Design against static loads and stress concentration
Static design with failure theories (Rankine, Tresca, von Mises) and stress concentration factors, including when Kt applies to ductile and brittle materials.
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Why it matters
Most machine parts carry combined stresses (tension plus shear, bending plus torsion), but material data come from a simple tension test. Failure theories bridge that gap, and stress concentration factors tell you how much a hole, keyway or shoulder raises the local stress. Together they decide whether a bracket, lever or shaft survives its static load and how much margin it really has.
Key ideas
Static load means a load applied slowly and held (or changing so few times in the life of the part that fatigue is not a concern). Design then compares a stress state with a static strength: yield strength σ_y for ductile materials, ultimate strength σ_ut (and σ_uc) for brittle ones.
Theories of failure predict failure under a multi-axial stress state from uniaxial test data. With principal stresses σ₁ ≥ σ₂ ≥ σ₃:
- Maximum principal stress theory (Rankine) - failure when the largest principal stress reaches the uniaxial strength. Suits brittle materials; unsafe for ductile materials under shear.
- Maximum shear stress theory (Tresca, Guest) - failure when τ_max reaches the τ_max at yield in a tension test, σ_y/2. Conservative for ductile materials; predicts shear yield strength τ_y = 0.5·σ_y.
- Distortion energy theory (von Mises, Hencky) - failure when the distortion strain energy reaches its value at tensile yield. Best agreement with tests on ductile metals; predicts τ_y = 0.577·σ_y.
- Maximum principal strain theory (Saint-Venant) and total strain energy theory (Haigh) are mainly of historical and exam interest. In pure shear, Tresca and von Mises differ the most (about 15%); for uniaxial stress all theories agree.
Stress concentration. At an abrupt change of geometry (hole, notch, groove, keyway, shoulder with a small fillet) the local peak stress is higher than the nominal stress predicted by elementary formulas. The theoretical (geometric) stress concentration factor is
Kt = σ_max / σ_nom.
Kt depends only on geometry and load type (tension, bending, torsion), not on material, and is read from charts (Peterson) or found by FEA or photoelasticity. Each chart defines σ_nom on a stated section - usually the net section at the notch - so always use the area the chart uses. Reference values: small circular hole in a wide plate in tension, Kt ≈ 3; elliptical hole with semi-axis a perpendicular to the load and b along it, Kt = 1 + 2a/b (a crack, b → 0, gives very large Kt).
When to apply Kt under static load:
- Brittle materials: apply Kt, because there is no yielding to relieve the peak. Exception: grey cast iron contains graphite flakes that already act as internal notches, so external notches add little and Kt is often taken as 1.
- Ductile materials: the peak region yields locally and load is shared by surrounding material, so Kt is usually ignored for static design (it must not be ignored for fatigue - next topic).
Reducing stress concentration: generous fillet radii at shoulders, gradual tapers, relief grooves or extra small holes beside a notch to smooth the stress flow, placing holes in low-stress regions, and avoiding sharp re-entrant corners.
Formulas
σ_nom = F / A_net (axial) · σ_nom = M·y / I (bending) · τ_nom = T·r / J (torsion)
- F: force (N); A_net: net cross-sectional area at the notch (m²); M: bending moment (N·m); y: distance from neutral axis (m); I: second moment of area (m⁴); T: torque (N·m); r: radius (m); J: polar moment of area (m⁴).
σ_max = Kt · σ_nom
- Kt: theoretical stress concentration factor (dimensionless, ≥ 1), from a chart for the same nominal-stress definition.
σ₁,₂ = (σx + σy)/2 ± √(((σx − σy)/2)² + τxy²)
- σx, σy: normal stresses (Pa); τxy: shear stress (Pa); plane stress.
Rankine: σ₁ = σ_ut / n
Tresca: max(|σ₁ − σ₂|, |σ₂ − σ₃|, |σ₃ − σ₁|) = σ_y / n (for plane stress with σ₃ = 0 this includes |σ₁| and |σ₂|)
von Mises (plane stress): σe = √(σ₁² − σ₁·σ₂ + σ₂²) = √(σx² − σx·σy + σy² + 3·τxy²) = σ_y / n
- σe: von Mises equivalent stress (Pa); n: factor of safety.
Kt (elliptical hole) = 1 + 2a/b
- a: semi-axis perpendicular to the load (m); b: semi-axis parallel to the load (m). Infinite plate, uniaxial tension.
Worked examples
Example 1 (standard). A flat bar 60 mm wide and 10 mm thick has a central transverse hole of 12 mm diameter and carries a static axial load of 40 kN. From the chart (net-section basis) Kt = 2.5. Find the peak stress and the FoS if the bar is (a) a brittle material with σ_ut = 250 MPa, (b) a ductile steel with σ_y = 300 MPa.
- Net area:
A_net = (w − d)·t= (60 − 12) × 10 = 480 mm². - Nominal stress:
σ_nom = F / A_net= 40,000 / 480 = 83.33 MPa. - Peak stress:
σ_max = Kt·σ_nom= 2.5 × 83.33 = 208.3 MPa. - (a) Brittle: n = σ_ut / σ_max = 250 / 208.3 = 1.20.
- (b) Ductile, static: Kt ignored, n = σ_y / σ_nom = 300 / 83.33 = 3.60.
Example 2 (GATE level). A point in a ductile steel part (σ_y = 300 MPa) has σx = 80 MPa, σy = −40 MPa, τxy = 30 MPa (plane stress). Find the FoS by Tresca and von Mises, and show why Rankine is unsafe here.
- Centre and radius: (σx + σy)/2 = 20 MPa; R = √(60² + 30²) = 67.08 MPa.
