Mohr's Circle: Stress States from Element to FEM

Push a steel beam and something happens inside it that your intuition quietly gets wrong. A fibre near the flange, the metal around a bolt hole, the throat of a weld, a slice of web carrying bending and shear at once — none of them feels a single "up-or-down" stress. Each one sits in a full two-dimensional stress state that changes depending on the angle of the plane you choose to look at. Turn the plane and the normal stress rises while the shear falls; turn it further and the shear vanishes entirely. The material does not care about your coordinate axes — it fails on whichever plane is working hardest, and that plane is almost never the one you drew first.

Mohr's circle is the 140-year-old picture that lets an engineer see that state and read the worst plane straight off it. Plot normal stress on one axis and shear on the other, and every plane through a point collapses onto a single circle. The two places where it crosses the horizontal axis are the principal stresses σ₁ and σ₂ — the planes with zero shear — and the radius is the maximum shear the point will ever feel. One drawing answers the whole question of "which direction is dangerous."

This is meant to be the definitive walkthrough, and it goes the whole distance: from the σ_x/σ_y/τ_xy definition of a stress element, through a hand-drawn circle in five steps, all the way to a real FEM member fibre and the code check that decides whether it passes. Every worked number here was produced by the CalcSteel finite-element engine — the same solver that ships in the app — and cross-checked against closed-form algebra to three decimals. There is a live, draggable Mohr's circle embedded further down the page so you can reproduce each result yourself.

We wrote it for three readers at once. If you are a student, treat it as the stress-transformation chapter your course races through in a single lecture. If you are a practising engineer, jump to the FEM hand-off and the real steel building, where the circle stops being a textbook toy and starts painting the verification colours on a 3-D model. And if you are simply curious how software knows a member is safe, follow the story — it ends inside a working portal frame.

The CalcSteel Mohr's circle calculator is free, runs in your browser with no login for the math, and draws every circle in this article as you read it.

Mohr's circle is a graphical form of the two-dimensional stress-transformation equations: plot normal stress σ on the horizontal axis and shear stress τ on the vertical axis, and every plane through a point maps to exactly one point on a single circle. Where the circle crosses the σ-axis — where the shear is zero — are the principal stresses, and its radius is the maximum in-plane shear.

That is the whole idea in one paragraph. The state of stress at a point is not one number; it is three — σ_x, σ_y and τ_xy — and those three completely determine what happens on every other plane through that point. The transformation equations tell you the normal and shear stress on a plane rotated by any angle θ. Mohr's insight was that if you feed those equations onto a σ-versus-τ chart, the locus of all possible planes is not some messy curve — it is a perfect circle, centred on the σ-axis at the average of σ_x and σ_y.

Once you have the circle, you have the answers by inspection. The two horizontal-axis crossings are the principal stresses σ₁ (the largest normal stress at the point) and σ₂ (the smallest), and on those two planes the shear is exactly zero. The top and bottom of the circle give the maximum in-plane shear τ_max, equal to the radius. The centre never moves as you rotate, because the sum σ_x + σ_y is an invariant — a fact we will lean on hard as a self-check.

The single most important thing to carry into the next section is the angle relationship. Physically rotating the inspection plane by an angle θ sweeps you twice as far — a full 2θ — around Mohr's circle. Rotate the real plane by 45° and you travel 90° around the circle, which is exactly why the plane of maximum shear sits 45° from the plane of maximum normal stress. Master that one doubling and the rest of the picture reads itself.

Mohr's circle feels timeless, but almost every idea inside it has a date and a name. It took the better part of a century to get from the first rigorous definition of stress on a plane to the elegant circle engineers now sketch from memory — and only a few decades more before a solver could build that circle at every fibre of a structure automatically.

• 1822 — Augustin-Louis Cauchy puts stress on a firm mathematical footing, introducing the stress tensor and the traction (force per area) acting on an arbitrary plane through a point. Everything that follows is a way of visualising Cauchy's tensor.
• 1858 — William Rankine names and formalises principal stress — the planes on which shear vanishes and the normal stress is extreme — giving the two σ-axis crossings their meaning.
• 1866 — Karl Culmann, working in graphical statics, draws the first stress circle: a geometric construction for transforming stress, and the direct ancestor of the diagram we use today.
• 1882 — Otto Mohr generalises Culmann's construction into the complete circle that now carries his name, showing how to read every plane, both principal stresses and the maximum shear from one drawing — and arguing that it is the shear stress that governs the failure of ductile materials.
• ~1900 — August Föppl and others popularise the method in the great German engineering textbooks, and it becomes standard fare in strength-of-materials teaching worldwide.
• 1950s–60s — matrix structural analysis and the finite element method turn stress analysis into linear algebra a computer can execute, evaluating Cauchy's tensor at every point of a mesh without a single hand construction.

