Dowel bar length and spacing numerical

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 $s$: $\Sigma=1+\sum_j(1-js/1.8l)$.

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.