Fixed supportThree reactions, and M starts at RMA

Why does the wall give a moment too? (30 seconds)

• There is no second support to hold the end → without a moment, the beam would rotate about the wall.
• ΣM about A: RAX and RAY pass through A, they have no lever arm → a single unknown remains: RMA.
• The free end: S and M close at 0 there.

A 3 m cantilever

• Resultant
  1 · 3 = 3 t, in the middle: x = 1.5
• ΣFx
  No horizontal forces → RAX = 0
• ΣFy
  RAY − 3 − 2 = 0 → RAY = 5 t
• ΣMA
  Assume RMA is clockwise: 3·1.5 + 2·3 + RMA = 0 → RMA = −10.5 t·m — the minus: in fact 10.5 counter-clockwise
• S
  5 → diagonal line 5 − 1·3 = 2 → the force at the end: 2 − 2 = 0 closes ✓
• M
  Starts at RMA = −10.5 → trapezoid (5 + 2) · 3 / 2 = 10.5 → −10.5 + 10.5 = 0 closes ✓

That's it. All of M is negative — and the largest moment in magnitude sits at the wall: 10.5 t·m.

M starts at RMA — with which sign? With the sign that came out of ΣMA: M(0) = RMA = −10.5. (In BST: M1 = RMA — M1 is the value of M at the left end.)

ΣFy as an equation? Here, yes: there is one support, and ΣFy is the only way to RAY. The check in its place — S and M closing at 0 at the free end.

4 pitfalls

• M starts at 0. At a fixed support — start at RMA.
• Looking for BY. There is no support B — the whole load goes to the wall.
• Counting 2 unknowns. Fixed support = 3: RAX, RAY, RMA.
• A diagonal line under a trapezoid. Trapezoid in S → parabola in M.