Strain transformation

A strain gauge glued to a machine part cannot feel stress directly — it can only feel the tiny stretch of the metal underneath it, and only along the one direction it is glued in. Yet engineers routinely report principal stresses from gauge readings. The bridge between "stretch along one line" and "the full state of deformation at a point" is strain transformation: the same rotation mathematics used for stress, applied to strain instead. Given the normal strains $\varepsilon_x$ , $\varepsilon_y$ and shear strain $\gamma_{xy}$ on one pair of reference axes, transformation equations hand you the normal and shear strain along any rotated direction, the principal strains, the maximum shear strain, and a Mohr's circle to visualize all of it. The one twist to remember: shear strain enters the formulas as $\gamma_{xy}/2$ , not $\gamma_{xy}$ .

The core insight

Strain transforms exactly like stress under a change of axes, provided you always substitute $\gamma_{xy}/2$ for $\tau_{xy}$ , which lets strain-gauge measurements along fixed directions be converted into principal strains via the same rotation equations and Mohr's circle used for stress.

The core idea

Stress and strain transformation are mathematically the same problem wearing different labels. Stress transformation rotates $(\sigma_x,\sigma_y,\tau_{xy})$ ; strain transformation rotates $(\varepsilon_x,\varepsilon_y,\gamma_{xy})$ , with the substitution $\sigma\to\varepsilon$ and $\tau\to\gamma/2$ . That halving is not cosmetic — it comes from how shear strain is defined (the total angular distortion between two originally perpendicular lines, split as two half-angle rotations), while the transformation algebra is naturally built around half of that total. For a plane-strain state at angle $\theta$ from the reference axes:

\varepsilon_{x'} = \frac{\varepsilon_x+\varepsilon_y}{2} + \frac{\varepsilon_x-\varepsilon_y}{2}\cos2\theta + \frac{\gamma_{xy}}{2}\sin2\theta

\frac{\gamma_{x'y'}}{2} = -\frac{\varepsilon_x-\varepsilon_y}{2}\sin2\theta + \frac{\gamma_{xy}}{2}\cos2\theta

where: $\varepsilon_x,\varepsilon_y$ are the normal strains on the reference axes, $\gamma_{xy}$ is the engineering shear strain, $\theta$ is the angle to the rotated axis $x'$ (measured the same sense as in the stress equations), and $\varepsilon_{x'},\gamma_{x'y'}$ are the transformed normal and shear strains. At the two special orientations where the shear strain vanishes, the normal strains take their extreme values — the principal strains:

\varepsilon_{1,2} = \frac{\varepsilon_x+\varepsilon_y}{2} \pm \sqrt{\left(\frac{\varepsilon_x-\varepsilon_y}{2}\right)^2+\left(\frac{\gamma_{xy}}{2}\right)^2}

where: $\varepsilon_1,\varepsilon_2$ are the principal (extreme) normal strains and the square-root term is the radius $R$ of the strain Mohr's circle. The maximum in-plane shear strain is simply $\gamma_{max}/2 = R$ , so $\gamma_{max}$ is twice that radius.

Mohr's circle for strain is the exact geometric twin of the stress circle: plot $\varepsilon$ on the horizontal axis and $\gamma/2$ on the vertical axis, centre it at $\left(\frac{\varepsilon_x+\varepsilon_y}{2},0\right)$ with radius $R$ , and a physical rotation of $\theta$ appears on the circle as a rotation of $2\theta$ — same doubling rule as stress. Why does this whole toolkit matter in practice? Stress cannot be measured directly at a point on a loaded part — there is no instrument that reads "pascals" off a surface. Strain, on the other hand, is a change in length, and a strain gauge (a thin resistive foil bonded to the surface) measures exactly that along one fixed direction, very precisely, through the tiny change in electrical resistance the stretch produces. Because a single gauge only reports strain along its own axis, engineers bond three gauges at known angles into a rosette and use the transformation equations, run in reverse, to solve for $\varepsilon_x,\varepsilon_y,\gamma_{xy}$ from the three readings. Only once that plane-strain state is known can the principal strains — and, via Hooke's law, the principal stresses — be recovered. This experimental chain, gauge readings to transformed strains to stresses, is the reason strain transformation is taught alongside stress transformation rather than as an afterthought. For the common 45° rosette, with gauges at $0°$ , $45°$ , $90°$ reading $\varepsilon_a,\varepsilon_b,\varepsilon_c$ respectively, the reduction to $\varepsilon_x,\varepsilon_y,\gamma_{xy}$ is direct rather than requiring simultaneous equations: $\varepsilon_x=\varepsilon_a$ , $\varepsilon_y=\varepsilon_c$ , and $\gamma_{xy}=2\varepsilon_b-(\varepsilon_a+\varepsilon_c)$ , obtained by substituting $\theta=45°$ into the $\varepsilon_{x'}$ transformation equation and solving for $\gamma_{xy}$ . A different arm spacing (60° rectangular or delta rosettes) needs its own reduction formula, re-derived the same way for that specific geometry.

