Discussion/comments of «Wave-induced uplift pressure on berm revetment with Seabee slope” by Zijun Zhou, Yongping Chen, Yi Pan, Yusheng Zhen & Min Gan

ABSTRACT

The purpose of these comments and discussion has been to demonstrate how wave statistics can be incorporated into future applications of the proposed empirical equations to estimate the maximum relative uplift pressure under positive and negative freeboards on berm revetment with Seabee slope. This is demonstrated by using a joint distribution of significant wave height and surf parameter as well as by giving examples of the results.

KEYWORDS:

• Berm revetment
• Seabee slope
• uplift pressure
• surf parameter
• significant wave height
• wave statistics

1. Discussion

First, the discussers wish to compliment the authors, Zhou et al. (2020) (hereafter referred to as Z20), on their results proposing empirical equations to estimate the maximum relative uplift pressure under positive and negative freeboards on berm revetment with Seabee slope. The empirical equations were obtained as best fit to data from laboratory tests in model scale 1:14 representing irregular wave conditions. Overall, the uneven distribution of the uplift pressure on this revetment structure would affect its stability and cause damage. These comments and discussion have been written to demonstrate how wave statistics can be incorporated into future applications of the authors’ empirical equations.

Z20 defined the dimensionless parameter

(1) $p_{1\mathrm{\%}}^{\ast }=\frac{p_{1\mathrm{\%}}}{\rho g{H}_s}$(1)

representing the 1% relative uplift pressure, where $p_{1\mathrm{\%}}$ is the uplift pressure level exceeded by 1% of the measured time series, $\rho$ is the fluid density, g is the acceleration due to gravity, and $H_s$ is the significant wave height. The following empirical equations to estimate the maximum relative freeboard on berm revetment with Seabee slope were proposed:

(2) $p_{1\mathrm{\%}}^{\ast }=\gamma a{\xi }_p^{b}exp(c{\xi }_p)$(2)

For negative freeboard:

(3) $\gamma =1.172-(3.623\frac{R_b}{H_s}+0.6836{)}^2;-0.36\le \frac{R_b}{H_s}\le -0.06$(3)

(4) $(a,b,c)=(0.5214,1.1,-0.2027);1.38\le {\xi }_p\le 2.78$(4)

For positive freeboard:

(5) $\gamma =0.97+0.009{\left(\frac{R_b}{H_s}\right)}^{-1};0.09\le \frac{R_b}{H_s}\le 0.56$(5)

(6) $(a,b,c)=(0.76,-0.78,0.31);1.3\le {\xi }_p\le 2.8$(6)

where $\gamma$ is the influence parameter of the elevation of the berm expressed in terms of the relative freeboard $R_b/H_s$, and $R_b$ is the freeboard of the berm. Moreover, ${\xi }_p$ is the surf parameter defined as ${\xi }_p=tan\theta /\sqrt{H_s/{\lambda }_p}$ where $tan\theta$ is the slope with an angle $\theta$ to the horizontal, ${\lambda }_p=(g/2\pi )T_p^2$ is the deep water spectral wave length, $T_p$ is the spectral period, and $H_s$ is the deep water significant wave height. Equations (1(1) $p_{1\mathrm{\%}}^{\ast }=\frac{p_{1\mathrm{\%}}}{\rho g{H}_s}$(1) ) to (6(6) $(a,b,c)=(0.76,-0.78,0.31);1.3\le {\xi }_p\le 2.8$(6) ) were obtained as best fit to data obtained in physical model tests performed in a wave flume using irregular waves in model scale 1:14 (see Z20 for more details).

The statistical features of $p_{1\mathrm{\%}}^{\ast }$ (and thus of $p_{1\mathrm{\%}}$) will be exemplified based on wave statistics obtained from the joint probability density function (pdf) of $H_s$ and ${\xi }_p$ provided in the Appendix, which is based on data from wave measurements made in the Northern North Sea during a 29 years period. Here, the conditional expected value of $p_{1\mathrm{\%}}^{\ast }$ given $H_s,E\left[p_{1\mathrm{\%}}^{\ast }|H_s\right]$ and the conditional variance of $p_{1\mathrm{\%}}^{\ast }$ given $H_s$, $Var\left[p_{1\mathrm{\%}}^{\ast }|H_s\right]$ are considered.

