von Mises and Failure Criteria Calculator

von Mises and Tresca side by side for a ductile material, maximum normal stress and Coulomb–Mohr for a brittle one — with the 0.577 shear yield derived, and the material class asked first because it decides which criterion is applicable at all.

Last updated 24 Sep 2026

von Mises and failure criteria

Principal stresses and material class → factor of safety

Material class — ASK THIS FIRSTvon Mises and Tresca are for DUCTILE materials: they are shear-based, they are indifferent to hydrostatic stress, and they treat tension and compression alike. A brittle material is none of those things — grey cast iron is three to four times stronger in compression than in tension — so a von Mises check on it is not conservative, it is the wrong check. This dropdown decides which criterion the headline reports.

Principal stress σ₁ (N/mm²)A principal stress, not a component. The Mohr’s circle page turns σ_x, σ_y and τ_xy into these three. The page sorts whatever you enter, so the order does not matter.

Principal stress σ₂ (N/mm²)The second principal stress. For plane stress one of the three is exactly zero — and it is a value, not an absence: the intermediate principal stress is what separates von Mises from Tresca, because Tresca ignores it entirely.

Principal stress σ₃ (N/mm²)The third. Leave it at zero for plane stress. Make it strongly negative for a contact patch or a constrained fit, and watch the von Mises stress FALL even as the individual stresses rise — that is hydrostatic pressure doing no damage, and it is the whole reason metal forming works.

Yield strength S_y (N/mm²)The 0.2 per cent proof stress from a tensile test. A36 structural steel is 250 minimum; 4140 quenched and tempered is around 655; 304 stainless annealed is 215; 6061-T6 is 276 typical and 240 minimum. A brittle material has no yield point at all, which is why this field is ignored for the brittle classes.

Ultimate TENSILE strength S_ut (N/mm²)The maximum engineering stress in a tensile test. For a ductile metal it is used for the factor against fracture rather than yield; for a brittle one it IS the strength, because there is nothing between first yield and fracture.

Ultimate COMPRESSIVE strength S_uc (N/mm²)Enter it as a positive magnitude. For a ductile metal it is about equal to the tensile value and is ignored. For grey cast iron it is three to four times higher — class 30 iron is roughly 214 tensile and 750 compressive — and that single asymmetry is the entire reason the Coulomb–Mohr criterion exists.

Not a circuit: the two ductile yield loci in principal stress space, drawn in units of the yield strength so that ONE picture serves every material. The hexagon is Tresca and the ellipse is von Mises, and the crosshairs are your own stress point — inside both envelopes is safe, outside either is yielding by that criterion. Read three things off it. The two loci TOUCH at six points, which are uniaxial tension and compression in each axis and the two equal-biaxial corners: at those states the criteria agree exactly, which is why a tensile test cannot distinguish them and why nobody argues about one. They are furthest apart along the marked pure-shear diagonal, where the hexagon crosses at 0.500 S_y and the ellipse at 0.577 — the two small ticks — a gap of 15.47 per cent and the largest it ever gets. And the hexagon is always INSIDE the ellipse, which is the geometric statement that Tresca is never less conservative than von Mises. Note that the figure is drawn for σ₃ = 0; a third principal stress moves both loci bodily along the hydrostatic diagonal without changing either shape, because neither criterion cares about hydrostatic stress at all.

1.766Example

A ductile steel at S_y = 250 N/mm², in a plane stress state with principal stresses 160, 51 and 0 N/mm²

Two ductile criteria, three brittle ones, and one question first

von Mises: σ′ = √(½[(σ₁−σ₂)² + (σ₂−σ₃)² + (σ₃−σ₁)²]) · Tresca: σ₁ − σ₃ · Coulomb–Mohr: σ₁/S_ut − σ₃/S_uc = 1/n

σ′: the von Mises or distortion-energy equivalent stress. It is √(3J₂), where J₂ is the second invariant of the DEVIATORIC stress — so it is blind to hydrostatic stress by construction, which is the right physics for a ductile metal
σ₁ − σ₃: the Tresca equivalent: twice the absolute maximum shear. It ignores the intermediate principal stress entirely, which is the only difference between the two criteria
0.577: 1/√3. The shear yield strength a ductile metal actually has, as a fraction of its tensile yield — derived below, not asserted. Tresca says 0.500, and the 15.47 per cent between them is the largest the two criteria ever differ
S_uc: the ultimate COMPRESSIVE strength, as a positive number. For a ductile metal it equals S_ut and drops out. For grey cast iron it is three to four times larger, and that asymmetry is what the brittle criteria exist to represent
n: factor of safety. On STRESS here. A factor on LOAD is the same number only when the stress is proportional to the load, which excludes contact pressure, a preloaded joint, anything geometrically nonlinear and anything past yield

