Static Failure Criteria and Design: Notes on Von Mises, Tresca, Centroids, and Example Problems
Static Failure Criteria and Design: Comprehensive Notes
Theme of the week:
Applying Solar Mechanics (statics and stress analysis) to design.
Static failure criteria as a stepping stone to dynamic failure (fatigue).
Recognize that real design may later involve dynamics, fluids, or thermodynamics, but this unit emphasizes static analysis and failure criteria.
Static analysis workflow (recap):
Start with internal vs external loading, reactions, and equilibrium (forces sum to zero).
Build internal loading diagrams (normal stresses, bending, shear).
Introduce stress concentration considerations (last week).
Use Mohr’s circle to visualize stresses, principal stresses, and planes of maximum shear.
Define a comparative stress metric (often Von Mises, sometimes maximum normal stress or maximum distortion energy).
Determine the critical location (the failure “point”), which is not always the location of maximum external load due to stress concentrations or geometry changes.
Stress concentration (context):
Real shafts/parts aren’t uniform along their length; steps, shoulders, holes, and attachments create localized stress risers.
Stress concentration factors (SCFs) and the need to account for geometry-driven peaks in stress.
Stress metrics used for static failure criteria (stress-based):
Maximum Normal Stress (brittle materials): failure when the peak normal stress reaches a material limit (e.g., concrete, cast iron).
Maximum Shear Stress (Tresca): failure when the maximum distortion due to shear reaches a limit in ductile materials.
Maximum Distortion Energy (Von Mises): energy-based criterion for ductile materials; often the most practical general-purpose criterion for metals.
Distortion energy (Von Mises) concept (high-level):
Von Mises stress,
Distortion energy is the energy associated with deviatoric (distortion) stresses after removing hydrostatic (isotropic) stress.
Intuition: removal of hydrostatic pressure; what remains drives yielding in many metals.
Simple physical example: in a high hydrostatic pressure environment (e.g., deep ocean), yield is driven by distortion, not simply the pressure.
Practical note: von Mises is widely used because it correlates well with yielding in many metals and is easy to implement in FEA.
Key equations and definitions (LaTeX-formatted):
Von Mises equivalent stress (3D):
c sigma{vm} = \sqrt{\sigmax^2 + \sigmay^2 + \sigmaz^2 - \sigmax\sigmay - \sigmay\sigmaz - \sigmaz\sigmax}In plane stress (two principal stresses and , with \sigma3 = 0): \sigma{vm} = \sqrt{\sigma1^2 + \sigma2^2 - \sigma1\sigma2}
Alternative (often used):
\sigma{vm}^2 = \frac{1}{2}[(\sigma1-\sigma2)^2 + (\sigma2-\sigma3)^2 + (\sigma3-\sigma_1)^2]Tresca (maximum shear stress) criterion is conceptually based on the maximum difference of principal stresses, but is usually used in its own form as a separate check for ductile materials.
Distortion energy and hydrostatic (mean) stress intuition: remove the average normal stress (hydrostatic pressure) to obtain the distortional stress state.
Stress notation and subscripts (recap):
Normal stresses: \SigmaX, \SigmaY, \SigmaZ (also sometimes written as \Sigma{XX}, \Sigma{YY}, \Sigma{ZZ}).
Shear stresses: \Tau{XY} = \Tau{YX}, \Tau{YZ} = \Tau{ZY}, \Tau{ZX} = \Tau{XZ}.
Subscripts denote the plane and the normal direction of the load:
For example, \Sigma{XX} is normal stress on the X-plane in the X-direction; \Tau{XY} is shear on the X-plane in the Y-direction.
In 2D analyses on the XY plane, terms like \Tau{XZ}, \Tau{YZ} often vanish (or are neglected) for simplicity.
Complementary (equal) shear stresses: \Tau{XY} = \Tau{YX}, etc.
Principal stresses and principal planes (concept):
Principal stresses are the normal stresses on planes where shear stress is zero.
In 2D, a single orientation exists where shear vanishes and stresses are purely normal.
Principal stress concept facilitates Mohr’s circle construction and simplifies failure criteria application.
Stress criteria and practical use in design:
All criteria ultimately compare a real, complex stress state to a known reference state, typically the uniaxial tensile yield point from a material test.
The comparison is usually against the yield strength or a proof stress; allowable stress is often a fraction of the yield strength (proof stress, etc.).
Von Mises is particularly convenient in modern design and FEA, but one must be mindful of cases where directionality and compression/tension asymmetries matter (e.g., pre-stressed concrete, anisotropic materials).
