Simple Beam — Uniformly Distributed Load

1

Geometry and Load

Define the span and uniformly distributed load

Span ℓ

m

Distributed load q

kN/m

2

R, M_max

R = V = qℓ2 = — = —kN

M_maxmidspan = qℓ²8 = — = —kN·m

3

Define E and I for δ_max

Elastic modulus E

MPa

Moment of inertia I

cm⁴

δ_maxmidspan = 5qℓ⁴384 EI = — = —mm

Deflected Shape Diagram

4

Compare σ_max with the allowable 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 at a given x

Distance x

m

0 ℓ/2 ℓ

V_x = q(ℓ2 − x) = — = —kN

M_x = qx2(ℓ − x) = — = —kN·m

δ_x = qx24 EI(ℓ³ − 2ℓx² + x³) = —mm