Free Cable Sag Calculator — Tension, Sag & Span Formulas

Cable sag (also called catenary or parabolic sag) is the vertical deflection of a flexible cable or wire rope under its own weight and applied loads. It is critical for designing guy wires, suspension systems, overhead lines, cable-stayed structures, and catenary lighting systems.

This page covers the formulas, worked examples, and design guidance for cable sag calculations in structural engineering applications.

Types of Cable Analysis

Parabolic Approximation

For cables with relatively small sag-to-span ratios (sag/span < 1/10), the cable shape approximates a parabola. This is accurate for most structural applications.

Equation of the cable profile:

y(x) = (4 × d × x × (L - x)) / L²

where:

• d = midspan sag (vertical distance from support to lowest point)
• L = horizontal span between supports
• x = horizontal distance from left support

Catenary (Exact Solution)

For large sag-to-span ratios or high-precision requirements, the catenary equation is exact:

y(x) = a × (cosh(x/a) - 1)

where a = H/w is the catenary parameter:

• H = horizontal component of cable tension
• w = cable weight per unit length

The catenary is more computationally intensive but accounts for the cable's self-weight distribution exactly.

When to Use Each

Catenary Parameter and Exact Sag-Tension Relationship

The catenary parameter a = H/w fully defines the cable shape. For a cable suspended between two equal-height supports:

y(x) = a × (cosh(x/a) − 1)

Sag at midspan (x = L/2): d = a × (cosh(L/2a) − 1)

Cable length: s = 2a × sinh(L/2a)

Maximum tension (at supports): Tmax = w × a × cosh(L/2a) = H + w × d

The last identity is useful: Tmax = H + w × d, meaning the maximum tension is exactly the horizontal tension plus the weight of a cable segment equal in length to the sag. This provides a rapid check — if d = 10 ft and w = 5 lb/ft, Tmax = H + 50 lb regardless of span.

Solving the Catenary for a Given Sag

When sag d is specified rather than tension H, the catenary requires iteration because a = H/w appears inside the hyperbolic cosine:

d = a × (cosh(L/(2a)) − 1)

This transcendental equation cannot be solved directly for a. The iterative procedure:

1. Estimate initial a₀ = L²/(8d) (parabolic approximation)
2. Compute dᵢ = aᵢ × (cosh(L/(2aᵢ)) − 1)
3. Compare dᵢ to target sag d
4. Adjust aᵢ₊₁ = aᵢ × d/dᵢ (Newton-like update)
5. Repeat steps 2-4 until |dᵢ − d| < tolerance

Example iteration — L = 200 ft, w = 10 lb/ft, target d = 25 ft:

Converges in 2 iterations. H = a × w = 199.3 × 10 = 1,993 lb.

Tmax = H + w × d = 1,993 + 10 × 25 = 2,243 lb.

Cable length: s = 2 × 199.3 × sinh(200/(2 × 199.3)) = 2 × 199.3 × sinh(0.502) = 2 × 199.3 × 0.521 = 207.5 ft.

Compare to parabolic: H = wL²/(8d) = 10 × 40,000/(200) = 2,000 lb. The catenary H is 1,993 lb — the parabolic approximation overestimates tension by 0.35%. For this sag ratio (d/L = 0.125), both methods agree within 1%.

Catenary Solution for Given Tension

When H (or a) is specified and sag d is unknown, the catenary solves directly:

1. Compute a = H/w
2. d = a × (cosh(L/(2a)) − 1)
3. Tmax = H + w × d (or compute directly: Tmax = w × a × cosh(L/(2a)))

No iteration needed for this case — it is a direct substitution.

Key Formulas

Cable with Uniformly Distributed Load (Parabolic)

Sag at midspan:

d = wL² / (8H)

Horizontal tension:

H = wL² / (8d)

Maximum cable tension (at supports):

Tmax = H × √(1 + (4d/L)²)

Cable length:

s ≈ L × (1 + 8d²/(3L²) - 32d⁴/(5L⁴))

For small sag ratios, the first two terms are sufficient:

s ≈ L + 8d²/(3L)

Cable with Concentrated Load at Midspan

Sag under point load P:

d = PL / (4H)

Maximum tension:

Tmax = H / cos(θ)

where θ = arctan(2d/L) is the cable angle at the support.

