Engineering · Strength of materials

Principal stresses in plane stress

Principal stresses are the extreme normal stresses obtained by rotating a plane stress element.

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Formula

σ_{1, 2} = \frac{σ _{x} + σ _{y}}{2} \pm (\frac{σ _{x} - σ _{y}}{2})^{2} + τ_{x y}^{2}

σ₁, σ₂, σx, σy and τxy in Pa

Variables & units

Variables & units
Symbol	Meaning	Unit
σ₁, σ₂	Maximum and minimum principal normal stress	Pa
σx, σy	Normal stresses in the selected coordinate system	Pa
$τxy$	Shear stress in the selected coordinate system	Pa

LaTeX

\sigma_{1,2}=\frac{\sigma_x+\sigma_y}{2}\pm\sqrt{\left(\frac{\sigma_x-\sigma_y}{2}\right)^2+\tau_{xy}^2}

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Derivation & background

Principal stresses are eigenvalues of the symmetric plane-stress tensor; shear stress vanishes in the corresponding principal directions. They do not depend on the original coordinate orientation. This expression assumes plane stress, not a general three-dimensional state. Use a consistent sign convention and units for all components.

Exam blueprint

Validity range

Derivation steps

Principal stresses are the extreme normal stresses obtained by rotating a plane stress element.

1State the model assumptions and list quantities with units.
2Substitute into σ₁,₂ = (σx+σy)/2 ± √[((σx−σy)/2)²+τxy²]; convert quantities to SI first.
3For σx=80 MPa, σy=20 MPa and τxy=30 MPa, the mean is 50 MPa and the square-root term is 42.4 MPa, giving σ₁≈92.4 MPa and σ₂≈7.6 MPa.

Rearrangements

Base equation for the target quantity

\sigma_{1,2}=\frac{\sigma_x+\sigma_y}{2}\pm\sqrt{\left(\frac{\sigma_x-\sigma_y}{2}\right)^2+\tau_{xy}^2}

Preserve signs, boundary conditions and units when rearranging.

Task variant

Work through the numerical example for “Principal stresses in plane stress” and check its unit and magnitude.

For σx=80 MPa, σy=20 MPa and τxy=30 MPa, the mean is 50 MPa and the square-root term is 42.4 MPa, giving σ₁≈92.4 MPa and σ₂≈7.6 MPa.

Common mistakes

Substituting a related quantity that shares the same symbol.

Use variable definitions and check: σ₁, σ₂, σx, σy and τxy in Pa.

Exam context

Find critical normal stresses and principal directions before material or component assessment.

Formula cluster

Strength of materials

Principal stresses are the extreme normal stresses obtained by rotating a plane stress element.

σv = √(σx² − σxσy + σy² + 3τxy²)

Von Mises equivalent stress in plane stress

σ_m = F/A

Average axial normal stress

σ_x(y) = −My/I

Bending stress in an Euler–Bernoulli beam

Worked example

For σx=80 MPa, σy=20 MPa and τxy=30 MPa, the mean is 50 MPa and the square-root term is 42.4 MPa, giving σ₁≈92.4 MPa and σ₂≈7.6 MPa.

Applications

Find critical normal stresses and principal directions before material or component assessment.

How do you calculate with Principal stresses in plane stress?

Here is how to work through a typical Principal stresses in plane stress (σ₁,₂ = (σx+σy)/2 ± √[((σx−σy)/2)²+τxy²]) task step by step:

1
Task
Work through the numerical example for “Principal stresses in plane stress” and check its unit and magnitude.
Solution path
For σx=80 MPa, σy=20 MPa and τxy=30 MPa, the mean is 50 MPa and the square-root term is 42.4 MPa, giving σ₁≈92.4 MPa and σ₂≈7.6 MPa.

Frequently asked questions about Principal stresses in plane stress

What are principal stresses?+

Principal stresses are normal stresses on planes oriented so that shear stress vanishes. In plane stress, σ₁ and σ₂ are the largest and smallest eigenvalues of the 2×2 stress tensor. Their magnitudes help identify critical tension or compression but do not replace an appropriate yield or fracture criterion. The equation gives the two normal-stress extrema of a plane stress tensor. All components must use the same sign convention and units. Shear stress vanishes in the principal axes. A general three-dimensional stress state requires the three eigenvalues of the full tensor. Finally check units, assumptions, measurement conditions and reference quantities against the problem before using the result.

When does the shear stress rotate to zero?+

The stress transformation gives an orientation where transformed shear stress is zero. The angle depends on τxy relative to σx−σy; if τxy=0, the current axes are already principal axes when aligned with the tensor. The principal-stress formula gives values, but not by itself the orientation angle. The equation gives the two normal-stress extrema of a plane stress tensor. All components must use the same sign convention and units. Shear stress vanishes in the principal axes. A general three-dimensional stress state requires the three eigenvalues of the full tensor. Finally check units, assumptions, measurement conditions and reference quantities against the problem before using the result.

