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



EXAMPLE PROBLEM 25************* 



(This example is taken from an actual situation which occurred at Seward, 

 Alaska, in 1964; see Wilson and T0rum, 1968.) 



GIVEN : A 104.5-metric ton (230,000 pounds) railroad locomotive was over- 

 turned by a tsunami surge. The surge was assumed to have a depth of 

 1.83 meters. The clear space under the locomotive was approximately 

 0.91 meter (3 feet) and the length of the locomotive body was 12.5 

 meters (41 feet). The width between the rails was 1.52 meters (5 feet) 

 and the width of the locomotive body was 3.05 meters (10 feet). The 

 surge was assumed to act normal to the side of the locomotive. 



FIND : The overturning force on the locomotive. 



SOLUTION : The buoyant force is given by equation (324) as 



F g = pg V = 1, 026(9. 81) (1. 83 - 0.91) (3.05) (12.5) 

 F D = 3.53 x 10 5 newtons (78,700 pounds) 



D 



As indicated previously, the coefficient of drag can be determined by 

 doubling the wetted height and assuming both underflow and overflow 

 for a flat surface 1.83 meters high and 12.5 meters long. Interpolating 

 in Table 6 for a flat plate for L/d = 6.8 gives 



C D = 1.24 



The velocity can be obtained from equation (318), so for h = 1.83 meters 



u = 2/gh = 2/9.81(1.83) = 8.47 meters per second 



From equation (336) , the drag force is 



v d - p S A T 



¥ D = 1, 026(1. 24) (1. 83 - 0.91)(12.5) ( 8 ' 47) 



F = 5.24 x 10 5 newtons (1.17 x 10 5 pounds) 



which will act against the side of the locomotive at a distance, Z, 

 above the ground, given as 



Z = 0.91 + C1 - 83 - °- 91 ) 



2 



Z = 1.37 meters (4.5 feet) 



181 



