Mohr's Circle and Principal Stress Calculator

Principal stresses, the principal angle and BOTH maximum shears — in-plane and absolute — from a plane stress state or a full three-dimensional one, with the third principal stress carried explicitly because in plane stress it is zero rather than absent.

Last updated 24 Sep 2026

Mohr's circle and principal stress

Stress components → principal stresses and both shears

Stress statePlane stress is the right model for a thin plate, a shaft surface, a sheet-metal part or anything where one face is free. It does NOT mean the third principal stress is absent — it means it is zero, which is a number and it matters. Choose the 3D state for a pressure vessel wall, a shrink fit, a contact patch or anywhere material is constrained on every side.

σ_x (N/mm²)Normal stress on the face whose outward normal is the x axis. Tension positive, compression negative — the sign convention is not optional here, because the whole construction depends on it.

σ_y (N/mm²)Normal stress on the y face. If you are analysing a shaft under bending and torsion, this is usually zero and σ_x is the bending stress.

σ_z (N/mm²)Normal stress on the z face. Locked to zero in plane stress, which is the definition of plane stress. For a thin-walled pressure vessel it is the radial stress and is small; for a contact patch or a constrained fit it is large and compressive and it changes the answer completely.

τ_xy (N/mm²)The in-plane shear. For a shaft in torsion this is T·r/J and it is usually the whole of the shear. Positive when it acts in the +y direction on the +x face.

τ_yz (N/mm²)Out-of-plane shear, locked to zero in plane stress.

τ_zx (N/mm²)The other out-of-plane shear, locked to zero in plane stress.

Plane angle θ to report the state on (°)Rotate the element by this much and the page prints the normal and shear stress on the rotated face. Useful for checking a weld line, a grain direction, a bonded joint or a gauge orientation — all of which care about the stress on a plane that is not the one you happened to set up axes on.

Not a circuit: Mohr's circles for your own stress state, drawn in NORMALISED coordinates — σ₃ at the left end of the axis and σ₁ at the right, whatever their values. That is why the big circle never changes shape: normalising makes the σ₁–σ₃ circle the same circle for every stress state there is, so what you are watching is everything else moving against it. The tick on the axis is σ₂, and the two smaller circles it sits between are the σ₁–σ₂ and σ₂–σ₃ circles. The crosshairs mark the point (σ_x, τ_xy) — your element as you set it up. Three things to read. Every admissible combination of normal and shear stress on every plane through the point lies in the region between the big circle and the two small ones, so the big circle's radius is the ABSOLUTE maximum shear and it is marked. In plane stress, σ₃ = 0 sits at one end of the axis whenever both in-plane principals share a sign — and then the in-plane circle is one of the SMALL ones, not the big one, which is the error the page exists to stop. And when σ₂ reaches either end of the axis, one small circle has collapsed to a point and you have a repeated root.

159.083N/mm²Example

A plane stress state: σ_x = 150, σ_y = 60, τ_xy = 30 N/mm²

The circle, the principals, and the third one

σ₁,₂ = (σ_x+σ_y)/2 ± √(((σ_x−σ_y)/2)² + τ_xy²) · tan 2θ_p = 2τ_xy/(σ_x−σ_y) · τ_in-plane = R · τ_absolute = (σ₁ − σ₃)/2

(σ_x+σ_y)/2: the centre of Mohr’s circle on the normal-stress axis. It is I₁/2 in plane stress and it does not move when you rotate the axes
R: the radius. It is the in-plane maximum shear stress, and the two principal stresses are the centre plus and minus it
θ_p: the principal angle. The factor of 2 is the whole geometry of the construction: rotating the element by θ moves you 2θ round the circle, which is why the maximum-shear planes are 45° from the principal ones and not 90°
σ₃: THE THIRD PRINCIPAL STRESS. In plane stress it is zero. Zero is a value, not an absence, and if σ₁ and σ₂ have the same sign then zero is outside them and the governing shear is on a plane the two-dimensional circle never draws
τ_absolute: (σ₁ − σ₃)/2 using all three principal stresses sorted by value. This is the number a shear yield criterion needs. It equals R only when σ₁ and σ₂ straddle zero

