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21-Mat-A5 Phase Transformations and Thermal Treatment · December 2014

Question 5 of 8: General Yield vs. Fast Fracture in a Marine Steel Plate; Cold Brittleness; HCP Ductility

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

Paper format. National Exams, December 2014 — 10-Met-A5, Mechanical Behaviour and Fracture of Materials. Three hours, closed book, any non-communicating calculator permitted. Eight questions of 20 marks each; the rubric states that five questions constitute a complete paper and that only the first five appearing in the answer book are marked. All eight are answered here, because this set is a study resource rather than an exam script. Several questions ask explicitly for essay-format answers, and the marking scheme rewards clarity and organisation, so the discursive answers below are written as structured prose rather than as note form.

Note on the exam title

The printed exam header reads 10-Met-A5, Mechanical Behaviour and Fracture of Materials. The paper examines fracture mechanics and fatigue-crack-growth life, strengthening and toughening of engineering materials, creep and fatigue testing, deformation processing selection, and elastic–plastic forming behaviour; it has no classical phase-transformation or heat-treatment (TTT/CCT diagram, hardenability, tempering-curve) questions. The answers below are written to the printed subject.

Reference texts. The answers below are keyed to the works normally recommended for this syllabus code:



Question 5: General Yield vs. Fast Fracture in a Marine Steel Plate; Cold Brittleness; HCP Ductility (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

5.1 — (a) Fast fracture or general yield at the ultrasonic detection limit?

Given.

QuantitySymbolValue
Fracture toughness$K_c$53 MPa$\sqrt{\text{m}}$ ($=53\ \text{MN\,m}^{-3/2}$)
Yield strength$\sigma_y$950 MPa
Smallest detectable surface crack$a$1.0 mm

Find. Whether a plate containing a crack at the detection limit fails by general yield or by fast fracture first.

Approach. Compare the crack size $a$ against the transition flaw size $a_t$ — the crack length at which the fast-fracture stress $\sigma_f=K_c/\sqrt{\pi a}$ exactly equals $\sigma_y$. If the actual flaw is larger than $a_t$, the plate reaches $K_c$ (fast fracture) at a nominal stress below $\sigma_y$; if smaller, the plate yields generally before fast fracture can occur.

  1. Compute the transition flaw size. $$a_t=\frac{1}{\pi}\left(\frac{K_c}{\sigma_y}\right)^2=\frac{1}{\pi}\left(\frac{53}{950}\right)^2=\frac{1}{\pi}(0.05579)^2=9.907\times10^{-4}\ \text{m}.$$ $$\boxed{a_t \approx 0.991\ \text{mm}}$$
  2. Compare to the actual (detectable) crack size. $a=1.0$ mm is larger than $a_t=0.991$ mm. Equivalently, the fast-fracture stress at $a=1.0$ mm is $$\sigma_f=\frac{K_c}{\sqrt{\pi a}}=\frac{53}{\sqrt{\pi(0.001)}}=945.6\ \text{MPa},$$ which is below $\sigma_y=950$ MPa.
  3. Conclusion. $$\boxed{\sigma_f = 945.6\ \text{MPa} \;\lt\; \sigma_y = 950\ \text{MPa} \;\Longrightarrow\; \text{fast fracture governs, not general yield.}}$$ A plate loaded from zero, containing a crack right at the smallest size the inspection method can find, fractures catastrophically at about 945.6 MPa — 0.5% below the stress that would cause general yield. The margin is thin under the simplest ($Y=1$, through-thickness crack) geometry model; the more conservative surface-crack correction $Y\approx1.12$ lowers $\sigma_f$ to about 844 MPa, a clearly sub-yield fast-fracture stress with no ambiguity.
QuantityResult
Transition flaw size, $a_t$ ($Y=1$)0.991 mm
Fast-fracture stress at $a=1.0$ mm, $Y=1$945.6 MPa
Fast-fracture stress at $a=1.0$ mm, $Y=1.12$844.3 MPa
Governing failure mode$\boxed{\text{Fast fracture (before general yield)}}$
Check: safety implication

Because $a\approx a_t$, this is a marginal case by design. In practice this means the plate should not be qualified on inspection alone at this $K_c/\sigma_y$ ratio: either a higher-resolution NDE method (reliably detecting flaws comfortably under $a_t$) or a design margin on stress (operating below $\sigma_f$ with a safety factor) is required, since a crack at the very limit of what ultrasonic inspection can find is already large enough to fail the plate by fast fracture rather than by the more forgiving, visibly-warned general-yield route.

