Cable Loads

Overview

Engineering reference data for Cable Loads in mechanics.

Key Formulas

Newton's Second Law

$F = ma$

Force = mass × acceleration.

Work

$W = Fd\cos\theta$

Work = force × displacement × cos(angle).

Kinetic Energy

$E_k = \frac{1}{2}mv^2$

Energy of motion.

Potential Energy

$E_p = mgh$

Gravitational potential energy.

Variables

Worked Examples

Example 1: Uniform Cable Load (Imperial Units)

A cable with length 100 ft and sag 30 ft has a uniform load of 850 lb/ft.

Horizontal support forces: $R_{1x} = R_{2x} = \frac{qL^2}{8h} = \frac{(850)(100)^2}{8(30)} = 35{,}417 \text{ lb}$

Vertical support forces: $R_{1y} = R_{2y} = \frac{qL}{2} = \frac{(850)(100)}{2} = 42{,}500 \text{ lb}$

Resultant forces at supports: $R_{1,2} = \sqrt{R_{1x}^2 + R_{1y}^2} = \sqrt{35417^2 + 42500^2} = 55{,}323 \text{ lb}$

Angle: $\theta = \tan^{-1}\left(\frac{R_{1y}}{R_{1x}}\right) = \tan^{-1}\left(\frac{42500}{35417}\right) = 50.2°$

Cable length: $s = L + \frac{8h^2}{3L} = 100 + \frac{8(30)^2}{3(100)} = 124 \text{ ft}$

Example 2: Uniform Cable Load (SI Units)

A cable with length 30 m and sag 10 m has a uniform load of 4 kN/m.

Horizontal support forces: $R_{1x} = R_{2x} = \frac{(4000)(30)^2}{8(10)} = 45{,}000 \text{ N} = 45 \text{ kN}$

Vertical support forces: $R_{1y} = R_{2y} = \frac{(4000)(30)}{2} = 60{,}000 \text{ N} = 60 \text{ kN}$

Resultant forces: $R_{1,2} = \sqrt{45^2 + 60^2} = 75 \text{ kN}$

Angle: $\theta = \tan^{-1}(60/45) = 53.1°$

Cable length: $s = 30 + \frac{8(10)^2}{3(30)} = 38.9 \text{ m}$

Example 3: Known Tension at Supports — Calculate Sag and Length

A 30 m cable with uniform load 4 kN/m has resultant tension at supports of 100 kN.

Vertical forces: $R_{1y} = R_{2y} = (4)(30)/2 = 60 \text{ kN}$

Horizontal forces: $R_{1x} = R_{2x} = \sqrt{100^2 - 60^2} = 80 \text{ kN}$

Angle: $\theta = \tan^{-1}(60/80) = 36.9°$

Sag: $h = \frac{qL^2}{8R_{1x}} = \frac{(4)(30)^2}{8(80)} = 5.6 \text{ m}$

Cable length: $s = 30 + \frac{8(5.6)^2}{3(30)} = 32.8 \text{ m}$

Inclined Cable Formulas

For cables spanning between supports at different heights with uniform horizontal load $q$, the horizontal support forces are equal:

$R_{1x} = R_{2x} = \frac{qa^2}{2h_1} = \frac{qb^2}{2h_2}$

When $b > a$, the maximum resultant forces are:

$R_2 = \sqrt{R_{2x}^2 + (qb)^2} \qquad R_1 = \sqrt{R_{1x}^2 + (qa)^2}$

The angles between horizontal and resultant forces:

$\theta_2 = \cos^{-1}\left(\frac{R_{2x}}{R_2}\right) \qquad \theta_1 = \cos^{-1}\left(\frac{R_{1x}}{R_1}\right)$

Cable segment lengths:

$s_b = b\left(1 + \frac{2}{3}\left(\frac{h_2}{b}\right)^2\right) \qquad s_a = a\left(1 + \frac{2}{3}\left(\frac{h_1}{a}\right)^2\right)$

$s = s_a + s_b$

Important Notes

• The parabolic cable equations are valid when the sag-to-span ratio $h/L < 0.1$. For larger ratios, use catenary equations instead.
• The cable length approximation (Eq. 1d) is not valid when $h > L/4$.
• For cables loaded only with self-weight, these parabolic approximations apply provided the sag ratio condition is met.
• Unit conversions: $1 \text{ kip} = 1000 \text{ lb}$ and $1 \text{ klf} = 1 \text{ kip/ft}$.
• The inclined chord equations use an iterative algorithm to fit the parabolic shape to the specified span $L$ and heights $h_1$, $h_2$.

