By now, everything should have been settled about modulus of subgrade reaction. We all know some typical statements about it:
However, let's think about a weathered rock with E=500 MPa and a foundation to be built on top of this rock with 20 x 50 m dimensions. You can say it depends on many factors as much as you like, everybody has a rough idea already: 100,000 kN/m3. So, if you are a fancy engineer and dare to make some calculations, you can find much lower values. Will they believe you or will they think that you are being too conservative (if lowering the subgrade reaction means being conservative)?
Let's start with simple terms. The equation that everybody knows and nobody wants to use:
$$ K=\frac{q}{s} $$
So, the subgrade reaction is equal to a spring stiffness distributed under the foundation. If you divide the pressure by the settlement, you will find the subgrade reaction, amount of deformation for unit pressure.
If we think about how we calculate settlement (how it depends on many factors), we can see actually how complex this modulus is.
Use any method you like. If you calculate the settlement of a foundation (say for 200 kPa uniform loading) with dimensions of 20 x 50 m on a weathered rock with elasticity modulus of 500 MPa, you will find something around 7-8 mm (or someting close) that will result in 25000 kN/m3 modulus of subgrade reaction
You know the easiest method without bothering with any influence factor:
$$ s=\frac{qB(1-v^2)}{E} $$
If you do that, you will find 28000 kN/m3.
If you use Settle 3D with flexible foundation, you will find something similar:
So, the question is: Where does this 100000 kN/m3 comes from?
Bowles was a great engineer and teacher. His books have guided most of us. Since he was a practicing engineer, his recommendations were pin-point to most of the problems we had. It was the first geotechnical book I have read from the cover to the end.