Chin's Method v4l2p
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Chin’s Method
The Chin’s method required the test pile result of load-settlement relationship. The Chin’s method just only to plot a graft which the settlement (∆) / load (P) against settlement (∆). Then the ultimate loads are deriving from the inverse slopes of the lines. Chin’s method assumes the form of the load or deformation curve is hyperbolic and it is an empirical method. It is useful in that if the result is plotted at the time of the test, deviation from the expected curve is easily seen and the test procedure can be checked
1. PILE A
1st Cycle Total Load, Ave. P Sett, ∆ (Tons) (mm)
Table 1:
∆/P
0
0
690
0.37
1323
0.98
1975
1.72
2755
2.75
2006
2.49
1191
2.04
721
1.47
0 0.00053 6 0.00074 1 0.00087 1 0.00099 8 0.00124 1 0.00171 3 0.00203 9
0
0.95
0
2nd Cycle 0
0.95
0 0.00175 4
741
1.3
1368
1.82
2056
2.39
2710
2.97
3409
4.02
4050
5.38
4750
7.03
5365
8.78
6310
10.78
6761
14.73
6119
14.44
5427
14.08
4555
13.36
4059
13.03
3330
12.26
2632
11.59
1993
10.91
1402
10.23
0.00321 0.00368 2 0.00440 3 0.00547 4 0.00729 7
795
9.27
0.01166
0
8.03
0
Minus Residual Sett. ∆/P (mm) 0 0.35
0.00133 0.00116 2 0.00109 6 0.00117 9 0.00132 8
0.87
0.00148 0.00163 7 0.00170 8 0.00217 9
6.08
0.00236 0.00259 4 0.00293 3
1.44 2.02 3.07 4.43
7.83 9.83 13.78
0 0.00047 2 0.00063 6 0.0007 0.00074 5 0.00090 1 0.00109 4 0.00128 0.00145 9 0.00155 8 0.00203 8
Calculation for Graph ∆/P against ∆ for Pile A
Graft 1: ∆/P against ∆ for Pile A
Estimation of Qu Using First Cycle Result Equation from graph: y =0.000187x + 0.000514 Slope of the plot: m =0.000187 kN-1 Know that the inverse slope of the plot gives the ultimate capacity, thus Ultimate Capacity of the pile, Qult = 1/0.000187 kN-1 = 5,347.59kN Estimation of Qu Using Minus Residual Result Equation from graph: y =0.000111x + 0.000535 Slope of the plot: m =0.000111 kN-1 Know that the inverse slope of the plot gives the ultimate capacity, thus Ultimate Capacity of the pile, Qult = 1/0.000111 kN-1 = 9,009.01 kN
2. PILE B Table 2: Calculation for Graft ∆/P against ∆ for Pile B 1st Cycle Total Load, P (Tons) 0 215 500 699 953 726 470 246 0
Avg. sett. ∆ (mm) 0 0.49 1.49 2.2 2.88 2.41 2.11 1.51 0.76
∆/P 0 0.002279 0.00298 0.003147 0.003022 0.00332 0.004489 0.006138 0
2nd Cycle 0 277 471 705 957 1163 1380 1622 1850 2076 2393 2200 1887 1655 1437
0.76 1.45 1.81 2.31 2.96 3.7 4.7 5.65 6.79 8.08 10.67 10.29 10.11 9.73 9.33
0 0.005235 0.003843 0.003277 0.003093 0.003181 0.003406 0.003483 0.00367 0.003892 0.004459 0.004677 0.005358 0.005879 0.006493
1156 968 733 504 281 0
8.62 8.1 7.53 6.71 5.83 3.98
0.007457 0.008368 0.010273 0.013313 0.020747 0
Minus Residual Sett. ∆ ∆/P (mm) 0 0 0.69 0.002491 1.05 0.002229 1.55 0.002199 2.2 0.002299 2.94 0.002528 3.94 0.002855 4.89 0.003015 6.03 0.003259 7.32 0.003526 9.91 0.004141
Graft 2: ∆/P against ∆ for Pile B Estimation of Ultimate Load, Qu Using First Cycle Result Equation from graft:
y
=0.000323x + 0.002286
Slope of the plot:
m
=0.000323 kN-1
