Chapter12

12.1 Show that the ultimate load for a strip footing under long-term conditions 1 using the two triangle failure surfaces shown in Fig. P12.1 is Pu  γB 2 N γ , 2 2 2 tan    tan    3 2sin 3   where N  1  tan 2     2 1  tan    cos    sin 3   

Solution 12.1

P B WB B NB

TAB

TAB NAB

WA

A TA

TB

NA Z X

From the free body diagram of A, we get

A  N A tan 

AB

  Fx 

 N AB tan 

 0 : N AB  N A sin 45  A cos 45  0 Putting equation (1) into (3) we get

(1) (2)

(3)

 N AB  NA 

 Fz  0 :

2 N A 1  tan  2 N AB 2 1  tan 

2



AB  WA  N A cos 45  A sin 45  0

(4)

(5)

(6)

Putting equation (1) and (2) into (6), we get

N AB tan   WA 

2 N A 1  tan  2

(7)

Substituting (5) into (7) we obtain

 1  tan      WA N AB  tan   1  tan     tan   tan 2   1  tan    = - WA N AB   1  tan    1  B2 1  tan   WA 1  tan  N AB   2 tan 2'  2tan'  1 tan 2'  2tan'  1 where WA 

(8)

1 2 B 2

Similarly summing forces in the x direction for B, we get

N AB 

2 N B 1  tan  2

(9)

Similarly summing forces in the z direction for B, after substituting (9) and simplifying we get

 1  2tan  tan 2  N AB   = WB  P 1  tan  

WB 

1 1 2 B and P  B 2 N  2 2

Substituting (8) and (11) in (10) we get

(10)

(11)

N 



 1+tan  1  2tan  tan 2

 1

 tan   2tan  1 1  tan 2





2tan tan 2  3 2tan 3  6tan  1  tan tan 2  2tan  1 1  tan 1  tan 2  2 2sin3 Simplifying weget N   cos  sin3 





12.2 A strip footing, 5 m wide, is founded on the surface of a deep deposit of clay. The undrained shear strength of the clay increases linearly from 3 kPa at the surface to 10 kPa at a depth of 5 m. Estimate the vertical ultimate load assuming that the load is applied at an eccentricity of 0.5 m from the center of the footing’s width. [Hint: Try a circular failure surface, determine the equation for the distribution of shear strength with depth, and integrate the shear strength over the radius to find the shear force.] Solution 12.2 Pa Pa -e e O 3 kPa



z

B/2

d dl

5m

Slope = 7/5

Radius = B

10 kPa The answer will vary according to the assumed failure mechanism. The eccentricity could be to the right (e) or left (-e) of the centerline. Assume a circular failure plane and consider an element at a depth z. The undrained shear strength at this depth is su = 7z/5 + 3 Now, consider an element d as shown in the figure above. The arc length is dl = B d and z = B sin  The force on this arc length = su B d = (7z/5 +3) B d = (7B sin  /5 +3) B d

 Mo  0 : / 2    B   7  Pu   e   2   Bsin   3  Bd B  0 2      0 5  / 2

B   7  Pu   e   2  Bcos   3 B2  0 2   5 0 (The 2 in the second term of the above equations comes from the fact we are integrating over a quadrant and we have two quadrants.) / 2

/ 2

 7   7  Now, 2   Bcos   3 B2  2    5  cos   3 52  585.7 kN.m  5 0  5 0

Pu 

585.7 5    0.75  2 

 180.2kN or 334.7kN

12.3

The centroid of a square foundation of sides 5 m is located 10 m away from the edge of a vertical cut of depth 4 m. The soil is a stiff clay with an undrained strength of 20 kPa and a unit weight of 16 kN/m3. Calculate the vertical ultimate load. Assume a circular failure surface for the footing and a planar surface for the cut.

Solution 12.3

P 1.5m 2.5m

r = 1.5m 3.5m

E cut

D

O

E

C

= 450

D

 C Section B

For undrained condition,   0 Assuming a planar failure plane for the cut. The slope of the failure plane (Chapter 10) is:    45   45 2 Assuming a circular failure mode for the footing. We find that the circular failure surface intersects with the failure plane of the cut. Therefore, we can assume that the surface CE has no shearing resistance.

