StructuralMind
Start free trial

Modelling curved members

Updated:

On this page

Results.

ExampleReferenceRecordsLargest deviation
Three-hinged parabolic arch, full-span loadClosed form3< 0.001%
Two-hinged parabolic arch, full-span loadClosed form55.51%
Two-hinged parabolic arch, local moment at mid-segmentClosed form10%

Problem. MindFEM has no curved frame element, and an arch is modelled with straight frame elements whose nodes lie on its axis. The effect of the number of segments on the thrust of a three-hinged and a two-hinged parabolic arch is examined. The local moment that forms at the middle of each straight segment is also given.

Reference document. For the three-hinged arch the thrust follows from statics. For the two-hinged arch the reference is computed with Castigliano's theorem from the bending and axial strain energy along the true parabolic axis.

For the three-hinged arch the horizontal thrust under a full-span load is given by this relation.

H=wL28fH = \dfrac{wL^2}{8f}

For the two-hinged arch the thrust is found by setting the derivative of the strain energy with respect to the thrust to zero. M0M_0 and V0V_0 are the moment and shear of a simple beam of the same span, and θ\theta is the angle of the axis to the horizontal.

H=∫M0 yEI ds−∫V0sin⁡θcos⁡θEA ds∫y2EI ds+∫cos⁡2θEA dsH = \dfrac{\displaystyle\int \frac{M_0\,y}{EI}\,ds - \int \frac{V_0 \sin\theta\cos\theta}{EA}\,ds}{\displaystyle\int \frac{y^2}{EI}\,ds + \int \frac{\cos^2\theta}{EA}\,ds}

The moment at the middle of a straight segment exceeds the mean of its two end moments by this amount. Here aa is the horizontal spacing of the nodes.

M=w a28M = \dfrac{w\,a^2}{8}

Numerical model. The arch is a parabola with a 24 m span and a 6 m rise, its nodes lie on the parabola and the load is 20 kN/m, uniform over the horizontal projection. The in-plane rotation is free at the supports, and the three-hinged arch has a further hinge at the crown node.

Three-hinged parabolic arch, full-span load

Closed form. Statics

H=wL2/(8f)H = wL^2/(8f)
QuantityConditionReferenceComputedDeviation
Horizontal thrust HHL = 24 · f = 6 · segments = 42402400%
Horizontal thrust HHL = 24 · f = 6 · segments = 162402400%
Horizontal thrust HHL = 24 · f = 6 · segments = 64240240< 0.001%

Two-hinged parabolic arch, full-span load

Closed form. Castigliano's theorem, bending and axial energy along the true curved axis

QuantityConditionReferenceComputedDeviation
Horizontal thrust HHL = 24 · f = 6 · segments = 4238.994252.1695.51%
Horizontal thrust HHL = 24 · f = 6 · segments = 8238.994242.1911.34%
Horizontal thrust HHL = 24 · f = 6 · segments = 16238.994239.7870.332%
Horizontal thrust HHL = 24 · f = 6 · segments = 32238.994239.1920.0828%
Horizontal thrust HHL = 24 · f = 6 · segments = 64238.994239.0440.0207%

Note. A coarse mesh gives a high thrust, because the chords lie below the true axis and the arch behaves as if it were flatter. The error falls with the square of the number of segments, and sixteen segments are enough for an error below one per cent. The value wL2/8fwL^2/8f = 240, which neglects axial shortening, is 0.4% above the reference. The tolerances are a regression band.

Two-hinged parabolic arch, local moment at mid-segment

Parameters. L = 24 · f = 6 · segments = 16

Closed form. Bending of a straight segment between its nodes

M=wa2/8M = wa^2/8
QuantityReferenceComputedDeviation
Local moment at mid-segment5.6255.6250%

Note. Under a full-span load the true parabolic arch works in pure compression and its moment is zero everywhere. This moment comes from representing the curved axis with straight segments, is the same in every segment and falls to a quarter when the number of segments is doubled.