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The BST binders · Beams binder · Reactions and loads · updated 27.9.2026

Reactions in a beamTwo moment equations — and you are done

Reaction = the force that the support applies to the beam. One equation for every reaction — and one more for the check.

  • ΣMA = 0 → BY
  • ΣMB = 0 → AY
  • ΣFy = 0 → check only

Distributed load? First the resultant: P = Q · L, at the middle of the segment — and for the reactions only.

  1. 01

    Why about the support? (30 seconds)

    • Moment about A: the reactions of A pass through A → their lever arm is 0 → one unknown is left.
    • Every force × its horizontal distance from the support (the lever arm). A force that rotates the beam about the point clockwise — plus, counter-clockwise — minus.
    • How do you know? A force down to the right of the point → clockwise (+). A force up to the right of the point — like BY about A → counter-clockwise (−). To the left of the point — the opposite: a force up on the left (like AY about B) → clockwise (+).
    • ΣFy at the end — if it does not come to zero, there is a mistake, and you find it now and not in the middle of the M diagram.
  2. 02

    A beam of 8 m — two equations

    2 t/m8 t6 tAY = 6.5 tBY = 7.5 t2 m2 m4 m
    The resultant (dashed blue) — for the reactions only · scroll the drawing sideways
    • Resultant
      2 · 4 = 8 t, at the middle of the segment → x = 6
    • ΣMA
      6·2 + 8·6 − BY·8 = 0 → BY = 7.5 t
    • ΣMB
      AY·8 − 6·6 − 8·2 = 0 → AY = 6.5 t
    • Check
      6.5 + 7.5 − 6 − 8 = 0 ✓

    That's it. Two equations, one check — and the reactions are in your pocket.

  3. 03

    4 pitfalls everyone falls into

    • A lever arm from the end of the beam instead of from the support. When the support is not at the end — the whole equation shifts.
    • A concentrated moment times a distance. M₀ enters ΣM at its full value, without multiplying by a distance.
    • The resultant stays in the drawing of S and M. It is for the reactions only — in the diagrams you work with the load itself.
    • ΣFy as the second equation instead of a check. A mistake in AY passes straight to BY, and nothing catches it.

    Before anything else: count the unknowns. Roller support = 1, pinned = 2 (horizontal and vertical), fixed = 3 → in a statically determinate beam the total must be 3. Here: AY, BY and AX — and AX = 0 because there are no horizontal forces (ΣFx = 0).

Beams binder · BST · beamsolvertool.com/en/learn/beams/support-reactions/

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