A three-story structure is to be constructed over an 8000-m2 site. The initial subsurface exploration indicates the presence of sinkholes and voids due to dissolution of the limestone formation. The predominant soil type is a silty fine sand grading to a fine sand with seams of sandy clay. The design indicates that shallow foundations can be used for this project provided the soils were made more homogeneous as far as load support and no voids were present within the depth up to 7.6 m below the

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A three-story structure is to be constructed over an 8000-m2 site. The initial subsurface exploration indicates the presence of sinkholes and voids due to dissolution of the limestone formation. The predominant soil type is a silty fine sand grading to a fine sand with seams of sandy clay. The design indicates that shallow foundations can be used for this project provided the soils were made more homogeneous as far as load support and no voids were present within the depth up to 7.6 m below the ground surface. Assume groundwater is not a concern. Dynamic compaction is proposed to improve the ground. The local contractor doing dynamic compaction has a 15-ton tamper with the diameter of 2.0 m and the height of 1.4 m. You are requested to conduct the preliminary design for this dynamic compaction project including drop height, spacing, number of drops, number of passes, estimated crater depth, and settlement

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Answer:a) 24.07 m b)  4 mc)  14 number of dropsd) p = number of passese)  Dcd = 2.27 0.69 m Explanation:Given data:Depth  ( D )= 7.6 m below ground surface dynamic compaction ( w )  = 15-ton , diameter of tamper = 2.0 m , thickness = 1.4 m Determine :A) drop height ( H )   D = n √wH  therefore H = 361 / 15 = 24.07 m where : D = 7.6 m ,  n = 0.4 , w = 15 B) Drop spacing drop spacing = average of ( 1.5 to 2.5 )  * diameter  of tamper                        = 2 * 2.0m =  4 mC) number of drops since the applied energy for fine grained soils and day fills range from 250 - 350 kj/m^2  the number of drops can be calculated using the relation below AE = [tex]\frac{NWHP}{SPACING ^2}[/tex]w = 15, H = 24.07 , Np = ?  , AE = 300 kj/m^2∴ Np = 4800 / 361.05  = 13.3 the number of drops at one pass =  14 D) number of passesp = number of passesE) estimated crater depth and settlement crater depth ( Dcd ) = 0.028 [tex]N_{d} ^{0.55} \sqrt{wtIt}[/tex]Nd = 14 ,  wt = 15, It = 24.07therefore : Dcd = 2.27 estimate settlement is within 3 to 5% therefore the improved settlement = 2.27 * 0.04 * 7.6 = 0.69 m

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