Yield stress, ultimate stress — material behavior: Exercises

Constants reused below (all "engineering" quantities — force over the original area $A_{0}$ ):

Level 1 — Recognition

L1.1

A rod carries $F = 6000 N$ across an original area $A_{0} = 40 mm^{2}$ . Which formula gives the stress, and what is it in MPa?

Recall Solution

WHAT: stress is force spread over area, $σ = F / A_{0}$ . WHY that formula: the parent note's first mistake ("stress is just force") is exactly what we avoid — the same force on a smaller area is more punishing, so we always divide by area. Unit care — the one place people slip: $1 mm^{2} = 1 0^{- 6} m^{2}$ , so $40 mm^{2} = 40 \times 1 0^{- 6} m^{2}$ . $σ = \frac{6000}{40 \times 1 0 ^{- 6}} = 1.5 \times 1 0^{8} Pa = 150 MPa .$

L1.2

State whether each quantity is a material property (fixed for the alloy) or depends on the part's shape/load: (a) $σ_{y}$ , (b) the force at yield $F_{y}$ , (c) $E$ , (d) $σ_{a pp}$ .

Recall Solution

WHAT/WHY: a material property is baked into the metal's atomic bonds and dislocations, so it does not care how thick you cut the bar.

(a) $σ_{y}$ — material property. Yielding is dislocation glide; that happens at a set stress, not a set force.
(b) $F_{y} = σ_{y} A_{0}$ — depends on shape. Thicker bar → same $σ_{y}$ but bigger $F_{y}$ .
(c) $E$ — material property (slope of the elastic line).
(d) $σ_{a pp}$ — depends on load and shape (it's $F / A_{0}$ of your loading).

Level 2 — Application

L2.1

A titanium strut has $A_{0} = 25 mm^{2}$ , $E = 114 GPa$ , $σ_{y} = 830 MPa$ . It carries $F = 10000 N$ . Find the stress, confirm it is still elastic, and find the strain.

Recall Solution

Step 1 — stress. $σ = F / A_{0} = 10000/ (25 \times 1 0^{- 6}) = 4.0 \times 1 0^{8} Pa = 400 MPa$ . Step 2 — elastic check (WHY first): Hooke's law $σ = E ε$ is only legal below $σ_{y}$ . Here $400 < 830$ , so we're in the straight-line region — allowed. Step 3 — strain. $ε = σ / E = 400 MPa /114000 MPa = 3.51 \times 1 0^{- 3}$ . (Keep both in MPa so units cancel to a pure number.)

L2.2

Same strut. Required $FoS_{y i e l d} = 1.25$ . Compute the margin of safety against yield.

Recall Solution

WHAT: margin of safety, $MoS = \frac{σ _{a l l o w}}{FoS _{r e q} σ _{a pp}} - 1$ , with $σ_{a l l o w} = σ_{y}$ for a yield check. $MoS = \frac{830}{1.25 \times 400} - 1 = \frac{830}{500} - 1 = 1.66 - 1 = 0.66.$ WHY it means "pass": $MoS = 0.66 \geq 0$ — after already dividing by the required 1.25, we still have 66% headroom.

Level 3 — Analysis

L3.1 (offset yield, geometric)

A stress–strain test gives elastic slope $E = 68 GPa$ . The 0.2% offset line is $σ = 68000 (ε - 0.002)$ (MPa). It intersects the measured curve at $ε = 0.0095$ . Find $σ_{y}$ , and read off the picture below what the offset construction means.

Figure — Yield stress, ultimate stress — material behavior

Recall Solution

WHAT the offset does (look at the figure): many alloys bend gradually, so there is no single obvious "yield point". We draw a line parallel to the elastic slope (same $E$ ) but shifted right to start at $ε = 0.002$ . Where the shifted line (teal, dashed) crosses the real curve (orange) defines $σ_{y}$ — that intersection is the plum dot. WHY 0.2%: it pins yield to a fixed permanent strain of 0.002, so every lab gets the same number. Compute: $σ_{y} = 68000 (0.0095 - 0.002) = 68000 \times 0.0075 = 510 MPa .$

L3.2 (engineering vs true stress reasoning)

A specimen reaches its ultimate point. The engineering stress there is $σ_{u} = 500 MPa$ (using original area $A_{0}$ ). Just after, a neck forms and the real load-bearing area drops to $0.80 A_{0}$ while the force momentarily holds at the ultimate force. Compute the true stress at that instant and explain, from the parent's third mistake, which number the handbook prints.

Recall Solution

Step 1 — recover the ultimate force. Engineering stress uses $A_{0}$ : $F_{u} = σ_{u} A_{0} = 500 MPa \times A_{0}$ . Step 2 — true stress uses the shrunken area $A_{in s t} = 0.80 A_{0}$ : $σ_{t r u e} = \frac{F _{u}}{A _{in s t}} = \frac{500 A _{0}}{0.80 A _{0}} = 625 MPa .$ WHY they diverge: after necking the real area collapses, so dividing by the smaller area makes true stress climb even as engineering stress falls. The handbook prints the engineering value ( $500 MPa$ ) — design allowables and $σ_{u}$ tables are all $F / A_{0}$ .

