------- | ----------------- | --------------- | ----------------------------------------- | | < 0.10 | Parabolic OK | < 1% | Sufficient for almost all structural work | | 0.10 - 0.15 | Parabolic usable | 1 - 3% | Acceptable for preliminary design | | 0.15 - 0.20 | Catenary advised | 3 - 5% | Parabolic underpredicts tension | | > 0.20 | Catenary required | > 5% | Parabolic unreliable; use full catenary |

Physical Interpretation of the Catenary Parameter

The catenary parameter a = H/w is the radius of curvature at the lowest point and represents the length of cable whose weight equals the horizontal tension. Large a means a highly tensioned, shallow cable; small a means a slack, deep cable. For reference, the main cables of the Akashi Kaikyo Bridge have a ~ 1,400 m for a 1,991 m span, giving d/L = 0.089. A typical overhead conductor has a ~ 1,000-3,000 m for a 300 m span, giving d/L = 0.004-0.011.

Solving the Catenary: Two Practical Scenarios

Scenario A: Given Sag d, Find Tension H

When clearance requirements specify a maximum sag, tension becomes the unknown. The equation d = a [cosh(L/(2a)) - 1] is transcendental in a and requires iteration:

1. Compute initial estimate: a_0 = L^2 / (8d) (parabolic approximation)
2. Compute d_i = a_i [cosh(L/(2 a_i)) - 1]
3. If |d_i - d| < tolerance (e.g., 0.1 mm), accept a_i
4. Otherwise, update: a_{i+1} = a_i * (d / d_i) (Newton step, typically converges in 2-4 iterations)
5. Return to step 2

Once a converges: H = a * w, T_max = H + w d, s = 2a sinh(L/(2a)).

Convergence example: L = 200 ft, w = 10 lb/ft, target d = 25 ft (d/L = 0.125):

Result: a = 199.35 ft, H = 1,993.5 lb, T*max = 1,993.5 + 10 * 25 = 2,243.5 lb, s = 2 _ 199.35 _ sinh(200/(2 _ 199.35)) = 207.5 ft.

For comparison, the parabolic gives H = 10 _ 200^2 / (8 _ 25) = 2,000 lb -- only 0.3% difference at d/L = 0.125.

Scenario B: Given Tension H, Find Sag d

When tension is limited by anchor capacity or cable working load, sag follows directly:

1. Compute a = H / w
2. d = a [cosh(L/(2a)) - 1]
3. T_max = H + w d
4. s = 2a sinh(L/(2a))

No iteration required. This is the direct case: specify the tension you can provide, and the sag is the outcome.

Worked Example 1: Transmission Line Conductor

Problem

A 300 m span overhead transmission line uses ACSR "Drake" conductor (diameter 28.1 mm, weight 1.628 kg/m, rated breaking strength 140 kN). Design tension is limited to 22% of breaking strength (31 kN) at 15 degrees C. Determine the sag, maximum tension, and conductor length.

Given Data

• Span L = 300 m
• Conductor weight w = 1.628 * 9.81 = 15.97 N/m = 0.01597 kN/m
• Horizontal tension H = 31 kN
• Catenary parameter a = H/w = 31 / 0.01597 = 1,941 m
• Ratio L/(2a) = 300 / (2 * 1,941) = 0.0773
• d/L expected approx: 0.005 - 0.01 (shallow transmission line sag)

Step 1: Midspan Sag (Catenary)

d = a [cosh(L/(2a)) - 1] = 1,941 * [cosh(0.0773) - 1]

cosh(0.0773) = 1 + 0.0773^2/2 + 0.0773^4/24 = 1 + 0.00299 + 0.000004 = 1.00299

d = 1,941 * 0.00299 = 5.80 m

Sag ratio: d/L = 5.80 / 300 = 0.0193 (1.93% of span). Well within parabolic range but catenary used for precision.

