Lecture 18

NPTEL - ADVANCED FOUNDATION ENGINEERING-1

Module 5

(Lecture 18) MAT FOUNDATIONS

Topics 1.1 FIELD SETTLEMENT OBSERVATIONS FOR MAT FOUNDATIONS 1.2 COMPENSATED FOUNDATIONS 1.3 Example

FIELD SETTLEMENT OBSERVATIONS FOR MAT FOUNDATIONS Several field settlement observations for mat foundations are currently available in the literature. In this section we compare the observed settlements for some mat foundations constructed over granular soil deposits with those obtained from equations (12 and 13). Meyerhof (1965) compiled the observed maximum settlements for mat foundations constructed on sand and gravel, as listed in table 1. In equation (13) if the depth factor, 1 + 0.33(𝐷𝐷𝑓𝑓 /𝐵𝐵), is assumed to be approximately 1, 𝑞𝑞 all (net )

𝑆𝑆𝑒𝑒 = 0.25𝑁𝑁

cor

[5.19]

Table 2 shows a comparison of the observed maximum settlements in table 1 and the settlements obtained from equation (19). For the cases considered the ratio of 𝑆𝑆𝑒𝑒 calculated /𝑆𝑆𝑒𝑒 observed varies from 0.84 to 3.6. Thus calculation of the net allowable bearing capacity with equation (12 or 13) will yield a safe and conservative value. Stuart and Graham (1975) reported the case history of the 13-story Ashby Institute building of Queens University, Belfast, Ireland, construction of which began in August 1960. It was supported by at foundation 180 ft (length)× 65 ft (width). Figure 5.5a shows a schematic diagram of the building cross section. The nature of the subsoil along with the field standard penetration resistance values at the south end of the building are shown in figure 5.5b. The base of the mat was constructed about 20 ft below the ground surface.

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Figure 5.5 Ashby Institute Building of Queens University, as reported by Stuart and Graham (1975); (a) building cross section; (b) subsoil conditions at south end The variation of the corrected standard penetration number with depth is shown in table 3. Note that the average 𝑁𝑁cor value between the bottom of the mat and a depth of 30 ft (≈ 𝐵𝐵/2) is about 17. The engineers estimated the average net dead and live load [equation (16)] at the level of the mat foundation to be about 3360 lb/ft 2 . From equation (13) 𝑆𝑆𝑒𝑒 =

𝑞𝑞 all (net )

𝐷𝐷 𝑓𝑓

0.25𝑁𝑁cor �1+0.33�

𝐵𝐵

��

[5.20]

Substituting appropriate values into equation (20) yield the settlement at the south end of the building: (3360 /1000 )

𝑆𝑆𝑒𝑒 = (0.25)(17)[1+0.33(20/65)] = 0.72 in.

The construction of the building was completed in February 1964. Figure 5.6 shows the variation of the mean settlement of the mat at the south end. In 1972 (eight years after completion of the building) the mean settlement was about 0.55 in. Thus the estimated settlement of 0.72 in. is about 30% higher than that actually observed.

NPTEL - ADVANCED FOUNDATION ENGINEERING-1 Table 1 Observed Maximum Settlement of Mat Foundations on Sand and Gravel Reference

𝐵𝐵(ft)

Case no.

Structure

1

T. Edison

Rios and Silva Sao Paulo, (1948) Brazil

60

2

Banco Brasil

do Rios and Silva (1948); Sao Paulo, Vargas Brazil (1961)

3

Iparanga

𝑁𝑁cor (𝑎𝑎𝑎𝑎𝑎𝑎) 𝑞𝑞all (net ) (kip/ Observed maximum settlement, 𝑆𝑆𝑒𝑒 (in. ) ft 2 ) 15

4.8

0.6

75

18

5.0

1.1

Vargas (1948)

30

9

6.4

1.4

C. B. I. Vargas Esplanada (1961)

48

22

8.0

1.1

Vargas (1948)

13

20

4.8

0.5

Schultze (1962)

74

25

5.0

0.95

Schultze (1962)

52

20

4.6

0.8

Schultze (1962)

67

10

3.6

0.4

Sao Paulo, Brazil 4

Sao Paulo, Brazil 5

Riscala Sao Paulo, Brazil

6

Thyssen Dusseldorf, Germany

7

Ministry Dusseldorf, Germany

8

Chimney Cologne, Germany

NPTEL - ADVANCED FOUNDATION ENGINEERING-1 Table 2 Comparison of Settlements Observed and Calculated Case 1

