Principal stresses are the maximum and minimum normal stresses at a point in a loaded body, and they act on the two planes where the shear stress is zero. For plane stress they are given by σ1,2 = (σx + σy)/2 ± √[((σx − σy)/2)² + τxy²]. The two principal planes are always 90° apart, and the maximum in-plane shear stress equals the square-root term, on planes 45° from the principal planes. Designers care because cracks and yielding start where the normal stress is largest, so a design check starts by finding σ1 and σ2.
Why principal stresses matter in design
A shaft under combined bending and torsion does not fail on the plane you happened to draw. At every point the stress you measure depends on the orientation of the cut you imagine through it. Turn the element and the normal stress rises and falls, and the shear stress rises and falls with it.
Two orientations are special:
- The principal planes, where shear stress is zero and normal stress reaches its largest and smallest values. A brittle material such as grey cast iron cracks on a plane roughly normal to the largest tensile principal stress, which is why a brittle shaft in pure torsion fails along a 45° helix.
- The maximum shear planes, 45° away, where shear stress peaks. A ductile material such as mild steel yields by slip on these planes, which is why a mild steel tensile specimen shows slip bands and a shear lip at about 45° to the load axis.
So the same point in the same part is checked with different numbers depending on the material. Either way, you first need σ1, σ2 and τmax.
The 2D stress element and sign conventions
Plane stress means the third direction carries nothing: σz = 0, τxz = 0, τyz = 0. This covers thin plates, shafts, beam surfaces and pressure-vessel walls, which is most of what a second-year strength of materials course handles.
The element is a small square carrying three independent quantities: σx, σy and τxy. Get the signs right before touching a formula.
| Quantity | Positive when | Negative when |
|---|---|---|
| σx, σy | Tensile, pulling the face outward | Compressive, pushing the face inward |
| τxy | On the face whose outward normal is +x, the shear acts along +y (the pair tends to rotate the element anticlockwise about the corner) | The opposite sense |
| θ | Measured anticlockwise from the x-axis to the outward normal of the inclined plane | Clockwise |
Complementary shear means τxy = τyx in magnitude, so one value describes both. Compressive stress goes into the formula with a minus sign, every time. Most wrong answers in this topic are sign errors, not algebra errors.
Stress transformation equations
Cut the element along a plane whose outward normal makes an angle θ anticlockwise with the x-axis, and write force equilibrium on the wedge. The normal and shear stresses on that plane are:
σθ = (σx + σy)/2 + ((σx − σy)/2) cos 2θ + τxy sin 2θ
τθ = −((σx − σy)/2) sin 2θ + τxy cos 2θ
Two facts fall straight out of these. First, everything repeats every 180° in θ, because the equations contain 2θ. Second, adding σθ and the stress on the perpendicular plane (θ + 90°) always gives σx + σy. That sum is the first stress invariant, and it is the cheapest check you have.
Derivation of the principal stress formula
Principal planes are where the normal stress is stationary, so differentiate σθ with respect to θ and set it to zero:
dσθ/dθ = −(σx − σy) sin 2θ + 2τxy cos 2θ = 0
Compare that expression with τθ: it is exactly 2τθ. So the planes of maximum and minimum normal stress are the same planes on which the shear stress is zero. That is not a coincidence to memorise, it is the definition falling out of the algebra. Rearranging gives the principal plane angle:
tan 2θp = 2τxy / (σx − σy)
The tangent function repeats every 180°, so 2θp has two solutions separated by 180°, which means θp and θp + 90°. The two principal planes are perpendicular to each other. Always.
From tan 2θp, build the right triangle with opposite side τxy and adjacent side (σx − σy)/2, so the hypotenuse is R = √[((σx − σy)/2)² + τxy²]. Then sin 2θp = τxy/R and cos 2θp = ((σx − σy)/2)/R. Substituting back into σθ:
σ = (σx + σy)/2 + [((σx − σy)/2)² + τxy²]/R = (σx + σy)/2 + R
The second root gives the minus sign, so:
σ1,2 = (σx + σy)/2 ± √[((σx − σy)/2)² + τxy²]
By convention σ1 is the algebraically larger value. Two invariant checks come free:
- σ1 + σ2 = σx + σy
- σ1σ2 = σxσy − τxy²
Maximum in-plane shear stress
Repeat the exercise on τθ. Setting dτθ/dθ = 0 gives tan 2θs = −(σx − σy)/(2τxy), which is the negative reciprocal of tan 2θp. Negative reciprocal means 2θs and 2θp differ by 90°, so:
θs = θp ± 45°
The maximum in-plane shear stress is the same square-root term, the radius R:
τmax(in-plane) = √[((σx − σy)/2)² + τxy²] = (σ1 − σ2)/2
One more result that students often miss: on the maximum shear planes the normal stress is not zero. It equals the average, (σx + σy)/2, on both of them.
