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

This is the toolbox page for Structural design process — load cases, FOS (factor of safety). We assume you have seen nothing. We start from a push.

1. Force — a push or a pull, with a size and a direction

The picture: draw an arrow. The length of the arrow is the size $F$ ; the way it points is the direction. The whole arrow together is $F$ .

Why the topic needs it: every load on a spacecraft — the rocket shoving it upward, vibration rattling a bracket, a tank pushing outward — is ultimately a collection of forces. If we cannot draw the arrow, we cannot design against it.

2. Mass, gravity, and the number " $g$ "

The size of the weight force from gravity is $F = m g$ . This is force number one that the object always feels.

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

Figure s01 — Alt text: a green mass block with two downward arrows. A short lavender arrow labelled "1g = m·9.81" shows normal gravity; a long coral arrow labelled "9g = m·9.81·9" shows the launch acceleration as nine times longer. The picture defines a "g-load" as a multiple of gravity's pull.

3. Adding forces — general vectors, and the perpendicular shortcut

Sometimes two forces act on the same part at the same time in different directions. Forces are arrows ( $F$ ), so they add the way arrows add: tip-to-tail, not by simply adding their sizes.

Where does the size formula come from? (a picture, not magic.) Put $F_{1}$ flat along the ground and stand $F_{2}$ on its tip at angle $θ$ . Drop the tip of $F_{2}$ straight down to the ground line: this splits $F_{2}$ into a flat part of length $F_{2} cos θ$ (along $F_{1}$ ) and an upright part of length $F_{2} sin θ$ (at right angles to $F_{1}$ ). Now the resultant is the long slanted arrow of a right triangle whose horizontal leg is $F_{1} + F_{2} cos θ$ and whose vertical leg is $F_{2} sin θ$ . Apply plain Pythagoras to that right triangle:

$F^{2} = (F_{1} + F_{2} cos θ)^{2} + (F_{2} sin θ)^{2} .$

Multiply out and use the identity $cos^{2} θ + sin^{2} θ = 1$ : the $F_{2}^{2} cos^{2} θ$ and $F_{2}^{2} sin^{2} θ$ collapse into a single $F_{2}^{2}$ , leaving

$F^{2} = F_{1}^{2} + F_{2}^{2} + 2 F_{1} F_{2} cos θ .$

So the "law of cosines" is nothing but Pythagoras applied after splitting the tilted arrow into flat and upright parts — see the figure.

Figure s02 — Alt text: tip-to-tail construction. A lavender arrow $F_{1}$ lies flat; a mint arrow $F_{2}$ stands on its tip at angle $θ$ and is split by a dashed line into a flat leg $F_{2} cos θ$ and an upright leg $F_{2} sin θ$ . The coral resultant $F$ is the hypotenuse of the right triangle with legs $F_{1} + F_{2} cos θ$ and $F_{2} sin θ$ — the geometric source of the law of cosines.

Now read off the three special cases:

Same line, same way ( $θ = 0^{\circ}$ , so $cos θ = 1$ ): $F = F_{1}^{2} + F_{2}^{2} + 2 F_{1} F_{2} = (F_{1} + F_{2})^{2} = F_{1} + F_{2}$ — you just add.
Opposite ways ( $θ = 18 0^{\circ}$ , $cos θ = - 1$ ): $F = (F_{1} - F_{2})^{2} = ∣ F_{1} - F_{2} ∣$ — you subtract.
Right angles ( $θ = 9 0^{\circ}$ , $cos θ = 0$ ): the middle term vanishes and we get the clean RSS rule below.

Why the topic needs it: the axial ( $9 g$ ) and lateral ( $4 g$ ) launch loads happen simultaneously and at right angles, so RSS is exact for them. Adding them as plain numbers would fake a bigger load than reality, forcing a heavier part — and mass is the enemy in space. See Quasi-static Loads and Launch Environment.

Figure s03 — Alt text: a right triangle of forces. The lavender horizontal leg is the axial force $F_{a} = 4415$ N, the mint vertical leg is the lateral force $F_{l} = 1962$ N, and the coral hypotenuse is the resultant $F = F_{a}^{2} + F_{l}^{2} = 4832$ N. Shows why perpendicular loads combine by Pythagoras, giving less than the scalar sum.

