A new mathematical model to calculate the equilibrium scour depth around a pier

Abstract

This paper sheds light on the formulation of a new equilibrium local scour depth equation around a pier. The total bed materials removed from the scour hole due to the force exerted by the flowing fluid after colliding with the pier in the flow field are estimated. At the equilibrium condition, the shape of the scour hole around the pier may take any form, viz. linear, circular, parabolic, triangular, or combination of different shapes. To consider that, two functions are assumed at the stoss and the lee sides of the pier. The total volume of bed materials removed from the scour hole of an arbitrary shape at the stoss and the lee sides of the pier is obtained by integrating the two functions. The equilibrium scour depth is formed by applying the energy balance theorem. An example problem is illustrated and the results are compared with the equations presented by Melville and Coleman (Bridge scour. Water Resources Publication, Colorado, 2000) and HEC-18 (Richardson and Davis in Evaluating scour at bridges, HEC-18. Technical report no. FHWA NHI, 2001).

Appendix: Calculation of scour depth

The detailed calculation procedure of the scour depth given by Melville and Coleman (2000) and HEC-18 (Richardson and Davis 2001) can be found in Dey (2014).

Melville and Coleman

Velocity of incoming flow, $u = \frac{q}{h} = \frac{10}{5} = 2\hbox { ms}^{-1}$

Threshold shear velocity, $u_{*c}$ , and incoming flow, $u_{cr}$ , are estimated as follows:

$u_{*c}$ $(0.1\le d_{50}<1 \hbox { mm}) = 0.0115 + 0.0125$ $d_{50}^{1.4} = 0.02 \hbox { ms}^{-1}$

$u_{cr} = u_{*c} 5.75 \log \left( 5.53 \frac{h}{d_{50}} \right) = 0.52\, \hbox {ms}^{-1}$

For uniform sediment, $u_{a} = u_{cr}$.

k-factor

For $\frac{b}{h} = \frac{1.5}{5} = 0.3 < 0.7, k_{h} = 2.4 b = 3.6\hbox { m}$

For $\frac{u - u_{a}-u_{cr}}{u_{cr}} = 3.84 > 1$, $k_{I} = 1$

For $\frac{b}{d_{50}} = 1875 > 1$, $k_{d} = 1$

For a circular pier, $k_\mathrm{s} = 1$

For the projected width, $b_\mathrm{p} = L \hbox {sin}\alpha + b \hbox {cos}\alpha = 3\hbox { m}$

$k_{\alpha } = \left( \frac{b_\mathrm{p}}{b}\right) ^{0.65} = 1.569$

For equilibrium scour depth, ($t = t_\mathrm{e}$), $k_{t} = 1$

Now, scour depth $d_\mathrm{s} = k_{h}k_{I}k_{d}k_\mathrm{s}k_{\alpha }k_{t} = 5.65\hbox { m}$

HEC-18

For a circular pier, $k_\mathrm{s} = 1$

For $\frac{L}{b} = 4$ and $\alpha = 15^{0}$, $k_{\alpha } = 1.5$

Assuming small dunal bed form and Froude number, $F_{r} = \frac{2}{\sqrt{9.8 \times 5}} = 0.28$, $k_\mathrm{bed} = 1.1$

For $d_{50} < 2\hbox { mm}$, $k_{a} = 1$

Now, scour depth

$d_{s} = bk_\mathrm{s}k_{\alpha }k_\mathrm{bed}k_{a}\left( \frac{h}{b}\right) ^{0.35} F_{r}^{0.43} = 2.2\hbox { m}$

Present equation

Considering one-sixth power law, $(m = 6)$

For p = 1, q = 1

$$\begin{aligned} A & = \left( \frac{\pi p}{2+p}\right) \left( 1-\theta \right) \left( \gamma _\mathrm{s} -\gamma \right) = 5771.32\\ B& = \left( \frac{\pi q}{2+q}\right) \left( 1-\theta \right) \left( \gamma _\mathrm{s} -\gamma \right) = 5771.32\\ C&= \frac{\rho u_\mathrm{c}^{2}hb }{2} = 15{,}000\\ E&= \left( \rho u_\mathrm{c}^{2}hb \right) \left( \frac{2+2m}{2+3m}h \right) = 105{,}000\\&5771.32d^{1.5}_\mathrm{s} + 5771.32d^{1.5}_{s} - 15{,}000d_{s} - 105{,}000 = 0 ,\\ d_\mathrm{s}&= 6.88\hbox { m} \end{aligned}$$

Similarly, for p = 1, q = 2

$$\begin{aligned} A &= 5771.32, B = 8656.98, C = 15{,}000,\hbox { and }E = 105{,}000;\\&d_\mathrm{s} = 3.72\hbox { m} \end{aligned}$$

For p = 2, q = 1

$$\begin{aligned} A&= 8656.98, B = 5771.32, C = 15{,}000,\hbox { and } E = 105{,}000; \\&d_\mathrm{s} = 3.72\hbox { m} \end{aligned}$$

For p = 2, q = 2

$$\begin{aligned} A&= 8656.98, B = 8656.98, C = 15{,}000,\hbox { and } E = 105{,}000; \\&d_\mathrm{s} = 2.93\hbox { m} \end{aligned}$$