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Mechanical

Normal stress on any plane (three dimensions)

Normal stress on a plane whose unit normal is n, found from all six components of the stress tensor.

Ref: MEC-58
Rev: v1
Field: Mechanical
Result: σ_n [MPa]

σ_{n} = σ_{x} {n_{x}}^{2} + σ_{y} {n_{y}}^{2} + σ_{z} {n_{z}}^{2} + 2 τ_{xy} n_{x} n_{y} + 2 τ_{yz} n_{y} n_{z} + 2 τ_{xz} n_{x} n_{z}

For a plane with unit normal n, the traction t = σn is the force per unit area on it. Its component along n is the normal stress σ_n = nᵀσn, written out above in terms of the six stress components and the direction components n_x, n_y, n_z. The direction you enter does not have to have unit length, because it is scaled for you. Positive σ_n is tension across the plane. The remaining part of the traction is the shear stress on the plane, in the next entry. The tensor convention: τ_xy acts in the +y direction on the +x face, with the equal complementary shear on the +y face.

Normal stress on a plane whose unit normal is n, found from all six components of the stress tensor. It is the first half of splitting the traction on the plane into a normal part and a shear part.

Calculator

Solve for

σ_nNormal stress on the planeComputed

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

n_xNormal direction component x

−11

n_yNormal direction component y

−11

n_zNormal direction component z

−11

Traction magnitude |t|109.1 MPa
Shear stress magnitude |τ_n| on the plane8.4 MPa
Length of the entered direction (scaled to 1)1

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.

Normal stress on the plane (MPa) current value 108.8 MPa

σ_yNormal stress in yKey driver

−70.4250

τ_xyShear stress xy

−64224

σ_xNormal stress in x

8188

n_yNormal direction component y

50110

n_xNormal direction component x

50110

n_zNormal direction component z

64.4

σ_zNormal stress in z

τ_yzShear stress yz

τ_xzShear stress xz

On the plane with normal (0.6, 0.8, 0) the lecture’s three-dimensional state carries 108.8 MPa. The largest value over every plane is the first principal stress, 110 MPa, on the plane with normal (1, 1, 0)/√2; entering that direction here gives 110 MPa.
Planes with normals along x, y and z give the three normal stresses directly: 80, 80 and 20 MPa, a quick check.

Chart data as tables

Ranked variables

Variable	Typical range	Lowest result	Highest result	Swing (MPa)	Local step	Change in result
σ_y Normal stress in y	−200 to 300 MPa	−70.4	249.6	320	+10 MPa	+5.88 %
τ_xy Shear stress xy	−150 to 150 MPa	−64	224	288	+6 MPa	+5.29 %
σ_x Normal stress in x	−200 to 300 MPa	8	188	180	+10 MPa	+3.31 %
n_y Normal direction component y	−1 to 1	50	110	60	+0.04	−0.388 %
n_x Normal direction component x	−1 to 1	50	110	60	+0.04	+0.43 %
n_z Normal direction component z	−1 to 1	64.4	108.8	44.4	+0.04	−0.13 %
σ_z Normal stress in z	−200 to 300 MPa	108.8	108.8	0	+10 MPa	0 %
τ_yz Shear stress yz	−150 to 150 MPa	108.8	108.8	0	+6 MPa	0 %
τ_xz Shear stress xz	−150 to 150 MPa	108.8	108.8	0	+6 MPa	0 %

Sweep of Normal stress in x

Normal stress in x (MPa)	Normal stress on the plane (MPa)
−200	8
−137.5	30.5
−75	53
−12.5	75.5
50	98
112.5	120.5
175	143
237.5	165.5
300	188

Sweep one variable

Vary

Normal stress on the plane (MPa)Hover, touch, or use the arrow keys on the chart

Variables and how each one contributes

σ_n

Normal stress on the plane

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 σ_n by n_x² times the change. Raising it by 10 MPa (to 90 MPa) moves the result by +3.31 %, from 108.8 to 112.4 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 σ_n by n_y² times the change. Raising it by 10 MPa (to 90 MPa) moves the result by +5.88 %, from 108.8 to 115.2 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 σ_n by n_z² times the change. Raising it by 10 MPa (to 30 MPa) moves the result by +0 %, from 108.8 to 108.8 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 σ_n by twice n_xn_y times the change. Raising it by 10 MPa (to 40 MPa) moves the result by +8.82 %, from 108.8 to 118.4 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 σ_n by twice n_yn_z times the change, which is nothing when the plane normal has no z component. Raising it by 10 MPa (to 10 MPa) moves the result by +0 %, from 108.8 to 108.8 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 σ_n by twice n_xn_z times the change, which is nothing when the plane normal has no z component. Raising it by 10 MPa (to 10 MPa) moves the result by +0 %, from 108.8 to 108.8 MPa.

n_x

Normal direction component x

· typical −1 to 1

The x component of the direction perpendicular to the plane. The direction can have any length and is scaled to a unit vector, so only the ratios of the three components matter. The slider runs from −1 to +1. Tilts the plane, which changes the normal stress through the direction cosines. Raising it by 0.1 (to 0.7) moves the result by +0.859 %, from 108.8 to 109.7 MPa.

n_y

Normal direction component y

· typical −1 to 1

The y component of the direction perpendicular to the plane. The direction is scaled to a unit vector, so only the ratios of the three components matter. Tilts the plane, which changes the normal stress through the direction cosines. Raising it by 0.1 (to 0.9) moves the result by −1.02 %, from 108.8 to 107.7 MPa.

n_z

Normal direction component z

· typical −1 to 1

The z component of the direction perpendicular to the plane. The direction is scaled to a unit vector, so only the ratios of the three components matter. Tilts the plane, which changes the normal stress through the direction cosines. Raising it by 0.1 (to 0.1) moves the result by −0.808 %, from 108.8 to 107.9 MPa.

Worked example with the values above

Scale the direction to unit length: n = (0.6, 0.8, 0).
Traction t = σn = (72, 82, 0) MPa.
σ_n = n·t = 108.8 MPa.

Assumptions and limits

A symmetric stress tensor with six independent components, all in one set of axes.
The direction is the normal to the plane, in the same axes.

Common mistakes

Entering the direction lying in the plane instead of the direction perpendicular to it.
Leaving out the factor 2 on the shear terms when working by hand. This entry includes it.

Statistics and uncertainty

The normal stress is least sensitive to the plane direction near a principal plane. A 1° tilt (0.0175 rad) changes σ_n by only about (σ₁ − σ₃) × 0.0175² ≈ 0.028 MPa there, but by up to (σ₁ − σ₃) × 0.0175 ≈ 1.6 MPa on planes 45° away from a principal plane, 57 times more.

Where it is used

Normal stress across a fracture, weld or bonded interface in a full three-dimensional state.
Checking a principal-stress calculation by evaluating the stress on the principal plane.

Maths you need

Direction cosines: the components of a unit vector
The stress tensor: six independent components

Related formulas

Practise this formula