Question 5 of 8: TTT Diagram Reading & Isothermal Anneal Design; Critical Crack Size

Question 5: TTT Diagram Reading & Isothermal Anneal Design; Critical Crack Size (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.

Given. (a) TTT diagram for 1035 steel (Fig 2), with hardness values correlated to hold temperature for fully transformed structures ($1100^\circ$F→240 HB, $1000^\circ$F→300 HB, $M_s=600^\circ$F, $M_f=490^\circ$F). (b) $\sigma_y=600$ MPa, $K_{IC}=120$ MPa$\cdot$m$^{1/2}$, applied stress $\le0.70\sigma_y$.

Find. (a)(i) Two candidate as-cast microstructures giving 400 HB. (a)(ii) A time-temperature path achieving ≤250 HB. (b) Minimum (critical) crack size and applicable NDT methods.

1. (a)(i) As-cast 400 HB structures. Reading Fig 2's hardness scale, 400 HB sits at the $M_s$ line ($600^\circ$F), i.e. in the hardest region the diagram documents. An as-cast part that air-cools moderately quickly (especially a thin section) can easily miss the pearlite "nose" and transform at low temperature without a controlled isothermal hold, so two plausible as-cast microstructures giving this hardness are (untempered) martensite (if the local cooling rate carried the section past the nose and down through $M_s$–$M_f$) and lower bainite (if the cooling rate was fast enough to avoid the pearlite nose but not fast enough to reach $M_s$, so the material isothermally-in-effect transforms in the low-temperature bainite range on the way down) — both are hard, brittle, difficult-to-machine products consistent with an uncontrolled as-cast cooling path.
2. (a)(ii) Choosing the isothermal hold temperature. Interpolating Fig 2's hardness correlation between $1100^\circ$F (240 HB) and $1000^\circ$F (300 HB), a target of 250 HB max falls just below $1100^\circ$F. Holding at $$\boxed{T_{hold}=1100^\circ\text{F}\ (\approx240\ \text{HB, well inside the }\le250\text{ HB spec})}$$ is the practical choice: it is also the temperature of the curve's nose (the fastest-transforming temperature), so it minimizes the required furnace/hold time and produces the softest, coarsest pearlite the diagram documents.
3. Reading the isothermal hold time. At $1100^\circ$F transformation starts at the nose of the start curve, $t\approx100$ s, and the finish curve is reached at its own nose, so transformation is complete by $$t_{hold}\approx600\ \text{s}\quad(\approx10\ \text{min})$$ (a chart reading, disclosed as approximate to the precision the reproduced curve supports).

4. (b) Critical crack size. With the page-1 fracture-toughness formula $K_{IC}=f\sigma\sqrt{\pi a}$ and assuming the standard geometry factor $f=1$ (no other geometry is specified), the applied stress is $$\sigma=0.70\times600=420\ \text{MPa},$$ and solving for the crack size $a$ at the point of failure ($K=K_{IC}$): $$a=\frac{1}{\pi}\left(\frac{K_{IC}}{f\sigma}\right)^2 =\frac{1}{\pi}\left(\frac{120}{420}\right)^2$$ $$\boxed{a\approx26\ \text{mm}}$$ (the half-length of an internal flaw, or the full depth of an edge/surface flaw, under this $f=1$ convention — i.e. a total internal crack length of ≈52 mm).
5. Detection methods for a ≈25–50 mm crack. A flaw of this size is macroscopic and well within the resolving power of standard NDT: radiography (X-ray or $\gamma$-ray, showing internal flaws as a density/absorption contrast on film or a digital detector), ultrasonic testing (a pulse-echo probe detects the crack as a reflected/back-scattered signal, and can also size and depth-locate it), and, if the crack reaches the surface, the faster and cheaper magnetic particle inspection (for this ferromagnetic low-alloy steel — iron filings/ink concentrate at the leakage flux over a surface-breaking crack) or dye penetrant inspection. Any of these would readily detect a crack well below the ≈25–50 mm critical size, giving a real inspection margin before the tank reaches the failure condition.