Fatigue strength and the S-N diagram - Design of Machine Elements - Mechanical Engineering

Fatigue Failure

Fatigue failure is the process by which a material fails under repeated or fluctuating stresses that are often much lower than its ultimate tensile strength and may even be below its yield strength. With repeated loading, the resistance to these fluctuating stresses decreases as the number of cycles increases. This progressive loss of resistance, culminating in crack initiation and propagation until sudden fracture, is called fatigue.

Difference Between Static Load Failure and Fatigue Failure

Failure under static loads and failure by fatigue differ fundamentally in mechanism, appearance and design implications.

Static Load Failure

• Illustration: Simple tensile test is a typical example.
• Load Application: Load is applied gradually, allowing time for material deformation.
• Material Behaviour: Ductile materials show significant plastic deformation before fracture.
• Fracture Surface: Fracture surface shows a silky, fibrous appearance from crystal stretching.

Fatigue Failure

• Initiation: Begins at a localised crack which originates from a stress concentrator.
• Likely Crack Origins: Oil holes, keyways, screw threads, surface scratches, stamp marks, inspection marks, internal defects such as blow holes and non‐metallic inclusions.
• Propagation: Under fluctuating stresses the crack progressively grows, reducing the effective cross‐section until sudden final fracture.

Characteristics of Fatigue Failure

1. Slow Crack Growth: The crack propagation region exhibits a fine fibrous appearance indicating gradual progression.
2. Sudden Fracture: The final failure region is abrupt and shows a coarse granular appearance.

Endurance Limit (Endurance Strength)

The endurance limit (also called fatigue limit) of a material is the maximum value of completely reversing (fully reversed) stress that a standard specimen can sustain for an effectively unlimited number of cycles without failure. It is denoted by σe′ (for a standard rotating beam specimen).

Endurance limits are determined experimentally (commonly with a rotating-beam machine; method developed by R. R. Moore). Results are plotted as an S-N curve, showing applied maximum stress σf versus the number of cycles N to failure on log-log or semi‐log axes. Each test gives one failure point; test points scatter and an average curve is drawn.

For many ferrous materials the S-N curve becomes asymptotic around 10⁶ cycles; the corresponding stress is taken as the endurance limit (infinite life criterion). Non‐ferrous materials usually show a continuing downward slope beyond 10⁶ cycles and do not possess a clear limiting endurance stress; their allowable stress is then expressed as a function of the required number of cycles.

Notch Sensitivity

The theoretical stress concentration factor is denoted K_t. In fatigue situations the actual reduction in endurance strength due to a notch is usually less severe than that predicted by K_t, so a separate fatigue stress concentration factor K_f is used for design. The fatigue stress concentration factor depends on material microstructure (grain size) and the notch geometry.

The notch sensitivity factor q quantifies how much of the theoretical stress concentration is effective in fatiguing the material. It is defined so that:

The relation between K_f, K_t and q is usually written as:

K_f = 1 + q (K_t − 1)

Here K_t depends on geometric ratios (for example d/w in shaft shoulders). A value of q close to 1 means the material is highly notch sensitive (notch has strong effect), while q close to 0 indicates little sensitivity.

Estimation of Endurance Limit for Practical Components

Approximate relationships between the rotating-beam endurance limit σe′ and ultimate tensile strength σut (50% reliability) are often used as starting values:

• Steels: σe′ ≈ 0.5 σut
• Cast iron and cast steels: σe′ ≈ 0.4 σut

The component endurance limit σe differs from the standard specimen value σe′ because of several modifying factors. A commonly used relationship is:

σe = k_a k_b k_c k_d σe′

• k_a = surface finish factor (depends on σut and surface condition)
• k_b = size factor (larger parts generally have lower endurance due to more surface defects)
• k_c = reliability factor (corrects σe′ from 50% test reliability to desired reliability)
• k_d = factor accounting for stress concentration (usually related to fatigue stress concentration factor)

The modifying factor k_d to account for stress concentration is sometimes expressed in terms of the effective fatigue stress concentration; common practical approximations are used in design tables.

Endurance Limit in Torsion and Relation to Bending

The endurance strength in torsion (τe) can be related to the reversed bending endurance (σe) using failure theories. According to the maximum shear stress theory (Tresca) and the distortion energy theory (von Mises), conversion formulae give τe in terms of σe.