Figure

Strain Mohr circle showing the centre point average strain the radius R and the two principal strain points where the circle crosses the horizontal axis

Derivation

Derivation APrincipal strains from a plane-strain state

The goal is to locate the orientation at which the shear strain $\gamma_{x'y'}$ vanishes and show that the corresponding normal strains equal the extreme values $\varepsilon_{1,2}$ given above, starting from nothing but the transformation equation for $\gamma_{x'y'}/2$ and ordinary calculus-free trigonometry — the same route used to find principal stresses.

Step 1. Set the shear-strain transformation equation to zero, since principal directions are defined as the orientations with no shear distortion: $0 = -\dfrac{\varepsilon_x-\varepsilon_y}{2}\sin2\theta_p + \dfrac{\gamma_{xy}}{2}\cos2\theta_p$ .

Step 2. Solve for the principal angle. Rearranging gives $\dfrac{\gamma_{xy}}{2}\cos2\theta_p = \dfrac{\varepsilon_x-\varepsilon_y}{2}\sin2\theta_p$ , so $\tan2\theta_p = \dfrac{\gamma_{xy}}{\varepsilon_x-\varepsilon_y}$ . This equation has two solutions $180°$ apart in $2\theta_p$ , i.e. $90°$ apart in $\theta_p$ — the two mutually perpendicular principal directions.

Step 3. Build a reference right triangle for $2\theta_p$ with opposite side $\gamma_{xy}/2$ , adjacent side $(\varepsilon_x-\varepsilon_y)/2$ , and hypotenuse $R=\sqrt{\left(\frac{\varepsilon_x-\varepsilon_y}{2}\right)^2+\left(\frac{\gamma_{xy}}{2}\right)^2}$ . Then $\cos2\theta_p = \dfrac{(\varepsilon_x-\varepsilon_y)/2}{R}$ and $\sin2\theta_p = \dfrac{\gamma_{xy}/2}{R}$ .

Step 4. Substitute these back into the normal-strain transformation equation $\varepsilon_{x'} = \dfrac{\varepsilon_x+\varepsilon_y}{2} + \dfrac{\varepsilon_x-\varepsilon_y}{2}\cos2\theta_p + \dfrac{\gamma_{xy}}{2}\sin2\theta_p$ . The two cosine and sine terms each become (something) $^2/R$ , and adding them gives $\dfrac{(\varepsilon_x-\varepsilon_y)^2/4+(\gamma_{xy})^2/4}{R} = \dfrac{R^2}{R} = R$ .

Step 5. So $\varepsilon_{x'} = \dfrac{\varepsilon_x+\varepsilon_y}{2} + R$ at this orientation; taking the other root ( $2\theta_p+180°$ , which flips both trig signs) gives $\dfrac{\varepsilon_x+\varepsilon_y}{2} - R$ . These are exactly $\varepsilon_1$ and $\varepsilon_2$ from the boxed formula — the shear-free directions genuinely do carry the maximum and minimum normal strain, confirming the principal-strain result algebraically rather than by analogy alone.

Worked example

Using the same state as the reference article — $\varepsilon_x = 800\,\mu\varepsilon$ , $\varepsilon_y = 200\,\mu\varepsilon$ , $\gamma_{xy} = 400\,\mu\text{rad}$ — find $\varepsilon_{x'}$ and $\gamma_{x'y'}$ at $\theta = 30°$ , then the principal strains and the principal direction.

Step 1. Compute the building blocks: average $= (800+200)/2 = 500\,\mu\varepsilon$ , half-difference $= (800-200)/2 = 300\,\mu\varepsilon$ , half-shear $= 400/2 = 200\,\mu\text{rad}$ .

Step 2. At $\theta=30°$ , $2\theta=60°$ , so $\cos60°=0.5$ and $\sin60°=0.866$ . Then $\varepsilon_{x'} = 500 + 300(0.5) + 200(0.866) = 500+150+173.2 = 823.2\,\mu\varepsilon$ .

Step 3. For the shear strain, $\gamma_{x'y'}/2 = -300(0.866)+200(0.5) = -259.8+100 = -159.8\,\mu\text{rad}$ , so $\gamma_{x'y'} = -319.6\,\mu\text{rad}$ . Check: $\varepsilon_{y'} = 1000-823.2 = 176.8\,\mu\varepsilon$ , consistent with the invariant $\varepsilon_x+\varepsilon_y=\varepsilon_{x'}+\varepsilon_{y'}$ .

