S diagramLeft to right, jump after jump

Why does it work? (30 seconds)

• S at a point = the sum of the vertical forces to the left of the point.
• Every force you pass → the sum changes by exactly its size → a jump.
• A distributed load adds a little in every centimetre → the graph falls gradually → a diagonal line.

A beam of 8 m — 6 steps

• x = 0
  AY up → jump to 6.5
• 0–2
  No load → straight line
• x = 2
  6 t down → 6.5 − 6 = 0.5
• 2–4
  No load → straight line
• 4–8
  2 t/m → diagonal line: 0.5 − 2·4 = −7.5
• x = 8
  BY up → −7.5 + 7.5 = 0 closed ✓

• Z
  S crosses 0 on the diagonal line → Z = S / Q = 0.5 / 2 = 0.25 m after x = 4 → x = 4.25

That's it. 6 steps, and the graph closes at 0 — a sign that the reactions are correct.

And if S crosses 0 in a jump? For example a point force that takes S down from +3.8 to −1.2. No diagonal line crosses — there is no Z. That is fine: the point of the force is the place of Mmax.

A concentrated moment? It does not change S at all — it only rotates.

4 pitfalls everyone falls into

• Starting from the end instead of from the reaction. The graph starts at the point where the first force meets the beam.
• The resultant instead of the load. You get a jump of 8 at x = 6 instead of a diagonal line.
• Q from another segment in Z. Take the Q of the segment where S crosses 0.
• It did not close at 0 — and you carry on. That is a mistake in the reactions. Go back there before M.