Structural design process — load cases, FOS (factor of safety): Worked examples

Before anything, recall the two formulas we lean on (from the parent note):

Read that again: $M S$ is a fraction of spare capacity. $M S = 0$ means "exactly enough, no cushion." $M S = 0.2$ means "20% stronger than the factored demand." $M S < 0$ means "fails."

The scenario matrix

Every FOS/load-case problem you will ever meet lands in one of these cells:

#	Case class	What makes it distinct	Example
A	Size a part (solve for area)	Unknown is geometry; set $M S = 0$	Ex 1
B	Check a part (compute $M S$ )	Geometry known; is $M S \geq 0$ ?	Ex 2
C	Two separate FOS (yield and ultimate)	Part must pass both checks	Ex 3
D	Combine simultaneous loads (RSS & law of cosines)	Vector add, not scalar sum	Ex 4
E	Zero / degenerate input	Zero load, zero margin, $S_{allow} = S_{design}$	Ex 5
F	Negative margin → redesign	$M S < 0$ ; find new geometry	Ex 6
G	Real-world word problem (force-based $S$ )	Pull numbers out of prose	Ex 7
H	Exam twist (which is limiting?)	Two candidate failure modes; pick worst	Ex 8

The eight examples below hit all eight cells. Each ends with a Verify step, and every number is machine-checked in the VERIFY block that accompanies this page.

Prerequisites you may want open: Stress and Strain — Yield vs Ultimate Strength, Quasi-static Loads and Launch Environment.

Ex 1 — [Cell A] Size a strut

Worked example Statement

A strut carries a limit tensile force $P_{limit} = 20 kN$ . Material yield stress $σ_{y} = 300 MPa$ , $F O S_{yield} = 1.25$ . Find the minimum cross-sectional area $A$ .

Forecast: guess before reading — will $A$ be bigger or smaller than $P_{limit} / σ_{y}$ ? (It must be bigger, because we size against the factored load, not the raw one.)

Step 1 — factor the demand. Here the checked quantity $S$ is a force, so let $P_{design} \equiv S_{design}$ (the general design load specialised to a force). Then $P_{design} = F O S \times P_{limit} = 1.25 \times 20 = 25 kN$ . Why this step? FOS inflates the load, never the material. We must survive the factored force.

Step 2 — write the stress limit. Stress = force / area, and we require it not to exceed yield: $σ = \frac{P _{design}}{A} \leq σ_{y}$ Why this step? Stress (force per unit area) is the quantity the material actually "feels." Below yield → no permanent bend.

Step 3 — solve for the smallest area (set $M S = 0$ ). $A \geq \frac{P _{design}}{σ _{y}} = \frac{25 000 N}{300 \times 1 0 ^{6} Pa} = 8.33 \times 1 0^{- 5} m^{2} = 83.3 mm^{2}$ Why this step? Equality is the knife-edge ( $M S = 0$ ); any larger area gives positive margin.

Verify: Put $A = 83.3 mm^{2}$ back: $σ = 25 000/ (83.3 \times 1 0^{- 6}) = 300 \times 1 0^{6}$ Pa $= σ_{y}$ . Exactly at yield → $M S = 0$ . ✓ Units: $N / m^{2} = Pa$ . ✓

Ex 2 — [Cell B] Check a bracket's margin

Worked example Statement

A bracket has allowable stress $S_{allow} = σ_{allow} = 180 MPa$ . Worst-case stress from the limit load is $120 MPa$ , with $F O S_{ult} = 1.4$ . Does it pass?

Forecast: $120$ is well below $180$ — feels safe. But we haven't factored yet. Guess the sign of $M S$ .

Step 1 — factor the limit stress to design. $S_{design} = 1.4 \times 120 = 168 MPa$ . Why this step? The mistake "positive stress = positive margin" (parent note) dies here: you must compare against the factored demand.

Step 2 — compute margin. $M S = \frac{180}{168} - 1 = 0.0714$ . Why this step? $M S$ is fractional spare capacity after factoring. Both numerator and denominator are stresses (MPa) → units cancel.

Result: $M S = + 0.07 \geq 0$ → passes, but only 7% cushion. Thin; an engineer may add material.

Verify: $180 = 1.0714 \times 168$ ? $1.0714 \times 168 = 180$ . ✓ And $168 < 180$ confirms $M S > 0$ . ✓

Ex 3 — [Cell C] Both yield AND ultimate

Worked example Statement

A fitting sees a limit stress $S_{limit} = σ_{limit} = 100 MPa$ . Material: yield $σ_{y} = 250 MPa$ , ultimate $σ_{u} = 400 MPa$ . Factors: $F O S_{yield} = 1.1$ , $F O S_{ult} = 1.4$ . Does it pass both checks?

