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Mechanical

Shear stress magnitude on any plane (three dimensions)

Size of the shear stress on a plane with unit normal n.

Ref: MEC-59
Rev: v1
Field: Mechanical
Result: |τ_n| [MPa]

| τ_{n} | = \sqrt{{t_{x}}^{2} + {t_{y}}^{2} + {t_{z}}^{2} - {σ_{n}}^{2}} t = σ n

The traction t = σn on a plane splits into a part along the normal, σ_nn, and a part in the plane, the shear stress vector τ_n = t − σ_nn. By Pythagoras its size is |τ_n| = √(tᵀt − σ_n²). Only the magnitude is given: the direction of the shear within the plane is a separate calculation. In a spreadsheet or code a plane with no shear can leave a tiny negative number under the root from round-off, which is set to zero. The direction can have any length and is scaled to a unit vector.

Size of the shear stress on a plane with unit normal n. The traction vector t is split into the part along the normal and the rest, and the rest is the shear; its direction in the plane is not given by the magnitude.

Calculator

Solve for

|τ_n|Shear stress magnitude 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

Normal stress σ_n on the plane108.8 MPa
Traction magnitude |t|109.1 MPa
Largest shear over all planes, (σ₁ − σ₃)/245 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.

Shear stress magnitude on the plane (MPa) current value 8.4 MPa

σ_yNormal stress in yKey driver

1.2143

σ_xNormal stress in x

2126

τ_yzShear stress yz

120

τ_xzShear stress xz

90.4

τ_xyShear stress xy

042

n_zNormal direction component z

44.8

n_xNormal direction component x

030

n_yNormal direction component y

030

σ_zNormal stress in z

The plane with normal (0.6, 0.8, 0) carries 8.4 MPa of shear, although it lies near a principal plane. On the principal plane with normal (1, 1, 0)/√2 the shear is 0 MPa, as it must be.
The largest shear over every plane is (σ₁ − σ₃)/2 = 45 MPa, on planes at 45° between the first and third principal directions. A plane with normal along x carries 30 MPa, the size of τ_xy when τ_xz is zero.

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	1.2	142.8	141.6	+10 MPa	−57.1 %
σ_x Normal stress in x	−200 to 300 MPa	2	126	124	+10 MPa	+57.1 %
τ_yz Shear stress yz	−150 to 150 MPa	8.4	120.3	111.9	+6 MPa	+15.2 %
τ_xz Shear stress xz	−150 to 150 MPa	8.4	90.39	81.99	+6 MPa	+8.8 %
τ_xy Shear stress xy	−150 to 150 MPa	0	42	42	+6 MPa	+20 %
n_z Normal direction component z	−1 to 1	8.4	44.8	36.4	+0.04	+8.47 %
n_x Normal direction component x	−1 to 1	0	30	30	+0.04	−21.6 %
n_y Normal direction component y	−1 to 1	0	30	30	+0.04	+15.8 %
σ_z Normal stress in z	−200 to 300 MPa	8.4	8.4	0	+10 MPa	0 %

Sweep of Shear stress xy

Shear stress xy (MPa)	Shear stress magnitude on the plane (MPa)
−150	42
−112.5	31.5
−75	21
−37.5	10.5
0	0
37.5	10.5
75	21
112.5	31.5
150	42

Sweep one variable

Vary

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

Variables and how each one contributes

|τ_n|

Shear stress magnitude 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. Changes the traction in the x direction, and so the shear, by an amount that depends on the plane’s direction. Raising it by 10 MPa (to 90 MPa) moves the result by +57.1 %, from 8.4 to 13.2 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. Changes the traction in the y direction, and so the shear, by an amount that depends on the plane’s direction. Raising it by 10 MPa (to 90 MPa) moves the result by −57.1 %, from 8.4 to 3.6 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. Changes the traction in the z direction, and so the shear, by an amount that depends on the plane’s direction. Raising it by 10 MPa (to 30 MPa) moves the result by +0 %, from 8.4 to 8.4 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. Changes the traction, and so the shear, through the product of the x and y direction components. Raising it by 10 MPa (to 40 MPa) moves the result by +33.3 %, from 8.4 to 11.2 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. Changes the shear only if the plane’s normal has a y or z component. Raising it by 10 MPa (to 10 MPa) moves the result by +38.1 %, from 8.4 to 11.6 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. Changes the shear only if the plane’s normal has an x or z component. Raising it by 10 MPa (to 10 MPa) moves the result by +22.9 %, from 8.4 to 10.32 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 shear through the direction cosines. Raising it by 0.1 (to 0.7) moves the result by −52.6 %, from 8.4 to 3.982 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 shear through the direction cosines. Raising it by 0.1 (to 0.9) moves the result by +37.4 %, from 8.4 to 11.54 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 shear through the direction cosines. Raising it by 0.1 (to 0.1) moves the result by +44.4 %, from 8.4 to 12.13 MPa.

Worked example with the values above

Traction t = σn = (72, 82, 0) MPa, so |t|² = 11,910.
Normal part: σ_n = 108.8 MPa, so σ_n² = 11,840.
|τ_n| = √(11,910 − 11,840) = 8.4 MPa.

Assumptions and limits

A symmetric stress tensor, six components in one set of axes.
The result is a magnitude. It does not say in which direction the shear acts in the plane.

Common mistakes

Taking the magnitude as a signed shear on a chosen direction.
Using τ_xy as the shear on every plane. It is the shear on the x face only.

Statistics and uncertainty

The shear on a plane is zero at a principal plane and varies linearly with a small tilt there, by about (σ₁ − σ₃) × tilt. A 1° error in locating a fracture plane therefore changes the computed shear by up to about 1.6 MPa for a 90 MPa stress spread, a substantial fraction of a small shear.

Where it is used

Shear on a fracture plane or a bonded interface, for a friction or bond check.
Verifying that principal planes carry no shear.

Maths you need

Vectors: dot products and lengths
Normal stress on a plane: the part of the traction along the normal

Related formulas

Practise this formula