Following is the variation of the field standard penetration number (№60) in a sand deposit: Depth (m) N60 1.5 6 3 8 4.5 9 6 8 7.5 9 13 14 The groundwater table is located at a depth of 6 m. Given: the dry unit weight of sand from 0 to a depth of 6 m is 19 kN/m³, and the saturated unit weight of sand for depth 6 to 12 m is 21.2 kN/m³. Use the relationship given in the equation 0.5 1 CN = σ'o/Pa to calculate the corrected penetration numbers. (Round your answers to the nearest whole number.) Depth (m) N60 (N1) 60 1.5 6 3 8 4.5 9 6 8 7.5 13 9 14

Following is the variation of the field standard penetration number (№60) in a sand deposit: Depth (m) N60 1.5 6 3 8 4.5 9 6 8 7.5 9 13 14 The groundwater table is located at a depth of 6 m. Given: the dry unit weight of sand from 0 to a depth of 6 m is 19 kN/m³, and the saturated unit weight of sand for depth 6 to 12 m is 21.2 kN/m³. Use the relationship given in the equation 0.5 1 CN = σ'o/Pa to calculate the corrected penetration numbers. (Round your answers to the nearest whole number.) Depth (m) N60 (N1) 60 1.5 6 3 8 4.5 9 6 8 7.5 13 9 14

Structural Analysis

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

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A standard penetration test is carried out in sand where the efficiency of the hammer nH =70%. If the measured N-value at 30 ft depth is 24, find N60 and (N1)60. The unit weight of the sand is 115.0 lb/ft3. Assume nB = nS = nR =1.
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