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Pile Design and Construction Practice, Fifth edition

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Thus the load to be carried by anchorage in shale � 1100 – 72 � 1028 kN. This load can be<br />

resisted by an anchor cable made up from seven 15.2 mm 7-wire Dyform str<strong>and</strong>s. The<br />

breaking load of each str<strong>and</strong> is 300 kN. Therefore<br />

working load � 1028<br />

� 0.49�breaking load of str<strong>and</strong><br />

7�300<br />

which is satisfactory. The approximate overall diameter of the 7-str<strong>and</strong> cable is 45 mm.<br />

Therefore for an allowable bond stress between steel <strong>and</strong> grout of 1.0 N/mm 2 :<br />

required bond length of cable � 1028�1000<br />

� 7.3 m<br />

��45�1.0�1000<br />

<strong>Pile</strong>s to resist uplift <strong>and</strong> lateral loading 359<br />

Drill the cable hole to 9 m <strong>and</strong> provide an unwrapped <strong>and</strong> cleaned bond length of 7 m with<br />

compression fittings swaged on to the lower end. The cable can be fed down a 150 mm<br />

borehole for which,<br />

1028�1000<br />

working bond stress between rock <strong>and</strong> grout �<br />

�0.31 N/mm2<br />

��150 �7.0 �1000<br />

which is satisfactory for a strong shale (Table 6.3). The stress is not excessive if the anchors<br />

are stressed to 1.5 times the working load during installation.<br />

From Figure 6.17a, the volume of a rock cone with a 30� half angle lifted by single anchor<br />

cable is 0.35 � 9 3 � 255 m 3 . The submerged weight of the rock cone � 1.3 � 9.81<br />

� 255/1000 � 3.25 MN.<br />

Factor of safety against uplift � 3.25<br />

� 3.1 (which is satisfactory)<br />

1.03<br />

The anchorage of the whole dock floor requires 70 lines of anchors (at right-angles to the<br />

centre-line of the dock) <strong>and</strong> 20 ranks of anchors ( parallel to the centre-line of the dock) to<br />

form the 3 m square grid. Therefore in Figure 6.17b, N � 70, M � 20, <strong>and</strong> P � S � 3 m.<br />

From Figure 6.17a, m/L � n/L � 0.57, <strong>and</strong> therefore, m � n � 0.57 � 9 � 5.1 m. Then<br />

P/n � s/m � 0.59 so that, from Figure 6.17b, �Vn�Vc ��Vm�Vc � 0.45.<br />

Because<br />

(P/n) 2 � (S/m) 2 � 2 � 0.59 2 � 0.7 is less than 4, there is composite overlapping of the rock<br />

cones, <strong>and</strong> the charts are not valid. The intersecting cones represent a rock volume roughly<br />

estimated to be 69 � 3 � 19 � 3 � 6 � 70794 m 3 .<br />

The sum giving the total force resisting uplift is thus as follows:<br />

210 � 60 � 0.8 � 2.4 � 9.81<br />

Weight of dock floor � � 237.3 MN<br />

1000<br />

210 � 60 � 8.0 � 0.8 � 9.81<br />

Submerged weight of soft clay � � 791.1 MN<br />

1000<br />

70 794 � 1.3 � 9.81<br />

Submerged weight of anchored rock � � 902.8 MN<br />

1000<br />

Total � 1931.2 MN

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