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<br />Fr.= V./(gYa)O.5 <br /> <br />v. = Q.lA. <br /> <br />Q. = flow cbstructed by the abutment and approach <br />embcnkment, <br /> <br />Y = depth of flow at the abutment <br />a <br /> <br />Problem 1. <br /> <br />Scour at right ba~k abutment, <br /> <br />a. Assume flow conditions in channel. <br /> <br />K, = O. 55 <br /> <br /> <br />K2 = (80/9(1)0.13 = 0.98 <br /> <br /> <br />a' = (52 X 2.0)/ 9,0 = 11.6 ft <br /> <br />v. = 70/(:,2 X 2,C) = 0,67 ft/s <br /> <br /> <br />Fr. = 0,67 / (32.2 X 9.0)0.5 = 0,04 <br /> <br />y.l9.0 = 2.27 :, 0,55 '? C.98 X <br /> <br />r> I. <br />( '1 ,. C " o. -~ " <br />~... . 0/ ... . ~ I <br /> <br />- .' <br />:'. ::;.,;, ~..." - 1 <br /> <br />Ys/9.0 = 1.19, Ys = 10,7 ft <br /> <br />This scour depth would place the total depth of scour at 18.0 ft <br />below the present stream bed (10,7+5,3+2,0), The last t'.vO terms <br />are the contraction scour and the long term degradation. <br /> <br />Comments <br />This is probably deE'per scour than will occur! The flow coming <br />around the abutment intersects the flow in the channel, causing <br />vortices, and will drill a hole but not this deep. The equations for <br />abutment scour give the worst case results. Also, the large depth is <br />caused by using the depth of flow of 9.0 ft at the toe of the <br />abutment. <br />What to do? <br />1. The scour depth \lould be between that calculated using the <br />overbank flow dept.h at the abutment (2 ft) and the channel flow <br />depth (9.0 ft) at the abutment. <br /> <br />2. To help in makinq a decision calculate abutment scour using the <br />overbank depth at the abutment. <br /> <br />68 <br />