--- title: "Rigid pavement thickness by IRC procedure" subject: "Transportation Engineering II" chapter: 2 key: "semester_7/transportation_engineering_ii:rigid-thickness-irc-procedure" source: https://www.bctengineer.com/q/semester_7/transportation_engineering_ii/rigid-thickness-irc-procedure site: BCT Engineering --- # Rigid pavement thickness by IRC procedure Transportation Engineering II · Chapter 2: Highway Pavement - Asked in: 2079 Bhadra, 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/rigid-thickness-irc-procedure - Whole chapter: https://www.bctengineer.com/md/semester_7/transportation_engineering_ii/chapter-2 ## Question Discuss the step by step procedure for the determining of rigid pavement thickness based on IRC guidelines. ## Short answer The IRC:58 procedure involves collecting design parameters, assuming a trial thickness, computing radius of relative stiffness, then calculating edge and corner load stresses, temperature warping stresses, and friction stress. Critical stress combinations are checked against permissible stress, thickness is revised by trial, and traffic correction and joint design are applied. ## Answer The design follows IRC:58 (Guidelines for the Design of Plain Jointed Rigid Pavements for Highways), based on the Westergaard/Bradbury/Pickett-Ray approach. 1. **Collect data** - design wheel load (5100 kg for highways, 4100 kg for small roads), tyre pressure, radius of contact area, flexural strength (modulus of rupture), E, Poisson's ratio and thermal coefficient of concrete; modulus of subgrade reaction k (on top of the sub-base); traffic (CVPD), growth rate and design life. 2. **Design traffic** - find the commercial vehicles at the end of construction, classify the traffic (A to G) and note the thickness adjustment. 3. **Assume a trial slab thickness h** and the joint spacing (transverse L, longitudinal width B). 4. **Radius of relative stiffness** $l=\left[\dfrac{Eh^3}{12(1-\mu^2)k}\right]^{1/4}$. 5. **Load stresses** - edge load stress (IRC chart based on the modified Westergaard equation) and corner load stress (Westergaard's formula as modified by Kelly, $\sigma_c=\frac{3P}{h^2}\left[1-\left(\frac{a\sqrt2}{l}\right)^{1.2}\right]$). 6. **Temperature (warping) stresses** - from the temperature differential between top and bottom: Bradbury's formula $\sigma_{te}=\dfrac{C\,E\,\alpha\,t}{2}$ with coefficient C from L/l and B/l; corner warping stress $\sigma_{tc}=\dfrac{E\alpha t}{3(1-\mu)}\sqrt{a/l}$. 7. **Frictional stress** $\sigma_f=\dfrac{WLf}{2\times10^4}$ (W in kg/m3, L in m). 8. **Check critical combinations** - (a) summer midday, edge: $\sigma_e+\sigma_{te}-\sigma_f$; (b) winter midday, edge: $\sigma_e+\sigma_{te}+\sigma_f$ (with the winter $\Delta T$); (c) night, corner: $\sigma_c+\sigma_{tc}$. Each must not exceed the permissible flexural stress (modulus of rupture divided by a factor of safety, or checked by the fatigue stress ratio in IRC:58-2002/2015). 9. **Revise the thickness** by trial until the stresses are safe; apply the traffic correction (A to G) to the thickness. 10. **Joints and steel** - design contraction, expansion and longitudinal joints, dowel bars at transverse joints and tie bars at longitudinal joints (IRC:58 / IRC:15). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What design wheel load does the procedure take for highways? A. 2000 kg B. 4100 kg C. 5100 kg D. 8160 kg Answer: C. 5100 kg is used for highways and 4100 kg for small roads. (Key point: Design wheel load) 2. After collecting data and finding design traffic, what is the next logical step? A. Design the dowel bars B. Assume a trial slab thickness and joint spacing C. Apply the thickness correction and finish D. Calculate frictional stress only Answer: B. A trial thickness h and joint spacing are assumed, then l and the stresses are computed and checked. (Key point: Order of procedure) 3. Which combination is checked for summer midday? A. Edge load stress + warping stress + frictional stress B. Edge load stress + warping stress - frictional stress C. Corner stress only D. Frictional stress - warping stress Answer: B. In summer friction is compressive and relieves tension, so sigma_e + sigma_te - sigma_f; winter adds sigma_f. (Key point: Critical stress combinations) 4. When is the trial thickness accepted? A. When the combined stresses do not exceed the permissible flexural stress B. When the slab is thicker than 30 cm C. When l is larger than 100 cm D. When the traffic class is A Answer: A. Each critical combination must not exceed the modulus of rupture divided by a factor of safety (or the fatigue stress ratio); otherwise thickness is revised. (Key point: Acceptance criterion) 5. Which expression gives the corner load stress in the procedure? A. Kelly's formula 3P/h^2 [1 - (a sqrt2 / l)^0.6] B. Bradbury's C E alpha t / 2 C. W L f / (2 x 10^4) D. 0.316 P/h^2 [4 log(l/b) + 1.069] Answer: A. Kelly's formula gives corner load stress; Bradbury's gives warping and WLf/2x10^4 gives friction. 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