Fatih Küçükkarakurt — Writing

What a Mohr circle actually tells you

fatih@fkkarakurt.com (Fatih Küçükkarakurt) — Tue, 11 Aug 2026 00:00:00 GMT

The Mohr circle is a drawing that replaces a calculation. Everything it shows can be obtained algebraically, so it earns its place only if it makes relationships visible that the algebra hides — which it does, provided you remember that every angle on it is doubled.

Every number in this article is recomputed from its inputs by a script that runs on each build of this site, and the diagram’s coordinates are checked against the same arithmetic.

The circlePermalink to “The circle”

Take a two-dimensional stress state: normal stresses $\sigma_x$ and $\sigma_y$ on two perpendicular planes, with shear stress $\tau_{xy}$ on both. Rotating the reference planes by an angle $\theta$ traces out a circle in the $(\sigma, \tau)$ plane with

\text{centre} = \left(\frac{\sigma_x + \sigma_y}{2},\ 0\right), \qquad R = \sqrt{\left(\frac{\sigma_x - \sigma_y}{2}\right)^{2} + \tau_{xy}^{2}}

The principal stresses are where the circle crosses the $\sigma$ axis, because that is where the shear stress vanishes:

\sigma_1 = \text{centre} + R, \qquad \sigma_3 = \text{centre} - R

and the radius is therefore the maximum shear stress in the plane, $R = (\sigma_1 - \sigma_3)/2$ . Take $\sigma_x = 260$ kPa, $\sigma_y = 140$ kPa, $\tau_{xy} = 60$ kPa:

\text{centre} = 200\ \text{kPa}, \qquad R = \sqrt{60^{2} + 60^{2}} = 60\sqrt{2} = 84.85\ \text{kPa}

giving $\sigma_1 = 284.85$ kPa and $\sigma_3 = 115.15$ kPa.

The doubled anglePermalink to “The doubled angle”

This is the fact that causes the errors. The angle subtended at the centre of the circle is twice the angle through which the physical plane has rotated. The principal planes are found from

\tan 2\theta_p = \frac{2\tau_{xy}}{\sigma_x - \sigma_y}

For the numbers above, $2\theta_p = \arctan(120/120) = 45^\circ$ , so the major principal plane is at $\theta_p = 22.5^\circ$ — not 45°.

The reason is structural, not a convention that could have been chosen differently. The stress transformation equations contain $\cos 2\theta$ and $\sin 2\theta$ , so a rotation of the plane by $\theta$ moves the stress point by $2\theta$ around the circle. Two planes 90° apart in the ground sit on opposite sides of the circle, 180° apart — which is exactly why they have the same shear stress magnitude and why the circle closes after a physical rotation of only 180°.

If an angle taken off a Mohr circle looks twice what it should be, it is.

Adding a failure envelopePermalink to “Adding a failure envelope”

Now give the soil a strength: a straight envelope

\tau = c + \sigma \tan\varphi

A stress state is safe while its circle stays clear of the line, and at failure the circle touches it. Tangency is a distance condition, and that is all the Mohr-Coulomb criterion is.

Write the envelope as $\sigma\tan\varphi - \tau + c = 0$ . The perpendicular distance from the circle centre $(s, 0)$ , where $s = (\sigma_1 + \sigma_3)/2$ , to that line is

d = \frac{\left|s\tan\varphi + c\right|}{\sqrt{\tan^{2}\varphi + 1}} = \left(s\tan\varphi + c\right)\cos\varphi = s\sin\varphi + c\cos\varphi

Setting $d = R = (\sigma_1 - \sigma_3)/2$ gives

\frac{\sigma_1 - \sigma_3}{2} = \frac{\sigma_1 + \sigma_3}{2}\sin\varphi + c\cos\varphi

Collecting terms in $\sigma_1$ ,

\sigma_1\left(1 - \sin\varphi\right) = \sigma_3\left(1 + \sin\varphi\right) + 2c\cos\varphi

and using $\cos\varphi = \sqrt{(1-\sin\varphi)(1+\sin\varphi)}$ , so that $\cos\varphi/(1-\sin\varphi) = \sqrt{K_p}$ , this becomes the form everyone quotes:

\sigma_1 = K_p\,\sigma_3 + 2c\sqrt{K_p}, \qquad K_p = \frac{1 + \sin\varphi}{1 - \sin\varphi}

The triaxial formula and the tangency condition are the same statement. Nothing was assumed about how the soil fails beyond “the circle touches the line”.