Step 2: Parabolic Check

d = w L^2 / (8 H) = 0.01597 _ 300^2 / (8 _ 31) = 1,437.3 / 248 = 5.80 m

Parabolic and catenary agree to 2 decimal places at this sag ratio.

Step 3: Maximum Tension at Supports

T_max = H + w d = 31 + 0.01597 * 5.80 = 31 + 0.093 = 31.09 kN

The sag adds less than 0.1 kN to the maximum tension. For overhead lines, T_max differs negligibly from H.

Step 4: Conductor Length

s = 2a sinh(L/(2a)) = 2 _ 1,941 _ sinh(0.0773)

sinh(0.0773) = 0.0773 + 0.0773^3/6 = 0.0773 + 0.000077 = 0.07738

s = 3,882 * 0.07738 = 300.4 m

The conductor is 0.4 m longer than the span. This slack is taken up entirely by the sag curvature.

Step 5: Ground Clearance Check

Required ground clearance at midspan: 7.5 m (AS/NZS 7000 for 132 kV over roads).

Attachment height = clearance + sag = 7.5 + 5.80 = 13.3 m minimum above ground. Specifying 15 m pole height provides 1.7 m margin for temperature sag increase.

Step 6: Tension Variation with Sag

If the same conductor were tensioned to 15 kN instead of 31 kN:

a = 15 / 0.01597 = 939 m

d = 939 * [cosh(300/(2*939)) - 1] = 939 _ [cosh(0.160) - 1] = 939 _ 0.0128 = 12.0 m

Halving the tension more than doubles the sag (5.80 to 12.0 m). This nonlinearity -- sag is inversely proportional to tension -- is the fundamental design constraint for overhead lines.

Worked Example 2: Guy Wire for a Communication Tower

Problem

A 60 m tall lattice communication tower requires guy wires at four levels. Design the top-level guy wire, attached at 55 m height and anchored 30 m from the tower base at ground level. The guy must resist a design wind load of 32 kN applied at the attachment point (from antenna and tower wind area). Use 19 mm 7x19 galvanized wire rope. Determine guy tension, sag, and check the working load.

Given Data

• Attachment height: h = 55 m
• Horizontal anchor distance: x = 30 m
• Cable straight-line length: L_straight = sqrt(55^2 + 30^2) = 62.65 m
• Angle from horizontal: theta = atan(55/30) = 61.4 degrees
• Wire rope: 19 mm 7x19 IWRC, w = 0.137 kg/m = 1.344 N/m, MBL = 223 kN, WLL (SF=5) = 44.6 kN
• Wind load at attachment: F_wind = 32 kN (horizontal, from ASCE 7-22 calculated wind pressure)

Step 1: Guy Tension from Wind Load

The guy wire resolves the horizontal wind force at the attachment point. Taking moments about the tower base, or directly from the force triangle:

Horizontal component of guy tension must equal wind force: H = F_wind = 32 kN

Guy tension (along the chord): T_chord = H / cos(theta) = 32 / cos(61.4) = 32 / 0.479 = 66.8 kN

Step 2: Check Against Working Load Limit

WLL = 44.6 kN. T_chord = 66.8 kN exceeds WLL. The single 19 mm rope is insufficient. Options:

1. Use two 19 mm guys at this level (each takes 33.4 kN < 44.6 kN, OK)
2. Use a single 26 mm 7x19 (w = 0.255 kg/m, MBL = 419 kN, WLL = 83.8 kN). T_chord = 66.8 < 83.8, OK.

Select option 2 for simplicity: 26 mm 7x19 IWRC galvanized wire rope.

Step 3: Sag Under Self-Weight (Inclined Cable)

For an inclined cable, resolve the weight into components normal and parallel to the chord. The sag normal to the chord is:

w*normal = w * cos(theta) = 0.255 _ 9.81 _ cos(61.4) = 2.50 _ 0.479 = 1.20 N/m

Effective horizontal tension for sag calculation: H = T*chord * cos(theta) = 66.8 _ 0.479 = 32.0 kN

dnormal = w_normal * Lstraight^2 / (8 * H) = 1.20 _ 62.65^2 / (8 _ 32,000) = 4,713 / 256,000 = 0.0184 m = 18.4 mm

Sag ratio normal to chord: 0.0184 / 62.65 = 0.00029 (negligible). The cable is essentially straight under this tension. Parabolic approximation is exact.

