Yield stress, ultimate stress — material behavior: Worked examples

The scenario matrix

Every question about yield/ultimate stress lives in one of these cells. The "Which cell" tag on each example tells you where it sits.

#	Cell class	What makes it special	Covered by
A	Elastic, safe	$σ_{a pp} < σ_{y}$ , margin positive	Ex 1
B	Exactly at yield	$σ_{a pp} = σ_{y}$ , $M o S = 0$ boundary	Ex 2
C	Past yield, below ultimate	plastic but not broken	Ex 3
D	Negative margin (fails)	$M o S < 0$ , part is under-designed	Ex 4
E	Zero / degenerate input	$F = 0$ , or $A \to 0$ (limiting stress $\to \infty$ )	Ex 5
F	0.2% offset geometry	reading $σ_{y}$ off a bending curve	Ex 6
G	Ultimate + necking	engineering vs true stress diverge	Ex 7
H	Real-world word problem	pick area to survive a load	Ex 8
I	Exam twist	doubling area / changing material; trap question	Ex 9

We'll use these symbols throughout (all defined in the parent note, repeated here so you never guess):

Definition Symbol cheat-sheet

$F$ = pulling force, in newtons ( $N$ ).
$A_{0}$ = original (undeformed) cross-section area, in $m^{2}$ .
$A_{in s t}$ = instantaneous (current) cross-section area during a test — smaller than $A_{0}$ once the specimen necks. Used only for true stress.
$F_{f r a c}$ = the force the specimen is carrying at the moment of fracture.
$σ = F / A_{0}$ = engineering stress, in pascals ( $Pa = N/m^{2}$ ). $1 MPa = 1 0^{6} Pa$ , $1 GPa = 1 0^{9} Pa$ .
$σ_{a pp}$ = the engineering stress actually applied to a part in service, i.e. $σ_{a pp} = F_{a ppl i e d} / A_{0}$ . (It is just $σ$ evaluated for the working load — we write $σ_{a pp}$ when comparing against a limit.)
$ε = Δ L / L_{0}$ = strain, a pure number (no units).
$E$ = Young's modulus (elastic slope), see Young's modulus and elasticity.
$σ_{y}$ = yield stress, $σ_{u}$ = ultimate stress.
$σ_{allow}$ = allowable stress = the limit ( $σ_{y}$ or $σ_{u}$ ) divided by the required factor of safety, i.e. $σ_{allow} = σ_{l imi t} / F o S_{r e q}$ . This is the single number every example compares the applied stress against — always form it explicitly.
$F o S$ = factor of safety; $F o S_{r e q}$ = the required factor of safety a design must meet (e.g. $1.25$ for yield, $1.5$ for ultimate). $M o S$ = margin of safety — see Factor of safety and margins.

The master formula we'll lean on, straight from the parent. Notice that once we've built $σ_{allow} = σ_{l imi t} / F o S_{r e q}$ , the margin is just "allowable over applied, minus one": $M o S = \frac{σ _{allow}}{σ _{a pp}} - 1 = \frac{σ _{l imi t}}{F o S _{r e q} \cdot σ _{a pp}} - 1 (\geq 0 means "passes")$

Every example below first forms $σ_{allow}$ , then divides by $σ_{a pp}$ , so the two written forms above are always the same computation.

Here is the whole matrix drawn on one stress–strain curve, so you can see where each cell lives before we compute anything.

Figure — Yield stress, ultimate stress — material behavior

The worked examples

Worked example Example 1 — Elastic and safe (Cell A)

A titanium bracket has area $A_{0} = 40 mm^{2}$ and carries $F = 6000 N$ . Material: $σ_{y} = 830 MPa$ , $E = 114 GPa$ . Required $F o S_{y i e l d} = 1.25$ . Is it safe, and how much does it stretch fractionally?

Forecast: $6000 N$ on $40 mm^{2}$ — guess: is $σ$ closer to $100$ or $1000$ MPa? Will the margin be big or tiny?