P

Consider section ECD as shown in the figure “Section B”

 Length CE = r () = 1.5  0.375  1.178m 4 Length CD = 1.5  cos(45 )  1.061 m

B/2

3.5m

D

su (CD)

O

C

B su

Moment about O (figure on right)

 Mo  0 : B  s u ( B  1.178 )  B  ( s u  1.061 cos 45 )  3.5  0 2 2.5P  20(  5  1.178 )  5  ( 20  1.061 cos 45 )  3.5

P

P

1400  560 kN / m 2.5

su (CD) cos 45

12.4

Calculate the ultimate net bearing capacity of (a) a strip footing 2 m wide, (b) a square footing 3 m  3 m, and (c) a circular footing 3 m in diameter. All footings are located on the ground surface and the groundwater level is at the ground surface. The soil is medium-dense coarse-grained with γsat = 17kN/m3 and  p' = 30 from direct shear tests.

Solution 12.4

1 BN  s  ,     = 17 – 9.8 = 7.2 kN/ m 3 2 N  0.1054 exp (9.6' )  16.06

qu 

B B B  0 (strip footing)  1 (square footing), , L L L s   0.6 (square or circular footing); s γ  1 (strip)

s   1  0.4

1  7.2  2  16.06  1 = 116 kPa 2 1  7.2  3  16.06  0.6 = 104 kPa (b) q u (square)= 2 1  7.2  3  16.06  0.6 = 104 kPa (c) q u (circular)= 2 (a) q u (strip)=

12.5

A strip footing, founded on dense sand (  'p = 35 from direct shear tests.and γsat = 17kN/m3 ), is to be designed to support a vertical load of 400 kN per

meter length. Determine a suitable width for this footing for FS = 3. The footing is located 1 m below the ground surface. The groundwater level is 10 m below the ground surface. Solution 12.5

Assume B = 1.2m and the test data are from plane strain tests. Use Davis & Booker expression for N  . Using the spreadsheet with B =1.1 m

    35 , Nq  33.3, Nq  1  32.3, N  37.1, sq  s  1 dq  1.23,d   1.0

q u   Df (N q  1) s q d q  0.5  B N  s  d  q u  (17 1 32.3 11.23)  (0.5  17  1.2  37.1 1 1)  1023kPa 400  363.6 1.11 1023 FS   3.0 363.6  17 1

a 

B = 1.1 m is suitable width for the footing

12.6

A square footing, 3 m wide, is located 1.5 m below the surface of a stiff clay. Determine the allowable bearing capacity for short-term condition if (su)p =100 kPa, and γsat = 20 kN/m3. If the footing were located on the surface, what would be the allowable bearing capacity? Use FS = 3. Comment on the use of the (su)p value for both the embedded and the surface footing.

Solution 12.6

Stiff clay s u  100kPa,  sat  20kN / m 3

q u  5.14s u sc d c sc  1.2,d c  1.17 Position (1): D f  1.5m

qu  5.14 100 1.2 1.17  719 kPa qu  Df 3 719   20 1.5  270 kPa 3

qa 

Position (2): Df  0

qu  5.14 100 1.2 1  617kPa

q u 617   206kPa 3 3 Comment on the use of the (su)p value for both the embedded and the surface footing. The (su)p may not be the same. The value of su depends on the void ratio (or confining pressure). The confining pressure changes with depth so su will change with depth. Also, soils tend to be overconsolidated near the surface with the possibility that a tensile rather than a general failure mode can occur. qa 

12.7

A column carrying a load of 750 kN is to be founded on a square footing at a depth of 2 m below the ground surface in a deep clay stratum. What will be the size of the footing for FS = 3 for TSA? The soil parameters are, γsat = 18.5kN/m3 , and su = 55 kPa. The groundwater level is at the base of the footing but it is expected to rise to the ground surface during rainy seasons.

Solution 12.7

Clay sat  18.5kN / m3 , su  55kPa , '  8.7kN / m3 ,

TSA Assume B =2 m

B  1.2 L = 5.14  55  1.2  1.33 = 451 kPa

q u  5.14 suscdc , sc  1  0.2 q ult

applied stress, = a  FS =

750  187.5 kPa 2 2

451 3 187.5  8.7  2

Use B =2 m

, dc  1.33

12.8

Repeat Exercise 12.7 with a moment of 250 kN.m about an axis parallel to the length in addition to the vertical load.

