Two-Hinged Parabolic Arch — Concentrated Load

1

Geometry and load

Span, rise, 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

Model: A and C are pinned, no crown hinge — one redundant (H), recovered from compatibility with the classical assumption I = I₀·secφ (so ds/I = dx/I₀; axial strain neglected). Arch axis: y(x) = 4f·(x/ℓ)·(1−x/ℓ). 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 compatibility (strain energy) — closed form

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

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

H∂U/∂H = 0 → H = ∫M₀·y·dx ÷ ∫y²·dx — with k = a/ℓ = 5Pℓ8f·(k − 2k³ + k⁴) = — = —kN

∫y²dx = 8f²ℓ/15 and ∫M₀·y·dx = (Pf ℓ²/3)·(k − 2k³ + k⁴) give the closed form. Symmetric in k ↔ 1−k; maximum 25Pℓ/(128f) at the crown — 22 % less than the three-hinged value Pℓ/4f. With constant EI instead of I₀secφ the thrust changes by well under 1 % (—).

3

Bending moment (M)

M(x) = M₀(x) − H·y(x) — no longer zero at the crown; extremes located along the axis

M(x)M₀ = simple-beam moment; y = arch height at x

= 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 only at A and C — the crown carries moment (—). The positive peak sits under the load; the negative extreme lies on the unloaded side, its position found by evaluating M(x) at 1000 stations.

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 = (4f/ℓ)(1−2x/ℓ)

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

N_Aat support A, φ_A = atan(4f/ℓ) = 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. The axial jump at the load is P·sinφ(a). V_crown = V₀(ℓ/2) is the same as for the three-hinged arch; N_crown = H is lower.

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