Beam Overhanging One Support — Concentrated Load at End of Overhang

1

Geometry and Load

Main span ℓ (R₁→R₂), overhang a (R₂→free end), P at the overhang tip

Main span ℓ

m

Overhang a

m

Concentrated load P

kN

R₁ is directed downward (holding the beam down). The moment is linear and entirely negative in the span; on the overhang, M = −Px₁ (x₁ from the free end).

2

Calculation details

R₁, R₂, V₂, M_max

R₁ = V₁downward= −Paℓ= —kN

R₂ = V₁+V₂= P(ℓ+a)ℓ= —kN

V₂overhang= P= —kN

M_maxat R₂= −Pa= —kN·m

M_xspan= −Paxℓ M_x₁overhang= −Px₁

3

Deflection calculation

Upward in the span, downward on the overhang — define E and I

Elastic modulus E

MPa

Moment of inertia I

cm⁴

Δ_maxspan, x = ℓ/√3, upward= Paℓ²9√3 EI= —mm

Δ_maxoverhang tip, downward= Pa²(ℓ+a)3EI= —mm

Deflected Shape Diagram

4

Stress check

σ_max = |M_max| / S

Section modulus S

cm³

Allow. bending F_b

MPa

σ_max=—MPa

—σ / F_b

5

Point analysis

V_x, M_x, δ_x — x measured from R₁

Distance x

m

0 (R₁)ℓ (R₂)ℓ+a

V_x=—kN

M_x=—kN·m

δ_x+ downward=—mm