Step 4: Anchor Block Design Load

Vertical component at anchor: V = T*chord * sin(theta) = 66.8 _ sin(61.4) = 66.8 * 0.878 = 58.6 kN (uplift)

The anchor block must resist 58.6 kN uplift plus 32 kN horizontal pull. A concrete deadman anchor with plan dimensions 1.5 m _ 1.5 m _ 1.2 m deep (mass ~ 6.5 tonnes, weight ~ 64 kN) provides a factor of safety of 64/58.6 = 1.09 against uplift by dead weight alone, before soil resistance. With soil friction and passive pressure, the overall factor of safety exceeds 2.0 per AS/NZS 1170.0 requirements.

Step 5: Pre-Tension Specification

The guy is installed with an initial tension of 10% MBL = 22.3 kN to prevent slackening under wind reversal and to maintain tower alignment. At this pre-tension:

a = H/w = (22.3 _ cos(61.4) _ 1000) / 1.20 = 10,680 / 1.20 = 8,900 m

d = 8,900 * [cosh(62.65/(2*8900)) - 1] = 8,900 _ [cosh(0.00352) - 1] = 8,900 _ 0.0000062 = 0.055 m = 55 mm

Pre-tension sag of 55 mm is visually imperceptible over 62 m. The cable appears straight when pre-tensioned.

Wire Rope Selection Data

MBL = Minimum Breaking Load. WLL = Working Load Limit at safety factor 5. Elastic modulus for IWRC ropes; fibre core ropes are approximately 15-20% lower. Values per AS 3569 and manufacturer data for 1770 MPa grade wire.

Wind and Ice Loading on Cables (ASCE 7-22)

Overhead cables accumulate ice and experience wind pressure, both of which increase the effective weight per unit length and therefore the sag.

Ice Loading

Per ASCE 7-22 Section 10.8 and AS/NZS 1170.2:

Ice weight: wice = rho_ice * g _ pi _ tice * (D + t_ice)

where:

• rho_ice = 900 kg/m^3 (glaze ice density)
• t_ice = radial ice thickness (mm), mapped by region in ASCE 7 Fig. 10.3-1
• D = cable diameter (mm)

Example: 26 mm guy wire with 12.5 mm radial ice (ASCE 7 ice region for Midwest USA):

w*ice = 900 * 9.81 _ pi _ 0.0125 _ (0.026 + 0.0125) = 27,730 _ 0.0125 _ 0.0385 = 13.35 N/m

The ice weight (13.35 N/m, 1.36 kg/m) is comparable to the cable self-weight (2.55 kg/m). Combined weight in iced condition: w_combined = 2.55 + 1.36 = 3.91 kg/m => 38.4 N/m -- a 53% increase.

Wind Pressure on Cables

Per ASCE 7-22 Eq. 29.4-1 for other structures:

F*wind = q_z * G _ C_f * A_f

where q_z = velocity pressure at height z, G = gust effect factor (0.85 for flexible structures), C_f = force coefficient (1.2 for circular sections, Re dependent), A_f = projected area = D * L.

Example: 26 mm guy wire, 62.65 m length, basic wind speed V = 45 m/s, exposure C, height 55 m:

qz = 0.613 * Kz * Kzt * Kd * V^2 = 0.613 _ 1.25 _ 1.0 _ 0.85 _ 45^2 = 0.613 _ 1.25 _ 0.85 * 2,025 = 1,319 Pa

F*wind = 1,319 * 0.85 _ 1.2 _ (0.026 _ 62.65) = 1,319 _ 0.85 _ 1.2 * 1.63 = 2,195 N = 2.2 kN distributed along the cable.

