Arch bridges operate according to the following (I am assuming arch dams are the same):

Tensional force in arch bridges are virtually negligible. The natural curve of the arch and its ability to dissipate the force outward greatly reduces the effects of tension on the underside of the arch.

The greater the degree of curvature (the larger the semicircle of the arch), the greater the effects of tension on the underside of the bridge. (Source)

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  1. WHY does a greater degree of curvature results in more tension? A diagram would be useful if possible.

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2. How does the changing the radius of curvature affect the tension and compression in the dam, and ultimately the strength of it. enter image description here


1 Answer 1


The presence of tension in an arch is not really dependent on the curvature but on how well the arch matches a catenary shape.

Robert Hooke famously stated:

As hangs a flexible cable so, inverted, stand the touching pieces of an arch.

So a catenary arch will have only compression stresses (since a flexible cable can only have tension stresses).

Catenary Arch

Notice how the line of action of the compression in the arch runs through the centre of the arch. If the arch shape is not a catenary, the line of action will still follow a catenary shape but will not follow the centre line of the arch:


We can see how this can cause tension by examining the distribution of stress as a result of the position of the line of action (green is tension):

Stress Distribution

In general the rule for arches is the cantenary line must be within the middle $\frac{1}{3}$ of the arch to avoid tension.

Form finding can be used to find the catenary shape for a particular loading. This can be done computationally, or even by the use of hanging models.

Arch dams behave in the same way. However, since the main loading on the dam (from water) is a pressure normal to the dam surface the equivalent of the catenary shape is a circle. In this case the compressive stress is calculated in the same way as the hoop stress of a pressure vessel (where I have assumed r >> t):

$$\sigma = \frac{Pr}{t}$$

where $\sigma$ is the stress P is the applied pressure, r is the radius and t is the thickness. Since the pressure P increases with the depth of water dams sometimes decrease radius with depth (variable-radius dams) or change the thickness (constant-radius dams).

Naturally the design of a dam has many other complicating factors which affect the design.


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