Types of Fatigue Design Problems

Fatigue design problems are commonly of two broad types:

• Components subjected to completely reversed stresses (mean stress σm = 0): the stress alternates equally in tension and compression. Design may be for infinite life (use endurance limit criterion) or finite life (use S-N curve).
• Components subjected to fluctuating stresses (mean stress σm = 0): stresses have a non‐zero mean and may be tensile, compressive or mixed; design uses mean‐stress correction relationships such as Gerber, Goodman or Soderberg.

Design for Completely Reversed Stresses

For components to have infinite life under completely reversed loading, the induced stress amplitudes must be less than the corrected endurance strengths. Typically we require:

For finite life design, the S-N curve corresponding to the required number of cycles is used.

Cumulative Fatigue Damage (Miner's Rule)

When a component experiences varying stress levels during service, the damage is cumulative. Miner's linear damage rule (Palmgren-Miner rule) gives an estimate of life consumption:

If the component experiences stress σ₁ for n₁ cycles and σ₁ would cause failure at N₁ cycles, then the proportion of life consumed by those cycles is n₁/N₁. Summing over all stress levels:

When the actual number of cycles at each level is unknown but proportions α_i of the total life are known, and N is total life, then n_i = α_i · N and

Substituting into Miner's equation gives:

With α₁ + α₂ + ... + α_x = 1. Miner's rule is a useful engineering approximation but has limitations (it ignores load sequence effects and changes in material properties during cycling).

Components Subjected to Fluctuating Stresses (Mean Stress Corrections)

When mean stress σm is not zero, fatigue strength is reduced. The three commonly used mean‐stress correction methods are:

• Gerber's Parabola
• Goodman's Straight Line
• Soderberg's Straight Line

Gerber's Parabola

Gerber proposed a parabolic relationship between alternating stress amplitude σa and mean stress σm. Gerber's curve assumes the limiting conditions σa = σe when σm = 0, and σm = σut when σa = 0. The parabolic relation (standard form) is:

σa = σe · [1 − (σm/σut)²]

Gerber's criterion is often closest to experimental results for ductile materials but is less conservative than Goodman.

Goodman's Straight Line

Goodman approximated the relation by a straight line between the points (σm = 0, σa = σe) and (σm = σut, σa = 0):

σa/σe + σm/σut = 1

Goodman is simpler and more conservative than Gerber in the positive mean stress region and is commonly used in design.

Soderberg's Straight Line

Soderberg replaced σut by the yield strength σy in the straight-line relation, giving a more conservative criterion:

σa/σe + σm/σy = 1

Soderberg's line yields lower allowable σa for a given σm and thus is the most conservative of the three.

Experimental Evidence and Practical Choice

Experimental fatigue data for ductile materials generally lie between Gerber's parabola and Goodman's straight line. Brittle materials' results often lie below Goodman's line and above a lower bound related to an inverse parabola. A generalised empirical relation is sometimes used (see figure):

where typical exponent values are approximately x = 1.6 for ductile materials and x = 0.6 for brittle materials. For design purposes, Goodman's line is frequently adopted because it is simple and conservative.

For negative mean stresses (compressive mean), ductile materials often sustain nearly the same alternating stress amplitude as for zero mean stress; brittle materials may exhibit linear increase in allowable σa with decreasing σm in compression.

Complex Fluctuating Stresses

If the state of stress at any point is described by then the principal fluctuating stresses σ_p1 σ_p2 and can be calculated from the results the mean principal stresses σ_p1m and σ_p2m may be calculated. Similarly, the variable components of two principal stresses σ_p1 σ_p2 and may be calculated.

Compute principal mean stresses and principal alternating components, then reduce to equivalent mean and equivalent alternating stresses using an appropriate failure criterion:

• For ductile steels, use the distortion energy (von Mises) criterion to obtain equivalent (von Mises) mean and alternating stresses.
• For example, if for steel,  and are equivalent Von Mises stresses (i.e. following distortion energy criterion), then;

  For cast iron, if  and are equivalent principal stresses, then

  

Equivalent Stress Method

Goodman's relation can be rewritten to define an equivalent normal stress σn at the critical point. In rearranged form:

Thus an equivalent direct or normal stress is defined as:

Similarly an equivalent shearing stress τ_n may be defined for fluctuating shear, with an analogous relation:

Once σn and τn are obtained they can be used in a chosen static failure theory (e.g., von Mises, Tresca) to determine dimensions or factor of safety under fluctuating multiaxial loads.

