Republic of the Philippines
NUEVA VIZCAYA STATE UNIVERSITY
Bayombong, Nueva Vizcaya
INSTRUCTIONAL MODULE
IM No.:IM-MECH1A-1STSEM-2020-2021
College of Engineering
Bayombong Campus
DEGREE PROGRAM
SPECIALIZATION
YEAR LEVEL
I.
BSCE
SE/CEM/TE/WRE
2nd
COURSE NO.
COURSE TITLE
TIME FRAME
N145/N155/N165/N177
Statics of Rigid Bodies
WK NO. 9-10 IM NO.
6
UNIT TITLE/CHAPTER TITLE
Cables and Catenary
II.
LESSON TITLE
Cables and Catenary
III.
LESSON OVERVIEW
1. Cables
a. Distributed Loads – Parabolic Cables
b. Cables under Concentrated Loads
2. Catenary
IV.
DESIRED LEARNING OUTCOMES
•
V.
To include analysis on cables due to the tensional force it applies.
LESSON CONTENT
Cables under Distributed Loads - Introduction
Flexible cables are used in numerous engineering applications. Common examples are power
transmission lines and suspension bridges. The term flexible means that the cables are incapable of
developing internal forces other than tension. In earlier chapters we treated cables as two-force
members; that is, the weights of the cables were neglected, and the loading consisted of end forces only.
Here we consider the effects of distributed forces, such as the weight of the cable or the weight of a
structure that is suspended from the cable.
Figure (a) shows a cable that is suspended from its endpoints A and B. In order to support the distributed
loading of intensity w, the cable must assume a curved shape. It turns out that the equation describing
this shape is simplified if we place the origin of the xy-coordinate system at the lowest point O of the
cable. We let s be the distance measured along the cable from O. The shape of the cable and the location
of point O are generally unknown at the beginning of the analysis.
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The units of the load intensity “w” are lb/ft or N/m. The length can be measured in two ways: along the
horizontal x-axis (w as a function of x) or along the cable (w as a function of s). Although these two cases
must be treated separately, we first consider the elements of the analyses that are common to both. The
free-body diagram (FBD) of a segment of the cable, extending from the lowest point O to an arbitrary
point C, is shown in (b). The tensile forces in the cable at O and C are denoted by To and T, respectively;
W is the resultant of the distributed loading; and θ represents the slope angle of the cable at C. The force
equilibrium equations of the cable segment are
Hence:
The first of equation shows that the horizontal component of the cable force, namely T cos θ, is constant
throughout the cable. The solution of equations for θ and T yields:
Parabolic Cables
Here we analyze the special case in which the loading is distributed uniformly along the horizontal; that
is, w(x) = wo, where wo is the constant load intensity. This case arises, for example, in the main cables
of a suspension bridge where wo represents the weight of the roadway per unit length. It is assumed that
the roadway is connected to the main cables by a large number of vertical cables and that the weights of
all cables are negligible compared to the weight of the roadway. Taking equations as the starting point,
we now derive several useful equations that describe the geometry of the cable and the variation of the
tensile force within the cable.
Substituting tan θ = dy/dx, the equation can be written as dy/dx = wox/To. Upon integration, we get where
the constant of integration was set equal to zero to satisfy the condition y = 0 when x = 0. Equation, which
represents a parabola with its vertex at O, could also be obtained from a moment equilibrium equation
using the FBD in Fig. (b).
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It is often necessary to compute the length s of the cable between points O and C in Fig. (b). The
infinitesimal length of the cable is:
Sample Problem 1
Figure (a) shows a cable that carries the uniformly distributed load wo = 80 lb/ft, where the distance is
measured along the horizontal. Determine the shortest cable for which the cable tension does not exceed
10 000 lb, and find the corresponding vertical distance H.
Method of Analysis
Because the loading is distributed uniformly over the horizontal distance, we know that the shape of the
cable is parabolic. It is also apparent that the location of the lowest point O of the cable and the cable
tension T0 at that point are not known. Therefore, the computation of these unknowns is addressed first.
A good starting point is the free-body diagram of the entire cable, shown in Fig. (b).
The forces appearing on this diagram are the cable tensions at the endpoints (TA and TB) and the
resultant of the distributed load: W = (80 lb/ft)(200 ft) = 16 000 lb. The tension in the cable increases with
x (x is measured from the vertex of the parabola). It follows that the maximum cable tension occurs at B;
that is, TB = 10 000 lbs, as shown in the figure. The FBD in Fig. (b) now contains three unknowns: the
slope angles θA and θB and the tension TA, all of which could be computed from the three available
equilibrium equations. It turns out that we need only θB, which can be obtained from the moment
equation.
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As the next step, we draw the FBD of the portion of the cable that lies to the right of point O, as shown in
Fig. (c). Assuming that θB has already been computed, this FBD contains three unknowns: LB (which
locates point O), T0, and H. Because there are also three equilibrium equations available, all the
unknowns can now be calculated. The final step is to calculate the length of the cable from the equations
above.