The wind force distributed along the cable adds to the point wind load at the attachment already accounted for in the guy tension calculation. This distributed wind further increases the catenary sag.

Combined Ice and Wind

The governing load combination for sag: 1.0 _ Dead + 1.0 _ Ice + 0.75 * Wind (ASCE 7 ASD load combination 4, or 1.2D + 1.0Ice + 1.0W for LRFD).

The combined load acts vertically (gravity) and horizontally (wind), producing a resultant inclined cable plane. For sag calculation, use the resultant distributed load w_r = sqrt((w_dead + w_ice)^2 + w_wind^2) and solve the catenary in the inclined plane.

Temperature Effects on Cable Sag

Governing Equation: Sag-Tension-Temperature Interaction

A change in temperature modifies both the cable length (thermal strain) and the tension (elastic strain). For a change from state 1 (T1, H1, d1) to state 2 (T2, H2, d2), the cable length compatibility equation is:

s2 - s_1 = (H_2 - H_1) * L / (E _ A) + alpha _ (T2 - T_1) * L

where si = L + 8 * di^2 / (3 * L), d*i = w * L^2 / (8 _ H_i), alpha = thermal expansion coefficient (12 x 10^-6 per degree C for steel), E = elastic modulus, A = metallic cross-sectional area.

Substituting d in terms of H:

(w^2 _ L^3 / 24) _ (1/H*2^2 - 1/H_1^2) = (H_2 - H_1) * L / (E _ A) + alpha _ DeltaT _ L

This is a cubic equation in H_2. Solve iteratively:

1. Start with H_1 at installation temperature T_1
2. For a new temperature T_2, estimate initial H_2 (use H_2 ~ H_1 * T_1 / T_2 as initial guess)
3. Compute d*2 = w * L^2 / (8 _ H_2)
4. Compute s2 = L + 8 * d2^2 / (3 * L)
5. Compute Delta_L_available = s_2 - s_1
6. Compute DeltaL_required = (H_2 - H_1) * L / (E _ A) + alpha _ (T2 - T_1) * L
7. If available and required lengths differ, adjust H_2 (increase H if available > required; decrease otherwise) and return to step 3
8. Converge when |Delta_L_available - Delta_L_required| < 0.1 mm

Worked Iteration: 300 m Overhead Line, Summer vs. Winter

ACSR Drake conductor (A = 469 mm^2, E = 77 GPa, w = 15.97 N/m, alpha = 20.9 x 10^-6 per degree C for ACSR). Installed at 15 deg C with H_1 = 31 kN, d_1 = 5.80 m.

Summer condition (+40 deg C, Delta_T = +25 deg C):

The conductor expands, tension decreases, sag increases. After iteration: H_2 = 22.4 kN, d_2 = 8.03 m.

Ground clearance at 8.03 m sag: 15.0 - 8.03 = 6.97 m. If required clearance is 7.5 m, this condition governs -- either raise the pole or increase the initial tension.

Winter condition (-10 deg C, Delta_T = -25 deg C):

The conductor contracts, tension increases, sag decreases. After iteration: H_2 = 43.6 kN, d_2 = 4.13 m.

Tension at 43.6 kN represents 43.6/140 = 31% of breaking strength. This is above the typical 25% everyday tension limit for ACSR but below the 60% extreme limit. Acceptable for occasional winter minimum.

Result table:

The summer sag governs the support height. The winter tension governs the conductor and hardware capacity check.

Cable Dynamics and Vibration

Natural Frequency

Cables have negligible flexural stiffness but significant geometric stiffness. The taut-string natural frequency is:

f_n = (n / 2L) * sqrt(H / m)

where n = mode number (1, 2, 3...), m = mass per unit length (kg/m).

