Question bank — Structural design process — load cases, FOS (factor of safety)

False — $FOS$ covers known ignorance (material scatter, analysis error, load uncertainty); an unmodelled load case or a manufacturing flaw can still cause failure below the factored load.

False — a higher $FOS$ adds mass, and mass stolen from payload or propellant can compromise the mission; safety is a balance, not "more is always better".

False — you must first multiply the limit stress by $FOS$; a stress comfortably under allowable can still give $MS<0$ once the factor is applied (see Figure 1's shifted danger line).

False — yield and ultimate are separate checks with different $FOS$ and different allowables; a part could survive rupture yet permanently deform, which is its own failure.

False — it is because the load changes slowly compared to the structure's response time, so the structure reaches near-equilibrium; rigidity is not the reason.

False — they act at different frequencies and times; you either envelope them or combine them by root-sum-square where they genuinely coincide, not by naïve addition.

False — $MS=0$ means $S_{\text{allow}}$ exactly equals the factored design load $FOS\cdot S_{\text{limit}}$, i.e. zero spare margin but still acceptable; failure is $MS<0$.

False — perpendicular loads combine as vectors (the diagonal in Figure 3, Pythagoras / RSS); scalar summing overstates the load and wastes mass, as covered in Quasi-static Loads and Launch Environment.

The error is inflating the supply; $FOS$ multiplies the demand ($S_{\text{limit}}$) or equivalently divides the allowable — we never boost the material's real strength.

The $FOS$ is missing from the denominator; the correct form is $MS=S_{\text{allow}}/(FOS\cdot S_{\text{limit}})-1$, so the unfactored version dangerously overstates the margin.

Load cases occur at different times/frequencies, so summing them all is over-conservative; you design to the worst-case envelope and RSS only truly simultaneous orthogonal loads.

A negative stress just means compression (our sign convention, Figure 2); it is the margin $MS$ that signals failure when negative, and margin is computed from magnitudes against the relevant allowable.

Qualification testing is typically done at or above ultimate to prove the factored design load $FOS\cdot S_{\text{limit}}$ is survivable — see Qualification vs Acceptance Testing; passing only at limit proves nothing about the margin.

You forgot the $FOS$; with $FO{S}_{\text{yield}}=1.25$ the design stress is $187.5$ MPa and the real margin is $300/187.5-1=0.6$, not $1.0$.

They are algebraically equivalent, but inflating the demand keeps a single, clearly-defined "design load" $FOS\cdot S_{\text{limit}}$ the structure is checked against (the shifted line in Figure 1), avoiding double-counting when several uncertainties stack.

Permanent deformation (yield) is a recoverable-in-design, less catastrophic outcome than rupture (ultimate), so we tolerate a smaller cushion against yield and a larger one against fracture — the two danger zones in Figure 1.

Their strength has more scatter and less validation — composites vary batch to batch, pressure vessels fail catastrophically — so a larger cushion absorbs that greater uncertainty.

Because low-frequency loads change slowly relative to the structure's response, the dynamic amplification is negligible and a constant-$g$ static analysis captures the peak stress far more cheaply.

You launch it only once and cannot recover it to redesign; the whole process is forecasting the loads and proving on paper and in ground tests, per Quasi-static Loads and Launch Environment.

It converts the capacity/demand ratio $S_{\text{allow}}/(FOS\cdot S_{\text{limit}})$ into fractional spare capacity, so $MS$ reads directly as "how much extra strength beyond the factored load", with the pass line landing cleanly at zero.

Extra structural mass reduces available payload and propellant, raising launch cost or shrinking the mission — in space, mass is the enemy, so margins are trimmed to just-enough.

Zero — $FOS\times 0=0$, so that case imposes no demand; the part is sized by whichever other load case dominates the envelope.

No — such a razor-thin margin leaves no room for the analysis uncertainty $FOS$ is supposed to smooth over locally, so most engineers add material or re-examine assumptions rather than ship it.

The part is massively over-strength for that load — a signal to lighten it (thinner walls, lighter material) to save mass, provided no other load case governs, per Mass Budget and Structural Efficiency.

The ultimate check, because a brittle material fractures with little prior yielding, so its higher $FOS$ against rupture drives the sizing and the yield margin becomes almost irrelevant.

Not necessarily — the compressive case can trigger buckling at a stress far below the yield allowable, so it may use a lower buckling allowable and yield a smaller margin, even though the magnitudes match.

Yes — thermal $\Delta T$ induces internal stress through constrained expansion, which counts as a static load case that must be enveloped with the mechanical ones, and can be checked in Finite Element Analysis of Spacecraft Structures.

Raise the limit load estimate $S_{\text{limit}}$ to a conservative value first (that is a modelling choice); $FOS$ covers residual scatter, not gross uncertainty in the load definition itself.

The static ultimate check missed a resonance that amplified stress dynamically; the fix is a proper dynamic analysis and requalification, tying into Random & Sine Vibration Testing.