Question 2 of 7: True Strain from Gauge Length and Diameter; Validity During Necking; Brinell Hardness

Question 2: True Strain from Gauge Length and Diameter; Validity During Necking; Brinell Hardness (2 of 5)

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 tensile specimen with initial gauge length $l_0$ (written $L_0$ below), initial diameter $d_0$; during the test the instantaneous gauge length $L$ and diameter $d$ change continuously and the specimen volume is assumed constant (fully plastic deformation).

Find. (a) Derive Eqs. (1) and (2) and identify the more valid one during necking, with reasoning. (b) Describe the Brinell test, explain its correlation with tensile strength for structural steels, why that correlation fails for heat-treated steels, and a better test for them.

Approach

True strain is defined as the integral of incremental engineering strain referenced to the instantaneous length, $d\epsilon_T = dL/L$; integrating from $L_0$ to $L$ gives Eq. (1). Constant volume ($A_0L_0=AL$, with $A\propto d^2$) converts the length ratio into a diameter ratio for Eq. (2). The necking question follows from which quantity (average length vs. local diameter) remains representative once deformation localises. Part (b) is descriptive, drawing on Callister’s hardness-testing chapter.

1. (a) Eq. (1): true strain from gauge length. By definition, an increment of true strain is the incremental elongation divided by the current (not original) length: $$d\epsilon_T = \frac{dL}{L}$$ Integrating from the original length $L_0$ to the current length $L$: $$\epsilon_T = \int_{L_0}^{L}\frac{dL}{L} = \big[\ln L\big]_{L_0}^{L}$$ $$\boxed{\epsilon_T = \ln\!\left(\frac{L}{L_0}\right)}$$
2. (a) Eq. (2): true strain from diameter, via constant volume. Constant volume during plastic deformation requires $A_0L_0 = AL$, i.e. $L/L_0 = A_0/A$. For a circular cross-section, $A=\tfrac{\pi}{4}d^2$, so $A_0/A=(d_0/d)^2$. Substituting into Eq. (1): $$\epsilon_T = \ln\!\left(\frac{L}{L_0}\right) = \ln\!\left(\frac{A_0}{A}\right) = \ln\!\left(\frac{d_0^2}{d^2}\right)$$ $$\boxed{\epsilon_T = 2\ln\!\left(\frac{d_0}{d}\right)}$$
3. (a) Which expression is valid during necking. The diameter-based expression, $\epsilon_T=2\ln(d_0/d)$, remains valid during necking; the length-based expression does not. Once necking begins, deformation localises into a small region of the gauge length — the diameter $d$ measured at the neck is a genuinely local measurement of the most-deformed cross-section, so the constant-volume substitution $A_0L_0=AL$ still applies locally. The gauge length $L$, by contrast, is an average over the whole (now non-uniformly strained) gauge section: most of the gauge length outside the neck has stopped straining once necking localises, so the average elongation $L/L_0$ increasingly under-represents the true local strain at the neck as deformation proceeds. $$\boxed{\text{diameter-based } \epsilon_T = 2\ln(d_0/d) \text{ remains valid post-necking}}$$
4. (b) The Brinell hardness test. A hardened steel (or tungsten-carbide) ball, standard diameter $D=10$ mm, is pressed into the flat, polished test surface under a fixed load $P$ (commonly $500$–$3000$ kgf, held $10$–$30$ s), and the diameter $d$ of the resulting circular indentation is measured (typically with a low-power microscope/scale). The Brinell hardness number follows from the load divided by the curved (spherical-cap) area of the indentation: $$HB = \frac{2P}{\pi D\!\left(D-\sqrt{D^2-d^2}\right)}$$
5. (b) Why Brinell correlates closely with tensile strength. The Brinell indenter is large (10 mm ball) and the resulting indentation is correspondingly large, so the test samples a substantial volume of material — averaging over grains, inclusions, and any local microstructural inhomogeneity, in much the same way a full-size tensile specimen does. This large sampled volume is what makes $HB$ correlate empirically closely with $\sigma_{UTS}$ for plain-carbon and low-alloy structural steels via the well-known relation $\sigma_{UTS}(\text{MPa})\approx3.45\,HB$. Sharper, small-indenter tests (Vickers diamond pyramid, Rockwell cones/small balls) probe a much smaller, more locally-sensitive volume, so they are more sensitive to local microstructural variation and correlate less consistently with the bulk tensile property.
6. (b) Why the correlation does not exist for heat-treated steels. The factor $3.45$ is empirical: it holds because, in a ductile ferrite–pearlite structural steel, the metal under the ball work-hardens and flows in the same way as the necking region of a tensile bar, so the mean indentation pressure tracks the tensile strength. A quenched (martensitic) or quenched-and-tempered steel behaves differently in three ways: (1) its work-hardening response and ductility are very different, so the ratio of indentation pressure to UTS changes with the heat-treatment condition; (2) hardening is often non-uniform through the section (hard case, softer core), so a surface indentation no longer represents the bulk tensile bar; and (3) above roughly $450$–$500$ HB the standard hardened-steel ball itself flattens under the $3000$ kgf load, so the measured impression is no longer a true Brinell reading at all.
7. (b) A better test for heat-treated (hardened) steels. Rockwell C (HRC), using a conical diamond (Brale) indenter under a $150$ kgf load, is the standard choice for heat-treated steels: the diamond does not deform, the small, shallow indentation makes the test effectively non-destructive on finished/thin hardened parts, and the reading is fast (direct-reading dial, no optical measurement needed). $$\boxed{\text{Better test for hardened steel: Rockwell C (HRC), diamond cone indenter}}$$