Two-Hinged Parabolic Arch

1

Geometry and load

Span, rise and full-span vertical UDL — section (A, I) only for the rib-shortening correction

Span ℓ

m

Rise f

m

UDL w (horizontal projection)

kN/m

Section properties for the rib-shortening correction (A, I)

Model: A and C are pinned, no crown hinge — one redundant (H), released and recovered from compatibility (zero horizontal displacement at C). Arch axis: y(x) = 4f·(x/ℓ)·(1−x/ℓ). w acts per unit horizontal length.

2

Reactions and horizontal thrust

Rv from equilibrium; H from compatibility (strain energy) — closed form

R_vat each support, vertical — same as a simple beam = wℓ2 = — = —kN

Hredundant — ∂U/∂H = 0 with M = M₀ − H·y = ∫M₀·y·ds/EI∫y²·ds/EI = wℓ²8f = — = —kN

H_rswith rib shortening — ρ = (I/A)·∫cosφ dx ÷ ∫y²·secφ dx = H1 + ρ = — = —kN

For this load M₀(x) = (wℓ²/8f)·y(x) exactly, so the compatibility ratio is wℓ²/8f whatever the EI distribution — the same thrust as the three-hinged arch. Only axial strain (rib shortening) lowers it, by the factor 1/(1+ρ), typically well under 1 %.

3

Bending moment and shear

Zero at every section with H = wℓ²/8f; rib shortening leaves a small residual

M(x) · V(x)the arch axis coincides with this load's funicular (thrust line) ≡ 0 (at every x, exactly)

M_rs(x)residual from rib shortening — (H − H_rs)·y(x), largest at the crown = —kN·m

The arch axis matches the funicular curve of the UDL — the arch carries this load in pure axial compression. The residual above is the price of axial strain: small, sagging, and usually neglected.

Bending Moment and Shear Diagrams (M ≡ 0, V ≡ 0 for H = wℓ²/8f) — along the arch axis

4

Axial force (N)

The only internal force — pure compression, minimum at the crown, maximum at the supports

N(x)φ = tangent angle, tanφ = (4f/ℓ)(1−2x/ℓ)

= H / cos φ(x)

N_crownx=ℓ/2, φ=0 — minimum, pure horizontal thrust = H = —kN

N_supportx=0 or ℓ — maximum = √(H² + R_v²) = —kN

The sign never changes (compression throughout); the horizontal component equals H at every section (N·cosφ ≡ H).

Axial Force Diagram (N) — along the arch axis

5

At a given section

Read the values at any x — marked in amber on the diagrams

Distance x (from A, horizontal)

m

0 ℓ/2 ℓ

y(x)arch height at this section = —m

N(x)= H / cos φ(x) — compression = —kN

M(x) · V(x)here too — as at every x = 0 · 0