Formulations for Poulos & Davis (1980) Method for Single Pile Settlement 1

[bdbd]

Hey everyone,

Most of you know the Poulos & Davis (1980) for calculation of settlement of single pile. There are four or five charts. Have you ever seen formulation of these charts? I want to digitize that process, however, I do not want to deal with that extensive effort if there is something similar. Maybe you have created some excel charts?

 

[Lomarandil]

Interested to follow this.

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The name is a long story -- just call me Lo.

 

[EngMan40]

I digitized a curve from a book before by reading the x and y values from the curve and then created an excel table. I then plotted the values in excel and fitted a trendline to the plot to get the equation. Alternative way to obtain the points is to insert a scaled curve image into CAD and trace it with spline or points and export those points into excel by a lisp. For log scale you can manipulate the date in excel before plotting them so the trendline gives you higher R^2 value.

 

[bdbd]

I know we can digitize it, there are many softwares and web based apps to digitize easily. But the point is if there is anyone to share their formulas, excel files.

 

[Phil1934]

I'll get you started
I0=0.97(L/d)^-0.80
RH for L/d=50=-1.63(h/L)^4+11.2(h/L)^3-28.67(h/L)^2-32.64(h/L)-13.11
RH for L/d=25=1.56(h/L)^3-7.88(h/L)^2+13.25(h/L)-6.63
RH for L/d=10=1.62(h/L)^3-8.2(h/L)^2+13.9(h/L)-7.18

 

[Phil1934]

I had to do a little research on input values before I could proceed. Poisson's ratio silt 0.3-0.5, sand 0.3-0.4, clay 0.1-0.3, sat. clay 0.4-0.5
and subgrade modulus from NAVFAC DM 7.1-219 Figure 6 is 11.36N-60.67 for clay and 5.70N+13.46 for sand in tons/ft^3. To convert to kPa multiply by 100.
E from Bowles Fdn Eng Table 5-6 sand E=500(N+15). Saturated sand E=250(N+15). Silt&clay=300(N+6) in kPa. To convert to ksf divide by 50.
So Rv=0.34v+0.83 as 500-2000 about same and for k=100 Rv=0.25v+0.88. Rk for L/d=50
Rk=-0.45ln(k)+4.78, for L/d=25 Rk=-0.30ln(k)+3.5, for L/d=10 Rk=-0.17ln(k)+2.4.
Since I was interested in an end bearing helical pile with a slender shaft 25d long in 50 blow silt
I0=0.97*25^-0.8=.074.
Rk=-0.30ln((11.36*50-60.67)*100)+3.5=1.63
Rb=1
Rv=0.34x0.4+0.83=0.97
I=.074x1.63x1x0.97=0.12
Ρ=50Kx0.12/(336K/SFx0.36 ft)=0.05 ftx12”/’=0.6 inches which is less than half expected. Let us know if you have better luck.