Know that the inverse slope of the plot gives the ultimate capacity, thus Ultimate Capacity of the pile,
Qult
= 1/0.000323 kN-1 = 3,095.98 kN
Estimation of Ultimate Load, Qu Using Minus Residual Result Equation from graft:
y
=0.000285x + 0.001544
Slope of the plot:
m
=0.000285 kN-1
Know that the inverse slope of the plot gives the ultimate capacity, thus Ultimate Capacity of the pile,
Qult
= 1/0.000285 kN-1 = 3,508.77 kN
3. Pile C Table 3: Calculation for Graft ∆/P against ∆ for Pile C
Total Load,P (Tons) 0
1st Cycle Avg sett. ∆ (mm) 0
2777
2.05
5082
4.18
2646 0
2.82 0.48
∆/P 0 0.0007 38 0.0008 23 0.0010 66 0
2nd Cycle 0
0.48
2751
2.4
4071
3.33
5078
4.18
6243
5.04
7479
7.23
6265
6.6
5105
5.94
4028
5.29
2785
4.44
1693 0
3.54 1.91
0 0.0008 72 0.0008 18 0.0008 23 0.0008 07 0.0009 67 0.0010 53 0.0011 64 0.0013 13 0.0015 94 0.0020 91 0
Minus Residual Sett. ∆ ∆/P (mm) 0 0 0.00069 1.92 8 2.85 3.7 4.56 6.75 6.12 5.46 4.81 3.96 3.06 1.43
0.0007 0.00072 9 0.00073 0.00090 3 0.00097 7 0.00107 0.00119 4 0.00142 2 0.00180 7 0
Graft 3: ∆/P against ∆ for Pile C Estimation of Qu Using First Cyle Result Equation from graft:
y
=0.000196x + 0.000114
Slope of the plot:
m
=0.000196 kN-1
Know that the inverse slope of the plot gives the ultimate capacity, thus Ultimate Capacity of the pile,
Qult
= 1/0.000196 kN-1 =5,102.04 kN
Estimation of Qu Using Minus Residual Result Equation from graft:
y
=0.000117x + 0.000245
Slope of the plot:
m
=0.000117 kN-1
Know that the inverse slope of the plot gives the ultimate capacity, thus Ultimate Capacity of the pile,
Qult
= 1/0.000117 kN-1 =8,547.01 kN
4. PILE D Table 4: Calculation for Minus Residual Graft ∆/P against ∆ for Pile D.
Total Load,P (Tons) 0
1st Cycle Avg sett. ∆ (mm) 0
460 839
1.01 2.19
1283
3.98
1707
6.65
2189
10.52
1577
9.69
1083
8.78
527 0
7.7 5.88
∆/P 0 0.00219 6 0.00261 0.00310 2 0.00389 6 0.00480 6 0.00614 5 0.00810 7 0.01461 1 0
2nd Cycle 0
5.88
571
7.26
1090
8.37
1669
9.45
2114
10.72
2636
14.08
3188
19.87
3711
26.77
4257
34.77
4801
43.9
5320
55.1
5821
69.88
4206
67.96
0 0.01271 5 0.00767 9 0.00566 2 0.00507 1 0.00534 1 0.00623 3 0.00721 4 0.00816 8 0.00914 4 0.01035 7 0.01200 5 0.01615 8
Minus Residual Sett. ∆ ∆/P (mm) 0 0 0.00241 1.38 7 0.00228 2.49 4 0.00213 3.57 9 0.00228 4.84 9 0.00311 8.2 1 0.00438 13.99 8 0.00562 20.89 9 0.00678 28.89 6 0.00791 38.02 9 0.00925 49.22 2 0.01099 64 5
2702
65.25
1121 0
60.99 55.5
0.02414 9 0.05440 7 0
Graft 4: ∆/P against ∆ for Pile D
Estimation of Qu Using First Cycle Result: Equation from graft:
y
=0.000274x + 0.00199
Slope of the plot:
m
=0.000274 kN-1
Know that the inverse slope of the plot gives the ultimate capacity, thus Ultimate Capacity of the pile,
Qult
= 1/0.000274 kN-1 = 3,649.64 kN
Estimation of Qu Using Minus Residual Result Equation from graft:
y
=0.000156x + 0.001714
Slope of the plot:
m
=0.000156 kN-1
Know that the inverse slope of the plot gives the ultimate capacity, thus Ultimate Capacity of the pile,
Qult
= 1/0.000156 kN-1 = 6,410.26 kN
PILE E Table 5: Calculation for Minus Residual Graft ∆/P against ∆ for Pile E.
