The vertical height to the center of submerged weight, 



z = ((3.42) * (135) + (0) * (905)) ^ 1040 = 0.44 feet 

 csw 



M = z * F, + z * F. = (3.42) * (410) + (0.44) * (109) = 

 did csw l 



1450 foot-pounds (7) 



For a crossed strip footing 



2 

 F = W , -r (2 * L - B) + (12 * M,) ■*• L = 128 pounds per foot (11) 

 sub d 



Allowable for B = 1.5 feet is only 48 pounds per foot (from Figure 3a); 



therefore, increase B while holding W , constant or use tiedowns. 



sub 



Solution 1 - Using Tiedowns 



A tiedown would be required on each end of both footings and would 

 need a capacity of, 



F = M, * r . =206 pounds (14) 

 a d mm 



For a sand bottom, a screw anchor with a 4-inch diameter blade, embedded 

 at least 2 feet would be adequate. 



Solution 2 - Increasing B and Using No Tiedowns 



The required F is 128 pounds per foot. From Figure 3a, F a -Q for 

 B = 2.8 feet is 130 pounds per foot for clean sand. 



Try B = 2.8 feet and plate thickness equals 5/16 inch with four 

 1-1/4 inch stiffners: 



W L = 135 + (2 * L - B) * 36 = 1045 pounds 

 sub 



Lateral resistance is satisfactory since W , has not changed significantly. 



sub 



Check overturning 



F = 1045 -: (2 * 14 - 2.8) + (12 * 1450) + (14) 2 = 130 pounds per 

 foot (11) 



Since F a -Q = 130 pounds per foot for clean sand, this is satisfactory. 

 Recalculate keying edge, 



d = 0.17 + 2.8 * 0.05 = 0.31 feet 



24 



