Find principal stresses, maximum shear stress, and plane angles for a 2D stress state. Transform stresses at a chosen angle with worked steps.
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Worked examples
These are the solver's own reference problems — the answers come from published sources or independent hand computation, never from the solver itself. The solution below is the live solver output for each.
Example 1
SI 3-4-5 state: σx=80, σy=20, τxy=40 MPa
Given
sigma x
80 MPa
sigma y
20 MPa
tau xy
40 MPa
Assumptions
Plane stress: the out-of-plane stress σz = 0 and is the third principal stress for the absolute maximum shear check.
Hibbeler sign convention: normal stresses tension-positive; positive τxy acts in the +y direction on the +x face. Angles are measured counterclockwise from the +x axis.
Solution steps
Center and radius of Mohr's circle
The circle is centered on the normal-stress axis at the average normal stress; its radius comes from the half-difference of normal stresses and the shear.
σavg = (σx + σy)/2; R = √(((σx − σy)/2)² + τxy²)
σavg = 50 MPa; R = 50 MPa
Principal stresses
The principal stresses sit at the two ends of the circle's horizontal diameter — the average stress plus and minus the radius. No shear acts on the principal planes.
σ1 = σavg + R and σ2 = σavg − R
σ1 = 100 MPa; σ2 = 0 MPa
= 100 MPa
Principal and maximum-shear angles
The half-angle arctangent has two solutions a right angle apart; the two-argument arctangent picks the rotation that carries the x-axis onto the σ1 direction (the other principal plane is 90° away). The maximum-shear planes bisect the principal planes.
2θp = atan2(2τxy, σx − σy); θs = θp1 − 45°
θp1 = 26.57 deg (σ1 direction, i.e. 2θp = 53.13 deg on the circle); θs = -18.43 deg or 71.57 deg
Maximum shear stresses
The in-plane maximum shear equals the circle radius. For the absolute maximum, the out-of-plane principal stress σz = 0 joins the comparison: when σ1 and σ2 have the same sign, the largest circle is the one through zero.
τmax(in-plane) = R; τabs = max(R, |σ1|/2, |σ2|/2)
τmax(in-plane) = 50 MPa; τabs = 50 MPa
= 50 MPa
Hibbeler convention · θp = 26.57°
Results
Major principal stress σ1
100MPa
Minor principal stress σ2
0MPa
Average normal stress (circle center)
50MPa
Maximum in-plane shear stress
50MPa
Absolute maximum shear stress (3D, σz = 0)
50MPa
Principal angle θp1 (x-axis → σ1 direction, ccw +)
26.57deg
Maximum-shear plane angle θs (ccw +)
-18.43deg
Where this answer was checked
source
Plane-stress transformation closed forms (NCEES FE Reference Handbook) — σ1,2 = σavg ± R with R = √(30² + 40²) = 50
US 5-12-13 state, both principals positive: σx=20, σy=10, τxy=12 ksi
Given
sigma x
20 ksi
sigma y
10 ksi
tau xy
12 ksi
Assumptions
Plane stress: the out-of-plane stress σz = 0 and is the third principal stress for the absolute maximum shear check.
Hibbeler sign convention: normal stresses tension-positive; positive τxy acts in the +y direction on the +x face. Angles are measured counterclockwise from the +x axis.
Solution steps
Center and radius of Mohr's circle
The circle is centered on the normal-stress axis at the average normal stress; its radius comes from the half-difference of normal stresses and the shear.
σavg = (σx + σy)/2; R = √(((σx − σy)/2)² + τxy²)
σavg = 15 ksi; R = 13 ksi
Principal stresses
The principal stresses sit at the two ends of the circle's horizontal diameter — the average stress plus and minus the radius. No shear acts on the principal planes.
σ1 = σavg + R and σ2 = σavg − R
σ1 = 28 ksi; σ2 = 2 ksi
= 28 ksi
Principal and maximum-shear angles
The half-angle arctangent has two solutions a right angle apart; the two-argument arctangent picks the rotation that carries the x-axis onto the σ1 direction (the other principal plane is 90° away). The maximum-shear planes bisect the principal planes.
