Mohr’s circle & stress transformation: practice problems with step-by-step solutions

Also called: principal stresses · plane stress

Free Mohr’s circle & stress transformation practice problems with diagrams, hints and full worked solutions. New values every time, at three levels: principal stresses, maximum shear and principal direction, stress on an inclined plane.

Free practice problemEasy
σx​ = 90σy​ = −20MPaprincipal element

At a point in a loaded part the in-plane principal stresses are σ₁ = 90 MPa and σ₂ = −20 MPa (tension positive). Find the normal stress σavg that acts on the planes of maximum in-plane shear stress.

σ1 = 90 MPaσ2 = −20 MPa
MPa
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What you practise

  1. Easyprincipal stresses
  2. Mediummaximum shear and principal direction
  3. Hardstress on an inclined plane

How to solve Mohr’s circle & stress transformation problems

  1. Calculate the centre of Mohr’s circle: C = (σx + σy)/2.
  2. Calculate the radius: R = √(((σx − σy)/2)² + τxy²).
  3. The principal stresses are C ± R and the largest in-plane shear stress is R.
  4. Angles on the circle are double the angles on the element: tan 2θp = 2τxy/(σx − σy).

Common mistakes

  • Using θ instead of 2θ in the transformation equations.
  • Adding (σx − σy)/2 and τxy instead of combining them with Pythagoras.
  • Getting the sign convention for shear stress wrong.

Key formulas

Centre
C = (σx + σy)/2
Radius
R = √(((σx − σy)/2)² + τxy²)
Principal stresses
σ1,2 = C ± R
Principal direction
tan 2θp = 2τxy/(σx − σy)
Inclined plane
σθ = C + (σx − σy)/2·cos 2θ + τxy·sin 2θ

Learn Mohr’s circle & stress transformation from scratch

Before this: Axial stress & strain

1Why stress depends on direction

In the axial lesson, stress was one number: force over area. But the stress at a point depends on the direction of the cut you look at.

Twist a piece of chalk until it breaks. It breaks along a spiral at 45°, not straight across. On that 45° plane the tension is largest, and chalk is weak in tension.

To describe the stress at one point, engineers draw a tiny square, the stress element, with the stresses on its faces:

σx​σy​τxy​xy
  • σx and σy are normal stresses, at right angles to the faces. Positive means tension (arrows pointing out), negative means compression.
  • τxy is the shear stress, along the faces. It tries to turn the square into a diamond.

Rotate the square, and the stresses on its faces change. This lesson shows how to find them for every direction, and which directions are the most dangerous.

Self-check At a point, σx = −40 MPa. What does this mean?
  1. A compression of 40 MPa in the x-direction
  2. A compression of 40 MPa in the y-direction
  3. A shear stress of 40 MPa
  4. A tension of 40 MPa in the x-direction

Answer: A compression of 40 MPa in the x-direction. With the usual sign rule tension is positive, so a negative σx is compression, acting on the faces normal to x.

2Mohr’s circle

Mohr’s circle is a picture of all the directions at once. Every point on the circle is one possible cut through the point in the part.

Draw normal stress σ to the right and shear stress τ upwards. Then plot two points:

  • point x at (σx, τxy): the stresses on the face normal to x,
  • point y at (σy, −τxy): the stresses on the face normal to y.

The circle goes through both points. Its centre C lies on the σ-axis, halfway between them. Its radius R follows from Pythagoras:

C = (σx + σy) / 2 R = √( ((σx − σy)/2)² + τxy² )
στxyCσ1​σ2​R

Where the circle crosses the σ-axis, the shear stress is zero. These two values are the principal stresses, the largest and smallest normal stress at the point. The top of the circle is the largest in-plane shear stress:

σ1 = C + R σ2 = C − R τmax = R
Self-check σx = 50 MPa, σy = 10 MPa and τxy = 15 MPa. What is σ1?
  1. 45 MPa
  2. 50 MPa
  3. 55 MPa
  4. 65 MPa

Answer: 55 MPa. C = 30 MPa and R = √(20² + 15²) = 25 MPa, so σ1 = 55 MPa. 65 adds 20 + 15 instead of using Pythagoras; 50 ignores the shear stress.

3Angles: double on the circle

Rotating the element by an angle θ moves its point on Mohr’s circle by 2θ, in the same direction. Counter-clockwise is positive.

The direction of σ1 is the angle θp from the x-axis, found from:

tan 2θp = 2τxy / (σx − σy)

Mind the quadrant. If σx − σy is negative, add 180° to 2θp when τxy > 0, or subtract 180° when τxy < 0. Then halve it.

The planes of maximum shear lie 45° away from the principal directions (90° on the circle).

For a plane whose normal is rotated θ from the x-axis, the stresses are:

σθ = C + (σx − σy)/2 · cos 2θ + τxy · sin 2θ
τθ = −(σx − σy)/2 · sin 2θ + τxy · cos 2θ
Self-check At a point, tan 2θp = 1 and σx − σy is positive. What is θp?
  1. 22.5°
  2. 45°
  3. 90°
  4. −67.5°

Answer: 22.5°. 2θp = 45°, so θp = 22.5°. Answering 45° forgets to halve the angle from the circle. −67.5° is the direction of σ2, 90° away.