A worked failure statePermalink to “A worked failure state”

Take $c = 20$ kPa, $\varphi = 30^\circ$ , $\sigma_3 = 100$ kPa. Then $K_p = 3$ exactly, and

\sigma_1 = 3 \times 100 + 2 \times 20 \times \sqrt{3} = 369.28\ \text{kPa}

s = \frac{369.28 + 100}{2} = 234.64\ \text{kPa}, \qquad R = \frac{369.28 - 100}{2} = 134.64\ \text{kPa}

Check the tangency independently, from the distance formula:

d = s\sin\varphi + c\cos\varphi = 234.64 \times 0.5 + 20 \times 0.8660 = 134.64\ \text{kPa}

which equals $R$ . The circle touches the envelope, as it must.

The failure circle for c = 20 kPa, φ = 30°, σ₃ = 100 kPa. Every coordinate in this drawing is computed from those three inputs.

Where the soil failsPermalink to “Where the soil fails”

The radius to the tangent point is perpendicular to the envelope. The envelope makes an angle $\varphi$ with the $\sigma$ axis, so that radius makes an angle $90^\circ + \varphi$ with it. That angle is measured on the circle, so it is twice the physical one:

2\theta = 90^\circ + \varphi \qquad\Longrightarrow\qquad \theta = 45^\circ + \frac{\varphi}{2}

The failure plane sits at $45^\circ + \varphi/2$ from the major principal plane. For $\varphi = 30^\circ$ that is 60°, and the tangent point is at

\sigma = s + R\cos\left(90^\circ + \varphi\right) = 167.32\ \text{kPa}, \qquad \tau = R\sin\left(90^\circ + \varphi\right) = 116.60\ \text{kPa}

Substituting back into the envelope: $20 + 167.32 \times \tan 30^\circ = 116.60$ kPa. The point lies on the line, which it must, and computing it two ways is the cheapest check available.

Note what this angle is not. It is not 45°, which is the answer only for $\varphi = 0$ — the undrained case. And it is not measured from the horizontal unless the major principal plane happens to be horizontal.

Total and effective stressPermalink to “Total and effective stress”

Pore pressure $u$ acts equally in all directions, so it subtracts equally from $\sigma_1$ and $\sigma_3$ :

\sigma'_1 = \sigma_1 - u, \qquad \sigma'_3 = \sigma_3 - u

The difference $\sigma_1 - \sigma_3$ is unchanged. The effective stress circle is the same size as the total stress circle, shifted left by $u$ . Nothing about its diameter changes; only where it sits.

Two consequences follow immediately from that picture. Strength is governed by the effective stress circle, so the total stress circle can sit anywhere and tell you nothing on its own. And an undrained test on a saturated soil, where the effective stresses are fixed by the initial state, produces total stress circles of constant diameter — which is drawn as a horizontal envelope, $\tau = s_u$ , with $\varphi_u = 0$ . That apparent zero friction angle is a consequence of the drainage condition, not a property of the soil.

What the circle does not tell youPermalink to “What the circle does not tell you”

It is a two-dimensional construction, and $\sigma_2$ does not appear anywhere in it. The Mohr-Coulomb criterion inherits that blindness: strength is made to depend only on $\sigma_1$ and $\sigma_3$ .

That is a real modelling decision with a visible consequence. Because the criterion ignores the intermediate principal stress, its surface in principal stress space is an irregular hexagonal pyramid rather than a cone — six flat faces, one for each ordering of the three principal stresses, meeting at edges where the gradient is undefined. Implementing the criterion means dealing with those edges, and it is why the strength measured in a triaxial compression test does not transfer unchanged to a plane strain problem. Whether that matters depends on the problem; whether it is true does not.