Based on Equations (4)(4) $(a,b,c)=(0.5214,1.1,-0.2027);1.38\le {\xi }_p\le 2.78$(4) and (6(6) $(a,b,c)=(0.76,-0.78,0.31);1.3\le {\xi }_p\le 2.8$(6) ), $p_{1\mathrm{\%}}^{\ast }$ is valid within a finite interval, and thus, the conditional pdf of ${\xi }_p$ given $H_s$, $f({\xi }_p|H_s)$ in Equation (A2)(A2) $f({\xi }_p|H_s)=\frac{1}{\sqrt{2\pi }\sigma {\xi }_p}exp\left[-\frac{{(ln{\xi }_p-\mu )}^2}{2{\sigma }^2}\right]$(A2) in the Appendix follows the truncated lognormal pdf

(7) $f_t({\xi }_p|H_s)=\frac{1}{N}f({\xi }_p|H_s);{\xi }_{P1}=1.3\le {\xi }_p\le {\xi }_{P2}=2.8$(7)

(8) $N=\mathrm{\Phi }\left[\frac{ln{\xi }_{p2}-\mu }{\sigma }\right]-\mathrm{\Phi }\left[\frac{ln{\xi }_{P1}-\mu }{\sigma }\right]$(8)

where $\mathrm{\Phi }$ is the standard Gaussian cumulative distribution function (cdf) (where $\mu$ and ${\sigma }^2$ are the mean value and the variance, respectively, of $ln{\xi }_p$) given by

(9) $\mathrm{\Phi }(x)=\frac{1}{\sqrt{2\pi }}\underset{-\mathrm{\infty }}{\overset{x}{\int }}{e}^{-t^2/2}dt$(9)

It should be noted that the interval limits ${\xi }_{P1}$ and ${\xi }_{P2}$ in Equation (7)(7) $f_t({\xi }_p|H_s)=\frac{1}{N}f({\xi }_p|H_s);{\xi }_{P1}=1.3\le {\xi }_p\le {\xi }_{P2}=2.8$(7) are based on the results depicted in Fig. 13 of Z20 (i.e. slightly extending the interval in Equation (4)(4) $(a,b,c)=(0.5214,1.1,-0.2027);1.38\le {\xi }_p\le 2.78$(4) ). Further details of $f({\xi }_p|H_s)$ are given in Equations (A2(A2) $f({\xi }_p|H_s)=\frac{1}{\sqrt{2\pi }\sigma {\xi }_p}exp\left[-\frac{{(ln{\xi }_p-\mu )}^2}{2{\sigma }^2}\right]$(A2) ) to (A6(A6) $(b_1,b_2,b_3)=(0.001,0.097,-0.255)$(A6) ) in the Appendix.

First, define

(10) $P({\xi }_p|H_s)\equiv \frac{p_{1\mathrm{\%}}^{\ast }}{\gamma }=a{\xi }_p^{b}exp(c{\xi }_p)$(10)

Then, the conditional expected value of P given $H_s$, $E\left[P|H_s\right]$ and the conditional variance of P given $H_s$, $Var\left[P|H_s\right]$ are calculated from the truncated pdf of ${\xi }_p$ given $H_s$ in Equations (7)(7) $f_t({\xi }_p|H_s)=\frac{1}{N}f({\xi }_p|H_s);{\xi }_{P1}=1.3\le {\xi }_p\le {\xi }_{P2}=2.8$(7) and (8(8) $N=\mathrm{\Phi }\left[\frac{ln{\xi }_{p2}-\mu }{\sigma }\right]-\mathrm{\Phi }\left[\frac{ln{\xi }_{P1}-\mu }{\sigma }\right]$(8) ) as (Bury 1975)

(11) $E\left[P|H_s\right]=\underset{{\xi }_{P1}}{\overset{{\xi }_{P2}}{\int }}P\left({\xi }_p|H_s\right)f_t\left({\xi }_p|H_s\right)d{\xi }_p$(11)

(12) $Var\left[P|H_s\right]=E\left[{(P({\xi }_p|H_s))}^2\right]-{\left(E\left[P|H_s\right]\right)}^2$(12)

where

(13) $E\left[{(P({\xi }_p|H_s))}^2\right]=\underset{{\xi }_{P1}}{\overset{{\xi }_{P2}}{\int }}{\left(P({\xi }_p|H_s)\right)}^2{f}_t\left({\xi }_p|H_s\right)d{\xi }_p$(13)