𝑆𝑆𝑒𝑒calculated 𝑆𝑆𝑒𝑒observed

1

Maximum observed settlement, 𝑆𝑆𝑒𝑒 (in. ) 0.6

Calculated settlement, 𝑆𝑆𝑒𝑒 [equation (19)]

2

1.1

1.11

1.0

3

1.4

2.84

2.03

4

1.1

1.45

1.32

5

0.5

0.96

1.92

6

0.95

0.8

0.84

7

0.8

0.92

1.15

8

0.4

1.44

3.6

1.28

2.1

Refer to table 1

Table 3 Determination of Corrected Standard Penetration Resistance Depth ground (ft) 20

below Field standard 𝜎𝜎𝑣𝑣′𝑎𝑎 (ton/ft 2 ) surface penetration number, 𝑁𝑁𝐹𝐹 21

1.2

25

22

30

1 𝐶𝐶𝑁𝑁 = � ′ 𝜎𝜎 𝛾𝛾

(𝑏𝑏)

𝑁𝑁𝑐𝑐𝑐𝑐𝑐𝑐 [equation chapter 2)]

0.91

19

1.5

0.82

18

21

1.8

0.75

16

35

15

2.1

0.69

10

40

20

2.4

0.65

13

45

18

2.7

0.61

11

50

50

3.0

0.58

29

𝜎𝜎′𝑣𝑣 = (depth); 𝛾𝛾 = 120 lb/ft 3 (assumed) Table 4 from chapter 2

7

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Figure 5.6 Mean settlement at the south end of the mat foundation, as reported by Stuart and Graham (1975) COMPENSATED FOUNDATIONS The settlement of a mat foundation can be reduced by decreasing the net pressure increase on soil, which can be done by increasing the depth of embedment, 𝐷𝐷𝑓𝑓 . This increase is particularly important for mats on soft clays, where large consolidation settlements are expected. From equation (16), the net average applied pressure on soil is 𝑄𝑄

𝑞𝑞 = 𝐴𝐴 − 𝛾𝛾𝐷𝐷𝑓𝑓

For no increase of the net soil pressure on soil below a raft foundation, 𝑞𝑞 should be zero. Thus 𝑄𝑄

𝐷𝐷𝑓𝑓 = 𝐴𝐴𝐴𝐴

[5.21]

This relation for 𝐷𝐷𝑓𝑓 is usually referred to as the depth of a fully compensated foundation.

The factor of safety against bearing capacity failure for partially compensated foundations (that is, 𝐷𝐷𝑓𝑓 < 𝑄𝑄/𝐴𝐴𝐴𝐴) may be given as 𝐹𝐹𝐹𝐹 =

𝑞𝑞 net (𝐮𝐮) 𝑞𝑞

=

𝑞𝑞 net (𝑢𝑢 ) 𝑄𝑄 −𝛾𝛾𝐷𝐷𝑓𝑓 𝐴𝐴

[5.22]

For saturated clays, the factor of safety against bearing capacity failure can thus be obtained by substituting equation (10) into equation (22): 𝐹𝐹𝐹𝐹 =

𝐷𝐷 𝑓𝑓 0.195 𝐵𝐵 ��1+0.4 � 𝐿𝐿 𝐵𝐵 𝑄𝑄 −𝛾𝛾𝐷𝐷𝑓𝑓 𝐴𝐴

5.14�1+

[5.23]

NPTEL - ADVANCED FOUNDATION ENGINEERING-1 Example 3 Refer to figure 5.4. The mat has dimensions of 60 ft × 100 ft. The total dead and live load on the mat is 25 × 103 kip. The mat is placed over a saturated clay having a unit weight of 120 lb/ft 3 and 𝑐𝑐𝑢𝑢 = 2800 lb/ft 2 . Given 𝐷𝐷𝑓𝑓 = 5 ft, determine the factor of safety against bearing capacity failure. Solution From equation (23), the factor of safety 𝐹𝐹𝐹𝐹 =

