Question

2GLE 418-12 runs A Standard penetration test taas been come in a museum statum having unit weight = 15 pef to a diet of 33 the lovel table is a below the sound level antais 121 . The hizo follows: 0-6°= 8.6-12* = 12. 12-08 = 15 h . Dercermine the c SPT value (N ) if the testing is assumed toamis te e . Further the were conducted using US sole dionut fame in asce m with a standard sampler Los AS

Answer 1

Ans) Given,

Unit weight of soil ($\gamma$) = 115 pcf

Depth (z) = 30 ft

We know,

N₆₀ = E_m C_B C_S C_R N / 0.60

where, E_m = hammer efficiency

C_B = bore hole diameter correction

C_S = Sampler correction

C_R = rod length correction

N = Recorded number of blows

For US style safety hammer, E_m = 0.60

For bore hole diameter of 6" , C_B = 1.05

For standard sampler , C_s = 1.00

For rod length of 30 ft, C_R = 0.95

=> N₆₀ = 0.6 x 1.05 x 1 x 0.95 N / 0.6

=> N₆₀ = 0.9975 N

For 0 -6 " , N₆₀ = 0.9975 x 8 = 7.98

For 6 - 12" , N₆₀ = 0.9975 x 12 = 11.97

For 12 - 18 " , N₆₀ = 0.9975 x 18 = 17.95

Now apply correction for overburden pressure, (N₆₀)₁ ,

(N₆₀)₁ =  C_N x (N₆₀)  

where, C_N = overburden correction factor = ( 2000 / $\sigma$ _v)^1/2

$\sigma$_v = Effective Vertical stress at 30 ft deoth

= $\gamma$ z = (115 x 4) + (115 - 62.4) 26 = 1827.6 psf

=> C_N = (2000 / 1827.6)^1/2 = 1.046

=> For 0 -6 " , (N₆₀)₁ = 1.046 x 7.98 = 8.34

For 6 - 12" , (N₆₀)₁ = 1.046 x 11.97 = 12.52

For 12 - 18 " , (N₆₀)₁ = 1.046 x 17.95 = 18.77

For SPT valve more than 15, apply Dilatancy correction (C_D),

C_D = 15 + 0.5 ( (N₆₀)₁ - 15)

= 15 + 0.5 ( 18.77 - 15)

= 16.89

    Hence, final corrected values are tabulated below :

Penetration Depth (in)	Corrected 'N' value
0 - 6	8.34
6 - 12	12.52
12 - 18	16.89