Worked examples — Verification methods — analysis, test, inspection, demonstration

This page is the practice arena for the four verification methods. The parent note told you what analysis, test, inspection, and demonstration are. Here we throw every kind of case at you — every method, the edge cases where a method fails, the degenerate "zero" inputs, the limiting values, a real word problem, and a nasty exam twist — and work each one from the ground up.

Before we compute anything, two words that will appear over and over must be earned:

Walk through the figure (s01) concretely. The mint green bar on top is the allowable — say a material can take $8$ units of stress before it yields. The coral bar below it is the applied — the launch only produces $5$ units. Because the mint bar is longer, its tip pokes out past the coral tip; that stick-out piece is drawn as the lavender double arrow labelled "margin". The dashed vertical line sits exactly at the coral (applied) tip — that is the "MoS zero line". Read the picture like this: if the mint tip lands to the right of the dashed line, you PASS; if it lands exactly on it, MoS $=0$; if the coral bar ever grows past the mint bar, applied overshoots and you FAIL. Every example below is just this same picture with different bar lengths.

Why subtract 1? Because $\frac{\text{allowable}}{\text{applied}}$ alone is a ratio: it equals 1 when the two are equal. Subtracting 1 shifts that "just barely equal" point to zero, so the sign of MoS instantly tells you pass/fail. That shift is why the dashed zero line in the figure sits right at the applied tip.

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Building the beam-stress tool from scratch

Several examples below need the bending stress of a beam. The parent note quoted the collapsed formula $\sigma =\frac{6M}{b{h}^2}$; let us earn every symbol so no example makes an unexplained leap.

Walk through figure s02. The slate rectangle is the beam's cross-section (width $b$ across, height $h$ up). The dashed horizontal line through its middle is the neutral axis — the fibres there are neither stretched nor squashed. The lavender arrow points from that middle line up to the top edge; its length is $c=h/2$, the farthest fibre. The little coral tick marks stacked up the height remind you that each sliver of area contributes $y^{2}dA$ to $I$ — the higher up (bigger $y$), the more it counts, which is why $I$ grows like $h^3$.

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The scenario matrix

Every problem this topic can throw at you falls into one of these cells. Each worked example below is tagged with the cell it hits, and together they cover the whole grid. The flow-schema below (figure s03) shows how the cells fan out from a single starting question, so you can see that we cover the whole space, not just a list.

Walk through figure s03. The butter box on the left is the single root question every verification starts from: "Which verification scenario am I in?" Eight arrows fan out from it to the right, one per row of the table below — the arrow colour just groups related cells (mint = passes, coral = fails/wrong-method, lavender = degenerate/limit, butter = statistical). Reading top-to-bottom the leaf boxes are cells A through H; each box names the case class and its one-line "trick". The point of the picture is coverage: no matter which node you land on, an example below works it out, so the fan has no missing branch.

#	Cell (case class)	What makes it tricky	Example
A	Analysis, positive margin	Straight number-crunch, PASS	Ex 1
B	Analysis, negative margin (FAIL)	Sign flips negative → redesign	Ex 2
C	Zero / degenerate input	Divide-by-zero, undefined MoS	Ex 3
D	Limiting value (MoS = 0)	Exactly on the boundary	Ex 4
E	Test, qualification vs acceptance	Two different levels, same unit	Ex 5
F	Wrong method chosen	Requirement can't be tested → must inspect	Ex 6
G	Real-world word problem	Extract numbers from prose	Ex 7
H	Exam twist: combined + statistical	Two uncertainties combine, then verify	Ex 8

We link the physics tools as they enter: Margin Philosophy, Finite Element Analysis, Thermal Math Modeling, Vibration Testing, Acceptance Testing, Traceability Matrix.

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Cell A — Analysis, positive margin

Forecast: guess before computing — will a 2 kg mass on a 1 cm-thick aluminium bar at 8g pass, or is it too thin? Write down PASS or FAIL now.