Convert area to SI. $40 mm^{2} = 40 \times 1 0^{- 6} m^{2}$ . Why this step? Stress needs $N/m^{2}$ ; $1 mm^{2} = 1 0^{- 6} m^{2}$ , so leaving it in mm² inflates the answer by a million.
Compute stress. $σ = \frac{6000}{40 \times 1 0 ^{- 6}} = 150 \times 1 0^{6} Pa = 150 MPa$ . Why this step? Stress is what we compare to $σ_{y}$ , not force (see Stress and strain fundamentals).
Compare to yield. $150 < 830$ → elastic, so Hooke's law $σ = E ε$ is valid. Why this step? Only below yield can we use $E$ to get strain.
Strain. $ε = \frac{σ}{E} = \frac{150}{114000} = 1.316 \times 1 0^{- 3}$ . Why this step? We rearrange Hooke's law $σ = E ε$ into $ε = σ / E$ to get the fractional stretch the question asks for; this is valid only because step 3 confirmed we are elastic.
Allowable stress and margin. $σ_{allow} = \frac{σ _{y}}{F o S _{r e q}} = \frac{830}{1.25} = 664 MPa$ , so $M o S = \frac{σ _{allow}}{σ _{a pp}} - 1 = \frac{664}{150} - 1 = 3.43$ . Why this step? We first build the de-rated allowable stress (the single limit the part is really permitted to reach), then compare the applied $150$ MPa against it — that ratio minus one is the margin.

Verify: $M o S = 3.43 > 0$ → passes, hugely. Sanity: $150$ MPa is $\approx 18%$ of yield, so a big margin is expected. Units: $MPa / MPa$ is dimensionless ✓; strain is dimensionless ✓.

Worked example Example 2 — Exactly at yield, the

$M o S = 0$ boundary (Cell B) An aluminium rod has $σ_{y} = 480 MPa$ . With no factor of safety demanded ( $F o S_{r e q} = 1$ ), what applied stress gives exactly $M o S = 0$ ? What does $M o S = 0$ physically mean?

Forecast: $M o S = 0$ is the razor's edge. Guess the applied stress before reading on.

Form the allowable and set $M o S = 0$ . With $F o S_{r e q} = 1$ , $σ_{allow} = σ_{y} /1 = 480 MPa$ , so $0 = \frac{σ _{allow}}{σ _{a pp}} - 1$ . Why this step? $M o S = 0$ is the exact pass/fail line, the boundary of cell A and cell D; writing it via $σ_{allow}$ keeps every example on the same footing.
Solve. $\frac{σ _{allow}}{σ _{a pp}} = 1 \Rightarrow σ_{a pp} = σ_{allow} = 480 MPa$ . Why this step? Rearranging isolates the applied stress that just touches the allowable.
Interpret. At $σ_{a pp} = σ_{allow} = σ_{y}$ (since $F o S_{r e q} = 1$ ), the applied stress exactly reaches the allowable: zero cushion. Why this step? We interpret the algebra physically because a design review needs the meaning ("no cushion left"), not just a number — this tells the engineer the part is at the very onset of permanent set.

Verify: Plug back: $M o S = 480/480 - 1 = 1 - 1 = 0$ ✓. Meaning: the part is right at the onset of permanent deformation — any more load and it starts to yield. Real spacecraft never fly here; that's why $F o S_{r e q} > 1$ .

Worked example Example 3 — Past yield but below ultimate (Cell C)

A test coupon is deliberately loaded to $σ_{a pp} = 520 MPa$ . Material: $σ_{y} = 480 MPa$ , $σ_{u} = 570 MPa$ , $E = 70 GPa$ . Has it yielded? Has it broken? Is the elastic strain formula still valid?

Forecast: $520$ sits between $480$ and $570$ . Predict which of the two limits it has crossed.

Compare to yield. $520 > 480$ → yielded: permanent (plastic) deformation has begun. Why this step? Crossing $σ_{y}$ is the definition of the onset of plastic set.
Compare to ultimate. $520 < 570$ → not fractured: still carrying load, past the yield knee but before the peak. Why this step? Ultimate is the load-carrying peak; below it the part is intact.
Check the strain rule. Since $520 > σ_{y}$ , Hooke's law $σ = E ε$ no longer gives total strain — part of the strain is now permanent and does not obey the elastic slope. Why this step? $E$ only describes the springy part; you'd underestimate total stretch if you used $ε = σ / E$ here.

Verify: Order check: $480 < 520 < 570$ , i.e. $σ_{y} < σ_{a pp} < σ_{u}$ ✓. So "yielded, not broken" is consistent. If you unloaded now, the coupon would spring back only partway — leaving a permanent set, the fingerprint of cell C.

Worked example Example 4 — Negative margin: the part fails (Cell D)

A strut sees $σ_{a pp} = 420 MPa$ . Material $σ_{y} = 480 MPa$ , required $F o S_{y i e l d} = 1.25$ . Compute $M o S$ and decide.

