Three-Hinged Parabolic Arch

1

Geometry and load

Span, rise and full-span vertical UDL

Span ℓ

m

Rise f

m

UDL w (horizontal projection)

kN/m

Model: A and C are pinned; a third hinge at the crown (B) makes the system statically determinate. Arch axis: y(x) = 4f·(x/ℓ)·(1−x/ℓ). w acts per unit horizontal length.

2

Reactions and horizontal thrust

Direct from equilibrium + crown condition (closed form)

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

Hhorizontal thrust, inward at both supports = wℓ²8f = — = —kN

H follows from zero moment at the crown hinge (ΣM_B=0) — no stiffness terms involved. H grows sharply as f decreases (∝ 1/f).

3

Bending moment and shear

Both are zero at every section, for this load case

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

The arch axis matches the funicular curve of the UDL — the arch carries this load in pure axial compression.

Bending Moment and Shear Diagrams (M ≡ 0, V ≡ 0) — 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