522
9 Variational Principles in Mechanics and Their Applications
the mass m relative to the reference point P; (2) The kinetic energy of motion of the
system relative to the reference point P; (3) The scalar product of the radius vector of
the reference point and the momentum of the system relative to the reference point.
Example 9.1.1 The problem of compression bar stability. The simple bar of length
L that the both ends are under the action of the compression force P occurs the
longitudinal bending, find the minimum compression force P (the Euler critical
compressive force) making the bar bend. The boundary is y(0) = 0.
Solution Let the deflection curve be y = y(x), EI is the flexural rigidity of the bar,
ϕ is the tangential direction angle of the bar at some point, k =
dϕ
ds
is the curvature
of the bar. The bending strain energy of the bar is
J 1 =
1
2
L
0
E I k
2 ds =
1
2
L
0
E I k
2
1 + y 2 dx
The potential energy of load caused by the deflection of the bar is
J 2 = −P
L
0
(1 − cos ϕ)ds = −P
L
0
(1 − cos ϕ)
1 + y 2 dx
In the case of small deformation, ds ≈ dx, y
≈ 0, 1 − cos ϕ ≈
ϕ
2
2
, thus J 2 can
be written as
J 2 = −
P
2
L
0
ϕ
2 ds ≈ −
P
2
L
0
ϕ
2 dx
The total potential energy is
J = J 1 + J 2 =
1
2
L
0
E I
dϕ
ds
2
− Pϕ
2
ds ≈
1
2
L
0
E I
dϕ
dx
2
− Pϕ
2
dx
The Euler equation of the functional is
ϕ
+ a
2
ϕ = 0
where, a
2
=
P
E I
. The general solution of the Euler equation is
ϕ = c 1 cos ax + c 2 sin ax
9 Variational Principles in Mechanics and Their Applications
the mass m relative to the reference point P; (2) The kinetic energy of motion of the
system relative to the reference point P; (3) The scalar product of the radius vector of
the reference point and the momentum of the system relative to the reference point.
Example 9.1.1 The problem of compression bar stability. The simple bar of length
L that the both ends are under the action of the compression force P occurs the
longitudinal bending, find the minimum compression force P (the Euler critical
compressive force) making the bar bend. The boundary is y(0) = 0.
Solution Let the deflection curve be y = y(x), EI is the flexural rigidity of the bar,
ϕ is the tangential direction angle of the bar at some point, k =
dϕ
ds
is the curvature
of the bar. The bending strain energy of the bar is
J 1 =
1
2
L
0
E I k
2 ds =
1
2
L
0
E I k
2
1 + y 2 dx
The potential energy of load caused by the deflection of the bar is
J 2 = −P
L
0
(1 − cos ϕ)ds = −P
L
0
(1 − cos ϕ)
1 + y 2 dx
In the case of small deformation, ds ≈ dx, y
≈ 0, 1 − cos ϕ ≈
ϕ
2
2
, thus J 2 can
be written as
J 2 = −
P
2
L
0
ϕ
2 ds ≈ −
P
2
L
0
ϕ
2 dx
The total potential energy is
J = J 1 + J 2 =
1
2
L
0
E I
dϕ
ds
2
− Pϕ
2
ds ≈
1
2
L
0
E I
dϕ
dx
2
− Pϕ
2
dx
The Euler equation of the functional is
ϕ
+ a
2
ϕ = 0
where, a
2
=
P
E I
. The general solution of the Euler equation is
ϕ = c 1 cos ax + c 2 sin ax