- Principal stresses: σ₁ = 20 + 67.08 = 87.08 MPa; σ₂ = 20 − 67.08 = −47.08 MPa; σ₃ = 0.
- Tresca: largest difference = σ₁ − σ₂ = 134.16 MPa → n = 300 / 134.16 = 2.24.
- von Mises: σe = √(80² − 80 × (−40) + (−40)² + 3 × 30²) = √13,900 = 117.9 MPa → n = 300 / 117.9 = 2.54.
- Rankine would give n = 300 / 87.08 = 3.45, overestimating the safety of a ductile part with opposite-sign principal stresses.
Common mistakes
- Using gross area when the Kt chart is based on the net section (or vice versa).
- Applying Kt to a ductile part under static load and grossly over-designing it - or ignoring it for a brittle part.
- Using Rankine for a ductile material, especially when principal stresses have opposite signs.
- Forgetting σ₃ = 0 in plane stress: if σ₁ and σ₂ have the same sign, Tresca's τ_max is σ₁/2, not (σ₁ − σ₂)/2.
- Thinking Kt depends on the material or the load magnitude - it depends only on geometry and load type.
- Quoting Tresca shear yield as 0.577·σ_y (that is von Mises; Tresca gives 0.5·σ_y).
For GATE ME
Frequent question types: choosing the correct failure theory for a material, computing the FoS or the permissible load by Tresca and von Mises from a given stress state, the ratio of shear yield strength to tensile yield strength under each theory, and peak stress at a hole or fillet from a given Kt. Practise principal-stress calculation quickly and remember to include σ₃ = 0.
Quick check
- Which failure theory best predicts yielding of ductile metals?
- Ratio τ_y/σ_y by maximum shear stress theory?
- Kt for a small circular hole in a wide plate in tension?
- Is Kt usually applied to a ductile part under static load?
- Elliptical hole with a = 10 mm (perpendicular to load), b = 5 mm: Kt? Answers: 1. Distortion energy (von Mises). 2. 0.5. 3. About 3. 4. No - local yielding redistributes the stress. 5. Kt = 1 + 2 × 10/5 = 5.
Interview questions
All Machine Design interview questionsTry answering each one aloud before you open it.
1.What is stress concentration and why is it important in machine design?Concept
Stress concentration is the local rise of stress above the nominal value at an abrupt change in geometry such as a hole, notch, keyway or sharp shoulder. It matters most for brittle materials under static load, which fracture when the peak reaches their strength, and for all materials under fluctuating load, where fatigue cracks start at the peak. Ductile parts under static load usually tolerate it because the peak region yields locally and redistributes load.
2.Explain the concept of a stress concentration factor (Kt).Concept
The stress concentration factor (Kt) is a dimensionless factor that quantifies the increase in stress at a point of discontinuity, such as holes, notches, or sharp corners, compared to the nominal stress. It is defined as the ratio of the maximum stress at the discontinuity to the nominal stress. Kt helps engineers predict where failures might occur and design components to mitigate these effects.
3.How can stress concentration be reduced in a mechanical component?Application
Use generous fillet radii at shoulders and gradual tapers instead of steps, avoid sharp re-entrant corners, and place holes and keyways in low-stress regions. Relief grooves or small auxiliary holes beside a notch smooth the flow of stress lines. For fatigue, surface treatments such as shot peening and good surface finish further reduce the effect of the notch.
4.Why are fillets used in the design of mechanical components?Application
Fillets are used in mechanical components to reduce stress concentration at sharp corners or edges. By providing a smooth transition between surfaces, fillets distribute the stress more evenly across the component, reducing the likelihood of failure. This is particularly important in components subjected to cyclic or static loads, where stress concentration can lead to fatigue or sudden failure.
5.What happens if a component with a high stress concentration is subjected to a static load?Application
In a brittle material the peak stress is not relieved, so a crack starts at the notch once σ_max = Kt·σ_nom reaches the ultimate strength, and the part fractures at a load well below what the nominal stress suggests. In a ductile material the peak region yields locally and load shifts to the surrounding material, so static strength is governed mainly by nominal stress on the net section. If the same load is repeated, however, the notch becomes the site of fatigue cracking.
6.Explain how material selection affects design against static loads and stress concentration.Application
The material decides both the failure criterion and how much the notch matters. Ductile materials are designed on yield strength and are relatively insensitive to stress concentration under static load because of local yielding. Brittle materials are designed on ultimate strength with Kt applied, except grey cast iron, whose graphite flakes already act as internal notches so external notches add little.
7.What is the role of finite element analysis (FEA) in addressing stress concentration issues?Application
Finite element analysis (FEA) is a computational tool used to simulate and analyze the behavior of components under various loads and conditions. In addressing stress concentration issues, FEA helps identify areas of high stress and evaluate the effectiveness of design modifications. By providing detailed insights into stress distribution, FEA allows engineers to optimize designs to minimize stress concentration and improve component performance.
8.Calculate the maximum stress in a flat plate with a hole, given that the nominal stress is 100 MPa and the stress concentration factor (Kt) is 3.Numerical
σ_max = Kt × σ_nom = 3 × 100 MPa = 300 MPa at the edge of the hole, on the diameter perpendicular to the load. Make sure the nominal stress uses the same section (gross or net) as the chart that gave Kt. For a ductile material under static load this peak would yield locally; for a brittle material or for fatigue it governs the design.
9.Discuss the impact of geometric discontinuities on the fatigue life of a component.Application
Geometric discontinuities, such as holes, notches, and sharp corners, can significantly impact the fatigue life of a component by creating areas of high stress concentration. These areas are more susceptible to crack initiation and propagation under cyclic loading, leading to premature failure. By addressing these discontinuities through design modifications, such as adding fillets or using smoother transitions, the fatigue life of the component can be improved.
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