Fixed-Fixed Parabolic Arch

1

Geometry and load

Span, rise and full-span vertical UDL — section (A, I) governs the rib-shortening moments

Span ℓ

m

Rise f

m

UDL w (horizontal projection)

kN/m

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

Area A

cm²

Moment of inertia I

cm⁴

Only the ratio I/A (radius of gyration²) enters — E cancels. With bending alone the end moments are exactly zero; A and I decide how much the fixed ends resist the axial shortening.

Model: A and C are fully fixed (no rotation, no translation) — three redundants; by symmetry two remain at the crown: the thrust H and the crown moment M_B, found from zero rotation and zero horizontal displacement of the crown relative to the fixed support. Arch axis: y(x) = 4f·(x/ℓ)·(1−x/ℓ). w acts per unit horizontal length.

2

Reactions, thrust and end moments

Rv from equilibrium; H and M_A from compatibility — closed form for bending, corrected for rib shortening

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

Hbending only — the funicular state satisfies both compatibility conditions = wℓ²8f = — = —kN

M_A = M_Cend moments, bending only = 0

H_rswith rib shortening — 2×2 compatibility system (θ_B = 0, u_B = 0) with the axial term ∫N·cosφ·ds/EA = —kN M_A,rs = —kN·m

With H = wℓ²/8f the arch is curvature-free (M ≡ 0), so the fixed ends have nothing to restrain — the same thrust as the two- and three-hinged arches, and zero end moments, for any ℓ, f, w. Axial shortening breaks this: the crown drops, the fixed ends resist, H falls slightly and end moments appear — scaled by I/A and f.

3

Bending moment and shear

Zero at every section for bending; rib shortening leaves hogging at the crown and sagging at the fixed ends

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 — M_B,rs − H_rs·(f − y) + w(x − ℓ/2)²/2

M_B,rscrown = —kN·m M_A,rsfixed ends = —kN·m

The arch axis matches the funicular curve of the UDL — the arch carries this load in pure axial compression whatever its rise or span. The residual above is the price of axial strain in a fixed arch: a few percent of wℓ²/8, larger than in the two-hinged case because the ends now resist the crown drop.

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