NPTEL β ADVANCED FOUNDATION ENGINEERING-I. Module 9. (Lecture 38).
DRILLED-SHAFT AND CAISSON FOUNDATIONS. Topics. 1.1 UPLIFTΒ ...
NPTEL β 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
NPTEL β ADVANCED FOUNDATION ENGINEERING-I
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, πππ’π’π’π’ , is πππ’π’π’π’ = πππ’π’π’π’ β ππ
[9.32]
Figure 9.22 Nature of failure surface in soil caused by uplifting force on drilled shaft with bell Where πππ’π’π’π’ = gross ultimate uplift capacity
ππ = effective weight of the drilled shaft
The magnitude of πππ’π’π’π’ drilled shafts in sand can be estimated by the procedure outlined by Meyerhof and Adams (1968) and Das and Seely (1975): πππ’π’π’π’ = π΅π΅ππ π΄π΄ππ πΎπΎπΎπΎ
[9.33]
NPTEL β ADVANCED FOUNDATION ENGINEERING-I Where π΅π΅ππ = breakout factor π΄π΄ππ = (ππ/4)π·π·ππ2
πΎπΎ = 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 πΏπΏ
πΏπΏ
π΅π΅ππ = 2 π·π· πΎπΎβ²π’π’ tan ππ οΏ½ππ π·π· + 1οΏ½ + 1 ππ
[9.34]
ππ
Where πΎπΎβ²π’π’ = nominal uplift coefficient ππ = soil friction angle
ππ = shape factor coefficient
The value of πΎπΎβ²π’π’ may be taken as 0.9 for all values of ππ from 30 β 45Β° . Meyerhof and Adams (1968) gave the variation of m as Soil friction angle, ππ (deg) 30
0.15
35
0.25
40
0.35
45
0.50
ππ
Experiments have shown that the value of π΅π΅ππ increases with the πΏπΏ/π·π·ππ ratio to a critical value, (πΏπΏ/π·π·ππ )ππππ , and remains constant thereafter. The critical embedment ratio, (πΏπΏ/π·π·ππ )ππππ , increases with the soil friction angle. The approximate ranges are Soil friction angle, ππ (deg) 30
(πΏπΏ/π·π·ππ )ππππ
35
5
4
NPTEL β ADVANCED FOUNDATION ENGINEERING-I
40
7
45
9
Hence drilled shafts with πΏπΏ/π·π·ππ β€ (πΏπΏ/π·π·ππ )ππππ are shallow foundations, and shafts with πΏπΏ/π·π·ππ β€ (πΏπΏ/π·π·ππ )ππππ 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 π΅π΅ππ with πΏπΏ/π·π·ππ is shown in figure 9.23.
Figure 9.23 Variation of the breakout factor, π΅π΅ππ , with πΏπΏ/π·π·ππ 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 πΏπΏ, π·π·ππ , and πΏπΏ/π·π·ππ . 2. Estimate (πΏπΏ/π·π·ππ )ππππ and hence πΏπΏcr . 3. If πΏπΏ/π·π·ππ β€ (πΏπΏ/π·π·ππ )ππππ , obtain π΅π΅ππ from figure 9. 23. Now,
NPTEL β ADVANCED FOUNDATION ENGINEERING-I πππ’π’π’π’ = π΅π΅ππ π΄π΄ππ πΎπΎπΎπΎ + ππ
4. If πΏπΏ/π·π·ππ > (πΏπΏ/π·π·ππ )ππππ ,
πΏπΏβπΏπΏππππ
πππ’π’π’π’ = π΅π΅ππ π΄π΄ππ πΎπΎπΎπΎ + ππ + β«0
(πππ·π·π π )ππ β² π£π£ πΎπΎ β² π’π’ tan πΏπΏ ππππ
[9.35]
The last term of equation (35) is for the frictional resistance developed along the soilshaft interface from π§π§ = 0 to π§π§ = πΏπΏ β πΏπΏππππ and is similar to equations (77 and 78 from chapter 9). The term ππβ²π£π£ is the effective stress at any depth π§π§, and πΎπΎπ’π’ and πΏπΏ 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): πππ’π’π’π’ = (πππ’π’ π΅π΅ππ + πΎπΎπΎπΎ)π΄π΄ππ
[9.36]
Where πππ’π’ = undrailed cohesion π΅π΅ππ = breakout factor
πΎπΎ = unit weight of clay soil above the bell
As in the case of π΅π΅ππ , the value of π΅π΅ππ increases with the embedment ratio to a critical value of πΏπΏ/π·π·ππ = (πΏπΏ/π·π·ππ )ππππ and remains constant thereafter. Beyond the critical depth, π΅π΅ππ β 9. The critical embedment ratio is related to the undrained cohesion by πΏπΏ
οΏ½π·π· οΏ½ ππ
ππππ
Where
= 0.107πππ’π’ + 2.5 β€ 7
[9.37]
πππ’π’ is in kN/m2
In English units, πΏπΏ
οΏ½π·π· οΏ½ ππ
ππππ
Where
= 0.738πππ’π’ + 2.5 β€ 7
πππ’π’ is in lb/in2
[9.38]
NPTEL β ADVANCED FOUNDATION ENGINEERING-I Following is a step-by-step procedure for determining the net ultimate uplift capacity of drilled shafts with bell in clay: 1. 2. 3. 4. 5.