Forecast: Two checks — which do you expect to be tighter, yield or ultimate?

Step 1 — yield margin. Design $= 1.1 \times 100 = 110 MPa$ ; $M S_{y} = \frac{250}{110} - 1 = 1.273$ . Why this step? Yield prevents permanent deformation; it uses the lower factor against the lower strength.

Step 2 — ultimate margin. Design $= 1.4 \times 100 = 140 MPa$ ; $M S_{u} = \frac{400}{140} - 1 = 1.857$ . Why this step? Ultimate prevents rupture; higher factor, higher strength.

Step 3 — the part must pass the worst of the two. $min (1.273, 1.857) = 1.273$ → yield governs, still positive → passes both. Why this step? Both are separate physical requirements; the smaller margin is the real one.

Verify: $250/110 = 2.2727 \Rightarrow M S_{y} = 1.2727$ . $400/140 = 2.857 \Rightarrow M S_{u} = 1.857$ . Yield $<$ ultimate margin, both $\geq 0$ . ✓ (Kills the parent's "same FOS for both" mistake.)

Ex 4 — [Cell D] Combine simultaneous loads — RSS and the general angle

Worked example Statement

A $50 kg$ box on a bracket sees QSL of $9 g$ axial and $4 g$ lateral at the same instant ( $g = 9.81 m/s^{2}$ ). (a) Find the resultant force when the two act at $9 0^{\circ}$ (perpendicular). (b) State how to combine them if instead they met at a general angle $θ$ .

Forecast: Should you add $9 g$ and $4 g$ to get $13 g$ , or something smaller? Look at the figure — the answer is the length of the red diagonal, not the two legs laid end-to-end.

Reading the figure: The blue arrow is the axial force $F_{a}$ pointing right; the orange arrow is the lateral force $F_{l}$ pointing up. They start from the same corner (the small black right-angle square shows they are exactly $9 0^{\circ}$ apart). The red arrow from the corner to the opposite corner of the rectangle is the resultant $F$ — its length is what we compute. The faint gray dotted line along the bottom is the (wrong) scalar sum $F_{a} + F_{l}$ , drawn to show how much longer, and therefore how over-conservative, it is.

Step 1 — axial force. $F_{a} = m (9 g) = 50 \times 9.81 \times 9 = 4414.5 N$ . Step 2 — lateral force. $F_{l} = m (4 g) = 50 \times 9.81 \times 4 = 1962 N$ . Why these steps? Force $=$ mass $\times$ acceleration; each $g$ -level is an acceleration.

Step 3a — resultant by Pythagoras (RSS), the $9 0^{\circ}$ case. Because the two forces are the perpendicular legs of the right triangle in the figure, the resultant (hypotenuse, red) is $F = F_{a}^{2} + F_{l}^{2} = 4414. 5^{2} + 196 2^{2} = 4831.6 N$ Why RSS, not $F_{a} + F_{l}$ ? Perpendicular vectors combine along the hypotenuse. The scalar sum $F_{a} + F_{l} = 6376.5 N$ (the gray dotted path) is physically impossible for orthogonal loads and wastes mass.

Step 3b — the general angle $θ$ (law of cosines). RSS is only the special case $θ = 9 0^{\circ}$ . If the two forces meet at an interior angle $θ$ between their directions, the resultant magnitude comes from the law of cosines: $F = F_{a}^{2} + F_{l}^{2} + 2 F_{a} F_{l} cos θ$ Why cosine here? $cos θ$ measures how much the two loads reinforce or oppose each other. Check the limits: at $θ = 9 0^{\circ}$ , $cos 9 0^{\circ} = 0$ and the middle term vanishes → we recover RSS ( $4831.6 N$ ). At $θ = 0^{\circ}$ (same direction), $cos 0^{\circ} = 1$ → $F = F_{a} + F_{l} = 6376.5 N$ , the full scalar sum. At $θ = 18 0^{\circ}$ (directly opposing), $cos 18 0^{\circ} = - 1$ → $F = ∣ F_{a} - F_{l} ∣ = 2452.5 N$ , the smallest possible. So RSS is the middle-ground case, and every real combination sits between $∣ F_{a} - F_{l} ∣$ and $F_{a} + F_{l}$ .

