and from Figure 21, for 



-2_ = 0.0038 , 

 gT 2 



The runup is 



R = (h) (H ^ = ( 3 -°)( 3 - 05 ) 



R = 9.15 meters (30.0 feet) 



(See Sec. VI for the appropriate scale-effect correction factor.) 

 ************************************ 



************* EXAMPLE PROBLEM 6************** 



GIVEN : Conditions are similar to example problem 5 with one exception. 

 An impermeable, smooth, 1 on 2 structure is fronted by a 1 on 10 

 beach slope. The beach slope extends seaward to a depth of 15.0 

 meters beyond which the slope is approximately 1 on 100. The 

 design wave approaches normal to the structure, and has a height 

 of H = 2.8 meters and period of T = 9.0 seconds, measured at a 

 depth of 16 meters. The exception is that the structure is located 

 at the waterline; i.e., d s = 0. 



FIND : Determine the height of wave runup. 



SOLUTION : From example problem 5, 



H^ =3.05 meters 



ti o 



-=V = 0.00384 . 



gT 2 



However, dg = 0; d^/H^ = 0. To enable determination of runup, the 

 depth at the toe of the beach slope (d = 15.0 meters) is used. 



d 15 = 4.92- 5.0. 



Because the slope length is longer than in example problem 5, 

 i.e., I = (15-0) 10 = 150.0 meters (492.0 feet), then 



I 



L > °' 5 - 



56 



