Design for static loading and factor of safety
Designing against yielding and fracture under steady loads: factor of safety, stress concentration, and the Rankine, Tresca and von Mises failure theories for combined stresses.
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Why it matters
Every machine element, from a tie rod to a crane hook or a shaft, is first sized so that it neither yields nor fractures under the steady load it carries. Choosing the right strength (yield or ultimate), the right failure theory for a combined stress state, and a sensible factor of safety is the starting point of all design calculations, including fatigue design later.
Key ideas
Static load. A load applied gradually that does not change in magnitude or direction during service (or changes only a few times in the part's life). Loads that fluctuate many times need fatigue design; suddenly applied and impact loads need dynamic factors.
Failure modes. A ductile material (steel, aluminium; elongation above about 5 %) is considered failed when it yields, because permanent deformation spoils the fit and function of a machine part. A brittle material (cast iron) fails by fracture at its ultimate strength, with little warning.
Factor of safety (FoS). The ratio of failure stress to working (allowable) stress:
- ductile:
N = S_yt/σ, - brittle:
N = S_ut/σ. It covers uncertainty in material properties, in the loads, in the stress analysis, in manufacturing (residual stresses, flaws), and the consequences of failure. Typical values for steel under steady load are around 1.5 to 2.5, larger for cast iron and for uncertain or shock loads; take the value your code or data book specifies.
Stress concentration. Holes, fillets, keyways and grooves raise the local stress by the theoretical factor K_t (from charts). Under static load a ductile material yields locally and redistributes stress, so K_t is usually ignored; for a brittle material it is applied, because there is no yielding to relieve the peak. (Under fatigue, K_t matters for both.)
Theories of failure (combined stresses). With principal stresses σ₁ ≥ σ₂ ≥ σ₃ at the critical point:
- Maximum principal stress (Rankine): failure when σ₁ reaches the strength. Good for brittle materials, unsafe for ductile ones in shear.
- Maximum shear stress (Tresca/Guest): failure when τ_max reaches S_yt/2. Conservative and simple for ductile materials.
- Distortion energy (von Mises–Hencky): failure when the von Mises stress reaches S_yt. Best match to tests on ductile materials; it predicts shear yield strength S_sy = 0.577 S_yt, against 0.5 S_yt from Tresca.
- Maximum principal strain (Saint-Venant) and total strain energy (Haigh) theories are of historical and exam interest. For a shaft with bending stress σ and torsional shear τ (the most common GATE case), the formulas below collapse to simple forms.
Design procedure. Identify the critical section → find the load and stresses there (direct, bending, torsion, shear) → combine with the right theory → set the equivalent stress equal to strength/N → solve for the dimension → round up to a standard size.
Formulas
σ = P/A, σ_b = M·y/I = 32M/(πd³) (solid round), τ = T·r/J = 16T/(πd³) (solid round). Stresses in Pa (N/m²), forces in N, moments in N·m, dimensions in m.
N = S_yt/σ_eq (ductile) or N = S_ut/σ₁ (brittle) — FoS, dimensionless.
Principal stresses (plane): σ₁,₂ = (σ_x + σ_y)/2 ± √[((σ_x − σ_y)/2)² + τ_xy²].
Max principal stress: σ₁ = S_ut/N.
Max shear stress: τ_max = (σ₁ − σ₃)/2 = S_yt/(2N); with σ and τ only, √(σ² + 4τ²) = S_yt/N.
Distortion energy: σ_vm = √(σ₁² − σ₁σ₂ + σ₂²) = S_yt/N (plane stress); with σ and τ only, √(σ² + 3τ²) = S_yt/N.
Equivalent loads for shafts: T_e = √(M² + T²) (Tresca), M_e = ½(M + √(M² + T²)) (Rankine).
Worked examples
Example 1 (standard). A round steel tie rod carries a steady tensile load of 50 kN. S_yt = 300 MPa and N = 2.5. Find the diameter.
- Allowable stress:
σ = S_yt/N = 300/2.5 = 120 MPa. - Area:
A = P/σ = 50 000/120 = 416.7 mm²(N and MPa give mm²). d = √(4A/π) = √(4 × 416.7/π) = 23.0 mm.- Use d = 24 mm (next standard size).
Example 2 (GATE level). At the critical point of a ductile steel shaft (S_yt = 300 MPa) the bending stress is 80 MPa and the torsional shear stress is 60 MPa. Find the factor of safety by the maximum principal stress, maximum shear stress and distortion energy theories.
τ_max = √((80/2)² + 60²) = √(1600 + 3600) = 72.11 MPa.σ₁ = 40 + 72.11 = 112.11 MPa(σ₂ = 40 − 72.11 = −32.11 MPa).- Rankine:
N = 300/112.11 = 2.68. - Tresca:
N = S_yt/(2τ_max) = 300/144.22 = 2.08. - Von Mises:
σ_vm = √(80² + 3 × 60²) = √17 200 = 131.15 MPa;N = 300/131.15 = 2.29. - N ≈ 2.68 (Rankine), 2.08 (Tresca), 2.29 (von Mises). For a ductile shaft, the Rankine value is unsafe; Tresca is the most conservative, von Mises the most realistic.