ReferencesPermalink to “References”

Mohr, O. (1900). Welche Umstände bedingen die Elastizitätsgrenze und den Bruch eines Materials? Zeitschrift des Vereines Deutscher Ingenieure, 44, 1524–1530 and 1572–1577. The origin of the graphical construction and of the failure hypothesis that bears his name.
Coulomb, C.A. (1776). Essai sur une application des règles de maximis et minimis à quelques problèmes de statique relatifs à l’architecture. Mémoires de Mathématique et de Physique, Académie Royale des Sciences, 7, 343–382. The origin of the linear strength relation $\tau = c + \sigma\tan\varphi$ .
Terzaghi, K. (1943). Theoretical Soil Mechanics. John Wiley & Sons, New York. The effective stress principle applied throughout soil mechanics.

The derivations above are carried out in full rather than cited, so that they can be checked rather than trusted. The circle geometry, the tangency condition and the failure plane angle follow from the two definitions at the top of the article and nothing else.

Classification decides which strength parameters are even applicable, and the arithmetic there has its own trap — see soil classification from a grading curve. The implementation of the criterion derived above, edges and all, is what KATAI 2D has to get right.

Soil classification from a grading curve

fatih@fkkarakurt.com (Fatih Küçükkarakurt) — Tue, 04 Aug 2026 00:00:00 GMT

Classification is the first thing done to a soil sample and the last thing anyone checks. It is also entirely mechanical: given five numbers from the laboratory, the group symbol follows from a table with no judgement required. This article works one sample all the way through, showing every step of the arithmetic.

Every figure below is recomputed from the input data by a script that runs on each build of this site. If a number here is wrong, the build fails.

What the classification needsPermalink to “What the classification needs”

The Unified Soil Classification System, ASTM D2487, sorts soil first by how much of it is fine. Two sieves do that work:

Sieve	Aperture	Separates
No. 4	4.75 mm	gravel from sand
No. 200	0.075 mm	sand from fines (silt and clay)

A soil with more than 50% retained on the No. 200 sieve is coarse-grained; 50% or more passing it is fine-grained. Within a coarse-grained soil, more than 50% of the coarse fraction retained on the No. 4 sieve makes it a gravel, and 50% or more passing makes it a sand.

Note the phrase coarse fraction. The No. 4 split is applied to the material coarser than the No. 200 sieve, not to the whole sample. Applying it to the whole sample is a common and silent error, and it changes the answer whenever the fines content is appreciable.

The samplePermalink to “The sample”

A sieve analysis, expressed as percentage passing:

Sieve aperture (mm)	19.0	9.5	4.75	2.36	1.18	0.600	0.300	0.150	0.075
Passing (%)	100	88	74	60	47	35	22	12	6

Take the fractions straight off the table:

\text{fines} = 6\%, \qquad \text{sand} = 74 - 6 = 68\%, \qquad \text{gravel} = 100 - 74 = 26\%

Fines are 6%, well under 50%, so the soil is coarse-grained. The coarse fraction is $100 - 6 = 94\%$ of the sample, and the part of it passing the No. 4 sieve is

\frac{68}{94} \times 100 = 72.3\%

which is 50% or more, so this is a sand.

D10, D30 and D60 — the logarithmic stepPermalink to “D10, D30 and D60 — the logarithmic step”

The three characteristic sizes are defined by the percentage passing: $D_{60}$ is the aperture through which 60% of the sample passes, and so on. Only $D_{60}$ falls exactly on a sieve here — the 2.36 mm sieve passes exactly 60%, so

D_{60} = 2.360\ \text{mm}

$D_{30}$ and $D_{10}$ have to be interpolated, and the interpolation is logarithmic, because a grading curve is plotted against $\log(\text{aperture})$ . Between two sieves with apertures $a_1 > a_2$ passing $p_1 > p_2$ , the size at a target percentage $p$ is

D = \exp\!\left[\ln a_2 + \frac{p - p_2}{p_1 - p_2}\left(\ln a_1 - \ln a_2\right)\right]

For $D_{30}$ , between the 0.600 mm sieve (35%) and the 0.300 mm sieve (22%):

