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

• Allowable $S_{\text{allow}}=300\text{MPa}$ (what the material can take — its yield).
• Limit $S_{\text{limit}}=150\text{MPa}$ (worst load expected, the demand).
• $FOS=1.25$.

$FOS$ multiplies the limit (the demand). So the design stress is $1.25\times 150=187.5\text{MPa}$. We inflate the demand, not the material strength. (This is exactly the "Fear inflates the FORCE, not the METAL" rule from the parent note.)

A structure is acceptable when $MS\ge 0$.

• Bracket A ($MS=+0.30$) passes — it has $30\%$ spare strength beyond the already-factored (design) load.
• Bracket B ($MS=-0.05$) fails — it is $5\%$ short of even the factored load. Redesign (thicken, add a rib, or change material).

Step 1 — design stress (WHAT: factor the limit; WHY: we must survive the inflated demand): $S_{\text{design}}=1.4\times 160=224\text{MPa}.$ Step 2 — margin (WHAT: fractional spare capacity; WHY: the "$-1$" makes "exactly enough" read as $0$): $MS=\frac{250}{224}-1=\mathrm{0.116.}$ $MS=+0.12\ge 0$ → passes, with about $12\%$ spare.

Step 1 — design load (WHAT: factor the demand): $P_{\text{design}}=1.25\times 30=37.5\text{kN}=37500\text{N}.$ Step 2 — stress condition (WHY: stress is force per area; keep it under yield to avoid permanent deformation, as in the figure): $\sigma =\frac{P_{\text{design}}}{A}\le {\sigma }_y.$ Step 3 — solve for the smallest $A$ ($MS=0$ gives the minimum): $A\ge \frac{P_{\text{design}}}{{\sigma }_y}=\frac{37500}{275\times 10^6}=1.364\times 10^{-4}{\text{m}}^2=136.4{\text{mm}}^2.$ Any larger area gives $MS>0$. Answer: $A_{\min }\approx 136{\text{mm}}^2$.

Two independent checks — the part must pass both.

Yield check: $M{S}_{\text{yield}}=\frac{240}{1.1\times 100}-1=\frac{240}{110}-1=1.182-1=\mathrm{0.182.}$ Ultimate check: $M{S}_{\text{ult}}=\frac{320}{1.4\times 100}-1=\frac{320}{140}-1=2.286-1=\mathrm{1.286.}$ Both $\ge 0$ → acceptable. The driving (smaller) margin is yield at $+0.18$; ultimate is comfortable. In practice the smaller margin is the one to watch.

Step 1 — design load: $P_{\text{design}}=1.4\times 12=16.8\text{kN}=16800\text{N}$. Step 2 — design stress (force over the given area): ${\sigma }_{\text{design}}=\frac{16800}{50{\text{mm}}^2}=\frac{16800}{50\times 10^{-6}{\text{m}}^2}=336\times 10^6\text{Pa}=336\text{MPa}.$ Step 3 — required allowable: For $MS=0$ we need $S_{\text{allow}}={\sigma }_{\text{design}}$. Answer: material must supply $\ge 336\text{MPa}$ ultimate.

Step 1 — axial force (WHAT: mass × acceleration; the $8g$ is $8\times g$): $F_a=m\cdot (8g)=40\times 8\times 9.81=3139.2\text{N}.$ Step 2 — lateral force: $F_l=m\cdot (3g)=40\times 3\times 9.81=1177.2\text{N}.$ Step 3 — resultant (WHY Pythagoras: axial and lateral are perpendicular, so they combine as the two legs of a right triangle — look at the blue triangle in the figure): $F=\sqrt{F_a^2+F_l^2}=\sqrt{3139.2^2+1177.2^2}=3352.7\text{N}.$ Compare to the naive scalar sum $3139.2+1177.2=4316.4\text{N}$. The scalar sum is $29\%$ larger — it pretends both loads point the same way. Designing to it wastes mass, and mass is the enemy in space. Answer: resultant $\approx 3353\text{N}$ (root-sum-square), not $4316\text{N}$.

The load cases occur at different times, so we design to the envelope (worst single case), not their sum.

• QSL: design stress $=1.4\times 95=133\text{MPa}$.
• Random: design stress $=1.4\times 130=182\text{MPa}$.
• Thermal: design stress $=1.4\times 60=84\text{MPa}$.

Random vibration is the largest → it drives the design. Its margin: $MS=\frac{200}{1.4\times 130}-1=\frac{200}{182}-1=\mathrm{0.099.}$ Answer: random vibration drives, $MS=+0.10$ (about $10\%$ spare — thin, worth watching).

Step 1 — allowed limit resultant (undo the FOS: the $5000\text{N}$ is the design load, so the limit is smaller): $F_{\text{limit}}=\frac{5000}{1.4}=3571.4\text{N}.$ Step 2 — lateral limit force: $F_l=60\times 2\times 9.81=1177.2\text{N}.$ Step 3 — solve the right triangle for the axial leg: $F_a=\sqrt{F_{\text{limit}}^2-F_l^2}=\sqrt{3571.4^2-1177.2^2}=3371.8\text{N}.$ Step 4 — convert to $g$: $n_a=\frac{F_a}{mg}=\frac{3371.8}{60\times 9.81}=\mathrm{5.73.}$ Answer: up to about $5.7g$ axial (with the $2g$ lateral present).

Step 1 — envelope: worst limit load $=\max (18,22)=22\text{kN}$ (sine drives). Step 2 — design loads:

• Yield: $1.1\times 22=24.2\text{kN}$.
• Ultimate: $1.4\times 22=30.8\text{kN}$.

