Interactive Lecture
Mohr's Circle, Principal Stresses, and Failure Criteria
by Dr Arnaldo Delli Carri
When analysing stress at a point in a material, we represent it using an infinitesimal element. However, this element can be oriented in different directions, and the normal and shear stresses acting on its faces depend on the chosen reference frame. This dependency on orientation is not ideal for understanding the true stress state at a point.
The stress state at a point is invariant, but its representation changes with the coordinate system. We need a way to describe stress that is independent of the reference frame.
Definition: Principal stresses are the normal stresses that act on planes where the shear stress is zero. These represent the maximum and minimum normal stresses at a point.
For 2D stress state:
\(\sigma_{1,2} = \frac{\sigma_{xx} + \sigma_{yy}}{2} \pm \sqrt{\left(\frac{\sigma_{xx} - \sigma_{yy}}{2}\right)^2 + \tau_{xy}^2}\)
where \(\sigma_1 \geq \sigma_2\) by convention.
The principal directions are the orientations of the planes on which these principal stresses act:
\(\tan(2\theta_p) = \frac{2\tau_{xy}}{\sigma_{xx} - \sigma_{yy}}\)
Definition: The maximum in-plane shear stress occurs on planes oriented at 45° to the principal planes.
\(\tau_{max} = \frac{\sigma_1 - \sigma_2}{2} = \sqrt{\left(\frac{\sigma_{xx} - \sigma_{yy}}{2}\right)^2 + \tau_{xy}^2}\)
The orientation of maximum shear planes:
\(\tan(2\theta_s) = -\frac{\sigma_{xx} - \sigma_{yy}}{2\tau_{xy}}\)
Mohr's circle is a graphical representation of the stress transformation equations. It provides a visual way to determine stresses on any plane and to find principal stresses and maximum shear stress.
Explore how stress components change with orientation. The left panel shows the stress element, while the right panel displays the corresponding Mohr's circle.
When testing materials, we typically obtain a single failure stress value (yield stress for ductile materials, ultimate stress for brittle materials) from a uniaxial tensile test. However, real components experience complex multiaxial stress states. Failure criteria allow us to convert these complex stress states into an equivalent stress that can be compared with the material's failure stress.
Definition: The factor of safety (FS) is the ratio of the material's failure stress to the equivalent stress:
\(FS = \frac{\sigma_{failure}}{\sigma_{equivalent}}\)
A component is considered safe if FS > 1 (typically FS ≥ 1.5-3 for design)
Application: Brittle materials
Theory: Failure occurs when the maximum principal stress reaches the material's ultimate tensile or compressive strength.
For tensile failure: \(\sigma_1 = \sigma_{ut}\)
For compressive failure: \(|\sigma_2| = \sigma_{uc}\)
Equivalent stress: \(\sigma_{eq} = \max(\sigma_1, |\sigma_2|)\)
Application: Ductile materials
Theory: Yielding occurs when the largest shear stress across all three principal planes reaches the shear yield strength, \(\tau_{max,abs} = \sigma_y/2\).
In plane stress the out-of-plane principal stress \(\sigma_3 = 0\) is itself one of the three principals, so it enters the comparison:
Equivalent stress: \(\sigma_{eq} = \max\big(|\sigma_1 - \sigma_2|,\ |\sigma_1|,\ |\sigma_2|\big)\)
When \(\sigma_1\) and \(\sigma_2\) have opposite signs the \(|\sigma_1 - \sigma_2|\) term governs and this reduces to \(\sigma_1 - \sigma_2\). When they share a sign, \(\sigma_3 = 0\) sets the largest gap, so \(\sigma_{eq} = \max(|\sigma_1|, |\sigma_2|)\) — using \(\sigma_1 - \sigma_2\) alone would then under-predict the shear and overstate safety.
Application: Ductile materials
Theory: Yielding occurs when the distortion energy reaches a critical value.
