# Load conversion for geotech problem

This is a basic question but for some reason it has me stumped. I am modeling a continuous load on a soil layer on a commercial FEA package. This is a plane strain problem where the layer is modelled 20m wide by 10m deep.

I am trying 2 variations with 0.2 m thickness and 1 m thickness. The load the software is applying is 1 kN/m/m for the 1m thickness variation. I would like to find an equivalent loading (i.e one which will give me equivalent stresses and strains throughout the layer) for the 0.2 m thickness case.

I am hesitating between using 0.2 kN/m/m and 5 kN/m/m. The reasoning behind these two numbers is either 1/0.2 = 5 or 1*0.2 = 0.2. Can anyone please explain what 1kN/m/m actually means? and how do we convert it for different thicknesses?

• I would expect that the last "/m" is per metre into your screen, and the first "/m" is per metre across your screen (left to right). So I would expect that the depth of the foundation doesn't affect the applied load. [I may have completely misunderstood your question. In which case either a) you need a geotech to answer your question rather than a structural like me, or b) you need to clarify (ideally with pictures). – AndyT Apr 19 '17 at 8:17
• There is no foundation. It is just a layer of soil with continuous load applied on top. You are right that the last "/m" is per meter into the screen. I believe the question is simple enough that it is not discipline-specific – user32882 Apr 19 '17 at 11:03
• If there is no foundation then what is changing thickness and can be either 0.2m thick or 1m thick? – AndyT Apr 19 '17 at 13:56

In the comments you seem to already understand that kN/m/m means "kN per meter along your screen, per meter into your screen". This can be simplified as kN/m2: kN per meter in each direction.

The beauty of units is that they follow all the same rules of algebra as numbers. That's because kN/m/m actually means:

$$\dfrac{1\text{ kN}}{1\text{ m}\cdot1\text{ m}}$$

So, we can do some cross-multiplication:

$$\dfrac{1\text{ kN}}{1\text{ m}\cdot1\text{ m}} = \dfrac{X\text{ kN}}{1\text{ m}\cdot0.2\text{ m}}$$

From this it is clear that $X = 0.2$. If you are reducing the area where the total force is applied, you need to reduce the force in order to maintain the same overall stress.

That being said, it depends on how exactly you are applying this load. Are you applying it already as concentrated nodal forces? In this case, the above description is valid. It is also valid if you are applying linearly distributed loads (kN/m) where the described distance is along the screen (meaning that you need to consider the "depth" of the applied load).

If you're using a software package that can handle distributed loads (and then internally transforms them into nodal forces), those are usually defined with area-load units (such as kN/m/m). In this case, you just keep applying the same 1 kN/m/m and let the program handle the math.