Module II : Analysis and design of concrete pavements
Lecture 2 : Analysis of Concrete Pavement
 
Solutions for temperature stress

The Westergaard's (1926) equation for maximum tensile stress at the top for cental area of infinite slab due to negative temperature differential ( $\triangle T$) is:
\begin{displaymath} \sigma_0=\frac{E\alpha \triangle T}{2(1-\mu)} \;\;\mbox{both in x and y direction, (Interior condition)}\\ \end{displaymath}
  
(17)

The Westergaard expression for deflection and maximum tensile stress of semi infinite slab ( $x=-\infty\;\; to\;\; \infty$ and   $y=0\;\; to\;\; \infty$)

$\displaystyle \omega_y = - \omega_0 \sqrt2 \cos( \frac{y}{l \sqrt2}+\pi/4) e^{-\frac{y}{l \sqrt2}}$
$\displaystyle \sigma_y = \sigma_0 \left[ 1 - \sqrt2 \sin( \frac{y}{l \sqrt2}+\pi/4) e^{-\frac{y}{l \sqrt2}} \right]$
(18)
$\displaystyle \sigma_x = \sigma_0 \left[ 1 - \mu \sqrt2 \sin( \frac{y}{l \sqrt2}+\pi/4) e^{-\frac{y}{l \sqrt2}} \right]$
(19)


where, $\omega_0=\frac{\alpha \triangle T (1+\mu)}{h}l^2$ and   $\sigma_0 $   as expressed above.

The $\displaystyle \sigma_y$ maximum occurs at $y=l\sqrt 2 \pi$ and the tensile stress at the edge location ( $y=0$), $\sigma_{e,x}=\frac{E \alpha \triangle T}{2}$ ; $\sigma_{e,y}=0$ (Edge condition)
For the slab with finite width ( $y=-B/2\;\;to\;\;B/2$) and infinite length ( $x=-\infty\;\; to\;\; \infty$), the deflection and stress equations are -

$\displaystyle \omega_y$ $\textstyle =$ $\displaystyle - \omega_0 \frac{2\cos \lambda_y \cosh \lambda_y}{\sin 2\lambda_y... ...bda_y+\tanh \lambda_y) \cos \frac{y}{l \sqrt2} \cosh \frac{y}{l \sqrt2} \right.$
$\displaystyle \left. + ( tan \lambda_y+\tanh \lambda_y)\sin \frac{y}{l \sqrt2} \sinh \frac{y}{l \sqrt2}\right]$
$\displaystyle \sigma_y$ $\textstyle =$ $\displaystyle \sigma_0 \left[ 1- \frac{2\cos \lambda_y \cosh \lambda_y}{\sin 2\... ...anh \lambda_y) \cos \frac{y}{l \sqrt2} \cosh \frac{y}{l \sqrt2} \right. \right.$
$\displaystyle \left. \left. + ( tan \lambda_y - \tanh \lambda_y)\sin \frac{y}{l \sqrt2} \sinh \frac{y}{l \sqrt2} \right) \right]$ (20)
$\displaystyle \sigma_x$ $\textstyle =$ $\displaystyle \sigma_0 \left[ 1- \mu \frac{2\cos \lambda_y \cosh \lambda_y}{\si... ...anh \lambda_y) \cos \frac{y}{l \sqrt2} \cosh \frac{y}{l \sqrt2} \right. \right.$
$\displaystyle \left. \left. + ( tan \lambda_y - \tanh \lambda_y)\sin \frac{y}{l \sqrt2} \sinh \frac{y}{l \sqrt2} \right) \right]$ (21)


where $\lambda_y=\frac{B}{l\sqrt 8}$. Similar solution can be obtained for the slab with finite length ( $y=-L/2\;\;to\;\;L/2$) and infinite width ( $x=-\infty\;\; to\;\; \infty$) and taking, $\lambda_x=\frac{L}{l\sqrt 8}$.

The Bradbury (1938) equation for maximum tensile stress at the edge ( $\sigma_e$) and interior ( $\sigma_i$) location with linear temperature gradient and finite slab with all edges free, over Winkler foundation are:

$\displaystyle \sigma_{e,x}=\frac{E \alpha \triangle T}{2}C_x \;\;\mbox{and}\;\; \sigma_{e,y}=\frac{E \alpha \triangle T}{2}C_y$ (22)
$\displaystyle \sigma_{i,x}=\frac{E \alpha \triangle T}{2(1-\mu^2)} \left[ C_x+ ... ...\sigma_{i,y}=\frac{E \alpha \triangle T}{2(1-\mu^2)}\left[ C_y+ \mu C_x \right]$ (23)

where,
$C_x= 1- \frac{2\cos \lambda_x \cosh \lambda_x}{\sin 2\lambda_x + \sinh 2 \lambda_x} (tan \lambda_x + \tanh \lambda_x)$ ; $C_y= 1- \frac{2\cos \lambda_y \cosh \lambda_y}{\sin 2\lambda_y + \sinh 2 \lambda_y} (tan \lambda_y + \tanh \lambda_y)$           and, $\lambda_x$   and $\lambda_y$ are same as above.

