foundation engineering eciv 4352 first term 2010site.iugaza.edu.ps/nhindi/files/chapter31.pdf · 2...

1

Foundation Engineering ECIV 4352

Chapter (3) Bearing Capacity of Shallow Foundation

First term

2010

Ultimate bearing capacity (qu)

ف انتشبة Shear failure أقم حم يقسو عهى انساحة ؤدي نحذث

Allowable bearing capacity (qall)

Shear failure عهى انساحة ك ا تتحه انتشبة بذ حذثحم يقسو

should be adequate to prevent excessive settlement and shear failure.

Types of shear failure:

1- General shear failure: For Dense sand.

2- Local shear failure: For Medium compaction soil.

3- Punching shear failure: For loose soil

2



First term

2010

Calculation of Ultimate Bearing Capacity of Shallow Foundations without

Eccentricity:

1. Terzaghi’s theory:

Assumption for Terzaghi’s theory:

i. The foundation is considered to be shallow if BD f , in recent studies the

foundation is considered to be shallow if 4/ BD f . Otherwise it is considered to

be deep foundation.

ii. Foundation is considered to be strip if 00.0/ LB .

iii. The soil from ground surface (سطح األزض انطثعح) to the bottom of the foundation

is replaced by stress (سطح انتأسس) fDq .

For General shear failure:

Type of foundation Ultimate bearing capacity qu

Strip Footing BNqNcNq qcu2

1

Square footing BNqNcNq qcu 4.03.1

Circular footing BNqNcNq qcu 3.03.1

c: Cohesive.

fDq

B: Foundation width (Diameter if circular).

NNN qc ,, : Bearing capacity factors given from Table 3.1 P.139 as function of angle of

friction

3



First term

2010

For Local shear failure:

Type of foundation Ultimate bearing capacity qu

Strip Footing '''

2

1

3

2BNqNcNq qcu

Square footing ''' 4.0867.0 BNqNcNq qcu

Circular footing ''' 3.0867.0 BNqNcNq qcu

''' ,, NNN qc : Factors for bearing capacity given from Table 3.2 P.140

2. Meyerhof’s equations (General bearing capacity equation):

Terzagi equations neglect:

i. Rectangular footings.

ii. Inclination of loads.

iii. Shear strength of soil above the foundation.

Meyerhof’s equation takes in consideration theses variables:

idsqiqdqsqcicdcscu FFFBNFFFqNFFFcNq 5.0

NNN qc ,, : Table 3.3 P.144

belowShown

factors.n Inclinatio ,,

factors.Depth ,,

factors. Shape ,,

iqici

dqdcd

sqscs

FFF

FFF

FFF

4



First term

2010

Shape Factors:

L

BF

L

BF

N

N

L

BF

s

qs

c

q

cs

4.01

tan1

1

'

Depth Factors: Case I: 1/ BD f

1

sin1tan21

4.01

2''

d

f

qd

f

cd

F

B

DF

B

DF

Depth Factors: Case II: 1/ BD f

1

tansin1tan21

tan4.01

12''

1

d

f

qd

f

cd

F

B

DF

B

DF

Inclination Factors

2

2

'1

901

i

qici

F

FF

The Term

B

D f1tan is in

radian

شاح يم انحم ع

انحز انسأس، هصو أ ك

انحم يائم ف يتصف انقاعدج

5



First term

2010

Effect of water table in bearing capacity equations

Case I) Water table is located at depth Dw1 so that 0 ≤ Dw1 ≤ Df:

1

'

1 wfw DDDq (انحد انثا ي انعادنح)

wsat ' (انحد األخس ي انعادنح)

Case II) Water table is located at depth Dw2 below the foundation so that 0 ≤ Dw2 ≤ B:

fDq .(انحد انثا ي انعادنح)

'2' B

Dw ( انحد األخس ي انعادنح )

Case III) Water table is located at depth Dw2 below the foundation so that Dw2 > B:

No changes in equations.

