Fatigue — S-N curves, Miner's rule: Exercises — Physics deep dive

Two constants recur, so let us name them once, in plain words:

Figure — Fatigue — S-N curves, Miner's rule

Level 1 — Recognition

L1.1

On an S–N curve, which axis is the number of cycles to failure, and is it linear or logarithmic?

Recall Solution

The horizontal (x) axis is $N$ , the number of cycles to failure, drawn on a logarithmic scale. The vertical (y) axis is the stress amplitude $S = σ_{a}$ . Why log? Because $N$ ranges over many powers of ten ( $1 0^{3}$ to $1 0^{9}$ ); a linear axis would squash everything against the left edge. Look at figure s01 — the equally-spaced ticks are $1 0^{3}, 1 0^{4}, 1 0^{5}, \dots$ , not $1, 2, 3$ .

L1.2

A part sees $σ_{ma x} = + 220$ MPa and $σ_{min} = - 60$ MPa each cycle. Find the mean stress $σ_{m}$ and the amplitude $σ_{a}$ .

Recall Solution

What we do: plug into the two cycle formulas. $σ_{m} = \frac{σ _{ma x} + σ _{min}}{2} = \frac{220 + ( - 60 )}{2} = \frac{160}{2} = 80 MPa$ $σ_{a} = \frac{σ _{ma x} - σ _{min}}{2} = \frac{220 - ( - 60 )}{2} = \frac{280}{2} = 140 MPa$ Why: the mean is the middle of the swing (holds the crack ajar); the amplitude is the half-swing (opens and closes it). See Stress and Strain for the underlying $σ$ definitions.

L1.3

True or false: aluminium and Ti-6Al-4V both have a true endurance limit like plain-carbon steel.

Recall Solution

False. A true endurance limit — a flat floor below which life is effectively infinite — is mainly a plain-carbon / low-alloy steel feature. Aluminium and common titanium alloys (Ti-6Al-4V) keep sloping down past $1 0^{7}$ cycles, so engineers quote a "fatigue strength at $N = 5 \times 1 0^{8}$ " instead.

Level 2 — Application

L2.1

A strut has $σ_{f}^{'} = 1800$ MPa and $b = 0.09$ . It is cycled at amplitude $S = 450$ MPa. How many cycles to failure?

Recall Solution

Step 1 — pick the tool. We know $S$ and want $N$ , so invert Basquin: $N = (S / σ_{f}^{'})^{- 1/ b}$ . Step 2 — the ratio. $S / σ_{f}^{'} = 450/1800 = 0.25$ . Step 3 — the exponent. $- 1/ b = - 1/0.09 = - 11.11$ . Step 4 — evaluate with logs (a power like $0.2 5^{- 11.11}$ is easiest through $ln$ ): $ln N = - 11.11 \times ln (0.25) = - 11.11 \times (- 1.3863) = 15.40$ $N = e^{15.40} \approx 4.9 \times 1 0^{6} cycles$ Why logs? They turn "raise to the $- 11.11$ power" into a single multiplication, then we exponentiate back.

L2.2

Same strut ( $σ_{f}^{'} = 1800$ MPa, $b = 0.09$ ). Design life is $N = 1 0^{7}$ cycles. What amplitude $S$ is allowed?

Recall Solution

Step 1 — pick the tool. We know $N$ , want $S$ : use Basquin forward, $S = σ_{f}^{'} N^{- b}$ . Step 2 — the power. $N^{- b} = (1 0^{7})^{- 0.09} = 1 0^{- 0.63}$ . $1 0^{- 0.63} = e^{- 0.63 \times l n 10} = e^{- 1.4506} = 0.2344$ Step 3 — multiply. $S = 1800 \times 0.2344 \approx 422 MPa$ . Why: at a longer required life the allowed stress is lower — the curve slopes down. Compare to L2.1: pushing life from $4.9 \times 1 0^{6}$ to $1 0^{7}$ cost us ~28 MPa. Consult Safety Factors & Margins of Safety before using this bare number.

L2.3

A bracket sees $n_{1} = 1.0 \times 1 0^{4}$ cycles at a level where $N_{1} = 1.0 \times 1 0^{5}$ , and $n_{2} = 3.0 \times 1 0^{4}$ cycles at a level where $N_{2} = 6.0 \times 1 0^{4}$ . Compute total Miner damage $D$ . Does it survive?