Today that whole lineage runs the instant you press a button. A modern solver computes the stress tensor at every fibre and builds the corresponding Mohr's circle in milliseconds — the same construction Mohr drew by hand in 1882, reproduced thousands of times over before you finish reading this sentence. When the CalcSteel calculator draws a circle, it is compressing a century of mechanics into a few milliseconds, and it is free.

Everything starts with a tiny square cut from the loaded body — the stress element. Imagine zooming in on a single point until the material around it looks like a little square (in three dimensions, a cube) aligned with your x and y axes. On its two vertical faces acts the normal stress σ_x; on its two horizontal faces, σ_y; and along its edges, the shear stress τ_xy. Those three numbers are the 2-D state of stress at that point. Choose any other orientation and the same physical state shows up as different σ and τ values — which is exactly what the circle organises.

Before drawing anything, fix a sign convention and never let go of it. Throughout this guide:

• Tension is positive, compression is negative for normal stress σ. A fibre being pulled apart reads to the right on the circle; a fibre being crushed reads to the left.
• Shear is plotted so a physical counter-clockwise rotation of the plane becomes a counter-clockwise sweep on the circle. Concretely, we plot the X face at the point (σ_x, −τ_xy) and the Y face at (σ_y, +τ_xy). Those two points are always diametrically opposite — they are the ends of a diameter — and the line joining them passes through the centre.

Which convention you choose matters far less than using it consistently from the first point to the last. Pick this one, keep it, and the angle you read off the circle will always turn the same way as the plane you are rotating in the real element.

The circle is just a picture of the stress-transformation equations. For a plane rotated by an angle θ from the x-axis, the normal and shear stresses are:

• σ_x'(θ) = (σ_x + σ_y)/2 + (σ_x − σ_y)/2 · cos 2θ + τ_xy · sin 2θ
• σ_y'(θ) = (σ_x + σ_y)/2 − (σ_x − σ_y)/2 · cos 2θ − τ_xy · sin 2θ
• τ_x'y'(θ) = −(σ_x − σ_y)/2 · sin 2θ + τ_xy · cos 2θ

Notice the 2θ everywhere: this is the algebra behind the "real angle θ, circle angle 2θ" doubling from the last section. The first equation is the horizontal coordinate of a point moving around a circle of centre (σ_x + σ_y)/2; the third is its vertical coordinate. That is all Mohr's circle is — those two equations, drawn.

Here is the canonical five-step recipe to build the circle from any (σ_x, σ_y, τ_xy):

1. Find the centre. It sits on the σ-axis at the average σ_avg = (σ_x + σ_y)/2 — no shear, because the centre is where the two normal-stress contributions balance.
2. Find the radius. R = √[ ((σ_x − σ_y)/2)² + τ_xy² ]. This is the maximum in-plane shear the point can feel.
3. Plot the two face points. X = (σ_x, −τ_xy) and Y = (σ_y, +τ_xy); the line through them is a diameter and confirms your centre and radius.
4. Read the principal stresses. σ₁ = σ_avg + R and σ₂ = σ_avg − R, the two crossings of the σ-axis, where τ = 0. The principal-plane angle is θ_p = ½·atan2(2τ_xy, σ_x − σ_y), measured to σ₁.
5. Read the maximum shear. τ_max = R, at the top and bottom of the circle, on a plane 45° from the principal plane (θ_s = θ_p − 45°).

Every worked example that follows uses exactly these five steps — and so does the calculator, just faster. If you want to see where the σ and τ that feed step 1 come from in a real member, the companion combined axial and bending and bending moment primers trace the internal forces back to the section.

Theory sticks the moment you draw one. Let's take the same general plane-stress state you'll meet in the live calculator a section from now and build its Mohr circle by hand, step by step. The inputs are the canonical case: σ_x = 80 MPa (tension), σ_y = 20 MPa (tension) and τ_xy = 30 MPa. This is exactly the calculator's default preset, so when you reach the embed you can reproduce every number below without typing a thing.

