Simple Beam — Partial UDL at One End

1

Geometry and Load

Define the span, load, and the loaded length

Span ℓ

m

Distributed load q

kN/m

Loaded length a

m

c = ℓ − a = 4 m unloaded length (right side)

2

R₁, R₂, M_max

R₁left support = qa2ℓ(2ℓ − a) = — = —kN

R₂right support = qa²2ℓ = — = —kN

M_maxx* = R₁/q = R₁²2q = — = —kN·m

3

Define E and I for δ_max

Elastic modulus E

MPa

Moment of inertia I

cm⁴

δ_maxx = x* = maximum of the δ_x curve ⇒ —mm

AWC gives a closed form for δ_x (separate for each region). δ_max is found numerically from the peak of the δ_x curve.

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 ℓ

x is in the loaded region (x < a)

V_xx < a = R₁ − qx = — = —kN

M_xx < a = R₁x − qx²/2 = — = —kN·m

δ_xx < a = qx24 EIℓ·[a²(2ℓ−a)² − 2ax²(2ℓ−a) + ℓx³] = —mm