Mohr's Circle & Stress on a Tilted Plane

Set the three stresses at a point and turn the element. The circle shows every plane at once: where the normal stress peaks, where the shear peaks, and the circle through the thickness that the in-plane picture leaves out. The table under the drawing works every figure out.

Worked example — Mohr's Circle & Stress on a Tilted Plane

Centre of the circle	20.00 MPa
Half-difference	30.00 MPa
Radius	50.00 MPa
Principal stress σ1	70.00 MPa
Principal stress σ2	−30.00 MPa
Product of principal stresses	−2,100.00 MPa²
Principal angle θp	26.57°
σ2 plane	−63.43°

These are the values the calculator shows when the page opens. Change any input above to recompute them.

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Lesson

The theory — Mohr's Circle & Stress on a Tilted Plane

The stress acting on a plane through a point depends on which plane you cut. In two dimensions three numbers describe it, σx, σy and τxy, and turning the plane by θ gives new values by the transformation equations. Mohr's circle draws all of them at once. Each plane is a point (σ, τ) on a circle centred at (σx + σy) ÷ 2 with radius R = √(((σx − σy) ÷ 2)² + τxy²), and turning the plane by θ moves its point 2θ round the circle. Where the circle crosses the σ axis are the principal stresses σ1 and σ2, on planes that carry no shear; its top and bottom are the maximum in-plane shear, R, on planes at 45° to them.

What each symbol means

σ1: the larger in-plane principal stress, C + R. The greatest normal stress on any plane square to the page
σ2: the smaller in-plane principal stress, C − R. Its plane is at right angles to σ1's
R: the circle's radius, and the largest shear on any plane square to the page
θp: the angle from the x axis to the plane of σ1. Half the angle between them on the circle
τ_abs: the absolute maximum shear, half the spread of σ1, σ2 and σ3 = 0. Larger than R whenever σ1 and σ2 share a sign

Breaks when: Thin-wall theory drops the radial stress, which is −p at the inside face. For the vessel preset σ3 is −2 MPa there rather than zero, and with the thin-wall hoop stress the largest shear at the inner surface is 51 MPa, not 50. σ3 = 0 holds exactly only at the outer surface, and the thin-wall stresses themselves are approximations, good to about 1% for a wall this thin.

Pull a bar and its worst shear is on a slant 🖖

Press Uniaxial tension. The bar is pulled at 100 MPa, and on its cross-section, at θ = 0, the shear is zero. Turn the element and shear appears, growing until θ = 45°, where it reaches 50 MPa in size, half the pull, and falling away again after. The table prints it as −50.00: under the sign convention, shear on a plane turned that way points the negative way. On the circle that is the top or bottom point, a quarter-turn from where you started, because every angle on the element is doubled on the circle. Metals like steel yield by dislocations gliding along slip planes, which is driven by shear, so they give way on the slant: a steel test bar pulled until it breaks ends in a cup and cone whose rim runs at roughly 45° to the axis. Tresca's criterion turns this into a rule. A ductile metal yields when its largest shear reaches half its yield strength in tension.

Twist chalk and it breaks on a 45° spiral 🖖

Press Pure shear (torsion). A twisted shaft carries 100 MPa of shear on its cross-section, and turning the element by 45° makes the shear vanish: one pair of faces is pulled with 100 MPa and the other pushed with 100 MPa. Torsion is tension and compression at right angles to each other, set at 45° to the shaft, and which of the two a material gives way to decides how it breaks. Press Cast iron in torsion. Cast iron is weak in tension, so it fails where the tension is, on the plane at +45°, and a twisted cast-iron bar cracks along a helix at 45° to its axis. A stick of blackboard chalk does the same between your fingers. A steel shaft fails in shear instead, and the shear is largest on the cross-section itself, θs = 0°, so steel twists off square across the bar.

A pressure vessel's worst shear is not on the circle you drew 🖖

Press Pressure vessel wall: 100 MPa round the hoop and 50 MPa along the axis. The circle through those two has a radius of 25 MPa, and 25 MPa is the largest shear on any plane square to the wall. The wall has a third direction, through its thickness, and at the outer surface nothing presses on it, so σ3 = 0. The circle from 100 down to 0 is twice as big, and the shear on it is 50 MPa, on planes slanting through the thickness. Check the vessel against the in-plane circle and Tresca's safety factor comes out as 250 ÷ (2 × 25) = 5.00; the page's 2.50 is the true one. The trap opens whenever the two principal stresses share a sign, and it is starkest at Equal biaxial, where the in-plane circle is a point and no plane across the sheet carries any shear, yet 30 MPa of it runs through the thickness.

Problem solved in full

Checking a pressure vessel against yield 5 steps
A steel cylinder of radius 500 mm with a 10 mm wall holds gas at 2 MPa. The steel yields at 250 MPa. Find the stresses in the wall, the largest shear, and the safety factor against yield by Tresca's criterion. Press Pressure vessel wall to check each figure.

Open this problem in the tool
1. Thin-wall theory gives the two stresses in the wall directly. The hoop stress is twice the axial one:
  
  $σ_{h} = \frac{p r}{t} = \frac{2 \cdot 500}{10} = 100 MPa, σ_{a} = \frac{p r}{2 t} = 50 MPa$
2. Neither direction carries shear, so these are already the principal stresses, and the in-plane circle follows:
  
  $σ_{1} = 100, σ_{2} = 50, C = \frac{100 + 50}{2} = 75, R = \frac{100 - 50}{2} = 25 MPa$
3. The largest shear on any plane square to the wall is the radius:
  
  $τ_{m a x, in} = R = 25 MPa$
4. At the outer surface nothing presses on the wall, so the third principal stress is zero, and the biggest of the three circles runs from σ1 to it:
  
  $σ_{3} = 0, τ_{abs} = \frac{σ _{1} - σ _{3}}{2} = \frac{100 - 0}{2} = 50 MPa$
5. Tresca compares twice the largest shear with the yield strength:
  
  $σ_{T} = 2 τ_{abs} = 100 MPa, n = \frac{250}{100} = 2.50$
Answer

The wall carries 100 MPa round the hoop and 50 MPa along the axis. The largest shear is 50 MPa, through the thickness, not the 25 MPa the in-plane circle shows, and the Tresca safety factor is 2.50. Worked from the in-plane circle alone it would have been 5.00, twice what the vessel has.

$n = 2.50, \frac{250}{2 \cdot 25} = 5.00$

Learning path

Where the load goes

References (2)

Where the circle comes from. Mohr published the graphical construction for the state of stress at a point in 1882: O. Mohr, "Über die Darstellung des Spannungszustandes und des Deformationszustandes eines Körperelementes." Zivilingenieur 28, 1882.
Where the two material values would be checked. THEY WERE NOT TAKEN FROM HERE — see the note: A materials handbook or the relevant specification: ASTM A36 or EN 10025 S275 for structural steel, ASTM A48 for grey cast iron.

By the 42Math team · Reviewed against published sources. Corrections welcome.