Cut a thread on a bar and you have taken metal away. So when a bolt is pulled in tension, the area resisting that pull is not the circle on the size stamped on its head — it is smaller, and every proof load, grade marking and tensile strength on a fastener is quoted against the smaller figure. That figure is the tensile stress area.

$A_t = \dfrac{\pi}{4}\left(d - k_t p\right)^{2}$ — read aloud A-t equals pi over four, times the quantity d minus k-t p, squared. $A_t$ is the tensile stress area in square millimetres. $d$ is the nominal major diameter in millimetres — the number after the M. $p$ is the thread pitch in millimetres, crest to crest; for an inch thread it is 1 divided by the threads per inch. And $k_t$ is the thread-form coefficient, a bare number with no units: 0.9382 for ISO metric threads, 0.9743 for Unified inch. It is an input, not a constant, because the two thread forms truncate their roots differently and applying a metric constant to an inch bolt is quietly wrong.

Notice what the construction is doing. It builds an EFFECTIVE diameter, $d - k_t p$, that sits between the pitch diameter and the minor diameter, then puts a plain circle on it. Not the shank — that over-rates the bolt by about a sixth. Not the root either: the thread is a helix, so a little of the crest material carries load, and the root area is too pessimistic. Check it once against the published table — M12 × 1.75 comes out 84.27 mm², and ISO prints 84.3 — and you will trust the formula for the rest of your career.

One caveat worth carrying. This is the right area for a bolt in TENSION. A bolt loaded in shear across an unthreaded shank uses the full shank area; sheared across its threads, the root area. Three areas, one fastener, and the load path decides which.