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The soil on both sides of the anchored sheet pile wall detailed in Figure 6.40 h

ID: 1827727 • Letter: T

Question

The soil on both sides of the anchored sheet pile wall detailed in Figure 6.40 has a saturated unit weight of 21 kN/m3 and a unit weight above the water table of 18kN/m3. Characteristic parameters for the soil are c0 ¼ 0, 0 ¼ 36 and ¼ 0. There is a lag of 1.5m between the water table behind the wall and tidal level in front. Determine (a) the factor of safety with respect to gross passive resistance and the force in each tie rod using the traditional method and (b) the required depth of embedment using the limit state method. The soil on both sides of the anchored sheet pile wall detailed in Figure 6.40 has a saturated unit weight of 21 kN/m3 and a unit weight above the water table of 18kN/m3. Characteristic parameters for the soil are c0 ¼ 0, 0 ¼ 36 and ¼ 0. There is a lag of 1.5m between the water table behind the wall and tidal level in front. Determine (a) the factor of safety with respect to gross passive resistance and the force in each tie rod using the traditional method and (b) the required depth of embedment using the limit state method. The soil on both sides of the anchored sheet pile wall detailed in Figure 6.40 has a saturated unit weight of 21 kN/m3 and a unit weight above the water table of 18kN/m3. Characteristic parameters for the soil are c0 ¼ 0, 0 ¼ 36 and ¼ 0. There is a lag of 1.5m between the water table behind the wall and tidal level in front. Determine (a) the factor of safety with respect to gross passive resistance and the force in each tie rod using the traditional method and (b) the required depth of embedment using the limit state method.

Explanation / Answer

(a) For

0 ¼ 36, Ka ¼ 0:26 and Kp ¼ 3:85;

0 ¼ 21 9:8 ¼ 11:2kN=m3

The pressure distribution is shown in Figure Q6.8. In this case the net water pressure at

C is given by

uC ¼

15:0

16:5 1:5 9:8 ¼ 13:4kN=m2

The average seepage pressure is

j ¼

1:5

16:5 9:8 ¼ 0:9kN=m3

Hence,

0 þ j ¼ 11:2 þ 0:9 ¼ 12:1kN=m3

0 j ¼ 11:2 0:9 ¼ 10:3kN=m3

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