Deformation and relative strain

The same extension can mean different deformation

A 1 mm extension is substantial for a 10 mm gauge length and smaller relative to a 100 mm gauge length. Engineering strain compares the length change with the original measured length: ε = ΔL/L₀. The gauge length is the interval over which extension is measured.

Choose two marks on the specimen and call their original separation L₀. When their separation becomes L, their extension is ΔL = L − L₀. The Greek letter ε, epsilon, denotes the relative change. A strain of 0.01 means the length changed by one hundredth of its original value.

Displacement describes movement of a point; deformation concerns a change of shape or size. If both gauge marks move 3 mm in the same direction and their separation stays 50 mm, the specimen has translated but that gauge has no axial extension. Tracking only one end would not establish strain.

A ratio with no length unit

Divide lengths expressed in matching units. The units cancel, so strain is dimensionless. Multiplying the ratio by 100 expresses percent strain. Thus ε = 0.002 is the same relative change as 0.2 %, not 0.002 %. The written form mm/mm can remind us what was divided; it does not give strain the dimension of length.

We use positive strain for an increase in gauge length and negative strain for a decrease. If a 40 mm gauge shortens to 39.8 mm, ΔL = −0.2 mm and ε = −0.2/40 = −0.005, or −0.5 %. Some reports give only a positive compression magnitude; check the sign convention before comparing results.

Keep the original length in the denominator when calculating engineering strain. Dividing by the final length answers a different question. Other strain measures are useful for large changes and are introduced later; the present formula is explicitly a relative change referred to L₀.

The gauge is a measurement interval

If extension is distributed uniformly, a gauge twice as long contains twice the extension, so ΔL/L₀ stays the same. Equal extension on different gauge lengths is a different comparison: 1 mm over 10 mm is 10 %, whereas 1 mm over 100 mm is 1 %.

A gauge can also contain regions that deform differently. Its overall strain then describes the average relative length change over that interval. It does not reveal the largest local strain. For adjoining collinear gauge segments, add their extensions and original lengths before taking the overall ratio; a plain average of the two strain numbers works only when their original lengths are equal.

An extensometer measures the change between defined specimen points. The testing machine’s moving crosshead can also move because grips settle, the load train deforms or material outside the selected gauge stretches. Its displacement is therefore not automatically the extension of that gauge. The measurement interval must match the quantity being calculated.

A strain value needs a loading state and a time

A length measurement while loaded can include recoverable and permanent deformation. Removing the load may allow immediate recovery, delayed recovery, or both. A residual strain is the relative change still present at a stated time after unloading; that word describes an observation, not its final cause.

For example, internal rearrangements in a time-dependent material can allow recovery to continue after the external load is removed. A residual measured immediately after unloading need not be permanent. Conversely, a specimen that eventually returns to L₀ still had nonzero strain while it was loaded.

Record the gauge definition, loading state, elapsed recovery time and relevant conditions such as temperature. This lesson calculates relative length changes; deciding which microscopic mechanism caused them needs further evidence. Elastic response, plasticity and time-dependent behaviour are developed in their own lessons.

Worked examples

Try the reasoning

1. A 2 mm extension is reported for one specimen. A second specimen has a 200 mm gauge and 0.5 % strain. What missing information prevents a fair strain comparison, and what is the second specimen’s extension?

   Show the answer and reasoning

   The first specimen’s original gauge length is missing. The second extends 1 mm.

   The first strain would be 2 mm/L₀, so extension alone does not rank the relative changes. For the second, 0.5 % = 0.005 and ΔL = 0.005 × 200 mm = 1 mm.

2. At constant temperature, an illustrative 100 mm gauge is 101 mm while loaded, 100.4 mm immediately after unloading and 100.1 mm after ten minutes. Calculate the three strains. What can and cannot be concluded about permanent deformation?

   Show the answer and reasoning

   Loaded strain is 1 %; immediate residual strain is 0.4 %; the ten-minute residual is 0.1 %. Some recovery continued after unloading, but the final permanent strain is undetermined.

   Subtract the same original 100 mm and divide each extension by 100 mm. The decline from 0.4 % to 0.1 % demonstrates delayed recovery in these stipulated readings. A later value or suitable recovery evidence is needed before calling the remaining value permanent; zero strain later would not erase the earlier loaded strain.

3. A 25.0 mm gauge shortens by 0.10 mm. A learner calculates 0.10/24.90 = +0.004016. Correct the engineering-strain calculation and identify both errors.

   Show the answer and reasoning

   ε = −0.10/25.0 = −0.004 = −0.4 %.

   The change is negative under the stated tension-positive convention, and engineering strain uses the original 25.0 mm denominator. Dividing by the final length gives a different ratio. A positive compression magnitude could be quoted only with that alternative convention stated.

4. During one test, the machine crosshead moves 0.30 mm while a 50 mm specimen gauge extends 0.10 mm. Compare the gauge strain with the value obtained by dividing crosshead travel by 50 mm. What does the discrepancy tell you?

   Show the answer and reasoning

   The gauge strain is 0.2 %. The crosshead-based calculation gives 0.6 %, three times as large.

   The matched measurement is 0.10/50 = 0.002. The other quotient, 0.30/50 = 0.006, assigns motion from a different measurement interval to this gauge. The difference may include machine, grip or out-of-gauge deformation; these two numbers do not isolate which contribution caused it.

What to carry forward

• Extension is a length change; engineering strain divides it by the original gauge length.
• Use matching length units, the stated sign convention and the correct percent conversion.
• A gauge average need not equal a local maximum, and machine motion need not equal gauge extension.
• Name the loading state and recovery time before interpreting a residual strain.