For the 300 m ACSR conductor at H = 31 kN, m = 1.628 kg/m:

f*1 = (1 / 600) * sqrt(31,000 / 1.628) = 0.00167 _ 138.0 = 0.23 Hz

f_2 = 0.46 Hz, f_3 = 0.69 Hz. Frequencies below 1 Hz are susceptible to wind-induced galloping.

Vortex Shedding Check

Shedding frequency: f_s = St * V / D where St ~ 0.2 (circular cylinder).

At V = 5 m/s (moderate breeze), D = 0.0281 m: f_s = 0.2 * 5 / 0.0281 = 35.6 Hz.

This is far above f_1 = 0.23 Hz -- no resonance. Vortex shedding is a high-frequency phenomenon for small-diameter cables and typically excites Aeolian vibration (50-100 Hz for conductors), not the low-frequency structural modes.

Galloping

Galloping is a low-frequency (0.1-3 Hz), large-amplitude (up to 10 * D) oscillation of ice-accreted cables. It occurs at wind speeds above 5-10 m/s when ice forms an aerodynamically unstable cross-section. The asymmetric shape creates negative aerodynamic damping -- energy is extracted from the wind and fed into the cable motion.

Galloping mitigation: increase cable tension to raise natural frequency above the aerodynamic excitation band; use detuning pendulums or interphase spacers on bundled conductors; select conductor orientations that minimise ice accretion asymmetry.

Stockbridge Damper Selection

Stockbridge dampers (tuned mass dampers) are the standard mitigation for Aeolian vibration. For the ACSR Drake example:

• Damper mass: 2-4 kg each
• Placement: at vibration anti-nodes for the first 3-4 modes
• Typical spacing: 1/4, 3/8, and 1/2 of the shortest anticipated vibration wavelength
• Number of dampers: 2 per span for spans under 400 m; 4 per span for 400-800 m

End Fittings and Anchorage

Fitting Types

Spelter sockets (ASTM A603) develop 100% of MBL by splaying the individual wires inside a conical cavity and potting with molten zinc or polyester resin. The cone angle (typically 10-16 degrees included) generates a wedging action that increases grip as tension increases.

Anchor Block Design

The anchor block must resist the vertical uplift component and the horizontal pull component of the cable tension. Design checks:

1. Dead weight resistance: W_block >= V_uplift * FS_overturning
2. Sliding resistance: mu _ (W_block - V_uplift) >= H_pull _ FS_sliding
3. Soil bearing: (W_block + soil overburden) / A_base <= q_allowable
4. Passive soil pressure: for shallow anchors, the passive wedge in front of the block contributes to horizontal resistance

Anchorage Fatigue

Stress concentrations at socket entries reduce the fatigue life of cable anchorages by 50% or more compared to the free cable length. For structures with cyclic loading (bridges, crane cables, vibrating equipment supports), specify socket geometries with generous entry radii (minimum 3 * cable diameter) and avoid abrupt stiffness transitions at the socket mouth.

Safety Factors by Application

Safety factor is applied to the minimum breaking load (MBL), not the yield load. The design tension must not exceed MBL / SF under any factored load combination.

Corrosion Protection

Galvanizing Classes (ASTM A603 / AS 3569)

For marine and heavy industrial environments, galvanized Class C wire with a polyethylene (PE) or nylon outer sheath is standard. The sheath is extruded directly over the galvanized wires and provides both a moisture barrier and mechanical protection.

Inspection Schedule

• Annual: Visual inspection of all accessible cables, end fittings, and corrosion protection. Check for broken wires (> 2% of visible wires in any strand requires investigation), corrosion products, and loose or deformed fittings.
• 5-year: Detailed inspection with NDT (magnetic flux leakage for internal wire breaks, acoustic emission for active cracking). Measure residual tension by lift-off or vibration frequency.
• 10-year: Full structural assessment. Replace cable if wire breaks exceed 5% in any 1 m length, or if measured tension differs from design by more than 15%.