Worked example

A plane stress state: σ_x = 150, σ_y = 60, τ_xy = 30 N/mm²
The centre of the circle is the average normal stress: (150 + 60)/2 = 105 N/mm². This does not move however you rotate the axes, which is the first invariant at work
The radius is the hypotenuse of the half-difference and the shear: R = √(((150−60)/2)² + 30²) = √(45² + 30²) = √(2,025 + 900) = √2,925 = 54.0833 N/mm²
So the two in-plane principal stresses are the centre plus and minus the radius: σ_A = 105 + 54.083 = 159.083 and σ_B = 105 − 54.083 = 50.917 N/mm². Check them against the invariants: they sum to 210, which is σ_x + σ_y, and their product is 8,100.0, which is σ_xσ_y − τ_xy² = 9,000 − 900
The principal angle: tan 2θ_p = 2 × 30/(150 − 60) = 60/90, so 2θ_p = 33.690° and θ_p = 16.845°. Rotate the element by that and the shear on its faces vanishes. The maximum-shear planes are 45° further round, at 61.845°
NOW THE STEP EVERYBODY SKIPS. Both in-plane principals are POSITIVE — 159.1 and 50.9. The third principal stress is σ₃ = 0, because this is plane stress, and zero is smaller than both of them. So sorted by value the three principal stresses are 159.08, 50.92 and 0
The in-plane maximum shear is R = 54.083 N/mm². The ABSOLUTE maximum shear is (σ₁ − σ₃)/2 = (159.083 − 0)/2 = 79.542 N/mm². That is 47 per cent higher, and it acts on a plane tilted out of the page. Read the two-dimensional circle alone and you have underestimated the governing shear by nearly half
When does this happen? Exactly when the two in-plane principals have the SAME sign, so that zero falls outside them rather than between them. If σ_A and σ_B straddle zero, the in-plane circle is the biggest of the three and the flat answer is right. Equal biaxial tension is the extreme case: the in-plane circle shrinks to a point, the in-plane maximum shear is zero, and the true maximum shear is half of σ₁
What to do with these three numbers: the von Mises equivalent stress is √(½[(σ₁−σ₂)² + (σ₂−σ₃)² + (σ₃−σ₁)²]) = 140.71 N/mm² and the Tresca equivalent is σ₁ − σ₃ = 159.08. Tresca is the more conservative of the two by 13.1 per cent here, and never by more than 15.47 per cent. The failure criteria page compares them against a material's yield and asks the question that has to come first — whether the material is ductile or brittle

Six plane-stress states, and which shear governs each one

Read the last two columns together. Whenever the two in-plane principal stresses have OPPOSITE signs, zero lies between them, σ₃ = 0 is the middle principal stress, and the in-plane circle is the largest of the three — the two-dimensional answer is the right answer. Whenever they have the SAME sign, zero lies outside them, it becomes σ₃ (or σ₁), and the governing circle is one that the two-dimensional construction never draws. The fifth row is the case that catches people: two modest tensile principals, and the real maximum shear is 47 per cent higher than the in-plane figure. Pure biaxial tension is the extreme of it — the in-plane circle collapses to a POINT and the in-plane maximum shear is zero, which would suggest a state that can never yield in shear, while the true maximum shear is half of σ₁.
State	σ_x	σ_y	τ_xy	σ_A	σ_B	In-plane τ_max	ABSOLUTE τ_max	Governing plane	Ratio
Pure tension	200	0	0	200.0	0.0	100.00	100.00	in-plane	1.000×
Pure shear (a shaft in torsion)	0	0	100	100.0	-100.0	100.00	100.00	in-plane	1.000×
Equal biaxial tension (a pressure vessel end)	150	150	0	150.0	150.0	0.00	75.00	OUT of plane	—
Bending + torsion on a shaft	150	0	60	171.0	-21.0	96.05	96.05	in-plane	1.000×
Both principals POSITIVE — the trap	150	60	30	159.1	50.9	54.08	79.54	OUT of plane	1.471×
Tension one way, compression the other	120	-80	0	120.0	-80.0	100.00	100.00	in-plane	1.000×

The same stress state on six different sets of axes

Every row is the SAME physical stress state, written on axes rotated by a different amount. The first three columns change completely. The next two do not change at all, and the last two do not change at all. That is the point of the whole construction: the individual stress components are an accident of where you drew your axes, and the invariants and the principal stresses are properties of the state. It is also why a failure criterion is always written in principal stresses or in invariants and never in σ_x and τ_xy — a criterion written in components would give a different answer depending on how you set up the problem, which would make it not a criterion. Notice the row at 16.845°: the shear column has reached zero, and the two normal stresses have become the two principal stresses. That rotation is θ_p.
Axes rotated by	σ_x′	σ_y′	τ_x′y′	σ_x′+σ_y′ (I₁)	σ_x′σ_y′−τ² (I₂)	σ₁	σ₂
0°	150.000	60.000	30.000	210.000	8,100.00	159.083	50.917
15°	158.971	51.029	3.481	210.000	8,100.00	159.083	50.917
30°	153.481	56.519	-23.971	210.000	8,100.00	159.083	50.917
45°	135.000	75.000	-45.000	210.000	8,100.00	159.083	50.917
60°	108.481	101.519	-53.971	210.000	8,100.00	159.083	50.917
90°	60.000	150.000	-30.000	210.000	8,100.00	159.083	50.917