The conditional coefficient of variation is

(14) $R\left[P|H_s\right]=\frac{{(Var\left[P|H_s\right])}^{1/2}}{E\left[P|H_s\right]}$(14)

Figures 1 and 2 depict $E\left[P|H_s\right]$ (Figure 1) and $R\left[P|H_s\right]$ (Figure 2) for the negative and positive freeboards versus $H_s$ in the range 2– 3 m. This interval of $H_s$ includes the full scale values $H_s$= 2.1 m, 2.5 m, 2.8 m corresponding to the model scale (1:14) values $H_s$= 0.15 m, 0.18 m, 0.20 m, respectively (see Z20). Here and in the following examples, the slope $tan\theta$ is taken as 1:3 (as in Z20).

Figure 1. $E\left[P|H_s\right]$ versus $H_s$ for negative and positive freeboards with slope 1:3.

The positive and negative freeboard ranges represent the Z20 data.

Figure 2. $R\left[P|H_s\right]$ versus $H_s$ for negative and positive freeboards with slope 1:3.

From Figure 1, it appears, as $H_s$ increases from 2 m to 3 m, that: $E\left[P|H_s\right]$ increases from about 0.84 to about 0.88 for positive freeboard; and decreases from about 0.71 to about 0.62 for negative freeboard. The Z20 data representing the full scale values $H_s$= 2.1 m, 2.5 m, 2.8 m corresponding to ${\xi }_p$ = 2.04, 1.86, 1.77, respectively, using $T_p=1.9\sqrt{14}s=7.1s$, are also shown for the positive and negative freeboard ranges taken from Fig. 13 in Z20. It appears that the two curves are within the ranges of the corresponding data.

From Figure 2, it appears, as $H_s$ increases from 2 m to 3 m, that: $R\left[P|H_s\right]$ varies slightly from about 0.038 to about 0.035 with a maximum of about 0.039 for $H_s$ = 2.3 m for positive freeboard; and decreases from about 0.133 to about 0.105 for negative freeboard.

Furthermore, examples of physical results are provided for negative and positive freeboards using the results in Figures 1 and 2. Combination of Equations (1)(1) $p_{1\mathrm{\%}}^{\ast }=\frac{p_{1\mathrm{\%}}}{\rho g{H}_s}$(1) , (10(10) $P({\xi }_p|H_s)\equiv \frac{p_{1\mathrm{\%}}^{\ast }}{\gamma }=a{\xi }_p^{b}exp(c{\xi }_p)$(10) ) and (11(11) $E\left[P|H_s\right]=\underset{{\xi }_{P1}}{\overset{{\xi }_{P2}}{\int }}P\left({\xi }_p|H_s\right)f_t\left({\xi }_p|H_s\right)d{\xi }_p$(11) ) yields

(15) $E\left[p_{1\mathrm{\%}}|H_s\right]=\rho g{H}_s\gamma E\left[P|H_s\right]$(15)

First, for negative freeboard:

Taking $R_b/H_s=-0.25$ in Equation (3)(3) $\gamma =1.172-(3.623\frac{R_b}{H_s}+0.6836{)}^2;-0.36\le \frac{R_b}{H_s}\le -0.06$(3) gives $\gamma =1.12$ which substituted in Equation (15)(15) $E\left[p_{1\mathrm{\%}}|H_s\right]=\rho g{H}_s\gamma E\left[P|H_s\right]$(15) for $H_s$= 2.5 m, $\rho$= 1025 kg/m³ and $E\left[P|H_s=2.5m\right]=0.656$ (see Figure 1) yields

(16) $E\left[p_{1\mathrm{\%}}|H_s=2.5m\right]=1025\times 9.81\times 2.5\times 1.12\times 0.656=1.85\times {10}^4\frac{N}{m^2}$(16)

Furthermore, now $R\left[p_{1\mathrm{\%}}|H_s=2.5m\right]=R\left[P|H_s=2.5m\right]=0.124$ (see Figure 2), and thus $E\left[p_{1\mathrm{\%}}|H_s=2.5m\right]$ plus and minus one standard deviation of 12.4% become $2.08\times {10}^{4}N/m^2$ and $1.62\times {10}^{4}N/m^2$, respectively.