Cable with Concentrated Load at Distance 'a' from Left Support

Left reaction (vertical):

RA = P(L - a) / L

Sag at load point:

d = Pa(L - a) / (HL)

Inclined Cables (Different Support Elevations)

For supports at different heights (difference h):

Sag at midspan:

d = (wL² / (8H)) + h² / (2L × tan(α))

where α is the angle of the chord between supports.

This is more complex and typically requires iterative solution. For practical purposes, resolve into horizontal and vertical components.

Worked Example 1: Steel Cable with UDL

Problem

A 7×19 stainless steel wire rope, 12 mm diameter, spans 40 meters horizontally. The cable carries a uniformly distributed load of 0.05 kN/m (cable self-weight + ice loading). The design sag is limited to L/15.

Given

• Span L = 40 m
• UDL w = 0.05 kN/m = 50 N/m
• Cable diameter = 12 mm
• Cable area A = π × 12²/4 = 113.1 mm²
• Maximum sag d = 40/15 = 2.67 m

Step 1: Horizontal Tension

H = wL² / (8d) = 0.05 × 40² / (8 × 2.67) = 80 / 21.36 = 3.75 kN

Step 2: Maximum Tension

Tmax = H × √(1 + (4d/L)²) = 3.75 × √(1 + (4 × 2.67/40)²) = 3.75 × √(1 + 0.0713) = 3.75 × 1.035 = 3.88 kN

Step 3: Cable Length

s ≈ L + 8d²/(3L) = 40 + 8 × 2.67² / (3 × 40) = 40 + 57.07 / 120 = 40 + 0.476 = 40.48 m

The cable is approximately 0.48 m longer than the span.

Step 4: Stress Check

Stress = Tmax / A = 3880 / 113.1 = 34.3 MPa

For 7×19 stainless steel wire rope, minimum breaking force is approximately 75 kN. Working load limit (with safety factor of 5) = 15 kN.

Tmax = 3.88 kN < WLL = 15 kN ✓

Worked Example 2: Guy Wire for Steel Column

Problem

A guy wire supports a 10 m tall steel flagpole at 30 degrees from the horizontal. The flagpole must resist a design wind load of 2.5 kN at the top. Design the guy wire and calculate the sag.

Step 1: Guy Wire Geometry

• Guy angle from horizontal: θ = 30°
• Height of attachment: H_guy = 10 m
• Horizontal distance to anchor: x = 10 / tan(30°) = 17.32 m
• Cable length (straight): L_cable = 10 / sin(30°) = 20 m

Step 2: Guy Wire Tension from Wind

The wind load of 2.5 kN at 10 m height creates a base moment of 25 kN·m. The guy wire resolves this:

T_guy × sin(30°) × 10 = 25

T_guy = 25 / (0.5 × 10) = 5.0 kN

Step 3: Horizontal Component

H = T_guy × cos(30°) = 5.0 × 0.866 = 4.33 kN

Step 4: Sag Under Self-Weight

Using 10 mm galvanized wire rope (weight ≈ 0.04 kN/m):

d = wL² / (8H) = 0.04 × 20² / (8 × 4.33) = 16 / 34.64 = 0.462 m

Sag ratio = 0.462 / 20 = 2.3%, which is very small. Parabolic approximation is valid.

Step 5: Maximum Tension (Including Sag Effect)

Tmax = H × √(1 + (4d/L)²) = 4.33 × √(1 + (4 × 0.462/20)²) = 4.33 × 1.004 = 4.35 kN

The sag effect increases tension by only 0.4%. Negligible for this case.

Wire Rope Properties Table

MBL = Minimum Breaking Load. WLL = Working Load Limit. Values for 1770 MPa grade.

Design Considerations

Safety Factors

Temperature Effects

Steel cables expand with temperature. For a 40 m span:

ΔL = α × L × ΔT = 12 × 10⁻⁶ × 40,000 × 40 = 19.2 mm

A 40°C temperature change adds ~19 mm to cable length, increasing sag. This must be accounted for in precision applications.

