CalcHandbook

Three-Hinged Parabolic Arch — UDL on Half Span

Arches · Three-Hinged Parabolic Arch · Pinned Bases + Crown Hinge · Statically Determinate · Uniform Load on the Left Half — Pattern Load
1

Geometry and load

Span, rise, and a vertical UDL over the left half (A to the crown)
m
m
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 on the left half only — the pattern-load case that puts an otherwise funicular arch into bending. Sign convention: M positive when the intrados (inside face) is in tension; N positive in compression.
2

Reactions and horizontal thrust

Resultant W = wℓ/2 at x = ℓ/4; H from the crown condition on the unloaded half
RAvertical, loaded side = 3wℓ8 = — = —kN
RCvertical, unloaded side = wℓ8 = — = —kN
Hhorizontal thrust, inward at both supports = wℓ²16f = — = —kN
H = M0(ℓ/2)/f = RC·(ℓ/2)/f, from ΣMB=0 on the unloaded half — exactly half the full-span value wℓ²/(8f).
3

Bending moment (M)

M(x) = M0(x) − H·y(x) — antisymmetric: sagging on the loaded half, hogging on the other
M(x)M0 = simple-beam moment; y = arch height at x
= RA·x − wx²/2 − H·y(x) = (wx/8)(ℓ − 2x)0 ≤ x ≤ ℓ/2 = RC·(ℓ − x) − H·y(x)ℓ/2 ≤ x ≤ ℓ
Mmax+loaded half, x = ℓ/4 = wℓ²64 = —kN·m
Mmax−unloaded half, x = 3ℓ/4 = −wℓ²64 = —kN·m
M = 0 at A, B and C; the diagram is antisymmetric about the crown, with equal extremes ±wℓ²/64 at the quarter points. The full-span case gives M ≡ 0 — moving the same load onto one half is what puts the arch in bending.
Bending Moment Diagram (M) — drawn on the tension side
4

Shear (V) and axial force (N)

Resolved along the arch tangent — continuous, with a kink where the load ends
V(x), N(x)V0 = simple-beam shear: RA − wx on the loaded half, −RC on the other; tanφ = (4f/ℓ)(1−2x/ℓ)
V = V0·cosφ − H·sinφshear, normal to the axis N = H·cosφ + V0·sinφaxial, compression positive
NAat support A, φA = atan(4f/ℓ) = H·cosφA + RA·sinφA = —kN
NCat support C = H·cosφA + RC·sinφA = —kN
Ncrownx = ℓ/2, φ = 0 = H = —kN
VAat support A = RA·cosφA − H·sinφA = —kN
Vcrownx = ℓ/2, φ = 0 — equals V0 = −RC = −wℓ8 = —kN
N stays compressive along the whole arch, largest at A (loaded side). V0 is continuous at the crown (RA − wℓ/2 = −RC), so V and N have no jump — only a slope change where the load ends.
Shear Force Diagram (V) — along the arch axis
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
m
0 ℓ/2 ℓ
y(x)arch height at this section = —m φ(x) = —°
M(x)= M0(x) − H·y(x) = —kN·m
V(x)= V0·cosφ − H·sinφ = —kN
N(x)= H·cosφ + V0·sinφ — compression = —kN