Verify: $4414. 5^{2} + 196 2^{2} = 1.9488 \times 1 0^{7} + 3.8494 \times 1 0^{6} = 2.3338 \times 1 0^{7}$ ; $= 4831.6 N$ . Cosine law at $θ = 9 0^{\circ}$ gives the same, at $θ = 0^{\circ}$ gives $6376.5$ , at $θ = 18 0^{\circ}$ gives $2452.5$ . And $2452.5 < 4831.6 < 6376.5$ confirms RSS lies between the extremes. ✓ Units: $kg \cdot m/s^{2} = N$ . ✓

Ex 5 — [Cell E] Degenerate inputs (zero & knife-edge)

Worked example Statement

Three quick sub-cases to nail the edges: (a) A part with zero applied load. (b) A part where $S_{allow}$ exactly equals the design load. (c) A part with a zero-thickness (degenerate) member.

Forecast: what is $M S$ when there's no load at all?

Step 1 — (a) zero load, treated as a limit. As the load shrinks, $S_{limit} \to 0^{+}$ so $S_{design} = F O S \times S_{limit} \to 0^{+}$ . We must not write $S_{allow} /0$ as raw arithmetic (division by zero is undefined). Instead read $M S$ as the limit $M S = lim_{S_{design} \to 0^{+}} (\frac{S _{allow}}{S _{design}} - 1) = + \infty$ Why this step? The ratio grows without bound as the denominator approaches zero from above. Physically: a part carrying (almost) nothing has (almost) unbounded spare capacity against that load — it can never fail from it. The limit language keeps the statement rigorous.

Step 2 — (b) knife-edge. Let $S_{allow} = 168 MPa$ and $S_{design} = 168 MPa$ . Then $M S = \frac{168}{168} - 1 = 0$ . Why this step? This is the exact boundary of acceptability — "just barely OK," zero cushion. Any scatter above the estimate now fails it.

Step 3 — (c) zero area. From Ex 1, $σ = P / A$ . As $A \to 0^{+}$ , $σ \to + \infty$ , so (holding $S_{allow}$ finite) $M S \to - 1$ (fully failed). Why this step? A vanishing cross-section concentrates all force into no material → infinite stress. This is why struts have a minimum area, never zero.

Verify: (b) $168/168 - 1 = 0$ . ✓ (a) $lim_{S_{design} \to 0^{+}} S_{allow} / S_{design} - 1 = + \infty$ . ✓ (c) $lim_{A \to 0^{+}} S_{allow} / (P / A) - 1 = - 1$ . ✓

Ex 6 — [Cell F] Negative margin → redesign

Worked example Statement

A plate has $S_{allow} = σ_{allow} = 200 MPa$ , limit stress $S_{limit} = 160 MPa$ , $F O S_{ult} = 1.4$ . (a) Show it fails. (b) By what factor must you increase the area (equivalently reduce stress) to reach $M S = 0$ ?

Forecast: $160 < 200$ , so it "looks" fine — but factor first. Will it fail?

Step 1 — factor & check. $S_{design} = 1.4 \times 160 = 224 MPa$ ; $M S = \frac{200}{224} - 1 = - 0.107$ . Why this step? $224 > 200$ → demand exceeds capacity → fails by ~11%. This is exactly the "positive stress but negative margin" trap.

Step 2 — target the fix. Stress scales as $1/ A$ (force fixed). To get $M S = 0$ we need the new stress $S_{limit}^{'}$ such that $1.4 S_{limit}^{'} = 200$ , i.e. $S_{limit}^{'} = 142.9 MPa$ . Why this step? We can't add strength (material fixed); we lower the stress by adding area.

Step 3 — area ratio. Since $σ \propto 1/ A$ : $\frac{A _{new}}{A _{old}} = \frac{S _{limit}}{S _{limit}^{'}} = \frac{160}{142.9} = 1.12$ . Why this step? Thicken the plate by 12% to reach the knife-edge; add a little more for real cushion.

Verify: New design stress $= 224/1.12 = 200 MPa = S_{allow}$ → $M S = 0$ . ✓ Original $M S = 200/224 - 1 = - 0.1071$ . ✓

Ex 7 — [Cell G] Real-world word problem (allowable is a force)

Worked example Statement

"During liftoff, a $12 kg$ avionics unit is bolted to a shelf. The launch vehicle user manual quotes a quasi-static limit acceleration of $6 g$ axial ( $g = 9.81 m/s^{2}$ ). Each of 4 identical bolts shares the axial load equally. A single bolt's proof (allowable) tensile force is $F_{allow} = 3000 N$ . With $F O S_{ult} = 1.4$ , are the bolts adequate?"

Note on notation: here the checked quantity is a force, so we set $S_{allow} \equiv F_{allow} = 3000 N$ and $S_{design} \equiv F_{design}$ . Both sides of the $M S$ ratio are forces (N) → units cancel, exactly as with stresses.