Step 1 — Plot the two face points

Using the plotting convention from the previous section, the X face goes at (σ_x, −τ_xy) = (80, −30) and the Y face at (σ_y, +τ_xy) = (20, +30). Those two points are diametrically opposite on the circle — draw the line between them and you already have the diameter.

Step 2 — Find the centre

The centre always sits on the σ-axis at the average normal stress:

• σ_avg = (σ_x + σ_y) / 2 = (80 + 20) / 2 = 50 MPa.

So the centre is C = (50, 0). Notice the shear coordinate of the centre is always zero — that's what makes the principal axis the σ-axis.

Step 3 — Find the radius

The radius is the distance from the centre to either face point. Horizontally that distance is (σ_x − σ_y)/2 = 30 MPa; vertically it is τ_xy = 30 MPa. So:

• R = √[((σ_x − σ_y)/2)² + τ_xy²] = √(30² + 30²) = √1800 = 42.43 MPa.

Step 4 — Read the principal stresses

The principal stresses are where the circle crosses the σ-axis — the centre plus or minus the radius:

• σ₁ = σ_avg + R = 50 + 42.43 = 92.43 MPa,
• σ₂ = σ_avg − R = 50 − 42.43 = 7.57 MPa.

The maximum in-plane shear is simply the radius, τ_max = R = 42.43 MPa, sitting at the very top and bottom of the circle.

Step 5 — Find the angle to the principal plane

The angle from the X face round to σ₁ — measured on the circle, so it's double the physical rotation — comes straight from the geometry:

• θ_p = ½ · atan2(2τ_xy, σ_x − σ_y) = ½ · atan2(60, 60) = ½ · 45° = 22.5°.

So rotating the physical element by 22.5° lands you on the principal plane where the shear vanishes and σ₁ = 92.43 MPa acts alone. A quick sanity check that never fails: σ₁ + σ₂ = 92.43 + 7.57 = 100 MPa = σ_x + σ_y — the sum of normal stresses is invariant at every rotation angle, so if it doesn't add up, a number is wrong.

Five steps, one circle, and every quantity a design check needs is now readable off the drawing. In the next section we'll walk the circle itself and see exactly where each of these values lives — and in the section after that you can rebuild this whole picture live in the CalcSteel Mohr's circle calculator.

Once the circle is drawn, it becomes a dashboard — every quantity a stress check needs has a fixed home on it. Keep the worked state in mind (σ_x = 80, σ_y = 20, τ_xy = 30 MPa, centre 50, radius 42.43) and let's tour the geometry.

The two σ-axis crossings are the principal stresses

Where the circle cuts the horizontal axis, the shear coordinate is zero — no shear on that plane — so those crossings are by definition the principal stresses. The right-hand crossing is σ₁ = 92.43 MPa, the left-hand one is σ₂ = 7.57 MPa. Physically, σ₁ is the largest tension any plane through the point feels, and it acts on a face rotated θ_p = 22.5° from the original X face.

The top and bottom are the maximum in-plane shear

The highest and lowest points of the circle sit a full radius above and below the axis, so the maximum in-plane shear equals the radius, τ_max = R = 42.43 MPa. Because a diameter subtends 180° on the circle and 180° on the circle is 90° in the real element, the top of the circle is 90° of arc from the principal crossing — a physical 45°. That is the rule worth burning in: the plane of maximum shear is always 45° from the principal plane, here at θ_s = θ_p − 45° = −22.5°. On that plane the shear is at its peak and the normal stress is exactly σ_avg = 50 MPa.

The invariant that checks your work

Spin the element to any angle and the two normal-stress readings still sum to the same number: σ_x' + σ_y' = σ_x + σ_y = σ₁ + σ₂ = 100 MPa. Geometrically that's obvious — the two points stay diametrically opposite, so their σ-coordinates always average to the centre. It's the fastest self-check there is.

The out-of-plane circle you must not forget

Plane stress isn't truly two-dimensional — the third principal stress is σ₃ = 0, out of the plane. Draw the circle between σ₁ = 92.43 and that σ₃ = 0 and its radius is larger than the in-plane one. The absolute maximum shear is therefore τ_abs,max = σ₁/2 = 46.21 MPa, not the 42.43 MPa in-plane value. This bigger number — not the in-plane τ_max — is what a Tresca check actually uses, and it's the single most common thing engineers miss when they treat a plane-stress problem as if the third circle didn't exist.

