Analysis and design of concrete pavements

Module II : Analysis and design of concrete pavements

Lecture 2 : Analysis of Concrete Pavement

Analysis of load stress

A concrete pavement is generally idealized as slab (or, plate) resting on elastic foundation. So as to develop the formulation for load stress calculation, the theory of plates is to be recapitulated first.

Basic theory of plates

For analysis purpose, a plate can be assumed as infinite in both the directions, semi-infinite, finite in one direction, or, finite in both the directions and accordingly the reference coordinates can be chosen. Figure 16 explains this configurations schematically.

For a basic plate analysis, it is assumed that the plate is homogenous, isotropic and elastic material and the cross section normal to the neutral axis remains plane before and after bending and, thus, there is no deformation along the thickness of plate. Therefore, the bending stress, $\sigma_{zz}^\omega(z) =0$ at any point across the depth () and the analysis becomes a plane stress case. For small deflection ( $\omega$ ) of plate (Timoshenko, and Kreiger 1959), it can be written,

$\begin{eqnarray*}\frac{1}{R_{xx}^\omega} = -\frac{\partial^2 \omega}{\partial x^... ...frac{1}{R_{yy}^\omega} = -\frac{\partial^2 \omega}{\partial y^2} \end{eqnarray*}$

where, $\omega(x,y)$ represents the deflection in Cartesian coordinate system along the z-direction (thickness) and $R^\omega$ is the radius of curvature in the respective direction. The negative sign represents that the upward curvature is due to downward deflection.

Figure 16: Various possible dimensions of a plate

Plate bending theory can be grossly grouped into two, thin plate theory and thick plate theory. For concrete pavement analysis purpose, it is generally assumed as thin plate. The assumption of thin plate (called as Kirchhoff plate) bending theory is that the thickness of the plate () is small as compared to the other dimensions. Thus, the effects of $\sigma_{xz}^\omega(z)$ and $\sigma_{yz}^\omega(z)$ on bending are neglected where, $\sigma^\omega(z)$ represent the bending stress due to external load in respective planes and represent the depth from the neutral axis (Timoshenko, and Kreiger 1959). Thus,
$\begin{eqnarray*} \epsilon_{xx}^\omega(z) = \frac{z}{R_{xx}^\omega} \;\;\;\mbox{and}\;\;\; \epsilon_{yy}^\omega(z) = \frac{z}{R_{yy}^\omega} \end{eqnarray*}$

where, $\epsilon^\omega(z)$ is the strain in the respective direction due to external load. So, the strain-deflection relationship due to bending may be presented as:

$\begin{eqnarray*} \epsilon_{xx}^\omega(z) = -z \frac{\partial^2 \omega}{\partial... ...^\omega(z) = -2z \frac{\partial^2 \omega}{\partial x \partial y} \end{eqnarray*}$

The strain-stress relationship due to bending may be represented as:

$\begin{eqnarray*} \epsilon_{xx}^\omega(z) &=& \frac{\sigma_{xx}^\omega(z)}{E}-\m... ...gma_{xy}^\omega(z)}{G}= \frac{2(1+\mu)}{E} \sigma_{xy}^\omega(z) \end{eqnarray*}$

where, and are the Young's modulus and shear modulus respectively (i.e. $G=\frac {E}{12(1+\mu)}$ ). The bending stress-strain relationship due to external load becomes:
$\begin{eqnarray*} \sigma_{xx}^\omega(z) &=& \frac{E}{1-\mu^2} \epsilon_{xx}^\ome... ...ma_{xy}^\omega(z) &=& \frac{E}{2(1+\mu)} \epsilon_{xy}^\omega(z) \end{eqnarray*}$

Therefore, the stress-deflection relationship may be expressed as:

$\displaystyle \sigma_{xx}^\omega(z)$	$\textstyle =$	$\displaystyle - \frac{Ez}{1-\mu^2} \left ( \frac{\partial^2 \omega}{\partial x^2}+ \mu \frac{\partial^2 \omega}{\partial y^2} \right )$	(4)
$\displaystyle \sigma_{yy}^\omega(z)$	$\textstyle =$	$\displaystyle - \frac{Ez}{1-\mu^2} \left ( \frac{\partial^2 \omega}{\partial y^2} +\mu \frac{\partial^2 \omega}{\partial x^2} \right )$	(5)
$\displaystyle \sigma_{xy}^\omega(z)$	$\textstyle =$	$\displaystyle -\frac{Ez}{1+\mu} \frac {\partial^2 \omega}{\partial x \partial y}$	(6)

If $M_{xx}^\omega$ , $M_{yy}^\omega$ are the bending moments per unit length due to load, parallel to and axis respectively, and $M_{xy}^\omega$ is the respective twisting moment in the plane, then (Timoshenko, and Kreiger 1959),