Safety factor and problem-solving strategy (design context):
Factor of safety (N) is defined as a comparison between the material yield and the actual equivalent stress:
N = \frac{S{yield}}{\sigma{vm}}If N < 1, the design is failing (Sigma' > S_yield). If N > 1, the design is safe.
In design problems, you may be asked to:
Determine the required thickness/geometry for a given N (e.g., a bracket thickness t for N = 2).
Determine the maximum allowable load for a given geometry and material with a required N.
Determine the cross-section shape or material choice for a target N.
Practical notes on rounding: when calculating allowable loads, rounding decisions should be conservative (round down for loads to ensure the safety factor remains above the required threshold).
Worked approach: typical two-step problem structure (as illustrated in the lecture):
Step 1: Define assumptions and problem statement.
Often assume elastic and homogeneous material, static (not dynamic) loading, and simple attachment without initial stress concentrations for the hand calc.
Step 2: Build load cases and free-body diagram (FBD).
Resolve external loads into components along the neutral axes.
Identify reaction forces and internal force components (axial, bending, shear).
Step 3: Compute internal stresses at critical cross-sections/elements.
Axial stress: \sigma_a = \frac{F}{A} with appropriate A (e.g., rectangular cross-section A = b t).
Bending stress: \sigma_b = \frac{M c}{I} with I = \frac{b h^3}{12},\quad c = \frac{h}{2} (for a rectangle, orientation dependent).
Transverse shear stress: for a rectangular section, a common estimate is proportional to the parabolic shear distribution, but the exact formula depends on the orientation and section.
Step 4: Combine stresses for the chosen criterion (Von Mises in the examples).
For each critical point, compute
\sigma{vm} = \sqrt{\sigmax^2 + \sigmay^2 + \sigmaz^2 - \sigmax\sigmay - \sigmay\sigmaz - \sigmaz\sigmax}In many cases, you may only have axial and bending (no shear), or a 2D reduction (plane stress).
Step 5: Solve for the design variable (e.g., thickness t) by setting \sigma{vm} = \frac{S{yield}}{N} and solving for the unknown.
Step 6: Compare multiple candidate critical locations; take the most conservative (largest required t or smallest allowable load).
Step 7: Discuss practical implications, such as manufacturability, availability of stock thicknesses, and realistic considerations like stress concentration regions (holes, joints), and pre-stressing effects.
Centroid and second moment of area (concepts and formulas):
Centroid (center of area) of a composite cross-section:
\bar{x} = \frac{\sumi Ai xi}{\sumi Ai}, \quad \bar{y} = \frac{\sumi Ai yi}{\sumi Ai}For a simple T-beam example, split into rectangles, compute each sub-area centroid and sum to get the overall centroid; this gives the neutral axis.
Parallel axis theorem for second moment of area:
I = \sumi \left( Ii^{centroid} + Ai di^2 \right)Rectangle second moment about its own centroid: I_{rect} = \frac{b h^3}{12}.
Quick check trick for second moment of area: use an approximate simple box, e.g., for a 5 mm by 5 mm square, I \, (approx) = \frac{b h^3}{12} = \frac{5^2 \cdot 5}{12} = \frac{625}{12} \text{ mm}^4 \approx 52.08 \text{ mm}^4. This is a sanity check against a more complex exact calculation.
The centroid and second moment calculations are critical inputs to bending stress and deflection analyses, and to Von Mises calculations when computing combined stress states.
Example 1 (design problem): cantilever bracket with a 50 kN load at 30°; solve for thickness t for a safety factor of 2 using Von Mises
Problem setup and assumptions:
Cable stay attached to a cantilever-like bracket; load P = 50 kN at angle 30° to the horizontal.
Cross-section: bracket has width 0.1 m (100 mm) and unknown thickness t (m); material: AISI 1020 steel; yield strength S_y = 200 MPa (in the example context).
Objective: determine thickness t so that Von Mises stress does not exceed S_y / N with N = 2.
Assumptions: weight of the stay negligible; static load (no dynamic fatigue); elastic and homogeneous material; perfect attachment (no stress concentration) for the hand calculation.
Free-body diagram and load decomposition:
Decompose P into components in the chosen plane (e.g., Y and Z or Y and horizontal, depending on orientation):
F{vertical} = P \sin(\theta),\quad F{horizontal} = P \cos(\theta)For P = 50 kN and a 30° angle: approximately, but exact components depend on chosen axes. Components are used to determine axial and bending actions in the bracket cross-section.