How the cubic is solved, and why it does not need iteration

This is Viète’s trigonometric solution of the cubic, and it is used here for a specific reason: this site’s expression engine has no loops, so anything iterative has to be unrolled a fixed number of times and proved to converge. A closed form needs neither. The solution was checked against a completely different algorithm — a cyclic Jacobi eigenvalue rotation of the same 3×3 matrix — over 400 random stress states, agreeing to better than 10⁻⁸ N/mm². One honest caveat: the arccosine is ill-conditioned at its endpoints, which is exactly where a REPEATED root puts it, so at a state like equal biaxial tension the roots lose about half their significant figures — three parts in 10⁸, which is invisible at any displayed precision and is stated here rather than hidden.
Step	What it is	Why it works
I₁, I₂, I₃	the three stress invariants, built from the six components	the characteristic equation of the stress tensor is σ³ − I₁σ² + I₂σ − I₃ = 0, and its roots are the principal stresses
substitute σ = s + I₁/3	the depressed cubic s³ + Ps + Q = 0	shifting by the mean removes the quadratic term, which is what makes a closed form possible at all
r = 2√(−P/3)	the amplitude of the trigonometric substitution s = r cosθ	P is never positive for a symmetric tensor, because the three roots are always real — so the square root is always defined and no complex arithmetic is needed
cos 3θ = −4Q/r³	one arccosine, and nothing else	substituting s = r cosθ and using cos3θ = 4cos³θ − 3cosθ makes the linear term vanish identically, leaving a single equation in 3θ
σ_k = I₁/3 + r cos(θ + 2πk/3), k = 0,1,2	the three roots, all of them, in closed form	the three branches of the arccosine give the three roots directly. No iteration, no convergence proof, no loop — which is why this page can do a full three-dimensional state rather than refusing it

The circle, the invariants, and why the third principal stress is not optional

The construction IS the page. Mohr’s circle is not a mnemonic or a graphical shortcut — it is the exact locus of (σ_n, τ) as you rotate the cutting plane, and it is a circle because both stress transformation equations are pure sinusoids in 2θ. That factor of two is the only thing about it that has to be memorised, and it is what makes the maximum-shear planes lie at 45° to the principal ones rather than at 90°. Read the chart on this page and you are reading the circle unrolled: where the normal-stress curve peaks, the shear curve crosses zero, which is what a principal plane is.

The third principal stress is zero in plane stress, and zero is a value. This is the one thing on this page that is worth being insistent about, because it is where a correct two-dimensional calculation gives a wrong answer. Plane stress does not mean there are two principal stresses; it means the third one is zero. Sort all three by value and the absolute maximum shear is (σ₁ − σ₃)/2. If the two in-plane principals straddle zero, then zero is the middle one, the in-plane circle is the largest of the three, and the flat answer is right. If they have the SAME sign, zero is an extreme, and the governing shear lives on a plane tilted out of the page that the two-dimensional construction never draws. On this page’s defaults — two modest tensile principals — the in-plane maximum shear is 54.1 N/mm² and the true maximum is 79.5, an error of 47 per cent. Equal biaxial tension is the pathological case: the in-plane circle shrinks to a point and the in-plane maximum shear is exactly zero.

A full three-dimensional state, without iteration. The three principal stresses are the roots of σ³ − I₁σ² + I₂σ − I₃ = 0, and this page solves that cubic in closed form using Viète’s trigonometric method: shift out the quadratic term, substitute s = r cos θ, and the linear term vanishes identically, leaving one arccosine whose three branches are the three roots. Because the stress tensor is symmetric, all three roots are real and the square root inside r is always defined, so no complex arithmetic is needed and no convergence has to be proved. The result was checked against a cyclic Jacobi eigenvalue solve of the same matrix over 400 random stress states. That is why this page takes a full 3D state rather than refusing one.

The invariants are why a failure criterion is written in principal stresses. Rotate your axes and σ_x, σ_y and τ_xy change completely — they are an accident of where you happened to draw the axes. What does not change is I₁, I₂, I₃ and the three principal stresses. A criterion written in stress components would give different answers depending on how you set the problem up, which would make it not a criterion; a criterion written in principal stresses or invariants cannot. The table on this page prints the same state on six different axis systems so you can watch the components move and the invariants sit still.

What to do with the principal stresses once you have them. Nothing on this page is a strength check — it is a transformation, and it knows nothing about the material. The von Mises and failure criteria page takes these three numbers and a material’s yield and ultimate strengths and asks the question that has to come first: is the material ductile or brittle? For a ductile metal, von Mises or Tresca. For a brittle one, maximum normal stress or Coulomb–Mohr, because a brittle material is much weaker in tension than in compression and no shear-based criterion sees that. If the stresses cycle, the fatigue and S-N page is the next stop, and the numbers it wants are the alternating and mean components rather than the peak.

