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Mechanical

Absolute maximum shear stress in plane stress

Largest shear stress over all planes through the point, taken from the biggest and smallest of the three principal stresses.

Ref: MEC-55
Rev: v1
Field: Mechanical
Result: τ_abs [MPa]

τ_{abs} = \frac{σ_{1} - σ_{3}}{2} = \frac{\max (σ_{a}, σ_{b}, 0) - \min (σ_{a}, σ_{b}, 0)}{2}

Mohr’s circles can be drawn between every pair of the three principal stresses, and the biggest of them has diameter σ₁ − σ₃. So the largest shear stress on any plane is τ_abs = (σ₁ − σ₃)/2. In plane stress, where the zero out-of-plane stress is one of the three, this equals the in-plane radius R only if the two in-plane principal stresses have opposite signs. If both are tensile, τ_abs = σ_a/2 (the larger one over two); if both are compressive, τ_abs = |σ_b|/2. Twice this value is the Tresca equivalent stress.

Largest shear stress over all planes through the point, taken from the biggest and smallest of the three principal stresses. It equals the Mohr circle radius only when the in-plane principal stresses have opposite signs.

Calculator

Solve for

τ_absAbsolute maximum 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

Mohr circle radius R (in-plane maximum shear)50 MPa
Tresca equivalent stress, 2τ_abs100 MPa
Difference τ_abs − R0 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.

Absolute 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)

41.3153

The lecture state gives 50 MPa, the same as the circle radius 50 MPa, because the in-plane values 70 and −30 MPa have opposite signs.
With both in-plane values tensile the two differ. Equal biaxial tension of 100 MPa has R = 0 MPa but τ_abs = 50 MPa. A state of 100, 20 and 30 MPa shear has R = 50 MPa but τ_abs = 55 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	41.26	153.1	111.9	+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)	Absolute 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

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

Variables and how each one contributes

τ_abs

Absolute 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. Raises τ_abs while σ_x is above σ_y, through the gap σ_x − σ_y that sets the circle radius. If both principal stresses are tensile, the larger one alone sets τ_abs. 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. Raising σ_y towards σ_x shrinks the circle and lowers τ_abs, until the zero out-of-plane stress takes over and keeps it at half the larger principal stress. 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 τ_abs whichever way the shear acts, because it only enters the radius 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

Principal stresses with the zero included: σ₁ = 70 MPa, σ₃ = −30 MPa.
Difference: σ₁ − σ₃ = 70 − (−30) = 100 MPa.
τ_abs = 100/2 = 50 MPa.

Assumptions and limits

Plane stress, so one principal stress is exactly zero.
Principal stresses ordered algebraically, with the zero included.

Common mistakes

Using the in-plane Mohr radius as the maximum shear when both in-plane values have the same sign.
Forgetting to double it again when comparing with the uniaxial yield strength: Tresca yields at τ_abs = S_y/2.

Statistics and uncertainty

τ_abs is the difference of two principal stresses, so it inherits the errors of both. If each principal stress has an independent error of ±3 MPa, the difference has about ±4.2 MPa and τ_abs about ±2.1 MPa. A systematic offset applied equally to all three principal stresses cancels completely.

Where it is used

Shear checks in biaxially loaded plates and pressure vessels.
Reading the Tresca equivalent stress as twice the largest shear stress.

Maths you need

Principal stresses sorted: σ₁ ≥ σ₂ ≥ σ₃
Mohr’s circle: one circle for each pair of principal stresses

Related formulas

Practise this formula