$\displaystyle M_{xx}^\omega$	$\textstyle =$	$\displaystyle \int_{-h/2}^{h/2} \sigma_{xx}^\omega(z)z dz = -\frac{Eh^3}{12(1-\... ...al^2 \omega} {\partial x^2}+\mu \frac{\partial^2 \omega}{\partial y^2} \right )$
	$\textstyle =$	$\displaystyle -D \left (\frac{\partial^2 \omega}{\partial x^2}+\mu \frac{\partial^2 \omega}{\partial y^2} \right )=\frac{h^2}{6} \sigma_{xx,max}^\omega$	(7)
$\displaystyle M_{yy}^\omega$	$\textstyle =$	$\displaystyle \int_{-h/2}^{h/2} \sigma_{yy}^\omega(z) z dz = -\frac{Eh^3}{12(1-... ...ial^2 \omega}{\partial y^2}+\mu \frac{\partial^2 \omega}{\partial x^2} \right )$
	$\textstyle =$	$\displaystyle -D \left (\frac{\partial^2 \omega}{\partial y^2}+\mu \frac{\partial^2 \omega}{\partial x^2} \right )=\frac{h^2}{6} \sigma_{yy,max}^\omega$	(8)
$\displaystyle M_{xy}^\omega$	$\textstyle =$	$\displaystyle - \int_{-h/2}^{h/2} \sigma_{xy}^\omega(z) z dz = \frac{Eh^3}{12(1+\mu)} \frac{\partial^2 \omega}{\partial x \partial y}$
	$\textstyle =$	$\displaystyle D(1-\mu) \frac{\partial^2 \omega}{\partial x \partial y}=-\frac{h^2}{6} \sigma_{xy,max}^\omega$	(9)

where, flexural rigidity, $D= \int_{-h/2}^{h/2} \frac{Ez^2}{1-\mu^2} dz = \frac{Eh^3}{12(1-\mu^2)}$ . Also, the stresses may be expressed as:

$\begin{eqnarray*} \sigma_{xx}^\omega(z)=\frac{Ez}{1-\mu^2}\frac{M_{xx}^\omega}{D... ...\sigma_{xy}^\omega(z)=-\frac{Ez}{1-\mu^2}\frac{M_{xy}^\omega}{D} \end{eqnarray*}$

Let, and are the shear forces per unit length parallel to and axis, then

$\textstyle =$ $\displaystyle \frac{\partial M_{xx}^\omega}{\partial x} - \frac{\partial M_{xy}^\omega}{\partial y}$

$\textstyle =$ $\displaystyle -D \frac {\partial}{\partial x}\left ( \frac{\partial^2 \omega}{\partial x^2}+ \frac{\partial^2 \omega}{\partial y^2} \right )$
(10)

$\textstyle =$ $\displaystyle \frac{\partial M_{yy}^\omega}{\partial y}-\frac{\partial M_{xy}^\omega}{\partial x}$

$\textstyle =$ $\displaystyle -D \frac {\partial}{\partial y}\left ( \frac{\partial^2 \omega}{\partial x^2}+ \frac{\partial^2 \omega}{\partial y^2} \right )$
(11)

If $q^\star$ is the net pressure (downward positive) over the plate surface which may include the external load, subgrade reaction as well as self weight of the plate, then from the equilibrium considerations:

$\displaystyle \frac{\partial Q_x}{\partial x}+\frac{\partial Q_y}{\partial y}$
$\textstyle =$ $\displaystyle -q^\star$
(12)

Therefore, differentiating equation (10) with respect to and equation (11) with respect to and, then, substituting in equation (12), the equilibrium equation becomes (Timoshenko, and Kreiger 1959):

$\displaystyle \frac{\partial^2 M_{xx}^\omega}{\partial x^2} + \frac{\partial^2 ... ...^\omega}{\partial y^2}-2 \frac{\partial^2 M_{xy}^\omega}{\partial x \partial y}$	$\textstyle =$	$\displaystyle -q^\star$
$\displaystyle \Rightarrow \frac{\partial^4 \omega}{\partial x^4}+2 \frac{\partial^4 \omega}{\partial x^2 \partial y^2} + \frac{\partial^4 \omega}{\partial y^4}$	$\textstyle =$	$\displaystyle \frac {q^\star}{D}$
$\displaystyle \Rightarrow \nabla^2 \nabla^2 \omega = \nabla^4 \omega$	$\textstyle =$	$\displaystyle \frac {q^\star}{D}$	(13)

where, $\nabla^2= \frac {\partial^2}{\partial x^2}+ \frac {\partial^2}{\partial y^2}$ (Laplace biharmonic operator).

When the curvatures in x and y directions are equal such as the case of symmetric interior loading, the deflection ( $\omega$ ) with free edges may be expressed as shown below:

$\begin{eqnarray*} \frac{1}{R_{xx}^\omega}&=& \frac{1}{R_{yy}^\omega} \;\;\;\Righ... ...ac{M}{2D(1+\mu)}(x^2+y^2)\;\;;\;(\mbox{since,}\;\;c_1=c_2=c_3=0) \end{eqnarray*}$