A tensile machine divides load by the ORIGINAL cross-section, because that is the only area it was ever told. Call that the engineering stress $\sigma_e$, in MPa. But the bar is getting thinner as you pull it, so the area actually carrying the load is smaller than the one on the calibration sheet. Divide by THAT area and you have the true stress $\sigma_t$, also in MPa — and it is always the larger of the two.

Constant volume does the rest of the work: $\sigma_t = \sigma_e (1 + e)$ — read aloud, sigma-t equals sigma-e times one plus e. The $e$ here is the engineering strain: the extension divided by the original gauge length, a length over a length, so it wears no units at all. Exam papers quote it as a percent and the formula wants it as a ratio, which is the single most reliable slip in this conversion — 15 % is 0.15, and $1 + e$ is 1.15, never 16.

Its twin handles the strain side: $\varepsilon_t = \ln(1 + e)$, the true or natural strain, which adds up honestly over successive stretches in a way engineering strain does not. Both conversions hold only while the deformation is uniform. Once the specimen necks there is no single strain in the bar to speak of, and every relation on this page politely stops.