Practical Measures to Combat Fatigue

• Material Selection: Choose homogeneous materials with fine grain structure and minimal inclusions or internal defects, since these act as crack initiation sites.
• Reduce Stress Concentration: Design transitions, grooves, and notches to minimise theoretical stress concentration K_t. Avoid sharp corners; use fillets, generous radii and improved transitions. Circumferential grooves can be split or relocated to reduce peak concentration.
• Machining and Fabrication: Ensure good surface finish; remove nicks and sharp machining marks. Design welded joints carefully as welding introduces residual stresses and stress raisers.
• Surface Treatments: Induce beneficial compressive residual surface stresses by shot peening, hammer peening or surface rolling. Surface hardening processes such as carburising, nitriding, cyaniding, flame or induction hardening improve fatigue resistance.

Solved Numericals

Q1. A cylindrical shaft is subjected to alternating stress of 100 MPa. Fatigue strength to sustain 1000 cycles is 490 MPa. If the corrected endurance strength is 70 MPa, What will be the estimated shaft life in revolutions? If the fatigue strength of the bar to withstand N cycles is 100 MPa then what will be the value of N (To the nearest integer)?
Ans: 281800 - 282000
The fatigue strength of the bar is calculated using the SN diagram where for 103 cycles the stress is 0.9Sut and for 106 cycles, the stress is endurance limit (S_e).

Calculation (Solution steps)

Given:

σ = 100 MPa → N = ?

For 1000 cycles → σ = 490 MPa = 0.9 S_ut

The corrected endurance limit, S_e = 70 MPa

At σ_e → N = 10⁶ cycles

Use the straight-line approximation on the log-log S-N curve between (N = 10³, σ = 490 MPa) and (N = 10⁶, σ = 70 MPa):

$\frac{\log }{10}=\frac{6}{-}$
⇒ N = 281914 cycle

Q2. The figure shows the relationship between fatigue strength (S) and fatigue life (N) of a material. The fatigue strength of the material for a life of 1000 cycles is 450 MPa, while its fatigue strength for a life of 10⁶ cycles is 150 MPa.

The life of a cylindrical shaft made of this material subjected to an alternating stress of 200 MPa will then be ____________ cycles (round off to the nearest integer).
Ans: 152000 - 165000

Solution:

In Δ APC

In Δ BQC

Calculation:
Given:
S₁ = 450 MPa, N₁ = 1000 cycles, S₂ = 150 MPa, N₂ = 10⁶ cycles.

6 - log N = 0.785578
log N = 5.214422
Number of cycle (N) equals 10^5.214422 = 163842 cycles.

Q3. A machine element has an ultimate strength (σ_u) of 600 N/mm², and endurance limit (σ_en) of 250 of 250 N/mm². The fatigue curve for the element on a log-log plot is shown below. If the element is to be designed for a finite life of 10000 cycles, the maximum amplitude of a completely reversed operating stress is_______N/mm².

Ans: 370 - 390

Solution:

Given:

σ_u = 600 N/mm²

σ_en = 250 N/mm²

Points on log-log fatigue curve:

A → (log(0.8 σ_u), 3)

B → (log S, 4)

C → (log σ_en, 6)

Slope of line AB equals slope of AC; solve for S (stress amplitude at N = 10⁴):

Computed value S ≈ 386.19 N/mm² (≈ 386, within the stated range 370-390).

Q4. A cylindrical shaft is subjected to an alternating stress of 100 MPa. Fatigue strength to sustain 1000 cycle is 490 MPa. If the corrected endurance strength is 70 MPa, estimated shaft life will be
Solution:

Equation of straight line through points (log₁₀N = 3, log₁₀490) and (log₁₀N = 6, log₁₀70):

Use the linear equation in x = log₁₀N and y = log₁₀σ to find x corresponding to y = log₁₀(100) = 2.

Solving gives x ≈ 5.449

N = 10^5.449 ≈ 281,190 cycles. The most appropriate answer close to interpolation is ≈ 281,914 cycles (option C in the original choices).