Mohr’s circle
Square and add the two transformation equations and the θ terms disappear:
(σθ − σavg)² + τθ² = R²
That is the equation of a circle in the (σ, τ) plane. Every possible orientation of the element is one point on it.
| Feature of the circle | Value | Meaning |
|---|---|---|
| Centre | C = ((σx + σy)/2, 0) | Average normal stress, always on the σ axis |
| Radius | R = √[((σx − σy)/2)² + τxy²] | Maximum in-plane shear stress |
| Right intercept | C + R | σ1 |
| Left intercept | C − R | σ2 |
| Top and bottom | (C, ±R) | Maximum shear planes, normal stress = C there |
How to construct it.
- Draw the σ axis horizontal, tension to the right. Draw the τ axis vertical. Take τ positive downward, so that a rotation on the circle runs in the same sense as the physical rotation of the element.
- Plot point X for the x-face at (σx, τxy) and point Y for the y-face at (σy, −τxy).
- Join XY. It cuts the σ axis at the centre C, because the midpoint of σx and σy is the average.
- Draw the circle on XY as diameter. Its radius is R.
- Read σ1 and σ2 where the circle crosses the σ axis, and τmax at the top and bottom of the circle.
- The angle from CX round to the σ axis is 2θp. Halve it to get the physical angle θp. All angles on Mohr’s circle are double the real angles.
Some books plot τ positive upward instead, which reverses the apparent sense of rotation. The magnitudes of σ1, σ2 and τmax are identical either way; only the sign of the angle changes, so pick one convention and stay in it for the whole paper.
Solved example 1: all stresses positive
Given: σx = 80 MPa, σy = 20 MPa, τxy = 30 MPa.
Average: (80 + 20)/2 = 50 MPa.
Radius: R = √[((80 − 20)/2)² + 30²] = √(30² + 30²) = √1800 = 42.43 MPa.
Principal stresses: σ1 = 50 + 42.43 = 92.43 MPa, σ2 = 50 − 42.43 = 7.57 MPa. Both tensile.
Invariant check: 92.43 + 7.57 = 100 MPa = 80 + 20. Correct. Second check: 92.43 × 7.57 = 700 = (80)(20) − 30² = 1600 − 900. Correct.
Principal plane: tan 2θp = 2(30)/(80 − 20) = 60/60 = 1, so 2θp = 45° and θp = 22.5°. The other principal plane is at 112.5°.
Which root goes with which plane? Substitute θ = 22.5° back: σ = 50 + 30 cos 45° + 30 sin 45° = 50 + 21.21 + 21.21 = 92.43 MPa. So 22.5° carries σ1, and 112.5° carries σ2. Never assume, always substitute one of them back.
Maximum in-plane shear: τmax = R = 42.43 MPa, on planes at 22.5° − 45° = −22.5° and at 67.5°, each carrying a normal stress of 50 MPa.
Solved example 2: negative normal stress and negative shear
Given: σx = −40 MPa (compressive), σy = 60 MPa (tensile), τxy = −30 MPa.
Average: (−40 + 60)/2 = 10 MPa.
Half-difference: (−40 − 60)/2 = −50 MPa. Keep the sign; it matters for the angle even though it is squared in R.
Radius: R = √[(−50)² + (−30)²] = √(2500 + 900) = √3400 = 58.31 MPa.
Principal stresses: σ1 = 10 + 58.31 = 68.31 MPa (tensile), σ2 = 10 − 58.31 = −48.31 MPa (compressive).
Invariant check: 68.31 + (−48.31) = 20 MPa = −40 + 60. Correct. Product check: 68.31 × (−48.31) = −3300 = (−40)(60) − (−30)² = −2400 − 900. Correct.
Principal plane: tan 2θp = 2(−30)/(−40 − 60) = −60/−100 = 0.6, so 2θp = 30.96° and θp = 15.48°, with the second plane at 105.48°.
Now substitute back, because this is where the trap sits. At θ = 15.48°: σ = 10 + (−50)(cos 30.96°) + (−30)(sin 30.96°) = 10 − 42.87 − 15.43 = −48.31 MPa. That is σ2, not σ1. The larger principal stress of 68.31 MPa acts on the plane at 105.48°. The tan formula gives you the pair of planes but never tells you which root sits on which, and with negative inputs the answer is often the one you did not expect.
Maximum in-plane shear: τmax = 58.31 MPa, on planes at 15.48° + 45° = 60.48° and at −29.52°, each carrying a normal stress of 10 MPa.
The third principal stress and the absolute maximum shear
A stress state always has three principal stresses, because stress is a second-order tensor in three dimensions. In plane stress the out-of-plane face carries nothing, so σ3 = 0. It is a genuine principal stress, not a missing one, and ignoring it is one of the most common errors in this topic.
