Simple Beam — Partial Uniformly Distributed Load

1

Geometry and Load

Define the span, load, and the load's position

Span ℓ

m

Distributed load q

kN/m

Left offset a

m

Loaded length b

m

c = ℓ − a − b = 2 m unloaded length, right

2

R₁, R₂, M_max

R₁left support = qb(2c + b)2ℓ = — = —kN

R₂right support = qb(2a + b)2ℓ = — = —kN

M_maxx* = a + R₁/q = R₁(a + R₁2q) = — = —kN·m

3

Define E and I for δ_max

Elastic modulus E

MPa

Moment of inertia I

cm⁴

δ_maxx = x* = EI · y(x) = ∫∫ M(x) dx² + boundary conditions ⇒ —mm

With a partial UDL, M(x) is piecewise; instead of an AWC closed form, δ is obtained by (numerical) double integration.

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 a a+b ℓ

x is in the loaded region (a ≤ x ≤ a+b)

V_xa ≤ x ≤ a+b = R₁ − q(x − a) = — = —kN

M_xa ≤ x ≤ a+b = R₁x − q(x − a)²/2 = — = —kN·m

δ_x = y(x) at this x ⇒ —mm