Parabolic arches

Application of parabolic arches

Parabolic arches are valued in construction not only for their functionality but also for their aesthetic appearance. Due to their ability to effectively transfer loads, they are chosen where a light and durable structure is important, especially in large open space projects. Thanks to their geometry, parabolic arches can evenly distribute loads and are more durable than circular arches under the same conditions.

Here are some places where they can be found:

• Bridges and viaducts: The characteristic shape of the parabolic arch makes it ideal for spanning rivers, valleys, or other terrain obstacles.
• Churches and sacred buildings: Parabolic arches play an important role in sacred architecture, where they are used to construct domes, vaults, and arcades. Their elegant form not only provides strength but also gives interiors a unique character.
• Roofs and halls: Parabolic arches are used in the construction of large roofs over halls, warehouses, or hangars, allowing for large spans without the need for numerous supports.
• Tunnels and galleries: In underground construction, parabolic arches are used as structural elements in tunnels and galleries.
• Stadiums and sports facilities: Roof structures in stadiums often rely on parabolic arches, which provide lightness to the structure and the ability to cover large areas without internal supports.

Internal forces in parabolic arches

Support reactions for arches are calculated the same way as for beams, frames, or trusses.

The greatest difficulty in calculating internal forces is the curvilinear shape of the rod. However, we can rely on trigonometry, which will allow us to calculate normal and shear forces. Remember that shear and normal forces will change their directions along the curvilinear line of the arch.

Functions of internal forces

To clarify the key elements of the analysis of internal forces in parabolic arches, let’s consider the partially below example:

Fig. 1. Thought cross-section for a parabolic arch

For parabolic arches, unlike circular arches, the coordinate system is assumed at the left or right end.

To determine the vertical coordinate of the point of force application, as well as to calculate \( \tan \alpha \), \( \sin \alpha \) and \( \cos \alpha \), we must first determine the equation of the parabolic arch (the equation of the parabola).

We will show two methods.

Equation of the axis of the parabolic arch

Method I

We know 3 points that belong to the arch - support A, B, and the vertex - we substitute them into the equation of the parabola, solve the system of equations, and determine the parameters of the quadratic function.

The equation of the parabola \(y=a \cdot x^2+b \cdot x+c\)

Coordinates of the points:

\( A(0,0), B(10,0), W(5,3) \)

We substitute into the equation:

Point A:

\[ 0=a \cdot 0^2+b \cdot 0+c \] \[ c=0 \]

Point B:

\[ \begin{aligned} & 0=a \cdot 10^2+b \cdot 10+c \\ & b=-10a \end{aligned} \]

Point W:

\[ 3=a \cdot 5^2+b \cdot 5+c \] \[ \begin{aligned} & 3=a \cdot 5^2+(-10a) \cdot 5 \\ & 3=-25a \\ & a=\frac{-3}{25}=-0.12 \end{aligned} \]

Therefore \( b=-10a=1.2 \)

So the function of the parabola: \(y=-0.12 \cdot x^2+1.2 \cdot x \)

Method II

Using the equation of the axis of the parabolic arch, with the origin of the coordinate system at support A: \( y=\frac{4f}{L^2} \cdot(L-x) \cdot x\), where:

• \(f\) - height of the arch,
• \(L\) - span between supports.

We have \(f=3, L=10\), therefore:

\[ y=\frac{4 \cdot 3}{10^2} \cdot(10-x) \cdot x=0.12 \cdot(10-x) \cdot x=1.2x-0.12x^2 \]

Knowing the equation of the axis of the parabolic arch, we can calculate the vertical coordinate of the force application.

For \(x=8 \text{ m} \quad y_P=-0.12x^2+1.2x=1.92 \text{ m}\)

Normal and shear direction

We make a thought cross-section, drawing a tangent to the arch passing through the point where we made the cross-section (in the above drawing in blue). This tangent defines the normal direction in this thought cross-section. Perpendicular to the normal direction is the shear direction. We need to project the forces acting on this cross-section from the left side onto the normal and shear directions.

Fig. 2. Projection of forces onto normal and shear components

Math recap - the tangent to the arch is inclined to the x-axis at an angle \( \alpha \), whose tangent:

\[ \tan \alpha=\frac{\partial y}{\partial x} \]

In the considered example:

\[ \tan \alpha=\frac{d(-0.12x^2+1.2x)}{dx}=-0.24 \cdot x+1.2 \]

From trigonometry, we have relationships that will allow us to determine the sine and cosine of this angle:

\[ \sin \alpha=\frac{\tan \alpha}{\sqrt{1+\tan^2 \alpha}} \quad, \quad \cos \alpha=\frac{1}{\sqrt{1+\tan^2 \alpha}} \]

Having done all this, we write the functions of internal forces classically.

We invite you to explore the topic and analyze internal forces in parabolic arches.

Good luck! 🛠️🔍