Question 17.8: Figure Ex. 17.8 shows a drilled pier penetrating an IGM: cla...

Figure Ex. 17.8 shows a drilled pier penetrating an IGM: clay-shale to a depth of 8 m. Joints exists within the IGM stratum. The following data are available: L_{s}=8 m \left(=z_{c}\right), d=1.5 m , q_{u}(\text { rock })=3 \times10^{3} kN / m ^{2}, E_{i}(\text { rock })=600 \times 10^{3} kN / m ^{2} concrete slump = 175 mm, unit weight of concrete \gamma_{c}=24 kN / m ^{3}, E_{c} \text { (concrete) }=435 \times 10^{6} kN / m ^{2}, \text { and } RQD =70 \text { percent, } q_{u} \text { (concrete) }=435 \times 10^{6} kN / m ^{2}. Determine the ultimate frictional load Q_{f}(\max ).

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(a) Determine \alpha in Eq. (17.26)

f_{a}=\alpha q_{u} (17.26)

f_{a}=\alpha q_{u} \text { where } q_{u}=3 MPa for rock

For the depth of socket L_{s}=8 m, and slump =175 mm

 

From Eq. (17.27)

 

\sigma_{n}=M \gamma_{c} z_{c} (17.27)

 

\sigma_{n}=M \gamma_{c} z_{c}=0.76 \times 24 \times 8=146 kN / m ^{2}

 

p_{a}=101 kN / m ^{2}, \sigma_{n} / p_{a}=146 / 101=1.45

 

From Fig. 17.16 for q_{u}=3 MPa \text { and } \sigma_{n} / p_{a}=1.45, \text { we have } \alpha=0.11

 

(b) Determination of f_{a}

 

f_{a}=0.11 \times 3=0.33 MPa

 

(c) Determination f_{a a} in Eq. (17.28)

 

f_{a a}=f_{a} R_{a} (17.28)

 

For RQD = 70%, E_{m} / E_{i}=0.1 from Table 17.5 for open joints, and f_{a a} / f_{a}\left(=R_{a}\right)=0.55 from Table 17.6

 

f_{\max }=f_{a a}=0.55 \times 0.33=0.182 MPa =182 kN / m ^{2}

 

(d) Ultimate friction load Q_{f}

 

Q_{f}=P L f_{a a}=3.14 \times 1.5 \times 8 \times 182=6,858 kN

 

Table 17.5 Estimation of E_{m} / E_{i} based on ROD (Modified after Carter and Kulhawy, 1988)
E_{m} / E_{i}
ROD (percent) Closed joints Open joints
100 1 0.6
70 0.7 0.1
50 0.15 0.1
20 0.05 0.05

 

Table 17.6 f_{a a} / f_{a} \text { based on } E_{m} / E_{i} (O’Neill et al., 1996)
E_{m} / E_{i} f_{aa} / f_{a}
1 1
0.5 0.8
0.3 0.7
0.1 0.55
0.05 0.45
17.8.

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