Frequently asked questions

Why is the absolute maximum shear sometimes bigger than the circle’s radius?

Because the circle you drew is only one of three. In plane stress the principal stresses are σ_A, σ_B and zero, and there are three Mohr circles — one for each pair. The in-plane circle is the σ_A–σ_B one. If σ_A and σ_B have the same sign, zero is outside them, so the σ_A–0 circle is bigger, and its radius — half of the larger principal stress — is the true maximum shear. It acts on a plane at 45° to the σ_A direction and to the free surface, tilted out of the plane you drew.

Is plane stress the same as plane strain?

No, and they give different answers. Plane stress sets σ_z = 0: the third principal stress is zero, which is right for a thin plate or a free surface. Plane STRAIN sets ε_z = 0: the material cannot contract in z, so a σ_z of ν(σ_x + σ_y) develops, which is right for a long thick member, a deep weld or the middle of a thick plate. That third stress changes the maximum shear and therefore the yield prediction. This page does plane stress and the full 3D case; for plane strain, compute σ_z = ν(σ_x + σ_y) yourself and enter it in the 3D mode.

Why does rotating the element by θ move me 2θ round the circle?

Because the stress transformation equations contain cos 2θ and sin 2θ, not cos θ and sin θ. That in turn is because a stress state on a plane repeats every 180° of rotation, not every 360° — the plane at 190° is the same plane as the one at 10°. So half a turn of the element is a full turn of the circle. The practical consequence is the useful one: the maximum-shear planes are 90° apart ON THE CIRCLE and therefore 45° apart in the material, which is why ductile shafts shear on a 45-degree surface.

Can I use this for second moments of area?

Yes, and it is the same circle. I_x, I_y and I_xy transform exactly as σ_x, σ_y and −τ_xy do, so the principal second moments and the principal axis rotation come from the identical construction. That is precisely what the section properties page does for an angle section, whose product of inertia is not zero and whose weakest axis is therefore rotated off its legs. Watch the sign on the shear term, which is the only difference.

What is the principal angle actually telling me?

The direction, measured from the x axis, of the plane on which there is no shear stress and the normal stress is at its maximum. Physically it is the direction a brittle crack will run perpendicular to, the direction a strain gauge should point to read the largest strain, and the direction a fibre should lie along in a composite. There is a 90-degree ambiguity in the arctangent that the two-argument form used here resolves, so the angle this page gives belongs to σ₁ and not to σ₂.

Does a hydrostatic stress state really produce no shear at all?

None, on any plane through the point. All three Mohr circles collapse to a single point on the normal-stress axis, and the von Mises equivalent stress is exactly zero. That is not a numerical quirk: the distortion-energy criterion predicts that a ductile metal under uniform pressure will never yield, however high the pressure, and experiments at tens of thousands of atmospheres agree. It does not mean the state is harmless. Hydrostatic TENSION drives brittle fracture, cavitation and void growth, none of which is yield and none of which any criterion on these pages covers.

My shaft is under bending and torsion. What do I enter?

σ_x = the bending stress M·c/I, σ_y = 0, τ_xy = the torsional shear T·r/J, and leave plane stress selected — a shaft surface is a free surface, so σ_z is genuinely zero. The two in-plane principals will have opposite signs for any real combination of bending and torsion, so the in-plane circle governs and the two-dimensional answer is the right one. If the shaft also has a fillet, a groove or a keyway, apply the stress concentration factor to the nominal stresses BEFORE entering them — the shaft fillet page computes it.

Related calculators

References

R. C. Hibbeler, Mechanics of Materials, on the superposition of axial and bending stress, on Mohr’s circle and the stress-transformation equations, and on the absolute maximum shear stress in plane stress — the point that the third principal stress is zero rather than absent, so the governing shear may be σ₁/2 and not (σ₁ − σ₂)/2.
Shigley, chapter 5, for the static failure criteria: the distortion-energy (von Mises) and maximum-shear-stress (Tresca) theories for ductile materials, the maximum-normal-stress theory, and the brittle Coulomb–Mohr and modified-Mohr theories with their quadrant conditions. The modified-Mohr form printed here was checked for continuity at both of its corners — at σ_B = −σ_A it must reduce to n = S_ut/σ_A and at σ_A = 0 to n = S_uc/|σ_B| — which is the only internal test a transcription of a piecewise criterion can be given, and it passes both.
ISO 6892-1, Metallic materials — Tensile testing — Part 1: Method of test at room temperature, and ASTM E8/E8M. Cited by number for what a yield strength and an ultimate tensile strength actually are: a 0.2 per cent proof stress and a maximum engineering stress from a standard specimen. Every criterion on these pages is only as good as the two numbers you feed it, and both are defined by these documents rather than by any calculator.