ENERGY METHODS.pdf

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17
Energy methods
17.1
Introduction
Energy methods are very useful for analysing structures, especially for those that are statically
indeterminate. This chapter introduces the principle of virtual work and applies it to statically
determinate and statically indeterminate frameworks. The chapter also shows how the method can
be used for the plastic design of beams and rigid-jointed plane frames.
The chapter then introduces strain energy and complementary strain energy, and through the
use of worked examples, shows how these methods can be used for analysing structures.
In Chapters
24
and
25,
energy methods are used for developing the finite element method,
which is one of the most powerful methods for analysing massive and complex structures with the
aid of digital computers.
17.2
Principle of virtual work
In its simplest form the
principle
o
virtual work
is that
f
For a system
of
forces acting
on
a particle, the particle is
in
statical equilibrium
if;
when it
i
given any virtual displacement, the net work done by theforces is zero.
s
A
virtual displacement is any arbitrary displacement of the particle. In the virtual displacement the
forces are assumed to remain constant and parallel to their original lines of actions. Consider a
particle under the action of three forces,
F,,
F2
and
F,,
Figure
17.1.
Figure
17.1
System
of
forces
in
statical
equilibrium
acting
on
a particle.
Imagine the particle to be given a virtual displacement of any magnitude in any direction.
Suppose the displacements of the particle along the lines of action of the forces
F , ,
F,
and
F,,
are
6,, 6,
and
6,,
respectively; these are known as
corresponding
displacements. Then the forces form
a system in statical equilibrium if
Deflections
of
beams
391
F,6,
+
F,6,
+
F363
=
0
(17.1)
On the basis of the principle of virtual work we can show that the resultant of the forces acting on
a particle
in
statical equilibrium is zero. Suppose the forces
F,,
F,
and
F,,
acting on the particle
of Figure 17.1, have a resultant of magnitude
R
in some direction; then by giving the particle a
suitable virtual displacement,
A,
say, in the direction of
R,
the net work is
RA
But by the principle of virtual work the net work is zero,
so
that
RA
=
0
(17.2)
As
A
can be non-zero,
R
must be zero. Hence, by adopting the principle of virtual work as a basic
concept, we can show that the resultant of a system of forces in statical equilibrium is zero.
17.3
Deflections of
beams
In a pin-jointed frame subjected to loads applied to the joints only the tensile load in any member
is constant throughout the length of that member. In the case of a beam under lateral loads the
bending moments and shearing forces may vary from one section to another,
so
that the state
of
stress is not uniform along the length of the beam. In applying the principle of virtual work to
problems of beams we must consider the loading actions on the virtual displacement of an
elemental length of the beam.
Figure 17.2 Deflections
of
a
straight beam.
Consider a straight beam
AB,
Figure 17.2, which is in statical equilibrium under the action of
a system
of
external forces and couples. The beam is divided into a number
of
short lengths; the
loading actions on a short length such as
6z
consist of bending moments
M
and
( M
+
JM),
an
external lateral load
W,
and lateral shearing forces at the ends of the short length. Now suppose
392
Energy
methods
the short lengths of the beam are given small virtual displacements,
8.
If the elements remain
,
connected to each other, then for given values of
8
the external forces, such as W suffer certain
displacements, such as
6.
Then the values of
8
and
6
form a compatible
system
of rotations and
displacements, and the virtual work of any system of forces and couples
in
statical in equilibrium
on
these rotations and displacements is zero. Then
y 6 M x 0 + y W x 6
=
0
(17.3)
because the net work of the internal shearing forces is zero. The summation
Z6M
x
8
is
carried out
for all short lengths
of
the beam, whereas the summation
2
W
x
6
is carried out for all external
loads, including couples and force reactions at points
of
support. If the virtual rotations
8
are
small, the virtual displacements
6
can be found easily. If the lengths
6z
of the beam are
infinitesimally small,
(1
7.4)
where the integration
is
carried out over the whole length
L
of the beam. But
I,
Now
z = L
=o
0dM
=
[MB
- , M d B I = L
=o
and is the work of the end couples
on
their respective virtual displacements;
t h ~ s
work has already
been taken account of in the summation
ZW
x
6,
so
that
equation (17.3) becomes
(17.5)
Now
(de/dz)
is the curvature of the beam when it
is
given the virtual rotations and displacements.
If we put
d
-
-
e -
-
1
dz
R
where
R
is
the radius of curvature
of
the beam, then
(17.6)
(17.7)
As
an example of the application
of
equation (17.7), consider the cantilever shown in Figure 17.3;
having a uniform flexural stiffness EI. The cantilever carries a vertical load
W
at the free end; the
Deflections
of
barns
393
bending moment at any section due to
W
is
Wz,
so
that, if the beam remains elastic, the
corresponding curvature at any section is
-
1
-
-
-
R
El
w
z
Suppose the corresponding deflection
of W
is
6,
Figure
17.3;
then the values of
1lR
and
6
form
a
system of compatible curvature and displacements.
Figure 17.3
Deflections
of
a
cantilever with an end
load.
We derive a simple system of forces and couples in statical equilibrium by applying a unit vertical
load at the end of the cantilever; the bending moment at any section due to this unit load is
M
=
l x z
=
z
Then, from equation
(17.7),
1
x
6
=
pfp)&
=
JoL
S
d
Z
Then
6 = -
WL
3
3EI
Problem
17.1
A
simply-supportedbeam, ofuniform flexural stiffness
EI,
carries a lateral load
W
at a distance
u
from the end
A.
Estimate the vertical deflection
of
W.
394
Energy methods
Solution
The bending moment a distance
zI
from
A ,
for the section
AB,
is
Wbz
I
-
L
The curvature for
A B
if
therefore
Wbz
1
-
1
-
-
-
Rl
EIL
Similarly, the curvature at any section in
B C
is
-
1
-
-
-
Waz,
EIL
R2
Now consider the beam with a
unit
vertical load at
B ;
the bending moments
at
sections
i
AB
and
n
B C
are, respectively,
MI
=
bZl
-,
L
M2
=
azz
-
L
Then, equation
( 1
7.7)
gives
=
+
l o
ZF
u
Wb2
2
Zl
4
+
6
=
lo0
$)
4
/b”
4
i ) h 2
MI[
1
s,
h
Wa2
2
k
2
2&- -

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