Forecast: $420$ is below $480$ , so it feels safe. But there's a $F o S$ of $1.25$ eating into the allowable. Guess the sign of $M o S$ .

Allowable stress. $σ_{allow} = \frac{σ _{y}}{F o S _{r e q}} = \frac{480}{1.25} = 384 MPa$ . Why this step? The factor of safety shrinks how much stress we're allowed to use, not the material's actual yield.
Compare. $σ_{a pp} = 420 > 384 = σ_{allow}$ → over the allowable. Why this step? Passing is $σ_{a pp} \leq σ_{allow}$ ; here it isn't.
Margin. $M o S = \frac{σ _{allow}}{σ _{a pp}} - 1 = \frac{384}{420} - 1 = 0.9143 - 1 = - 0.0857$ . Why this step? Using the already-formed $σ_{allow}$ keeps the arithmetic identical to $σ_{y} / (F o S_{r e q} \cdot σ_{a pp}) - 1$ and makes the negative sign obvious.

Verify: $M o S = - 0.086 < 0$ → fails. Note the trap: the raw stress ( $420$ ) is below yield ( $480$ ), so the part won't yet deform permanently — but it violates the required safety margin, which is what design demands. Cell D is exactly this "looks fine, fails the margin" case.

Worked example Example 5 — Degenerate inputs: zero force, vanishing area (Cell E)

Take $σ_{y} = 480 MPa$ and required $F o S_{r e q} = 1.25$ throughout. (a) A strut carries $F = 0 N$ on $A_{0} = 50 mm^{2}$ . What is its stress and margin? (b) A tie-rod of fixed force $F = 4000 N$ is machined thinner and thinner. What happens to the stress as $A_{0} \to 0$ ?

Forecast: Guess: with zero force is the margin zero, or infinite? As the rod gets razor-thin, does stress settle to a number or blow up?

Zero-force stress. $σ = \frac{0}{50 \times 1 0 ^{- 6}} = 0 Pa$ . Why this step? Stress is force per area; no force means no stress, whatever the geometry.
Zero-force margin. With $σ_{allow} = σ_{y} / F o S_{r e q} = 480/1.25 = 384 MPa$ , the margin is $M o S = \frac{σ _{allow}}{σ _{a pp}} - 1 = \frac{384}{0} - 1$ : dividing by $σ_{a pp} = 0$ is undefined (a number over zero has no value). Why this step? We must plug in honestly to expose that the formula breaks at $σ_{a pp} = 0$ — you cannot report a finite margin here.
Interpret the undefined result as $+ \infty$ . Watch the limit: keep the load tiny but nonzero and shrink it, $σ_{a pp} \to 0^{+}$ . Then $\frac{σ _{allow}}{σ _{a pp}} \to + \infty$ , so $M o S \to + \infty$ . Because the margin grows without bound as the load vanishes, we define the zero-load case as $M o S = + \infty$ ("infinitely safe"). Why this step? An undefined algebraic expression only becomes meaningful through the limiting process; the limit shows the margin heads to $+ \infty$ , which is the physical statement "a part carrying no load can never fail," so we adopt that value by convention.
Vanishing area, fixed force. $σ (A_{0}) = \frac{4000}{A _{0}}$ . As $A_{0} \to 0^{+}$ , $σ \to + \infty$ . Why this step? Force squeezed through ever-less area is the physical reason a nick, crack, or notch (tiny local area) is where things fail — see Fatigue and fracture mechanics.

Verify: (a) $σ = 0$ , and $F = σ_{y} A$ at yield with $F = 0$ input gives $0$ ✓; the limit argument gives $M o S = + \infty$ . (b) Numerically: $A_{0} = 10 mm^{2} \Rightarrow σ = 400 MPa$ ; $A_{0} = 1 mm^{2} \Rightarrow 4000 MPa$ ; $A_{0} = 0.1 mm^{2} \Rightarrow 40000 MPa$ — growing without bound ✓. This limiting behaviour is the whole reason stress concentrations are dangerous.

Worked example Example 6 — 0.2% offset geometry (Cell F)

An alloy has elastic slope $E = 68 GPa$ and no sharp yield point. The 0.2% offset line is $σ = 68000 (ε - 0.002)$ MPa. It intersects the measured curve at $ε = 0.0090$ . Find $σ_{y}$ .