6



First term

2010

Factor of safety:

Ultimate bearing capacity ي خالل انعادالت انسابقة حسبا

ز انقة تثم االجاد انزي ارا أثش عهى انتشبة تاس عذ، نزا ال جب استخذاي عذ انتصى بم ستخذو

.قة أقم ي زا ي خالل قسة ز انقة عهى يعايم أيا

load. Ultimate

capacity bearing allowableNet

capacity bearing allowable Gross

capacity bearing ultimateNet

capacity bearing ultimate Gross

GrossQ

q

q

qqq

q

u

netall

all

unetu

u

FS

qq

FS

qq

FS

qq

FS

qq

all

unetu

netall

u

all

FS = (3 – 4) for bearing capacity

7



First term

2010

Example 1: Determine the size of square footings to carry gross allowable load(295KN)

given that:

Use Terzagi equations assuming general shear failure.

Solution

295KN

Df=

1.0

0

B

C=0.00

=35

=18.15KN/m3

92cmB :error and By trial

684.22814.2

41.4515.184.044.41115.180885

41.45 , 44.41 , 75.5735At

4.03.1

:footing squareFor

.885

3295

.295

23

2

22

2

BB

BB

NNN

BNqNcNq

BBFSqq

BA

Qq

AreaqQ

qc

qcu

allu

all

all

allall

00.0

.35

./15.18

00.1

3

3

C

mKN

D

FS

f

8



First term

2010

Example 2: Determine the net allowable load that foundation can carry (no

inclination), Use Meyerhof equation given that:

./50

.25

./10

00.2

4

2

3

mKNC

mKN

D

FS

w

f

Solution:

idsqiqdqsqcicdcscu FFFBNFFFqNFFFcNq 5.0

88.10 , 66.10 , 72.2025At NNN qc

The water table is at depth = 1m <Df :

./4.9104.19

./2.261)104.19(18.16

3'

2

mKN

mKNq

Shape factors:

733.03

24.014.01

311.125tan3

21tan1

343.172.20

66.10

3

211

L

BF

L

BF

N

N

L

BF

s

qs

c

q

cs

9



First term

2010

Depth factors:

1

313.12

225sin125tan21

4.12

24.01

12/2/

2

d

qd

cd

f

F

F

F

BD

Inclination factors:

Due to absence of inclined load, the inclination factor is 1 everywhere.

KNAqQ

mKNFS

qq

mKNqqq

mKNq

q

netallnetall

netu

netall

unetu

u

u

1.3716635.619

/35.6194

4.2477

/4.24772.266.2503

./6.2503

11733.088.1024.95.01313.1311.166.102.2614.1343.172.2050

2

2

2

10



First term

2010 Example 3: Find the ultimate bearing capacity of square footing (2.5m x 2.5m) which is

placed 2.5m below the ground surface of soil having the shown properties:

2.5 m

2.5

md=18KN/m3

C=0.00

=30

h

The water table is located at distance (h) below the ground surface; if sat=19KN/m2

Find the ultimate bearing capacity using Terzagi equation for the following cases:

h = 7m.

h = 4m.

h = 1m.

h = 0m.

Comment on the results.

Solution case I) h=7m

d = 4.5m>B No effect of water table on bearing capacity.

./94.1355

18.195.2184.046.22450

4.03.1

18.19,46.22

30

/455.218

2

2

mKNq

q

BNqNcNq

NN

mKNq

u

u

qcu

q

o

11



First term

2010

Case II) h = 4m

d=1.5m<B

./892.1286

18.195.24.144.046.22450

4.03.1

/4.149185.2

5.11019

/455.218

2

3

2

mKNq

q

BNqNcNq

mKN

mKNq

u

u

qcu

Case III) h = 1m

D1 = 1m, D2 = 1.5m

./11.880

18.195.294.046.225.310

4.03.1

/91019

/5.315.11019118

2

3

2

mKNq

q

BNqNcNq

mKN

mKNq

u

u

qcu

Case IV) h = 0.00

D1 = 0m, D2 = 2.5m

./97.677

18.195.294.046.225.220

4.03.1

/91019

/5.225.21019

2

3

2

mKNq

q

BNqNcNq

mKN

mKNq

u

u

qcu

Comment; as the water table is being near to the foundation, the bearing capacity

decreases.