Total Load,P (Tons) 0
1st Cycle Avg sett. ∆ (mm) 0
724
0.43
984
0.69
1496
1.24
2014
1.92
2506
2.55
2984
3.25
3491
3.89
4013
4.6
4511
5.34
5222
6.57
3835
5.5
2957
4.28
1723 0
3.61 2.42
∆/P 0 0.00059 4 0.00070 1 0.00082 9 0.00095 3 0.00101 8 0.00108 9 0.00111 4 0.00114 6 0.00118 4 0.00125 8 0.00143 4 0.00144 7 0.00209 5 0
2nd Cycle
Graft for
0
2.42
2.42
1366
3.37
3.37
2618
4.54
4.54
3765
5.65
5.65
5031
6.93
6.93
6286
8.54
8.54
7444
10.58
10.58
8751
13.64
13.64
9975
18.31
18.31
11056
106.46
106.46
9664
105.72
105.72
6566
103.48
103.48
3581
100.21
100.21
0
94.04
94.04
Minus Residual Sett. ∆ ∆/P (mm) 0 0 0.0006 0.95 95 0.0008 2.12 10 0.0008 3.23 58 0.0008 4.51 96 0.0009 6.12 74 0.0010 8.16 96 0.0012 11.22 82 0.0015 15.89 93 0.0094 104.04 10 0.0106 103.3 89 0.0153 101.06 91 0.0273 97.79 08 0.0000 91.62 00
5: ∆/P against ∆ Pile E
Estimation of Ultimate Load, Qu Using First Cycle Result Equation from graft:
y
=0.000100x + 0.000685
Slope of the plot:
m
=0.0001 kN-1
Know that the inverse slope of the plot gives the ultimate capacity, thus Ultimate Capacity of the pile,
Qult
= 1/0.0001 kN-1 = 10,000 kN
Estimation of Ultimate Load, Qu Using Minus Residual Result Equation from graft:
y
=0.000085x + 0.000475
Slope of the plot:
m
=0.000085 kN-1
Know that the inverse slope of the plot gives the ultimate capacity, thus Ultimate Capacity of the pile,