4. Stress — spreading a force over an area (and its sign)

A force alone does not tell you whether a part breaks. A 20 kN pull snaps a thin wire but barely stretches a thick bar. What matters is how concentrated the force is.

The picture: imagine the same pull carried by a fat rope versus a thin thread. Same force $F$ , but the thread has a small $A$ , so $σ = F / A$ is huge — it snaps. Stress is force concentrated.

Figure s04 — Alt text: two bars side by side. Left, a lavender bar pulled by two outward coral arrows labelled "σ > 0 (tension)". Right, a butter-yellow bar squeezed by two inward slate arrows labelled "σ < 0 (compression)". Defines the sign convention of stress by picture: pulling is positive, squashing is negative.

5. Strength — yield and ultimate, the two limits of a material

The material fights back only up to a point. There are two important points, not one.

The picture: stretch a metal bar and watch a stress-vs-stretch graph. It rises as a straight line (springy), then bends over at yield, keeps climbing, and finally reaches a peak — ultimate — where it snaps. Two different heights, two different failures. Full detail lives in Stress and Strain — Yield vs Ultimate Strength.

Figure s05 — Alt text: a stress-vs-strain curve in lavender. It rises straight, bends over at a coral dot marked "yield σ_y = 300 MPa (bends permanently)", climbs further, and peaks at a mint dot marked "ultimate σ_ult (ruptures)". Shows the two distinct failure heights of a metal.

6. Force vs stress — the one letter "S", and how to convert

The parent note uses the symbol $S$ for the design quantities — $S_{allow}$ , $S_{limit}$ , $S_{design}$ . Here is the crucial rule that keeps you out of trouble.

7. Limit load — the worst push we expect

The picture: back to the Lego-bridge story — it is your honest guess of how heavy the toy car is. Real, but a guess.

Why the topic needs it: it is the number we will multiply by the safety factor. Everything downstream is built on this one "worst expected" value.

8. A ratio bigger than 1 — the factor of safety, and why we multiply demand

Why a ratio and not an added amount? A fixed added margin (say "+50 N") means nothing without knowing the scale — 50 N is huge for a wire, trivial for a beam. A ratio scales with the load automatically, so one number ( $1.25$ ) works everywhere.

9. Margin of safety — the "how much spare?" number

We now have supply ( $S_{allow}$ ) and inflated demand ( $F O S \cdot S_{limit}$ ), both on the same basis and units. The final question: is supply enough, and how much is left over?

How the foundations feed the topic

Equipment checklist

What does bold

F

mean versus plain

F

Bold

F

is the whole force arrow (size and direction); plain

F = ∣ F ∣

is only its size, a positive number.

How do you turn "

9 g

" acting on a mass

m

into a real force?

F = m g \times 9

, with

g = 9.81 m/s^{2}

General rule for adding two forces at any angle?

Add tip-to-tail; the resultant size is

F = F_{1}^{2} + F_{2}^{2} + 2 F_{1} F_{2} cos θ

, which is just Pythagoras after splitting the tilted arrow into flat and upright parts.

Two forces at right angles — how do you combine them?

RSS:

F = F_{a}^{2} + F_{l}^{2}

(the

cos 9 0^{\circ}

term is zero), where

F_{a}

is axial size and

F_{l}

is lateral size.

What is stress and its unit?

Force per area,

σ = F / A

; measured in pascals, usually MPa

= N/mm^{2}

What does the sign of

σ

mean?

Positive = tension (being pulled apart); negative = compression (being squashed).

Besides crushing, how else can a compressed part fail?

By buckling — a slender column bows sideways and collapses at a stress well below the material's compressive strength; it is a shape failure, not a material one.

Difference between yield and ultimate strength?

Yield = stress where it permanently bends; ultimate = higher stress where it ruptures.

Can you compare a force to a stress directly?

No — both sides must be the same basis and unit; convert with

σ = F / A

(or

F = σ A

) first.

What is the limit load?

The worst load (best estimate of demand) expected in the real mission, as a force or the stress it produces.

What does FOS multiply, and why?

The applied/limit load (demand), because we are uncertain about loads — we inflate the push, not the metal.

Write the margin of safety and its pass condition.

M S = \frac{S _{allow}}{F O S \cdot S _{limit}} - 1

; passes when

M S \geq 0

Why "

- 1

" in the margin formula?

So that supply exactly equal to inflated demand gives

M S = 0

— the zero-spare pass/fail line.