Everything so far has been ingredients. Here they meet. A beam under a bending moment is being stretched along one face and squashed along the other, with a surface somewhere in between that is doing neither — the neutral axis. The stress grows in a straight line from zero at that axis to a maximum at the outermost fibres, and the relation that prices it is the flexure formula: $\sigma = \dfrac{M c}{I}$, read aloud sigma equals M c over I.

Every letter, named: $\sigma$ — the Greek letter sigma — is the bending stress in MPa, which is one newton per square millimetre. $M$ is the bending moment at the section, in kN·m. $c$ is the distance from the neutral axis to the fibre you are asking about, in mm — and since the worst fibre is the outermost one, $c$ is usually half the depth. $I$ is the second moment of area in mm⁴. The $c$ upstairs is the reason the outside of a beam works hardest; the $I$ downstairs is the reason a deeper beam works less hard.

Since $S = I/c$, the same statement collapses to $\sigma = \dfrac{M}{S}$ — sigma equals M over S — and that is how the job is done in practice, because handbooks tabulate $S$. Read backwards it becomes the design equation: $S = \dfrac{M}{\sigma_{allow}}$ gives the section you must buy. Demand in, geometry out.

Keep the working units and both forms come out in MPa with no bridging arithmetic at all: a moment in kN·m times a $c$ in mm, divided by an $I$ in millions of mm⁴, is MPa exactly. A moment in kN·m divided by an $S$ in thousands of mm³ needs one factor of a thousand and nothing else. Mixed scales are where this lesson's answers go to die, so decide your units before you decide your arithmetic.