Solution 12.8

Clay sat  18.5kN / m3 , su  55kPa , '  8.7kN / m3 ,

TSA

eB 

250  0.33 m 750

Assume B = 2.8 m max 

750  6  0.33  1  164 kPa 2.8  2.82 

B'  B  2e B  3  2  0.33  2.33m sc  1.2,dc  1.31 ,

TSA qu  5.14  55  1.2  1.31  444 kPa

FS =

444 3 164  2  8.7

Use a footing of width 2.8 m

12.9

A square footing located on a dense sand is required to carry a dead load of 200 kN and a live load of 300 kN, both inclined at 15 to the vertical plane along the width. The building code requires an embedment depth of 1.2 m. Groundwater level is at 1 m below the ground surface. Calculate the size of the footing using ASD and LRFD for  p'  35 from direct shear tests, γsat = 18.5kN/m3 and FS = 3. Assume the soil above the groundwater level

to be saturated. Solution 12.9 Nq  33.3, Nq  1  32.3, N  37.13 iq  0.63 i   0.46,   18.5  9.8  8.7 kN/ m 3

ASD: Total load + 300 + 200 = 500 kN LRFD: Pu = 1.25 x 200 + 1.75 x 300 = 775 kN Assume B =1.6 m

D f  1  18.5  8.7  0.2  20.2kPa qu  ( 32.3  20.2  0.63 )  ( 0.5  8.7  1.6  27.13  0.46 )  528 kPa

ASD: Vertical load = 500 cos (15) = 483 kN Applied stress =

FS =

483  188.7 kPa 1.62

528 3 188.7  20.2

Use a square footing of width 1.6 m

LRFD Vertical load = 775 cos (15) = 748.6 kN qult  528  20.2  548.2kPa

i qult  0.8  548.2  438.6 kPa Applied stress =

748.6  292.4 kPa < 438.6 1.62

kPa

The footing size is more than adequate for LRFD. You can use a smaller footing ( B = 1.2 m can be used)

12.10 The footing for a bridge pier is to be founded in sand, as shown in Fig. P12.10. The clay layer is normally consolidated with Cc = 0.25. Determine the factor of safety against bearing capacity failure and the total settlement (elastic compression and primary consolidation) of the pier. The shear strength parameters were obtained from direct simple shear tests.

Cc = 0.25

Solution 12.10 B = 3m, Df  5m  B, GWL below B from base. No effect from GWL

q applied 

12  103  400 kPa 3  10

Dense sand:

 2.7  0.52  3  sat   9.8  20.8 kN m  1  0.52     20.8  9.8  11.0 kN m 3

Clay:

 2.7  0.92  3  sat   9.8  18.5 kN m  1  0.92 

   18.5  9.8  8.7 kN m 3 Bearing capacity of sand Nq  37.8, Nq 1  36.8, N  43.9, sq  1.22, dq  1.25,s  0.88,d   1.0

qu   20.8  5  36.8  1.22  1.25   0.5  20.8  3  43.9  0.88  1.0   7042 kPa

7042  23.8 400  20.8  5 Bearing capacity of clay FS =

H cr =

B   2 cos  45  p  2  '

  

exp  Atan ' p 

 ' p A   45   2 

   36    radians      0.47  4 2 180   3 H cr = exp  0.47tan36  4.64m > (9 m – 5 m = 4 m) 36   2 cos  45   2   Failure surface would penetrate clay Short term:

qu = 5.14 x 40 x (1 +0.2 x 3/10) = 218 kPa 12000 Vertical stress at top of clay =  70.5kPa 10  17 FS = 218/70.5 = 3. ; OK Elastic settlement of sand

Ab 2

4L



B 3   0.3 L 10

 s  0.45(.3) 0.38 =0.71

 emb  1  0.04

5 3 2

 4  1  3  0.3  0.81  

A w  23  5  10  5  130m 2 ;

A w 130   4.33 Ab 30

 wall  1  0.164.330.54  0.65 e 

12  10 3  10  55  10 3     2

1  0.35  0.71  0.81  0.65 2

= 0.0143m = 14.3mm

Elastic settlement of clay Assume a load dispersion of 2:1. The equivalent size of footing on the top of the clay is (3+4)  (10+4) = 7m  14m.