Worked example

A ductile steel at S_y = 250 N/mm², in a plane stress state with principal stresses 160, 51 and 0 N/mm²
ASK THE MATERIAL QUESTION FIRST, because it decides which criterion is even applicable. This is a ductile steel, so von Mises and Tresca apply and the brittle criteria do not. Had it been grey cast iron, both would have been the wrong check and optimistic rather than conservative
von Mises: σ′ = √(½[(160−51)² + (51−0)² + (0−160)²]) = √(½[11,881 + 2,601 + 25,600]) = √20,041 = 141.57 N/mm²
Tresca is simply the spread: σ₁ − σ₃ = 160 − 0 = 160 N/mm². Note that it never looked at σ₂ = 51 at all — that is the entire difference between the two criteria
So the two factors of safety against yield are 250/141.57 = 1.766 on von Mises and 250/160 = 1.5625 on Tresca. Tresca is 13.0 per cent more conservative here, and it is never more than 15.47 per cent more conservative anywhere
WHERE THE 0.577 COMES FROM, which is the most useful thing on this page. Put the material in pure shear: the principal stresses are +τ, 0 and −τ. von Mises gives σ′ = √(½[τ² + τ² + 4τ²]) = √(3τ²) = √3·τ. Setting that equal to S_y gives shear yield at τ = S_y/√3 = 0.5774·S_y = 144.3 N/mm² for this steel
Tresca on the same state gives σ₁ − σ₃ = 2τ, so shear yield at 0.500·S_y = 125 N/mm². The ratio is 2/√3 = 1.1547, and that is exactly where the 15.47 per cent comes from. It is a maximum, not a rule of thumb: holding σ₁ − σ₃ fixed, the von Mises value is smallest when σ₂ sits exactly half way between the other two, which is pure shear plus a hydrostatic term. Everywhere else the two criteria are closer, and in uniaxial or equal-biaxial tension they are identical
So a shear yield is NOT half a tensile yield, and the difference matters. A shaft, a key, a pin, a weld throat and a bolt in shear are all sized on shear yield; using half instead of 0.577 throws away 15 per cent of the material's capacity. Published practice usually uses 0.577 (it is in most machine design texts and in most pressure vessel codes) precisely because von Mises fits experimental data for ductile metals better than Tresca does
AND FINALLY: a factor of safety on STRESS is not a factor on LOAD unless the stress is proportional to the load. It is here, so 1.766 times the load is where this part yields. It is NOT for a Hertz contact pressure, which goes as the cube root of load — a factor of 1.77 on stress there is a factor of 5.5 on load. It is not for a preloaded bolt, where part of the stress does not scale with the external load at all. And it is not for anything past yield, where a small load increase produces a large deformation and the stress barely moves

Eight stress states, both criteria, and a 250 N/mm² yield

Three rows carry the whole argument. The first and fourth — uniaxial and equal biaxial tension — have a ratio of exactly 1.0000: the criteria agree, which is why a tensile test cannot distinguish them. The pure shear row has a ratio of 1.1547, which is 2/√3 and is the largest the ratio can ever be, at the only state where it reaches it. And the last two rows are the ones that look wrong and are not. Hydrostatic tension at 250 N/mm² in all three directions has a von Mises stress of EXACTLY ZERO and an infinite factor of safety against yield, because there is no distortion in it at all. The contact patch has principal stresses up to 1,400 N/mm² and a von Mises stress of only 917 — which is why a ball bearing survives contact pressures four times its material’s yield strength, and why the bearing life page rates contact by a fatigue law rather than a yield criterion. Hydrostatic pressure does no damage in a ductile metal, and both criteria on this page know that. This is a first-pass calculation on an idealised part — uniform section, static load, no stress raisers beyond those stated, and room-temperature material properties. Real parts have fillets, keyways, surface finish and duty cycles that a closed-form answer cannot see.
State	σ₁	σ₂	σ₃	von Mises σ′	Tresca	Tresca ÷ von Mises	Hydrostatic	n on von Mises
Uniaxial tension	250.0	0.0	0.0	250.00	250.00	1.0000×	83.3	1.000
Pure shear (shaft in torsion)	144.3	0.0	-144.3	250.00	288.68	1.1547×	0.0	1.000
… the same, at Tresca’s limit	125.0	0.0	-125.0	216.51	250.00	1.1547×	0.0	1.155
Equal biaxial tension	250.0	250.0	0.0	250.00	250.00	1.0000×	166.7	1.000
Biaxial 2:1	250.0	125.0	0.0	216.51	250.00	1.1547×	125.0	1.155
Triaxial: pressure vessel wall	200.0	100.0	-20.0	190.79	220.00	1.1531×	93.3	1.310
Hydrostatic tension	250.0	250.0	250.0	0.00	0.00	—	250.0	∞
Contact patch: triaxial compression	-400.0	-600.0	-1,400.0	916.52	1,000.00	1.0911×	-800.0	0.273