\frac{30 - 22}{35 - 22} = \frac{8}{13} = 0.6154, \qquad D_{30} = 0.300 \times \left(\frac{0.600}{0.300}\right)^{0.6154} = 0.300 \times 2^{0.6154} = 0.4596\ \text{mm}

For $D_{10}$ , between 0.150 mm (12%) and 0.075 mm (6%):

\frac{10 - 6}{12 - 6} = \frac{2}{3}, \qquad D_{10} = 0.075 \times 2^{2/3} = 0.1191\ \text{mm}

What linear interpolation costsPermalink to “What linear interpolation costs”

Interpolating linearly on aperture instead — reading the curve as though the axis were linear — gives $D_{30} = 0.4846$ mm and $D_{10} = 0.1250$ mm. Carried through:

	$D_{10}$ (mm)	$D_{30}$ (mm)	$C_u$	$C_c$
Logarithmic (correct)	0.1191	0.4596	19.82	0.75
Linear (wrong)	0.1250	0.4846	18.88	0.80

The error in $C_c$ is 6%. Here it changes nothing, because both values are far from the threshold. Near a threshold it decides the group symbol, and nothing in the output would tell you which interpolation produced it.

Uniformity and curvaturePermalink to “Uniformity and curvature”

Two ratios summarise the shape of the grading curve. ERDC/CRREL TR-15-4 gives them as

C_u = \frac{D_{60}}{D_{10}}, \qquad C_c = \frac{\left(D_{30}\right)^2}{D_{60} \times D_{10}}

For this sample:

C_u = \frac{2.360}{0.1191} = 19.82, \qquad C_c = \frac{0.4596^2}{2.360 \times 0.1191} = 0.75

The criteria for a clean sand are $C_u \ge 6$ and $1 \le C_c \le 3$ . This sample passes the first comfortably and fails the second.

So it is poorly graded, SP, despite $C_u \approx 20$ . That is the point of having two ratios. A large $C_u$ says only that the sample spans a wide range of sizes; it says nothing about whether the range is filled. $C_c$ below 1 means the intermediate sizes are under-represented — a gap-graded soil, which spans a wide range and still packs badly. One number cannot express both, and quoting $C_u$ alone is how gap-graded material gets described as well graded.

Fines between 5% and 12%: a dual symbolPermalink to “Fines between 5% and 12%: a dual symbol”

The fines content decides how much further the classification has to go:

Below 5% — a clean sand or gravel. The group symbol is settled by $C_u$ and $C_c$ .
Above 12% — the fines govern, and the symbol becomes SM, SC, GM or GC depending on where the fines plot on the plasticity chart.
Between 5% and 12% — both matter, and ASTM requires a dual symbol: SW-SM, SW-SC, SP-SM or SP-SC for sands, and the equivalent GW-GM, GW-GC, GP-GM, GP-GC for gravels.

This sample has 6% fines, so it lands in the middle case. It is SP from the grading, and the second half of the symbol depends on whether those fines are silty or clayey.

The plasticity chartPermalink to “The plasticity chart”

Fines are classified by Atterberg limits, plotted as plasticity index against liquid limit. Two straight lines govern the chart:

\text{A-line:}\quad PI = 0.73\left(LL - 20\right) \qquad\qquad \text{U-line:}\quad PI = 0.9\left(LL - 8\right)

The plasticity chart. Line positions are computed from the two equations; the plotted point is the worked sample.

The A-line is the one that classifies. Above it a fine-grained soil is a clay, below it a silt. The U-line classifies nothing — it is an upper bound on what natural soils have been observed to do. A result plotting above the U-line is not an unusual soil; it is almost certainly a test that needs repeating.

That is a genuinely useful thing to know at the bench, and it is why the U-line belongs on the chart even though no group symbol depends on it.

The sample’s finesPermalink to “The sample’s fines”

The fines from this sample have $LL = 42$ and $PI = 22$ . Evaluate both lines at $LL = 42$ :

PI_{\text{A}} = 0.73\left(42 - 20\right) = 16.06, \qquad PI_{\text{U}} = 0.9\left(42 - 8\right) = 30.60

The measured $PI = 22$ is above 16.06 and below 30.60: above the A-line, so the fines are clayey; below the U-line, so the result is plausible. With $LL = 42 < 50$ the fines are CL, a lean clay.