4The method step by step

  1. Write down σx, σy and τxy with their signs: tension positive, compression negative.
  2. Find the centre C = (σx + σy)/2.
  3. Find the radius R = √(((σx − σy)/2)² + τxy²).
  4. Principal stresses: σ1 = C + R and σ2 = C − R. Largest in-plane shear: τmax = R.
  5. Directions: tan 2θp = 2τxy/(σx − σy), check the quadrant, then halve.
  6. For a given plane θ: use the σθ and τθ equations with 2θ.

A quick sketch of the circle, even a rough one, shows you the signs and which point is σ1.

5A worked example

Follow every step. Try to predict the next one before you open it.

Worked example

A point has σx = 80 MPa, σy = 20 MPa and τxy = 40 MPa. Find the principal stresses, τmax, the direction θp of σ1, and the normal stress on a plane rotated 45° counter-clockwise.

1

Look at the element

xy

Both normal stresses are tension. The shear stress is positive.

2

Centre and radius

C = (80 + 20) / 2 = 50 MPa
R = √(30² + 40²) = 50 MPa
3

Principal stresses and maximum shear

στxyCσ1​σ2​τmax​ = R
σ1 = 50 + 50 = 100 MPa σ2 = 50 − 50 = 0 MPa
τmax = R = 50 MPa
4

Principal direction

tan 2θp = 2 · 40 / (80 − 20) = 1.333

σx − σy is positive, so no correction is needed: 2θp = 53.1° and θp = 26.6° counter-clockwise from x.

5

Plane rotated 45°

With θ = 45°, 2θ = 90°: cos 2θ = 0 and sin 2θ = 1.

σθ = 50 + 30 · 0 + 40 · 1 = 90 MPa
6

Check the answer

The sum σx + σy = 100 MPa equals σ1 + σ2 = 100 MPa. This sum is the same for every rotation, so it is a quick check.

Also, 90 MPa lies between σ2 = 0 and σ1 = 100 MPa, as every normal stress must.

Answer

σ1 = 100 MPa, σ2 = 0, τmax = 50 MPa, θp = 26.6° and σ45° = 90 MPa.

6Common mistakes

  • Adding (σx − σy)/2 and τxy directly. The radius combines them with Pythagoras.
  • Using θ instead of 2θ in the transformation equations, or forgetting to halve the angle read from the circle.
  • Dropping a minus sign: a compressive stress must go in as a negative number.
  • Plotting point y at (σy, +τxy). It is at −τxy, on the opposite side of the centre from x.
  • Missing the quadrant of 2θp when σx < σy.
Self-check σx = 60 MPa, σy = 0 and τxy = 40 MPa. What is the maximum in-plane shear stress?
  1. 30 MPa
  2. 40 MPa
  3. 50 MPa
  4. 70 MPa

Answer: 50 MPa. τmax = R = √(30² + 40²) = 50 MPa. 70 adds 30 + 40; 30 ignores the shear stress; 40 is only τxy.

7How exam questions look

  • Easy: principal stresses. Given σx, σy and τxy, find σ1 or σ2. Variants: pure shear, or an axial bar cut at an angle.
  • Medium: maximum shear and principal direction. Find τmax = R or the angle θp, between −90° and 90°.
  • Hard: stress on an inclined plane. Use the σθ or τθ equation with 2θ. Variants combine axial load and torsion, or use the principal stresses in a failure check such as von Mises or Tresca.

Write C and R down first. Almost every question in this topic is answered from those two numbers.

Learn it step by step in the appLearn it step by step in the app

Worked examples

Example 1Easy
σx​ = 30σy​ = 50τxy​MPa

A point has σₓ = 30 MPa and σᵧ = 50 MPa, plus some shear stress. On an element rotated by 30° the normal stress in the rotated x′ direction is 87 MPa. Find the normal stress σy′ in the rotated y′ direction.

σx = 30 MPaσy = 50 MPaσx′ = 87 MPaθ = 30°

Find σy′ (MPa)

Show hints
  1. You do not need τₓᵧ or the transformation equations.
  2. The sum of the normal stresses on two perpendicular planes is the same for every orientation (the centre of Mohr’s circle does not move).
  3. σₓ + σᵧ = 80 MPa.
Show the answer and the step-by-step solution

Plane stress transformation: on a plane rotated by θ, σ = (σₓ + σᵧ)/2 + (σₓ − σᵧ)/2·cos 2θ + τₓᵧ·sin 2θ. Mohr’s circle shows all planes at once: its centre is at C = (σₓ + σᵧ)/2 and its radius is R = √(((σₓ − σᵧ)/2)² + τₓᵧ²). The principal stresses are C ± R, and the largest in-plane shear stress is R.

First stress invariant

Points x′ and y′ are at opposite ends of a diameter, so their average is always the centre C.

