Bölüm 7 Çubuk ve Kablolarda Kuvvetler
Consider a straight two-force member AB subjected at A
and B to equal and opposite forces F and -F directed along AB.
Cutting the member AB at C and drawing the free-body
diagram of portion AC, we
conclude that the internal forces
-F
B
which existed at C in member AB
are equivalent to an axial force -F
equal and opposite to F.
C
C
-F
A
For a two-force member which is
A
not straight, the internal forces
reduce to a force-couple system
F
and not to a single force.
F
D
D
T
T
J
J
C
Cx
Cy
V
M
F
FBE
B
Ax
A
Ay
Considering multiforce member AD,
cutting it at J, and drawing the freebody diagram of portion JD, we
conclude that the internal forces at J
are equivalent to a force couple system
consisting of the axial force F, the
shearing force V, and a couple M.
D
D
T
J
C
Cx
Cy
FBE
B
Ax
A
Ay
T
The magnitude of
the shearing force
V
J
measures the shear
at point J, and the
M
moment of the couple
F
is referred to as the
bending moment at J.
Since an equal and opposite force-couple
system would have been obtained by
considering the free-body diagram
of portion AJ, it is necessary to specify
which portion of member
AD was used when recording the
answers.
Beams are usually long, straight prismatic members
designed to support loads applied at various points along
the member. In general, the loads are perpendicular to the
axis of the beam and produce only shear and bending in
the beam. The loads may be either concentrated at specific
points, or distributed along the entire length or a portion of
the beam. The beam itself may be supported in various
ways.
Since only statically determinate beams are considered in
this text, we limit our analysis to that of simply supported
beams, overhanging beams, and cantilevered beams.
To obtain the shear V and bending moment M at a given
point C of a beam, we first determine the reactions at the
supports by considering the entire beam as a free body.
We then cut the beam at C and use the free-body diagram
of one of the two portions obtained in this fashion to
determine V and M.
C
C
M
V
M’
V’
The sign convention for positive shear force and bending
moment is as shown. Once the values of shear and bending
moment are established at several select points along the
beam, it is usually possible to draw the shear diagram and
bending-moment diagram for the entire beam.
M M’
V
V’
Constructing these diagrams is generally facilitated by using
the relationships
dV
= -w
dx
dM
=V
dx
Integrating these, we find
VD-VC = -(area under the load curve between C and D )
MD-MC = area under the shear curve between C and D
L
Ay
A
Ax
By
d
C1
D
C2
P1
P2
x1
B
C3
Bx
P3
x2
x
3
For a flexible cable with negligible weight supporting
concentrated loads, and using the entire cable AB as a
free body, the three available equations of equilibrium are
not sufficient to determine the four unknown reactions at
supports A and B.
L
Ay
A
Ax
d
C1
By
D
C2
x1 P1
P2
x2
B Bx
C3
P3
Knowing the coordinates
of point D, an additional
equation can be obtained
by constructing the freebody diagram of portions
AD or DB of the cable.
x3
Once the reactions are known, the elevation of any point of
the cable and the tension in any section of the cable can be
determined.
T
D
TO
q
T
W
C
q
TO
W
For a cable carrying a distributed load we observe that the
horizontal component of the tension T at D is constant and
must be equal to the tension TO at C. The magnitude and
direction of T are obtained from the force triangle.
T = TO2+ W 2
tan q =
W
TO
y
B
A
D(x,y)
C
x
w
In a suspension
bridge, the load
is uniformly
distributed along
the horizontal.
The load supported
by portion CD of
the cable is
W = wx.
The curve formed by the cable is a parabola of equation
wx2
y=
2To
y
B
D(x,y)
A
s
C
c
O
x
For a cable hanging
under its own weight,
the load is uniformly
distributed along
the cable itself.
The load supported
by portion CD of
the cable is
W = ws.
Choosing the origin O at a distance c = To /w below C , the
relations between the geometry of the cantenary and
cable tension are
x
s = c sinh c
TO = wc
x
y = c cosh c
y2-s2=c2
W = ws
T = wy