--- title: "Plain CC pavement thickness by IRC, 7 m highway" subject: "Transportation Engineering II" chapter: 2 key: "semester_7/transportation_engineering_ii:plain-cc-pavement-thickness-irc-7m" source: https://www.bctengineer.com/q/semester_7/transportation_engineering_ii/plain-cc-pavement-thickness-irc-7m site: BCT Engineering --- # Plain CC pavement thickness by IRC, 7 m highway Transportation Engineering II · Chapter 2: Highway Pavement - Asked in: 2075 Chaitra, 8 marks - Video: [Lec-36_Rigid Pavement Slab Thickness Design | PDHC | Civil Engineering](https://www.youtube.com/watch?v=EOHGUdtxrqU) — Civil Engineering Department_LJIET - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_7/transportation_engineering_ii/plain-cc-pavement-thickness-irc-7m - Whole chapter: https://www.bctengineer.com/md/semester_7/transportation_engineering_ii/chapter-2 ## Question Estimate the thickness of a plain cement concrete pavement for a 7m wide highway following the design procedure recommended by Indian Roads Congress (IRC) wherever applicable. Use given data, IRC load stress charts for edge and corner regions. Design wheel load = 5100 kg Traffic growth rate = 7.5% Present traffic intensity = 1050 cvpd Design life = 20 years Construction period = 3 years Radius of contact area = 15 cm Modulus of elasticity of concrete = 3.0 × 10⁵ kg/cm² Poisson's ratio of concrete = 0.15 Modulus of rupture of concrete = 40 kg/cm² Thermal expansion of concrete = 10 × 10⁻⁶/°C Modulus of subgrade reaction = 6 kg/cm³ Maximum temperature in summer = 50° (as printed, unit cut off) Maximum temperature in winter = 15°C The temperature differential in slab in the region is 17.3, 19.0 and 20.3 degree Celsius for thickness of 15, 20 and 25 cm respectively. Tables attached: traffic classification vs adjustment in design thickness of CC pavement (A: 0-15 cvpd, -5 cm; B: 15-45, -5; C: 45-150, -2; D: 150-450, -2; E: 450-1500, 0; F: 1500-4500, 0; G: >4500, +2 cm); coefficient C versus L/l or B/l (1: 0.000, 2: 0.040, 3: 0.175, 4: 0.440, 5: 0.720, 6: 0.920, 7: 1.030, 8: 1.077, 9: 1.080, 10: 1.075, 11: 1.050, 12: 1.000); and IRC edge and corner load stress charts. ## Short answer Trial thickness 25 cm satisfies edge load + warping stress (39.4 kg/cm²) and corner stress (29.9 kg/cm²) below 40 kg/cm² modulus of rupture. Traffic class E requires no adjustment. Provide 25 cm thick plain cement concrete slab. ## Answer #### Method IRC:58 (jointed plain concrete pavement) - trial thickness is checked for the critical stress combinations by Westergaard, Bradbury and Pickett-Ray analysis, then the traffic adjustment is applied. #### Assumptions Pavement 7 m wide = two lanes of 3.5 m with a central longitudinal joint ($B=3.5$ m); transverse contraction joint spacing $L=4.5$ m; unit weight of concrete 2400 kg/m3; friction coefficient 1.5; permissible (design) flexural stress = modulus of rupture 40 kg/cm2; the IRC edge and corner load stress charts are replaced by the formulas they are based on (Westergaard's edge formula as modified by Teller and Sutherland, and Kelly's corner formula). The summer and winter air temperatures (50 and 15 degrees C) affect the joint gap only, while