Forecast: The offset line starts at $ε = 0.002$ on the strain axis. Guess whether $σ_{y}$ is near $E \times 0.002 = 136$ MPa or much larger.

Write the offset line. Slope $E$ , shifted right by $0.002$ : $σ = E (ε - 0.002)$ . Why this step? Same stiffness $E$ as the elastic region (a parallel line), but pinned to $0.2%$ permanent strain — this is why the offset is repeatable across labs (parent note).
Evaluate at the intersection strain. $σ_{y} = 68000 (0.0090 - 0.002) = 68000 \times 0.0070$ . Why this step? By definition $σ_{y}$ is where the offset line crosses the material curve — the intersection strain feeds straight in.
Compute. $68000 \times 0.0070 = 476 MPa$ .

Verify: Units: $MPa \times$ (dimensionless strain) $= MPa$ ✓. Geometry check (see figure below): the elastic part reaches $\approx E \times 0.002 = 136$ MPa at $ε = 0.002$ — but the yield point sits at a much larger strain ( $0.009$ ) and higher stress ( $476$ MPa), on the bent-over part of the curve, exactly as expected.

The figure below builds this construction step by step: the dashed white line is the elastic slope $E$ ; the yellow line is that same slope slid right to start at $ε = 0.002$ (the green dotted marker); where yellow meets the blue material curve is the red dot that defines $σ_{y} = 476$ MPa. Trace yellow from the green marker up to the red dot to see the "which angle has this steepness, shifted by 0.2%?" idea made visible.

Worked example Example 7 — Ultimate + necking: engineering vs true stress (Cell G)

At the ultimate point a specimen carries peak force $F = 28500 N$ on original area $A_{0} = 50 mm^{2}$ . Just before fracture the necked region has shrunk to instantaneous area $A_{in s t} = 30 mm^{2}$ while carrying fracture force $F_{f r a c} = 21000 N$ . Find engineering ultimate stress $σ_{u}$ , the engineering fracture stress, and the true fracture stress. Which is bigger?

Forecast: Engineering uses $A_{0}$ (the original area from the cheat-sheet); true uses the shrunk area $A_{in s t}$ . Guess: does engineering stress rise or fall from ultimate to fracture? Does true stress?

Engineering ultimate. $σ_{u} = \frac{28500}{50 \times 1 0 ^{- 6}} = 570 \times 1 0^{6} = 570 MPa$ . Why this step? $σ_{u}$ is the peak of the engineering curve, always over $A_{0}$ (design handbooks use engineering stress).
Engineering fracture stress. $σ_{f r a c, e n g} = \frac{F _{f r a c}}{A _{0}} = \frac{21000}{50 \times 1 0 ^{- 6}} = 420 MPa$ . Why this step? After the neck forms, force drops but we keep dividing by the original area $A_{0}$ , so engineering stress falls below $σ_{u}$ .
True fracture stress. $σ_{f r a c, t r u e} = \frac{F _{f r a c}}{A _{in s t}} = \frac{21000}{30 \times 1 0 ^{- 6}} = 700 MPa$ . Why this step? True stress divides by the real shrinking area $A_{in s t}$ ; the neck concentrates the same force through less material, so true stress rises.

Verify: $σ_{f r a c, e n g} = 420 < σ_{u} = 570$ ✓ (engineering drops after necking). $σ_{f r a c, t r u e} = 700 > 570$ ✓ (true stress keeps climbing). This is precisely the parent's warning: never mix engineering and true stress — they diverge after necking.

Worked example Example 8 — Real-world word problem: size the strut (Cell H)

A lander leg must carry $F = 9000 N$ at touchdown (see Structural load cases and launch loads). Material $σ_{y} = 480 MPa$ , required $F o S_{y i e l d} = 1.25$ . What is the minimum cross-sectional area, and what standard round-bar diameter (nearest mm) works?

Forecast: More force needs more area. Guess whether the answer is a few mm² or tens of mm².