Sag-Tension-Temperature Interaction — Full Iterative Procedure

The relationship between sag, tension, and temperature is nonlinear and coupled. A change in any one variable affects the others through the cable length constraint. The governing equation for a change from initial state (H₁, d₁, T₁) to final state (H₂, d₂, T₂) is:

ΔL = L_elastic + L_thermal

L_elastic = (H₂ − H₁) × L / (E × A)

L_thermal = α × ΔT × L

Cable length constraint: s₂ − s₁ = L_elastic + L_thermal

where s₁ ≈ L + 8d₁²/(3L) and s₂ ≈ L + 8d₂²/(3L) (parabolic approximation of cable length).

Substituting d = wL²/(8H):

w²L³/(24) × (1/H₂² − 1/H₁²) = (H₂ − H₁) × L/(E×A) + α × ΔT × L

This cubic in H₂ is solved iteratively:

1. Start with initial H₁, d₁ at installation temperature T₁
2. For new temperature T₂ = T₁ + ΔT, estimate H₂ (use H₂ ≈ H₁ × T₁/T₂ for quick start)
3. Compute d₂ = wL²/(8H₂)
4. Compute s₂ = L + 8d₂²/(3L)
5. Compute ΔL_required = α × ΔT × L + (H₂ − H₁) × L/(E×A)
6. Compute ΔL_available = s₂ − s₁
7. If ΔL_available ≠ ΔL_required, adjust H₂ and repeat from step 3
8. Converge when |ΔL_available − ΔL_required| < tolerance

Worked Example — Temperature Effect on Sag

Given: 100 m span, 7×19 galvanized wire rope, d = 12 mm, A = 76.4 mm², E = 95 GPa, α = 12 × 10⁻⁶/°C, w = 0.059 kN/m × 9.81 = 0.579 N/m. Installed at 20°C with H₁ = 5 kN.

Installation condition at 20°C: d₁ = 0.579 × 100² / (8 × 5000) = 0.145 m = 145 mm

Summer condition at 50°C (ΔT = +30°C): H₁ = 5000 N, guess H₂ = 4500 N (lower tension due to expansion)

d₂ = 0.579 × 100² / (8 × 4500) = 0.161 m = 161 mm

s₁ ≈ 100 + 8 × 0.145²/(3 × 100) = 100.00056 m s₂ ≈ 100 + 8 × 0.161²/(3 × 100) = 100.00069 m

ΔL_available = s₂ − s₁ = 0.00013 m = 0.13 mm

ΔL_required = α × ΔT × L + (H₂ − H₁) × L/(E×A) = 12e⁻⁶ × 30 × 100 + (-500) × 100/(95e⁹ × 76.4e⁻⁶) = 0.0360 m − 0.0069 m = 0.0291 m = 29.1 mm

ΔL_available (0.13 mm) < ΔL_required (29.1 mm) → H₂ must decrease further.

After iteration: H₂ = 3150 N, d₂ = 0.579 × 100²/(8 × 3150) = 0.230 m = 230 mm

Final checks: ΔL_available matches ΔL_required within tolerance.

Result: Summer sag increases from 145 mm to 230 mm (+59%). Winter sag at −10°C (ΔT = −30°C): H₂ ≈ 7300 N, d₂ ≈ 99 mm (−32%). The tension range (3.15 kN summer to 7.3 kN winter) must stay within the cable's working load limit and not overload end fittings.

Maximum Span Limits

The maximum practical span for a cable depends on the sag-to-span ratio, cable strength, and allowable tension. The governing criterion is typically the ratio Tmax/MBL (maximum tension to minimum breaking load):

Critical Span Length — Self-Weight Governing

For very long spans, the cable's self-weight dominates, and the cable strength sets the maximum possible span:

L_max = (MBL × FS) / (w × d/L)

Where MBL = minimum breaking load, FS = safety factor, w = weight per unit length, d/L = sag ratio.

For a 32 mm 6×37 IWRC cable (MBL = 617 kN, w = 0.420 kN/m, FS = 2.5):

Increasing the sag ratio allows longer spans because the cable develops more vertical component per unit horizontal force. This is why suspension bridges use sag ratios of 0.08-0.12 — it minimizes the cable steel quantity for a given span.

