Inputs
Example data table
| Material | L | d | Load | E | ΔT | ΔL (approx.) |
|---|---|---|---|---|---|---|
| Steel | 2.5 m | 2.0 mm | 1200 N | 200 GPa | 25 °C | ~2.37 mm |
| Aluminum | 1.8 m | 3.0 mm | 0.9 kN | 69 GPa | 40 °C | ~2.74 mm |
| Copper | 4.0 m | 1.6 mm | 800 N | 110 GPa | 15 °C | ~6.06 mm |
| Brass | 3.0 m | 2.5 mm | 65 MPa | 100 GPa | 0 °C | ~1.95 mm |
Formulas used
How to use this calculator
- Choose a calculation mode: force-based or stress-based.
- Enter wire length and either diameter or cross-sectional area.
- Select a material preset, or keep custom values.
- Provide Young’s modulus and your load input.
- Optionally enable thermal effects and add ΔT and α.
- Press Calculate to view results above the form.
- Use the CSV or PDF buttons to export results.
Wire elongation guide
1) What elongation means
Elongation is the change in wire length under load. For small elastic stretches, the wire returns to its original length after unloading. This calculator reports mechanical elongation (from load), thermal elongation (from temperature), total elongation, and final length using consistent units.
2) Key elastic inputs
The mechanical part follows Hooke’s law, so stiffness matters most. Typical Young’s modulus values are about 200 GPa for steel, 69 GPa for aluminum, 110 GPa for copper, and 100 GPa for brass. Higher E produces smaller elongation for the same stress.
3) Area, diameter, and stress
Stress is σ = F/A, so cross-sectional area strongly controls results. If you enter diameter, the tool computes area for a round wire with A = πd²/4. Doubling diameter increases area fourfold, reducing stress and elastic elongation by roughly 75% at the same force.
4) Thermal contribution
Thermal elongation uses ΔLth = αLΔT. Typical α values are ~12×10⁻⁶/°C for steel, ~23×10⁻⁶/°C for aluminum, and ~16.5×10⁻⁶/°C for copper. A 2.5 m steel wire with ΔT = 25°C expands about 0.75 mm from heat alone.
5) Choosing force or stress mode
Use force mode when you know the applied load (N, kN, or lbf). Use stress mode when loading is specified as a stress limit (MPa, GPa, psi). The calculator converts between force and stress internally and still reports both values for quick verification.
6) Interpreting strain
Strain is ε = ΔL/L (dimensionless). In many design checks, strains above about 0.002 (0.2%) indicate yielding risk for common metals, depending on alloy and heat treatment. If strain is high, confirm the load and consider using a stronger or larger wire.
7) Units and sanity checks
The tool converts all entries to SI for calculation, then returns outputs in your selected display units. As a quick check, wires with small diameters under kN loads can produce tens to hundreds of MPa stress. If your stress looks too low, verify area units (mm² vs m²).
8) Practical accuracy notes
Results assume a straight, uniform, isotropic wire in the linear elastic range. Real systems may include grips, temperature gradients, creep, and plastic deformation. For high temperatures or long-term loading, consult material data sheets and apply safety factors beyond the calculator’s scope.
FAQs
1) Can I use this for cables or strands?
Yes, if you know the effective cross-sectional area and elastic modulus. For stranded cables, E can be lower than solid wire due to lay angle and construction, so use measured or manufacturer values when possible.
2) What if I only know gauge size?
Convert gauge to diameter first, then enter diameter. If you already have the conductor area from a datasheet, enter area directly to avoid gauge conversion errors.
3) Why does diameter matter so much?
Area scales with d². A small diameter increase can greatly reduce stress and elongation because stress is force divided by area, and elastic stretch is proportional to stress.
4) Does temperature always increase length?
For positive α and positive ΔT, yes. If the wire cools (negative ΔT), thermal elongation becomes negative, meaning the wire contracts. The calculator handles either sign if you enter a negative ΔT.
5) Which elongation should I use for clearance checks?
Use total elongation when both load and temperature change apply. For purely mechanical deflection under constant temperature, use mechanical elongation only.
6) Is the calculated stress the same as tensile strength?
No. Stress here is the applied working stress. Tensile strength is a material limit. Compare the calculated stress against allowable stress or yield strength with an appropriate safety factor.
7) Why does my result differ from a handbook example?
Differences usually come from unit mix-ups, different E or α values, or area assumptions. Confirm your area units, verify modulus for the exact alloy, and check whether the reference includes thermal expansion.