--- title: "Temperature stresses in rigid pavement; dowel bar design" subject: "Transportation Engineering II" chapter: 2 key: "semester_7/transportation_engineering_ii:temperature-stress-rigid-dowel-bar-procedure" source: https://www.bctengineer.com/q/semester_7/transportation_engineering_ii/temperature-stress-rigid-dowel-bar-procedure site: BCT Engineering --- # Temperature stresses in rigid pavement; dowel bar design Transportation Engineering II · Chapter 2: Highway Pavement - Asked in: 2078 Bhadra, 8 marks - Video: [Temperature stresses in rigid pavement](https://www.youtube.com/watch?v=ZcagKyhc4-Y) — Online Study - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_7/transportation_engineering_ii/temperature-stress-rigid-dowel-bar-procedure - Whole chapter: https://www.bctengineer.com/md/semester_7/transportation_engineering_ii/chapter-2 ## Question Explain the temperature stresses in rigid pavement. Write down the design procedure of dowel bar in expansion joints. ## Short answer Temperature stresses include warping stress from top–bottom temperature differential, frictional stress from slab movement restrained by sub-base, and expansion stress if joints are blocked. Dowel bar design finds the least of shear, bending and bearing capacities; the embedded length is set by equating bending and bearing, total length is Ld+δ, and spacing ensures sum of (1 – x/1.8l) contributions ≥ 0.4P with one end greased or capped. ## Answer #### Temperature stresses in rigid pavement Temperature stresses in a concrete slab are of three kinds: 1. **Warping stress** - arises from the difference of temperature between the top and bottom of the slab (day: top hotter, bottom in tension; night: top in tension). The curling is resisted by the self-weight of the slab. Bradbury's formula for the edge is $$ \sigma_{te}=\frac{C\,E\,\alpha\,\Delta T}{2} $$ where $C$ is the coefficient from $L/l$ (chart), $E$ modulus of concrete, $\alpha$ coefficient of thermal expansion ($10\times10^{-6}$ per $^\circ$C) and $\Delta T$ the temperature differential across the slab. At the corner, $\sigma_{tc}=\dfrac{E\alpha\Delta T}{3(1-\mu)}\sqrt{\dfrac{a}{l}}$. 2. **Frictional stress** - due to overall expansion or contraction of the slab restrained by the sub-base: $$ \sigma_f=\frac{WLf}{2\times10^4} $$ 3. **Expansion stress** - if free movement is blocked at joints (for example by sand or stones in joints), a compressive stress $E\alpha\,\Delta t$ develops and the slab may buckle or blow up, so expansion joints and sealants are provided. Critical combinations: at the edge in summer (midday) $\sigma_e+\sigma_{te}-\sigma_f$; at the edge in winter (midday) $\sigma_e+\sigma_{te}+\sigma_f$; and at the corner at night $\sigma_c+\sigma_{tc}$ (friction is negligible at the corner). #### Design procedure of dowel bars at expansion joints Dowel bars transfer load across the expansion joint (smooth round mild steel bars, one end greased or sleeved with a cap). Design procedure (Bradbury, as in Khanna and Justo, with IRC:58 detailing): 1. **Given data** - wheel load $P$, joint width $\delta$, dowel diameter $d$, permissible stresses in shear ($F_s$), bending ($F_f$) and bearing ($F_b$), slab thickness, $E$, $k$, $\mu$. 2. **Radius of relative stiffness** $l=\left[\dfrac{Eh^3}{12(1-\mu^2)k}\right]^{1/4}$. 3. **Load to be transferred** by the dowel system is taken as 40 percent of the design wheel load. 4. **Capacity of a single dowel** (least of the three governs): - Shear: $P_s=0.785\,d^2F_s$ - Bending: $P_f=\dfrac{2d^3F_f}{L_d+8.8\delta}$ - Bearing: $P_b=\dfrac{F_b\,L_d^2\,d}{12.5\,(L_d+1.5\delta)}$ 5. **Length of dowel**: the embedded length $L_d$ is found by equating the bending and bearing capacities (by trial); the total bar length is $L_d+\delta$, rounded up to a practical value. 6. **Spacing** - the dowel under the load transfers its full capacity; a dowel at distance $x$ transfers the proportion $(1-x/1.8l)$ and dowels beyond $1.8l$ transfer nothing. For a trial spacing $s$ add the proportions on both sides: $\Sigma = 1+2\sum(1-ks/1.8l)$. 7. **Check** $P_d\times\Sigma\ge 0.4P$; adjust spacing (usually 30-45 cm) until satisfied. 8. **Detail** the bar: one half coated with bitumen or greased, and a cap leaving an expansion gap at the free end. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. During the day, where does tension develop in the slab because of warping? A. At the top B. At the bottom C. At the neutral axis only D. Nowhere, warping gives compression only Answer: B. By day the top is hotter, edges curl down, giving compression at the top and tension at the bottom; at night tension is at the top. (Key point: Daytime warping) 2. Which set lists the three temperature stresses in a rigid pavement? A. Warping, frictional and expansion stresses B. Shear, bearing and bending stresses C. Interior, edge and corner stresses D. Tensile, rupture and fatigue stresses Answer: A. The answer names warping, frictional and expansion stresses. (Key point: Kinds of temperature stress) 3. Why are expansion joints provided? A. To increase the modulus of rupture B. To avoid compressive stress E alpha dt when expansion is blocked, which could cause buckling or blow-up C. To reduce the CBR requirement D. To carry the tyre pressure Answer: B. If joints are blocked by sand or stones, expansion builds compression and the slab may buckle. (Key point: Blow-up) 4. In the dowel design, what part of the design wheel load must the dowel system transfer? A. 10 percent B. 40 percent C. 75 percent D. 100 percent Answer: B. The dowel system is designed for 40 percent of the wheel load. (Key point: Load to be transferred) 5. How is the capacity of a single dowel fixed? A. Sum of shear, bending and bearing capacities B. The least of the shear, bending and bearing capacities C. The shear capacity only D. The average of the three Answer: B. The weakest mode governs, so the least of the three is used. (Key point: Dowel capacity) 6. A dowel at distance x from the load transfers what proportion, and what is the limit? A. (1 - x/1.8 l); dowels beyond 1.8 l transfer nothing B. x/l; no limit C. (1 - x/l); dowels beyond l transfer nothing D. Always 50 percent Answer: A. Load transfer falls linearly to zero at 1.8 l; the sum of proportions times single-dowel capacity must be at least 0.4P. 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