Lecture 38

NPTEL ADVANCED FOUNDATION ENGINEERING-I

Module 9

(Lecture 38)

DRILLED-SHAFT AND CAISSON FOUNDATIONS

Topics

1.1 UPLIFT CAPACITY OF DRILLED SHAFTS 1.2 EXAMPLES & SOLUTIONS


UPLIFT CAPACITY OF DRILLED SHAFTS

Sometimes drilled shafts must resist uplifting loads. Field observations of drilled-shaft uplift capacity are relatively scarce. The procedure for determining the ultimate uplifting load for drilled shafts without bells is similar to that for piles described in chapter 9 and will not be repeated here. When a short drilled shaft with a bell is subjected to an uplifting load, the nature of the failure surface in the soil will be like that shown in figure 9. 22. The net ultimate uplift capacity, , is

= [9.32]

Figure 9.22 Nature of failure surface in soil caused by uplifting force on drilled shaft with bell

Where

= gross ultimate uplift capacity = effective weight of the drilled shaft The magnitude of drilled shafts in sand can be estimated by the procedure outlined by Meyerhof and Adams (1968) and Das and Seely (1975):

= [9.33]


Where

= breakout factor = (/4)2 = Unit weight of soil above the bell (Note: If the soil is submerged, the effective unit weight should be used).

The breakout factor may be expressed as

= 2 tan + 1 + 1 [9.34]

Where

= nominal uplift coefficient = soil friction angle = shape factor coefficient The value of may be taken as 0.9 for all values of from 30 45. Meyerhof and Adams (1968) gave the variation of m as

Soil friction angle, (deg) 30 0.15

35 0.25

40 0.35

45 0.50

Experiments have shown that the value of increases with the / ratio to a critical value, (/) , and remains constant thereafter. The critical embedment ratio, (/) , increases with the soil friction angle. The approximate ranges are

Soil friction angle, (deg) (/) 30 4

35 5


40 7

45 9

Hence drilled shafts with / (/) are shallow foundations, and shafts with / (/) are deep foundations with regard to the uplift. The failure surface in soil at ultimate load as shown in figure 9.22 is for shallow foundations. For deep foundations, local shear failure takes place, and the failure surface in soil does not extend up to the ground surface. Based on the preceding considerations, the variation of with / is shown in figure 9.23.

Figure 9.23 Variation of the breakout factor, , with / and soil friction angle Following is a step-by-step procedure for the calculation of the net ultimate uplift capacity of drilled shafts with bells in sand:

1. Determine , , and / . 2. Estimate (/) and hence cr. 3. If / (/) , obtain from figure 9. 23. Now,


= + 4. If / > (/) ,

= + + () tan 0 [9.35]

The last term of equation (35) is for the frictional resistance developed along the soil-shaft interface from = 0 to = and is similar to equations (77 and 78 from chapter 9). The term is the effective stress at any depth , and and are taken from figure 9. 36b and 9.36c from chapter 9, respectively.

The net ultimate uplift capacity of drilled shafts with bell in clay can be estimated according to the procedure outlined by Das (1980):

= ( + ) [9.36]

Where

= undrailed cohesion = breakout factor = unit weight of clay soil above the bell As in the case of , the value of increases with the embedment ratio to a critical value of / = (/) and remains constant thereafter. Beyond the critical depth, 9. The critical embedment ratio is related to the undrained cohesion by

= 0.107 + 2.5 7 [9.37] Where

is in kN/m2 In English units,

= 0.738 + 2.5 7 [9.38] Where

is in lb/in2


Following is a step-by-step procedure for determining the net ultimate uplift capacity of drilled shafts with bell in clay:

1. Determine , , , and / . 2. Obtain (/) from equation (37) or equation (38) and obtain . 3. If / < (/) , obtain the value of from figure 9.24. 4. Use equation (36) to obtain . 5. If / > (/) , = 9. The magnitude of may then be obtained from

(9 + ) + ()( ) [9.39]

Figure 9. 9.24 Nondimensional plot of the breakout factor,

The last term of equation (39) is the skin resistance obtained from the adhesion along the soil-shaft interface and is similar to equation (70 from chapter 9). The magnitude of can be obtained from equations 71, 72, 73, and 74 from chapter 9).

Example 6

Refer to figure 9.22. A drilled shaft with bell has a shaft diameter of 0.76m, a bell diameter of 1.85 m, and a length of 9.5 m. the bell is supported by a dense sand ( 9.5 m) layer. However, a fine, loose sand layer exists above the bell (z = 0-9.5 m). For


this sand, = 16.4 kN/m3, = 32, and the approximate relative density is 30%. The entire structure is located above the water table. Determine the net allowable uplift capacity of the drilled shaft with a factor of safety of 3.

Solution

We begin with = 9.5 m, = 1.85 m, and / = 9.5/1.85 = 5.14. For =30, (/ ) = 4; and for = 35, (/) = 5. By interpretation, (/) 4.2 for = 32. So = (4.2)() = 7.77. Because/ = 5.13 > (/) = 4.2, it is a deep foundation.

According to equation (34),

= 2 tan + 1 + 1 Note that

rather than / was used in the preceding equation because it is a deep foundation. For = 32, 0.17. Hence = (2)(4.2)(0.9)(tan 32)[(0.17)(4.2) + 1] + 1 = 0.09 From equation 35,

= = ()( tan )0 = + 2 tan( )2 = 4 ()2 = 4 (1.85)2 = 2.687 m2 = 9.5 7.77 = 1.73 m Also, from figure 9. 36b and 36c from chapter 9, for = 32 and relative density = 30%, = 1.5 and / 0.73. Hence = (9.09)(2.687)(16.4)(9.5) + 2 (16.4)(0.76)(1.5) [tan(0.73 32)](1.73)2 = 3805.4 + 37.96 = 3843.36 3843 So, the net allowable capacity = 3843/ = 3843/3 = 1281kN. Example 7

Consider the drilled shaft described in example 6. If the soil above the bell is clay with an average value of the undrianed shear strength of 95 kN/m2, calculate the net ultimate uplift capacity. For clay, = 17.9 kN/m3.


Solution

From equation (37).

= 0.107 + 2.5 = (0.107)(95) + 2.5 = 12.67 This quantity is more than 7, so use (/) = 7. Hence, = (7)(1.85) = 12.95 m. = 12.95 is greater than = 9.5 m, so this drilled shaft is a shallow foundation for uplift consideration. For shallow foundations [equation (36)],

= ( + ) The magnitude of the breakout factor, , is determined from figure 9. 24:

= 9.51.857

= 0.734 So, /9 = 0.92, or = 8.82, and = (/4)2 = (/4)(1.85)2 = 2.688 m2. Thus = [(95)(8.28) + (17.9)(9.5)]2.688 = 2571.5 kN.

Lecture 38

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