Three-Hinged Circular Arch — Concentrated Load

1

Geometry and load

Span, rise (→ radius R, half-angle θ), and a vertical point load at distance a from A

Span ℓ

m

Rise f

m

Point load P

kN

Load position a (from A)

m

0 ℓ/2 ℓ

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

Model: A and C are pinned; a third hinge at the crown (B) makes the system statically determinate. Circular axis: y(x) = √(R² − (x − ℓ/2)²) − (R − f). Sign convention: M positive when the intrados (inside face) is in tension; N positive in compression.

2

Reactions and horizontal thrust

Vertical reactions as for a simple beam; H from the crown condition

R_Avertical, at the near support = P·bℓ = — = —kN

R_Cvertical, at the far support (b = ℓ − a) = P·aℓ = — = —kN

Hhorizontal thrust, inward at both supports — c = min(a, b) = P·c2f = — = —kN

H = M₀(ℓ/2)/f — the simple-beam moment at the crown divided by the rise, from ΣM_B=0. Reactions and thrust depend only on ℓ, f, P and a — identical to the parabolic arch; the axis shape only changes M, V, N along the arch.

3

Bending moment (M)

M(x) = M₀(x) − H·y(x) with the circular y(x) — extremes located numerically along the axis

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

= R_A·x − H·y(x)0 ≤ x ≤ a = R_C·(ℓ − x) − H·y(x)a ≤ x ≤ ℓ

M_max⁺at x = … = max M(x) = —kN·m

M_max⁻at x = … = min M(x) = —kN·m

M = 0 at A, B and C. Unlike the parabolic arch, the circular axis is not the funicular of any uniform load, so the extremes have no simple closed form: M(x) is evaluated along the axis (1000 stations) and the extremes are read off. The positive peak usually sits under the load; the negative one on the unloaded half, but not at its quarter point.

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

4

Shear (V) and axial force (N)

Resolved along the arch tangent — both jump at the load

V(x), N(x)V₀ = simple-beam shear (R_A left of the load, R_A − P right of it); 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

ΔVjump in shear at the load, x = a = P·cosφ(a) = —kN

V_crownx = ℓ/2 — equals V₀ there, since φ = 0 = —kN

N stays compressive along the whole arch; its largest value is at the support on the loaded side (φ_A = θ). The axial jump at the load is P·sinφ(a).

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