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Mechanical

Principal stresses in three dimensions

Largest principal stress of a general three-dimensional stress state, from all six components.

Ref: MEC-63
Rev: v1
Field: Mechanical
Result: σ₁ [MPa]

σ_{1} = q + 2 p \cos φ σ_{3} = q + 2 p \cos (φ + \frac{2 π}{3}) σ_{2} = 3 q - σ_{1} - σ_{3}

The principal stresses are the eigenvalues of the symmetric stress tensor, ordered σ₁ ≥ σ₂ ≥ σ₃. The closed-form route removes the mean stress q = I₁/3 first, then uses p = √[(a² + b² + c² + 2(τ_xy² + τ_yz² + τ_xz²))/6] with a = σ_x − q, b = σ_y − q, c = σ_z − q, and the angle φ = ⅓ arccos(d/(2p³)) where d = abc + 2τ_xyτ_yzτ_xz − aτ_yz² − bτ_xz² − cτ_xy². The argument of the arccos is clamped to between −1 and 1 to absorb round-off. If p = 0 the state is hydrostatic and all three principal stresses equal q. For nearly hydrostatic states, where tiny differences matter, use a numerical eigenvalue solver instead. The calculator gives σ₁.

Largest principal stress of a general three-dimensional stress state, from all six components. The three roots of the characteristic cubic are found with a closed trigonometric formula, so no eigenvalue solver is needed.

Calculator

Solve for

σ₁Largest principal stressComputed

MPa

σ_xNormal stress in x

MPa

−200300

σ_yNormal stress in y

MPa

−200300

σ_zNormal stress in z

MPa

−200300

τ_xyShear stress xy

MPa

−150150

τ_yzShear stress yz

MPa

−150150

τ_xzShear stress xz

MPa

−150150

Middle principal stress σ₂50 MPa
Smallest principal stress σ₃20 MPa
Sum σ₁ + σ₂ + σ₃ (equals I₁)180 MPa
Tresca equivalent stress σ₁ − σ₃90 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.

Largest principal stress (MPa) current value 110 MPa

σ_xNormal stress in xKey driver

83.2304

σ_yNormal stress in y

83.2304

σ_zNormal stress in z

110300

τ_xyShear stress xy

80230

τ_yzShear stress yz

207

τ_xzShear stress xz

207

The lecture state (80, 80, 20 MPa with 30 MPa shear) gives principal stresses 110, 50 and 20 MPa. The trace is 180 MPa before and 180 MPa after, a quick check.
Tresca is 90 MPa and von Mises √6300 = 79.373 MPa. With a 250 MPa yield strength the factors of safety are 2.778 and 3.15. Setting σ_z to zero (plane stress) leaves σ₁ and σ₂ at 110 and 50 MPa and brings σ₃ down from 20 to 0 MPa, because the z direction is a principal direction here.

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	−200 to 300 MPa	83.18	304	220.8	+10 MPa	+4.92 %
σ_y Normal stress in y	−200 to 300 MPa	83.18	304	220.8	+10 MPa	+4.92 %
σ_z Normal stress in z	−200 to 300 MPa	110	300	190	+10 MPa	0 %
τ_xy Shear stress xy	−150 to 150 MPa	80	230	150	+6 MPa	+5.45 %
τ_yz Shear stress yz	−150 to 150 MPa	110	207.2	97.24	+6 MPa	+0.182 %
τ_xz Shear stress xz	−150 to 150 MPa	110	207.2	97.24	+6 MPa	+0.182 %

Sweep of Shear stress xy

Shear stress xy (MPa)	Largest principal stress (MPa)
−150	230
−112.5	192.5
−75	155
−37.5	117.5
0	80
37.5	117.5
75	155
112.5	192.5
150	230

Sweep one variable

Vary

Largest principal stress (MPa)Hover, touch, or use the arrow keys on the chart

Variables and how each one contributes

σ₁

Largest principal stress

Result · MPa

What the equation gives you. Everything else feeds it.

σ_x

Normal stress in x

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 σ₁ by a share of the change that depends on how much of the largest principal direction lies along x. Raising it by 10 MPa (to 90 MPa) moves the result by +4.92 %, from 110 to 115.4 MPa.

σ_y

Normal stress in y

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 σ₁ by a share of the change that depends on how much of the largest principal direction lies along y. Raising it by 10 MPa (to 90 MPa) moves the result by +4.92 %, from 110 to 115.4 MPa.

σ_z

Normal stress in z

MPa · typical −200 to 300

Normal stress on the z face, in MPa. Tension is positive and compression negative, so the slider runs from −200 to +300 MPa. Raises σ₁ by a share of the change that depends on how much of the largest principal direction lies along z. Raising it by 10 MPa (to 30 MPa) moves the result by +0 %, from 110 to 110 MPa.

τ_xy

Shear stress xy

MPa · typical −150 to 150

Shear stress τ_xy, in MPa, acting on the x face in the +y direction. Shear is signed here, so the slider runs from −150 to +150 MPa. Raises σ₁ whichever way the shear acts, because the principal values depend on its square. Raising it by 10 MPa (to 40 MPa) moves the result by +9.09 %, from 110 to 120 MPa.

τ_yz

Shear stress yz

MPa · typical −150 to 150

Shear stress τ_yz, in MPa, acting on the y face in the +z direction. Shear is signed here, so the slider runs from −150 to +150 MPa. Raises σ₁ whichever way the shear acts, because the principal values depend on its square. Raising it by 10 MPa (to 10 MPa) moves the result by +0.507 %, from 110 to 110.6 MPa.

τ_xz

Shear stress xz

MPa · typical −150 to 150

Shear stress τ_xz, in MPa, acting on the x face in the +z direction. Shear is signed here, so the slider runs from −150 to +150 MPa. Raises σ₁ whichever way the shear acts, because the principal values depend on its square. Raising it by 10 MPa (to 10 MPa) moves the result by +0.507 %, from 110 to 110.6 MPa.

Worked example with the values above

Mean stress q = 60 MPa. Scale p = 26.46 MPa, and the angle parameter r = 0.5399, so φ = ⅓ arccos(r) = 19.11°.
σ₁ = q + 2p cos φ = 60 + 2 × 26.46 × cos 19.11° = 110 MPa.
σ₃ = q + 2p cos(φ + 120°) = 20 MPa, so σ₂ = 3q − σ₁ − σ₃ = 50 MPa. The first is σ₁ = 110 MPa.

Assumptions and limits

A symmetric stress tensor in one set of axes, tension positive.
Round-off protected only by clamping the arccos argument. Near-hydrostatic states need a numerical eigensolver.

Common mistakes

Dropping the out-of-plane stresses in a state that has them.
Using the plane-stress formula σ_1,2 = C ± R when σ_z or the out-of-plane shears are not zero.

Statistics and uncertainty

Principal stresses of a symmetric tensor are well-conditioned: an error in the components moves each principal stress by no more than the size of the error tensor, so the errors are not amplified. With errors of ±2 MPa in each component, expect errors of about ±2 to ±4 MPa in each principal stress. The largest difference between principal stresses, used in Tresca, carries the errors of two of them.

Where it is used

Principal stresses from finite-element output or a measured stress tensor.
The input to Tresca and von Mises when the state is fully three-dimensional.

Maths you need

Stress invariants: the coefficients of the characteristic equation
Cosine and arccosine: angles in radians or degrees

Related formulas

Practise this formula