Now

Df  9m emb  1  0.04

9  4  1  x .5  0.83, A w  2  7  9  14  9   432m 2 , A b  7 14  98m 2  7 3  2  

 s  0.45(.5) 0.38  0.59 Neglect wall effect.

e 

12  103 1  0.452 14   15  103     2





 0.59  0.83  1  0.0446m  44.6 mm

Primary consolidation:  zo   20.8  8  11  1  8.7  0.5 ,  181.8 kPa :At center of clay:  z  143.4 kPa( surface stresses utility computer program)

final  181.8  143.4  325.2 kPa  c 

1000  325.2   0.25 log   32.4mm 1  0.95  181.8 

Total settlement = 14.3 + 44.6 + 32.4 = 91.3 mm

12.11 A multilevel building is supported on a footing 58 m wide  75 m long × 3 m thick resting on a very stiff deposit of saturated clay. The footing is located at 3 m below ground level. The average stress at the base of the footing is 350 kPa. Groundwater level is at 12 m below the surface. Field and laboratory tests gave the following results: Depth (m) su (kPa)

0.5 58

6 122

25 156

eo = 0.57, Cc = 0.16, Cr = 0.035, OCR = 10,  'p = 28 , cs' = 24 , Eu = 100 MPa, νu = 0.45, E  90 MPa, and v  0.3 . Determine the total settlement and the safety factor against bearing capacity failure. The shear strength parameters were obtained from direct simple shear tests. Solution 12.11

Short Term: (TSA) qu  5.14susc , sc  1.15 , dc = 1 (shear strength above footing depth lower than below) su at base of footing is 87kPa; su at a depth B below footing is 156kPa Take average su = 121.5kPa qult  5.14  121.5  1.15  721kPa Assume soil is saturated. G  eo 2.7  0.57  sat  w  9.8  20.4kN / m3 1  eo 1  0.57 Df = 3 x 20.4 = 61.2 kPa 721 FS   2.5 350  61.2 Long Term: (ESA) Use bearing capacity program utility to find N q , N  , shape and depth factors.

Nq  14.7, Nq  1  13.7, N  11.5 for 'p  28o

   20.4  9.8  10.6 kN/ m 3 ======= Short side failure sq  1.41 s  0.69,dq  1.02,d   1 ,

qu  20.4  3 13.7  1.41 1.02+0.5 11.5  ( 9  20.4  10.6  49 )  0.69 1 = 3994 kPa Long side failure sq  1.69 s   0.48,dq  1.01,d   1 qu  20.4  3 13.7  1.69 1.01+0.5 11.5  ( 9  20.4  10.6  66 ) 0.48 1 lk= 3869 kPa Long side failure governs FS =

3869  13.4 350  61.2

Long Term Settlement Assume effective depth=2 B. For consolidation settlement consider 10 layers, each 14.5 m, and compute Depth (z/B) z (kPa) 0.25 0.5 0.75 1 1.25 1.5 1.75 2

331.2 263.6 192 139.6 100 76 59.6 47.6

Then use the harmonic mean value (see Chapter 9). Elastic settlement

Ab 4L2



 z at center of each layer.

 z  210kPa

B 58   0.77 L 75

 s  0.45(.77) 0.38 =0.5

 emb  1  0.04

3 58 2

 4  1  3  0.77  0.98  

A w  2  358  75  798m 2 ;

Aw 798   0.18 Ab 58  75

 wall  1  0.16.180.54  0.94

e 

58  75  350  10 3 1  0.3 2  0.5  0.98  0.94 = 170 mm 75   100  10 3     2





Primary consolidation: The depth corresponding to a stress increase of 210 kPa is 40 m from the bottom of the footing  zo   20.4  12   10.6 31  573.4 kPa :  zc  10  573.4  5734kPa  zo   z  573.4  210  783.4kPa   zc

 c 

2  58  1000  783.4   0.035 log   351mm 1  0.57  573.4 

Note the thickness of the layer is 2B = 2 x 58 Total settlement = 170 + 351 = 520 mm This settlement is large. The actual settlement could be smaller because of the many simplifying assumptions made in this type of calculations.

Solution 12.12 You can set up a spreadsheet to do the calculations as shown below. ASD Assume zero eccentricity

