XYLENE POWER LTD.

FNR BASEPLATE

By Charles Rhodes, P.Eng., Ph.D.

BASEPLATE:
The FNR baseplate supports the entire weight of the nuclear island. In an ideal world the baseplate would rest on a single chunk of granite in dry soil. The granite would absorb local force concentrations. However, in the real world the FNR is surrounded by soil that sooner or later may become saturated with water, in which case the entire nuclear island in effect becmes a concrete boat.

In a boat, at any particular depth below grade, the hydraulic pressure is constant. However, the weight of the FNR has downward force concentrations directly underneath the sodium pool, the inner structural wall and the outer structural wall.

These force concentrations in combination with uniform external hydraulic pressure at a particular depth, lead to baseplate shear force concentrations at the bases of the loaded walls. To absorb these shear force concentrations it is necessary to make the concrete baseplate thicker and to increase the amount of reinforcement in the proximity of these walls. The reinforcement must be particularly dense at locations where the concrete is under maximum tensile stress.

In order to evenly distribute the downward forces related to the structural walls over the bottom of the baseplate, the baseplate may need to be as much as 6 m thick. The perimeter of the baseplate should extend about 3 m beyond the outer structural wall, indicating a baseplate outside diameter of:
54 m + 6 m = 60 m.
 

BELOW GRADE WALL STABILIZATION:
Further, when behaving as a boat there is substantial inward hydraulic radial force on the outer structural wall, immediately above the baseplate. From a theoretical perspective concrete is strong in compression. However, the distance between the major radial shear walls at the air locks is large. To provide additional structural stability it is necesary to introduce two stabilization floors below grade at 10 m and at 20 m m above the top surface of the baseplate.

There must be enough outside vertical reinforcement near the bottom of the outer structural sidewall to safely absorb the inward radial hydraulic force between successive radial shearwalls.
 

BASEPLATE SHEAR FORCE:
One of the circumstnces that must be faced is water logging of the ground underneath the nuclear island. Under these circumstances the nuclear island will tend to float like a boat. Then the external water pressure is evenly distributed over the bottom of the nuclar island, but the perimeter weight is concentrated at the outer structual wall and the inner weight is inside the inner structural wall. Between these two walls is about a 10 m width where the external water pressure will develop considerable shear stress. Refer to the profile diagram.

The torque produced by the external water pressure is about:
Integral from R = Ra to R = Rb of:
P 2 Pi R dR (R - Ra)

= P 2 Pi {[(R^3 / 3) - Ra (R^2 / 2)]|R = Rb - [(R^3 / 3) - Ra (R^2 / 2)]|R = Ra }
= P 2 Pi {[(Rb^3 / 3) - Ra (Rb^2 / 2)] - [(Ra^3 / 3) - Ra (Ra2 / 2)] }
= P 2 Pi {(Rb^3 / 3) - Ra Rb^2 / 2 + Ra^3 / 6}

For Ra = 17 m Rb = 23 m:
T = P 2 Pi (4055.7 - 4496.5 + 818.83) m^3
= P 2 Pi 378 m^3
= 200,000 Pa (6.28) (378 m^3)
= 474,768 kPa m^3

Now assume that the thickness of the bottom plate is 2 Zo:
S(Z) = S(Zo)[(Z / Zo)]

The torque resistqnce is:
2 Integral from Z = 0 to Z = Zo of:
s(Z)2 Pi Ra dZ Z
= Integral from Z = 0 to Z = Zo of:
S(zo)(Z / Zo) 4 Pi Ra Z dZ
= [S(Zo)4 Pi Ra / Zo] Zo^3 / 3
= S(Zo) 4 Pi Ra Zo^2 / 3

Assume that for concrete S(Zo) = 30 MPa = tensile strength

then torque resistance is:
30 MPa 4 Pi 17 m Zo^2 / 3

Hence:
30 MPa 4 Pi 17 m Zo^2 / 3 = 474,768 kPa m^3
or
Zo^2 = 3 (474.768 MPa m^3 / (30 MPa 4 Pi 17 m )
= 0.222 m^2

Hence Zo = .47 m, implying a bottom plate thickness of 1.0 m is sufficient to resist hydraulic pressure.

A much more concerning circumstance is one where due to perimeter ground errosion the inner structural wall has to support the outer structural wall.
That requires a much thicker base plate.
 

This web page last updated October 25, 2026.