What does a negative σ₂ mean?+

Under the usual tension-positive convention, a negative result represents a compressive principal stress; it is not an arithmetic error. For component assessment, compare tensile and compressive limits separately when the material behaves asymmetrically. Also check that plane stress is appropriate for the thin plate or surface considered. The equation gives the two normal-stress extrema of a plane stress tensor. All components must use the same sign convention and units. Shear stress vanishes in the principal axes. A general three-dimensional stress state requires the three eigenvalues of the full tensor. Finally check units, assumptions, measurement conditions and reference quantities against the problem before using the result.

How do I check the result and units for Principal stresses in plane stress?+

Write the quantity and unit from the variable list next to every value: σ₁, σ₂, σx, σy and τxy in Pa. Substitute only after converting to a consistent unit system, and state sign or reference direction explicitly. The example For σx=80 MPa, σy=20 MPa and τxy=30 MPa, the mean is 50 MPa and the square-root term is 42.4 MPa, giving σ₁≈92.4 MPa and σ₂≈7.6 MPa. is a calculation check: result, unit and order of magnitude must fit the described situation. Round after intermediate calculations so small contributions are not lost prematurely. The equation gives the two normal-stress extrema of a plane stress tensor. All components must use the same sign convention and units. Shear stress vanishes in the principal axes. A general three-dimensional stress state requires the three eigenvalues of the full tensor. Finally check units, assumptions, measurement conditions and reference quantities against the problem before using the result.

Which model limit matters for Principal stresses in plane stress?+

The background states this starting condition: Principal stresses are eigenvalues of the symmetric plane-stress tensor; shear stress vanishes in the corresponding principal directions. They. That limits when the result is a valid model prediction. Compare the problem’s boundary conditions with this assumption before substituting values; real components, material properties or loads can introduce additional effects. The cited reference supports the relation used here but does not remove the need to check whether the specific case fits the model. The equation gives the two normal-stress extrema of a plane stress tensor. All components must use the same sign convention and units. Shear stress vanishes in the principal axes. A general three-dimensional stress state requires the three eigenvalues of the full tensor. Finally check units, assumptions, measurement conditions and reference quantities against the problem before using the result.

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Subject references and further reading

This list names subject references and further reading. A page or section number appears when it is part of the reference; these are not verbatim quotations.

[1]MIT OpenCourseWare, Unified Engineering: Materials and Structures, stress transformation and principal stresses.

More formulas: Mechanics, strength and materials

E = \frac{\sigma}{\varepsilon} = \frac{F L_0}{A_0 \Delta L}

Young's modulus

\Delta\vec p=\int_{t_1}^{t_2}\vec F_{\mathrm{ext}}dt, \qquad \Delta\vec L=\int_{t_1}^{t_2}\vec\tau_{\mathrm{ext}}dt

Linear and angular impulse–momentum balance

M(x)=EI\,\kappa(x)\approx EI\frac{d^2w}{dx^2}

Euler–Bernoulli beam moment–curvature relation

\sum_i \vec F_i=\vec 0,\quad \sum_i M_{i,O}=0

Static equilibrium conditions

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Symbol

Meaning

Unit

σ₁, σ₂

Maximum and minimum principal normal stress

σx, σy

Normal stresses in the selected coordinate system

τxy

Shear stress in the selected coordinate system

Derivation & background

Exam blueprint

Validity range

Derivation steps

Principal stresses are the extreme normal stresses obtained by rotating a plane stress element.

1State the model assumptions and list quantities with units.
2Substitute into σ₁,₂ = (σx+σy)/2 ± √[((σx−σy)/2)²+τxy²]; convert quantities to SI first.
3For σx=80 MPa, σy=20 MPa and τxy=30 MPa, the mean is 50 MPa and the square-root term is 42.4 MPa, giving σ₁≈92.4 MPa and σ₂≈7.6 MPa.

How do you calculate with Principal stresses in plane stress?

Here is how to work through a typical Principal stresses in plane stress (σ₁,₂ = (σx+σy)/2 ± √[((σx−σy)/2)²+τxy²]) task step by step:

Task

Work through the numerical example for “Principal stresses in plane stress” and check its unit and magnitude.

Solution path

For σx=80 MPa, σy=20 MPa and τxy=30 MPa, the mean is 50 MPa and the square-root term is 42.4 MPa, giving σ₁≈92.4 MPa and σ₂≈7.6 MPa.

Frequently asked questions about Principal stresses in plane stress

What are principal stresses?+

When does the shear stress rotate to zero?+

What does a negative σ₂ mean?+

How do I check the result and units for Principal stresses in plane stress?+

Which model limit matters for Principal stresses in plane stress?+

Retain Principal stresses in plane stress for exams

Create a curated FSRS exam set for σ₁,₂ = (σx+σy)/2 ± √[((σx−σy)/2)²+τxy²]: formula recall, variables, derivation, rearrangement, worked example, common mistakes and exam context.

Curated formula set · LaTeX · FSRS spaced repetition · import from Essential