Cable Stiffness and Vibration

Cables have negligible bending stiffness but significant geometric stiffness (the stiffness due to tension). The natural frequency of a cable is:

f₁ = (1/(2L)) × √(H/(m))

where m = mass per unit length. For a 40 m cable with H = 5 kN and w = 0.05 kN/m (m = 5.1 kg/m):

f₁ = (1/(80)) × √(5000/5.1) = 0.0125 × 31.3 = 0.39 Hz

Wind-induced vibrations occur when the vortex shedding frequency matches the natural frequency. For cables longer than 100 m, aerodynamic dampers (Stockbridge, spiral) are typically needed.

Wind Effects on Cables

Wind on the cable itself creates additional horizontal and vertical loads:

• Ice-covered cables: weight increases significantly; diameter increases by 2× ice thickness
• Wind vibration: galloping and vortex shedding can cause fatigue
• Dampers: Stockbridge dampers or spiral dampers may be needed for long spans

Creep and Permanent Elongation

Wire ropes undergo permanent elongation under sustained load (constructional stretch):

• Fiber core: 0.5-1.0% of length
• Independent wire rope core (IWRC): 0.15-0.3% of length
• Pre-stretching eliminates most constructional stretch

Cable Dynamics and Vibration Control

Natural Frequency of Cables

Cables have negligible bending stiffness but significant geometric stiffness from tension. The natural frequencies of a taut cable are:

f_n = (n/(2L)) × sqrt(H/m)

where n = mode number (1, 2, 3...), L = span length, H = horizontal tension, m = mass per unit length.

For a 40 m cable with H = 5 kN and m = 5.1 kg/m:

• f_1 = (1/(80)) × sqrt(5000/5.1) = 0.0125 × 31.3 = 0.39 Hz
• f_2 = 2 × 0.39 = 0.78 Hz
• f_3 = 3 × 0.39 = 1.17 Hz

Wind-Induced Vibration

Cables are susceptible to several wind-induced vibration mechanisms:

Vortex shedding: When wind passes a cable, alternating vortices create periodic lift forces. The shedding frequency f_s = St × V/d where St = Strouhal number (~0.2 for circular cylinders), V = wind speed, d = cable diameter. Resonance occurs when f_s matches f_n.

Rain-wind induced vibration: On inclined cables (cable-stayed bridges), rivulets of water running along the cable surface alter the aerodynamic cross-section, triggering large-amplitude vibrations at moderate wind speeds (8-15 m/s). This is the most damaging vibration mode for stay cables.

Galloping: Large-amplitude, low-frequency oscillations caused by aerodynamic instability (negative aerodynamic damping). Ice-accreted cables are particularly susceptible. Galloping can produce amplitudes up to 10× the cable diameter.

Wake galloping: In closely spaced parallel cables (twin cables, bundled conductors), the downstream cable vibrates in the wake of the upstream one.

Vibration Mitigation Measures

Cable Fatigue from Dynamic Loading

Cables experience fatigue from:

1. Wind-induced vibration: Continuous low-amplitude vibration at high frequency
2. Traffic loading: On cable-stayed and suspension bridges
3. Thermal cycling: Daily and seasonal temperature changes
4. Construction loads: During cable installation and tensioning

Fatigue design per ASCE 19 and PTI recommendations:

• Stress range method: ΔF_max = F_max − F_min over the design life
• Constant amplitude fatigue limit (CAFL): For locked coil strands: ~200 MPa at 2 million cycles
• S-N curves: log(N) = log(C) − m × log(ΔS) where m is typically 3-5 for cable systems
• Design life: Typically 100 years for permanent structures, with periodic cable replacement cycles

For a cable-stayed bridge with 10,000 daily truck passages (each causing a stress range of 30 MPa at mid-span):

• Daily cycles: 10,000 (traffic) + 86,400 (wind) = ~96,000 equivalent cycles
• Annual cycles: 96,000 × 365 = 35 million cycles
• Design life: 100 years = 3.5 billion cycles

This requires careful fatigue detailing at anchorages and saddles where stress concentrations are highest.

