Two-Hinged Circular Arch

1

Geometry and load

Span, rise (→ radius R, half-angle θ) and full-span vertical UDL

Span ℓ

m

Rise f

m

UDL w (horizontal projection)

kN/m

Rradius of the circular axis = ℓ²8f + f2 = —m θhalf-angle, sinθ = ℓ/2R = —°

Model: A and C are pinned, no crown hinge — one redundant (H), recovered from compatibility with constant EI (axial strain neglected; E cancels). Circular axis: y(x) = √(R² − (x − ℓ/2)²) − (R − f). w acts per unit horizontal length. Sign convention: M positive when the intrados is in tension; N positive in compression.

2

Reactions and horizontal thrust

Rv from symmetry; H from compatibility (strain energy), integrated numerically along the arc

R_Avertical — by symmetry = wℓ2 = — = —kN

R_Cvertical — by symmetry = wℓ2 = — = —kN

H∂U/∂H = 0 with M = M₀ − H·y — Simpson along the arc, ds = secφ·dx = ∫M₀·y·ds∫y²·ds = — = —kN

The circular axis is not the funicular of a uniform load, so H no longer equals wℓ²/(8f): here H = — of that value, and a modest moment remains along the arch. Axial strain (rib shortening) would lower H by a further fraction of a percent.

3

Bending moment (M)

M(x) = M₀(x) − H·y(x) — small, symmetric, sagging at the crown, hogging near the haunches

M(x)M₀ = simple-beam moment; y(x) = √(R² − (x − ℓ/2)²) − (R − f)

= (wℓ/2)·x − wx²/2 − H·y(x)0 ≤ x ≤ ℓ

M_max⁺at x = … = max M(x) = —kN·m

M_max⁻at x = … = min M(x) = —kN·m

M = 0 only at A and C; the crown carries —. Compare the parabolic two-hinged arch under the same load: M ≡ 0. The residual here is the "shape error" of the circle against the parabolic thrust line — evaluated at 1000 stations to locate the extremes.

Bending Moment Diagram (M) — drawn on the tension side

4

Shear (V) and axial force (N)

Resolved along the arch tangent — continuous; V small, N close to H/cosφ

V(x), N(x)V₀ = wℓ/2 − wx (simple-beam shear); tanφ = −(x − ℓ/2)/√(R² − (x − ℓ/2)²)

V = V₀·cosφ − H·sinφshear, normal to the axis N = H·cosφ + V₀·sinφaxial, compression positive

N_Aat support A, φ_A = θ = H·cosφ_A + R_A·sinφ_A = —kN

N_Cat support C = H·cosφ_A + R_C·sinφ_A = —kN

N_crownx = ℓ/2, φ = 0 = H = —kN

V_Aat support A (mirror at C) = R_A·cosθ − H·sinθ = —kN

V_maxlargest |V| along the arch = —kN

N stays compressive along the whole arch, largest at the supports (φ = θ) and minimum H at the crown; V is a small residual of the same origin as M.

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

Distance x (from A, horizontal)

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