Grey cast iron, class 30: 214 tensile and 750 compressive

Read the last two columns. For the compression-dominated rows, applying von Mises to a brittle material gives a factor of safety far HIGHER than the brittle criteria do — which sounds backwards until you remember why. von Mises is blind to hydrostatic stress and treats tension and compression alike, so it cannot see that grey iron is three and a half times stronger one way than the other. Applied to a brittle material it is optimistic in exactly the states where the material is weak. The three brittle criteria differ from each other only in the mixed quadrant, where one principal stress is tensile and the other compressive: maximum normal stress ignores the compression entirely, Coulomb–Mohr draws a straight line between the two strengths, and modified Mohr follows the tensile limit until the compression exceeds the tension in magnitude and only then starts to bend. Published brittle data sits closest to modified Mohr, Coulomb–Mohr is the conservative choice, and both reduce to the same thing in pure tension and pure compression. This page sizes a part; it does not certify one. Where the answer carries a consequence — a load path, a lifting duty, a pressure boundary, a fastener holding something that can fall — confirm it against the design code that governs the application, and against the manufacturer’s own rating, before relying on it.
State	σ₁	σ₃	Max normal stress	Coulomb–Mohr	Modified Mohr	von Mises (wrongly applied)
Pure tension	200	0	1.070	1.070	1.070	1.250	von Mises is UNSAFE here
Tension with a little compression	200	-100	1.070	0.936	1.070	0.945
Pure shear	150	-150	1.427	1.110	1.427	0.962
Compression with a little tension	100	-400	0.535	0.999	1.153	0.546
Pure compression	0	-700	0.306	1.071	1.071	0.357

What a static criterion does not cover, and where each one lives

This is the honest boundary of the page. Every criterion here answers exactly one question — will this material yield or fracture under this stress state, right now, at room temperature, with no crack in it and no cyclic loading. That question is worth answering and it is not the only one. The first row is the one that matters most in machinery, because most machine parts see cyclic load and most machine failures are fatigue failures; a part that passes every criterion on this page with a factor of two is not thereby safe for ten million cycles. This page sizes a part; it does not certify one. Where the answer carries a consequence — a load path, a lifting duty, a pressure boundary, a fastener holding something that can fall — confirm it against the design code that governs the application, and against the manufacturer’s own rating, before relying on it.
Failure mode	Why no static criterion sees it	Where it belongs
FATIGUE	the stress never reaches the static limit at all. A part cycling at a third of its yield strength will fail after enough cycles, and no criterion evaluated on the peak stress predicts that	the fatigue and S-N page, which needs the alternating and mean stresses rather than the peak
BUCKLING	it is a stability problem, not a strength one. A slender strut fails at a load that depends on E, the length and the radius of gyration, and not on the yield strength at all	a structural code. The radius of gyration it needs is on the section properties page
CREEP	time is not in any of these equations. Above roughly 0.3 to 0.4 of the absolute melting temperature a metal deforms steadily under a constant stress well below yield	a creep-rupture curve for the specific alloy, at the specific temperature, for the specific design life
FRACTURE FROM AN EXISTING CRACK	a criterion evaluated on a nominal stress cannot see a crack. A cracked part fails when K = K_Ic, which depends on the crack LENGTH, and the nominal stress may be a small fraction of yield	linear elastic fracture mechanics. Not on this site
BRITTLE FRACTURE FROM TRIAXIALITY	a ductile metal in a strongly triaxial tensile state can fracture without yielding, because the hydrostatic part suppresses plastic flow while driving void growth. von Mises reports a low equivalent stress and is wrong	the triaxiality figure on this page is the warning flag; the assessment is a fracture-mechanics problem
STRESS CORROSION AND HYDROGEN EMBRITTLEMENT	the environment is not a stress. A high-strength steel in the wrong environment cracks at a stress far below any static limit, and higher strength makes it worse rather than better	the material and environment pairing. A published threshold stress intensity, not a yield criterion