Recall Solution

Step 1 — level-1 fraction. $D_{1} = n_{1} / N_{1} = 1 0^{4} /1 0^{5} = 0.10$ . Step 2 — level-2 fraction. $D_{2} = n_{2} / N_{2} = 3 \times 1 0^{4} /6 \times 1 0^{4} = 0.50$ . Step 3 — sum. $D = 0.10 + 0.50 = 0.60$ . Verdict: $0.60 < 1$ ⇒ survives, having spent $60%$ of its fatigue life.

Level 3 — Analysis

L3.1

A part has already accumulated $D = 0.60$ (from L2.3). A new load level appears with $N_{3} = 8.0 \times 1 0^{4}$ . How many cycles $n_{3}$ at this level until failure?

Recall Solution

Step 1 — remaining budget. $1 - 0.60 = 0.40$ of the life is left. Step 2 — set fraction = budget. $n_{3} / N_{3} = 0.40 \Rightarrow n_{3} = 0.40 \times 8.0 \times 1 0^{4} = 3.2 \times 1 0^{4}$ cycles. Why: Miner treats the budget as a single pool; whatever level empties the last $0.40$ triggers failure.

L3.2

Two nominally identical Ti struts are tested. Strut A fails at $S = 600$ MPa after $N = 1.0 \times 1 0^{4}$ cycles; strut B fails at $S = 300$ MPa after $N = 4.0 \times 1 0^{6}$ cycles. Find $b$ and $σ_{f}^{'}$ from these two points.

Recall Solution

Step 1 — pick the tool. Two points on a log–log straight line fix its slope. Take logs of Basquin $S = σ_{f}^{'} N^{- b}$ : $lo g S = lo g σ_{f}^{'} - b lo g N$ Step 2 — slope $b$ = (drop in $lo g S$ )/(rise in $lo g N$ ): $b = - \frac{l o g S _{B} - l o g S _{A}}{l o g N _{B} - l o g N _{A}} = - \frac{l o g 300 - l o g 600}{l o g ( 4 \times 1 0 ^{6} ) - l o g ( 1 0 ^{4} )}$ $= - \frac{- 0.3010}{6.6021 - 4} = - \frac{- 0.3010}{2.6021} = 0.1157$ Step 3 — coefficient $σ_{f}^{'}$ from point A: $σ_{f}^{'} = S_{A} N_{A}^{b} = 600 \times (1 0^{4})^{0.1157}$ . $(1 0^{4})^{0.1157} = 1 0^{0.4628} = 2.903 \Rightarrow σ_{f}^{'} = 600 \times 2.903 \approx 1742 MPa$ Why two points suffice: a straight line needs exactly two anchors; slope gives $b$ , either point back-solves the intercept $σ_{f}^{'}$ .

L3.3

Using the fit from L3.2 ( $b = 0.1157$ , $σ_{f}^{'} = 1742$ MPa), predict the life at $S = 400$ MPa.

Recall Solution

$N = (S / σ_{f}^{'})^{- 1/ b} = (400/1742)^{- 1/0.1157} = (0.22962)^{- 8.643}$ $ln N = - 8.643 \times ln (0.22962) = - 8.643 \times (- 1.4713) = 12.716$ $N = e^{12.716} \approx 3.33 \times 1 0^{5} cycles$ Sanity check: $400$ MPa sits between the two test stresses ( $300$ and $600$ ), and $3.3 \times 1 0^{5}$ sits between their lives ( $4 \times 1 0^{6}$ and $1 0^{4}$ ) — the interpolation is self-consistent.

Level 4 — Synthesis

L4.1

A launch profile hits a bracket with three vibration levels. From the S–N curve (fit $σ_{f}^{'} = 1742$ MPa, $b = 0.1157$ ):

$n_{1} = 2.0 \times 1 0^{3}$ cycles at $S_{1} = 500$ MPa
$n_{2} = 1.5 \times 1 0^{4}$ cycles at $S_{2} = 400$ MPa
$n_{3} = 8.0 \times 1 0^{4}$ cycles at $S_{3} = 300$ MPa

Compute the total Miner damage from one launch. How many launches can the bracket survive?

