Erratum for “Settlement Estimation of Piled Rafts for Initial Design” by Priyanka Bhartiya, Tanusree Chakraborty, and Dipanjan Basu

Bhartiya, Priyanka; Chakraborty, Tanusree; Basu, Dipanjan

doi:10.1061/(ASCE)GT.1943-5606.0002312

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Jun 4, 2020

Erratum for “Settlement Estimation of Piled Rafts for Initial Design” by Priyanka Bhartiya, Tanusree Chakraborty, and Dipanjan Basu

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Authors: Priyanka Bhartiya [email protected], Tanusree Chakraborty [email protected], and Dipanjan Basu, Ph.D., M.ASCE [email protected]Author Affiliations

Publication: Journal of Geotechnical and Geoenvironmental Engineering

Volume 146, Issue 8

https://doi.org/10.1061/(ASCE)GT.1943-5606.0002312

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The following corrections should be made to the original paper:

1.

In the section “Unpiled Raft Stiffness,” there is a typographical error in Eq. (3). The correct form of Eq. (3) is

k_{r s} = 1.73 C_{s} \times {\log (\frac{L_{r}}{T_{r}})}^{0.5} \times (\frac{1 - μ_{s}^{2}}{1 - μ_{r}^{2}}) \times {1 - {(\frac{T_{r}}{B_{r}})}^{3}} \times e^{(- 30 (E_{s} / E_{r}))}

2.

There are multiple typographical errors in the calculation steps in the section “Numerical Examples.” Therefore, the calculation steps of both the numerical examples with correct equations and numerical values are given as follows:

Example 1: Case Study of Rectangular PRF

1.

Given

E_{s} = 40 MPa

,

μ_{s} = 0.3

, and

A_{r} = 83 \times 46.5 = 3859.5 m^{2}

,

K_{s}

is calculated as

K_{s} = \frac{E_{s} (1 - μ_{s})}{(1 + μ_{s}) (1 - 2 μ_{s}) \sqrt{A_{r}}} = {40,000 \times (1 - 0.3)} / {(1 + 0.3) (1 - 2 \times 0.3) \sqrt 3,859.5} = 866.74 kN / m^{3}

Noting

C_{s} = {(B_{r} / L_{r})}^{0.25} = {(46.5 / 83)}^{0.25} = 0.865

,

k_{r s}

is calculated as

k_{r s} = 1.73 C_{s} \times {\log (\frac{L_{r}}{T_{r}})}^{0.5} \times (\frac{1 - μ_{s}^{2}}{1 - μ_{r}^{2}}) \times {1 - {(\frac{T_{r}}{B_{r}})}^{3}} \times e^{(- 30 (E_{s} / E_{r}))} = 1.73 \times 0.865 \times {\log (83 / 1)}^{0.5} \times {(1 - {0.3}^{2}) / (1 - {0.2}^{2}) \times {1 - {(1 / 46.5)}^{3}} \times e^{{- 30 \times (40) / 30,000}} = 1.89

This gives the raft stiffness

K_{r} = K_{s} k_{r s} = 866.74 \times 1.89 = 1638.14 kN / m^{3}

.

2.

Considering the pile group,

A_{p} = π \times {1.5}^{2} / 4 = 1.767 m^{2}

;

A_{g p} = π \times {1.5}^{2} / 4 \times 40 = 70.686 m^{2}

I_{p s} = 0.0084 \times {e^{- 0.001 (E_{p} / E_{s}) \cdot {(L_{p} / D_{p})}^{0.034}}} \times (\frac{μ_{p}}{μ_{s}}) \times (\frac{L_{p}}{D_{p}}) = 0.0084 \times {e^{- 0.001 (30,000 / 40) \cdot {(27.5 / 1.5)}^{0.034}}} \times (\frac{0.2}{0.3}) \times (\frac{27.5}{1.5}) = 0.045

Therefore,

K_{p 1} = A_{p} E_{p} I_{p s} / L_{p} = 1.767 \times 30,000 \times 1,000 \times 0.045 / 27.5 = 86,743.64 kN / m

. Now,

η = 1 - 0.4 \log (1 - A_{g p} / A_{r} - D_{p} / (2 s_{r})) = 1 - 0.4 \times \log {1 - (70.686 / 3,859.5) - (1.5 / 18)} = 1.019

. This gives

K_{p} = K_{p 1} n_{p} (e^{η} - 2) / A_{g p} = 86,743.64 \times 40 \times (e^{1.019} - 2) / 70.686 = 37,817.55 kN / m^{3}

.