Internal force components:
Axial force on the bracket section: \sigma_a = \frac{F}{A} = \frac{F}{b t} with b = 0.1 m and A = b t.
Bending moment at the critical cross-section: M = F d, where d is the perpendicular distance from the load to the fixed end (the location where the stress is maximum). For a rectangular cross-section, bending stress about the extreme fiber is
\sigma_b = \frac{M c}{I}, \quad I = \frac{b h^3}{12}, \quad c = \frac{h}{2}.Transverse shear stress can be estimated; for a rectangular section, a simplified form is sometimes used: but the specific value depends on the load path and orientation.
Von Mises at a critical point (example points 1 and 2):
At a chosen point 1 (with its local axial stress \sigmaa1 and bending stress \sigmab1, plus possible shear):
\sigma{vm,1} = \sqrt{\sigmaa1^2 + \sigma_b1^2 + (\text{shear term})^2\; - \; \text{cross-terms}}At a chosen point 2, similarly, with its local stresses.
Solve for t from the two critical points:
Point 1: find t such that \sigma{vm,1} = \frac{Sy}{N} = \frac{200\,\text{MPa}}{2} = 100\,\text{MPa} (convert all units consistently; often, stresses are in MPa, lengths in meters).
Example outcome reported: t ≈ 7.69 mm.
Point 2: find t such that \sigma_{vm,2} = 100\,\text{MPa}.
Example outcome reported: t ≈ 19 mm.
Critical case and final design choice:
The larger required thickness is the governing one (the more conservative design). In this example, Point 2 demanded t ≈ 19 mm, which is the thickness used to satisfy the N = 2 safety factor for both points.
Important practical note on interpretation:
If you were instead asked for the maximum allowable load with N = 2 for a fixed t, you would solve for the load that makes the von Mises stress reach 100 MPa and choose the smaller allowable load among the candidate critical points.
If you were asked to choose a real-world sheet thickness, you’d consider stock sizes and possibly round to a standard thickness, with a conservative approach to ensure the safety factor remains above the requirement (e.g., round down the allowable load or round up the thickness if necessary for manufacturability and safety).
Practical considerations highlighted in the example:
The critical location is not always the location of the largest external load due to geometry and stress concentrations.
A hole or fastener could dominate the local stress state (bearing stress, tear-out risk, fatigue considerations). In real design, you would re-check near stress concentrators.
The need to communicate results to manufacturability: if the required thickness is 19.0 mm, but stock comes in 19.05 mm or 20 mm, you need to decide on a feasible choice and justify conservatism.
Summary outcome (Example 1):
For a safety factor of 2, the minimum thickness t is governed by the more critical location (outer edge or a point near a hole, depending on the geometry). In the discussed calculation, the critical case required t ≈ 19 mm, ensuring Von Mises stress stays below 100 MPa under the given loading.
Example 2 (design problem): PFC (parallel flange channel) beam with a bolted plate; find the load F for which the factor of safety is 2
Problem setup and assumptions:
A 400 mm cantilever plate attaches to a PSC (parallel flange channel) column via bolts; a plate carries a load transferred to the PSC through the bolted connection.
Section properties: PSC cross-section about a given axis; yield strength S_y = 200 MPa (as used in the example).
Objective: determine the force F in the plate such that the safety factor with respect to Von Mises is 2.
Assumptions: neglected weight of the plate, load is static; material is elastic and homogeneous; the bolt hole stress concentrations are ignored for this hand calc; loads transferred through the centroid of the plate to the PSC cross-section.
Free-body and equivalent load approach:
The plate is treated as a cantilever delivering a shear force (compressive axial load on the PSC) and a bending moment (about the PSC centroid) at the connection.
The moment about the PSC centroid is M = F × distancetocentroidofbolts.
The column/PSC cross-section is treated as symmetric about its centroid (neutral axis at center) if the bending is about the symmetry axis.
Second moment of area (I) for the PSC cross-section:
Break the cross-section into sub-rectangles (or use an equivalent single-beam representation) and apply the parallel axis theorem:
I = \sumi \left( Ii^{centroid} + Ai di^2 \right)For a rectangular member, $I_{rect, about its centroid} = \frac{b h^3}{12}$.
Distances $d_i$ are the vertical/horizontal distances from the local centroid of each sub-area to the overall neutral axis of the entire cross-section.
Example calculation steps (brief outline):
Compute $I_y$ for the PSC section by decomposing into sub-areas (e.g., two flanges and web) and applying the parallel axis theorem.
Determine the centroid of the cross-section (neutral axis) and confirm symmetry so that $d_i$ can be taken accordingly.