Recall Solution

Step 1 — life at each level via $N = (S / σ_{f}^{'})^{- 1/ b}$ , with $- 1/ b = - 8.643$ :

$N_{1} = (500/1742)^{- 8.643} = (0.28703)^{- 8.643}$ : $ln N_{1} = - 8.643 (- 1.2481) = 10.788 \Rightarrow N_{1} \approx 4.85 \times 1 0^{4}$
$N_{2} = (400/1742)^{- 8.643} = 3.33 \times 1 0^{5}$ (from L3.3)
$N_{3} = (300/1742)^{- 8.643} = (0.17222)^{- 8.643}$ : $ln N_{3} = - 8.643 (- 1.7588) = 15.202 \Rightarrow N_{3} \approx 4.00 \times 1 0^{6}$ Step 2 — damage fractions: $D_{1} = \frac{2 \times 1 0 ^{3}}{4.85 \times 1 0 ^{4}} = 0.04124, D_{2} = \frac{1.5 \times 1 0 ^{4}}{3.33 \times 1 0 ^{5}} = 0.04505, D_{3} = \frac{8 \times 1 0 ^{4}}{4.00 \times 1 0 ^{6}} = 0.02000$ Step 3 — damage per launch: $D_{launch} = 0.04124 + 0.04505 + 0.02000 = 0.10629$ . Step 4 — launches to failure: $D = 1$ when $# launches = \frac{1}{0.10629} \approx 9.4 \Rightarrow 9 full launches$ (round down — the 10th would push $D$ past 1). Why round down: you may only fly a launch that finishes with $D \leq 1$ ; $9 \times 0.10629 = 0.957 < 1$ , but $10 \times = 1.063 > 1$ . Loads come from Launch Loads & Environments; the amplitude spectrum itself is derived in Random Vibration & PSD.

Level 5 — Mastery

L5.1

A structure sees two damage sources per orbit-day: (a) launch is a one-time event depositing $D_{launch} = 0.10629$ (from L4.1); (b) on orbit, each ~90-min thermal cycle imposes $S_{th} = 180$ MPa. Using the same fit ( $σ_{f}^{'} = 1742$ MPa, $b = 0.1157$ ), the mission runs for 5 years. With one thermal cycle every 90 minutes, does the structure survive? Apply a fatigue safety factor requiring $D \leq 0.5$ .

Recall Solution

Step 1 — thermal-cycle life. $N_{th} = (180/1742)^{- 8.643} = (0.10333)^{- 8.643}$ . $ln N_{th} = - 8.643 \times ln (0.10333) = - 8.643 \times (- 2.2694) = 19.615 \Rightarrow N_{th} \approx 3.30 \times 1 0^{8}$ Step 2 — count thermal cycles in 5 years. One every 90 min = 16 per day. $n_{th} = 16 \times 365.25 \times 5 = 29, 220 cycles$ Step 3 — thermal damage. $D_{th} = \frac{29 , 220}{3.30 \times 1 0 ^{8}} = 8.85 \times 1 0^{- 5}$ Step 4 — total mission damage. $D_{total} = D_{launch} + D_{th} = 0.10629 + 0.0000885 = 0.10638$ Step 5 — apply the safety factor. Requirement: $D \leq 0.5$ . Since $0.10638 \leq 0.5$ ⇒ survives with margin; the used-up fraction of the allowed half-budget is $0.10638/0.5 = 0.213$ , i.e. only $21%$ of the derated life. Physical insight: launch dominates by three orders of magnitude — a few minutes of violent vibration outweigh 5 years of gentle thermal breathing, because damage scales like $S^{1/ b}$ and launch stresses are far higher. Thermal cycling matters most for low-cycle, high-strain joints; see Thermal Cycling on Orbit. If a crack were already present, you would switch models entirely to Fracture Mechanics & Crack Growth (Paris' Law).

L5.2

For the same structure, how many extra thermal cycles alone (no more launches) would it take to reach the derated failure budget $D = 0.5$ , starting from $D_{total} = 0.10638$ ?

Recall Solution

Step 1 — remaining budget to the derated limit: $0.5 - 0.10638 = 0.39362$ . Step 2 — cycles at thermal level: $n = 0.39362 \times N_{th} = 0.39362 \times 3.30 \times 1 0^{8} = 1.30 \times 1 0^{8}$ cycles. Step 3 — convert to years at 16 cycles/day: $1.30 \times 1 0^{8} / (16 \times 365.25) \approx 2.22 \times 1 0^{4}$ years. Why the huge number: thermal amplitude sits far down the shallow tail of the S–N curve, so each cycle spends a vanishingly small fraction. Thermal fatigue is not the life-limiter here — launch is.