Cantilever Beam — Uniformly Distributed Load

1

Geometry and load

Define the span and the UDL (fixed at the left, free at the right)

Span ℓ

m

Distributed load q

kN/m

2

R, M_max

R = V_maxat the fixed end = qℓ = — = —kN

M_maxat the fixed end (negative: tension on top fibre) = qℓ²2 = — = —kN·m

3

Define E and I for δ_max

Modulus of elasticity E

MPa

Moment of inertia I

cm⁴

δ_maxat the free end = qℓ⁴8 EI = — = —mm

Deflected Shape Diagram

4

Compare σ_max with the allowable bending stress

Section modulus S

cm³

Allow. bending F_b

MPa

σ_max = M_maxS = — = —MPa

— σ / F_b

5

V_x, M_x, δ_x values at a given point x

Distance x

m

0 (fixed end) ℓ/2 ℓ (free end)

V_x = q(ℓ − x) = — = —kN

M_xnegative (tension on top fibre) = −q(ℓ − x)²2 = — = —kN·m

δ_x = q24 EI·(6ℓ²x² − 4ℓx³ + x⁴) = —mm