2θp = atan2(2τxy, σx − σy); θs = θp1 − 45°
θp1 = 33.69 deg (σ1 direction, i.e. 2θp = 67.38 deg on the circle); θs = -11.31 deg or 78.69 deg
Maximum shear stresses
The in-plane maximum shear equals the circle radius. For the absolute maximum, the out-of-plane principal stress σz = 0 joins the comparison: when σ1 and σ2 have the same sign, the largest circle is the one through zero.
τmax(in-plane) = R; τabs = max(R, |σ1|/2, |σ2|/2)
τmax(in-plane) = 13 ksi; τabs = 14 ksi
= 14 ksi
Hibbeler convention · θp = 33.69°
Results
Major principal stress σ1
28ksi
Minor principal stress σ2
2ksi
Average normal stress (circle center)
15ksi
Maximum in-plane shear stress
13ksi
Absolute maximum shear stress (3D, σz = 0)
14ksi
Principal angle θp1 (x-axis → σ1 direction, ccw +)
33.69deg
Maximum-shear plane angle θs (ccw +)
-11.31deg
Where this answer was checked
source
Plane-stress transformation + 3D absolute shear with σz=0 (NCEES FE Reference Handbook) — τabs = σ1/2 when σ1, σ2 share a sign
verified by
hand-recomputed
derivation
σavg = 15, half-difference = 5, R = √(25+144) = 13. σ1 = 28, σ2 = 2 ksi (both tension). τmax,in = 13 but τabs = max(13, 14, 1) = 14 ksi — the out-of-plane circle through σz = 0 governs. 2θp1 = atan2(24, 10) = 67.3801°, θp1 = 33.690°.
Example 3
SI transformed state at 30°: σx=−8, σy=12, τxy=−6 MPa
Given
sigma x
-8 MPa
sigma y
12 MPa
tau xy
-6 MPa
theta deg
30
Assumptions
Plane stress: the out-of-plane stress σz = 0 and is the third principal stress for the absolute maximum shear check.
Hibbeler sign convention: normal stresses tension-positive; positive τxy acts in the +y direction on the +x face. Angles are measured counterclockwise from the +x axis.
Solution steps
Center and radius of Mohr's circle
The circle is centered on the normal-stress axis at the average normal stress; its radius comes from the half-difference of normal stresses and the shear.
σavg = (σx + σy)/2; R = √(((σx − σy)/2)² + τxy²)
σavg = 2 MPa; R = 11.66 MPa
Principal stresses
The principal stresses sit at the two ends of the circle's horizontal diameter — the average stress plus and minus the radius. No shear acts on the principal planes.
σ1 = σavg + R and σ2 = σavg − R
σ1 = 13.66 MPa; σ2 = -9.662 MPa
= 13.66 MPa
Principal and maximum-shear angles
The half-angle arctangent has two solutions a right angle apart; the two-argument arctangent picks the rotation that carries the x-axis onto the σ1 direction (the other principal plane is 90° away). The maximum-shear planes bisect the principal planes.
2θp = atan2(2τxy, σx − σy); θs = θp1 − 45°
θp1 = -74.52 deg (σ1 direction, i.e. 2θp = -149 deg on the circle); θs = -119.5 deg or -29.52 deg
Maximum shear stresses
The in-plane maximum shear equals the circle radius. For the absolute maximum, the out-of-plane principal stress σz = 0 joins the comparison: when σ1 and σ2 have the same sign, the largest circle is the one through zero.
τmax(in-plane) = R; τabs = max(R, |σ1|/2, |σ2|/2)
τmax(in-plane) = 11.66 MPa; τabs = 11.66 MPa
= 11.66 MPa
Transformed stresses at the requested angle
Rotating the element corresponds to walking twice the angle around Mohr's circle; the companion face carries the remainder of the normal-stress sum.