Allowable stress. $σ_{allow} = \frac{σ _{y}}{F o S _{r e q}} = \frac{480}{1.25} = 384 MPa$ . Why this step? The leg is allowed to reach only the de-rated stress, not raw yield — $σ_{allow}$ is the ceiling every subsequent step must respect.
Minimum area from $σ = F / A$ . Require $σ_{a pp} = \frac{F}{A} \leq σ_{allow}$ , so $A_{min} = \frac{F}{σ _{allow}} = \frac{9000}{384 \times 1 0 ^{6}}$ . Why this step? Rearranging the stress definition turns a stress limit into a geometry requirement — this is how engineers pick sizes.
Compute. $A_{min} = 2.344 \times 1 0^{- 5} m^{2} = 23.44 mm^{2}$ . Why this step? Converting back to mm² gives a number we can match to real bar stock.
Round bar diameter. $A = \frac{π}{4} d^{2} \Rightarrow d = \frac{4 A}{π} = \frac{4 \times 23.44}{π} = 5.46 mm$ . Choose $d = 6 mm$ (round up). Why this step? You must round area/diameter up, never down — a smaller bar would exceed the allowable stress.

Verify: A $6 mm$ bar has $A = \frac{π}{4} (6)^{2} = 28.27 mm^{2} > 23.44$ ✓. Actual stress $= 9000/ (28.27 \times 1 0^{- 6}) = 318 MPa < 384$ ✓, so $M o S = σ_{allow} / σ_{a pp} - 1 = 384/318 - 1 = 0.21 > 0$ passes. Rounding down to $5 mm$ ( $19.6 mm^{2}$ ) would give $459$ MPa $>$ allowable — fails, confirming we must round up. Concluding insight: always solve for area first, then round the physical dimension up, because area is what stress actually depends on.

Worked example Example 9 — Exam twist: double the area, change the material (Cell I)

Original design: bar of area $A$ carries force $F$ at exactly its yield stress $σ_{y} = 480 MPa$ (so $σ_{a pp} = 480$ MPa). Take $F o S_{r e q} = 1$ . Now (a) double the area to $2 A$ keeping the same force $F$ ; (b) swap to a material with $σ_{y} = 480$ but twice the Young's modulus $E$ . For each, what happens to (i) the stress, (ii) the yield stress, (iii) the margin?

Forecast: The classic trap — people confuse changing $σ$ (from geometry) with changing $σ_{y}$ (a material property) and confuse $E$ (stiffness) with strength. Predict all three before reading.

(a) Double area, same force — stress. $σ_{n e w} = \frac{F}{2 A} = \frac{1}{2} \frac{F}{A} = \frac{σ _{y}}{2} = 240 MPa$ . Why this step? Stress is $F / A$ ; doubling the denominator halves the stress. Geometry changes stress.
(a) Yield stress. $σ_{y}$ is a material property → unchanged at $480 MPa$ , so $σ_{allow} = σ_{y} / F o S_{r e q} = 480$ MPa. Why this step? No new material was introduced; only shape changed. Yield stress — and hence the allowable — can't move from a geometry change.
(a) Margin. $M o S = \frac{σ _{allow}}{σ _{n e w}} - 1 = \frac{480}{240} - 1 = 2 - 1 = 1$ . Why this step? We went from $M o S = 0$ (at yield) to a comfortable $+ 1$ just by fattening the bar — the same $σ_{allow}$ over a halved applied stress.
(b) Double $E$ — stress, yield, margin. Stress $σ = F / A$ is unchanged (geometry same, force same). $σ_{y} = 480$ unchanged, so $σ_{allow} = 480$ unchanged, hence $M o S$ unchanged. Why this step? $E$ governs stiffness/how much it stretches, not strength/when it yields — the parent's "higher $E$ ≠ stronger" mistake. A stiffer bar deforms less elastically but yields at the same stress and has the same margin.

Verify: (a) $σ_{n e w} = 240$ MPa, $M o S = 1$ — plug back: $480/240 = 2$ , minus $1$ is $1$ ✓. (b) Doubling $E$ halves the elastic strain ( $ε = σ / E$ ) but leaves $σ$ , $σ_{y}$ , $σ_{allow}$ and $M o S$ untouched ✓. Trap defused: geometry moves stress, material moves strength, and $E$ moves neither strength nor stress — only elastic strain.

Wrap-up: matrix coverage

Recall Did we hit every cell?

A (Ex 1), B (Ex 2), C (Ex 3), D (Ex 4), E (Ex 5), F (Ex 6), G (Ex 7), H (Ex 8), I (Ex 9) — all nine cells worked. Every sign of margin (positive, zero, negative), the degenerate zero/vanishing inputs, the necking limit, a sizing word problem, and the double-area/double- $E$ exam trap are all covered.

Prerequisite links to revisit if a step felt shaky: Stress and strain fundamentals, Young's modulus and elasticity, Factor of safety and margins, Material selection for spacecraft.