# How does an arch transfer vertical load? [duplicate]

I am hoping that someone can explain to me how, physically, an arch transfers a purely vertical load (Fy) into a diagonal force to the abutments of the arch.

I understand that the y and x components of the reaction force support the load, but what creates the horizontal forces Rx1 and Rx2? I assume that they have to exist, as without an x component surely Fy could not otherwise transfer diagonally.

Many thanks

• Did you not already ask that here: engineering.stackexchange.com/questions/45139/…? Jul 22 at 21:28
• Start by thinking about the keystone at the top of of the arch. (The linked thread is irrelevant, because the triangle is upside down. The keystone is wider at the top, not at the bottom). It acts like a wedge. If you try to push it down, it has to push the adjacent stones sideways to make more room. The same thing happens to all the other stone segments around the arch. Jul 22 at 22:00
• I have got my head around the triangle truss type forces but in an arch I don't see how the arch does the same as there is no contraint as such. Jul 22 at 23:11
• With regards to the keystone - again this does make sense however arches can and do exist without keystones, or as a single structure, and still impart some element of a horizontal force presumably in order to transfer a downwards pressure into a diagonal pressure down and out. Jul 22 at 23:12
• @NickCory the principle is the same with the truss. The main difference is that you have additionally internal moments not only forces. The bending moment creates the curved shape that you see on the example of the no joints. Jul 22 at 23:57

The reason why forces are transmitted in arches is similar to the joint structure in the other question. Additionally there are the internal bending moments that also change the undeformed shape of the element (be it beam or arch).

# arches build with stone

Although not in your original post, I will focus on the forces in stone arches. In order to understand it, from another perspective have a look at this image, from the "The Geometrical Design of Masonry Arches".

Figure: Left, line of thrust in a semi-arch. (source The Geometrical Design of Masonry Arches

In the image above the arch is designed as part of individually cut stones which each is cut in a specific way, and are placed in a specific order. The general shape of all the stones is similar to a keystone

Figure 2: keystone basic shape

The forces acting on the voussoir are like in the image below:

Figure 3: Forces acting on a voussoir

Where

• $$T_1$$ is upward, because the stone on the left develops friction which stops the stone to move downwards (due to gravity).
• $$T_2$$ is the downward force applied from the stone on the right (which due to gravity is downwards).

It is noteworthy, that due to the angled shape of the stone and the added weight of the stone:

• $$N_1$$ is greater than $$N_2$$
• the friction forces which develop are also greater (because the normal forces are greater).

the above differences in forces results in a bending moment, which in turn results in an (sort of) triangular distribution of the normal Forces ($$N_1, N_2$$).

It is interesting to note that the stones become larger. There is a very good reason for that. The reason is that the added weight acts to stabilise the column (the resulting force vector is more vertical). See the following figure:

Figure: use of gargoyles to increase stability of columns from lateral weight (source: how-gargoyles-and-pinnacles-saved-gothic-architecture)

The arch works by being deformed and compressed under the vertical loads.

When it deforms it wants to settle down and its geometry changes trying to trace a shorter path. But that shorter path means it strains the material creating compression stress in the material of arch, being it pieces of rock or concrete or masonry or even bags of dirt.

That is the basics of the story. But the shape of the arch should be in a way that the moment is kept to a minimum or even zero.

For example, the optimal shape for an arch with uniform loading is a parabola.

Below is the free body diagram of a 3 hinged arch. The forces are passed through the internal stresses. See this article for examples.