3.

Now

r_{m} = 2.5 + L_{p} {2.5 (1 - μ_{p}) - 0.25} = 2.5 + 27.5 \times {2.5 \times (1 - 0.2) - 0.25} = 50.625

and

r_{r} = \sqrt{A_{r} / (π n_{p})} = \sqrt [3,859.5 / (π \times 40)] = 5.542

, so

α_{p r} = 1 - \frac{\ln (r_{r} / r_{p})}{\ln (r_{m} / r_{p})} = 1 - \frac{\ln (5.542 / 0.75)}{\ln (50.625 / 0.75)} = 0.525 and 0.38 e^{2.4 (A_{g p} / A_{r})} = 0.38 e^{2.4 (70.686 / 3,859.5)} = 0.397

This gives

K_{p r} = \frac{0.38 e^{(2.4 (A_{g p} / A_{r}))} K_{p} + K_{r} (1 - 2 α_{p r})}{1 + α_{p r}^{2} K_{r} / K_{p}} = \frac{0.397 \times 37,817.55 + 1,638.14 \times (1 - 2 \times 0.525)}{1 + {0.525}^{2} \times 1,638.14 / 37,817.55} = 14,755.49 kN / m^{3}

4.

Based on the estimated

K_{p r}

, the average PRF settlement

S_{p r, a v} = (P / A_{r}) / K_{p r} = 181 / 14,755.49 = 0.0123 m = 12.3 mm

. Therefore, the expected maximum PRF settlement

\approx 1.2 \times 12.3 = 14.75 mm

. The maximum settlement reported in the literature (Yamashita et al. 2011) is 18 mm.

Example 2: Circular PRF

1.

Given

E_{s} = 20 MPa

,

μ_{s} = 0.3

, and

A_{r} = π \times {4.37}^{2} = 60 m^{2}

K_{s} = \frac{E_{s} (1 - μ_{s})}{(1 + μ_{s}) (1 - 2 μ_{s}) \sqrt{A_{r}}} = {20,000 \times (1 - 0.3)} / {(1 + 0.3) (1 - 2 \times 0.3) \sqrt 60} = 3,475.75 kN / m^{3}

For a circular PRF,

C_{s} = 0.85

; therefore

k_{r s} = 1.73 C_{s} \times {\log (\frac{L_{r}}{T_{r}})}^{0.5} \times (\frac{1 - μ_{s}^{2}}{1 - μ_{r}^{2}}) \times {1 - {(\frac{T_{r}}{B_{r}})}^{3}} \times e^{(- 30 (E_{s} / E_{r}))} = 1.73 \times 0.85 \times {\log (\frac{8.74}{0.5})}^{0.5} \times (\frac{1 - {0.3}^{2}}{1 - {0.2}^{2}}) \times {1 - {(\frac{0.5}{8.74})}^{3}} \times e^{(- 30 (20 / 30,000))} = 1.52

This gives

K_{r} = K_{s} k_{r s} = 3,475.75 \times 1.52 = 5,283.14 kN / m^{3}

.

2.