Common mistakes
- Using S_ut for a ductile part under static load; design against yield.
- Applying K_t to a ductile part under static load, or ignoring it for cast iron.
- Using √(σ² + 4τ²) with the distortion energy theory (that is Tresca; von Mises uses 3τ²).
- Forgetting that τ_max = S_yt/2 in Tresca, not S_yt.
- Mixing units: N and mm give MPa directly; N and m give Pa.
- Reporting a calculated diameter without rounding up to a standard size.
For GATE PI
Expect FoS calculations for a given stress state using a named failure theory, equivalent stress for combined bending and torsion, comparison of Tresca and von Mises (including shear yield strength ratios 0.5 and 0.577), and simple sizing of tie rods and pins. Practise drawing Mohr's circle quickly and recognising which theory suits which material.
Quick check
- S_yt = 250 MPa, working stress 125 MPa. FoS?
- What shear yield strength does von Mises predict for S_yt = 300 MPa?
- For pure shear τ, what equivalent stress does Tresca give?
- Which theory would you use for a grey cast iron bracket?
Answers: 1. 2; 2. 173 MPa; 3. 2τ; 4. maximum principal stress (Rankine).
Interview questions
All Theory of Machines and Machine Design interview questionsTry answering each one aloud before you open it.
1.What is static loading in the context of machine design?Concept
Static loading refers to loads that are applied slowly to a structure or component and remain constant or change very slowly over time. These loads do not cause significant dynamic effects, and the structure is assumed to be in equilibrium under these loads. Examples include the weight of a stationary object or a constant force applied by a hydraulic press.
2.Explain the concept of factor of safety in machine design.Concept
The factor of safety is the ratio of the stress at which the material fails to the working stress the part is designed for. For ductile materials under static load the failure stress is the yield strength, N = S_yt/σ; for brittle materials it is the ultimate strength, N = S_ut/σ. It covers uncertainty in material properties, loads, stress analysis and manufacturing, and is larger where failure would be costly or dangerous.
3.Why is the factor of safety important in engineering design?Application
The factor of safety is crucial because it accounts for uncertainties in material properties, loading conditions, and potential flaws in manufacturing. It ensures that even if the actual conditions differ from the design assumptions, the component will not fail. This is especially important in critical applications where failure could lead to significant economic loss or safety hazards.
4.What happens if the factor of safety is too low in a design?Application
If the factor of safety is too low, the component may not be able to withstand unexpected loads or variations in material properties, leading to premature failure. This can result in costly repairs, downtime, or even catastrophic failure, especially in safety-critical applications. Therefore, selecting an appropriate factor of safety is essential for reliable and safe design.
5.How do you determine the appropriate factor of safety for a given application?Application
The appropriate factor of safety depends on several factors, including the material properties, the nature of the load, the consequences of failure, and industry standards or regulations. Engineers often refer to design codes and standards, which provide guidelines for selecting the factor of safety based on these considerations. Experience and judgment also play a role in determining the appropriate value.
6.Explain how material properties influence the design for static loading.Concept
Material properties such as yield strength, ultimate tensile strength, and modulus of elasticity are critical in designing for static loading. These properties determine how a material will respond to applied loads. For static loading, the design must ensure that the stresses do not exceed the material's yield strength to prevent permanent deformation. The choice of material directly affects the component's ability to withstand static loads safely.
7.What is the difference between yield strength and ultimate tensile strength?Concept
Yield strength is the stress at which a material begins to deform plastically, meaning it will not return to its original shape when the load is removed. Ultimate tensile strength is the maximum stress a material can withstand while being stretched before breaking. In design, yield strength is often more critical for ensuring that components do not undergo permanent deformation under static loads.
8.Why might a designer choose a higher factor of safety for a bridge compared to a household appliance?Application
A higher factor of safety might be chosen for a bridge because the consequences of failure are much more severe, potentially leading to loss of life and significant economic impact. Bridges are also subject to variable and unpredictable loads, such as traffic and environmental conditions. In contrast, household appliances typically have more predictable and less critical loading conditions, allowing for a lower factor of safety.
9.Calculate the factor of safety if a component is subjected to a maximum stress of 150 MPa and the material has a yield strength of 300 MPa.Numerical
The factor of safety (FoS) is calculated as the ratio of the material's yield strength to the maximum applied stress. FoS = Yield Strength / Maximum Stress = 300 MPa / 150 MPa = 2. Therefore, the factor of safety is 2.
10.A tie bar must carry a steady axial load of 5000 N. If the allowable tensile stress is 250 MPa, what is the minimum cross-sectional area?Numerical
Area = force / allowable stress = 5000 N / 250 N/mm² = 20 mm², i.e. 2 × 10⁻⁵ m². Working in N and mm gives MPa directly and avoids power-of-ten slips. In practice the bar would be rounded up to a standard size, and the allowable stress itself already includes the factor of safety.
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