the given differential $\Delta T$ is used for warping. #### Step 1 - Traffic classification Traffic at completion: $A=1050(1.075)^3=1304$ CVPD, which is in class E (450-1500 CVPD) and the thickness adjustment is **0 cm**. (Even the present 1050 CVPD is class E.) #### Step 2 - Frictional stress $\sigma_f=\dfrac{WLf}{2\times10^4}=\dfrac{2400\times4.5\times1.5}{2\times10^4}=0.81$ kg/cm2. #### Step 3 - Trial thickness: edge load and warping stress $l=\left[\dfrac{Eh^3}{12(1-\mu^2)k}\right]^{1/4}$, edge load stress $\sigma_e=\dfrac{0.529P}{h^2}(1+0.54\mu)\left[4\log_{10}\dfrac{l}{b}+\log_{10}b-0.4048\right]$ with $b=\sqrt{1.6a^2+h^2}-0.675h$ (since $a<1.724h$; $b=14.07$ cm for h = 20 cm and 14.51 cm for h = 25 cm), warping $\sigma_{te}=\dfrac{CE\alpha\Delta T}{2}$ with $C$ from the given table at $L/l$ (interpolated). | h (cm) | $\Delta T$ | $l$ (cm) | $\sigma_e$ | $L/l$ / $C$ | $\sigma_{te}$ | $\sigma_e+\sigma_{te}-\sigma_f$ | vs 40 | |---|---|---|---|---|---|---|---| | 20 | 19.0 | 76.4 | 26.9 | 5.89 / 0.898 | 25.6 | 51.7 | not safe | | 25 | 20.3 | 90.3 | 18.4 | 4.98 / 0.715 | 21.8 | 39.4 | safe | All stresses in kg/cm2. For h = 20 cm the combined stress is too high; for **h = 25 cm** it is within the 40 kg/cm2 limit. #### Step 4 - Corner check (h = 25 cm) Kelly's corner load stress: $\sigma_c=\dfrac{3P}{h^2}\left[1-\left(\dfrac{a\sqrt2}{l}\right)^{1.2}\right]=\dfrac{3\times5100}{25^2}\left[1-\left(\dfrac{15\sqrt2}{90.3}\right)^{1.2}\right]=20.2$ kg/cm2. Corner warping: $\sigma_{tc}=\dfrac{E\alpha\Delta T}{3(1-\mu)}\sqrt{a/l}=\dfrac{3\times10^5\times10\times10^{-6}\times20.3}{3(0.85)}\sqrt{15/90.3}=9.7$ kg/cm2. Total corner stress $=20.2+9.7=29.9$ kg/cm2 $<40$, safe. #### Step 5 - Result Thickness from analysis = 25 cm; traffic adjustment (class E) = 0 cm. **Answer: provide a 25 cm thick plain cement concrete slab** (modulus of rupture 40 kg/cm2) over a granular/DLC sub-base, with 4.5 m contraction joints, a longitudinal joint with tie bars at the 3.5 m lane edge and dowel bars at the transverse joints as per IRC:58. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Traffic at completion is 1050(1.075)^3 = 1304 CVPD. What thickness adjustment applies? A. Class E (450-1500), 0 cm adjustment B. Class D, -2 cm C. Class G, +2 cm D. Class B, -5 cm Answer: A. 1304 CVPD lies in class E which has no adjustment. (Key point: Traffic classification) 2. What is sigma_f for W = 2400 kg/m3, L = 4.5 m, f = 1.5? A. 0.81 kg/cm2 B. 8.1 kg/cm2 C. 0.081 kg/cm2 D. 16.2 kg/cm2 Answer: A. 2400x4.5x1.5/(2x10^4) = 0.81. (Key point: Frictional stress) 3. For h = 20 cm the combined stress is 52.0 kg/cm2 versus 40 allowed. What is concluded? A. Safe B. Not safe, so thickness is increased; 25 cm gives 39.5 and is safe C. Reduce thickness to 15 cm D. Increase modulus of rupture Answer: B. 52.0 exceeds 40, so a thicker slab is tried; 25 cm passes. (Key point: Trial thickness check) 4. In the corner check, which stresses are added? A. Kelly's corner load stress and corner warping stress B. Edge stress and friction C. Interior stress and friction D. Only friction Answer: A. 14.2 + 9.7 = 23.9 kg/cm2, less than 40. 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