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Mechanical

Principal stresses in plane stress

The two in-plane principal stresses at a point, found from the two normal stresses and the shear stress on perpendicular faces.

Ref: MEC-24
Rev: v1
Field: Mechanical
Result: σ_a [MPa]

σ_{a,b} = \frac{σ_{x} + σ_{y}}{2} \pm \sqrt{{(\frac{σ_{x} - σ_{y}}{2})}^{2} + {τ_{xy}}^{2}}

σ_x and σ_y are normal stresses on two perpendicular faces and τ_xy the shear stress between them, at the same point, with tension positive. The principal stresses are the ends of the horizontal diameter of Mohr’s circle: σ_a = C + R and σ_b = C − R, where C = (σ_x + σ_y)/2 is the centre and R = √[((σ_x − σ_y)/2)² + τ_xy²] the radius. These are the two in-plane values only. In plane stress the out-of-plane stress is zero and counts as a third principal stress, so the next entry sorts all three into σ₁ ≥ σ₂ ≥ σ₃ before any yield criterion is applied. The calculator gives σ_a, the larger in-plane value.

The two in-plane principal stresses at a point, found from the two normal stresses and the shear stress on perpendicular faces. On the planes where they act the shear stress is zero.

Calculator

Solve for

σ_aLarger in-plane principal stressComputed

MPa

σ_xNormal stress in x (tension positive)

MPa

−200300

σ_yNormal stress in y (tension positive)

MPa

−200300

τ_xyShear stress on the x face in the y direction

MPa

−150150

Smaller in-plane principal stress σ_b = C − R−30 MPa
Principal angle θ_p to the larger one26.57 °
Mohr circle radius R (maximum in-plane shear)50 MPa

What drives the result

Each variable is moved across its typical range while the others stay at the values above. The longest bar is the key driver.

Larger in-plane principal stress (MPa) current value 70 MPa

σ_xNormal stress in x (tension positive)Key driver

−1.92305

σ_yNormal stress in y (tension positive)

56.2306

τ_xyShear stress on the x face in the y direction

50173

The lecture state (σ_x = 50, σ_y = −10, τ_xy = 40 MPa) has a circle with centre 20 MPa and radius 50 MPa, so the principal stresses are 70 and −30 MPa. Their sum, 40 MPa, equals σ_x + σ_y, which is unchanged by rotation.
Shear raises the largest normal stress above either input: σ_x was only 50 MPa, but the shear lifts σ_a to 70 MPa. With the shear removed, σ_a would be just 50 MPa.

Chart data as tables

Ranked variables

Variable	Typical range	Lowest result	Highest result	Swing (MPa)	Local step	Change in result
σ_x Normal stress in x (tension positive)	−200 to 300 MPa	−1.922	305.1	307	+10 MPa	+11.6 %
σ_y Normal stress in y (tension positive)	−200 to 300 MPa	56.24	306.2	250	+10 MPa	+3.1 %
τ_xy Shear stress on the x face in the y direction	−150 to 150 MPa	50	173	123	+6 MPa	+7.03 %

Sweep of Shear stress on the x face in the y direction

Shear stress on the x face in the y direction (MPa)	Larger in-plane principal stress (MPa)
−150	173
−112.5	136.4
−75	100.8
−37.5	68.02
0	50
37.5	68.02
75	100.8
112.5	136.4
150	173

Sweep one variable

Vary

Larger in-plane principal stress (MPa)Hover, touch, or use the arrow keys on the chart

Variables and how each one contributes

σ_a

Larger in-plane principal stress

Result · MPa

What the equation gives you. Everything else feeds it.

σ_x

Normal stress in x (tension positive)

MPa · typical −200 to 300

Normal stress on the x face, in MPa. Tension is positive and compression negative, so the slider runs from −200 to +300 MPa. Raises σ_a by less than the full change, because the shear stress shares the load. The share is ½ + (σ_x − σ_y)/(4R), which is 80 % at the default state. Raising it by 10 MPa (to 60 MPa) moves the result by +11.6 %, from 70 to 78.15 MPa.

σ_y

Normal stress in y (tension positive)

MPa · typical −200 to 300

Normal stress on the y face, in MPa. Tension is positive and compression negative, so the slider runs from −200 to +300 MPa. Raises σ_a by a smaller share, ½ − (σ_x − σ_y)/(4R), which is 20 % at the default state where σ_x is the larger stress. Raising it by 10 MPa (to 0 MPa) moves the result by +3.1 %, from 70 to 72.17 MPa.

τ_xy

Shear stress on the x face in the y direction

MPa · typical −150 to 150

Shear stress on the x face acting in the +y direction, in MPa. The sign matters for directions and rotated stresses, so the slider runs from −150 to +150 MPa. Raises σ_a whichever way the shear acts, because it enters the radius squared. The sign of the shear changes the direction of σ_a, not its size. Raising it by 10 MPa (to 50 MPa) moves the result by +11.9 %, from 70 to 78.31 MPa.

Worked example with the values above

Centre: C = (50 + (−10))/2 = 20 MPa. Radius: R = √[(30)² + (40)²] = 50 MPa.
Larger: σ_a = C + R = 20 + 50.
σ_a = 70 MPa, and σ_b = C − R = −30 MPa.

Assumptions and limits

Plane stress at one point (out-of-plane stress zero).
Tension positive, compression negative, tensor shear convention.

Common mistakes

Mixing stresses from different points.
Treating σ_a and σ_b as the complete set of principal stresses. A yield check needs the zero out-of-plane value as well.

Statistics and uncertainty

Strain-gauge rosettes give stress components with about 2–5 % error, and the principal stresses inherit that. Because the radius combines two terms in quadrature, a 5 % error in each of the half-difference and the shear gives roughly a 5 % error in the radius, and slightly less in σ_a where the centre is exact.

Where it is used

Strength checks in thin plates and shells.
Strain-gauge interpretation, and the input to the Tresca and von Mises checks.

Maths you need

Mohr’s circle: centre and radius of the stress circle
Pythagoras: combining two components

Related formulas

Practise this formula