Mohr’s Circle
Step-by-Step Construction — Principal Stresses, Max Shear Stress & Stress Transformation
Last Updated: March 2026
📌 Key Takeaways
- Mohr’s circle is a graphical method for finding principal stresses, maximum shear stress, and stresses on any inclined plane.
- Principal stresses: σ₁, σ₂ = (σx+σy)/2 ± √[((σx−σy)/2)² + τxy²]
- Maximum shear stress: τmax = (σ₁ − σ₂)/2 = radius of Mohr’s circle.
- Principal planes have zero shear stress. Maximum shear planes are at 45° to principal planes.
- On the circle: the horizontal axis is normal stress (σ), the vertical axis is shear stress (τ).
- Mohr’s circle is one of the most frequently tested topics in GATE ME — 1–2 questions almost every year.
1. What is Mohr’s Circle?
When a structural element is subjected to combined loading (for example, a shaft under both bending and torsion), the stresses on different planes through a point vary. Mohr’s circle is a graphical technique that shows all possible combinations of normal stress (σ) and shear stress (τ) on planes passing through a point, plotted as a circle on the σ-τ plane.
It was developed by German engineer Christian Otto Mohr in 1882 and remains the most intuitive and rapid method for solving 2D stress transformation problems. While the same results can be obtained algebraically using transformation equations, Mohr’s circle provides visual insight and is faster for exam problems.
2. Stress Transformation Formulas
Given a state of stress (σx, σy, τxy) on an element, the stresses on a plane inclined at angle θ from the x-axis are:
Transformation Equations
σθ = (σx+σy)/2 + (σx−σy)/2 × cos2θ + τxysin2θ
τθ = −(σx−σy)/2 × sin2θ + τxycos2θ
Principal Stresses
σ₁, σ₂ = (σx+σy)/2 ± √[((σx−σy)/2)² + τxy²]
σ₁ = maximum principal stress (with + sign)
σ₂ = minimum principal stress (with − sign)
Maximum In-Plane Shear Stress
τmax = √[((σx−σy)/2)² + τxy²] = (σ₁ − σ₂)/2
This equals the radius of Mohr’s circle.
Principal Plane Orientation
tan(2θp) = 2τxy / (σx − σy)
θp gives two values 90° apart — the orientations of the two principal planes.
Maximum shear planes are at 45° to principal planes.
3. Step-by-Step Construction
Given: σx, σy, τxy at a point.
- Set up axes: Horizontal axis = normal stress σ (positive to the right). Vertical axis = shear stress τ (positive downward in some conventions, positive upward in others — be consistent).
- Plot point A: Represents the x-face → coordinates (σx, τxy).
- Plot point B: Represents the y-face → coordinates (σy, −τxy). Note: shear on the y-face has opposite sign.
- Find the centre C: C is the midpoint of line AB on the σ-axis → C = ((σx+σy)/2, 0).
- Draw the circle: The circle passes through both A and B, with centre C. The radius R = distance from C to A = √[((σx−σy)/2)² + τxy²].
- Read results: The rightmost point gives σ₁ = C + R. The leftmost point gives σ₂ = C − R. The topmost/bottommost points give τmax = R.
Quick Formulas from the Circle
Centre: C = (σx + σy)/2
Radius: R = √[((σx − σy)/2)² + τxy²]
σ₁ = C + R, σ₂ = C − R
τmax = R
4. Reading Results from the Circle
| What You Want | Where on Mohr’s Circle |
|---|---|
| Maximum principal stress σ₁ | Rightmost point of the circle |
| Minimum principal stress σ₂ | Leftmost point of the circle |
| Maximum shear stress τmax | Top (or bottom) of the circle = radius R |
| Normal stress on max shear plane | σavg = C (at top/bottom of circle) |
| Principal plane angle (2θp) | Angle from point A to the σ-axis (rightmost point), measured on the circle |
| Stresses on any inclined plane (angle θ) | Rotate 2θ from point A on the circle — the new point gives (σθ, τθ) |
Important: Angles on Mohr’s circle are doubled. A physical rotation of θ corresponds to 2θ on the circle. Principal planes 90° apart in reality are 180° apart on the circle (diametrically opposite).