For plane stress: \(\sigma_{eq} = \sqrt{\sigma_1^2 - \sigma_1\sigma_2 + \sigma_2^2}\)
Or in terms of stress components:
\(\sigma_{eq} = \sqrt{\sigma_{xx}^2 - \sigma_{xx}\sigma_{yy} + \sigma_{yy}^2 + 3\tau_{xy}^2}\)
This diagram shows the failure envelopes for different criteria on the principal stress plane. Points inside the envelope are safe; points on or outside indicate failure.
For a three-dimensional stress state with principal stresses σ₁ ≥ σ₂ ≥ σ₃, we have three Mohr's circles:
Circle 1: Between σ₂ and σ₃ (centre at \(\frac{\sigma_2 + \sigma_3}{2}\), radius \(\frac{\sigma_2 - \sigma_3}{2}\))
Circle 2: Between σ₁ and σ₃ (centre at \(\frac{\sigma_1 + \sigma_3}{2}\), radius \(\frac{\sigma_1 - \sigma_3}{2}\))
Circle 3: Between σ₁ and σ₂ (centre at \(\frac{\sigma_1 + \sigma_2}{2}\), radius \(\frac{\sigma_1 - \sigma_2}{2}\))
The absolute maximum shear stress is:
\(\tau_{max,abs} = \frac{\sigma_1 - \sigma_3}{2}\)
This occurs on planes inclined at 45° to the σ₁ and σ₃ directions.
The Von Mises criterion is based on the distortion energy theory. Here's the mathematical derivation:
The total strain energy density in a material is:
\(U = \frac{1}{2E}[\sigma_1^2 + \sigma_2^2 + \sigma_3^2 - 2\nu(\sigma_1\sigma_2 + \sigma_2\sigma_3 + \sigma_3\sigma_1)]\)
Decompose stress into hydrostatic (volumetric) and deviatoric (distortional) parts:
Hydrostatic stress: \(\sigma_h = \frac{\sigma_1 + \sigma_2 + \sigma_3}{3}\)
Deviatoric stresses: \(\sigma'_i = \sigma_i - \sigma_h\)
The distortion energy density is:
\(U_d = \frac{1+\nu}{3E}[(\sigma_1 - \sigma_2)^2 + (\sigma_2 - \sigma_3)^2 + (\sigma_3 - \sigma_1)^2]\)
For uniaxial tension at yield: \(\sigma_1 = \sigma_y\), \(\sigma_2 = \sigma_3 = 0\)
Therefore: \(U_{d,yield} = \frac{1+\nu}{3E} \cdot 2\sigma_y^2\)
Setting \(U_d = U_{d,yield}\) gives the Von Mises criterion:
\(\sqrt{\frac{1}{2}[(\sigma_1 - \sigma_2)^2 + (\sigma_2 - \sigma_3)^2 + (\sigma_3 - \sigma_1)^2]} = \sigma_y\)
For plane stress (\(\sigma_3 = 0\)):
\(\sigma_{eq} = \sqrt{\sigma_1^2 - \sigma_1\sigma_2 + \sigma_2^2}\)
Principal stress analysis is standard practice wherever a component sees more than one stress at once:
| Concept | Equation | Description |
|---|---|---|
| Stress Transformation (Normal) | \(\sigma_{x'} = \frac{\sigma_{xx} + \sigma_{yy}}{2} + \frac{\sigma_{xx} - \sigma_{yy}}{2}\cos(2\theta) + \tau_{xy}\sin(2\theta)\) | Normal stress on rotated plane |
| Stress Transformation (Shear) | \(\tau_{x'y'} = -\frac{\sigma_{xx} - \sigma_{yy}}{2}\sin(2\theta) + \tau_{xy}\cos(2\theta)\) | Shear stress on rotated plane |
