HANDOUT 12 Example Calculation of p-y Curve in Sand for Laterally Loaded Pile Reference: "COM624P - Laterally Loaded Pile Analysis Program for the Microcomputer, Version 2.0" by S.-T. Wang and L. C. Reese, Report No. FHWA-SA-91-048, FHWA, June 1993. Determine p-y curve for sand at depth of 720 in. shown in Figs. 42 and 43 on the next page Given: Static loading Pile diameter, b = Constant of subgrade reaction, nh = Depth below bearing level, x = From Fig. 42: φ = Effective unit weight below g.w.t, γ ' =
48 80 720 29 57.5
in. pci in. degrees = 0.506145 radians pcf = 0.033275 pci
Calculations: Follow steps shown on pp. 341-347, referring to Fig. 3.11 on p. 342 for definition of terminology and general shape of p-y curves. α=φ/2= β = 45o + φ / 2 = Ko = Ka = tan2(45o - φ / 2) =
14.5 59.5 0.4
degrees = 0.253073 radians degrees = 1.038471 radians
0.34697403
Compute p s as the smaller of p st and p sd (Eqs. 3.31 and 3.32 on p. 341) Calculate pst as follows: ⎡ K 0 x tan φ sin β ⎤ tan β ( b + x tan β tan α ) + K 0 x tan β ( tan φ sin β − tan α ) − K a b ⎥ p st = γx ⎢ + tan( β − φ ) ⎣ tan( β − φ ) cosα ⎦
Term1 = Term2 = Term3 = Term4 = γ'x= pst =
241.19859 1049.3949 107.07073 -16.65475 23.958333 33,087 lb/in.
Calculate psd as follows:
psd = Ka bγx( tan 8 β − 1) + K0bγx tan φ tan 4 β
Term1 = 27131.025 Term2 = 2117.9508 psd = 29,249 lb/in. Use lower of two values for ps, so ps =
© 2001, 2008 Evert C. Lawton
29,249
lb/in.
Example Calculation of p-y Curve in Sand for Laterally Loaded Pile.xls
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© 2001, 2008 Evert C. Lawton
Example Calculation of p-y Curve in Sand for Laterally Loaded Pile.xls
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Calculate ultimate values of p and y at point u x/b= From Fig. 3.12, p. 344: As =
15 0.88
pu = As ps =
25,739
yu = 3b / 80 =
1.80
lb/in. in.
Calculate ultimate values of p and y at point m From Fig. 3.13, p. 345: Bs = 0.50 pm = Bs ps = 14,624 lb/in. ym = b / 60 =
0.80
in.
Establish the initial straight line portion of the p-y curve p = (nh x) y nh x =
57,600
psi
Establish the parabolic section of the p-y curve p = C y1/n Slope of line between points m and u = (pu - pm) / (yu - ym) =
11,115
Power of the parabolic section, n = pm / (m ym) =
1.6447
Coefficient, C = pm / ym
1/n
=
psi
16,750
Determine point k where initial straight line portion intersects parabola yk = (C / (nh x))n/n-1 =
0.04281 in. From straight line, pk = (nh x) yk = 2,466 From parabola, pk = C yk1/n = 2,466
© 2001, 2008 Evert C. Lawton
lb/in. lb/in. - checks
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All required information is now available to draw the p-y curve as shown in Fig. 43. The following data points will be used to draw the p-y curve (shown on the next sheet): y (in.) 0 0.00856 0.0171 0.0257 0.0343 0.0385 0.0428 0.119 0.194 0.346 0.497 0.649 0.724 0.800 0.9 1.0 1.2 1.4 1.6 1.7 1.75 1.76 1.77 1.78 1.79 1.80 1.81 1.82 1.83 1.84 1.85 1.90 2.00 2.20 2.40 2.60 2.80 3.00
p (lb/in.) 0 493.2 986.4 1,480 1,973 2,219 2,466 4,580 6,185 8,781 10,951 12,873 13,767 14,624 15,736 16,847 19,070 21,293 23,516 24,628 25,183 25,295 25,406 25,517 25,628 25,739 25,739 25,739 25,739 25,739 25,739 25,739 25,739 25,739 25,739 25,739 25,739 25,739
© 2001, 2008 Evert C. Lawton
Location
(yk, pk)
(ym, pm)
(yu, pu)
Example Calculation of p-y Curve in Sand for Laterally Loaded Pile.xls
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30000
(yu, pu)
25000
p (lb/in.)
20000
15000
(ym, pm)
10000
5000 (yk, pk) 0 0.0
0.5
1.0
1.5
2.0
2.5
3.0
y (in.)
© 2001, 2008 Evert C. Lawton
Example Calculation of p-y Curve in Sand for Laterally Loaded Pile.xls
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