CC slab worst edge stress numerical

Given data and assumptions

$h=20$ cm, $k=15$ kg/cm3 (the unit given as kg/cm2 is taken as kg/cm3), $\Delta T=18^\circ$C, $L=4.5$ m (transverse joint spacing), $B=3.5$ m, $P=5100$ kg, $a=15$ cm, $E=3\times10^5$ kg/cm2, $\mu=0.15$, $\alpha=10\times10^{-6}$/$^\circ$C, $f=1.5$, unit weight of concrete $W=2400$ kg/m3. Method: IRC:58 (Westergaard, Bradbury).

Step 1 - Radius of relative stiffness

$$l=\left[\frac{3\times10^5\times20^3}{12(1-0.15^2)\times15}\right]^{1/4}=60.8\ \text{cm}$$

Step 2 - Edge load stress

Since $a=15 < 1.724h=34.5$ cm: $b=\sqrt{1.6a^2+h^2}-0.675h=\sqrt{1.6\times15^2+20^2}-13.5=14.07$ cm.

The IRC edge load stress chart is based on Westergaard's edge equation as modified by Teller and Sutherland, which is used here to get the value:

$$\begin{aligned} \sigma_e&=\frac{0.529P}{h^2}(1+0.54\mu)\left[4\log_{10}\frac{l}{b}+\log_{10}b-0.4048\right]\\ &=\frac{0.529\times5100}{20^2}(1.081)\left[4\log_{10}\frac{60.8}{14.07}+\log_{10}14.07-0.4048\right]=7.29\times3.286=24.0\ \text{kg/cm}^2 \end{aligned}$$

Step 3 - Warping stress at the edge (Bradbury)

$L/l=450/60.8=7.40$; from the Bradbury chart/table, $C=1.049$.

$$\sigma_{te}=\frac{C E\alpha\Delta T}{2}=\frac{1.049\times3\times10^5\times10\times10^{-6}\times18}{2}=28.3\ \text{kg/cm}^2$$

Step 4 - Frictional stress

$$\sigma_f=\frac{WLf}{2\times10^4}=\frac{2400\times4.5\times1.5}{2\times10^4}=0.81\ \text{kg/cm}^2$$

Step 5 - Critical combinations at the edge

• Summer midday (top hotter; warping tension at the bottom adds to the load tension; friction is compressive as the slab expands): $\sigma=\sigma_e+\sigma_{te}-\sigma_f=24.0+28.3-0.81=51.5$ kg/cm2
• Winter midday would be $\sigma_e+\sigma_{te}+\sigma_f$ with the smaller winter temperature difference (not given), so it is normally less critical.
• Night (top cooler, tension at the top; the edge load stress is not additive): $\sigma_{te}+\sigma_f=28.3+0.81=29.1$ kg/cm2

Answer: the worst combination is the summer-midday case, with a total edge stress of about 51.5 kg/cm2 (load 24.0 + warping 28.3 - friction 0.81). This must be compared with the flexural strength of concrete; a stress this high shows that the 20 cm slab is not adequate for 5100 kg unless the concrete is very strong, so a thicker slab is needed.