Level 4 — Synthesis

L4.1 (two checks, pick the governing margin)

An aluminium 7075 fitting: $A_{0} = 80 mm^{2}$ , $σ_{y} = 480 MPa$ , $σ_{u} = 570 MPa$ . A launch load case applies $F = 24000 N$ (see Structural load cases and launch loads). Using $FoS_{y i e l d} = 1.25$ and $FoS_{u l t} = 1.5$ , compute both margins and state which one governs (the smaller one wins).

Recall Solution

Step 1 — applied stress. $σ_{a pp} = 24000/ (80 \times 1 0^{- 6}) = 3.0 \times 1 0^{8} Pa = 300 MPa$ . Step 2 — yield margin. $MoS_{y} = \frac{480}{1.25 \times 300} - 1 = \frac{480}{375} - 1 = 1.28 - 1 = 0.28.$ Step 3 — ultimate margin. $MoS_{u} = \frac{570}{1.5 \times 300} - 1 = \frac{570}{450} - 1 = 1.2667 - 1 = 0.267.$ Step 4 — which governs (see figure): the smaller margin is the binding one — it's the first to fail as load grows. $MoS_{u} = 0.267 < MoS_{y} = 0.28$ , so ultimate governs here (barely). Both are $\geq 0$ , so the fitting passes.

L4.2 (design a thickness backwards)

Same alloy and load ( $F = 24000 N$ ), but now you must guarantee $MoS_{y} \geq 0$ and $MoS_{u} \geq 0$ with the same required factors. What is the minimum original area $A_{0}$ , and hence the minimum side of a square cross-section?

Recall Solution

WHY work backwards: $MoS \geq 0$ means $σ_{a l l o w} \geq FoS_{r e q} σ_{a pp}$ , i.e. the applied stress must not exceed the allowable. Turn each into a required area. Yield need: $σ_{a pp} \leq σ_{y} / FoS_{y} = 480/1.25 = 384 MPa$ $\Rightarrow A_{0} \geq F /384 MPa = 24000/ (384 \times 1 0^{6}) = 6.25 \times 1 0^{- 5} m^{2} = 62.5 mm^{2}$ . Ultimate need: $σ_{a pp} \leq σ_{u} / FoS_{u} = 570/1.5 = 380 MPa$ $\Rightarrow A_{0} \geq 24000/ (380 \times 1 0^{6}) = 6.316 \times 1 0^{- 5} m^{2} = 63.16 mm^{2}$ . Governing (bigger) area wins: $A_{0} = 63.16 mm^{2}$ . Square side: $s = 63.16 = 7.95 mm$ .

Level 5 — Mastery

L5.1 (full elastic-to-permanent story)

A steel tie-rod: $L_{0} = 500 mm$ , $A_{0} = 100 mm^{2}$ , $E = 200 GPa$ , $σ_{y} = 350 MPa$ , $σ_{u} = 500 MPa$ . (a) What force first causes yield? (b) At that force, how much has the rod stretched (elastic elongation, in mm)? (c) A fault case slams $F = 45000 N$ through it — is the rod still elastic, permanently bent, or broken? (d) What force would break it?

Recall Solution

(a) Force at first yield. Yield is a stress limit, so the force is $F_{y} = σ_{y} A_{0}$ : $F_{y} = 350 \times 1 0^{6} \times 100 \times 1 0^{- 6} = 35000 N .$ (b) Elastic elongation at that force. Right at yield the strain is (still on the elastic line) $ε = σ_{y} / E = 350/200000 = 1.75 \times 1 0^{- 3}$ , so $Δ L = ε L_{0} = 1.75 \times 1 0^{- 3} \times 500 mm = 0.875 mm .$ Why $L_{0}$ not $A_{0}$ here: elongation is strain × length, area already did its job in the stress. (c) Fault case $F = 45000 N$ . Stress $= 45000/ (100 \times 1 0^{- 6}) = 450 MPa$ . Compare to the landmarks: $350 (σ_{y}) < 450 < 500 (σ_{u})$ . So it is past yield but below ultimate — the rod is permanently deformed (plastically bent) but not broken. (d) Breaking force. Fracture (engineering) tracks $σ_{u}$ over $A_{0}$ : $F_{b r e ak} = σ_{u} A_{0} = 500 \times 1 0^{6} \times 100 \times 1 0^{- 6} = 50000 N .$

L5.2 (choose the alloy)

For a mass-critical bracket you may pick either alloy for the same load $σ_{a pp} = 250 MPa$ : Alloy A ( $σ_{y} = 300$ , $σ_{u} = 360 MPa$ ) or Alloy B ( $σ_{y} = 380$ , $σ_{u} = 430 MPa$ ). Required $FoS_{y i e l d} = 1.25$ , $FoS_{u l t} = 1.5$ . For each alloy compute both margins, decide pass/fail, and pick the alloy. (Ties into Material selection for spacecraft.)

Recall Solution

Alloy A: $MoS_{y} = 300/ (1.25 \times 250) - 1 = 300/312.5 - 1 = 0.96 - 1 = - 0.04$ → negative, FAILS yield. (No need to bother with ultimate: one negative margin condemns the design.) Alloy B: $MoS_{y} = 380/ (1.25 \times 250) - 1 = 380/312.5 - 1 = 1.216 - 1 = 0.216$ → pass. $MoS_{u} = 430/ (1.5 \times 250) - 1 = 430/375 - 1 = 1.1467 - 1 = 0.147$ → pass. Decision: Alloy A fails (its yield margin is below zero), so you must choose Alloy B, whose governing margin is $MoS_{u} = 0.147 \geq 0$ .