SHALLOW FOUNDATION DESIGN USING CSM DENSE SAND RESULTS CSM: BEARING CAPACITY, FS 1.32 CSM: SETTLEMENT 3.1

B = Width

Square 1.35 m

Depth of embedment

L l k M gsat

5.8 mm

DATA INPUT FOR FOOTING Select Footing type

Ip

0.218

p'o

33.2 kPa

Iq

0.388

qo

-12.4 kPa

24.9 kPa

m

Dp

59.8 kPa

p'k

149.3 kPa

1 m

Dq

106.4 kPa

qk

111.9 kPa

124.1 kPa

p'c

207.6

Ro

6.3

qyH

Footing Live Load

300 kN

Iz

Eccentricity

0

FS

Total - ASD

500 kN

DATA INPUT FOR LAB TEST

Ductility

0.7 1.32 OK

s'zc

223.9 kPa

s'xo

37.3 kPa

s'yo

37.3 kPa

30 degrees 0.08

Cr

0.016 9 18.8 KN/m3

Long term

3.1

f'cs

0.52 radian

f'cs

30.00 degrees

Total

3.1

mm

total

Ko nc

1-D

5.8

mm

max

Ko

oc

0.50 1.50

eo

0.851111

ec

0.838

Sample depth

1.675 m

ek

0.836

Depth of groundwater Poisson's ratio LOADING CONDITION Select

1m 0.35

tc m Rt

0.14

DATA INPUT FOR SAMPLE

axisymmetric

kPa

1.0544 OK

SETTLEMENT

Cc

Unit weight

0.8 0.03 0.007 1.20 18.8

s'zo

200 kN

OCR

274.3 kPa

Bearing Capacity Calculations

Footing Dead Load

Critical state friction angle

Surface stress

Help

nt

no mu E G rz

0.51 7.25 3 1.8 0.35 49363.266 kPa 18282.6911 kPa 3.1107252 mm

For eccentricity of a maximum of 10% of width SHALLOW FOUNDATION DESIGN USING CSM DENSE SAND RESULTS CSM: BEARING CAPACITY, FS 1.27 CSM: SETTLEMENT 2.6

B = Width

Square 1.5 m

Depth of embedment

L l k M gsat

3.3 mm

DATA INPUT FOR FOOTING Select Footing type

Ip

0.218

p'o

34.1 kPa

Iq

0.388

qo

-12.8 kPa

m

Dp

53.3 kPa

p'k

153.3 kPa

Dq

94.8 kPa

qk

115.0 kPa

104.2 kPa

p'c

268.3

Ro

7.9

Footing Live Load

300 kN

Iz

0.025 500 kN

DATA INPUT FOR LAB TEST

Ductility

0.08

Cr

0.016 9

Long term

kPa

1.27 OK

s'zc

230.0 kPa

1.0761 OK

s'xo

38.3 kPa

s'yo

38.3 kPa

30 degrees

Cc

Unit weight

FS

0.4

SETTLEMENT

OCR

25.6 kPa

1 m

qyH

Critical state friction angle

0.8 0.03 0.007 0.87 18.8

s'zo

200 kN

Total - ASD

244.4 kPa

Bearing Capacity Calculations

Footing Dead Load Eccentricity

Surface stress

Help

2.6

f'cs

0.52 radian

f'cs

30.00 degrees

Total

2.6

mm

total

Ko nc

1-D

3.3

mm

max

Ko

18.8 KN/m3

oc

0.50 1.50

eo

0.851111

ec

0.837

Sample depth

1.75 m

ek

0.836

Depth of groundwater Poisson's ratio LOADING CONDITION Select

1m 0.35

tc m Rt

0.20

nt

1.73205081

DATA INPUT FOR SAMPLE

plane strain

no mu E G rz

Minimum footing size with eccentricity requirement satisfied is 1.5m x 1.5 m Expected settlement: 2.6 mm (or say 3 mm) to 3.3 mm (or say 3.5 mm)

0.29 5.00 1.8 0.35 63750.2355 kPa 23611.1983 kPa 2.58191326 mm

LRFD at maximum allowable eccentricity SHALLOW FOUNDATION DESIGN USING CSM DENSE SAND RESULTS