Qult
= 1/0.000085 kN-1 = 11,764.71 kN
5. PILE F Table 6: Calculation for Minus Residual Graft ∆/P against ∆ for Pile F.
Total Load,P (Tons)
1st Cycle Avg sett. ∆ (mm)
0
0
288
0.59
613
1.18
832
1.58
1008
2.21
1349
2.75
1512
3.15
1649
3.42
1847
3.89
2064
5.61
2323
5.89
∆/P 0.00000 0 0.00204 9 0.00192 5 0.00189 9 0.00219 2 0.00203 9 0.00208 3 0.00207 4 0.00210 6 0.00271 8 0.00253 6
2613
6.34
1917
5.74
1290
4.72
641
3.58
0
2.16
0.00242 6 0.00299 4 0.00365 9 0.00558 5 0.00000 0
Minus Residual Sett. ∆ (mm)
2 nd Cycle 0
2.16
2.16
0
822
3.5
3.5
1.34
1090
3.78
3.78
1.62
1497
4.64
4.64
2.48
2002
5.34
5.34
3.18
2572
6.32
6.32
4.16
2737
6.81
6.81
4.65
2985
7.15
7.15
4.99
3270
7.8
7.8
5.64
3459
8.4
8.4
6.24
3708
9.14
9.14
6.98
3952
10.49
10.49
8.33
4192
11.34
11.34
9.18
4482
13.32
13.32
11.16
4737
14.99
14.99
12.83
4981
17.51
17.51
15.35
5217
21.29
21.29
19.13
5463
29.35
29.35
27.19
5696
38.68
38.68
36.52
5955 6237
51.83 75.2
51.83 75.2
49.67 73.04
∆/P 0.00000 0 0.00163 0 0.00148 6 0.00165 7 0.00158 8 0.00161 7 0.00169 9 0.00167 2 0.00172 5 0.00180 4 0.00188 2 0.00210 8 0.00219 0 0.00249 0 0.00270 8 0.00308 2 0.00366 7 0.00497 7 0.00641 2 0.00834 1 0.01171
1 4809 3206 1664 0
73.38 70.57 67.13 62.12
73.38 70.57 67.13 62.12
Graft 6: ∆/P against ∆ for Pile F
Estimation of Ultimate Load, Qu Using First Cycle Result Equation from graft:
y
=0.000126x + 0.001781
Slope of the plot:
m
=0.000126 kN-1
Know that the inverse slope of the plot gives the ultimate capacity, thus Ultimate Capacity of the pile,
Qult
= 1/0.000126 kN-1 = 7,936.51 kN
Estimation of Ultimate Load, Qu Using Minus Residual Result Equation from graft:
y
=0.000146x + 0.001009
Slope of the plot:
m
=0.000146 kN-1
Know that the inverse slope of the plot gives the ultimate capacity, thus Ultimate Capacity of the pile,
Qult
= 1/0.000146kN-1 = 6,849.32 kN
The Chin’s method required the test pile result of load-settlement relationship. The Chin’s method just only to plot a graft which the settlement (∆) / load (P) against settlement (∆). Then the ultimate loads are deriving from the inverse slopes of the lines. Chin’s method assumes the form of the load or deformation curve is hyperbolic and it is an empirical method. It is useful in that if the result is plotted at the time of the test, deviation from the expected curve is easily seen and the test procedure can be checked
1. PILE A
1st Cycle Total Load, Ave. P Sett, ∆ (Tons) (mm)
Table 1:
∆/P
0
0
690
0.37
1323
0.98
1975
1.72
2755
2.75
2006
2.49
1191
2.04
721
1.47
0 0.00053 6 0.00074 1 0.00087 1 0.00099 8 0.00124 1 0.00171 3 0.00203 9
0
0.95
0
2nd Cycle 0
0.95
0 0.00175 4
741
1.3
1368
1.82
2056
2.39
2710
2.97
3409
4.02
4050
5.38
4750
7.03
5365
8.78
6310
10.78
6761
14.73
6119
14.44
5427
14.08
4555
13.36
4059
13.03
3330
12.26
2632
11.59
1993
10.91
1402
10.23
0.00321 0.00368 2 0.00440 3 0.00547 4 0.00729 7
795
9.27
0.01166
0
8.03
0
Minus Residual Sett. ∆/P (mm) 0 0.35
0.00133 0.00116 2 0.00109 6 0.00117 9 0.00132 8
0.87
0.00148 0.00163 7 0.00170 8 0.00217 9
6.08
0.00236 0.00259 4 0.00293 3
1.44 2.02 3.07 4.43
7.83 9.83 13.78
0 0.00047 2 0.00063 6 0.0007 0.00074 5 0.00090 1 0.00109 4 0.00128 0.00145 9 0.00155 8 0.00203 8
