Continuous Beam — Two Equal Spans — Concentrated Load at Any Point

1

Geometry and load

a + b = ℓ; P at distance a in the left span

Span ℓ

m

Load P

kN

Distance a

m

b = ℓ − a = —

2

R₁, R₂, R₃, V₁, V₂, V₃, M_max, M₁

R₁ = V₁= Pb4ℓ³(4ℓ²−a(ℓ+a)) = —kN

R₂= Pa2ℓ³(2ℓ²+b(ℓ+a)) = —kN

R₃ = V₃= −Pab4ℓ³(ℓ+a) = —kN

V₂just left of R₂= Pa4ℓ³(4ℓ²+b(ℓ+a)) = —kN

M_maxat x=a= Pab4ℓ³(4ℓ²−a(ℓ+a)) = —kN·m

M₁at R₂= Pab4ℓ²(ℓ+a) = —kN·m

3

σ_max = M_krit / S

Section modulus S

cm³

Allow. bending F_b

MPa

σ_max=—MPa

M_krit = max(M_max, M₁)

—σ / F_b

4

V_x and M_x at the given location

Distance x

m

0ℓ2ℓ

V_x=—kN

M_x=—kN·m