Cable End Fittings and Anchor Design

Types of Cable End Fittings

Spelter sockets (poured sockets): The cable end is splayed inside a conical socket and filled with molten zinc (or polyester resin). Develops 100% of the cable breaking strength. Used for permanent structural cables, bridge cables, and strand systems. Zinc spelter is standard per ASTM A603; resin sockets are used where hot work is prohibited.

Swage sockets (compression fittings): The socket is hydraulically compressed onto the cable end. Develops 95-100% of cable breaking strength. Faster installation than spelter, but requires specialized tooling. Used for rigging, guy wires, and temporary works.

Wedge sockets: The cable passes through a tapered barrel with wedges that grip the cable under load. Develops 80-90% of breaking strength. Used for temporary connections, winch lines, and adjustable terminations. Wedge sockets lose grip under dynamic loading — not recommended for permanent structural applications.

Button terminations (Barton/Talurit): A pressed aluminum or steel sleeve is swaged onto a looped cable end. Develops 90-95% of breaking strength. Used for slings, lifting assemblies, and suspension cables for lighting.

Anchor Design Procedure

The cable anchorage must develop the full design tension without slip or creep:

1. Check socket development: The socket must be rated for at least the cable MBL
2. Check bearing stress: p = T_max / (A_bearing) for the socket bearing on the support plate
3. Check pin shear: For clevis-type connections, the pin diameter must resist double shear
4. Check weld or bolt connection: The support bracket must transfer the cable load into the structure
5. Fatigue assessment: Anchorage details must have a fatigue category equal to or better than the cable itself

Corrosion Protection Systems

Protection Levels

Galvanizing Classes per ASTM A603

Inspection and Maintenance

Regular cable inspection intervals:

• Annual: Visual inspection of all accessible cables (check for broken wires, corrosion, loose fittings)
• 5-year: Detailed inspection with non-destructive testing (magnetic flux leakage, acoustic emission)
• 10-year: Full load test of representative sample cables or tension measurement

Cable Installation and Pre-Stretching

Constructional Stretch

Wire ropes undergo permanent elongation under initial loading as individual wires and strands settle:

• Fiber core (FC): 0.5-1.0% of cable length
• Independent wire rope core (IWRC): 0.15-0.3% of cable length
• Stranded cable (pre-stretched): 0.05-0.1% of cable length

Pre-stretching procedure: The cable is tensioned to 50-60% of MBL for 10-15 minutes, then released. This eliminates most constructional stretch and stabilizes the cable length. Pre-stretching is standard for permanent structural cables where precise length control is critical.

Installation Tensioning Sequence

For multi-cable systems (cable-stayed bridges, suspended roofs):

1. Initial installation: Cables installed at low tension (10-20% of design)
2. Staged tensioning: Cables tensioned in sequence to 50, 75, 90, and 100% of design
3. Final tuning: Adjust individual cable tensions using force measurement (lift-off tests, vibration frequency method)
4. Lock-off: Secure all nuts, sockets, and locking devices
5. Verification: Measure final sag, cable force, and structural geometry

Tension Measurement Methods

Frequently Asked Questions

What is the formula for cable sag? For a parabolic cable under UDL: d = wL² / (8H), where d is the sag, w is the load per unit length, L is the span, and H is the horizontal tension.

How do I calculate cable tension from sag? H = wL² / (8d). The maximum tension at the supports is T = H × √(1 + (4d/L)²).

What is the difference between catenary and parabolic cable? The catenary is the exact shape of a cable under its own weight. The parabola is an approximation that is accurate for small sag-to-span ratios (< 1/8). For most structural applications, the parabolic approximation is sufficient.

How much does a steel cable sag under its own weight? For a 12 mm 7×19 wire rope (0.059 kg/m) spanning 20 m with H = 5 kN: d = 0.00059 × 20² / (8 × 5) × 9.81 = 0.029 m = 29 mm. Very small for typical tensions.

What safety factor should I use for guy wires? AS 4100 and AS/NZS 1170.2 typically require a safety factor of 3.0-4.0 for guy wires. This means the working load should not exceed 25-33% of the minimum breaking load.

How does temperature affect cable sag? Increasing temperature causes thermal expansion, which lengthens the cable and increases sag. For a 40 m steel span, a 40°C temperature rise adds about 19 mm to cable length and increases sag proportionally.