θB = 62.98◦
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Cables under Concentrated Loads
Sometimes a cable is called on to carry a number of concentrated vertical loads. If the weight of the cable
is negligible compared to the applied loads, then each segment of the cable is a two-force member and
the shape of the cable consists of a series of straight lines. The analysis of a cable loaded in this manner
is similar to truss analysis, except that with cables the locations of the joints (i.e., points where the loads
are applied) are sometimes unknown. As in the case of truss analysis, we can use the method of joints
and/or the method of sections to determine the equilibrium equations. However, it is often necessary to
include equations of geometric constraints in order to have enough equations to find all the unknowns.
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If a cable has n segments, then there are (n − 1) joints. For example, the cable in Fig. 6.12(a) has n = 3
segments, and (n − 1) = 2 joints, labeled 1 and 2. We use the following notation: Si is the segment length;
Li is the horizontal spacing of the loads; and θi is the angle between a segment and the horizontal, where
i = 1, 2, . . . , n is the segment number. The vertical position of the ith joint, measured downward from
end B, is denoted by Hi, i = 1, 2, …, n − 1.
Observe that equation represent (n − 1) equations that contain the (n + 1) unknowns To, θ1, θ2,..., θn.
Therefore, we must obtain two additional independent equations before we can calculate all of the
unknowns. The source of the additional equations depends primarily on the nature of the problem. It is
convenient to divide problems into two categories depending on whether the horizontal spacings of the
loads (Li) or the lengths of the cable segments (Si) are given and to discuss each category separately.
Because Equations have been derived from the equation in Parabolic Cables, the figure also defines the
sign conventions that have been used in the derivations: tensile forces and counterclockwise angles
measured from the horizontal are positive, and h is the vertical distance measured downward from the
right-hand support B. These conventions also apply to the equations that are derived in the remainder of
this article.
Consider a cable with n segments for which the horizontal spacings of the loads (L1, L2,..., Ln) are given.
For this case, the following geometric relationship can be obtained from (a):
Consider next a cable with n segments for which the lengths of the segments s1, s2,..., sn are known.
For this case, (a) yields two independent geometric relationships:
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Sample Problem 2
Method of Analysis:
The free-body diagram of the entire cable is shown in Fig. (b), where the labeling of the variables is
consistent with the notation used in Fig. 6.12 (recall that the positive direction for θ1, θ2, and θ3 is
counterclockwise from the horizontal, and positive h is measured downward from end B). We note that
the information given in Fig. (a) includes the horizontal spacing of the loads and the angle β3. Therefore,
according to the discussion, the problem is statically determinate, and a solution can be obtained by
writing and solving Equations.
In this problem, however, the difficulty of solving these simultaneous, nonlinear equations can be avoided.
Examination of the FBD in Fig. (b) reveals that T3 can be calculated from the equation Ma = 0. Equilibrium
equations for joints 2 and 1 will then determine the other unknowns without having to solve the equations
simultaneously.
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Catenary
Consider a homogeneous cable that carries no load except its own weight. In this case, the loading is
uniformly distributed along the length of the cable; that is, w(s) = wo, where wo is the weight of the cable
per unit length, and the distance s is measured along the cable. Therefore, the resultant of the loading is
W = w0s. The following useful relationships can now be derived from the previous equations in Parabolic
Cables.
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Sample Problem 3
Method of Analysis
Because the loading is distributed along the cable, the shape of the cable is a catenary. The cable is
obviously symmetric about the midpoint of AB, which means that the location of the lowest point O of the
cable is known.
From the Equations, we note that the maximum cable tension occurs at the endpoints, where s is a
maximum. We now draw the free-body diagram of the right half of the cable, shown in Fig. (b).
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VI.
LEARNING ACTIVITIES
Answer the following questions:
1. The cable of the suspension bridge spans L = 120 m with a sag H = 18 m. The cable supports a
uniformly distributed load of wo N/m along the horizontal. If the maximum allowable force in the cable
is 4 MN, determine the largest permissible value of wo. (See Figure P6.63-6.64)
2. Cable AB supports the uniformly distributed load of 2 kN/m. If the slope of the cable at A is zero,
compute the maximum tensile force in the cable and the length of the cable. (Figure 6.65)
3. Determine the angles B2 and B3, and the force in each cable segment if B1 = 40 deg. (Figure P6.82)
VII.
ASSIGNMENT
Answer the following questions in a clean sheet of bond paper.
1. The two main cables of the Akashi Kaikyo suspension bridge in Japan have a span L = 1990 m and
a sag H = 233 m. The loading on each cable is wo = 444.7 KN/m along the horizontal. Determine the
corresponding maximum tensional force in one of the cables. (See Figure P6.63-6.64)
2. Determine the maximum uniform loading w, measured in lb/ft, that the cable can support if it is capable
of sustaining a maximum tension of 3000 lb before it will break. (Fig 1.)
3. For the cable loaded as shown in the Figure (a), calculate the angles in each cable with the horizontal
and the force in each segment of the cable.
VIII.
REFERENCES
1.
2.
3.
4.
Singer, Ferdinand L., Engineering Mechanics: Statics and Dynamics.
Beer, Ferdinand P. and Johnson, E. Russel Jr., Mechanics for Engineers.
Hibbeler R.C., Engineering Mechanics: Statics
Pytel, Andrew and Kiusalaas, Jaan, Understanding Engineering Mechanics: Statics.
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