Calculation for Graph ∆/P against ∆ for Pile A
Graft 1: ∆/P against ∆ for Pile A
Estimation of Qu Using First Cycle Result Equation from graph: y =0.000187x + 0.000514 Slope of the plot: m =0.000187 kN-1 Know that the inverse slope of the plot gives the ultimate capacity, thus Ultimate Capacity of the pile, Qult = 1/0.000187 kN-1 = 5,347.59kN Estimation of Qu Using Minus Residual Result Equation from graph: y =0.000111x + 0.000535 Slope of the plot: m =0.000111 kN-1 Know that the inverse slope of the plot gives the ultimate capacity, thus Ultimate Capacity of the pile, Qult = 1/0.000111 kN-1 = 9,009.01 kN
2. PILE B Table 2: Calculation for Graft ∆/P against ∆ for Pile B 1st Cycle Total Load, P (Tons) 0 215 500 699 953 726 470 246 0
Avg. sett. ∆ (mm) 0 0.49 1.49 2.2 2.88 2.41 2.11 1.51 0.76
∆/P 0 0.002279 0.00298 0.003147 0.003022 0.00332 0.004489 0.006138 0
2nd Cycle 0 277 471 705 957 1163 1380 1622 1850 2076 2393 2200 1887 1655 1437
0.76 1.45 1.81 2.31 2.96 3.7 4.7 5.65 6.79 8.08 10.67 10.29 10.11 9.73 9.33
0 0.005235 0.003843 0.003277 0.003093 0.003181 0.003406 0.003483 0.00367 0.003892 0.004459 0.004677 0.005358 0.005879 0.006493
1156 968 733 504 281 0
8.62 8.1 7.53 6.71 5.83 3.98
0.007457 0.008368 0.010273 0.013313 0.020747 0
Minus Residual Sett. ∆ ∆/P (mm) 0 0 0.69 0.002491 1.05 0.002229 1.55 0.002199 2.2 0.002299 2.94 0.002528 3.94 0.002855 4.89 0.003015 6.03 0.003259 7.32 0.003526 9.91 0.004141
Graft 2: ∆/P against ∆ for Pile B Estimation of Ultimate Load, Qu Using First Cycle Result Equation from graft:
y
=0.000323x + 0.002286
Slope of the plot:
m
=0.000323 kN-1
Know that the inverse slope of the plot gives the ultimate capacity, thus Ultimate Capacity of the pile,
Qult
= 1/0.000323 kN-1 = 3,095.98 kN
Estimation of Ultimate Load, Qu Using Minus Residual Result Equation from graft:
y
=0.000285x + 0.001544
Slope of the plot:
m
=0.000285 kN-1
Know that the inverse slope of the plot gives the ultimate capacity, thus Ultimate Capacity of the pile,
Qult
= 1/0.000285 kN-1 = 3,508.77 kN
3. Pile C Table 3: Calculation for Graft ∆/P against ∆ for Pile C
Total Load,P (Tons) 0
1st Cycle Avg sett. ∆ (mm) 0
2777
2.05
5082
4.18
2646 0
2.82 0.48
∆/P 0 0.0007 38 0.0008 23 0.0010 66 0
2nd Cycle 0
0.48
2751
2.4
4071
3.33
5078
4.18
6243
5.04
7479
7.23
6265
6.6
5105
5.94
4028
5.29
2785
4.44
1693 0
3.54 1.91
0 0.0008 72 0.0008 18 0.0008 23 0.0008 07 0.0009 67 0.0010 53 0.0011 64 0.0013 13 0.0015 94 0.0020 91 0
Minus Residual Sett. ∆ ∆/P (mm) 0 0 0.00069 1.92 8 2.85 3.7 4.56 6.75 6.12 5.46 4.81 3.96 3.06 1.43
0.0007 0.00072 9 0.00073 0.00090 3 0.00097 7 0.00107 0.00119 4 0.00142 2 0.00180 7 0
Graft 3: ∆/P against ∆ for Pile C Estimation of Qu Using First Cyle Result Equation from graft:
y
=0.000196x + 0.000114
Slope of the plot:
m
=0.000196 kN-1
Know that the inverse slope of the plot gives the ultimate capacity, thus Ultimate Capacity of the pile,
Qult
= 1/0.000196 kN-1 =5,102.04 kN
Estimation of Qu Using Minus Residual Result Equation from graft:
y
=0.000117x + 0.000245
Slope of the plot:
m
=0.000117 kN-1
Know that the inverse slope of the plot gives the ultimate capacity, thus Ultimate Capacity of the pile,