Frequently Asked Questions

What is the difference between catenary and parabolic sag formulas? The catenary equation y = a [cosh(x/a) - 1] is the exact solution for a flexible cable under uniform self-weight. The parabolic equation y = 4d x (L-x) / L^2 assumes the load is uniform per horizontal foot rather than per arc length. For sag ratios d/L < 0.1, the difference is under 1%. For d/L > 0.2, only the catenary is reliable.

How do I calculate cable tension from a measured sag? Measure the sag d at midspan and the total span L between supports. For sag ratios under 0.1, use H = w L^2 / (8 d) and T_max = H sqrt(1 + (4d/L)^2). For deeper sags, iterate the catenary: guess a, compute d_predicted = a[cosh(L/(2a))-1], adjust a until d_predicted matches measured d, then H = a w and T_max = H + w d.

How does ice loading affect cable sag? Ice adds weight per unit length, increasing the distributed load w and therefore the sag approximately linearly with the weight increase. A cable with 12.5 mm radial ice on a 26 mm diameter sees a 50-60% weight increase, producing a corresponding sag increase. The design must verify that increased sag does not violate ground clearance or create interference with adjacent cables.

What safety factor applies to structural guy wires? AS/NZS 1170.0 and ASCE 7 specify a minimum safety factor of 3.0 on the minimum breaking load for permanent structural guys. For temporary construction guys, the factor may reduce to 2.0-2.5 per AS 3610. The working load must not exceed MBL / SF under the governing factored load combination.

How do temperature changes affect cable sag and tension? A temperature increase expands the cable, reducing tension and increasing sag. A temperature decrease contracts the cable, increasing tension and reducing sag. The sag-tension-temperature relationship must be solved iteratively because the three variables are coupled through the cable length constraint equation. For a typical 300 m ACSR overhead conductor, a +25 deg C temperature change can increase sag by 40% (from 5.8 m to 8.0 m) while decreasing tension by 28%.

What is the maximum practical span for a steel cable? Practical span limits are set by cable strength-to-weight ratio and allowable sag. For 32 mm 6x37 IWRC at SF = 2.5, d/L = 0.10: L_max ~ 590 m. For 50 mm locked coil strand at higher strength grades, spans of 1,000 m are achievable. Suspension bridges use parallel wire bundles (not preformed rope) to achieve main spans exceeding 2,000 m. Material strength limits are not the practical constraint -- erection, dynamics, and cost govern.

When should I use the levelling nut method? The levelling nut method applies to base plate connections, not cables. For cable anchorages, the equivalent is pre-tensioning: the cable is installed with a specified initial tension (typically 10-20% of MBL) to remove slack and define the sag geometry, then secured at the socket. The socket and anchor block design must accommodate the full range of tension from pre-tension to maximum design value.

What are Stockbridge dampers and when are they needed? Stockbridge dampers are tuned mass dampers that suppress Aeolian vibration (50-100 Hz) in overhead conductors and guy wires. They consist of two masses connected by a messenger cable, clamped to the conductor at a vibration anti-node. Needed when: (a) span exceeds 100 m, (b) terrain is open and windy, (c) the conductor experiences visible vibration for more than 100 hours per year, or (d) fatigue strand failures have been observed on similar installations.

Try it now: Check your cable sag with our free Cable Sag calculator →

Related Pages

• Cable Sag Calculator Tool -- Interactive cable sag analysis for any span and cable type
• Wind Load Calculator -- Wind pressure calculation per ASCE 7-22
• Steel Weight Calculator -- Mass and weight of steel sections, plates, and cables
• Bolt Grades and Strengths -- Mechanical properties of structural bolts and anchor rods
• Beam Serviceability Limits Calculator -- Flexural member deflection analysis

Disclaimer

This is a calculation reference, not a substitute for professional engineering certification. All results must be independently verified by a licensed Professional Engineer (PE), Chartered Professional Engineer (CPEng), or Registered Structural Engineer before use in construction, fabrication, or permit documents. The user bears full responsibility for the accuracy of all inputs and the independent verification of all outputs against applicable codes and standards.