Beam with a cantileverNegative M over the support — and Mmin

Why is M negative over the support? (30 seconds)

• The load on the cantilever rotates it about the support downward → over the support the beam bends 'upward' → M is negative.
• There is no support at the free end → S and M must reach 0 there.
• In BST, positive M is drawn below the axis — so negative M is drawn above it.

Beam 8 m, cantilever 2 m — everything

• Resultant
  2 · 6 = 12 t, at the middle of the segment: x = 3
• ΣMA
  12·3 + 4·8 − BY·6 = 0 → BY = 11.33 t
• ΣMB
  AY·6 − 12·3 + 4·2 = 0 → AY = 4.67 t — the 4 t on the cantilever rotates about B clockwise
• Check
  4.67 + 11.33 − 12 − 4 = 0 ✓
• S
  4.67 → diagonal line 4.67 − 2·6 = −7.33 → BY: −7.33 + 11.33 = 4 → straight along the cantilever → 4 − 4 = 0 closed ✓
• Z
  Z = S / Q = 4.67 / 2 = 2.33 m
• 0–Z
  Triangle above the axis 2.33 · 4.67 / 2 = 5.44 → Mmax = 5.44 t·m at Z
• Z–B
  Triangle below the axis: base 6 − 2.33 = 3.67, height 7.33 → 3.67 · 7.33 / 2 = 13.44 → 5.44 − 13.44 = −8 → Mmin = −8 t·m over B
• Cantilever
  Rectangle above the axis 4 · 2 = 8 → −8 + 8 = 0 closed ✓

Here |Mmin| = 8 is larger than Mmax = 5.44 — the largest value in magnitude sits over the support, not at Z.

And what about 'Mmax at Z'? The rule holds for the positive M. With a cantilever there is also a negative M — and you look for it over the support, where S jumps above the axis.

4 pitfalls

• S 'ends' at B. Continue along the cantilever — S reaches 0 only at the free end.
• The sign of the cantilever force in ΣMB. It is on the other side of B → it rotates opposite to the load between the supports.
• Looking only for Mmax at Z. Check Mmin over the support as well.
• M at the end of the cantilever is not 0. A sign of a mistake in the reactions or in the areas.