𝐷𝐷 𝑓𝑓 0.195 𝐵𝐵 ��1+0.4 � 𝐿𝐿 𝐵𝐵 𝑄𝑄 −𝛾𝛾𝐷𝐷𝑓𝑓 𝐴𝐴

5.14𝑐𝑐𝑢𝑢 �1+

Given: 𝑐𝑐𝑢𝑢 = 2800 lb/ft 2 , 𝐷𝐷𝑓𝑓 = 5 ft, 𝐵𝐵 = 60 ft, 𝐿𝐿 = 100 ft, and 𝛾𝛾 = 120 lb/ft 3 . Hence 𝐹𝐹𝐹𝐹 =

(0.195 )(60) 5 ��1+0.4� �� 100 60 25 ×10 6 lb � �−(120)(5) 60 ×100

(5.14)(2800 )�1+

Example 4

= 4.66

Consider a mat foundation 90 ft × 120 ft in plan, as shown in figure 5.7. The total dead load and live load on the raft is 45 × 103 kip. Estimate the consolidation settlement at the center of the foundation.

Figure 5.7

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Solution From equation (64 from chapter 1) 𝐶𝐶 𝐻𝐻

𝑝𝑝 𝑜𝑜 +∆𝑝𝑝 𝑎𝑎𝑎𝑎

𝑐𝑐 𝑐𝑐 𝑆𝑆𝑐𝑐 = 1+𝑒𝑒 log � 𝑜𝑜

𝑝𝑝 𝑜𝑜

�

𝑝𝑝𝑜𝑜 = (11)(100) + (40)(121.5 − 62.4) + 𝐻𝐻𝑐𝑐 = 18 × 12 in.

18 2

(118 − 62.4) ≈ 3964lb/ft 2

𝐶𝐶𝑐𝑐 = 0.28 𝑒𝑒𝑜𝑜 = 0.9

For 𝑄𝑄 = 45 × 106 lb, the net load per unit area 45×10 6

𝑄𝑄

𝑞𝑞 = 𝐴𝐴 − 𝛾𝛾𝐷𝐷𝑓𝑓 = 90×120 − (100)(6) ≈ 3567lb/ft 2

In order to calculate ∆𝑝𝑝𝑎𝑎𝑎𝑎 , we refer to section 5. The loaded area can be divided into four areas, each measuring45 ft × 60 ft. Now using equation (19 from chapter 4), we can calculate the average stress increase in the clay layer below the corner of each rectangular area, or ∆𝑝𝑝𝑎𝑎𝑎𝑎(𝐻𝐻2 /𝐻𝐻1 ) = 𝑞𝑞 � = 3567 �

𝐻𝐻2 𝐼𝐼𝑎𝑎 (𝐻𝐻 2 ) −𝐻𝐻1 𝐼𝐼𝑎𝑎 (𝐻𝐻 1 ) 𝐻𝐻2 −𝐻𝐻1

�

(5+40+18)𝐼𝐼𝑎𝑎 (𝐻𝐻 2 ) −(5+40)𝐼𝐼𝑎𝑎 (𝐻𝐻 1 )

For 𝐼𝐼𝑎𝑎(𝐻𝐻2 ) , 𝐵𝐵

18

�

45

𝑚𝑚 = 𝐻𝐻 = 5+40+18 = 0.71 𝐿𝐿

2

60

𝑛𝑛 = 𝐻𝐻 = 63 = 0.95 2

From figure 5.8, for 𝑚𝑚 = 0.71 and 𝑛𝑛 = 0.95, the value of 𝐼𝐼𝑎𝑎(𝐻𝐻2 ) is 0.21. Again, for 𝐼𝐼𝑎𝑎(𝐻𝐻1 ) ,

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Figure 5.8 Conventional rigid mat foundation design

Fig 5.9 (Continued)

NPTEL - ADVANCED FOUNDATION ENGINEERING-1 𝐵𝐵

45

𝑚𝑚 = 𝐻𝐻 = 45 = 1 𝐿𝐿

1

60

𝑛𝑛 = 𝐻𝐻 = 45 = 1.33 1

From figure 5.8, 𝐼𝐼𝑎𝑎(𝐻𝐻1 ) = 0.225, so ∆𝑝𝑝𝑎𝑎𝑎𝑎(𝐻𝐻2 /𝐻𝐻1 ) = 3567 �

(63)(0.21)−(45)(0.225) 18

� = 615.3 lb/ft 2

So, the stress increase below the center of the 90 ft × 120 ft area is (4)(615.3) = 2461.2 lb/ft 2 . Thus 𝑆𝑆𝑐𝑐 =

(0.28)(18×12) 1+0.9

3964+2461 .2

log �

3964

� = 6.68 in.