The absolute maximum shear stress at the point uses the largest and smallest of all three:
τabs max = (σmax − σmin)/2, taken over σ1, σ2 and σ3 = 0
Two cases follow:
- σ1 and σ2 have opposite signs. Zero lies between them, so the in-plane pair already spans the full range and τabs max = (σ1 − σ2)/2 = the in-plane value.
- σ1 and σ2 have the same sign. Zero now lies outside the pair, the range widens, and τabs max = σ1/2 if both are positive, or |σ2|/2 if both are negative. This is larger than the in-plane maximum.
Apply it to the two examples. In example 1 both principal stresses are tensile, so τabs max = 92.43/2 = 46.21 MPa, against an in-plane maximum of 42.43 MPa. Quoting 42.43 MPa as “the maximum shear stress” there understates the real peak by about 9%. In example 2 the signs are opposite, so τabs max = (68.31 + 48.31)/2 = 58.31 MPa, which is the in-plane value again.
On Mohr’s circle this is easy to see: draw all three circles for the pairs (σ1, σ2), (σ2, σ3) and (σ1, σ3). The largest of the three radii is τabs max. When both in-plane principal stresses share a sign, the biggest circle is one of the two that involve σ3 = 0.
Using principal stresses in failure theories
Once σ1, σ2 and σ3 are known, the design check is a one-line comparison against a material property.
| Theory | Criterion (plane stress) | Use for |
|---|---|---|
| Maximum principal stress (Rankine) | σ1 = σut in tension, or |σmin| = σuc in compression | Brittle materials: cast iron, concrete, ceramics |
| Maximum shear stress (Tresca / Guest) | σmax − σmin = σy, using all three principal stresses | Ductile materials, conservative and simple |
| Distortion energy (von Mises) | √(σ1² − σ1σ2 + σ2²) = σy | Ductile materials, closest to test data |
For plane stress the von Mises equivalent can also be written straight from the components as √(σx² − σxσy + σy² + 3τxy²), which is a useful independent check on your principal stresses.
Take example 1 again. Tresca equivalent stress = 92.43 − 0 = 92.43 MPa. Von Mises from the principal stresses = √(92.43² − 92.43 × 7.57 + 7.57²) = 88.88 MPa, and from the components = √(6400 − 1600 + 400 + 2700) = √7900 = 88.88 MPa. The two routes agree, and Tresca comes out higher, which is the general rule: the Tresca equivalent stress is never lower than the von Mises one, so Tresca predicts yielding a little sooner.
Example 2 gives a Tresca equivalent of 68.31 + 48.31 = 116.62 MPa and a von Mises equivalent of √10300 = 101.49 MPa. For a steel with a yield strength of 250 MPa, the factor of safety is 250/116.62 = 2.14 by Tresca and 250/101.49 = 2.46 by von Mises. Tresca is the safer number to quote when the question does not say which theory to use.
A last practical note: these checks assume a static load. Under fluctuating load you go on to a fatigue calculation, and under stress concentration you multiply by a factor Kt first. Principal stresses feed those methods rather than replacing them, in the same way that strain relations feed deflection calculations.
References
- AICTE Model Curriculum, Strength of Materials / Mechanics of Solids.
- NPTEL, Strength of Materials lecture series on stress transformation and Mohr’s circle, IIT.
FAQs
What is the principal stress formula?
For plane stress, σ1,2 = (σx + σy)/2 ± √[((σx − σy)/2)² + τxy²], where σ1 is the larger value. Compressive stresses enter as negative numbers. Check the answer with σ1 + σ2 = σx + σy.
What is a principal plane?
A principal plane is a plane through the point on which the shear stress is zero and the normal stress reaches a maximum or minimum. Its orientation comes from tan 2θp = 2τxy/(σx − σy), and the two principal planes at a point are always 90° apart.
How many principal stresses are there at a point?
Three, because stress is a three-dimensional tensor. In plane stress the third one is σ3 = 0, which still counts. Including it matters for the absolute maximum shear stress and for the Tresca yield check.
What is the maximum shear stress in terms of principal stresses?
The maximum in-plane shear stress is (σ1 − σ2)/2, equal to the radius of Mohr’s circle, on planes 45° from the principal planes. The absolute maximum is (σmax − σmin)/2 taken over all three principal stresses including σ3 = 0, so if σ1 and σ2 have the same sign it exceeds the in-plane value.
How do you find principal stresses using Mohr’s circle?
Plot (σx, τxy) and (σy, −τxy), join them to find the centre at ((σx + σy)/2, 0), and draw the circle of radius √[((σx − σy)/2)² + τxy²]. The principal stresses are the two points where the circle cuts the σ axis, and τmax is the radius. Angles on the circle are twice the physical angles.