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Mechanical

Maximum shear stress (plane stress)

Radius of Mohr’s circle: the largest shear stress on any plane perpendicular to the plane of the stresses.

Ref: MEC-37
Rev: v1
Field: Mechanical
Result: τ_max [MPa]

τ_{\max} = \sqrt{{(\frac{σ_{x} - σ_{y}}{2})}^{2} + {τ_{xy}}^{2}}

R is the radius of Mohr’s circle, drawn about the centre C = (σ_x + σ_y)/2. It is the largest shear stress on any plane perpendicular to the plane of the stresses, and it acts on faces turned 45° from the principal directions. It is the hypotenuse of a right triangle whose sides are the half-difference of the normal stresses and the shear stress. Step 2 of the stress route: the principal stresses are the ends of the horizontal diameter, C ± R. When both in-plane principal stresses have the same sign, the true maximum shear over all planes is larger than R; use the absolute maximum shear stress entry for that.

Radius of Mohr’s circle: the largest shear stress on any plane perpendicular to the plane of the stresses. It is the middle step between the circle centre and the principal stresses.

Calculator

Solve for

τ_maxMaximum shear 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

Centre of Mohr’s circle, (σ_x + σ_y)/220 MPa
Larger in-plane principal stress σ_a = C + R70 MPa
Smaller in-plane principal stress σ_b = C − R−30 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.

Maximum shear stress (MPa) current value 50 MPa

τ_xyShear stress on the x face in the y directionKey driver

30153

σ_xNormal stress in x (tension positive)

40.1160

σ_yNormal stress in y (tension positive)

40131

The lecture state gives a 30-40-50 right triangle: half-difference 30, shear 40, radius 50 MPa. With centre 20 MPa the principal stresses are 20 ± 50, that is 70 and −30 MPa.
R is the largest shear stress in the plane of the stresses, but not always the largest over all planes. Here the third principal stress is zero and lies between the two in-plane values, so R = 50 MPa is also the absolute maximum shear. With σ_x = σ_y = 100 MPa and no shear, R is zero, yet the largest shear over all planes is 50 MPa.

Chart data as tables

Ranked variables

Variable	Typical range	Lowest result	Highest result	Swing (MPa)	Local step	Change in result
τ_xy Shear stress on the x face in the y direction	−150 to 150 MPa	30	153	123	+6 MPa	+9.84 %
σ_x Normal stress in x (tension positive)	−200 to 300 MPa	40.14	160.1	119.9	+10 MPa	+6.3 %
σ_y Normal stress in y (tension positive)	−200 to 300 MPa	40	131.2	91.24	+10 MPa	−5.66 %

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

Shear stress on the x face in the y direction (MPa)	Maximum shear stress (MPa)
−150	153
−112.5	116.4
−75	80.78
−37.5	48.02
0	30
37.5	48.02
75	80.78
112.5	116.4
150	153

Sweep one variable

Vary

Maximum shear stress (MPa)Hover, touch, or use the arrow keys on the chart

Variables and how each one contributes

τ_max

Maximum shear 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. Changes R only through the difference σ_x − σ_y, so the effect depends on which stress is larger. Raising it by 10 MPa (to 60 MPa) moves the result by +6.3 %, from 50 to 53.15 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. Changes R only through the difference σ_x − σ_y, so raising σ_y towards σ_x shrinks the circle. Raising it by 10 MPa (to 0 MPa) moves the result by −5.66 %, from 50 to 47.17 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 R in quadrature with the half-difference. The sign of the shear does not matter here, because it is squared. Raising it by 10 MPa (to 50 MPa) moves the result by +16.6 %, from 50 to 58.31 MPa.

Worked example with the values above

Half the difference: (σ_x − σ_y)/2 = (50 − (−10))/2 = 30 MPa.
Squares added: 900 + 1,600 = 2,500.
R = √2,500 = 50 MPa.

Assumptions and limits

Plane stress: the stress through the thickness is zero.
All stresses act at one point, with the same sign convention throughout.

Common mistakes

Calling R the maximum shear stress when both principal stresses are tensile or both are compressive. The larger value is σ₁/2 or |σ₃|/2, which comes from the zero out-of-plane stress.
Dropping the sign of a compressive stress inside the difference σ_x − σ_y.

Statistics and uncertainty

R is a root-sum-square of two terms, so the larger term dominates and errors in the smaller one barely register. In the defaults the shear stress (40 MPa) dominates the half-difference (30 MPa). Stresses found from strain gauges carry the gauge angle error, which matters most when the shear term is small.

Where it is used

Drawing Mohr’s circle by hand and checking it.
Finding the principal stresses as C ± R, before a Tresca or von Mises check.

Maths you need

Pythagoras: √(a² + b²)
Normal and shear stress: perpendicular and parallel to a face

Related formulas

Practise this formula