Circular Arches

In this course, you will learn:

• how to determine support reactions in arches,
• how to calculate internal forces in arches,
• how to draw diagrams of normal, shear, and bending moment forces

Support reactions for arches are calculated the same way as for beams, frames, or trusses; there is no difference here.

The greatest difficulty in calculating internal forces is the curvilinear shape of the member. However, trigonometry comes to our aid, allowing us to calculate normal and shear forces. Remember that shear and normal forces will change their position along the curvilinear line of the arch.

Functions of internal forces

Let's see what the function of internal forces for a circular arch will look like as below.

We will skip the stage of calculating support reactions. For this example, we have three characteristic segments, looking from the left side:

1. from the support to the center of the arch,
2. from the center of the arch to the concentrated force,
3. from the concentrated force to the support.

Let's analyze the thought section in the third segment from the right side view.



In the above drawing, the calculated support reaction at the support on the right side is marked. A thought section is placed "somewhere" between the support and the concentrated force, with the coordinate system x,y marked. The variable with circular arches is the angle between the horizontal and the thought section, marked in the drawing as alpha. The positive directions of internal forces N,T,M are also marked in the right side view.

The point where we placed the section is projected onto the marked x and y axes. The distance "y" can be transferred from the vertical axis to the section, creating a key to understanding the problem being considered right triangle with legs x and y and the hypotenuse... indeed, the hypotenuse is the radius of our arch.

Before we deal with this triangle, let's add that from the center of the coordinate system to the support on the right side, we have a distance equal to the radius, which is R=4m. If from the center to the projection of the section on the horizontal axis we have a distance "x", then the remaining distance - from the projection of the section on the horizontal axis to the support will be "R-x", or "4-x", as marked in the drawing - this distance will be needed to determine the bending moment.

Triangle x-y-R which has variable lengths of sides x and y depending on angle alpha. In problems with circular arches, we must relate the lengths of sides x and y of this triangle to angle alpha. From trigonometric functions:

\( \frac{x}{R}=cos\alpha \Rightarrow x=R\cdot cos\alpha \)

\( \frac{y}{R}=sin\alpha \Rightarrow y=R\cdot sin\alpha \)

The only force we see from the right side is the reaction of 2.248kN. We need to project this force onto the normal and shear directions to the marked thought section of the arch. When we connect the center of the arch with the point of the thought section, this line represents the shear direction for the arch in this section. This line is inclined at angle alpha to the horizontal. Perpendicular to this line (and thus in the tangential direction to the arch) is the normal force (axial) to the considered thought section.



When it comes to projecting forces onto components, a good idea is to draw a "web" - make a drawing where we mark the vertical direction, horizontal direction, shear direction, and perpendicular to it the normal direction. The angle alpha between the horizontal and the shear direction. In such a drawing, every second angle is angle alpha.

We use the web with marked angles alpha - we apply the force to the web drawing, draw the two nearest components, and see where angle alpha is located.

The component at which angle alpha is marked is multiplied by the cosine of that angle, and the other by the sine of that angle.



Considering this triangle and projecting horizontal and vertical forces onto normal and shear components to the thought section of the arch is a key skill for circular arches. Now we write down the functions of internal forces. The variable is angle alpha; the question remains in what range does angle alpha change for the considered thought section? The angle is measured from the x-axis to the thought section, the maximum angle we will reach is marked in the drawing below.



In the topic, the horizontal dimensions of the position of the concentrated force were given. After constructing the triangle, we know that its hypotenuse is the radius R=4m. Knowing the sides - the leg 2m and the hypotenuse 4m, we can write that \(cos \alpha = 2/4 = 0,5 \). Therefore, \( \alpha = acos(0,5)= 60^o \).

Thus:

\begin{aligned} &0^o < \alpha < 60^o \\ &N(\alpha)=-2,248\cdot cos\alpha\\ &T(\alpha)=-2,248\cdot sin\alpha\\ &M(\alpha)=2,248\cdot (4-x)=2,248\cdot (4-4\cdot cos\alpha)\\ \end{aligned}

Now, depending on how often we want to have the result, we check the values of internal forces, e.g., every 10, 15, or 30 degrees. The results can be compiled in a table, and at the end, draw a diagram of internal forces.

The diagram of internal forces can be drawn on the arch or on a horizontal line.

The complete solution to this example can be found HERE

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

Good luck! 🛠️🔍