Uniformly distributed loadOne force in the middle — for reactions only

Why is it allowed — and why for reactions only? (30 seconds)

• The resultant gives the supports the same moment as all the small forces together → the reactions come out exact.
• But inside the beam the load is spread out and does not land at one point → in S it is a sloped line, not a jump.
• Where does S cross 0? Z = S / Q — how many metres the load needs to "eat" S.

A 6 m beam — from the load to Mmax

• Resultant
  2 · 6 = 12 t in the middle → x = 3
• Reactions
  ΣMA: 12·3 − BY·6 = 0 → BY = 6 t, and the beam is symmetric → also AY = 6 t
• S
  Starts at 6, goes down along a sloped line → 6 − 2·6 = −6 → BY closes it to 0 ✓
• Z
  Z = 6 / 2 = 3 m
• Mmax
  Area of the triangle up to Z: 3 · 6 / 2 → Mmax = 9 t·m, at Z

That's it. A resultant for the reactions, a sloped line in S, one triangle for Mmax.

One-second check: a simple beam with a uniform load along the whole length: Mmax = Q·L² / 8 = 2·36 / 8 = 9 ✓

4 pitfalls everyone falls into

• The resultant goes into S. You get a jump of 12 in the middle instead of a sloped line — and all of M is wrong.
• L of the whole beam. When the load is only on part of it — L is the length of the segment.
• The resultant in the middle of the beam. It is in the middle of the load's segment.
• Q from another segment in Z. Take the Q of the segment where S crosses 0.