1. Inertial force. $F=m(ng)=2\times (8\times 9.81)=156.96\text{ N}$. Why this step? During launch the acceleration feels like extra gravity (a quasi-static load); Newton's second law converts the load factor "$n=8$" into a real force the bracket must carry.
2. Bending moment. $M=FL=156.96\times 0.10=15.696\text{ N⋅m}$. Why this step? A force at a distance from the fixed root bends the bar. The moment $M$ (defined above) measures that bending leverage — the farther out the mass, the harder the twist at the root.
3. Bending stress using the transparent form $\sigma =\frac{6M}{b{h}^2}$ (derived above, with $M$ the bending moment from step 2): $\sigma =\frac{6\times 15.696}{0.02\times (0.01{)}^2}=\frac{94.176}{2\times 10^{-6}}=4.7088\times 10^7\text{ Pa}=47.088\text{ MPa}.$ Why this step? Stress is bending effort concentrated at the farthest fibre (distance $c=h/2$ from the neutral axis); the top fibre stretches most, so we check there — if the worst fibre survives, all do.
4. Margin. $\text{MoS}=\frac{270}{47.088}-1=5.734-1=4.734$. Why this step? This is the pass/fail verdict. Positive and large → the bracket is way over-strong.

Verify: Units — Pa = N/m². $47.088\text{ MPa}$ is far below $270\text{ MPa}$ (only about one-sixth of the allowable), so MoS $\approx 4.73>0$ → PASS. In the s01 picture, the coral "applied" bar is tiny and the mint "allowable" bar towers over it. Sanity: a chunky 1 cm bar carrying only 2 kg should be easily strong, matching a big positive margin.

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Cell B — Analysis, negative margin (FAIL)

Forecast: thinner and higher g — do you think it still passes? Guess.

1. New force. $F=m(ng)=2\times (12\times 9.81)=235.44\text{ N}$. Why this step? Higher load factor means more inertial force — same Newton's-law conversion, now with $n=12$.
2. New bending moment then stress. $M=FL=235.44\times 0.10=23.544\text{ N⋅m}$, so $\sigma =\frac{6M}{b{h}^2}=\frac{6\times 23.544}{0.02\times (0.004{)}^2}=\frac{141.264}{3.2\times 10^{-7}}=4.4145\times 10^8\text{ Pa}=441.45\text{ MPa}.$ Why this step? Notice $h$ appears squared in the denominator — shrinking $h$ from 10 mm to 4 mm ($2.5\times$ thinner) blows up stress by $2.5^2=6.25\times$. This is why thickness matters so much.
3. Margin. $\text{MoS}=\frac{270}{441.45}-1=0.6116-1=-0.3884$. Why this step? Negative! In the s01 picture the coral "applied" bar has now grown past the mint "allowable" bar — the tip lands left of the mint tip — so the part yields.

Verify: $441.45\text{ MPa}$ is larger than $270\text{ MPa}$ → MoS $<0$ → FAIL. The Finite Element Analysis would flag this region red. Fix: thicken $h$, add ribs, or pick a stronger alloy. Sanity check: ratio $\frac{441.45}{47.088}=9.375=1.5\times 6.25$ — the 1.5 is the load-factor increase ($12/8$), the 6.25 is the thickness effect. The physics is self-consistent.

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Cell C — Zero / degenerate input

Forecast: MoS $=\frac{{\sigma }_y}{\sigma }-1$. Plug in $\sigma =0$… what happens? Guess the trap.

1. Try the formula directly. $\text{MoS}=\frac{270}{0}-1$. Why this step? Dividing by zero is mathematically undefined — it is not a real number, and calling it "infinity" is only informal shorthand for "grows without bound as the denominator shrinks". You cannot write a finite margin here.
2. Interpret physically. The stress being exactly zero means the member is not loaded at all in this case. Why this step? A picture: the coral "applied" bar in figure s01 has zero length, so the mint "allowable" bar sticks out past it with no upper limit. The ratio $\frac{270}{\sigma }$ has no finite value — it just keeps rising as $\sigma$ approaches zero.
3. Correct engineering statement. Report: "MoS not applicable (denominator zero — undefined); member unloaded in this load case; verification satisfied by inspection of the load path, not by margin computation." Why this step? Never write a literal "$\infty$" or a made-up big number on a Traceability Matrix. Zero applied load is a different verification story — you inspect that the load path genuinely bypasses this member.