Ductile or brittle first, then the 0.577, then what a static criterion cannot see

Ductile or brittle is the first question and it is not a formality. von Mises and Tresca are both shear-based criteria. Both are completely indifferent to hydrostatic stress, and both treat tension and compression as equivalent. For a ductile metal that is the right physics — yielding is dislocation motion, dislocations move under shear, and uniform pressure does not move them, which is why a steel can sit at tens of thousands of atmospheres without yielding. For a brittle material it is the wrong physics entirely. Grey cast iron is three to four times stronger in compression than in tension because its failure is crack propagation from graphite flakes, and a criterion that cannot tell tension from compression cannot see that. Applied to grey iron, von Mises is not conservative — in a compression-dominated state it is optimistic, which is the wrong direction to be wrong in.

Where the 0.577 comes from, and why a shear yield is not half a tensile yield. Put a material in pure shear: the principal stresses are +τ, 0 and −τ. Substitute into the von Mises expression and you get σ′ = √3·τ, so yield at τ = S_y/√3 = 0.5774·S_y. Tresca on the same state gives σ₁ − σ₃ = 2τ, so yield at 0.500·S_y. The ratio is 2/√3 = 1.1547 — and that 15.47 per cent is not a rule of thumb, it is a MAXIMUM. Hold σ₁ − σ₃ fixed and vary σ₂: the von Mises value is √(½[(1−s)² + s² + 1]) with s the normalised intermediate stress, which is minimised at s = ½ — the midpoint, which is pure shear plus a hydrostatic term. Anywhere else the criteria are closer, and in uniaxial and equal-biaxial tension they agree exactly. The chart on this page traces both loci so you can watch the gap open and close.

The brittle family, and what distinguishes its three members. MAXIMUM NORMAL STRESS says failure occurs when the largest principal stress reaches the tensile strength (or the smallest reaches the compressive one). It is the simplest and it is right for a brittle material with equal strengths both ways. BRITTLE COULOMB–MOHR draws a straight line in the mixed quadrant between S_ut on the tension axis and S_uc on the compression axis: σ₁/S_ut − σ₃/S_uc = 1/n. MODIFIED MOHR keeps the tensile limit until the compressive principal exceeds the tensile one in magnitude, and only then starts to bend towards the compressive limit. Published brittle data sits closest to modified Mohr; Coulomb–Mohr is the conservative choice; all three agree in pure tension and pure compression, and they differ only in the mixed quadrant. This page computes all three whatever you select, so you can see the spread.

A factor on stress is not a factor on load unless the stress is linear in the load. This page’s factor of safety is on stress: it is the strength divided by the equivalent stress. When stress is proportional to load — which covers most direct tension, bending and torsion — that same number is the load multiplier at which the part yields, and the distinction does not arise. It arises in three common cases. A Hertz CONTACT pressure goes as the cube root of load, so a factor of 1.5 on stress is a factor of 3.4 on load — which is why a bearing can be loaded far past its apparent stress margin. A PRELOADED joint has a stress component that does not scale with the external load at all, so doubling the external load does not double the bolt stress; the bolted joint stiffness page has that arithmetic. And PAST YIELD nothing is linear: a small load increase produces a large deformation at almost constant stress, so a factor on stress becomes meaningless and the honest measure is a factor on load or on displacement. This page prints the load multiplier for the linear case and for a square and a cube-root relationship beside it.

What a static criterion does not cover, said plainly. FATIGUE: a part cycling at a third of its yield strength will fail, and no criterion evaluated on a peak stress predicts that. That is the fatigue and S-N page, and it needs the alternating and mean stresses rather than the peak. BUCKLING: a stability problem in which the yield strength does not appear at all. CREEP: time is not in any of these equations, and above roughly a third of the absolute melting temperature a metal deforms steadily under a stress well below yield. FRACTURE FROM AN EXISTING CRACK: a cracked part fails when K reaches K_Ic, which depends on the crack length, and the nominal stress can be a small fraction of yield. And BRITTLE FRACTURE OF A DUCTILE METAL under high triaxial tension, where the hydrostatic part suppresses plastic flow while driving void growth — von Mises reports a comfortable equivalent stress and is wrong. The triaxiality figure on this page is the warning flag for that last one.