The downward vertical displacement due to weight of slab over a dense liquid foundation may  be represented as:
\begin{displaymath} \omega_{\rho}=\frac{\rho h}{k} \nonumber \end{displaymath}

The Westerdard solution is valid when $\omega_0 \le \omega_{\rho}$ (slab in full contact). For large negative temperature differential                 ($\omega_0 > \omega_{\rho}$) the slab may curled up and considering the gap due to curling, a closed form solution for semi-infinite slab can be presented as (Tang et al. 1993):
$\displaystyle \omega_y$
 $\textstyle =$ $\displaystyle - \frac{\omega_{\rho}}{\cos \phi} \cos(\frac{y}{l \sqrt2}+\phi) e^{-\frac{y}{l \sqrt2}}$
$\displaystyle \sigma_y$
 $\textstyle =$ $\displaystyle \sigma_0 \left[ 1- \frac{\omega_{\rho}}{\omega_0 \cos \phi} \sin (\frac{y}{l \sqrt2}+\phi) e^{-\frac{y}{l \sqrt2}} \right]$
(24)

where, $\cos \phi= \frac{1}{\sqrt{2(2\gamma^2-2\gamma+1}}$ and $\gamma=\sqrt{\frac{\omega_0}{\omega_{\rho}}} \ge 1$.

The $\displaystyle \sigma_y$ maximum occurs at $y=l\sqrt 2\left( \frac{5\pi}{4}-\phi \right)$. When $\omega_0=\omega_{\rho}$, then, $\gamma = 1 \;\;or\;\; \phi=\pi/4$ and thus it becomes the Westergaard's solution. For slab with finite width and infinite length, Westergaard's solution is valid when $\omega_{y=\pm B/2}\le \omega_{\rho}$ and $\omega_{y=\pm B/2}=\omega_0 \frac{-\sin2\lambda_y+\sinh2\lambda_y}{\sin2\lambda_y+\sinh2\lambda_y}$. The general solution proposed by Tang et al. (1993)is:
$\displaystyle \omega_y$ $\textstyle =$ $\displaystyle a_1 \cos \frac{y}{l\sqrt2}\cosh \frac{y}{l\sqrt2}+b_1 \sin \frac{y}{l\sqrt2}\sinh \frac{y}{l\sqrt2}$
$\displaystyle \sigma_y$ $\textstyle =$ $\displaystyle \frac{a_1}{l\sqrt2} \sin \frac{y}{l\sqrt2}\sinh \frac{y}{l\sqrt2}-\frac{b_1}{l\sqrt2} \cos \frac{y}{l\sqrt2}\cosh \frac{y}{l\sqrt2}$ (25)
$\displaystyle \mbox{where,}$
$\displaystyle a_1$ $\textstyle =$ $\displaystyle \frac{l\sqrt2}{\sin \lambda_y \cos \lambda_y + \sinh \lambda_y \c... ..._y \sinh\lambda_y) (\frac{\omega_{\rho}}{l\sqrt2})(\frac{s}{l\sqrt2})^2 \right.$
$\displaystyle \left. +2 \cos\lambda_y \cosh\lambda_y (\frac{\omega_{\rho}}{l\sq... ...cosh \lambda_y-\cos\lambda_y \sinh\lambda_y )(\frac{\omega_0}{l\sqrt2}) \right]$
$\displaystyle b_1$ $\textstyle =$ $\displaystyle \frac{l\sqrt2}{\sin \lambda_y \cos \lambda_y + \sinh \lambda_y \c... ...a_y \sinh\lambda_y)(\frac{\omega_{\rho}}{l\sqrt2})(\frac{s}{l\sqrt2})^2 \right.$
$\displaystyle \left. +2 \sin\lambda_y \sinh\lambda_y (\frac{\omega_{\rho}}{l\sq... ...cosh \lambda_y+\cos\lambda_y \sinh\lambda_y )(\frac{\omega_0}{l\sqrt2}) \right]$
$\displaystyle s$ $\textstyle =$ $\displaystyle l\sqrt2(\gamma-1)\;\;\;\mbox{and ;}\;\;\ \gamma \ge 1 \;\; \mbox{(as above).}$

 

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