So from one drawing you read the worst tension (σ₁), the worst compression (σ₂), the worst in-plane shear (R), the worst shear of all (τ_abs,max), and the angles to each. That's the whole point of the circle — and next you'll do it yourself with a draggable version.

You've built the circle by hand and toured every landmark on it. The fastest way to make it permanent is to grab it and move it — so here it is, live and embedded right on this page.

Load the worked state — σ_x = 80, σ_y = 20, τ_xy = 30 MPa (it's the default, so it may already be there) — and confirm the numbers you just derived: centre 50, radius 42.43, σ₁ = 92.43, σ₂ = 7.57, θ_p = 22.5°. Then start dragging. Push τ_xy up and watch the circle swell, σ₁ and σ₂ spread apart, and θ_p rotate toward 45°; pull it to zero and the circle collapses onto the axis until σ_x and σ_y are the principal stresses. Rotating the plane sweeps twice as far around the circle — you can watch the 2θ relationship happen in real time.

It's the real tool, not a teaser: unlimited runs, completely free, and no login required for the math. When you're ready to go further you can load a real solved FEM section, switch from plane stress to a full 3-D triaxial state with all three circles, and share a permalink to the exact state you're looking at. If it opens in its own tab, here's the direct link to the CalcSteel Mohr's circle calculator.

Play with the textbook state here, because in the sections that follow the stresses stop coming from tidy inputs and start coming from a real member — a beam-column and then a full steel building — where the finite-element engine hands the circle its σ and τ automatically. If you want to see where those internal forces come from first, the companion guide on shear force and bending moment diagrams is the perfect warm-up.

Once you have drawn a few circles by hand, a handful of stress states start to look like old friends. Their circles have such distinctive shapes that you can read the answer before you finish the arithmetic. Memorise these three and you will sanity-check almost any real result in seconds.

Uniaxial tension (or compression)

Pull a bar straight along its axis and only one normal stress survives: σ_x = σ, σ_y = 0, τ_xy = 0. The circle is centred at σ/2 with radius σ/2, so it passes cleanly through the origin. The two σ-axis crossings are the principal stresses: σ₁ = σ and σ₂ = 0. The top of the circle sits at τ_max = σ/2, on a plane 45° to the bar axis — which is exactly why a ductile tensile coupon necks and shears along 45° slip planes, and why a brittle chalk stick in tension breaks square. This is not just a textbook toy: the extreme bending fibre of a real member is a uniaxial state, and you will meet it again as a genuine FEM result in the next section.

Pure shear

Now load a bolt group or a beam web so there is only shear: σ_x = σ_y = 0, τ_xy = τ. The centre lands at the origin and the radius is τ, so the circle is centred on zero and symmetric. The principal stresses come out equal and opposite: σ₁ = +τ and σ₂ = −τ. In other words, a state of pure shear is also a state of equal tension and compression rotated 45° away. That 45° is the reason a shaft in torsion made of brittle material fractures along a helical 45° surface rather than straight across — the tension diagonal, not the shear itself, is what tears it.

Equal biaxial and hydrostatic states

Finally, load a point equally in both directions: σ_x = σ_y = σ, τ_xy = 0. The radius R = √[((σ − σ)/2)² + 0²] is exactly zero, so the circle collapses to a single point on the σ-axis. A point-circle carries a blunt message: there is no shear stress on any plane whatsoever, and every direction is a principal direction. A thin spherical pressure-vessel wall approaches this equal-biaxial state; a truly hydrostatic (equal in all three directions) state is what a fluid feels, and it produces zero distortion — the fact that underlies the whole von Mises criterion we reach shortly.

These three silhouettes — a circle through the origin, a circle straddling the origin, and a circle shrunk to a dot — are the vocabulary of everything that follows. Load any of them into the free Mohr's circle calculator and watch the shape snap into place before you trust a single number.

Textbook circles start with σ_x, σ_y and τ_xy already handed to you. Real design never does. A finite-element solver gives you internal forces at a section — an axial force N, a shear V and a bending moment M — and it is up to you (or the engine) to turn those into the stresses that build the circle. The bridge is elementary mechanics of materials:

• Normal stress from axial plus bending: σ = N/A ± M/S_x, largest at the extreme fibre furthest from the neutral axis.
• Shear stress from the transverse force: τ ≈ V/A_w, largest at the neutral axis where bending stress vanishes.

The sign twist is the whole point: the two stresses peak in different places. So a single cross-section does not have one Mohr's circle — it has a different circle at every fibre, and the two that matter sit at opposite ends of the web.

