Small deformation shear strain $\gamma$ in engineering can be expressed as

$$\gamma_{xy}=\alpha +\beta ={\frac {\partial u_{y}}{\partial x}}+{\frac {\partial u_{x}}{\partial y}}$$

where $u$ represents the displacement of the edges of the infinetesimal element. But in the strain tensor we have defined

$$\epsilon_{xy} = 1/2 \gamma_{xy}$$

The shear strain represents the change in the angle between the lines of the infinitesimal element. So why do we define the tensorial component as half of this angle? Why not just use the angle itself?

  • $\begingroup$ Search for engineering shear strain on here... $\endgroup$
    – Solar Mike
    Commented Jun 7, 2019 at 19:08
  • $\begingroup$ @SolarMike I did that before asking this question. $\endgroup$
    – S. Rotos
    Commented Jun 7, 2019 at 19:14
  • $\begingroup$ So you must have found at least one then... $\endgroup$
    – Solar Mike
    Commented Jun 7, 2019 at 19:15
  • $\begingroup$ Have a look at engineering.stackexchange.com/q/6020/10902 $\endgroup$
    – Solar Mike
    Commented Jun 7, 2019 at 19:17
  • $\begingroup$ @SolarMike I don't see how that's relevant to my question. I'm asking why there is a factor of one half in the definition of shear strain, not why we use engineering strain. $\endgroup$
    – S. Rotos
    Commented Jun 7, 2019 at 19:24

2 Answers 2


A tensor is a mathematical object which has to obey certain rules about how to transform it from one coordinate system to another.

Engineers started using and measuring strains a century or more before tensors were invented (by Ricci, in around 1900, and not in the context of continuum mechanics). They just happened to make a bad choice of how to measure shear strains (though they got it right for shear stresses).

You can create an "engineering formulation" of continuum mechanics that doesn't use tensors at all, by pretending the 6 stress components are a "vector," using matrix arithmetic in an ad hoc way, and inventing complicated-looking formulas for transforming stresses and strains into different coordinate systems.

Alternatively, you can change the definition of shear strain by a factor of two, and use mathematics that doesn't need any "special" definitions, just standard vector calculus.

The second way has obvious advantages if you want to combine continuum mechanics with other phenomena such as fluid dynamics, or with special or general relativity.

Really, this is no more of an issue than measuring angles in degrees in real life, or in radians for mathematical purposes.


Consider a displacement map $u : A \subset\mathbb{R}^3 \rightarrow \mathbb{R}^3$. This map displaces a reference material point $x \in A$ by the action $x \mapsto x + u(x)$. We have $\gamma_{xy} = \frac{du}{dx} + \frac{du}{dy}$ as you stated in your question when small angle approximation applies.

Now we will consider a separate, seemingly unrelated problem. Consider some reference point $x_0 \in A$, and choose some other point $x$ very close by. We may call $\Delta a \equiv x - x_0$. The displacement map $u$ sends $x$ to $u(x) + x$ and $x_0$ to $u(x_0) + x_0$ and so gives rise to a $\Delta b = (u(x) + x) - (u(x_0) - x_0)$. We seek to characterize the difference in squared length, after distorting through the map $u$, i.e. approximate $|\Delta b|^2 - |\Delta a|^2$ for all $x$ very close to $x_0$.

$\Delta b = (u(x) - u(x_0)) + (x - x_0)$
$\Delta b = (u(x) - u(x_0)) + \Delta a$

Since $x$ is very close to $x_0$, we may Taylor expand $u$ about $x_0$.
$\Delta b \approx \nabla u|_{x_0}\cdot (x - x_0) + \Delta a$
$\Delta b \approx \nabla u|_{x_0} \cdot \Delta a + \Delta a$
$\Delta b \approx (\nabla u|_{x_0} + I)\cdot \Delta a$

And since $|\Delta b|^2 = \Delta b^T \Delta b$,
$|\Delta b|^2 \approx \Delta a^T(\nabla u|_{x_0} +I)^T(\nabla u|_{x_0} + I)\cdot \Delta a$
$|\Delta b|^2 \approx \Delta a^T(\nabla u|_{x_0}^T\nabla u|_{x_0} + \nabla u|_{x_0} + \nabla u|_{x_0}^T + I)\cdot \Delta a$

For infinitesimal strains, $\nabla u|_{x_0}^T\nabla u|_{x_0}$ is quadratic in the entries of $\nabla u|_{x_0}$ and therefore we can approximate it as $0$.

$|\Delta b|^2 \approx \Delta a^T(\nabla u|_{x_0} + \nabla u|_{x_0}^T + I)\cdot \Delta a$
$|\Delta b|^2 \approx \Delta a^T(\nabla u|_{x_0} + \nabla u|_{x_0}^T) \Delta a + |\Delta a|^2$

$|\Delta b|^2 - |\Delta a|^2 \approx \Delta a^T(\nabla u|_{x_0} + \nabla u|_{x_0}^T) \Delta a$

This is the sought-after approximation. Call the tensor $\nabla u|_{x_0} + \nabla u|_{x_0}^T \equiv T_{x_0}$. You can see that $T_{x_0}$ captures completely how infinitesimal segments (with one endpoint attached to $x_0$) change lengths under $u$. Therefore we cannot do better than $T_{x_0}$ when trying to formulate the concept of a strain tensor. And since the derivation does not depend on coordinate system, it is invariant to choice of $x$, $y$, and $z$ directions, which makes it practical to manipulate mathematically, as @alephzero's answer indicated.

You can see that any multiple of $T_{x_0}$, say $kT_{x_0}$, will provide exactly the same information, but will introduce a scale factor when using it to compute difference of squared lengths. $\Delta a^T \cdot k T_{x_0} \cdot \Delta a = k|b|^2 - k|a|^2$. This scaled $kT$ is less natural for computing the difference of squared lengths, but may be more natural for some other purpose.

We may want to seek the most natural and intuitive scaling parameter $k$, by somehow examining the entries of $T$ and comparing them to simpler engineering notions of strain. This would completely resolve the motivation of your post. However, look at the entries of $T_{x_0}$ in some coordinate system.

$[T_{x_0}]_{ij} = \frac{du_j}{dx_i}(x_0) + \frac{du_i}{dx_j}(x_0)$.

For $i \neq j$, the entries are $\gamma_{ij}(x_0)$, the engineering shear strains at reference point $x_0$.
For $i = j$ the entries are $2 \epsilon_i(x_0)$, twice the engineering direct strains at reference point $x_0$.

So we cannot at once scale the entries of $T_{x_0}$ to seem natural for both direct strains and shear strains. One of them will end up getting "unnaturalled" by a factor of 2. But now remember that engineering shear strain is formulated somewhat arbitrarily. You might even argue that it's more arbitrary than how direct engineering strain is formulated. (It's defined by taking the outer angles formed by one corner of a deformed quadrilateral. Why not take the opposite corner as well into the definition?)

Therefore, choose the scaling $k$ to be natural in terms of the direct engineering strain, i.e. set $k = \frac{1}{2}$.

  • 1
    $\begingroup$ Excellent first answer. Welcome to Eng.SE. $\endgroup$
    – Rob
    Commented Nov 3, 2022 at 23:53

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