Second, for positive freeboard:

Taking $R_b/H_s=0.25$ in Equation (5)(5) $\gamma =0.97+0.009{\left(\frac{R_b}{H_s}\right)}^{-1};0.09\le \frac{R_b}{H_s}\le 0.56$(5) gives $\gamma =1.01$ which substituted in Equation (15)(15) $E\left[p_{1\mathrm{\%}}|H_s\right]=\rho g{H}_s\gamma E\left[P|H_s\right]$(15) for $E\left[P|H_s=2.5m\right]=0.862$ (see Figure 1) yields

(17) $E\left[p_{1\mathrm{\%}}|H_s=2.5m\right]=1025\times 9.81\times 2.5\times 1.01\times 0.862=2.19\times {10}^4\frac{N}{m^2}$(17)

Furthermore, now $R\left[p_{1\mathrm{\%}}|H_s=2.5m\right]=R\left[P|H_s=2.5m\right]=0.039$ (see Figure 2), and thus $E\left[p_{1\mathrm{\%}}|H_s=2.5m\right]$ plus and minus one standard deviation of 3.9% become $2.28\times {10}^{4}N/m^2$ and $2.10\times {10}^{4}N/m^2$, respectively.

Thus, as expected, these examples demonstrate that the wave-induced uplift pressure on berm revetment with Seabee slope is larger under positive freeboard than under negative freeboard.

In future applications of the empirical equations to estimate the maximum relative uplift pressure under positive and negative freeboards on berm revetment with Seabee slope proposed by Zhou et al. (2020), it should be considered to implement the statistical properties of waves as demonstrated in this discussion, although the proposed equations and parameters may be subject to regional limitations. The utilization of these empirical equations jointly with wave statistics should represent an important component in, for instance, risk analysis used as part of revetment structure design. It should be noted, however, that implementation of wave statistics is not limited to the joint distribution of significant wave height and surf parameter applied here.

Disclosure statement

No potential conflict of interest was reported by the author(s).

Appendix.

Joint pdf of $H_s$ and ${\xi }_P$

Here, the joint pdf of $H_s$ and ${\xi }_P$ provided by Myrhaug and Fouques (2010) (hereafter referred to as MF10) is used as an example. This $(H_s,{\xi }_p)$ distribution was deduced based on a joint pdf of $H_s$ and $T_p$ obtained as a best fit to data from wave measurements made in the Northern North Sea during a 29-year period. The joint pdf of $H_s$ and ${\xi }_p$ is given as

(A1) $f(H_s,{\xi }_p)=f({\xi }_p|H_s)p(H_s)$(A1)

where $f(H_s)$ is the marginal pdf of $H_s$ given by the combined lognormal and Weibull distribution given in Equations (2(2) $p_{1\mathrm{\%}}^{\ast }=\gamma a{\xi }_p^{b}exp(c{\xi }_p)$(2) –4(4) $(a,b,c)=(0.5214,1.1,-0.2027);1.38\le {\xi }_p\le 2.78$(4) ) in MF10. The conditional pdf of ${\xi }_p$ given $H_s$ is given by the following lognormal distribution

(A2) $f({\xi }_p|H_s)=\frac{1}{\sqrt{2\pi }\sigma {\xi }_p}exp\left[-\frac{{(ln{\xi }_p-\mu )}^2}{2{\sigma }^2}\right]$(A2)

where $\mu$ and ${\sigma }^2$ are the mean value and the variance, respectively, of $ln{\xi }_p$, given by

(A3) $\mu =ln\left(\frac{tan\theta }{{(H_s/(g/2\pi ))}^{1/2}}\right)+a_1+a_2{H}_s^{a_3}$(A3)

(A4) ${\sigma }^2=b_1+b_2{e}^{b_3{H}_s}$(A4)

with

(A5) $(a_1,a_2,a_3)=(1.780,0.288,0.474)$(A5)

(A6) $(b_1,b_2,b_3)=(0.001,0.097,-0.255)$(A6)

where $H_s$ is in meters in Equations (A3)(A3) $\mu =ln\left(\frac{tan\theta }{{(H_s/(g/2\pi ))}^{1/2}}\right)+a_1+a_2{H}_s^{a_3}$(A3) and (A4(A4) ${\sigma }^2=b_1+b_2{e}^{b_3{H}_s}$(A4) ) (see MF10 for more details).