Pwerhitungan struktur kabel listrick
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1. The document discusses the shapes of cables that hang under their own weight. A catenary cable has uniform density along its length and forms the shape y = 1/acos(ax) + C. 2. A parabolic cable has uniform density along the x-axis, supporting a horizontal span. Its shape is a parabola described by y = ax^2/2 + C. 3. Historically, Galileo incorrectly thought the catenary was parabolic, but Huygens, Bernoulli, and Leibniz later derived the correct formula for the catenary curve.
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The Catenary and Parabolic Cables
The Catenary
Assume that one has a cable which
hangs under its own weight with only
tangential forces acting along the length of
the cable. The cable is assumed to have
uniform density, w0 , along its length.
Arrange the coordinate system so that the y-
axis passes through the lowest point of the
cable. Consider a free-body diagram of that
portion of the cable above the interval [0,x].
Suppose that the length of this portion of the
cable is s. Then its weight is w0s . In the
free-body diagram this is a force acting
downward on that portion of the cable. There is also a tangential force T0 acting
horizontally to the left in the free-body diagram. This force is due to the cable to the left
of the figure. Similarly, there is a force T acting tangentially at the right in the free-body
diagram. This force is due to the cable to the right of the figure. Since the cable is
assumed static, the sum of the horizontal and vertical forces must be zero. This leads to
two equations.
T cos(!) = T0 and T sin(!) = w0s
Dividing we eliminate T and get the following.
tan(!) =
w0s
T0
We rewrite this in terms of the derivative.
dy
dx
= a ! s where a =
w0
T0
We take the derivative of this expression with respect to x to get the following
expression.
d2
y
dx
2 = a !
ds
dx
= a ! 1 +
dy
dx( )
2
because
ds
dx
= 1+
dy
dx( )
2
Now let p =
dy
dx
in the above expression to get
dp
dx
= a ! 1+ p
2
.
w0s
T0
T
x ! axis
y ! axis
Figure 1
"](https://image.slidesharecdn.com/pwerhitunganstrukturkabellistrick-140416233745-phpapp02/85/Pwerhitungan-struktur-kabel-listrick-1-320.jpg)
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Pwerhitungan struktur kabel listrick
- 1. 1 The Catenary and Parabolic Cables The Catenary Assume that one has a cable which hangs under its own weight with only tangential forces acting along the length of the cable. The cable is assumed to have uniform density, w0 , along its length. Arrange the coordinate system so that the y- axis passes through the lowest point of the cable. Consider a free-body diagram of that portion of the cable above the interval [0,x]. Suppose that the length of this portion of the cable is s. Then its weight is w0s . In the free-body diagram this is a force acting downward on that portion of the cable. There is also a tangential force T0 acting horizontally to the left in the free-body diagram. This force is due to the cable to the left of the figure. Similarly, there is a force T acting tangentially at the right in the free-body diagram. This force is due to the cable to the right of the figure. Since the cable is assumed static, the sum of the horizontal and vertical forces must be zero. This leads to two equations. T cos(!) = T0 and T sin(!) = w0s Dividing we eliminate T and get the following. tan(!) = w0s T0 We rewrite this in terms of the derivative. dy dx = a ! s where a = w0 T0 We take the derivative of this expression with respect to x to get the following expression. d2 y dx 2 = a ! ds dx = a ! 1 + dy dx( ) 2 because ds dx = 1+ dy dx( ) 2 Now let p = dy dx in the above expression to get dp dx = a ! 1+ p 2 . w0s T0 T x ! axis y ! axis Figure 1 "
- 2. 2 This can be solved by separating variables. dp 1+ p2! = a " dx! and thus arcsinh(p) = a ! x + C1 By the choice of axes, when x = 0 , p = 0 . Thus C1 = 0 and we get the following. p = sinh(a ! x) and thus dy dx = sinh(a ! x) We now find y by integrating. y = 1 a cosh(a ! x) + C2 Notice that the constant is determined by the y-intercept. C2 = y(0) ! 1 a This material is adapted from Calculus Gems by George Simmons (McGraw-Hill, 1992), pp. 256-259. The Parabolic Cable Suppose now that the cable does not have uniform density along its length, but has uniform density along the x-axis. This would happen if it were supporting a heavy horizontal span of a bridge, for instance. The free-body diagram for this is given in Figure 2. We go through the calculations to see what shape this curve has. The explanation for each calculation is the same as for the catenary replacing s with x. However, the integration is easier and can be done in one step. T sin(!) = w0 x and tan(!) = w0 x T0 dy dx = a ! x where a = w0 T0 T0 T x ! axis y ! axis Figure 2 " w0 x
- 3. 3 y = a ! x2 2 + C Thus, this curve is a parabola. In 1645 Galileo thought that the catenary was a parabola. Christiaan Huygens showed this false in 1646. James Bernoulli asked in 1690 what was the formula for the curve of the catenary. John Bernoulli showed in 1691 that the catenary had the form y = 1 a cosh(a ! x) + C2 . Christian Huyhgens and Gottfried Leibniz also derived correct formulas for the catenary.