--- title: "Dowel bar length and spacing numerical" subject: "Transportation Engineering II" chapter: 2 key: "semester_7/transportation_engineering_ii:dowel-bar-design-expansion-joint-25mm" source: https://www.bctengineer.com/q/semester_7/transportation_engineering_ii/dowel-bar-design-expansion-joint-25mm site: BCT Engineering --- # Dowel bar length and spacing numerical Transportation Engineering II · Chapter 2: Highway Pavement - Asked in: 2081 Bhadra, 8 marks - Video: [Design of Dowel Bar || Pavement Engineering || Transportation Engineering](https://www.youtube.com/watch?v=sXqccs-FcRE) — Reinforce Engineering - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_7/transportation_engineering_ii/dowel-bar-design-expansion-joint-25mm - Whole chapter: https://www.bctengineer.com/md/semester_7/transportation_engineering_ii/chapter-2 ## Question Calculate the length and spacing of dowel bars of dia 25mm at an expansion joint of concrete pavement of thickness 25 cm. Design wheel load 5000 kg. Load capacity of the dowel system is 40 percent of design wheel load. Joint width is 2.0 cm and the permissible stress in shear, bending and bearing stress in dowel bars are 1000, 1400 and 100 kg/cm² respectively. Modulus of elasticity of concrete 2.45 x 10⁵ kg/cm², modulus of subgrade reaction = 8 kg/cm³, Poisson ratio = 0.15. ## Short answer Bradbury's method gives balanced embedment 40.5 cm, so dowel length 45 cm. Capacity per dowel is 699 kg (bending governs). With group factor 2.86, safe spacing is 30 cm c/c (capacity 2037 kg > needed 2000 kg). First dowel 15 cm from edge. ## Answer #### Given data Dowel diameter $d=2.5$ cm, slab $h=25$ cm, joint width $\delta=2.0$ cm, wheel load $P=5000$ kg, load transferred by dowels $=0.4P=2000$ kg, $F_s=1000$, $F_f=1400$, $F_b=100$ kg/cm2, $E=2.45\times10^5$ kg/cm2, $k=8$ kg/cm3, $\mu=0.15$. (Bradbury's method as in Khanna and Justo; IRC:58 detailing.) #### Step 1 - Radius of relative stiffness $$ l=\left[\frac{Eh^3}{12(1-\mu^2)k}\right]^{1/4}=\left[\frac{2.45\times10^5\times25^3}{12(1-0.15^2)\times8}\right]^{1/4}=79.9\ \text{cm} $$ #### Step 2 - Capacity of one dowel - Shear: $P_s=0.785\,d^2F_s=0.785\times2.5^2\times1000=4906$ kg - Bending: $P_f=\dfrac{2d^3F_f}{L_d+8.8\delta}=\dfrac{2\times2.5^3\times1400}{L_d+17.6}=\dfrac{43750}{L_d+17.6}$ - Bearing: $P_b=\dfrac{F_bL_d^2d}{12.5(L_d+1.5\delta)}=\dfrac{100\times2.5\,L_d^2}{12.5(L_d+3)}=\dfrac{20L_d^2}{L_d+3}$ #### Step 3 - Length of dowel Equating $P_f=P_b$ and solving by trial gives the balanced embedment length $L_d=40.5$ cm. Total length $= L_d+\delta = 40.5+2.0 = 42.5$ cm, so adopt a **dowel length of 45 cm** (within the usual 45-50 cm). Checking with $L_d=45$ cm (conservative, since bending governs and $P_f$ falls as $L_d$ rises): $P_f=699$ kg, $P_b=844$ kg, $P_s=4906$ kg. The least value governs: $P_d=699$ kg per dowel. #### Step 4 - Spacing Required capacity of the dowel group $=2000$ kg, so the capacity factor (number of dowels in terms of the dowel under the load) must be $\ge 2000/699=2.86$. The critical position is the wheel load at the slab edge, where dowels lie on one side only. The dowel under the load takes its full capacity, and one at distance $x$ takes $(1-x/1.8l)$ of it, up to $1.8l=143.9$ cm. For spacing $s$: $\Sigma=1+\sum_j(1-js/1.8l)$. | Spacing s (cm) | $\Sigma$ | $P_d\Sigma$ (kg) | Check vs 2000 kg | |---|---|---|---| | 20 | 4.11 | 2871 | safe | | 25 | 3.39 | 2372 | safe | | 30 | 2.92 | 2037 | safe | | 35 | 2.57 | 1795 | not safe | | 40 | 2.33 | 1631 | not safe | The largest safe spacing is 30 cm, which also matches IRC:58 practice; adopt **30 cm centre to centre** ($\Sigma=2.92$, capacity $=2037$ kg $>2000$ kg). **Answer:** 25 mm diameter dowels, length 45 cm, spacing 30 cm c/c (first dowel 15 cm from the slab edge), smooth bars with one half greased and capped at the free end. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Which dowel capacity governs the load transfer per dowel? A. The largest of shear, bending and bearing B. The least of shear, bending and bearing C. Always shear D. The average of the three Answer: B. The weakest mode governs; here bending gives 699 kg. (Key point: Governing capacity) 2. How is dowel length found? A. Equating bending and bearing capacities and solving by trial (about 40.5 cm), then rounding up to 45 cm B. Equal to the slab thickness C. Taking 10 times the diameter D. Equal to joint width Answer: A. Pf = Pb gives about 40.5 cm; 45 cm is adopted. (Key point: Dowel length) 3. The shear capacity of one 2.5 cm dowel with Fs = 1000 kg/cm2 is: A. 0.785 x 2.5^2 x 1000 = 4906 kg B. 2.5 x 1000 = 2500 kg C. 0.4 x 5000 kg D. 699 kg Answer: A. Ps = 0.785 d^2 Fs = 4906 kg. (Key point: Shear capacity) 4. How much load must the dowel group carry? A. 5000 kg B. 40 percent of 5000 = 2000 kg C. 100 kg D. 699 kg Answer: B. The dowel system capacity is 40 percent of the design wheel load. (Key point: Load to be transferred) 5. The dowel at distance x from the load takes what fraction of the dowel under the load? A. x / 1.8l B. 1 - x/1.8l C. 1 + x/1.8l D. Equal share Answer: B. Dowels share load linearly reducing to zero at 1.8 l, so the factor is (1 - x/1.8l). 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