1. Introduction
It is well known that the foundations of buildings are subject to loads which can result in deformation and subsidence. Hence, the analysis of foundation deformation must be conducted at the design stage. This article describes the computer simulation of foundation deformation. We propose an approach based on numerical solution of the stationary differential equation in partial derivatives. This equation describes the transversal deflection of a thin plate (foundation slab), taking elasticity into account, due to an external orthogonal force.
2. Plate Deflection Equation
Let the Cartesian coordinate system 
be the plate plane.By 
we define the domain of the plate in this plane. Let 
be the boundary of the domain. The function for plate deflection is given by 
. At small transversal (vertical) deflections, the function 
satisfies the following equation [1]:
(1) 
![\[ = f(x,y) + r(x,y,u), (x,y)\in \Omega \]](http://blog.simmakers.com/wp-content/ql-cache/quicklatex.com-513c7bedc0d35ed20fdd9e89c19182c7_l3.png)
The coefficients included in equation (1) are described in Table 1.
Table 1: Description of the Coefficients
| Coefficients | Description | Units of Measurement |
 |
Flexural rigidity of a plate at the point |
N∙m |
 |
The plate thickness at the point |
m |
 |
Young's modulus at the point |
Pa |
 |
Poisson's Ratio at the point |
– |
 |
Plate normal reaction pressure |
Pa |
 |
Normal pressure at the point |
Pa |
Since equation (1) defines a whole family of functions that satisfy it, the boundary conditions must be applied to obtain a unique solution. These boundary conditions depend on which of the three following phenomena apply to a particular slab: fixed and hinged boundaries, where the foundation is fastened to the raft, and free boundaries, where the slab rests on the ground surface or piles (the edges are not fixed). 
denotes the outward vector of the normal to the boundary at the point 
. If the boundary of the slab is fixed, then the boundary condition is:
|
|
(2) |
![\[ \begin{cases} u(x,y)|_{(x,y)\in\Gamma}=0, \\ \frac{\partial u(x,y)}{\partial \vec{n}}|_{(x,y)\in\Gamma}=0. \end{cases} \]](http://blog.simmakers.com/wp-content/ql-cache/quicklatex.com-5a15db5d7a5a71e6622d52cd6ad4bbd0_l3.png)
If it is hinged, then
|
|
(3) |
![\[ \begin{cases} u(x,y)|_{(x,y)\in\Gamma}=0, \\ G(x,y)|_{(x,y)\in\Gamma}=0. \end{cases} \]](http://blog.simmakers.com/wp-content/ql-cache/quicklatex.com-66515f730e85212518e92bbc3e05847c_l3.png)
where 
is the bending moment at point . Finally, if the boundary 
is free, then the boundary conditions are stated for no external forces on the border.
When modeling the deformation and subsidence of the slab, the reaction at point of slab equals zero or coincides with the reaction of the soil or piles beneath the slab. Experimental studies [2] show that the dependence between the soil reaction and transversal downward displacement of the slab is linear. This linear dependence holds as long as the reaction of the soil is less than the load-bearing capacity, otherwise the reaction of the soil is constant and does not depend on the displacement . Hence, when resting on soil, the reaction is calculated thus:
 |
(4) |
where 
denotes the coefficient of soil reaction [Н/m3] and 
is the bearing capacity [Pa].
If the support of the slab at point is a pile, then the reaction is assumed to depend linearly on the displacement until 
. When pile reaction exceeds load capacity there is no longer any resistance, resulting in 
. Therefore, when the slab rest on a pile the reaction is calculated thus:
 |
(5) |
where 
is the reaction coefficient of the pile [Н/m3].
Since the area of the plate is arbitrary, there is no general formula for the analytical solution of equation (1) and so numerical methods, in this case based on finite difference approximation of differential operators, are applied.
Due to the finite-difference approximation of fourth-order equation (1), there is a 9-point cross stencil. Given the problems with the approximation of this equation at the nodes adjacent to the boundary nodes, we reformulate fourth-order equation (1) into an equivalent set of three second-order equations:
|
|
(6) |
![\[ \begin{cases} D(x,y)\times (\Delta\nu(x,y)+\Delta w(x,y))=f(x,y)+r(x,y,u), \\ \\ \frac{\partial ^2u(x,y)}{\partial x^2}=\nu,\\ \\ \frac{\partial ^2u(x,y)}{\partial y^2}=w. \end{cases} \]](http://blog.simmakers.com/wp-content/ql-cache/quicklatex.com-5b1760c11af624ad09fdd8fcabfe7e9f_l3.png)
Despite the set of equations (6) having three times more unknowns than equation (1), its finite-difference approximation is straightforward because the stencil for the approximation of the second-order derivative contains only 3 points.
Due to the nonlinearity of set of equations (6), we use iteration methods to solve it numerically.
3. Sample Calculation of Foundation Deformation
We have provided a sample calculation for foundation deformation. Here, we assume that the material of the foundation is concrete having the physical properties shown in the Table 2.
Table 2: Physical Properties of Materials
| Physical Properties | Value |
| The Young's Modulus | E = 10.0 hPa |
| Poisson's Ratio |  |
| Density | ρ = 2000.0 kg/m3 |
Let the slab have a circle shape with the center at point 
m, 
m, and radius 
m. The slab thickness is equal to 
m. We consider two cases of slab connection with piles. In the first calculation, the slab stands on 60 piles: 40 piles are arranged in a circle with the center at the point 
and radius 
m, and the remaining 20 piles in a concentric circle with radius 
m. The coordinates along the Oz-axis of top points is 
m. The pile reaction coefficient is 
Н/m3; the load capacity of the piles is 
hPa.
External pressure (perpendicular to the plane of the slab) is 
kPa. Thus, with respect to the pressure created by the slab load, the total pressure at each point of the slab is equal to 
kPa. The boundary of the plate is free.
The foundation deformation results are shown in Figure 1.
Figure 1: Deflection of the Foundation Slab (Case 1)
Figure 1 shows the foundation slab vertical deflection by color distribution, with the black dots indicating the piles. Note that the load on any pile did not exceed its load capacity . The maximum deflection of the slab was 
m.
In the second computation, the inner circle of piles is removed and the results of the computation are shown in Figure 2.
Figure 2: Deflection of the Foundation Slab (Case 2)
As in the first calculation, the load on any one pile never exceeded its load capacity , but the maximum deflection of the foundation slab rose to as high as 0.62 m, which is sufficient to destroy the foundation.
4. Conclusion
The article demonstrates the application of numerical simulation for the evaluation of non-uniform deflection of foundation slabs under various loads with specific load-bearing pile arrangements. The same approach can be also used to calculate the deformation of foundation slabs on subsiding permafrost soil due to thawing.
5. References
- Timoshenko, S. and Woinowsky-Krieger, S. Theory of plates and shells. McGraw–Hill New York, 1959.
- Aynbinder A.B. Raschet magistralynykh i promyslovykh truboprovodov na prochnosty i ustoychivosty: spravochnoye posobiye (Main/Field Pipeline Strength and Stability Calculation: Reference Book). Moscow: Nedra, 1991. 287 p.