Simple Beam — Partial UDL at Each End

1

Geometry and Loads

Define the span and the loads at each end

Span ℓ

m

Left load w₁

kN/m

Left length a

m

Right load w₂

kN/m

Right length c

m

b = ℓ − a − c = 5 m unloaded middle region

2

Show calculation details

R₁, R₂, M_max

R₁left support = w₁a(2ℓ − a) + w₂c²2ℓ = — = —kN

R₂right support = w₂c(2ℓ − c) + w₁a²2ℓ = — = —kN

M_maxx* = R₁/w₁ = R₁²2w₁ = — = —kN·m

3

Deflection calculation

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 loading at both ends, M(x) is piecewise in three parts; AWC gives no closed form, so numerical double integration is used.

Deflected Shape Diagram

4

Bending stress check

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

Point analysis

V_x, M_x, δ_x at a given x

Distance x

m

0 a a+b ℓ

x is in the left loaded region (x < a)

V_xx < a = R₁ − w₁x = — = —kN

M_xx < a = R₁x − w₁x²/2 = — = —kN·m

δ_x = y(x) at this x (numerical) ⇒ —mm