| Principal Stresses | \(\sigma_{1,2} = \frac{\sigma_{xx} + \sigma_{yy}}{2} \pm \sqrt{\left(\frac{\sigma_{xx} - \sigma_{yy}}{2}\right)^2 + \tau_{xy}^2}\) | Maximum and minimum normal stresses |
| Principal Angle | \(\tan(2\theta_p) = \frac{2\tau_{xy}}{\sigma_{xx} - \sigma_{yy}}\) | Orientation of principal planes |
| Maximum Shear Stress | \(\tau_{max} = \frac{\sigma_1 - \sigma_2}{2} = \sqrt{\left(\frac{\sigma_{xx} - \sigma_{yy}}{2}\right)^2 + \tau_{xy}^2}\) | Maximum in-plane shear stress |
| Max Shear Angle | \(\tan(2\theta_s) = -\frac{\sigma_{xx} - \sigma_{yy}}{2\tau_{xy}}\) | Orientation of max shear planes (45° from principal) |
| Mohr's Circle Centre | \(\sigma_{avg} = \frac{\sigma_{xx} + \sigma_{yy}}{2}\) | Centre of Mohr's circle on σ axis |
| Mohr's Circle Radius | \(R = \sqrt{\left(\frac{\sigma_{xx} - \sigma_{yy}}{2}\right)^2 + \tau_{xy}^2}\) | Radius of Mohr's circle |
| Rankine Criterion | \(\sigma_{eq} = \max(\sigma_1, |\sigma_2|)\) | Maximum principal stress (brittle materials) |
| Tresca Criterion | \(\sigma_{eq} = \max(|\sigma_1 - \sigma_2|,\ |\sigma_1|,\ |\sigma_2|)\) | Maximum shear stress with \(\sigma_3 = 0\) (ductile materials) |
| Von Mises Criterion | \(\sigma_{eq} = \sqrt{\sigma_1^2 - \sigma_1\sigma_2 + \sigma_2^2}\) | Distortion energy (ductile materials) |
| Von Mises (Components) | \(\sigma_{eq} = \sqrt{\sigma_{xx}^2 - \sigma_{xx}\sigma_{yy} + \sigma_{yy}^2 + 3\tau_{xy}^2}\) | Von Mises in stress components |
| Factor of Safety | \(FS = \frac{\sigma_{failure}}{\sigma_{equivalent}}\) | Safety margin (FS > 1 for safe design) |
A stress element has \(\sigma_{xx} = 80\,\text{MPa}\), \(\sigma_{yy} = 20\,\text{MPa}\), and \(\tau_{xy} = 30\,\text{MPa}\). Calculate:
Show Solution
Solution:
Average stress: \(\sigma_{avg} = \dfrac{80 + 20}{2} = 50\,\text{MPa}\)
Radius: \(R = \sqrt{\left(\dfrac{80 - 20}{2}\right)^2 + 30^2} = \sqrt{900 + 900} = 42.43\,\text{MPa}\)
\(\sigma_1 = 50 + 42.43 = 92.43\,\text{MPa}\)
\(\sigma_2 = 50 - 42.43 = 7.57\,\text{MPa}\)
\(\tan(2\theta_p) = \dfrac{2(30)}{80 - 20} = 1 \;\Rightarrow\; \theta_p = 22.5°\)
\(\tau_{max} = R = 42.43\,\text{MPa}\)
Answer: \(\sigma_1 = 92.43\,\text{MPa}\), \(\sigma_2 = 7.57\,\text{MPa}\), \(\theta_p = 22.5°\), \(\tau_{max} = 42.43\,\text{MPa}\).
A component made of steel (\(\sigma_y = 300\,\text{MPa}\)) has principal stresses \(\sigma_1 = 200\,\text{MPa}\) and \(\sigma_2 = -100\,\text{MPa}\). Calculate the factor of safety using:
Show Solution
Solution:
Tresca:
\(\sigma_{eq} = \max\big(|\sigma_1 - \sigma_2|,\ |\sigma_1|,\ |\sigma_2|\big) = \max(300, 200, 100) = 300\,\text{MPa}\)
Here \(\sigma_1\) and \(\sigma_2\) have opposite signs, so \(|\sigma_1 - \sigma_2| = 300\,\text{MPa}\) governs.
