Beam Overhanging One Support — Uniformly Distributed Load

1

Geometry and Load

Main span ℓ (A to B), overhang c (B to C), distributed load w

Main span ℓ

m

Overhang c

m

Distributed load w

kN/m

A: pin support; B: roller support; C: free end. Total length = ℓ + c.

2

R_A, R_B, M_B, M_max+

R_A = w(ℓ²−c²)2ℓ = — = —kN

R_B = w(ℓ+c)²2ℓ = — = —kN

M_Bmoment at B = −wc²2 = —kN·m

M_max+at x₀ = R_A/w = R_A²2w = —kN·m

V_max = max(|R_A|, |V(ℓ⁻)|) = —kN

3

Define E and I for δ_max

Elastic modulus E

MPa

Moment of inertia I

cm⁴

δ_maxnumerical = —mm

δ_Cfree end = —mm

Deflected Shape Diagram

4

Compare σ_max with the allowable stress

Section modulus S

cm³

Allow. bending F_b

MPa

σ_max = M_kritS = — = —MPa

M_krit = max(|M_B|, M_max+)

— σ / F_b

5

V_x, M_x, δ_x at a specific point x

Distance x

m

0 (A) ℓ (B) ℓ+c (C)

V_x = —kN

M_x = —kN·m

δ_x = —mm