True Stress and True Strain Calculator

Analyze a material’s behavior more accurately by calculating true stress and true strain, which account for the actual, changing dimensions under load.

What is True Stress and True Strain?

True stress and true strain are fundamental concepts in materials science and engineering that describe a material’s response to an applied load. Unlike their “engineering” counterparts, they are based on the instantaneous (actual) dimensions of the material during deformation. Engineering stress and strain are calculated using the original, undeformed dimensions, which can lead to inaccuracies, especially after significant plastic deformation occurs.

True stress (σ_t) is defined as the applied load divided by the actual cross-sectional area of the specimen at that load. As a material is stretched (in tension), its cross-sectional area decreases. By using this smaller, instantaneous area, true stress provides a more accurate measure of the stress the material is actually experiencing. This is why, in a tensile test, true stress values are higher than engineering stress values after the material starts to yield.

True strain (ε_t), also known as logarithmic strain, is the natural logarithm of the ratio of the instantaneous length to the original length. It accounts for the cumulative deformation the material has undergone. This logarithmic approach provides a more consistent measure of strain, particularly for large deformations where engineering strain can be misleading.

The Formulas for True Stress and Strain

The calculations for true stress and strain are straightforward, relying on the instantaneous measurements of the material under load.

True Stress Formula

True stress (σ_t) is calculated by dividing the applied force (F) by the instantaneous cross-sectional area (A):

σt = F / A

True Strain Formula

True strain (ε_t) is the natural logarithm (ln) of the ratio of the instantaneous length (L) to the original length (L₀):

εt = ln(L / L₀)

Variables Used in Stress and Strain Calculations

Variable	Meaning	Common Units	Typical Range
F	Applied Force/Load	Newtons (N), Pounds-force (lbf)	0 – 1,000,000+
A₀	Original Area	mm², in²	1 – 10,000+
A	Instantaneous Area	mm², in²	Slightly less than A₀ in tension
L₀	Original Length	mm, in	10 – 1000+
L	Instantaneous Length	mm, in	Slightly more than L₀ in tension
σt / σe	Stress	Pascals (Pa), Megapascals (MPa), PSI	Dependent on material strength
εt / εe	Strain	Unitless (mm/mm or in/in)	0.001 – 2.0+

Practical Examples

Example 1: Steel Rod in Tension (Metric)

An engineer is testing a steel rod to understand its properties.

• Inputs:
  • Applied Force (F): 50,000 N
  • Original Area (A₀): 150 mm²
  • Instantaneous Area (A): 145 mm²
  • Original Length (L₀): 100 mm
  • Instantaneous Length (L): 105 mm
• Results:
  • Engineering Stress (σe) = 50,000 N / 150 mm² = 333.33 MPa
  • Engineering Strain (εe) = (105 mm – 100 mm) / 100 mm = 0.05
  • True Stress (σt) = 50,000 N / 145 mm² ≈ 344.83 MPa
  • True Strain (εt) = ln(105 mm / 100 mm) ≈ 0.0488

Example 2: Aluminum Specimen (Imperial)

A lab technician measures the deformation of an aluminum sample.

• Inputs:
  • Applied Force (F): 2,000 lbf
  • Original Area (A₀): 0.5 in²
  • Instantaneous Area (A): 0.48 in²
  • Original Length (L₀): 4 in
  • Instantaneous Length (L): 4.1 in
• Results:
  • Engineering Stress (σe) = 2,000 lbf / 0.5 in² = 4,000 PSI
  • Engineering Strain (εe) = (4.1 in – 4.0 in) / 4.0 in = 0.025
  • True Stress (σt) = 2,000 lbf / 0.48 in² ≈ 4,166.67 PSI
  • True Strain (εt) = ln(4.1 in / 4.0 in) ≈ 0.0247

For more advanced simulations, consider exploring resources on finite element analysis.

