R , 

 ri - -4^ tan 6~ 



I, = 



1 ' \l 



1 + — tan 6" 



1 -—■ (0.105) 



£ = TT-R 76 = 0.81(76) = 61.6 m (202 ft) 



^ ' 1+^(0.105) 



^2 = 0.81 i^ = 0.81(61.6) = 50 m (164 ft) 



£3 = 0.81 i^ = 0.81(50) = 40.5 m (132 ft) 



Using equation (5-13) 



R„ 



Si = I -v^ ;- Un = 1.81(76) = 137.6 m (451 ft) 



^ M + I tan 6 ' '^ 



S2 = 1.81 £j = 1.81(61.6) = 111.5 m (366 ft) 



and 



S3 = 1.81 ^2 = 1.81(50) = 90.5 m (297 ft) 



Using equations (5-6) and (5-9) as a check, on the above calculations, the 

 following is obtained 



Jlj^ = £^ - Sj^ tan 6 = 76 - 137.6 = 61.6 m 



and 



n - (H-'-^j «,1 - {^^^^) 2.0 - 137.6 „ 

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



f. Beach P rofiles Adjacent to Groins . Estimating the adjusted beach 

 alinement and determining the shape of the beach profiles adjacent to the 

 groin will permit the calculation of the differential soil loads on the 

 groin. The updrift side of the groin will have a higher sediment level than 

 the downdrift side. The profile, which is illustrated in Figure 5-20, is 

 drawn by the following steps: 



5-48 



