--- title: "Westergaard theory; warping and friction stresses" subject: "Transportation Engineering II" chapter: 2 key: "semester_7/transportation_engineering_ii:westergaard-theory-warping-friction-stress" source: https://www.bctengineer.com/q/semester_7/transportation_engineering_ii/westergaard-theory-warping-friction-stress site: BCT Engineering --- # Westergaard theory; warping and friction stresses Transportation Engineering II · Chapter 2: Highway Pavement - Asked in: 2076 Asoj, 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/westergaard-theory-warping-friction-stress - Whole chapter: https://www.bctengineer.com/md/semester_7/transportation_engineering_ii/chapter-2 ## Question Write down the assumption and analysis of Westergard Theory. How the warping stress and friction stress are developed in the Rigid pavement? ## Short answer Westergaard assumes a homogeneous isotropic elastic slab on a liquid foundation with reaction proportional to deflection, giving stress formulas for interior, edge and corner loads using radius of relative stiffness. Warping stress comes from top–bottom temperature difference causing curling, resisted by slab weight; frictional stress arises from overall expansion or contraction restrained by sub-base friction (σf = WLf/2×10^4). ## Answer #### Westergaard theory: assumptions and analysis Westergaard (1926) analysed a concrete slab resting on an elastic subgrade, treating the slab as a thin plate on a liquid (Winkler) foundation. **Assumptions** 1. The slab acts as a homogeneous, isotropic elastic plate of uniform thickness. 2. The reaction of the subgrade is vertical and proportional to deflection ($p=k\,\Delta$), where $k$ is the modulus of subgrade reaction (a "dense liquid"; no shear transfer in the subgrade). 3. The slab is in full contact with the subgrade at all times. 4. The load is applied on a circular area of radius $a$ with uniform pressure; the thickness is small compared with the other dimensions and the slab is treated as infinite or semi-infinite for the three cases. 5. The load is static; temperature, joint load transfer and friction are not included in the basic analysis. **Analysis - three loading positions** - **Interior load** (far from edges): $$ \sigma_i=\frac{0.316\,P}{h^2}\left[4\log_{10}\frac{l}{b}+1.069\right] $$ - **Edge load** (at the free edge): $$ \sigma_e=\frac{0.572\,P}{h^2}\left[4\log_{10}\frac{l}{b}+0.359\right] $$ Teller and Sutherland's modification (the basis of the IRC edge chart) is $\sigma_e=\dfrac{0.529P}{h^2}(1+0.54\mu)\left[4\log_{10}\dfrac{l}{b}+\log_{10}b-0.4048\right]$. - **Corner load**: $$ \sigma_c=\frac{3P}{h^2}\left[1-\left(\frac{a\sqrt2}{l}\right)^{0.6}\right] $$ Kelly's modification (used by IRC) changes the exponent from 0.6 to 1.2. Here $P$ is the wheel load, $h$ thickness, $a$ contact radius, $b=\sqrt{1.6a^2+h^2}-0.675h$ for $a<1.724h$ (else $b=a$) and $l=\left[\dfrac{Eh^3}{12(1-\mu^2)k}\right]^{1/4}$ the radius of relative stiffness. For the same load the edge stress is the largest bending stress and the interior stress is the smallest; the corner load gives the largest deflection. #### Development of warping and friction stresses **Warping stress** Warping stress comes from the temperature difference between the top and bottom of the slab. - By day, the top is hotter than the bottom, so the top tries to expand more and the slab edges curl downwards. The weight of the slab and the subgrade resist this, so compression develops at the top and tension at the bottom. - At night the top is cooler, the edges curl upwards and tension develops at the top. Bradbury gave the edge stress for a finite slab: $$ \sigma_{te}=\frac{C\,E\,\alpha\,\Delta T}{2} $$ and for the interior $\sigma_{ti}=\dfrac{E\alpha\Delta T}{2}\left(\dfrac{C_x+\mu C_y}{1-\mu^2}\right)$. The coefficient $C$ depends on $L/l$ and $B/l$ (Bradbury chart), $E$ is the modulus of concrete, $\alpha$ the thermal coefficient and $\Delta T$ the top-bottom temperature difference. **Frictional stress** When the whole slab expands or contracts with an average temperature change, the movement is resisted by friction between the slab and the sub-base. At the centre of the slab the tension from contraction is $$ \sigma_f=\frac{W\,L\,f}{2\times10^{4}}\ \ (\text{kg/cm}^2) $$ with $W$ the unit weight of concrete (2400 kg/m3), $L$ the joint spacing (m) and $f$ the coefficient of friction (about 1.5). It is tensile in winter or at night (contraction) and compressive in summer. It is generally small and fixes the spacing of contraction joints. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How does Westergaard's theory model the subgrade? A. As a dense liquid (Winkler) foundation with reaction proportional to deflection, p = k Delta B. As an elastic solid with shear transfer C. As a rigid base D. As a layered system of three elastic layers Answer: A. The subgrade is a liquid foundation: vertical reaction proportional to deflection, no shear. (Key point: Subgrade model) 2. Which ranking of load stresses for the same load is correct? A. Interior is largest B. Edge stress is largest bending stress, interior smallest C. Corner stress is smallest D. All are equal Answer: B. Edge loading gives the largest bending stress and interior the smallest; the corner gives the largest deflection. (Key point: Loading positions) 3. At night, how does the slab curl and where is tension? A. Edges curl down, tension at the bottom B. Edges curl up, tension at the top C. No curling D. Edges curl down, tension at the top Answer: B. At night the top is cooler, edges curl upward, and the slab weight causes tension at the top. (Key point: Night warping) 4. How does frictional stress develop? A. From top-bottom temperature difference B. From average temperature change causing slab movement that is resisted by friction with the sub-base C. From tyre pressure D. From moisture in the subgrade Answer: B. Expansion or contraction of the whole slab is restrained by the sub-base, giving sigma_f = WLf/(2x10^4). (Key point: Cause of friction stress) 5. What is the nature and use of frictional stress? A. Large; fixes slab thickness B. Generally small; tensile in winter or at night, and it fixes spacing of contraction joints C. Always compressive; ignored D. Larger than load stress Answer: B. Friction is tensile on contraction and compressive in summer, generally small, and governs joint spacing. 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