Continuous Beam — Two Equal Spans — Two Symmetric Point Loads

1

Geometry and Load

P at distance a from R₁ in the left span, and from R₃ in the right span (symmetric)

Span ℓ

m

Load P

kN

Distance a

m

b = ℓ − a = —

2

R₁=R₃, R₂, V₁=V₃, V₂=Vmax, M₁, M₂

R₁=R₃= Pb(2ℓb+a²)2ℓ³ = —kN

R₂ = 2V₂= 2(P − R₁) = —kN

V₂=V_max= P − R₁ = —kN

M₁at R₂= −Pab(ℓ+b)2ℓ² = —kN·m

M₂at x = a= R₁ · a = —kN·m

For a = ℓ/2: R₁=5P/16, V₂=11P/16, M₁=3Pℓ/16, M₂=5Pℓ/32

3

σ_max = M_cr / S

Section modulus S

cm³

Allow. bending F_b

MPa

σ_max=—MPa

M_cr = max(|M₁|, M₂)

—σ / F_b

4

V_x and M_x at a given location

Distance x

m

0ℓ2ℓ

V_x=—kN

M_x=—kN·m