Forecast: total factored load spread over 4 bolts — will each bolt's share stay under $3000 N$ ?

Step 1 — total limit force. $F_{limit} = m (6 g) = 12 \times 9.81 \times 6 = 706.3 N$ . Why this step? Convert the $g$ -level into an actual force via $F = ma$ ; the acceleration is $6 g$ .

Step 2 — factor it. $F_{design} = F O S_{ult} \times F_{limit} = 1.4 \times 706.3 = 988.8 N$ . Why this step? Bolts must survive the factored ultimate demand, not the raw estimate.

Step 3 — per-bolt share. $F_{bolt} = F_{design} /4 = 988.8/4 = 247.2 N$ . Why this step? A symmetric 4-bolt pattern splits the axial load equally, so each bolt sees a quarter.

Step 4 — margin per bolt. $M S = \frac{F _{allow}}{F _{bolt}} - 1 = \frac{3000}{247.2} - 1 = 11.14$ . Why this step? Fractional spare capacity, force over force — units (N) cancel inside the ratio.

Result: $M S \approx + 11 \geq 0$ → hugely adequate against axial QSL. In practice bolts are rarely sized by axial QSL alone; shear and preload usually govern. But this axial check passes cleanly with enormous margin.

Verify: $12 \times 9.81 \times 6 = 706.32$ ; $\times 1.4 = 988.85$ ; $/4 = 247.21$ ; $3000/247.21 - 1 = 11.14$ . ✓ Units: N throughout, ratio dimensionless. ✓

Ex 8 — [Cell H] Exam twist — which mode governs?

Worked example Statement

A pressurized tank wall sees a limit stress $S_{limit} = σ_{limit} = 150 MPa$ . Two candidate checks must both be satisfied:

Yield: $σ_{y} = 280 MPa$ , $F O S_{yield} = 1.1$ .
Ultimate: $σ_{u} = 350 MPa$ , but because the part is pressurized the user manual demands a higher $F O S_{ult} = 2.0$ .

Which check governs, and does the tank pass?

Forecast: the ultimate strength ( $350$ ) is bigger than the yield strength ( $280$ ) — but the ultimate FOS ( $2.0$ ) is also much bigger than the yield FOS ( $1.1$ ). Which effect wins?

Step 1 — yield margin. $S_{design, y} = 1.1 \times 150 = 165 MPa$ ; $M S_{y} = \frac{280}{165} - 1 = 0.697$ . Why this step? Yield prevents permanent deformation; it uses its own (lower) factor against its own strength.

Step 2 — ultimate margin. $S_{design, u} = 2.0 \times 150 = 300 MPa$ ; $M S_{u} = \frac{350}{300} - 1 = 0.167$ . Why this step? Ultimate prevents rupture; the elevated pressurized factor makes the factored demand large.

Step 3 — pick the worst (smallest) margin. $min (0.697, 0.167) = 0.167$ → ultimate governs. The big pressurized factor overwhelms the higher raw strength. Why this step? A part must pass every mode; the tightest margin is the real one. Never reuse one FOS for both (parent's "same FOS" mistake).

Result: both margins are $\geq 0$ → passes, but it is ultimate-limited at $M S = + 0.17$ . The trap is assuming the bigger raw strength ( $350$ ) means ultimate is safest — the factor flips it.

Verify: $280/165 - 1 = 0.6970$ ; $350/300 - 1 = 0.1667$ . Ultimate margin $<$ yield margin → ultimate governs; both $\geq 0$ → pass. ✓

Recall Which cell does each example hit?

Ex1 sizing (A) ::: solve for area, set $M S = 0$ Ex2 check margin (B) ::: compute $M S$ , positive but thin Ex3 two FOS (C) ::: yield and ultimate both checked, take min Ex4 combine loads (D) ::: RSS for $9 0^{\circ}$ , law of cosines for general angle Ex5 degenerate (E) ::: zero load → $M S \to + \infty$ (as a limit); zero area → $M S \to - 1$ Ex6 redesign (F) ::: negative $M S$ , thicken by area ratio Ex7 word problem (G) ::: force-based $S$ , split over bolts Ex8 exam twist (H) ::: higher FOS on pressurized part makes ultimate govern

Related deep dives to explore next: Finite Element Analysis of Spacecraft Structures (where $S_{allow}$ actually comes from), Mass Budget and Structural Efficiency (why we hate over-design), Qualification vs Acceptance Testing and Random & Sine Vibration Testing (proving the margin in hardware).