Step 3 — area of the $8\text{mm}$ rod: $A=\frac{\pi d^2}{4}=\frac{\pi (8{)}^2}{4}=50.27{\text{mm}}^2.$ Step 4 — check both stresses:

• Yield stress $=\frac{24200}{50.27{\text{mm}}^2}=481.4\text{MPa}>250\text{MPa}$ → fails (negative margin).
• Ultimate stress $=\frac{30800}{50.27}=612.7\text{MPa}>350\text{MPa}$ → fails.

Step 5 — minimum area for each check ($MS=0$):

• Yield needs $A\ge \frac{24200}{250}=96.8{\text{mm}}^2$.
• Ultimate needs $A\ge \frac{30800}{350}=88.0{\text{mm}}^2$.

The larger requirement (yield, $96.8{\text{mm}}^2$) governs. Step 6 — minimum diameter: $d=\sqrt{\frac{4A}{\pi }}=\sqrt{\frac{4\times 96.8}{\pi }}=11.10\text{mm}.$ Answer: the $8\text{mm}$ rod fails both; minimum diameter $\approx 11.1\text{mm}$ (yield-driven).

We need $\frac{{\sigma }_y}{FO{S}_{\text{yield}}{S}_{\text{limit-stress}}}-1\ge 0.20$, i.e. the actual stress must be $\le \frac{{\sigma }_y}{1.20\times FO{S}_{\text{yield}}}$ of the way... work in areas directly:

Step 1 — required area. From $MS=\frac{{\sigma }_y}{P_{\text{design,yield}}/A}-1=0.20$: $A\ge \frac{P_{\text{design,yield}}}{{\sigma }_y}\times (1+0.20)=96.8\times 1.20=116.16{\text{mm}}^2.$ Step 2 — diameter: $d=\sqrt{\frac{4\times 116.16}{\pi }}=12.16\text{mm}\Rightarrow \text{13 mm (rounded up)}.$ Check with $d=13\text{mm}$: $A=\frac{\pi (13{)}^2}{4}=132.7{\text{mm}}^2$, yield stress $=\frac{24200}{132.7}=182.4\text{MPa}$, so $M{S}_{\text{yield}}=\frac{250}{182.4}-1=0.371\ge 0.20✓.$ Answer: $13\text{mm}$ diameter (gives $M{S}_{\text{yield}}\approx +0.37$, comfortably above the $+0.20$ target).

(a) Design stress $=1.4\times 120=168\text{MPa}$. Then $MS=\frac{168}{168}-1=0.$ Exactly zero — allowable equals the design load, just barely OK, no spare. (b) New design stress $=1.4\times 126=176.4\text{MPa}$: $MS=\frac{168}{176.4}-1=-\mathrm{0.0476.}$ A mere $5\%$ load re-estimate flipped the part from "pass" to fail ($-4.8\%$). Lesson: $MS$ near $0$ is fragile — any load-estimate scatter (exactly the ignorance $FOS$ is meant to cover) can push it negative. Engineers keep a working margin (e.g. $\ge +0.10$–$0.20$) precisely so small updates don't cause a redesign.

No. RSS is only for loads that act simultaneously. Here the two $6g$ events happen at different times, so you take the envelope: each event alone is a $6g$ resultant (since the other axis is $0g$). The structure must survive a $6g$ case, not $8.49g$. $F_{\text{event 1}}=m(6g),F_{\text{event 2}}=m(6g),\text{design to }6g,\text{ not }8.49g.$ Blindly RSS-ing non-simultaneous cases inflates the load by $\sqrt{2}\approx 1.41\times$ — a large, unjustified mass penalty. The correct governing load is $6g$.

(B), the untested composite tank, gets the higher $FOS$. $FOS$ is a paid-for cushion against ignorance: manufacturing scatter, analysis error, material variability, and untested failure modes. Ground testing to qualification (part A) directly removes ignorance — you have physically demonstrated the metallic bracket holds, so a smaller cushion is justified ($FO{S}_{\text{ult}}\approx 1.25$). Part B stacks more unknowns: composites scatter more than metals, delamination is hard to inspect, and no full proof test means the demonstrated capability is uncertain. More unknowns → larger cushion (pressurized/composite hardware often uses $FO{S}_{\text{ult}}\approx 1.5$–$2.0$). The number tracks how much you don't know, not how strong the part looks on paper.

Step 1 — resultant limit force (RSS, simultaneous & perpendicular): $F=\sqrt{3000^2+4000^2}=\sqrt{9000000+16000000}=5000\text{N}.$ (The famous $3$-$4$-$5$ triangle.) Step 2 — limit stress: $S_{\text{limit}}=\frac{5000}{30{\text{mm}}^2}=\frac{5000}{30\times 10^{-6}{\text{m}}^2}=166.7\text{MPa}.$ Step 3 — margin: $MS=\frac{200}{1.4\times 166.7}-1=\frac{200}{233.3}-1=-\mathrm{0.143.}$ $MS=-0.14<0$ → fails. The lug must be enlarged: minimum area for $MS=0$ is $A\ge \frac{1.4\times 5000}{200}=35{\text{mm}}^2$. Answer: $S_{\text{limit}}=166.7\text{MPa}$, $MS=-0.14$ (fails); needs $\ge 35{\text{mm}}^2$.

L1 knows which number is which. L2 sizes a part through the FOS chain. L3 combines/compares load cases (RSS vs envelope). L4 runs the whole loop and redesigns to a target margin. L5 reasons about why — thin-margin fragility, degenerate combinations, and choosing $FOS$ from ignorance.