--- title: "CC slab worst edge stress numerical" subject: "Transportation Engineering II" chapter: 2 key: "semester_7/transportation_engineering_ii:cc-pavement-edge-stress-20cm-slab" source: https://www.bctengineer.com/q/semester_7/transportation_engineering_ii/cc-pavement-edge-stress-20cm-slab site: BCT Engineering --- # CC slab worst edge stress numerical Transportation Engineering II · Chapter 2: Highway Pavement - Asked in: 2081 Baisakh, 8 marks - Video: [Stresses in Rigid Pavement Slab, HM Westergaard Expressions](https://www.youtube.com/watch?v=UlFz4pesC-0) — talk on civil engineering (TOCE) - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_7/transportation_engineering_ii/cc-pavement-edge-stress-20cm-slab - Whole chapter: https://www.bctengineer.com/md/semester_7/transportation_engineering_ii/chapter-2 ## Question A CC pavement slab of thickness 20cm is constructed over a granular subbase having modular of reaction 15 kg/cm². The maximum temperature difference between the top and bottom of the slab during summer day and night is found to be 18°C. The spacing between the transverse contraction joint is 4.5 m and that between longitudinal joints is 3.5 m. The design wheel load is 5100 kg, radius of contact area is 15 cm, E value of CC is 3×10⁵ kg/cm², Poisson's ratio is 0.15 and coefficient of thermal expansion is 10×10⁻⁶ per °C and friction coefficient is 1.5. Find the worst combination of stresses at the edge. (The paper attaches the IRC edge load stress chart and Bradbury's warping stress coefficient chart.) ## Short answer Worst combination is summer midday edge: load stress 24.0 kg/cm2 (Westergaard) + warping stress 28.3 kg/cm2 (Bradbury C=1.049) minus friction 0.81 kg/cm2 = 51.5 kg/cm2. This exceeds typical flexural strength, so slab is inadequate. ## Answer #### 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. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is the radius of relative stiffness l computed here? A. About 60.8 cm B. About 6 cm C. About 600 cm D. 20 cm Answer: A. l = [3x10^5 x 20^3 / (12(1 - 0.15^2) x 15)]^(1/4) = 60.8 cm. (Key point: Radius of relative stiffness) 2. The Bradbury coefficient C is read using which ratio? A. L/l = 450/60.8 = 7.40 B. h/l C. a/h D. P/h Answer: A. C depends on slab length over radius of relative stiffness; here L/l = 7.40 and C = 1.049. (Key point: Bradbury coefficient) 3. What is the edge warping stress? A. C E alpha dT / 2 = about 28.3 kg/cm2 B. E alpha dT = 54 kg/cm2 C. 0.81 kg/cm2 D. 24.3 kg/cm2 Answer: A. 1.049 x 3x10^5 x 10x10^-6 x 18 / 2 = 28.3 kg/cm2. (Key point: Warping stress) 4. In the summer midday case, the edge stress is combined as: A. Load + warping - friction B. Load - warping + friction C. Warping + friction only D. Load only Answer: A. Top hotter gives bottom tension adding to load tension, while friction is compressive: 24.3 + 28.3 - 0.81 = 51.8 kg/cm2. (Key point: Worst combination) 5. What does the high total edge stress of 51.8 kg/cm2 indicate? A. The slab is adequate B. The 20 cm slab is not adequate unless concrete is very strong, so a thicker slab is needed C. Friction is critical D. Joints should be removed Answer: B. The stress must be below the flexural strength; this high value shows the slab is too thin. 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