Qult
= 1/0.000117 kN-1 =8,547.01 kN
4. PILE D Table 4: Calculation for Minus Residual Graft ∆/P against ∆ for Pile D.
Total Load,P (Tons) 0
1st Cycle Avg sett. ∆ (mm) 0
460 839
1.01 2.19
1283
3.98
1707
6.65
2189
10.52
1577
9.69
1083
8.78
527 0
7.7 5.88
∆/P 0 0.00219 6 0.00261 0.00310 2 0.00389 6 0.00480 6 0.00614 5 0.00810 7 0.01461 1 0
2nd Cycle 0
5.88
571
7.26
1090
8.37
1669
9.45
2114
10.72
2636
14.08
3188
19.87
3711
26.77
4257
34.77
4801
43.9
5320
55.1
5821
69.88
4206
67.96
0 0.01271 5 0.00767 9 0.00566 2 0.00507 1 0.00534 1 0.00623 3 0.00721 4 0.00816 8 0.00914 4 0.01035 7 0.01200 5 0.01615 8
Minus Residual Sett. ∆ ∆/P (mm) 0 0 0.00241 1.38 7 0.00228 2.49 4 0.00213 3.57 9 0.00228 4.84 9 0.00311 8.2 1 0.00438 13.99 8 0.00562 20.89 9 0.00678 28.89 6 0.00791 38.02 9 0.00925 49.22 2 0.01099 64 5
2702
65.25
1121 0
60.99 55.5
0.02414 9 0.05440 7 0
Graft 4: ∆/P against ∆ for Pile D
Estimation of Qu Using First Cycle Result: Equation from graft:
y
=0.000274x + 0.00199
Slope of the plot:
m
=0.000274 kN-1
Know that the inverse slope of the plot gives the ultimate capacity, thus Ultimate Capacity of the pile,
Qult
= 1/0.000274 kN-1 = 3,649.64 kN
Estimation of Qu Using Minus Residual Result Equation from graft:
y
=0.000156x + 0.001714
Slope of the plot:
m
=0.000156 kN-1
Know that the inverse slope of the plot gives the ultimate capacity, thus Ultimate Capacity of the pile,
Qult
= 1/0.000156 kN-1 = 6,410.26 kN
PILE E Table 5: Calculation for Minus Residual Graft ∆/P against ∆ for Pile E.