5.2 — (b) Why steel structures are more failure-prone in cold winter conditions

Body-centred-cubic steels undergo a ductile-to-brittle transition as temperature falls, because the lattice (Peierls–Nabarro) resistance to dislocation glide is strongly thermally activated in the BCC structure — unlike FCC metals, whose glide resistance is comparatively insensitive to temperature. As temperature drops, the stress needed to move dislocations (the effective yield stress) rises steeply, while the stress needed to trigger cleavage fracture on a low-index plane changes comparatively little. Above the ductile-to-brittle transition temperature (DBTT), the yield stress is low enough that the material yields and blunts a crack tip plastically before the local stress reaches the cleavage fracture stress, so failure is ductile, high-energy, and gives warning through visible deformation. Below the DBTT the required yield stress exceeds the cleavage stress, so the material fractures by low-energy transgranular cleavage before it can yield — often suddenly, at nominal stresses well below the room-temperature yield strength, frequently initiating at a stress concentrator such as a weld defect, notch, or pre-existing crack (exactly the fast-fracture mechanism quantified in part (a)). Structural steels typically show a DBTT in the range of roughly $-20$ to $+20\,{}^{\circ}\text{C}$ depending on composition, grain size and thickness, placing ordinary winter service temperatures for a marine vessel, bridge or offshore platform close to or below the transition, while summer temperatures sit safely in the ductile regime. Large section thickness compounds the effect by promoting a triaxial (plane-strain) stress state at any flaw, further raising the local stress needed to trigger cleavage relative to yield, and high loading rates (wave slam, impact) push the transition temperature upward — both effects stacking unfavourably in exactly the cold, dynamically loaded service this question describes. The historical case most often cited is the brittle fracture of welded Liberty-ship hulls in the cold North Atlantic during World War II, which motivated the fracture-mechanics and Charpy-impact-testing framework used to qualify structural steels today.

5.3 — (c) Why HCP metals are more brittle than FCC or BCC metals

Ductility in a polycrystal requires each grain to accommodate an arbitrary shape change imposed by its neighbours without cracking at the grain boundaries; the von Mises criterion establishes that this requires at least five independent slip systems operating simultaneously. FCC metals satisfy this easily with 12 slip systems on the close-packed $\{111\}\langle110\rangle$ family, and BCC metals, while lacking a single close-packed plane, compensate with a large number of nearly-close-packed systems ($\{110\}$, $\{112\}$, $\{123\}$, all $\langle111\rangle$) that together provide more than enough independent systems. HCP metals, by contrast, have only one close-packed plane, the basal $(0001)$, which offers just 2 independent slip systems on $\langle11\bar{2}0\rangle$ directions — nowhere near the required five. The remaining non-basal systems (prismatic $\{10\bar{1}0\}$, pyramidal $\{10\bar{1}1\}$) exist crystallographically but typically have a critical resolved shear stress several times that of basal slip, so they only activate at high stress or elevated temperature; deformation twinning is often needed to supply additional strain modes, but twinning itself is a much less flexible accommodation mechanism than slip and can itself nucleate cracks at twin–boundary or twin–twin intersections. The practical result is that polycrystalline HCP metals (magnesium, zinc, titanium at room temperature, zirconium) cannot easily accommodate the compatibility strains imposed by neighbouring, differently-oriented grains; strain concentrates at grain boundaries, and the metal fails at comparatively low overall ductility, with strong crystallographic texture and pronounced anisotropy compounding the effect. FCC and BCC metals, having an excess of available independent systems, distribute strain smoothly through a polycrystal and remain ductile to much larger overall strains.