Consider a straight two-force member AB subjected at A and B to equal and opposite forces F and -F...
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r a straight two-force member AB subjected at A to equal and opposite forces F and -F directed alon the member AB at C A B C F -F F -F C A and drawing the free-body diagram of portion AC, we conclude that the internal forces which existed at C in member AB are equivalent to an axial force -F equal and opposite to F. For a two-force member which is not straight, the internal forces reduce to a force-couple system and not to a Chapter 7 FORCES IN BEAMS AND CABLES Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
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Transcript of Consider a straight two-force member AB subjected at A and B to equal and opposite forces F and -F...
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
A
B
C
F
-F
F
-FC
A
and drawing the free-body diagram of portion AC, we conclude that the internal forces which existed at C in member AB are equivalent to an axial force -F equal and opposite to F.
For a two-force member which is not straight, the internal forces reduce to a force-couple system and not to a single force.
Chapter 7 FORCES IN BEAMS AND CABLES
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
D
J
T
MF
V
Considering multiforce member AD, cutting it at J, and drawing the free-body 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. A
B
C
D
J
Cx
Cy
Ax
Ay
T
FBE
The magnitude of the shearing force measures the shear at point J, and the moment of the couple is referred to as thebending moment at J. Since an equal and opposite force-couple system would have been obtained by considering the free-body diagramof portion AJ, it is necessary to specify which portion of memberAD was used when recording the answers.
D
J
T
MF
V
A
B
C
D
J
Cx
Cy
Ax
Ay
T
FBE
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
V
M
V’
M’
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.
V
M
V’
M’
Constructing these diagrams is generally facilitated by using the relationships
dVdx
= -wdMdx = V
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
Ay
Ax
P1P2 P3x1
Bx
Byd
L
A
BD
C1
C2 C3
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.
x2
x3
Ay
Ax
P1P2 P3
x1
x2
x3
Bx
Byd
L
A
BD
C1
C2 C3
Knowing the coordinates of point D, an additional equation can be obtained by constructing the free-body diagram of portions AD or DB of the cable.
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.
C
TO
TD
T
TO
W
W
For a cable carrying a distributed load we observe that the horizontal component of the tension T at D is constant andmust be equal to the tension TO at C. The magnitude and direction of T are obtained from the force triangle.
T = TO + W 22 tan = WTO
x
y
A
B
C
D(x,y)
In a suspension bridge, the loadis uniformlydistributed alongthe horizontal.The load supportedby portion CD of the cable isW = wx.
w
The curve formed by the cable is a parabola of equation
y =wx2
2To
x
y
A
B
C
D(x,y)
For a cable hanging under its own weight, the load is uniformlydistributed alongthe cable itself.The load supportedby portion CD of the cable isW = ws.
Choosing the origin O at a distance c = To /w below C , the relations between the geometry of the cantenary andcable tension are
s
O
c
s = c sinh y = c cosh y 2 - s 2 = c 2 xc
xc
TO = wc W = ws T = wy
