Module II : Analysis and design of concrete pavements
Lecture 2 : Analysis of Concrete Pavement
 

Slab resting on Winkler foundation

In this type of foundation, the subgrade is assumed to be composed of closely spaced linear springs and subgrade shear strength is neglected. The subgrade reaction is proportional to the deflection (equivalent to dense liquid foundation) and the proportionality (spring) constant is called as modulus of subgrade reaction     (k). So, the upward subgrade pressure ($p$) is $k \omega$. The equilibrium equation of plate resting on Winkler foundation becomes:
$\displaystyle D\nabla^4 \omega$
 $\textstyle =$ $\displaystyle q-p=q -k \omega$
$\displaystyle \Rightarrow l^4 \nabla^4 \omega + \omega$
 $\textstyle =$ $\displaystyle \frac {q}{k}$
(14)


where, radius of relative stiffness, $l=(D/k)^{1/4}$.

The closed form solution for deflection due to slab wight (unit wight= $\rho$) for finite dimensions ($L \times B$) supported over an area in the middle by subgrade reaction may expressed as (Timoshenko, and Kreiger 1959):

 \begin{eqnarray*} \omega &=& \frac{\rho h B^4}{384D} \left( 1-4\frac{y^2}{B^2} \... ...{1}{m^3}\left( a_m - \frac{a_m + \tanh a_m}{\tanh^2 a_m} \right) \end{eqnarray*}

where, $a_m=\frac{m \pi B}{2L}$.

 

Slab resting on Pasternak foundation

In this type of foundation, the shear interaction between the Winkler spring element is incorporated and subgrade shear force is assumed to be proportional to the variation in deflection in the soil layer (Cauwelaert et al. 2002). Therefore,
                                             \begin{eqnarray*} Q_{x,s}=G_s \frac {\partial \omega}{\partial x} \;\;\;\mbox{and}\;\;\; Q_{y,s}=G_s \frac {\partial \omega}{\partial y} \end{eqnarray*}

where, $Q_{x,s}$   and $Q_{y,s}$   are the subgrade shear forces per unit length acting on the plane orthogonal to $x$ and $y$ direction respectively and  $G_s$   is called as subgrade shear modulus. So, the net upward vertical pressure ($p$) is:

  \begin{eqnarray*} p= k \omega - \left ( \frac {\partial Q_{x,s}}{\partial x}+ \f... ...al Q_{y,s}}{\partial y} \right )= k \omega - G_s \nabla^2 \omega \end{eqnarray*}

Therefore, the equilibrium equation for plate resting on Pasternak foundation become:

$\displaystyle D\nabla^4 \omega = q-p$
 $\textstyle =$ $\displaystyle q - (k \omega - G_s \nabla^2 \omega)$
$\displaystyle \Rightarrow D\nabla^4 \omega - G_s \nabla^2 \omega + k \omega$
 $\textstyle =$ $\displaystyle q$
$\displaystyle \Rightarrow l^4 \nabla^4 \omega - \frac {G_s}{k} \nabla^2 \omega + \omega$
 $\textstyle =$ $\displaystyle \frac {q}{k}$
(15)


The Winkler foundation is a special case of Pasternak foundation (i.e when, $G_s=0$).

 

 

Slab resting on Kerr foundation

In Kerr foundation, the subgrade is characterized with two spring constants ( $k, k_0$) along with presence of subgrade shear interaction. The relationship between the upward subgrade reaction ($p$) and deflection ($\omega$) of the slab as follows (Cauwelaert et al. 2002):

                                                          \begin{eqnarray*} \left ( 1+ \frac {k}{k_0} \right ) p - \frac{G_s}{k_0}\nabla^2 p = k \omega -G_s \nabla^2 \omega \end{eqnarray*}

Therefore, the equilibrium equation (13) for Pasternak foundation becomes,

$\displaystyle D \nabla^6\omega =\nabla^2 q^\star=\nabla^2 q-\nabla^2 p \;\;\mbox{(since, $q^\star=p-q$)}$
$\displaystyle \Rightarrow D\nabla^6\omega = \nabla^2 q + \frac{kk_0}{G_s} \omega - k_0 \nabla^2 \omega - \frac{k_0}{G_s} \left ( 1+ \frac {k}{k_0} \right) p$
$\displaystyle \Rightarrow \frac{G_s}{kk_0}D\nabla^6\omega = \frac{G_s}{kk_0}\na... ...^2 \omega - \frac{1}{k} \left ( 1+ \frac {k}{k_0} \right ) ( q-D\nabla^4\omega)$
$\displaystyle \Rightarrow -\frac{G_sD}{k k_0}\nabla^6 \omega+\left (1+ \frac{k}... ...\omega = \left (1+ \frac{k}{k_0}\right )\frac{q}{k}-\frac{G_s}{k k_0}\nabla^2 q$ (16)

The Pasternak foundation is a special case of Kerr foundation (i.e when, $k_0=\infty$).

 

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