σx′ + σy′ = σx + σy
σy′ = 30 + 50 − 87 = −7 MPa

Answer

σy′ = −7 MPa
Example 2Medium
σx​ = 0σy​ = 20τxy​ = −30MPa

A point in a loaded part has σₓ = 0 MPa, σᵧ = 20 MPa and τₓᵧ = −30 MPa (tension positive). Find the angle θp from the x-axis to the direction of the largest principal stress, between −90° and 90°.

σx = 0 MPaσy = 20 MPaτxy = −30 MPa

Find θp (°)

Show hints
  1. Draw Mohr’s circle: point x at (σₓ, τₓᵧ), point y at (σᵧ, −τₓᵧ), centre C halfway.
  2. tan 2θp = 2τₓᵧ/(σₓ − σᵧ). Mind the quadrant.
  3. C = 10 MPa, R = 31.62 MPa.
Show the answer and the step-by-step solution

Plane stress transformation: on a plane rotated by θ, σ = (σₓ + σᵧ)/2 + (σₓ − σᵧ)/2·cos 2θ + τₓᵧ·sin 2θ. Mohr’s circle shows all planes at once: its centre is at C = (σₓ + σᵧ)/2 and its radius is R = √(((σₓ − σᵧ)/2)² + τₓᵧ²). The principal stresses are C ± R, and the largest in-plane shear stress is R.

στxyCσ1​σ2​

Angle on the circle

Point x sits at an angle 2θp from the σ-axis on Mohr’s circle; the quadrant follows from the signs of σₓ − σᵧ and τₓᵧ.

2θp = atan2(2·−30, 0 − 20) = −108.4°

Angle on the element

θp = −54.22°

Answer

θp = −54.22°

Positive angles are counter-clockwise.

Example 3Hard
60°D = 500 mm, t = 5 mm, p = 2.5 MPap

A thin-walled cylindrical tank of inner diameter 500 mm and wall thickness 5 mm holds gas at a gauge pressure of 2.5 MPa. It is made from plate welded along a helical seam at 60° to the axis of the cylinder. Find the magnitude of the shear stress |τn| acting along the weld seam.

D = 500 mmt = 5 mmp = 2.5 MPaα = 60°

Find |τn| (MPa)

Show hints
  1. Hoop stress σh = pr/t, axial stress σa = pr/(2t). There is no shear on the axial and hoop planes.
  2. Shear along the seam: |τn| = (σh − σa)·sin α·cos α.
  3. σh = 125 MPa, σa = 62.5 MPa.
Show the answer and the step-by-step solution

Plane stress transformation: on a plane rotated by θ, σ = (σₓ + σᵧ)/2 + (σₓ − σᵧ)/2·cos 2θ + τₓᵧ·sin 2θ. Mohr’s circle shows all planes at once: its centre is at C = (σₓ + σᵧ)/2 and its radius is R = √(((σₓ − σᵧ)/2)² + τₓᵧ²). The principal stresses are C ± R, and the largest in-plane shear stress is R. Thin-walled cylinder: hoop stress σh = pr/t and axial stress σa = pr/(2t); these are the principal stresses in the wall.

Wall stresses

r = 500/2 = 250 mm
σh = pr/t = 2.5·250/5 = 125 MPa
σa = pr/(2t) = 62.5 MPa

Transform to the seam

Take x along the axis (σa) and y around the hoop (σh). The seam’s normal makes 90° − α with the axis.

|τn| = (125 − 62.5)·sin 60°·cos 60° = 27.06 MPa

Answer

|τn| = 27.06 MPa

Questions and answers

What is Mohr’s circle & stress transformation?

In the axial lesson, stress was one number: force over area. But the stress at a point depends on the direction of the cut you look at.

Twist a piece of chalk until it breaks. It breaks along a spiral at 45°, not straight across. On that 45° plane the tension is largest, and chalk is weak in tension.

How do I solve Mohr’s circle & stress transformation problems step by step?
  1. Calculate the centre of Mohr’s circle: C = (σx + σy)/2.
  2. Calculate the radius: R = √(((σx − σy)/2)² + τxy²).
  3. The principal stresses are C ± R and the largest in-plane shear stress is R.
  4. Angles on the circle are double the angles on the element: tan 2θp = 2τxy/(σx − σy).
What are common mistakes in Mohr’s circle & stress transformation?
  • Using θ instead of 2θ in the transformation equations.
  • Adding (σx − σy)/2 and τxy instead of combining them with Pythagoras.
  • Getting the sign convention for shear stress wrong.
What do the Mohr’s circle & stress transformation exercises cover?
  • Easy: principal stresses
  • Medium: maximum shear and principal direction
  • Hard: stress on an inclined plane

Every problem has new values, so you learn the method instead of an answer. Stuck? Open a small hint first, and the full explanation only when you need it.

How practising works

  • Every problem has new values, so you learn the method instead of an answer.
  • Stuck? Open a small hint first, and the full explanation only when you need it.
  • Your progress, mistakes and saved problems are kept, so you can review before the exam.

Topics before and after this one

Learn first

Axial stress & strain

Calculators

Mohr’s circle calculator

Courses with this topic

Mechanics of MaterialsMachine DesignFE Mechanical exam practice

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Printable worksheet

Mohr’s circle & stress transformation worksheet with answers
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