The member: an IPE 330 beam-column

Take a 3 m cantilever in IPE 330 (area A = 62.6 cm², elastic modulus S_x = 685.9 cm³, web shear area A_w = 24.75 cm²), grade S355. Load the free tip with an axial F_x = −150 kN (compression) and a transverse F_y = 20 kN. Running it through the shipping CalcSteel FEM engine (the same run_analysis that drives the 3-D editor), the fixed base returns N = −150 kN, V = 20 kN, M = 60 kN·m. Now inspect two fibres of that one section.

Fibre 1 — the extreme fibre (bending governs, τ = 0)

At the outermost fibre the bending stress is maximal and the transverse shear is zero. Stacking axial and bending: σ_x = N/A − M/S_x = (−150/62.6 − 6000/685.9) × 10 = −111.4 MPa, with τ = 0. That is a uniaxial state, so the circle is the special case from the previous section — it passes through the origin: σ₁ = 0, σ₂ = −111.4 MPa, τ_max = 55.7 MPa, principal angle θ_p = 90°, and von Mises σ_vM = 111.4 MPa (utilisation η = 0.35 to NBR 8800). Pure compression, degenerate circle, one number to check.

Fibre 2 — the neutral axis (a genuine 2-D state)

Slide to the neutral axis and the story flips: the bending stress is zero there, but the axial stress still acts and the transverse shear is now at its maximum. So σ_x = N/A = −23.96 MPa and τ_xy = V/A_w = 8.08 MPa act together — a genuine combined state. The circle is centred at −11.98 MPa with radius 14.45 MPa, giving σ₁ = +2.47 MPa, σ₂ = −26.43 MPa, τ_max = 14.45 MPa and θ_p = 73.0°. Notice something a uniaxial view would have missed entirely: even under net compression this fibre carries a small tension, σ₁ = +2.47 MPa, because the shear tilts the principal directions.

The punchline: the same IPE 330 section produced two completely different Mohr circles — a degenerate uniaxial one at the extreme fibre and a real combined one at the neutral axis. A correct verification looks at both, because you cannot know in advance which fibre governs the member. This is exactly the combined axial-plus-bending interaction unpacked in combined axial & bending, and it flows directly from the internal forces you read off a shear and bending moment diagram. Load a solved section into the Mohr's circle calculator and you can flip between fibres yourself.

The Mohr's circle tells you the worst stresses and the planes they act on — but on its own it does not tell you whether the steel yields. For that you feed its principal stresses into a yield criterion, and the two that matter for structural steel both read straight off the circle.

Tresca — the maximum-shear criterion

Tresca is the honest, conservative one: yielding begins when the maximum shear stress reaches the shear yield of the material. On the circle that maximum shear is simply the radius of the largest of the three principal circles, so the criterion is τ_max = (σ₁ − σ₃)/2, i.e. the failure measure is σ₁ − σ₃. It is geometrically literal — you are reading the top of the biggest circle — which is exactly why it errs on the safe side.

von Mises — the distortion-energy criterion

Steel design almost always uses von Mises, because it matches ductile-metal test data more closely. For a plane-stress state it collapses to a clean formula in the two in-plane principal stresses:

σ_vM = √(σ₁² − σ₁σ₂ + σ₂²)

It captures the idea that distortion, not pure volume change, drives yielding — which is why a hydrostatic point-circle (all principal stresses equal) has von Mises zero no matter how large the pressure. One caution that trips people up: von Mises is a three-dimensional criterion. In genuine plane stress the third principal stress is σ₃ = 0, and it still belongs in the picture — it is the out-of-plane σ₃ = 0 circle that sets the true τ_abs,max, and dropping it silently is one of the most common Mohr's-circle mistakes.

The code check

Design turns that stress into a single utilisation ratio η, comparing the demand against the factored yield strength f_yd:

• NBR 8800: f_yd = f_y/γ_a1, with γ_a1 = 1.10.
• AISC 360 (LRFD): f_yd = φ·f_y, with φ = 0.90.

Put the canonical worked state through it — σ_x = 80, σ_y = 20, τ_xy = 30 MPa, giving σ₁ = 92.43 MPa and σ₂ = 7.57 MPa. Its von Mises stress is σ_vM = √(92.43² − 92.43·7.57 + 7.57²) = 88.88 MPa. For grade S355, the NBR 8800 check is η = 88.88 / (355/1.10) = 88.88 / 322.7 = 0.28. The member is at 28% of yield — comfortably safe. Notice how far apart the numbers are: σ₁ = 92.43 MPa but σ_vM = 88.88 MPa. Principal stress is not von Mises — the criterion blends σ₁ and σ₂, and only the von Mises value belongs in the code check.

