CalcHandbook

Three-Hinged Parabolic Arch — Concentrated Load

Arches · Three-Hinged Parabolic Arch · Pinned Bases + Crown Hinge · Statically Determinate · Vertical Point Load at Any x
1

Geometry and load

Span, rise, and a vertical point load at distance a from A
m
m
kN
m
0 ℓ/2 ℓ
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). 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
RAvertical, at the near support = P·bℓ = — = —kN
RCvertical, 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 ΣMB=0. It peaks at H = Pℓ/(4f) when the load sits at the crown.
3

Bending moment (M)

M(x) = M0(x) − H·y(x) — positive under the load, negative on the unloaded half
M(x)M0 = simple-beam moment; y = arch height at x
= RA·x − H·y(x)0 ≤ x ≤ a = RC·(ℓ − x) − H·y(x)a ≤ x ≤ ℓ
Mmax+under the load, x = a = P·a·b·|ℓ − 2a|ℓ² = —kN·m
Mmax−quarter point of the unloaded half, x = 3ℓ/4 = −P·c8 = —kN·m
M = 0 at A, B and C. The negative extreme always sits at the quarter point of the unloaded half, whatever a is — the stationary point of M0 − H·y there is independent of the load position.
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 (RA left of the load, RA − 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
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
ΔVjump in shear at the load, x = a = P·cosφ(a) = —kN
Vcrownx = 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).
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