The answerPermalink to “The answer”

Coarse-grained; sand, because 72.3% of the coarse fraction passes the No. 4 sieve. Poorly graded, because $C_c = 0.75$ falls outside 1 to 3. Fines at 6%, so a dual symbol, and the fines are clayey:

\boxed{\textbf{SP-SC}}

Poorly graded sand with clay.

Every step above is arithmetic on measured numbers. Which is exactly why it is worth automating — the calculation is short enough that nobody checks it, and long enough that a transposed digit survives. The same argument applies to the Mohr circle, where the standing error is a factor of two rather than a logarithm.

Classification is where the chain starts: a constitutive model in KATAI 2D is only as good as the parameters, and the parameters are only as good as the group symbol that decided which correlations were allowed to apply.

ReferencesPermalink to “References”

ASTM D2487, Standard Practice for Classification of Soils for Engineering Purposes (Unified Soil Classification System). ASTM International. The governing standard; the criteria applied above are its criteria.
ASTM D6913, Standard Test Methods for Particle-Size Distribution (Gradation) of Soils Using Sieve Analysis.
ASTM D4318, Standard Test Methods for Liquid Limit, Plastic Limit, and Plasticity Index of Soils.
García-Gaines, R.A. and Frankenstein, S. (2015). USCS and the USDA Soil Classification System: Development of a Mapping Scheme. ERDC/CRREL TR-15-4, U.S. Army Engineer Research and Development Center. Approved for public release. Its Table 4 and Figure 3 state the classification criteria, the definitions of $C_u$ and $C_c$ , and the A-line and U-line equations used above.
Casagrande, A. (1948). Classification and identification of soils. Transactions of the American Society of Civil Engineers, 113(1), 901–930. The origin of the plasticity chart.

The sieve data and Atterberg limits above are a worked illustration, not measurements from a real site. Everything derived from them is computed, not quoted.

Fatih Küçükkarakurt — Writing

What a Mohr circle actually tells you

The circlePermalink to “The circle”#

The doubled anglePermalink to “The doubled angle”#

Adding a failure envelopePermalink to “Adding a failure envelope”#

A worked failure statePermalink to “A worked failure state”#

Where the soil failsPermalink to “Where the soil fails”#

Total and effective stressPermalink to “Total and effective stress”#

What the circle does not tell youPermalink to “What the circle does not tell you”#

ReferencesPermalink to “References”#

Soil classification from a grading curve

What the classification needsPermalink to “What the classification needs”#

The samplePermalink to “The sample”#

D10, D30 and D60 — the logarithmic stepPermalink to “D10, D30 and D60 — the logarithmic step”#

What linear interpolation costsPermalink to “What linear interpolation costs”#

Uniformity and curvaturePermalink to “Uniformity and curvature”#

Fines between 5% and 12%: a dual symbolPermalink to “Fines between 5% and 12%: a dual symbol”#

The plasticity chartPermalink to “The plasticity chart”#

The sample’s finesPermalink to “The sample’s fines”#

The answerPermalink to “The answer”#

ReferencesPermalink to “References”#

The circlePermalink to “The circle”

The doubled anglePermalink to “The doubled angle”

Adding a failure envelopePermalink to “Adding a failure envelope”

A worked failure statePermalink to “A worked failure state”

Where the soil failsPermalink to “Where the soil fails”

Total and effective stressPermalink to “Total and effective stress”

What the circle does not tell youPermalink to “What the circle does not tell you”

ReferencesPermalink to “References”

What the classification needsPermalink to “What the classification needs”

The samplePermalink to “The sample”

D10, D30 and D60 — the logarithmic stepPermalink to “D10, D30 and D60 — the logarithmic step”

What linear interpolation costsPermalink to “What linear interpolation costs”

Uniformity and curvaturePermalink to “Uniformity and curvature”

Fines between 5% and 12%: a dual symbolPermalink to “Fines between 5% and 12%: a dual symbol”

The plasticity chartPermalink to “The plasticity chart”

The sample’s finesPermalink to “The sample’s fines”

The answerPermalink to “The answer”

ReferencesPermalink to “References”