This is the exact number the CalcSteel engine computes at every fibre, and in the next section you will watch that same von Mises value paint the verification colours on a real 3-D steel building. You can reproduce the 88.88 MPa yourself in the free Mohr's circle calculator — it reports both σ₁/σ₂ and the von Mises utilisation side by side.

The beam-column of the last section was one member. Now scale up to a building — because the whole reason a solver builds a Mohr's circle at every fibre is so it can tell you which fibre, in which member, of a whole frame, is closest to yielding. That is the number that decides whether the steel is safe.

Take a real portal frame, IPE 330 throughout: a 12 m span, 6 m eave columns, carrying gravity w = 12 kN/m on the rafter plus a 15 kN lateral wind push, on fixed bases. This is a statically indeterminate sway frame — the rigid knees and built-in feet give more restraints than equilibrium alone can resolve, so there is no closed-form textbook formula. Only a finite-element solution gets the answer, and the CalcSteel engine returns clean base reactions as a first sanity check: vertical ΣF_y = 144 kN (exactly w·L = 12 × 12) and horizontal ΣF_x = −15 kN (balancing the wind precisely).

The governing fibre

The engine scans every member and finds the worst-loaded fibre at the rafter mid-span: an axial N = −15.3 kN combined with a bending moment M = 184.7 kN·m. Fed through σ = N/A − M/S_x on the IPE 330 (A = 62.6 cm², S_x = 685.9 cm³), that extreme fibre reaches:

• σ_x = −271.7 MPa — a nearly uniaxial compression, so the Mohr circle passes essentially through the origin.
• von Mises σ_vM = 271.7 MPa — for uniaxial stress the von Mises value equals the stress magnitude itself.
• Utilization η = 271.7 / (355 / 1.10) = 0.842 to NBR 8800 (γ_a1 = 1.10), or η = 271.7 / (0.90 × 355) = 0.850 to AISC 360 (φ = 0.90).

The member passes, but only just — at roughly 84% utilised it is an amber, well-worked section, exactly the sort you want in an efficient design: no wasted steel, real margin still in hand.

Why frame action matters

Here is the story the circle tells. If that same 12 m rafter were a simply-supported beam under w = 12 kN/m, its mid-span moment would be the textbook wL²/8 = 12 × 12² / 8 = 216 kN·m. Bent into a rigid frame, the stiff knees pull moment out of the span and share it around the corners, so the governing rafter moment drops to 184.7 kN·m. That redistribution — the reason portal frames are so efficient, and the reason they need a solver — is unpacked fibre by fibre in our shear & bending moment pillar.

The punchline that ties this whole guide together: the engine turns that single governing fibre into a Mohr's circle, reads its principal stresses, computes σ_vM = 271.7 MPa, and that exact von Mises value is what paints the verification colours on the 3-D model in the CalcSteel editor — green for comfortable, amber for well-utilised, red for overstressed. The circle you learned to draw by hand is the same one colouring every bar of a real steel building.

The construction is clean, but the same handful of slips catch students and practising engineers alike. Run through this checklist before you trust any circle you draw — or read off a solver.

• Confusing the angle: it's 2θ on the circle, not θ. Rotating the physical plane by an angle θ sweeps twice that angle around the circle. A 45° rotation of the element is a 180° trip across the diagram — which is exactly why the maximum-shear plane and the principal plane sit 45° apart in space but 90° apart on the circle.
• Getting the shear-plotting sign wrong. To make a physical counter-clockwise rotation read as a counter-clockwise sweep on the circle, plot the X face at (σ_x, −τ_xy) and the Y face at (σ_y, +τ_xy). Flip that sign and your angles come out with the wrong handedness even though σ₁ and σ₂ still look right.
• Putting τ_max on the principal plane. The maximum in-plane shear acts on a plane 45° from the principal plane (θ_s = θ_p − 45°), at the top and bottom of the circle — never where σ₁ and σ₂ live. On the principal planes the shear is exactly zero.
• Forgetting the out-of-plane σ₃ = 0 circle. For a plane-stress state the true absolute maximum shear is often set by the third circle through σ₃ = 0, not the in-plane one. In our worked state τ_max in-plane is 42.43 MPa, but τ_abs,max = σ₁/2 = 46.21 MPa — and that larger value is what a Tresca check actually uses.
• Treating principal stress as the failure number. σ₁ is the largest normal stress, but ductile steel yields on distortion energy: the check runs on von Mises σ_vM = √(σ₁² − σ₁σ₂ + σ₂²), not σ₁ alone. In our example σ₁ = 92.43 MPa but σ_vM = 88.88 MPa — they are simply different quantities.
• Expecting shear somewhere on a point-circle. An equal-biaxial (hydrostatic in-plane) state collapses the circle to a single point of radius zero. Every plane then carries the same normal stress and zero shear — there is no orientation that produces any shear at all.

