Fixed-Fixed 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 fully fixed — three redundants (H, R_C, M_C), recovered from compatibility of the released end C 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, thrust and end moments

Three compatibility equations at C — closed form in k = a/ℓ

R_Avertical — same as a fixed-ended beam = P·(1 − k)²·(1 + 2k) = — = —kN

R_Cvertical — k = a/ℓ = P·k²·(3 − 2k) = — = —kN

Hhorizontal thrust, inward at both ends = 15Pℓ4f·k²(1 − k)² = — = —kN

M_Aend moment at A (sagging +) = Pℓ2·k(1 − k)²(5k − 2) = —kN·m

M_Cend moment at C = Pℓ2·k²(1 − k)(3 − 5k) = —kN·m

Released at C, the three conditions (u_C = v_C = θ_C = 0) with M = R_C(ℓ−x) − H·y + M_C − P⟨a−x⟩ integrate to these polynomials in k. The vertical reactions are those of a fixed-ended beam; H peaks at 15Pℓ/(64f) for a load at the crown — 20 % below the three-hinged Pℓ/4f, 20 % above the two-hinged value. The end moment changes sign at k = 0.4 (M_A) and k = 0.6 (M_C). With constant EI instead of I₀secφ: —.

3

Bending moment (M)

M(x) = R_C(ℓ − x) − H·y(x) + M_C − P⟨a − x⟩ — nonzero at both ends and at the crown

M(x)forces to the right of the section; y = arch height at x

= M_A + R_A·x − H·y(x)0 ≤ x ≤ a = M_C + 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

The end moments and the crown moment (—) are all nonzero; the diagram has two or three zero crossings. The positive peak sits under the load; the negative extreme is usually the end moment nearest the load or a point on the unloaded side — 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). V₀ now uses the fixed-beam reactions, so V_crown differs from the hinged arches.

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