Calculator Implementation Notes

The inclined cable calculator uses an iterative algorithm to adapt a parabola-shaped cable to the specified span L and heights h1 and h2. The resulting parabolic equation can be exported for use in spreadsheets or CAD systems.

Sag-to-Span Ratio Validity

The parabolic cable equations have two important validity constraints:

Unit Definitions

For imperial calculations:

Inclined Cable Parabolic Equation

For inclined cables with supports at different heights, the cable shape follows:

$y = ax^2$

where the coefficient $a$ is determined iteratively to satisfy the boundary conditions for span $L$ and heights $h_1$, $h_2$.

The total cable length is approximated by summing the segments:

$s = s_a + s_b$

$s_a = a\left(1 + \frac{2}{3}\left(\frac{h_1}{a}\right)^2\right)$

$s_b = b\left(1 + \frac{2}{3}\left(\frac{h_2}{b}\right)^2\right)$

Inclined Cable Formulas (Continued)

For inclined cables with uniformly distributed horizontal loads, the relationship between the parameters defines the parabolic shape. The following derived equations are useful for analysis:

The lengths of each cable segment from the lowest point to the supports can be approximated as:

$s_b = b \left(1 + \frac{2}{3} \left(\frac{h_2}{b}\right)^2\right)$

$s_a = a \left(1 + \frac{2}{3} \left(\frac{h_1}{a}\right)^2\right)$

The total length of the sagged cable is the sum of these segments:

$s = s_a + s_b$

The maximum cable forces occur at the supports. For the support at the lower point of the incline (Support 1), the resultant force is:

$R_1 = \sqrt{R_{1x}^2 + (q a)^2}$

And for the higher support (Support 2):

$R_2 = \sqrt{R_{2x}^2 + (q b)^2}$

The vertical components of the support forces are:

$R_{1y} = \sqrt{R_1^2 - R_{1x}^2} = q a$

$R_{2y} = \sqrt{R_2^2 - R_{2x}^2} = q b$

The angles the resultant forces make with the horizontal are:

$\theta_1 = \cos^{-1}\left(\frac{R_{1x}}{R_1}\right)$

$\theta_2 = \cos^{-1}\left(\frac{R_{2x}}{R_2}\right)$

Worked Examples (Continued)

Example: Known Support Tension - Calculate Sag and Length (SI Units)

For a 30 m long cable with a uniform load of 4 kN/m, the resultant tension at the supports is measured as 100 kN.

1. Vertical Forces: Since the load is uniform and horizontal, the vertical forces at the supports are equal to the load carried on that segment. $R_{1y} = R_{2y} = \frac{q \cdot L}{2} = \frac{4 \text{ kN/m} \cdot 30 \text{ m}}{2} = 60 \text{ kN}$

2. Horizontal Forces: Using the Pythagorean theorem with the known resultant tension and calculated vertical force. $R_{1x} = R_{2x} = \sqrt{R_{\text{support}}^2 - R_{1y}^2} = \sqrt{(100 \text{ kN})^2 - (60 \text{ kN})^2} = 80 \text{ kN}$

3. Sagging: Rearrange the horizontal force formula to solve for sag. $h = \frac{q \cdot L^2}{8 \cdot R_{1x}} = \frac{4 \text{ kN/m} \cdot (30 \text{ m})^2}{8 \cdot 80 \text{ kN}} = 5.625 \text{ m}$

4. Cable Length: Estimate the sagged cable length using the approximate formula. $s = L + \frac{8h^2}{3L} = 30 \text{ m} + \frac{8 \cdot (5.625 \text{ m})^2}{3 \cdot 30 \text{ m}} \approx 32.81 \text{ m}$

Inclined Cable Example Parameters

References

• Original Source