Verify: Limit check — as $\sigma \to 0^{+}$, the ratio $\frac{{\sigma }_y}{\sigma }$ increases without bound (no finite limit exists). The formula therefore confirms the trap: MoS is undefined at $\sigma =0$, hence the special reporting rule. Never a finite lie, and never a literal $\infty$ in the paperwork.

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Cell D — Limiting value (MoS exactly zero)

Forecast: applied equals allowable exactly. Pass or fail? Think about the "$>0$" in the requirement.

1. Compute MoS. $\text{MoS}=\frac{200}{200}-1=1-1=0$. Why this step? In figure s01 the coral and mint bars are now identical length; the lavender extra-length arrow has zero length; the mint tip lands exactly on the dashed zero line.
2. Read the requirement carefully. It said MoS $>0$ (strictly greater), not $\ge 0$. Why this step? The boundary case turns on one inequality symbol. $\text{MoS}=0$ means the part yields at precisely the design load — zero protection against any manufacturing scatter, temperature effect, or model error.
3. Verdict: MoS $=0$ does not satisfy "MoS $>0$". FAIL (marginal). Why this step? This is exactly why Margin Philosophy demands positive margin — reality always has scatter, and a zero-margin part is a coin flip.

Verify: $\frac{200}{200}-1=0$ exactly; $0>0$ is false → correctly fails the strict requirement. Limiting behaviour confirmed: this is the razor's edge between Ex 1 (positive) and Ex 2 (negative).

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Cell E — Test: qualification vs acceptance levels

Forecast: which level is higher, and why would you ever test a flight unit at a lower level than the design proof? Guess before reading.

1. Qualification level $=\text{Flight Limit}\times 1.25=8\times 1.25=10g$ RMS. Why this step? Qualification proves the design survives worse-than-flight, covering all future units built to that design. The $1.25$ buys margin against manufacturing + environment scatter.
2. Acceptance level $=\text{Flight Limit}\times 1.0=8\times 1.0=8g$ RMS. Why this step? Acceptance Testing confirms this specific unit has no build defects (cold solder, loose fastener) without over-stressing flight hardware — you already proved the design at 10 g on the qual unit, so you never re-punish the flight unit to 10 g.
3. What each verifies. Qual → the design margin. Acceptance → workmanship of one article. Why this step? Same requirement, two questions: "is the design good?" vs "is this copy built right?" Different levels answer different questions.

Verify: $8\times 1.25=10$; $8\times 1.0=8$. Qual ($10$) $>$ Acceptance ($8$) and Acceptance equals Flight ($8$) — the correct ordering. Ratio qual/acceptance $=10/8=1.25$, matching the design margin factor.

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Cell F — Wrong method chosen (method-selection twist)

Forecast: can you test a mass? What does testing even mean here? Guess the right method.

1. Reject the wrong method. Mass is a static property — it doesn't emerge from operation or environment. A mission simulation is a demonstration and reveals nothing about mass. Why this step? The parent note's intuition: you can't "test" a mass — you just weigh it, and weighing is inspection, not test.
2. Apply the correct method — inspection. Place the assembled, dry spacecraft on a calibrated scale: it reads $174.3\text{ kg}$. Why this step? Inspection directly observes the physical property against the limit — no model, no environment needed. Weighing is the textbook example of inspection.
3. Verify against the limit. Is $174.3\le 180$? Yes. For mass we report the remaining allocation rather than a stress-style MoS: spare $=180-174.3=5.7\text{ kg}$, which is $\frac{5.7}{180}=3.17\%$ under budget. Why this step? Mass margin is tracked in kilograms of spare allocation, which flows straight into mass budget tracking — not as the ratio-minus-one MoS used for stress.
4. State the verdict. $174.3\text{ kg}<180\text{ kg}$ → PASS by inspection, with $5.7\text{ kg}$ ($3.17\%$) of margin remaining. Why this step? The requirement is met and the method is now the correct one; both the number and the method must be right for a valid verification entry.

Verify: $174.3<180$ → PASS. Method-fit check: static scalar property → inspection ✓ (not test, not analysis, not demonstration). Spare $180-174.3=5.7$ kg; $5.7/180=0.03167\Rightarrow 3.17\%$. Both the numeric answer and the method selection are confirmed.

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Cell G — Real-world word problem

Forecast: which limit is the battery closer to violating — the hot ceiling or the cold floor? Guess.