With

A_{p} = π \times {0.5}^{2} / 4 = 0.196 m^{2}

and

A_{g p} = π \times {0.5}^{2} / 4 \times 9 = 1.764 m^{2}

I_{p s} = 0.0084 \times {e^{- 0.001 (E_{p} / E_{s}) \cdot {(L_{p} / D_{p})}^{0.034}}} \times (\frac{μ_{p}}{μ_{s}}) \times (\frac{L_{p}}{D_{p}}) = 0.0084 \times {e^{- 0.001 (30,000 / 20) \cdot {(10 / 0.5)}^{0.034}}} \times (\frac{0.2}{0.3}) \times (\frac{10}{0.5}) = 0.021

Thus,

K_{p 1} = A_{p} E_{p} I_{p s} / L_{p} = 0.196 \times 30,000 \times 1,000 \times 0.021 / 10 = 12,348 kN / m

. Now

η = 1 - 0.4 \log (1 - \frac{A_{g p}}{A_{r}} - \frac{D_{p}}{s_{r}}) = 1 - 0.4 \times \log {1 - (1.764 / 60) - (0.5 / 3)} = 1.038

Thus,

K_{p} = K_{p 1} n_{p} (e^{η} - 2) / A_{g p} = 12,348 \times 9 \times (e^{1.038} - 2) / 1.764 = 51,884.55 kN / m^{3}

.

3.

Now

r_{m} = 2.5 + L_{p} {2.5 (1 - μ_{p}) - 0.25} = 2.5 + 10 \times {2.5 \times (1 - 0.2) - 0.25} = 20

,

r_{r} = \sqrt{A_{r} / (π n_{p})} = \sqrt [60 / (π \times 9)] = 1.457

α_{p r} = 1 - \frac{\ln (r_{r} / r_{p})}{\ln (r_{m} / r_{p})} = 1 - \frac{\ln (1.457 / 0.25)}{\ln (20 / 0.25)} = 0.60, and 0.038 e^{(2.4 (A_{g p} / A_{r}))} = 0.038 \times e^{(2.4 \times (1.764 / 60))} = 0.04078

This results in the PRF stiffness

K_{p r} = \frac{0.038 e^{(2.4 (A_{g p} / A_{r}))} K_{p} + K_{r} (1 + 2 α_{p r})}{1 + α_{p r}^{2} K_{r} / K_{p}} = \frac{0.04078 \times 51,884.55 + 5,283.14 \times (1 + 2 \times 0.6)}{1 + {0.6}^{2} \times 5,283.14 / 51,884.55} = 13,252.9 kN / m^{3}

4.

The estimated average PRF settlement

S_{p r, a v (equation)} = 12 \times 1,000 / (13,252.9 \times 60) = 15 mm

. The average elastic settlement

S_{p r, a v (FEM)}

obtained directly from FE analysis is 19.5 mm. The expected maximum PRF settlement

\approx 1.2 \times 15 = 18 mm

. The maximum PRF settlement

S_{p r, \max (FEM)}

obtained directly from FE analysis is 20.4 mm.

The authors regret these errors.

References

Yamashita, K., T. Yamada, and J. Hamada. 2011. “Investigation of settlement and load sharing on piled rafts by monitoring full scale structures.” Soils Found. 51 (3): 513–532.

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Journal of Geotechnical and Geoenvironmental Engineering

Volume 146 • Issue 8 • August 2020

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Received: Mar 13, 2020

Accepted: Mar 16, 2020

Published online: Jun 4, 2020

Published in print: Aug 1, 2020

Discussion open until: Nov 4, 2020

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Priyanka Bhartiya [email protected]

Project Scientist, Dept. of Civil Engineering, Indian Institute of Technology, New Delhi 110016, India. Email: [email protected]

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Tanusree Chakraborty [email protected]

Associate Professor, Dept. of Civil Engineering, Indian Institute of Technology, New Delhi 110016, India. Email: [email protected]

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Dipanjan Basu, Ph.D., M.ASCE [email protected]

Associate Professor, Dept. of Civil and Environmental Engineering, Univ. of Waterloo, Waterloo, ON, Canada N2L 3G1 (corresponding author). Email: [email protected]

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Erratum for “Settlement Estimation of Piled Rafts for Initial Design” by Priyanka Bhartiya, Tanusree Chakraborty, and Dipanjan Basu

Example 1: Case Study of Rectangular PRF

Example 2: Circular PRF

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