Square

B = Width

1.66 m

Depth of embedment

s'zo

Ip

0.218

p'o

35.0 kPa

Iq

0.388

qo

-13.1 kPa

m

Dp

67.4 kPa

p'k

157.6 kPa

Dq

120.0 kPa

qk

118.2 kPa

107.1 kPa

p'c

275.8

Ro

7.9

qyH

Footing Live Load

300 kN

Iz

0.027666667 775 kN

DATA INPUT FOR LAB TEST

FS Ductility

0.4 1.00 OK 0.8494

SETTLEMENT

Critical state friction angle

30 degrees

Cc

0.08

Cr

0.016

OCR

9

Long term

3.5

236.4 kPa

s'xo

39.4 kPa

s'yo

39.4 kPa

f'cs

0.52 radian

f'cs

30.00 degrees

3.5

mm

total

Ko nc

1-D

4.5

mm

max

Ko oc

0.50 1.50

eo

0.851111

ec

0.837

Sample depth

1.83 m

ek

0.836

Depth of groundwater Poisson's ratio LOADING CONDITION Select

1m 0.35

tc m Rt

0.20

nt

1.73205081

DATA INPUT FOR SAMPLE

plane strain

kPa

s'zc

Total

18.8 KN/m3

Unit weight

26.3 kPa

1 m 200 kN

LRFD

0.8 0.03 0.007 0.87 18.8

Bearing Capacity Calculations

Footing Dead Load Eccentricity

309.4 kPa

L l k M gsat

DATA INPUT FOR FOOTING Select Footing type

Surface stress

Help

no mu E G

0.29 5.00 1.8 0.35 65546.7196 kPa 24276.5628 kPa

For LRFD the minimum footing size to satisfy the requirements is 1.66 m x 1.66 m Conventional method Ng  0.1054 exp (9.6' cs )  16.06; Ng  18.4; Ng  1  17.4

sg  1  0.4

B  0.6 , dq = 1.2, all other geometric factors are equal to 1 L

B = 1.5 m , e = 0.1B, B’ = B – 2e = 0.8B = 1.2 m q u  gDf ( Nq  1 )sq d q 

1 gBN g s g  18.8 117.4 1.58 12  0.5  ( 18.8  9.8 ) 1.2 16.06  0.6  672kPa 2

FS = 667/(244.4 – 18.8) = 3

12.13 A circular foundation of diameter 8 m supports a tank. The base of the foundation is at 1 m from the ground surface. The vertical load is 20 MN. The tank foundation was designed for short-term loading conditions (su = 80 kPa and γsat = 19 kN/m3). The groundwater level when the tank was initially designed was at 4 m below the ground surface. It was assumed that the groundwater level was stable. Fourteen months after the tank was constructed and during a week of intense rainfall, the tank foundation failed. It was speculated that failure occurred by bearing capacity failure. Establish whether this is so or not. The friction angle is

 p'  25

from simple shear tests

Solution 12.13 Applied vertical stress =

20 103  398kPa  2 8 4

Short term

sc  1.2,dc  1 qu  5.14  80  1.2  1  493 kPa  398 kPa Long Term

(no failure)

  25

sq  1.47 ,s   0.6, d q  1.04,d   1.0 N q  10.7 , N q  1  9.7 , N   6.95   19  9.8  9.2kN m3

qu   9.2  1  9.7  1.47  1.04    0.5  9.2  8  6.95  0.6  1  289 kPa  398 kPa

 Failure occurs under effective stress condition.

Solution 12.14

M 150 B  0.6  0.6  2.1  3.3 6.6   0.28m;    1.1m P 535 6 6 6 B'  6.6  2(0.28)=6.04m

e

  36 , Nq  37.75, Nq  1  36.75, N  43.9, sq  s  1 d   1.0 Depth of groundwater level is greater than B below base of the footing. No effects of groundwater. Neglect depth of embedment

q u  0.5  B' N  s  d  q u  (0.5  18  6.04  43.9 11)  2386 kPa

applied(max)  FS 

535  6  0.28  1  102 kPa 6.6 1  6.6 

2386  23.4 102

The settlement is not expected to be uniform because the vertical stresses at the base are non-uniform. B  However, the eccentricity is small  e   , so any tilting of the base would be within tolerable limits. 6 

Solution 12.15 Assume a square footing of width B Assume that the groundwater is more than B below the footing base and that B < 4 m. From Table A.11, the estimated unit weight is 18.5 kN/m3 Therefore use the N value for the top layer 0 to 5 m. Assume B = 1 m

q ult  32N1B; N1  c n N; c n  95.8 / (18.5 1.5) 

1/2

 1.86  2; use 1.9

N1  1.9  28  53 q ult  32  53  B  1696 B kPa ASD q ult  (FS)  P / B2  1696 B  3  700 / B2  1696 B B  1.07 m Use a footing of size 1.1 m x 1.1 m

LRFD:

Pu  1.25DL  1.75LL  1.25  200  1.75  500  1125kN Pu  i (32  53  B  B2 ) Table 12.1: i  0.45 for SPT 1

