Are material strains additive or multiplicative?

Question

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1. What the question is asking (in plain language)

A 1‑m aluminium rod is first cooled so that it becomes 10 ppm shorter (a strain of –10 × 10⁻⁶).
After it is warmed back to room temperature it is loaded in tension so that it becomes 10 ppm longer (a strain of +10 × 10⁻⁶).

If the two actions are performed together, what is the total change in length?

The student wonders whether the two strains add (‑10 ppm + +10 ppm = 0) or multiply (1–10 ppm × 1+10 ppm ≈ 0.9999 → a 0.01 % shortening).
The same question is raised for a much larger strain (±10 %).

In other words: Are strains additive or multiplicative?

2. Detailed solution

2.1. Definitions

Symbol	Meaning
(L_0)	Original length (1 m)
(\lambda)	Stretch ratio = final length / original length
(\varepsilon)	Engineering strain = ((L-L_0)/L_0) (dimensionless)
(\alpha)	Coefficient of linear thermal expansion (≈ 23 × 10⁻⁶ K⁻¹ for Al)
(\Delta T)	Temperature change
(\sigma)	Axial stress
(E)	Young’s modulus of aluminium (≈ 70 GPa)

The engineering strain (\varepsilon) is additive for small deformations (the usual linear‑elastic regime).
For finite deformations the true (logarithmic) strain or the stretch ratio (\lambda) is used and the multiplicative decomposition of deformation gradients becomes appropriate.

2.2. Small‑strain case (≈ 10 ppm)

Thermal strain
[ \varepsilon_{\text{th}} = \alpha\,\Delta T ] The cooling was chosen so that (\varepsilon_{\text{th}} = -10\times10^{-6}).
Mechanical strain (tension)
[ \varepsilon_{\text{mech}} = \frac{\sigma}{E} ] The weight was chosen so that (\varepsilon_{\text{mech}} = +10\times10^{-6}).
Superposition (additivity)
In the linear‑elastic regime the total engineering strain is the sum of the individual strains: [ \varepsilon_{\text{total}} = \varepsilon_{\text{th}} + \varepsilon_{\text{mech}} = (-10\times10^{-6}) + ( +10\times10^{-6}) = 0 . ]
Resulting length
[ L = L_0\,(1+\varepsilon_{\text{total}}) = 1\ \text{m}\times(1+0)=1.000\ \text{m}. ]

Conclusion: For the realistic 10 ppm values the two effects cancel exactly (to the precision of the linear model). The cold‑plus‑loaded rod is the same length as the original.

2.3. Large‑strain case (≈ 10 % change)

A 10 % strain is not “small” – the linear approximation is no longer accurate.
We must work with the stretch ratio (\lambda = 1+\varepsilon) (or with true/logarithmic strain).

Individual stretches
- Thermal shrinkage: (\lambda_{\text{th}} = 0.90) (10 % shorter).
- Mechanical elongation: (\lambda_{\text{mech}} = 1.10) (10 % longer).
Multiplicative combination

The total deformation gradient for a body subjected to two sequential deformations is the product of the individual stretch ratios: [ \lambda_{\text{total}} = \lambda_{\text{mech}}\,\lambda_{\text{th}} = 1.10 \times 0.90 = 0.99 . ]
Corresponding engineering strain

[ \varepsilon_{\text{total}} = \lambda_{\text{total}} - 1 = 0.99 - 1 = -0.01 = -1\%. ]
Resulting length

[ L = L_0\,\lambda_{\text{total}} = 1\ \text{m}\times 0.99 = 0.99\ \text{m}. ]

Conclusion: When the strain magnitude is large (10 %), the combined effect is multiplicative, giving a net 1 % shortening. The cold‑plus‑loaded rod is shorter than the original.

2.4. Why the two regimes differ

Regime	Governing relation	Reason
Small strains ((	\varepsilon	\lesssim 10^{-3}) i.e. < 0.1 %)	(\varepsilon_{\text{total}} = \varepsilon_1 + \varepsilon_2)	Linearised kinematics → superposition holds.
Finite strains (≥ ~1 %)	(\lambda_{\text{total}} = \lambda_1\lambda_2) (or (\ln\lambda_{\text{total}} = \ln\lambda_1 + \ln\lambda_2))	Exact geometry of deformation; multiplication of deformation gradients is required.

For engineering practice (most structural problems) strains are far below 0.1 %, so the additive rule is used.
When dealing with large thermal expansions, soft polymers, rubber, shape‑memory alloys, or high‑precision metrology where strains approach a few percent, the multiplicative treatment is necessary.

3. Final answer

For realistic small strains (10 ppm, i.e. 10⁻⁵) the thermal and mechanical contributions add linearly. The cold‑plus‑loaded rod returns to exactly its original length (to the accuracy of the linear model).
For large strains (e.g., ±10 %) the correct description is multiplicative: the total stretch is the product of the individual stretches, giving a net shortening of 1 % (the rod ends up 0.99 m long).

Thus, strains are additive only in the small‑strain (linear‑elastic) regime; for finite strains they combine multiplicatively.

4. Common mistakes

Mistake	Why it’s wrong	How to avoid it
Assuming linear superposition for any size strain	The linear relationship (\varepsilon = \sigma/E) is derived by neglecting higher‑order terms in the deformation gradient. It fails when (	\varepsilon	) approaches a few percent.	Check the magnitude of the strain. If (	\varepsilon	< 10^{-3}) (≈ 0.1 %), additive treatment is fine; otherwise use stretch ratios or true strain.
Confusing engineering strain with true (logarithmic) strain	Engineering strain (\varepsilon = \Delta L/L_0) is not additive for large deformations; true strain (\varepsilon_{\text{true}} = \ln(1+\varepsilon)) is additive.	For finite deformations convert to true strain, add, then exponentiate back to get the final stretch.
Neglecting the order of operations (e.g., heating then loading vs loading then heating)	In the finite‑strain regime the product (\lambda_{\text{mech}}\lambda_{\text{th}}) is commutative for pure axial stretches, but for more general deformations the order matters because deformation gradients do not commute.	Remember that the simple product works only for coaxial (same direction) stretches; for general cases use full deformation‑gradient multiplication.
Using the coefficient of thermal expansion beyond its linear range	(\Delta L = \alpha L_0 \Delta T) is linear only for modest (\Delta T). At very large temperature changes the material’s (\alpha) can vary with temperature.	Verify that (\Delta T) is small enough, or integrate (\alpha(T)) over the temperature interval.
Treating Poisson’s effect as irrelevant	Axial loading also causes lateral contraction; thermal expansion may be constrained, affecting the axial strain.	Include Poisson’s ratio if the rod is not free to expand laterally or if lateral constraints are present.

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