Determine πππ’π’ , πΏπΏ, π·π·ππ , and πΏπΏ/π·π·ππ . Obtain (πΏπΏ/π·π·ππ )ππππ from equation (37) or equation (38) and obtain πΏπΏππππ . If πΏπΏ/π·π· < (πΏπΏ/π·π·ππ )ππππ , obtain the value of π΅π΅ππ from figure 9.24. Use equation (36) to obtain πππ’π’π’π’ . If πΏπΏ/π·π· > (πΏπΏ/π·π·ππ )ππππ , π΅π΅ππ = 9. The magnitude of πππ’π’π’π’ may then be obtained from
πππ’π’π’π’ (9πππ’π’ + πΎπΎπΎπΎ) + Ξ£ (πππ·π·π π )(πΏπΏ β πΏπΏππππ )πΌπΌ β² πππ’π’
[9.39]
Figure 9. 9.24 Nondimensional plot of the breakout factor, π΅π΅ππ 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 πΌπΌβ² 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 (π§π§ β₯ 9.5 m) layer. However, a fine, loose sand layer exists above the bell (z = 0-9.5 m). For
NPTEL β ADVANCED FOUNDATION ENGINEERING-I this sand, πΎπΎ = 16.4 kN/m3 , ππ = 32Β° , 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 πΏπΏ = 9.5 m, π·π·ππ = 1.85 m, and πΏπΏ/π·π·ππ = 9.5/1.85 = 5.14. Forππ = 30Β° , (πΏπΏ/π·π·ππ )ππππ = 4; and for ππ = 35Β° , (πΏπΏ/π·π·ππ )ππππ = 5. By interpretation, (πΏπΏ/π·π·ππ )ππππ β 4.2 for ππ = 32Β° . So πΏπΏππππ = (4.2)(π·π·ππ ) = 7.77. BecauseπΏπΏ/π·π·ππ = 5.13 > (πΏπΏ/π·π·ππ )ππππ = 4.2, it is a deep foundation. According to equation (34), πΏπΏ
πΏπΏ
π΅π΅ππ = 2 οΏ½π·π· οΏ½ πΎπΎβ²π’π’ tan ππ οΏ½ππ οΏ½π·π· οΏ½ ππ
ππ
ππππ
πΏπΏ
ππππ
+ 1οΏ½ + 1
Note that οΏ½π·π· οΏ½ rather than πΏπΏ/π·π·ππ was used in the preceding equation because it is a deep ππ
ππππ
foundation. For ππ = 32Β° , ππ β 0.17. Hence
π΅π΅ππ = (2)(4.2)(0.9)(tan 32Β° )[(0.17)(4.2) + 1] + 1 = 0.09 From equation 35,
πΏπΏβπΏπΏππππ
πππ’π’π’π’ = πππ’π’π’π’ β ππ = π΅π΅ππ π΄π΄ππ πΎπΎπΎπΎ β«0 ππ
(πππ·π·π π )(ππ β² π£π£ πΎπΎ β² π’π’ tan ππ)ππππ
= π΅π΅ππ π΄π΄ππ πΎπΎπΎπΎ + 2 πΎπΎπ·π·π π πΎπΎβ²π’π’ tanπΏπΏ(πΏπΏ β πΏπΏππππ )2 ππ
ππ
π΄π΄ππ = οΏ½ 4 οΏ½ (π·π·ππ )2 = οΏ½ 4 οΏ½ (1.85)2 = 2.687 m2 πΏπΏ β πΏπΏππππ = 9.5 β 7.77 = 1.73 m
Also, from figure 9. 36b and 36c from chapter 9, for ππ = 32Β° and relative density = 30%, πΎπΎβ²π’π’ = 1.5 and πΏπΏ/ππ β 0.73. Hence ππ
πππ’π’π’π’ = (9.09)(2.687)(16.4)(9.5) + οΏ½ 2 οΏ½ (16.4)(0.76)(1.5) Γ [tan(0.73 Γ 32)](1.73)2 = 3805.4 + 37.96 = 3843.36 β 3843
So, the net allowable capacity = 3843/πΉπΉπΉπΉ = 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, πΎπΎ = 17.9 kN/m3 .
NPTEL β ADVANCED FOUNDATION ENGINEERING-I Solution From equation (37). πΏπΏ
οΏ½π·π· οΏ½ ππ
ππππ
= 0.107πππ’π’ + 2.5 = (0.107)(95) + 2.5 = 12.67
This quantity is more than 7, so use (πΏπΏ/π·π·ππ )ππππ = 7. Hence, πΏπΏππππ = (7)(1.85) = 12.95 m. πΏπΏππππ = 12.95 is greater than πΏπΏ = 9.5 m, so this drilled shaft is a shallow foundation for uplift consideration. For shallow foundations [equation (36)], πππ’π’π’π’ = (πππ’π’ π΅π΅ππ + πΎπΎπΎπΎ)π΄π΄ππ
The magnitude of the breakout factor, π΅π΅ππ , is determined from figure 9. 24: πΏπΏ οΏ½ π·π· ππ πΏπΏ οΏ½ οΏ½ π·π· ππ ππππ
οΏ½
=
οΏ½
9.5 οΏ½ 1.85
7
= 0.734
So, π΅π΅ππ /9 = 0.92, or π΅π΅ππ = 8.82, and π΄π΄ππ = (ππ/4)π·π·ππ2 = (ππ/4)(1.85)2 = 2.688 m2. Thus πππ’π’π’π’ = [(95)(8.28) + (17.9)(9.5)]2.688 = 2571.5 kN.