 1125  3 B   1.14 m  1696  .45  Use a footing of size 1.2 m x 1.2 m

12.16 The column load for an office building consists of a dead load of 200 kN and a live load of 250 kN. The soil at the site for the office building is a fairly homogeneous clay. Soil samples at a depth of 2 m gave the following average results. Triaxial tests: Isotropic consolidated CU tests on saturated samples, su = 36 kPa, confining stress = 100 kPa and average water content of 40%; One-dimensional consolidation tests: Cc = 0.16, Cr = 0.04 and OCR = 9. The minimum embedment depth of the footing is 1 m. Groundwater level is at the surface. Check the suitability of a 3.0 m square footing using the conventional ASD method with an FS = 3. Compare the results of the conventional method with CSM using an FS = 1.25. Assume  = 0.35.The tolerable settlement is less than 20 mm. assume the samples represent the soil at a depth 0.5B below the bottom of the footing. Solution 12.16 Calculate initial values. G e   2.7  1.08  3 γsat =  s o   w    9.8  17.8 kN / m  1  1.08   1  eo 

γ = γsat   w  17.8  9.8  8 kN / m3

( )

  su  f  '  po

   su  f   '  ic   zo

0.25 

3sin ' cs  1    3  sin ' cs  2 

Λ Λ  3sin ' cs  1  M 1  =      2 2 3  sin ' cs  2   ic 0.75

 ' cs  21.6o Mc =

6sin ' cs 6 sin 21.6o   0.84 3  sin ' cs 3  sin 21.6o

[(

)

(

)

(

]

[(

)

(

)

)

( (

) )

(

]

)

The current and past consolidation stresses in the field are: Current: po' =

1  2 Kooc ' 1  2 1.9  zo  16  25.6 kPa 3 3

(

)

(

)

a

Calculate past mean effective and deviatoric stresses. 1  2 Konc ' 1  2  0.63 Past: pk' =  zc  144  108.5 kPa 3 3

(

)

(

)

The preconsolidation mean effective stress on the ICL is p p ' c

' k

q   k

2

M 2 p'k

= 108.5 

53.32  145.6 kPa 0.842 108.5

Calculate increase in stresses from the surface load.

Table 12.7: Table 12.7: Slope of TSP is

[This is a not equal to 3 as in the standard

triaxial test because 3  0.) =

Calculate the deviatoric stress on the HV surface. For axisymmetric condition, which approximates the stress condition under the center of the footing, nt = 3.

tc 

1 1   0.07 2  nt  1  9  1  2   0.842    M  

Rt =1/0.07 = 14.3 > 5.7; failure would not occur from tension [ ( [ (

)

] )

]

Determine the factor of safety. ; Therefore, acceptable. Check if the imposed state is within the ductile region. (

)

(

)

Therefore, the imposed stress state in the soil will be in ductile region. The footing size for bearing capacity requirement can be reduced. However, we need to check that serviceability is satisfactory. Calculate the settlement.  = 0.35. ec  eo  ln R = 1.08 -0.017 ln (5.7) =1.05 z 

q s κB  10  Iq  I p   pc 1  ec   3 

50  0.017  3 103  145.6 1  1.05 

10   0.55  3  0.33  14 mm

Conservative settlement z  1.88

qs Cr BIz 50  0.04  3000  0.7  1.88  26.7mm zc 1  ec  144 1  1.05

Note: Iz is found from Chapter 7. Determine if the footing designed according to CSM is satisfactory. Settlement range = 14 mm to 26.7 mm; FS > 1.25. The footing satisfies both serviceability and ultimate limit state requirements. Conventional method

The conventional method required the peak undrained shear strength. We need to make an estimate this. However, we have to use CSM to do so. Calculate (su)f for OCR = 9 (Ro* = 5.7)   su  f    su  f  0.75 * Λ  '    '   R o   0.25   5.5   0.92  po   po  ic  su  f  0.92  25.6  23.6kPa

Calculate the initial yield value to check whether the intact soil would show a peak shear strength response Use  su  f  23.6kPa in the conventional bearing capacity method sc = 1.2 qu = 5.14 x 23.6 x 1.2 = 146 kPa FS = 146/(50 – 1 x 17.8) = 4.5 We do not have enough information to calculate settlement unless we use CSM to estimate E.

12.17 The results of a representative field vane shear test at a site are shown in Fig.12.22a. Previous studies reveal that  = 0.8 and  = 0.12 for the gray clay. A building with different column loads is to be erected on the site. Estimate the maximum centric load that a 2 m square footing can support using CSM. The minimum factor of safety is 1.25 and the settlement should not exceed 25 mm. Assume  = 0.35.