Frequently asked questions

Which should I use, von Mises or Tresca?

von Mises for a ductile metal, unless a code you are working to specifies otherwise. It fits experimental yield data for ductile metals better, it is what most machine design texts and most pressure vessel codes use, and it is smooth rather than faceted, which makes it better behaved numerically. Tresca is always the more conservative of the two and never by more than 15.47 per cent, so choosing it is a defensible conservatism rather than an error — it simply costs you up to a sixth of the material. Where the two matter most is shear and torsion; in tension or biaxial tension they agree exactly.

Why is the shear yield strength 0.577 of the tensile yield and not half?

Because yielding is driven by distortion energy rather than by maximum shear, and the two criteria give different answers for pure shear. Put ±τ and zero into the von Mises expression and it returns √3·τ, so yield arrives at τ = S_y/√3 = 0.5774·S_y. Tresca gives 2τ and therefore 0.500·S_y. Experiment on ductile metals sits closer to 0.577, which is why it is the value in general use. The practical consequence is that sizing a shaft, a key or a weld throat on half the tensile yield throws away fifteen per cent of the capacity.

Can I use von Mises on grey cast iron?

No, and the error is in the unsafe direction for the states cast iron is usually in. von Mises is blind to hydrostatic stress and symmetric in tension and compression; grey iron is three to four times stronger in compression than in tension. Use maximum normal stress if you only have one strength figure, and Coulomb–Mohr or modified Mohr if you have both. This page computes all of them side by side and the table shows how far apart they get.

How can the von Mises stress be zero when the principal stresses are 250 each?

Because that state is purely hydrostatic and there is no distortion in it at all — the material is being uniformly compressed or expanded, not sheared on any plane. The distortion-energy criterion therefore predicts no yield at any magnitude, and experiment agrees to remarkable pressures. It is correct and it is not the whole story: hydrostatic TENSION is the most dangerous state for brittle fracture and void growth, which no criterion on this page covers, and the triaxiality figure is the flag for it.

What factor of safety should I use?

That is a judgement about uncertainty rather than a calculation, and it depends on how well you know the load, the material and the geometry. General machine practice against yield under a well-known static load runs 1.5 to 2; 2 to 3 where the load is less certain or the material is a casting; 3 to 4 for a brittle material because there is no yielding to redistribute a local peak; and more again where a failure has consequences. Two reminders: a factor of safety does not cover a stress raiser you forgot to apply, and it does not cover fatigue at all.

Should the factor be against yield or against ultimate?

Both, and the governing one is whichever gives the lower answer — which for a ductile metal is nearly always yield, since the ultimate is 1.5 to 2 times the yield. Yield is the right limit when permanent deformation is a failure, which for a machine part holding a tolerance it usually is. Ultimate is the right limit when only separation matters, and it is the limit brittle criteria use because a brittle material has essentially nothing between yield and fracture. This page prints both for the ductile case.

Is a factor of safety of 2 on stress the same as being able to double the load?

Only when the stress is proportional to the load. For direct tension, bending and torsion it is, and the distinction never comes up. It fails in three familiar places. A Hertz contact pressure goes as the cube root of load, so a factor of 2 on stress is a factor of 8 on load. A preloaded bolt has a stress component that does not scale with the external load at all, so doubling the external load changes the bolt stress by much less than double. And once anything has yielded, load and stress stop tracking each other entirely. This page prints the load multiplier for the linear case and for two nonlinear ones beside it.

Related calculators

References

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.
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.
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.
ASTM A36/A36M, Standard Specification for Carbon Structural Steel. Cited by number, not reproduced. The two figures this page uses from it — a minimum yield of 250 MPa (36 ksi) for plate, bar and shapes under 200 mm and a tensile range of 400–550 MPa — are the specification’s headline values and are quoted here from the public summary of the standard rather than from the document.
Aluminium 6061-T6 is taken at a minimum yield of 240 MPa with typical values near 270, a minimum tensile of 290 MPa with typical values near 310, and E = 68–69 GPa. Stainless 304 and 316 in the annealed condition are taken at 215 and 205 MPa yield. Grey cast iron is carried with NO yield strength at all, deliberately: it is a brittle material with no yield point, its tensile strength is a fraction of its compressive strength, and a von Mises check on it is the wrong check. That is the whole reason the failure-criteria page asks for the material class before it asks for anything else.