UNDERSTAND IT. WORK IT OUT.

Learn: Maximum in-plane shear stress

The stresses on a tiny element change when you rotate the axes, even though its physical loading stays the same. Mohr’s circle turns the three plane-stress components into a centre and a radius so you can identify the important extremes.

Beginner-friendlyFree · No accountTwo worked examples + separate practice
01

What the formula is saying

Principal directions have zero in-plane shear; their normal stresses are the mean plus and minus the radius. The maximum in-plane shear equals that radius, obtained from the normal-stress half-difference and shear component.

τmax = √[((σx−σy)/2)²+τxy²]

Read the symbols in plain language

σx
Normal stress x

Normal stress x. The average is the centre of the in-plane Mohr circle.

Pa

One pascal is one newton per square metre. 1 MPa = 10⁶ Pa.

σy
Normal stress y

Normal stress y. The average is the centre of the in-plane Mohr circle.

Pa

One pascal is one newton per square metre. 1 MPa = 10⁶ Pa.

τxy
Shear stress

Shear stress. Use the square root of the sum of squares; the radius is nonnegative.

Pa

One pascal is one newton per square metre. 1 MPa = 10⁶ Pa.

τmax
Result to find

Maximum in-plane shear stress. The radius is the largest shear stress reachable by rotating axes within this plane.

Pa

Sort out the units first

Use the same stress unit for σx, σy and τxy; tension is positive and compression negative. Every intermediate and final stress is in Pa after conversion, and squared stresses are square-rooted back to stress units.

Assumptions before calculating

The inputs represent a symmetric plane-stress tensor at one point, with out-of-plane normal stress and shear taken as zero. This is a stress transformation, not a material failure criterion.

02

Let’s solve one together

Read the given values, follow each operation, then check what the result means.

Read the supplied values as one complete study case. Find τmax and explain the result in the stated output unit.

σx · Normal stress x
10000000 Pa
σy · Normal stress y
3000000 Pa
τxy · Shear stress
2000000 Pa
  1. Find the mean normal stress

    The average is the centre of the in-plane Mohr circle.

    ((10000000) + (3000000)) ÷ 2 = 6500000 Pa
  2. Find half the normal-stress difference

    The departure from the mean is combined with the original shear stress.

    ((10000000)-(3000000)) ÷ 2 = 3500000 Pa
  3. Find the Mohr-circle radius

    Use the square root of the sum of squares; the radius is nonnegative.

    √((3500000)^2 + (2000000)^2) ≈ 4031128.874 Pa
  4. Report maximum in-plane shear

    The radius is the largest shear stress reachable by rotating axes within this plane.

    (4031128.874) ≈ 4031128.874 Pa
Answer4031128.874 Pa

Does this worked answer make sense?

For pure shear, the in-plane principal stresses are +|τ| and −|τ|. Changing only the sign of τxy changes orientation but not the two principal values or the radius.

A second worked example — different values

A second case uses different data. Predict which way the answer will change, then calculate it without reusing the first answer.

σx · Normal stress x
0 Pa
σy · Normal stress y
0 Pa
τxy · Shear stress
5000000 Pa
  1. Find the mean normal stress

    The average is the centre of the in-plane Mohr circle.

    ((0) + (0)) ÷ 2 = 0 Pa
  2. Find half the normal-stress difference

    The departure from the mean is combined with the original shear stress.

    ((0)-(0)) ÷ 2 = 0 Pa
  3. Find the Mohr-circle radius

    Use the square root of the sum of squares; the radius is nonnegative.

    √((0)^2 + (5000000)^2) = 5000000 Pa
  4. Report maximum in-plane shear

    The radius is the largest shear stress reachable by rotating axes within this plane.

    (5000000) = 5000000 Pa
Answer5000000 Pa
03

Now try your own values

Change a value or its unit. The same method will show your calculation, step by step.

Normal stress x. The average is the centre of the in-plane Mohr circle.

Normal stress y. The average is the centre of the in-plane Mohr circle.

Shear stress. Use the square root of the sum of squares; the radius is nonnegative.

English, Arabic and Persian digits are supported. The steps convert inputs to the formula’s base units.

Results update only when you calculate. The lesson example above stays unchanged.

04

Your turn — check your understanding

Solve this separate case yourself. Use only the values below; the two worked examples use different data. Give the requested result in the selected unit.

σx · Normal stress x
8000000 Pa
σy · Normal stress y
-2000000 Pa
τxy · Shear stress
3000000 Pa

Find: Learn: Maximum in-plane shear stress

For repeating decimals, use at least four significant figures. Accepted rounding tolerance: 0.05% of the expected value; zero uses an absolute tolerance of 10⁻¹².

A hint, not the answer

Principal directions have zero in-plane shear; their normal stresses are the mean plus and minus the radius. The maximum in-plane shear equals that radius, obtained from the normal-stress half-difference and shear component.

Use the same stress unit for σx, σy and τxy; tension is positive and compression negative. Every intermediate and final stress is in Pa after conversion, and squared stresses are square-rooted back to stress units.

Show the full practice solution

Compare the steps with your work; revealing a solution does not mark the lesson complete.

  1. Find the mean normal stress

    The average is the centre of the in-plane Mohr circle.

    ((8000000) + (-2000000)) ÷ 2 = 3000000 Pa
  2. Find half the normal-stress difference

    The departure from the mean is combined with the original shear stress.

    ((8000000)-(-2000000)) ÷ 2 = 5000000 Pa
  3. Find the Mohr-circle radius

    Use the square root of the sum of squares; the radius is nonnegative.

    √((5000000)^2 + (3000000)^2) ≈ 5830951.895 Pa
  4. Report maximum in-plane shear

    The radius is the largest shear stress reachable by rotating axes within this plane.

    (5830951.895) ≈ 5830951.895 Pa
Answer5830951.895 Pa

Avoid the common trap

Do not omit the factor 1/2 from the normal-stress difference. Square shear before adding it. Do not confuse maximum in-plane shear with the absolute maximum shear in three dimensions.

When this method applies — and when it does not

The two returned principal values are in-plane values. For absolute three-dimensional maximum shear, include the third principal stress 0 and use half the difference between the largest and smallest of all three; the in-plane radius need not equal that value.

For study and understanding, not approval of a real structure, site operation or design. Apply the correct standard, National Annex and professional review to actual engineering work.

One idea understood. Keep going.

This optional checkmark is saved only in this browser. It is your own progress note, not a certificate.

Sources & further reading

References open in a new tab and explain the underlying principles. The teaching text and examples here are SimpleFlick’s own; the source organisations have not endorsed this calculator.

Reading focus: Maximum in-plane shear stress. Use the relevant elastic-response, equilibrium, constitutive-relations or shear-and-torsion module; the lesson states its particular sign convention.

Lesson updated: · Both examples and the separate practice case are checked against an independent high-precision numerical implementation. This verifies arithmetic for the stated model, not engineering certification.

Menu