Solution 12.17 Step 1: Inspect and interpret vane shear test data. Inspection of the vane shear test data shows that the soil is overconsolidated above 7 m and normally consolidated below 7 m. Recall that normally consolidated soils tend to show linear increase of shear strength with depth. In other words, the normalized shear strength is constant with depth.

Step 2: Calculate the critical state friction angle. (su)f at 7 m = 14 kPa Vertical effective stress at 7 m is (

)

From Eq. (11.74), the normalized undrained shear strength for normally consolidated fine-grained soils is   su  f  '   zo

 '   0.5sincs  DSS

14  0.5 sin ' cs 56.8 sin ' cs  0.493



 ' cs  29.5o M c = 3 sin' cs  3  0.493 = 0.85 Step 3: Calculate the initial stresses, overconsolidation ratio and preconsolidation stress at B/2 below the footing. The calculations will be done for 3 m x 3 m. A spreadsheet will be used for other footing sizes. The depth from the surface at B/2 below the maximum size footing is 2 + (3/2) = 3.5 m Vertical effective stress at 3.5 m is (

)

(su)f at 3.5 m = 28 kPa From Eq. (11.73) 0.8   su  f  3 sinc' s  OCR    '    2  2    zo  DSS

0.85  OCR   28   32.3   2  2  DSS

0.8

 OCR = 4.9 From Fig. 11.29, Ro = 4.2

The current and past consolidation stresses in the field are: 1  2 Kooc ' 1  2 1.13 Current: p =  zo   32.3  35 kPa 3 3 ' o

(

)

(

)

a

In Fig.12.19, point O represents ( po' , qo). [(

)

( (

) )

[(

)

] (

(

G e  γsat =  s o   w  1  eo   2.7  eo  16.8    9.8; eo  1.38  1  eo 

(

)

(

)

Step 4: Check if the soil element will fail in tension.

) )

]

1 1   0.194 2 2   nt   3 1  2  1    M   0.852  1 1 Rt    5.2  4.2 tc 0.194

tc 

Soil will not fail by tension. Step 5: Calculate the deviatoric stress on the HV surface. √ [ ( (

[

)

] )

√

]

Step 6: Estimate the load to width ratio to satisfy ultimate limit state. Since the eccentricity is 10% of the width then qs 

P  6e  P  6  0.1B  1.6P 1    2 1   B2  B B  B  B2

At the edge of the footing, A (Fig. E12.22a), Ip =

Step 7: Estimate the load-width ratio to satisfy settlement. ec = eo  ln Ro  1.38  0.024 ln(4.2)  1.35 Since  = 0.35, then z

qs κB pc 1 ec

3 Iq 2

1.54Ip

1.6P B  2    q s B 3 3 B z   Iq  1.54I p    Iq  1.54I p  pc 1  ec o   2  pc 1  ec   2  1.6P 1.6P    0.024    3 3 B B   Iq  1.54I p     0.49  1.54  0.15 Note: pc 1  ec   2  147.2 1  1.35   2   7.27 105

P m B

The settlement from the above equation is dependent only on P/B ratio because all the other parameters are constant. The maximum allowable settlement is 25 mm. Therefore 0.025  7.27 105

P B

P  343; P  343  2  686 kN B Since the load for bearing capacity consideration is lower than for settlement, bearing capacity governs the design. The allowable load is 230.8 kN ( say 231 kN)

12.18 Fig. P12.18 shows a proposed canal near a 5 story apartment building 30 m wide x 50 m long. The building is founded on a mat foundation. Describe and justify some of the concerns you may have regarding the stability of the mat foundation with constructing such a canal. If the owner insists on constructing the canal, research methods that you would consider so that the canal can be designed and constructed safely.

Solution 12.18 Concerns: 1. The top layer (sand mixed with silt and clay) can slide along the silt layer leading to instability. 2. Sliding is also possible at the interface of the fine sand and the silty clay. 3. General slope stability failure. 4. Seepage of water into the canal can lower the groundwater leading to additional settlement. The groundwater level is unlikely to be lowered uniformly so additional differential settlement could occur. Even for uniform decreases in groundwater level, the settlement would be nonuniform. 5. The additional differential settlement will cause addition bending moment and rotation of the mat. A possible alternative design method is to use a retaining wall rather that cutting a slope.

Proposed retaining wall 3m Center line GWL Canal

Bedrock