How to Use This True Stress and Strain Calculator

1. Enter Applied Force: Input the total force or load applied to the material.
2. Select Force Units: Choose the appropriate unit for your force measurement (Newtons, kilonewtons, or pounds-force).
3. Enter Original and Instantaneous Area: Provide the material’s cross-sectional area before deformation (Original Area) and the area at the moment of measurement (Instantaneous Area).
4. Select Area Units: Choose between square millimeters (mm²) or square inches (in²). Ensure this is consistent for both area inputs.
5. Enter Original and Instantaneous Length: Input the material’s gauge length before deformation and its length at the moment of measurement.
6. Calculate: Click the “Calculate” button to see the results. The calculator will display the primary results for True Stress and True Strain, along with the intermediate values for Engineering Stress and Strain.
7. Interpret Results: The displayed values help you understand the material’s true behavior under its current load, which is critical for advanced analysis and failure prediction. Understanding this can be a key part of material failure analysis.

Key Factors That Affect True Stress and Strain

Several factors influence the true stress-strain relationship of a material. Understanding these is vital for accurate material modeling and prediction.

• Material Composition: The inherent atomic structure and alloying elements dictate a material’s strength and ductility. For example, high-carbon steel behaves differently from aluminum.
• Strain Hardening: As a ductile material deforms, dislocations in its crystal structure move and multiply, making it stronger and harder. This is why true stress continues to increase even as the material necks. A related concept is work hardening.
• Temperature: Higher temperatures generally increase ductility and decrease strength. The material becomes softer and can deform more easily, affecting both stress and strain values.
• Strain Rate: The speed at which the material is deformed can have a significant impact. Many materials exhibit higher strength when pulled apart quickly compared to when they are pulled apart slowly.
• Necking: In ductile materials under tension, deformation eventually localizes in a small region, causing a significant reduction in cross-sectional area. This phenomenon, known as necking, marks the point where the engineering stress curve begins to fall, while the true stress curve continues to rise sharply.
• Volume Constancy: The calculation of true stress from engineering stress often relies on the assumption that the material’s volume remains constant during plastic deformation. While a good approximation, it’s not perfectly accurate for all conditions. You can learn more about this with a Poisson’s Ratio calculator.

Frequently Asked Questions (FAQ)

1. Why is true stress generally higher than engineering stress?

True stress uses the instantaneous, smaller cross-sectional area of a material under tension. Since you are dividing the same force by a smaller number, the resulting stress value is higher. Engineering stress uses the larger, original area, underestimating the actual stress after yielding.

2. What is the unit of strain?

Both true strain and engineering strain are unitless. They are a ratio of length to length (e.g., mm/mm or in/in), so the units cancel out.

3. Can true strain be negative?

Yes. In a compression test, the instantaneous length (L) is shorter than the original length (L₀). The natural logarithm of a number less than 1 is negative, so the true strain will be negative, indicating compressive deformation.

4. When is it critical to use true stress and strain?

It’s most critical in situations involving large plastic deformation, such as in finite element analysis (FEA) simulations, forming processes (like forging or stamping), and advanced material failure analysis. For simple elastic analysis, engineering stress is often sufficient.

5. How is the instantaneous area (A) measured in a real test?

In modern testing labs, non-contact methods like laser extensometers or digital image correlation (DIC) are used to continuously measure the change in the specimen’s width and calculate the instantaneous area throughout the test.

6. What is the relationship between true strain and engineering strain?

True strain can be calculated from engineering strain (εe) using the formula: εt = ln(1 + εe). This relationship holds true up until the onset of necking.

7. Why does the engineering stress curve dip after the ultimate tensile strength (UTS)?

The dip occurs because of necking. The load required to continue stretching the material decreases as the cross-sectional area rapidly shrinks in the necked region. Since engineering stress is based on the original area, it falsely appears as if the material is getting weaker. True stress, however, accounts for the shrinking area and continues to rise.

8. Does this calculator account for necking?

This calculator performs the calculation based on the instantaneous area you provide. If you input the area from a necked region, the resulting true stress will accurately reflect the stress in that specific location. The formulas themselves are valid before and after necking begins.