the structure is a rigid, thin barrier, calculate L in a depth 

 dg = 15 ft. using Equation 7-10. 



L = ^— = 5.12(8)2 ^ 328 ft., 

 " 27r 

 and 



^5 15 



— = = 0.0458 . 



Lc 328 



From Table C-1, Appendix C, setting dg = d, 



d 47rd , /47rd\ 



- = 0.0896, = 1.127, sinh = 1.378, 



L L \ L / 



and calculate, 



47rh h /47rd\ 10 



= - = — (1.127) = 0.751 . 



L d \ L / 15 



Find 0.751 in Column 9 of Table C-1, Appendix C and read sinh (4Trd/L) 

 from Column 10. Hence, 



. /47rh\ 

 sinh I = 0.825 . 



From Equation 7-10, 



Hj I (47rh/L) + sinh (47rh/L) ' 



"^^V " sinh(47rd,^^)+(47rd,/^)' 



_ (0.751) + (0.825) 

 H. M (1.378) + (1.127) ' 



1.576 



, 1 = x/0.370 = 0.609 



H. V 2.505 ^ 



The transmitted wave height is 



H^ = 0.609(6) = 3.67 ft., 

 say 



H^ = 3.5 ft. 



The calculated value can be assumed to exceed the true value, since the 

 finite structure crest width will decrease the transmitted wave height 

 and some energy dissipation will occur. 



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7-61 