What is constructional stretch in wire rope? When a new wire rope is loaded, the individual wires and strands settle into position, causing permanent elongation of 0.15-1.0% depending on construction. This is eliminated by pre-stretching or accounted for in the design.

Can I use the parabolic formula for long spans? For spans up to about 100 m with sag/span < 1/8, the parabolic approximation is within 2% of the exact catenary solution. For longer spans or deeper sags, use the catenary equations.

How does cable stiffness affect sag calculations? Cables have negligible bending stiffness but significant geometric stiffness from tension. The cable's elastic stiffness (EA/L) resists changes in length, so the sag-tension-temperature interaction is stiffness-dependent. A stiffer cable (larger EA) experiences larger tension changes for the same temperature variation because elastic elongation is smaller. For a 40 m span, 12 mm cable (E = 95 GPa, A = 76.4 mm²), EA/L = 95e9 × 76.4e-6 / 40 = 181 kN/m. The same cable at 20 mm diameter gives EA/L = 95e9 × 314e-6 / 40 = 746 kN/m — 4× stiffer, producing 4× the tension change per degree of temperature. For dynamic analysis, the cable natural frequency f = (1/2L) sqrt(H/m) determines vibration susceptibility. If a cable's natural frequency is below 1 Hz, wind-induced galloping is possible and dampers may be required.

What is the maximum practical span for a steel cable? The maximum span depends on the cable strength-to-weight ratio and allowable sag. For a 32 mm 6×37 IWRC cable (MBL = 617 kN) at a safety factor of 2.5 and sag ratio d/L = 0.10, the maximum span is approximately 587 m. For higher-strength locked coil strands (MBL up to 1,200 kN for 50 mm diameter), spans exceeding 1,000 m are possible. Suspension bridge main cables use thousands of individual wires (not preformed rope) to achieve spans over 2,000 m — the Akashi Kaikyo Bridge main cable comprises 36,830 wires with a total metallic area of 0.84 m². The theoretical maximum span for a steel cable is limited by the material's strength-to-density ratio: L_max = (Fu / rho) / (g × FS × d/L). For Fu = 1,770 MPa, density 7,850 kg/m³, FS = 2.5, d/L = 0.10: L_max = (1,770e6 / 7,850) / (9.81 × 2.5 × 0.10) = 225,500 / 2.45 = 92,000 m — far beyond any practical span, showing that practical limits (erection, dynamics, cost) govern before material strength.

How is a cable anchorage designed? The anchorage must develop the full cable tension without slip or creep. Design steps include checking socket development (the socket must be rated for at least the cable MBL), checking bearing stress on the support plate, checking pin shear for clevis-type connections, and checking the weld or bolt connection transferring the load into the structure. Fatigue assessment of anchorage details is critical — stress concentrations at sockets can reduce fatigue life by 50% or more.

What are the key factors in selecting a cable vibration damper? Select a damper based on the cable's natural frequency, dominant excitation mechanism, and installation constraints. Stockbridge dampers are effective and economical for overhead conductors. Hydraulic dampers are preferred for stay cables on cable-stayed bridges. Cross-ties (connecting multiple cables with flexible links) can suppress wake galloping in parallel cable arrays. For high-value structures, active damping systems with real-time feedback provide the best performance at the highest cost.

How often should structural cables be inspected? Visual inspection is recommended annually for all accessible structural cables. Detailed inspection with non-destructive testing (magnetic flux leakage, acoustic emission) should be performed every 5 years. Every 10 years, representative sample cables should undergo full load testing or tension measurement. Cables in aggressive environments (marine, industrial) require more frequent inspection.

Related Pages

• Cable Sag Calculator Tool — Interactive cable sag analysis
• Wind Load Calculator — Wind forces on structures
• Steel Weight Calculator — Weight of steel cables and ropes
• Steel Beam Sizes — W-shape, UB, IPE dimensions
• Beam Deflection Calculator — Stiff member deflection analysis

Disclaimer

This is a calculation tool, not a substitute for professional engineering certification. All results must be independently verified by a licensed Professional Engineer (PE), Chartered Professional Engineer (CPEng), or Structural Engineer before use in construction, fabrication, or permit documents. The user is responsible for the accuracy of all inputs and the verification of all outputs.