\(FS = \dfrac{300}{300} = 1.00\)
Von Mises:
\(\sigma_{eq} = \sqrt{200^2 - 200(-100) + (-100)^2}\)
\(\sigma_{eq} = \sqrt{40000 + 20000 + 10000} = \sqrt{70000} = 264.58\,\text{MPa}\)
\(FS = \dfrac{300}{264.58} = 1.13\)
Answer: Tresca \(FS = 1.00\); von Mises \(FS = 1.13\).
A stress element is subjected to the following stresses:
Using Mohr's Circle principles, determine:
Show Solution
(a) Centre and radius of Mohr's Circle:
\[C = \frac{\sigma_{xx} + \sigma_{yy}}{2} = \frac{100 + 40}{2} = 70 \text{ MPa}\]
\[R = \sqrt{\left(\frac{\sigma_{xx} - \sigma_{yy}}{2}\right)^2 + \tau_{xy}^2}\]
\[R = \sqrt{\left(\frac{100 - 40}{2}\right)^2 + (-50)^2} = \sqrt{30^2 + 50^2} = \sqrt{900 + 2500} = 58.31 \text{ MPa}\]
(b) Principal stresses:
\[\sigma_1 = C + R = 70 + 58.31 = 128.31 \text{ MPa}\]
\[\sigma_2 = C - R = 70 - 58.31 = 11.69 \text{ MPa}\]
(c) Principal angle:
\[\tan(2\theta_p) = \frac{2\tau_{xy}}{\sigma_{xx} - \sigma_{yy}} = \frac{2(-50)}{100 - 40} = \frac{-100}{60} = -1.667\]
\[2\theta_p = \arctan(-1.667) = -59.04°\]
\[\theta_p = -29.52° \text{ (or } 60.48° \text{ from the other solution)}\]
(d) Maximum in-plane shear stress:
\[\tau_{max} = R = 58.31 \text{ MPa}\]
This occurs at planes 45° from the principal planes:
\[\theta_s = \theta_p + 45° = -29.52° + 45° = 15.48°\]
(e) Stress components at θ = 30°:
Using stress transformation equations (or Mohr's Circle with 2θ = 60°):
\[\sigma_{x'} = \frac{\sigma_{xx} + \sigma_{yy}}{2} + \frac{\sigma_{xx} - \sigma_{yy}}{2}\cos(2\theta) + \tau_{xy}\sin(2\theta)\]
\[\sigma_{x'} = 70 + 30\cos(60°) + (-50)\sin(60°)\]
\[\sigma_{x'} = 70 + 30(0.5) + (-50)(0.866) = 70 + 15 - 43.3 = 41.7 \text{ MPa}\]
\[\sigma_{y'} = \frac{\sigma_{xx} + \sigma_{yy}}{2} - \frac{\sigma_{xx} - \sigma_{yy}}{2}\cos(2\theta) - \tau_{xy}\sin(2\theta)\]
\[\sigma_{y'} = 70 - 15 + 43.3 = 98.3 \text{ MPa}\]
\[\tau_{x'y'} = -\frac{\sigma_{xx} - \sigma_{yy}}{2}\sin(2\theta) + \tau_{xy}\cos(2\theta)\]
\[\tau_{x'y'} = -30\sin(60°) + (-50)\cos(60°) = -30(0.866) + (-50)(0.5)\]
\[\tau_{x'y'} = -26.0 - 25.0 = -51.0 \text{ MPa}\]
Answer: (a) \(C = 70\,\text{MPa}\), \(R = 58.31\,\text{MPa}\); (b) \(\sigma_1 = 128.31\,\text{MPa}\), \(\sigma_2 = 11.69\,\text{MPa}\); (c) \(\theta_p = -29.52°\) (or \(60.48°\)); (d) \(\tau_{max} = 58.31\,\text{MPa}\) at \(\theta_s = 15.48°\); (e) \(\sigma_{x'} = 41.7\,\text{MPa}\), \(\sigma_{y'} = 98.3\,\text{MPa}\), \(\tau_{x'y'} = -51.0\,\text{MPa}\).