Is Mohr's circle still used today?

Yes — constantly, though rarely with pencil and compass. The construction is the visual grammar behind every stress check: a solver computes the stress tensor at each fibre and reduces it exactly the way the circle does, to principal stresses and a von Mises value. Engineers still reach for the circle to reason about combined states, to sanity-check software, and to teach the geometry of stress. Draw one yourself in the free Mohr's circle calculator and the 140-year-old picture appears in real time.

What are principal stresses, physically?

They are the largest and smallest normal stresses at a point, and they act on the two planes where the shear stress vanishes completely. Rotate your imaginary cut until nothing is trying to slide the faces past each other — pure push or pull remains — and those pure-normal values are σ₁ and σ₂. On the circle they are simply the two points where it crosses the σ-axis. They matter because cracks tend to open perpendicular to σ₁, and yield criteria are built from them.

Why is the angle doubled on the circle?

Because the stress-transformation equations for σ and τ are functions of 2θ — they contain cos 2θ and sin 2θ, not cos θ and sin θ. Mohr's circle is just the geometric picture of those equations, so a physical rotation of θ necessarily maps to an arc of 2θ. It isn't a convention you could choose differently; it falls straight out of the algebra of rotating a symmetric tensor.

Is there a Mohr's circle for strain?

Yes. The same construction works with normal strain ε on the horizontal axis and half the shear strain (γ/2) on the vertical. It gives principal strains and the maximum shear strain in exactly the same way, which is precisely how a strain-gauge rosette is reduced to principal values. The geometry is identical — only the labels on the axes change.

You have travelled the whole road — from a little square element with σ_x, σ_y and τ_xy on its faces, to the governing fibre of a real steel frame and the code check that keeps it standing. Here is what to carry away.

• A point feels a 2-D state, not a single stress. Every plane through it maps to one point on a circle centred at σ_avg with radius R = τ_max. Reading the circle tells you the worst plane at a glance.
• The circle hands you the design numbers directly. The two σ-axis crossings are the principal stresses (σ₁ = 92.43, σ₂ = 7.57 MPa in our worked state); the top of the circle is the in-plane τ_max = 42.43 MPa; and the sum σ₁ + σ₂ = σ_x + σ_y is invariant at every rotation.
• Angles double, and the shear peak is offset. Rotating the plane by θ sweeps 2θ on the circle, and the maximum-shear plane sits 45° from the principal plane — never on it. Don't forget the out-of-plane σ₃ = 0 circle when you need τ_abs,max.
• One section can give two different circles. In a beam-column the extreme fibre gives a degenerate uniaxial circle (σ₂ = −111.4 MPa, von Mises 111.4) while the neutral axis gives a genuine combined circle (σ₁ = 2.47, σ₂ = −26.43 MPa). A correct check looks at both.
• The circle ends in a failure check. Its principal stresses feed von Mises σ_vM = √(σ₁² − σ₁σ₂ + σ₂²), and η = σ_vM / f_yd decides the member — 0.842 to NBR 8800, 0.850 to AISC 360 for the real portal frame's governing fibre.

Now stop reading and start drawing. Load your own σ_x, σ_y and τ_xy into the free Mohr's circle calculator — unlimited, no login for the math — drag the shear and watch σ₁, σ₂ and θ_p respond, then load a real solved FEM section and go triaxial. When you are ready for the whole member, the CalcSteel solution runs the same real FEM engine on complete beam-columns and portal frames right in your browser, on a genuinely free plan, colouring every bar by the exact von Mises value the circle produces. Students get everything unlocked through CalcSteel Education, free for verified accounts. The proof isn't a countdown — it's the tool itself, in your hands today.