Many things in civil engineering are expressed in terms of stress, rather than forces (loads).

The stress (pressure) is found by dividing the force (load) by area.

So,

Stress = Force / Area

So it has units of Newtons per milimeter square, N/mm^{2} (Mega Pascal – MPa), or pounds per inch square, lbs/inch^{2} (psi) for instance. Other conversions are also used depending on industry.

For example, the compressive or tensile stresses on the beam that you saw before in Figures 1a and 1b would be the magnitude of the applied force, load, divided by the area of the cross section of the beam. The shear stress in Figure 1c, is calculated by the shear force acting on that cross section divided by the area of the entire horizontal fiber face, so shear stress is also force / area.

To illustrate again with example:

Let’s assume the compressive force was 10 kN (10000N), and
the cross section area of the beam was 250cm^{2} (25000mm^{2}).

Based on this, the compressive stress acting on the beam is: 10000/25000 = 0.4 MPa

We show axial stress with the symbol “σ” (sigma).

So, here,

σ = 0.4 MPa

What does this mean? It means, if our beam material’s strength is more than 0.4 MPa, it will not fail and withstand this stress.

To visualize, concrete that we use to build our buildings and infrastructure usually has strength anywhere from 20 Mpa to around 40 Mpa, or in the US from 3000 psi to 5000 psi, altough obviously there are exceptions for specific applications. Note that when we talk about strength of concrete, we almost always talk about compressive strength, and ignore its tensile strength, because concrete is strong in compression but very weak in tension, therefore it is ignored in our structural calculations, to be on the conservative side.

Structural steel has much higher strength. Its values range between 230 to 350 MPa for most applications (and in the US about 30000 – 50000 psi), and in comparison to concrete which is only strong in compression, steel is strong in both tension and compression.

This stress example you just saw was in the direction of the long axis of the beam, therefore it is called as axial stress.

If we applied tensile force instead of compressive force, we would have calculated the stress in exactly the same way, except we would call it as “axial tensile stress” instead of “axial compressive stress”.

Now let’s see an example of shear stress. See figure below, which is the same as Figure 1c you saw before. Let’s assume this time the shear stress is 100kN, and the surface area it acts on, is 2500 cm2.

So we can calculate the shear stress as:

100000 / 250000 = 0.4 Mpa

We show shear stress with the symbol “τ” (tao). So, here,

τ = 0.4 MPa

Again we found the same result, because although we increased the force 10 times here, from 10 to 100 kN, the area that the force acts on, also happened to increase by 10 times. So in other words, if you apply higher load, it doesn’t necessarily mean higher stress. You must consider the area too. And what is exactly the area here? The area in this particular case here is the surface that these two beams touch and try to slide over each other.

So many things in civil engineering are expressed in terms of stress. Just a few examples are below, to give an idea… The list goes on and on:

- Strength of concrete
- Strength of steel
- Bearing capacity of soil
- Lateral earth pressure
- Hydrostatic pressure of water and liquids

Note that some of these can only be compressive, tensile or shear, some of them can be both compressive and tension or some of them can be for all three. But regardless of that, all are stress units, which is Force / Area, be it σ or τ.

As far as pressure and stress, you might have noticed that we used them as if having equal meaning. Yes they have both the same characteristic, except one important difference: Pressure is the external effect on something. Stress is how much a material withstands internally, when a pressure is applied. So you can consider pressure as action, and stress as reaction. For example, the load on a concrete column creates a pressure for that column coming on that column as an external effect. This pressure is countered by the column body, which generates a stress in that column. As long as this generated stress is below the allowable limit (yield strength) of that column, the column will not fail (assuming no other failure occurred first, such as due to geometric instability – see Structural Stability).

In the next post of this series, we will introduce the term “Mohr’s Circle”

You must be logged in to post a comment Login