Total Load,P (Tons) 0
1st Cycle Avg sett. ∆ (mm) 0
724
0.43
984
0.69
1496
1.24
2014
1.92
2506
2.55
2984
3.25
3491
3.89
4013
4.6
4511
5.34
5222
6.57
3835
5.5
2957
4.28
1723 0
3.61 2.42
∆/P 0 0.00059 4 0.00070 1 0.00082 9 0.00095 3 0.00101 8 0.00108 9 0.00111 4 0.00114 6 0.00118 4 0.00125 8 0.00143 4 0.00144 7 0.00209 5 0
2nd Cycle
Graft for
0
2.42
2.42
1366
3.37
3.37
2618
4.54
4.54
3765
5.65
5.65
5031
6.93
6.93
6286
8.54
8.54
7444
10.58
10.58
8751
13.64
13.64
9975
18.31
18.31
11056
106.46
106.46
9664
105.72
105.72
6566
103.48
103.48
3581
100.21
100.21
0
94.04
94.04
Minus Residual Sett. ∆ ∆/P (mm) 0 0 0.0006 0.95 95 0.0008 2.12 10 0.0008 3.23 58 0.0008 4.51 96 0.0009 6.12 74 0.0010 8.16 96 0.0012 11.22 82 0.0015 15.89 93 0.0094 104.04 10 0.0106 103.3 89 0.0153 101.06 91 0.0273 97.79 08 0.0000 91.62 00
5: ∆/P against ∆ Pile E
Estimation of Ultimate Load, Qu Using First Cycle Result Equation from graft:
y
=0.000100x + 0.000685
Slope of the plot:
m
=0.0001 kN-1
Know that the inverse slope of the plot gives the ultimate capacity, thus Ultimate Capacity of the pile,
Qult
= 1/0.0001 kN-1 = 10,000 kN
Estimation of Ultimate Load, Qu Using Minus Residual Result Equation from graft:
y
=0.000085x + 0.000475
Slope of the plot:
m
=0.000085 kN-1
Know that the inverse slope of the plot gives the ultimate capacity, thus Ultimate Capacity of the pile,
Qult
= 1/0.000085 kN-1 = 11,764.71 kN
5. PILE F Table 6: Calculation for Minus Residual Graft ∆/P against ∆ for Pile F.
Total Load,P (Tons)
1st Cycle Avg sett. ∆ (mm)
0
0
288
0.59
613
1.18
832
1.58
1008
2.21
1349
2.75
1512
3.15
1649
3.42
1847
3.89
2064
5.61
2323
5.89
∆/P 0.00000 0 0.00204 9 0.00192 5 0.00189 9 0.00219 2 0.00203 9 0.00208 3 0.00207 4 0.00210 6 0.00271 8 0.00253 6
2613
6.34
1917
5.74
1290
4.72
641
3.58
0
2.16
0.00242 6 0.00299 4 0.00365 9 0.00558 5 0.00000 0
Minus Residual Sett. ∆ (mm)
2 nd Cycle 0
2.16
2.16
0
822
3.5
3.5
1.34
1090
3.78
3.78
1.62
1497
4.64
4.64
2.48
2002
5.34
5.34
3.18
2572
6.32
6.32
4.16
2737
6.81
6.81
4.65
2985
7.15
7.15
4.99
3270
7.8
7.8
5.64
3459
8.4
8.4
6.24
3708
9.14
9.14
6.98
3952
10.49
10.49
8.33
4192
11.34
11.34
9.18
4482
13.32
13.32
11.16
4737
14.99
14.99
12.83
4981
17.51
17.51
15.35
5217
21.29
21.29
19.13
5463
29.35
29.35
27.19
5696
38.68
38.68
36.52
5955 6237
51.83 75.2
51.83 75.2
49.67 73.04
∆/P 0.00000 0 0.00163 0 0.00148 6 0.00165 7 0.00158 8 0.00161 7 0.00169 9 0.00167 2 0.00172 5 0.00180 4 0.00188 2 0.00210 8 0.00219 0 0.00249 0 0.00270 8 0.00308 2 0.00366 7 0.00497 7 0.00641 2 0.00834 1 0.01171
1 4809 3206 1664 0
73.38 70.57 67.13 62.12
73.38 70.57 67.13 62.12
Graft 6: ∆/P against ∆ for Pile F
Estimation of Ultimate Load, Qu Using First Cycle Result Equation from graft:
y
=0.000126x + 0.001781
Slope of the plot:
m
=0.000126 kN-1
Know that the inverse slope of the plot gives the ultimate capacity, thus Ultimate Capacity of the pile,
Qult
= 1/0.000126 kN-1 = 7,936.51 kN
Estimation of Ultimate Load, Qu Using Minus Residual Result Equation from graft:
y
=0.000146x + 0.001009
Slope of the plot:
m
=0.000146 kN-1
Know that the inverse slope of the plot gives the ultimate capacity, thus Ultimate Capacity of the pile,
Qult
= 1/0.000146kN-1 = 6,849.32 kN
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