Beam formulas/Overhanging

Overhanging Beam with Uniform Distributed Load

Reactions, shear, bending moment and deflection formulas, with a live calculator.

wABLa

A beam rests on a pin at A and a roller at B, a span L apart, and extends a distance a past B. A uniform load w acts over the full length L + a.

Formulas

Reactions

Reactions

Shear

Shear

Bending moment

Maximum positive moment

Moment over support B

Deflection

Deflection between supports (0 ≤ x ≤ L)

Deflection at the free end

  • A short overhang lifts the tip upward while the main span sags. A long overhang (a > 0.6L or so) makes the tip drop and can govern.
  • If a > L, R_A becomes negative: the left support must hold the beam down.

Calculator

Loads act downward as drawn. Steel: E = 200 GPa. I in millions of mm⁴ (100 means 100 × 10⁶ mm⁴).

Reactions

  • RA = 26.67 kN
  • RB = 53.33 kN

Forces: + up. Moments: − hogging.

Maximum values

  • |V|max = 33.33 kN at x = 6 m
  • Mmax = 35.56 kN·m at x = 2.667 m
  • δmax = 6.205 mm at x = 2.858 m
  • Span/δ = 966.9

Shear, moment and deflection diagrams

Shear force V (kN)
-33.33 kN
Bending moment M (kN·m)
35.56 kN·m
Deflection δ (downward) (mm)
6.205 mm

x is measured from the left end. Positive moment (sagging) plots above the axis. Need more loads or supports? Solve any beam with the full beam calculator.

How these formulas were checked

Every formula on this page is evaluated by the calculator above and was cross-checked numerically against PolyCalc's stiffness-method beam solver: reactions, shear and moment at dozens of points, and deflection by integrating the solver's moment diagram twice with the support conditions.

More beam formulas

FAQ

What are the reactions for an overhanging beam with uniform distributed load?

RA = w(L² − a²)/(2L) and RB = w(L + a)²/(2L), where L is the span between supports and a is the overhang.

What is the maximum bending moment of an overhanging beam with uniform distributed load?

The positive moment in the span is w(L + a)²(L − a)²/(8L²) and the negative moment over the support next to the overhang is wa²/2. Design for whichever is larger.

What is the maximum deflection of an overhanging beam with uniform distributed load?

Between the supports δ(x) = wx(L⁴ − 2L²x² + Lx³ − 2a²L² + 2a²x²)/(24EIL). At the tip δ = wa(4a²L − L³ + 3a³)/(24EI), where a negative value means the tip moves up.

What sign convention do these formulas use?

x is measured from the left end (from the fixed end for cantilevers and propped cantilevers). Loads act downward, reactions are positive upward, sagging moment is positive and deflection is positive downward. E is the elastic modulus and I the second moment of area of the section.

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