solution of
a
k
(t)=c
k
cos(
kπf
L
t).
The fundamental solutions of the problem become
y
k
(x, t)=c
k
cos(
kπf
L
t)sin(
L
x); k =1, 2,....
For any specific value of k the solution is a periodic function with period
2L
kf
and frequency
kπf
L
.
The quantities
λ
k
=
L
for a specific k value are the natural frequencies of the string. The correspond-
ing fundamental solutions are natural vibration modes shapes, or the normal
modes. The first three normal modes are shown in Figure 11.1 for an elastic
spring of unit tension force, mass density, and span. The figure demonstrates
that the half-period decreases along with increasing mode number.
The motion is initiated by displacing the string and releasing it. Let us
define this initial enforced amplitude as
y(x
m
, 0) = y
m
,

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