Determine the axial load on the PSC due to the plate, and the bending moment at the bolt line: the axial stress is compressive, the bending stress arises from the moment.
Von Mises stress at the critical point combines axial stress and bending stress (no shear term in the simplified case).
Von Mises stress for the two principal load components (when no shear at the critical point):
\sigma{vm} = \sqrt{ \sigmay^2 + \sigma_m^2 }Here, $\sigmay$ is the axial (compressive) stress, and $\sigmam$ is the bending stress at the critical point.
If shear is present, include the shear term: $\sqrt{ \sigmay^2 + \sigmam^2 + \tau^2 }$ with the appropriate combination per von Mises formulation.
Solve for F given safety factor 2:
Set $\sigma{vm} = Sy / N = 200 \text{ MPa} / 2 = 100 \text{ MPa}$ and solve for F, using the computed $I$, $d$, and cross-sectional areas.
Example outcome: $F \approx 11.4 \text{ kN}$.
Rounding and conservatism:
When converting to a safe design value, you may round down the allowable force to ensure the factor of safety remains above the required threshold (e.g., rounding a calculated 11.476 kN to 11.4 kN to maintain ~2.0 safety).
Practical notes about this example:
The problem illustrates translating a plate-side load into an equivalent PSC-side load via reactions and moments.
It demonstrates how to compute $I$ for a non-simple cross-section via decomposition and parallel axis theorem.
It emphasizes that the control variable can be a force (F) or a dimension (thickness, i). Here, the problem solved for F; for a different prompt you could solve for required thickness or a different geometry to meet the same safety target.
Example 3 (contextual, practical design scenario): analyzing a shifting spanner or gear-like component
The lecturer mentions a third example (shifting spanner) to practice bending stress and safety factor evaluation on a non-rectangular cross-section.
The idea is to move from idealized shafts/beams to components with more realistic cross-sections to determine loads, dimensions, and safety factors.
Practical insights and design philosophy emphasized in the session:
Von Mises is easy to apply in FEA (easy to display and compare to yield with a safety factor). However, it loses directional information about whether the critical stress is tensile or compressive. Always check the actual stress components and directions, especially when a material behaves differently under tension vs compression (e.g., concrete under tension is weak).
In pre-stressed concrete, the pre-compression can help keep the material in compression under service loads, delaying tensile failure.
Always consider the effect of geometry changes and discontinuities (holes, joints, fillets) on the location of the critical stress (stress concentration). Do not automatically assume the peak external load location is the critical point.
When performing hand calculations, use checks (e.g., a simplified box section to estimate $I$) to ensure the result is reasonable before proceeding with more complex calculations.
When communicating designs to machinists or manufacturing, consider standard stock sizes and manufacturability in the rounding/selection process to avoid infeasible dimensions.
Summary of key takeaways for exam prep:
Know the three major static failure criteria and what materials they best apply to:
Maximum Normal Stress (brittle materials)
Maximum Shear Stress (Tresca – typically for ductile metals)
Maximum Distortion Energy (Von Mises – most common for metals)
Be able to compute Von Mises stress for a general 3D state and for common 2D plane stress cases, including how to incorporate axial, bending, and shear components.
Be comfortable with centroids and second moments of area, and applying the parallel axis theorem for nonuniform cross-sections.
Be able to set up a design problem with target safety factor, identify the governing case, and solve for geometry (thickness, diameter) or load limits.
Recognize the practical limitations of hand calculations and the role of conservative design, manufacturability, and stress concentrations.
Quick recap of formulas to memorize (LaTeX):
Von Mises (3D):
\sigma{vm} = \sqrt{\sigmax^2 + \sigmay^2 + \sigmaz^2 - \sigmax\sigmay - \sigmay\sigmaz - \sigmaz\sigmax}Von Mises (plane stress 2D):
\sigma{vm} = \sqrt{\sigmax^2 - \sigmax\sigmay + \sigma_y^2}Safety factor:
N = \frac{S{yield}}{\sigma{vm}}Second moment of area (rectangle):
I_{rect} = \frac{b h^3}{12}Parallel axis theorem:
I = \sumi \left( Ii^{centroid} + Ai di^2 \right)Centroid position (example form):
\bar{x} = \frac{\sumi Ai xi}{\sumi A_i}
If you want these notes in a quick reference format for study, you can skim the sections above and pull out the formulas and the key decision rules (when to use Von Mises, how to interpret N, how to determine the critical location, and how to perform the parallel axis theorem for common cross-sections).