Enter the original length of the steel, the coefficient of thermal expansion, and the temperature change into the calculator to determine the shrinkage of the steel.
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Steel Shrinkage Formula
Steel shrinkage from cooling is calculated with the standard linear thermal expansion relationship. For contraction problems, the temperature change is negative, which produces a negative change in length. In practice, the absolute value is the physical amount of shrinkage.
S = L \cdot C \cdot \Delta T
Variable Definitions
| Symbol | Description | Common Units |
|---|---|---|
| S | Change in length of the steel member. | in, ft, mm, cm, m |
| L | Original length before the temperature change. | in, ft, mm, cm, m |
| C | Coefficient of thermal expansion for the steel being analyzed. | in/in/°F, mm/mm/°C |
| ΔT | Temperature change from initial condition to final condition. | °F, °C |
Rearranged Forms
If you know any three inputs, the missing value can be solved directly.
L = \frac{S}{C \cdot \Delta T}C = \frac{S}{L \cdot \Delta T}\Delta T = \frac{S}{L \cdot C}Relationship to Thermal Strain
The calculation can also be viewed as thermal strain multiplied by original length. This is helpful when comparing shrinkage across members of different sizes.
\epsilon = C \cdot \Delta T
S = L \cdot \epsilon
How to Use the Calculator
- Enter the original steel length.
- Enter the coefficient of thermal expansion that matches your material data and temperature unit system.
- Enter the temperature change from starting temperature to ending temperature.
- Keep all units consistent so the output length is meaningful.
- Interpret a negative result as contraction and use the magnitude when you need the physical amount of shortening.
Sign Convention
- A cooling event produces a negative temperature change.
- A negative result means the steel became shorter.
- If you only need the amount of contraction, use the absolute value.
\left| S \right| = L \cdot C \cdot \left| \Delta T \right|
Example Calculation
For a steel member with an original length of 100 inches, a coefficient of 0.00000645 in/in/°F, and a temperature drop of 50°F, the change in length is:
S = 100 \cdot 0.00000645 \cdot (-50) = -0.03225
The member shortens by 0.03225 inches.
Why This Calculation Matters
- Checking fit-up tolerances in fabrication and erection.
- Estimating thermal movement in frames, rails, beams, and pipe runs.
- Sizing gaps, slots, shims, and expansion allowances.
- Reviewing cold-weather dimensional changes before assembly.
- Comparing expected movement between members of different lengths.
Common Input Mistakes
- Mixing Fahrenheit-based coefficients with Celsius temperature changes.
- Using feet for length while expecting an output in inches without converting.
- Entering the final temperature instead of the temperature change.
- Forgetting that cooling should produce a negative temperature change.
- Using a generic coefficient when a project-specific alloy value is required.
Engineering Notes
- This method assumes uniform temperature change along the member.
- It is a linear approximation and is most appropriate when the coefficient can be treated as constant over the temperature range.
- If the steel is restrained, thermal stress may develop even when free shrinkage is small.
- For large gradients, complex restraint, or high-temperature applications, a more detailed analysis may be needed.