5. Special Cases
| State of Stress | Mohr’s Circle | Key Result |
|---|---|---|
| Uniaxial tension (σx only) | Circle passes through origin and (σx, 0) | τmax = σx/2 at 45° |
| Pure shear (τxy only) | Circle centred at origin | σ₁ = τxy, σ₂ = −τxy, at 45° to shear planes |
| Biaxial, equal tension (σx = σy) | Circle shrinks to a point | τmax = 0 (hydrostatic state in 2D) |
| Biaxial, equal and opposite (σx = −σy) | Circle centred at origin | Same as pure shear rotated 45° |
6. Worked Numerical Examples
Example 1: Find Principal Stresses
Problem: At a point in a stressed body: σx = 80 MPa, σy = −40 MPa, τxy = 30 MPa. Find the principal stresses and maximum shear stress.
Solution
C = (80 + (−40))/2 = 20 MPa
R = √[((80−(−40))/2)² + 30²] = √[(60)² + 30²] = √[3600 + 900] = √4500 = 67.08 MPa
σ₁ = C + R = 20 + 67.08 = 87.08 MPa
σ₂ = C − R = 20 − 67.08 = −47.08 MPa
τmax = R = 67.08 MPa
Principal plane: tan(2θp) = 2(30)/(80−(−40)) = 60/120 = 0.5 → 2θp = 26.57° → θp = 13.28°
Example 2: Pure Shear
Problem: A shaft surface has σx = 0, σy = 0, τxy = 50 MPa (pure torsion). Find the principal stresses.
Solution
C = 0, R = √[0 + 50²] = 50 MPa
σ₁ = +50 MPa, σ₂ = −50 MPa (equal and opposite)
τmax = 50 MPa
θp = 45° — principal stresses act at 45° to the shaft axis.
This is why torsional failures in brittle materials (like chalk) produce a 45° helical fracture surface.
Example 3: Combined Bending and Torsion
Problem: A shaft has bending stress σb = 100 MPa and shear stress (from torsion) τ = 60 MPa at the outer surface. Find the principal stresses.
Solution
At the critical point: σx = 100 MPa, σy = 0, τxy = 60 MPa
C = 100/2 = 50 MPa
R = √[(50)² + 60²] = √[2500 + 3600] = √6100 = 78.1 MPa
σ₁ = 50 + 78.1 = 128.1 MPa
σ₂ = 50 − 78.1 = −28.1 MPa
τmax = 78.1 MPa
7. Common Mistakes Students Make
- Confusing physical angle and circle angle: A rotation of θ on the physical element corresponds to 2θ on Mohr’s circle. This 2:1 relationship is the most common source of errors.
- Wrong sign for shear on the y-face: When plotting the two points on Mohr’s circle, the shear stress on the y-face has the opposite sign from the x-face. Point A = (σx, τxy), Point B = (σy, −τxy).
- Forgetting the normal stress on the maximum shear plane: The plane of maximum shear stress also carries a normal stress equal to σavg = (σ₁+σ₂)/2. It is NOT a plane of pure shear (unless σavg = 0).
- Using in-plane τmax when absolute τmax is needed: For a 3D stress state (σ₁, σ₂, σ₃), the absolute maximum shear stress is (σmax − σmin)/2, which may involve the out-of-plane stress σ₃ = 0.
- Not labelling signs correctly: Tensile stresses are positive, compressive are negative. Getting the signs wrong at the start propagates through the entire solution.
8. Frequently Asked Questions
What is Mohr’s circle used for?
Mohr’s circle is used to find principal stresses, maximum shear stress, and the stresses on any inclined plane through a point in a stressed body. It converts the algebraic stress transformation equations into a simple graphical method. It is essential for analysing combined loading (bending + torsion, axial + shear) and for applying failure theories.
What are principal stresses?
Principal stresses (σ₁, σ₂) are the maximum and minimum normal stresses that occur on planes where the shear stress is zero. These planes are called principal planes. Principal stresses are the most important values for failure analysis — failure theories compare principal stresses against material strength to predict whether a component will fail.