1. Extract the numbers from the prose. Allowable band $[0,40{]}^{\circ }\text{C}$; predicted band $[5,38{]}^{\circ }\text{C}$. Why this step? A word problem hides the numbers in sentences; the first job is always to name the applied values (predicted extremes) and the allowable values (the two limits).
2. Hot-side margin. Distance to the ceiling $=40-38=2^{\circ }\text{C}$ of headroom. Why this step? For a two-sided limit you must check both ends; a temperature can fail high or low. The hot case tests the ceiling.
3. Cold-side margin. Distance to the floor $=5-0=5^{\circ }\text{C}$ of headroom. Why this step? The cold case tests the floor. Both margins must be positive for the requirement to hold across the whole orbit.
4. Overall verdict. Both $2^{\circ }\text{C}$ and $5^{\circ }\text{C}$ are $>0$ → PASS by analysis. The tighter margin is the hot side ($2^{\circ }\text{C}$). Why this step? Reporting the tightest margin tells the model-validation team where a small model error would matter most — the hot ceiling is the risk driver.

Verify: $40-38=2>0$ ✓ and $5-0=5>0$ ✓. Predicted band $[5,38]\subset [0,40]$ → requirement met. Tighter margin correctly identified as the hot side ($2<5$).

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Cell H — Exam twist: combine uncertainties, then verify

Forecast: two 10% scatters — do they add to 20%? Guess before you combine them.

1. Combine independent uncertainties in quadrature. ${\sigma }_{\text{total}}=\sqrt{{\sigma }_{\text{mfg}}^2+{\sigma }_{\text{env}}^2}=\sqrt{0.10^2+0.10^2}=\sqrt{0.02}\approx \mathrm{0.1414.}$ Why this step? Independent random errors do not add directly (that would be $0.20$); they add as squares under a root — the Pythagorean rule for uncertainties. Picture two perpendicular arrows of length 0.10; their resultant is $0.10\sqrt{2}$, not 0.20.
2. Go out to $3\sigma$ for high ($99.7\%$) coverage. $\text{Test Level}=\text{Mean}+3{\sigma }_{\text{total}}=1.0+3(0.1414)=1.0+0.4243=\mathrm{1.4243.}$ Why this step? A $3\sigma$ band on a normal distribution captures $99.7\%$ of all units — testing to this level means almost every real flight unit is bounded by your test.
3. Compare to the practical $1.25\times$ factor. Why this step? The pure statistics demands $\approx 1.42\times$, but the industry standard is $1.25\times$ (Ex 5). So $1.25$ is less conservative than full $3\sigma$ — a deliberate cost/risk compromise, accepting slightly under $3\sigma$ coverage to avoid over-testing and over-building hardware. This is Margin Philosophy in one number.

Verify: $\sqrt{0.10^2+0.10^2}=\sqrt{0.02}=0.14142$; $1+3(0.14142)=1.42426$. And $1.25<1.4243$, confirming the practical factor is the less-conservative compromise. Not $0.20$: quadrature ≠ linear sum ✓.

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Recall Quick self-test

Ex 1 bracket MoS (2 kg, 10 mm, 8g) ::: $+4.73$ → PASS Ex 2 bracket MoS (4 mm, 12g) ::: $-0.39$ → FAIL (stress 441 MPa > 270) MoS when applied stress is exactly zero ::: undefined (division by zero) — report "member unloaded", verify by inspection Does MoS $=0$ satisfy "MoS $>0$"? ::: No — strict inequality, marginal FAIL Qual vs acceptance level for 8g flight limit ::: 10g (design) vs 8g (workmanship) Right method to verify a mass requirement ::: Inspection (weigh it), not test Combine two independent 10% scatters ::: $\sqrt{0.1^2+0.1^2}=0.1414$, not 0.20 $3\sigma$ statistical test level vs industry $1.25\times$ ::: $1.42\times$ (statistics) vs $1.25\times$ (practical, less conservative) What does the load factor $n$ mean? ::: a plain dimensionless number of "g's" — the acceleration feels like $n$ times ordinary gravity How does $\sigma =Mc/I$ become $6M/(b{h}^2)$? ::: put $c=h/2$, $I=b{h}^3/12$; the algebra collapses to $6M/(b{h}^2)$