Hall A.S., Archer F.E., Gilbert R.I. Engineering Statics
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10
kN
kN
2
M"-----
3
kN
2
kN
10
kN
(4
Q
3
ICN
2
kN
re
7.
This equilibrant
Q
is the force exerted on body
AC
by body
C
opposite force
Q'
must be exerted upon
CD
by
AC,
and this force m
e~uilibrium of freebody
CD
is considered (Figure 7.2~).
The fact that the equilibrant force does not actually pass through the cross-section
C
is
of
little importance since it may be replaced by a force at
C
together with a couple
(Figure 7.3a).
It
is
convenient in practice to resolve the force into its
two
components,
one parallel to the axis of the bar and one perpendicu~ar to it (Figure
733).
Thus in
Figure 7.2b the equilibrant
Q
is expressed by the three quantities: magnitude, direction
and position of a single force, whereas in Figure 7.3b it is replaced by a statically
equivalent system comprising
two
forces
S
and
N
and a couple
M,
all
acting at the
centroid of the cross-section
C.
It is more convenient to determine these components
directly rather than to determine the equilibrant as a single force.
(4
S
3
kN
This topic deals with beams whose cross-sectional dimensions are small compared
with their length. In many examples such beams will be represented by a single line
which is the longitudinal axis of the bar. Where this axis
is
curved, the component forces
of the internal action at any section are taken parallel and perpendicular to the tangent
to the curve at that section.
The component parallel to the axis is called the
~~~Z~~ce.
This is usually abbre~iat~d
to
AA?
and the force is denoted by
N
The component perpendicular to the axis is called
the
s~e~~~~ce.
This is abbreviated to
S,E:
and is denoted by
S
(or sometim~s
V).
The
couple
is
called the
~en~~ng
~u~en~.
This is abbreviated to
~.M.
and is denoted by
M.
oss-section
of
a beam there may exist an internal action.
.
In
this
ill~~t~atio~,
if
we
find that
M
is
anti
small
element
of
length
Cl
c2
The small element in Figure
7.4
is subjected to a pair of forces
NI
and
NZ
which in
this instance tend to increase its length (i.e. the element is in tension). The axial force
IV
is said to be positive if it puts the element in tension, and ne
compression.
The element is subjected to
a
pair of forces
S,
and
Sz
which tend to cause
a
shearing
type of de~ormation* If the forces
SI
and
S,
are in the directions shown in Figure
7.4,
then the shear force
S
is said to be positve.
bending moment
M
is said to be positive if the element bends concavely upward (or
concavely towards a specified positive direction if the bar
is
not horizontal).
The element is subjected to a pair of couples
M,
and
M’
which tend to ben
These sign conventions are summarized in Figure
7.5.
+
ve
+
ve
+
ve
-
ve
-
ve
-
ve
AXIAL
FORCE
SHEAR
FORCE
The above definition of the sign conventions indicates the physical significance of
these signs. In many practical problems the signs may be determined bp i~a~ining the
nature of the deformation. For instance, a simply supported beam carrying downward
loads will bend
so
that it becomes concave on the top, Provided the
x
axis
is
taken from
left to right and the
y
axis upward, such bending will be denoted as positive.
In more general problems, the physical determination of signs is less simple and
analytical rules are more convenient. This
is
important also for computer calculations,
since the computer cannot imagine the deformed shape of the beam se
an
x
axis
as
running along the beam as in Figure
7.6.
Then,
if
the beam is cut at
a
section
C,
the
x
axis is directed outward on one cut face, the left-hand face in Figure
7.6,
and
we may call this the
face
o~~os~~~~e
~nc~~ence.
If the forces
N,
S
and
M
on this face
with the direction of the
x
and y axes, then they are defined as positive.
face, the
x
axis is directed inward as on the right-hand side of Figure
7.6.
This face
is
called
~~e~ce
o~ne~ative ~~ci~ence
and on this face
N,
S
and Mare positive if they
~~sa~ee
with the
x
and y axes,
y
Face
of
+ve incidence Face
of
-ve
incidence
ure
'7.7
shows the beam of Figure
'7.2a
(page
90)
cut at
C.
On each freebody, the
al
actions are shown acting in their positive sense as defined above. On the left-
hand freebody, which has the positive cut face, Nand
S
are shown in the directions of
x
and y and ~anticloc~wise. These indicate the positive directions. In Figure
7.7%
the cut
ative,
so
the positive directions of
N
and
S
are opposed to
x
and y and
ositive
M is
clockwise.
(a)
y
!
6)
5
kN
S
kN
3
kN
2
kN
Either freebody may now be used, and the equilibrium equations will give
N,
S
and
Mwith the correct signs.
INTERNAL
i4c:TroN.S
IN
n the
axis
of
the beam is curved at the section where it
is
require
axes x and y are taken at this particular section such that
x
is
ta
beam. This
is
ill~strate$
in
Example
7.3.
~istributed loading was discussed in Section
4.4.
&"hen a beam supports
a
floor, the
weight of the floor,
as
well
as
the weight
of
the beam itself, constitutes a distributed
loading.
If
each unit length of beam supports the same load, the load is
~~~0~~
~~s~~~~~e~
In order to determine the reactions, the e~uilibrium of the whole beam
is
considered, and the distributed load is replaced by its resultant. However, to determine
internal actions at a section, the beam is cut at that section. The freebody to one side
of