Step 4. Radius of the strain circle: $R=\sqrt{300^2+200^2}=\sqrt{130000}=360.6\,\mu\varepsilon$ . Principal strains: $\varepsilon_1 = 500+360.6=860.6\,\mu\varepsilon$ , $\varepsilon_2 = 500-360.6=139.4\,\mu\varepsilon$ , and maximum shear strain $\gamma_{max}=2R=721.2\,\mu\text{rad}$ .

Step 5. Principal direction: $\tan2\theta_p = \gamma_{xy}/(\varepsilon_x-\varepsilon_y) = 400/600 = 0.667$ , giving $2\theta_p = 33.7°$ so $\theta_p = 16.85°$ .

Final answer: $\varepsilon_{x'}=823.2\,\mu\varepsilon$ , $\gamma_{x'y'}=-319.6\,\mu\text{rad}$ at $\theta=30°$ ; principal strains $\varepsilon_1=860.6\,\mu\varepsilon$ , $\varepsilon_2=139.4\,\mu\varepsilon$ at $\theta_p=16.85°$ , with $\gamma_{max}=721.2\,\mu\text{rad}$ .

Competitive stretch

A 45° strain-gauge rosette (arms at $0°,45°,90°$ ) on the same part reads $\varepsilon_a=800\,\mu\varepsilon$ , $\varepsilon_b=700\,\mu\varepsilon$ , $\varepsilon_c=200\,\mu\varepsilon$ . Recover $\varepsilon_x,\varepsilon_y,\gamma_{xy}$ and confirm they match the state used above. Directly, $\varepsilon_x=\varepsilon_a=800\,\mu\varepsilon$ and $\varepsilon_y=\varepsilon_c=200\,\mu\varepsilon$ . Then $\gamma_{xy}=2\varepsilon_b-(\varepsilon_a+\varepsilon_c)=2(700)-(800+200)=1400-1000=400\,\mu\text{rad}$ — exactly the state analysed in the main example, showing how a rosette's three readings alone reconstruct the full strain state needed for the principal-strain formulas.

Common traps

Forgetting the $\gamma/2$ substitution. Plugging $\gamma_{xy}$ directly where $\tau_{xy}$ sits in the stress equations (without halving) is the single most common strain-transformation error.
Mixing microstrain and strain. Strain-gauge data is usually quoted in microstrain ( $\mu\varepsilon=10^{-6}$ ); carry the units consistently rather than converting partway through a calculation.
Assuming principal strain and stress magnitudes coincide. For isotropic linear-elastic materials the principal directions of stress and strain do coincide, but the magnitudes differ through $E$ and $\nu$ — never equate $\varepsilon_1$ numerically with $\sigma_1$ .
Applying the 45° rosette shortcut to a non-45° rosette. The formula $\gamma_{xy}=2\varepsilon_b-(\varepsilon_a+\varepsilon_c)$ only holds for the standard 0°–45°–90° layout; a 60° rectangular or delta rosette needs its own re-derived reduction equations.

Ask the tutor: if you would like a walkthrough converting a full strain-gauge rosette reading into principal strains and principal stresses, or a side-by-side comparison sheet of the stress and strain transformation equations.

Why does shear strain need the factor of 2 but normal strain does not?

Engineering shear strain $\gamma_{xy}$ is defined as the total change in the right angle between two originally perpendicular lines — effectively the sum of two half-angle rotations. The rotation mathematics that produces the transformation equations naturally works with one of those half-angles at a time, which is exactly $\gamma_{xy}/2$ . Normal strain has no such split: it is a single stretch ratio along one line, so it transforms with no extra factor, just like $\sigma$ does in the stress equations.

Can I use the stress Mohr's circle procedure directly for strain, or do I need a separate method?

The procedure is identical — same circle construction, same $2\theta$ doubling rule, same reading of principal values off the horizontal-axis crossings. The only change is what you plot on the axes: strain $\varepsilon$ instead of stress $\sigma$ , and $\gamma/2$ instead of $\tau$ . If you already know how to draw a stress Mohr's circle, you already know how to draw a strain one — just relabel the axes and halve the shear value before plotting.

Once I have principal strains from a rosette, how do I get principal stresses?

You substitute $\varepsilon_1$ and $\varepsilon_2$ into the plane-stress Hooke's-law relations (using $E$ and $\nu$ for the material) to solve for $\sigma_1$ and $\sigma_2$ — this is why gauges are used at all, since stress cannot be measured directly but strain can. That substitution step belongs to the Hooke's-law/axial-loading topic rather than to transformation itself; transformation only gets you as far as the principal strains and their directions, which then feed into the stress calculation as a separate step.