Analysis of Cables in Engineering Mechanics

Analysis of Cables

Cables are classified into parabolic, catenary, and segmented cables.

Parabolic Cable - is one in which the load is distributed uniformly in a horizontal length. An example of a parabolic cable is a wire suspended at its end. Consider a wire supported at A and B as shown with its lowest point at C. Note that tensile forces act tangent to any point along the curve.

Cutting the cable at the lowest point, the free-body diagram below shows the right side of the cut. Concept

Then, at the left side, we have the following free-body diagram. Concept

$$\sum F_x = 0,\quad \sum M_{anypoint} = 0,\quad \sum F_z = 0 \quad$$

$$T = \sqrt{F_x^2 + F_y^2} $$

$$S_A = \int_0^{L_1} \sqrt{1+\left(\frac{dy}{dx} \right)^2}\,dx $$

$$S_B = \int_0^{L_2} \sqrt{1+\left(\frac{dy}{dx} \right)^2}\,dx $$

$$y=kx^2 $$

$$\frac{dy}{dx}=2kx$$

²

$$k=\frac{y}{x^2}$$

Catenary Cables

A catenary cable is one whose load is distributed uniformly along the cable itself. Examples of this are transmission lines, telephone lines, and guy wires on radio and television towers. More generally, when cables are subjected to their own weight only, they are analyzed as catenaries (sag/span>0.10). When the cable is supporting uniformly distributed loads aside from its own weight and when the weight of the cable is very small compared to the load, the cable is assume to form a parabolic curve.

The general equation of a catenary curve (with origin at the lowest point of the cable) is:

$$ y(x) = a \cosh\left( \frac{x}{a} \right) - a $$

Where:

$ x $ is measured horizontally from the lowest point of the cable (point C)
$ a = \frac{H}{w} $ is the catenary parameter

For a non-symmetrical cable where supports A and B are at different elevations:

$ x = -L_1 $ at point A
$ x = +L_2 $ at point B
$ y_A = a \cosh\left( \frac{L_1}{a} \right) - a $
$ y_B = a \cosh\left( \frac{L_2}{a} \right) - a $

The vertical difference between supports is then:

$$ y_B - y_A = a \left[ \cosh\left( \frac{L_2}{a} \right) - \cosh\left( \frac{L_1}{a} \right) \right] $$

Horizontal tension is given by:

$$ H = wa $$

Total tension at a point $ x $ is:

$$ T(x) = w \sqrt{a^2 + x^2} $$

Arc length from the lowest point C to support A is:

$$ s_A = a \sinh\left( \frac{L_1}{a} \right) $$

Arc length from the lowest point C to support B is:

$$ s_B = a \sinh\left( \frac{L_2}{a} \right) $$

Total cable length from A to B is:

$$ S = s_A + s_B = a \left[ \sinh\left( \frac{L_1}{a} \right) + \sinh\left( \frac{L_2}{a} \right) \right] $$

The integral form of the cable length between two points (e.g. from A to B) is:

$$ S = \int_{-L_1}^{L_2} \sqrt{1 + \left( \frac{dy}{dx} \right)^2} \, dx $$

For the specific catenary equation $ y(x) = a \cosh\left( \frac{x}{a} \right) - a $, this becomes:

$$ S = \int_{-L_1}^{L_2} \cosh\left( \frac{x}{a} \right) \, dx = a \left[ \sinh\left( \frac{x}{a} \right) \right]_{-L_1}^{L_2} $$

Segmented Cables

A cable of negligible self-weight that supports multiple concentrated loads forms a series of straight-line segments. Each segment experiences a constant tensile force, and the shape of the cable adjusts to satisfy the equilibrium conditions at each point of loading. Analyzing such systems involves applying the principles of equilibrium to both concurrent and non-concurrent force systems at the joints and supports.

$$\sum F_x = 0,\quad \sum M_{anypoint} = 0,\quad \sum F_z = 0 \quad$$

Problem 1 (Symmetrical Parabolic Cable):

The cable shown carries a uniform load of 5kN/m. The lowest point of the cable is 36m below the supports.

Determine the minimum tension in the cable
A. 600kN
B. 1000kN
C. 1166kN
D. 1616kN

Determine the maximum tension in the cable
A.600kN
B. 1000kN
C. 1166kN
D. 1616kN

Determine the angle that the maximum tension makes with the horizontal
A. 59º
B. 31º
C. 42º
D. 48º

Determine the length of the cable
A. 240m
B. 245m
C. 254m
D. 258m

Cut the cable at the lowest point and let us consider the left side of the cut.

The system consists of a force triangle. The minimum tension occurs at the lowest point, while the maximum tension occurs at the point where the slope is the largest (at the support). The lowest point of the cable is 36m below the support, as stated in the problem.

The integral expression below only solve the arc length up to the vertex, considering the left side. Just multiply the result by 2 for symmetrical parabolic cables to obtain the total length.

Problem 2 (Unsymmetrical Parabolic Cable):

A cable carries a uniformly distributed load of 500lb per foot of horizontal length. The supports are 500ft apart and the left support is 30ft lower than the right support. The lowest point on the cable is 25ft below the left support. Determine:

The minimum tension in the cable.
A. 405k
B. 432k
C. 443k
D. 418k

The maximum tension in the cable.
A. 405k
B. 432k
C. 443k
D. 418k

The angle between the cable and the horizontal at the right support.
A. 20.2º
B. 22.2º
C. 21.2º
D. 23.2º

The length of the cable
A. 500ft
B. 523ft
C. 509ft
D. 515ft

To obtain the value of x, we use the squared property of a parabola (SPP):

Since this is an unsymmetrical parabolic cable, we will analyze the left side and the right side separately after cutting the cable at the lowest point.

Generally, because the maximum tension occurs at the point wherein the slope is the largest, it naturally means that the maximum tension occurs at the side where the horizontal distance is also the largest. In this case, the right side has a horizontal span of 298.65ft, so the maximum tension will occur at the support at B.

To obtain the total cable length, we need to add s_A and s_B.

Problem 3 (Unsymmetrical Parabolic Cable with Unknown W):

The maximum tension of the cable shown is 400kN. It carries a uniform load, w, in kN/m. If the distance between A and B is 10m and the distance from point B to the lowest point of the cable is 13m, determine:

Determine the angle between the cable and the horizontal at the right support.
A. 21.1º
B. 68.9º
C. 30.4º
D. 59.6º

Determine the uniform load, w
A. 2.127kN/m
B. 3.127kN/m
C. 4.127kN/m
D. 1.127kN/m

Determine the length of the cable
A. 50.91m
B. 101.82m
C. 69.2m
D. 138.4m