12



First term

2010

P

e

BL

Pe

B

Eccentrically loaded foundation:

:عديا تتعسض انقاعدج نحم يسكص تد احساف فإ ضغط انتستح ك يتظا كا يضح تانشكم انتان

P

q=P/(BxL)

BxL

انقاعدج ذنك حسة عد جد احساف ف حم انقاعدج ع انسكص تسثة ذنك ف تند ضغط غس يتظى أسفم

.قح االحساف

One way eccentricity

13



First term

2010

P

B

M=Pe

qmax

qmin

1. For e<B/6:

B

e

LB

P

LB

eP

LB

P

LB

eP

LB

P

LB

BeP

LB

Pq

Bc

LBI

PeM

LBA

I

Mc

A

Pq

61

6

6

1

12

1

2/

2/

12

1

223

3

B انقا جاء تر انصزج أل انعصو انتند كا ف يست انضهع

:L تا ن كا انعصو ف اتجا انضهع

L

e

LB

Pq

L

e

LB

Pq

61

61

min

max

1. For e = B/6:

qmax

Remember, for circular foundation, the moment of inertia is given by:

4

64DI

, D is diameter of the

circle.

B

e

LB

Pq

B

e

LB

Pq

61

61

min

max

14



First term

2010

B

L

e

B/2 - eB/2 - e

B-2e2e

2. For e > B/6: There will be tension stresses on the foundation which is prohibited in

design, so we will neglect the tension stress and calculate qmax as follow:

qmax qmax

3(B/2 - e)

)2(3

4max

eBL

Pq

ألحال أقم ي انساحح األصهح حت حصم عه جد احساف ف األحال جعم يساحح انقاعدج انت تتحم ا

:يساحح أخس ك انحم فا ف انتصف كا يضح فا ه

BB

eLL

LBB

LBA

LL

eBB

used

'

'

'''

'''

'

'

2

:(L) of plane in the ismoment theIf

,.min

2

The equation used to calculate the bearing capacity is Meyerhof's or Terzagi's equation:

idsqiqdqsqcicdcscu FFFNBFFFqNFFFcNq '' 5.0

To find shape factors: use '' , LB

To find depth factors: use B, L

in last term is related to soil below the foundation.

To find the gross ultimate load Qu:

'' AqQ uu

15



First term

2010

max

'

max :qfor adequate issafety offactor

capacity) bearingAgainst (

q

qFS

thatCheck

Q

QFS

u

all

u

Example 4: Determine the size of square footing (B x B) that subjected to vertical load

(100,000Ib) and moment (25,000Ib.ft), the soil profile is given below:

Use FS=6, w=62.4 Ib/ft3

Qall=100,000Ib

B

M=25,000Ib.ftd=100Ib/ft3

C=0.00

=30

sat=120Ib/ft3

C=0.00

=30

4 ft

Solution:

1- Find the eccentricity (e):

ftP

Me 25.0

000,100

000,25

2- Find the effective area A':

BBBBA

BL

BLBB

BL

BB

LBA

used

5.05.0

.

5.0),.(min

5.0

2'

'

'''

'

'

'''

16



First term

2010

3- Find '

uq :


4.22 , 4.18 03At NNq

./6.574.62120

./4001004

3'

2

ftIb

ftIbq

Shape factors:

BB

BF

BB

BF

s

qs

2.06.0

5.04.01

2886.0577.130tan

5.01

Depth factors:

1

154.11

4)30sin1(30tan21

1/D Assume

2

f

d

qd

F

BBF

B



.5.0

000,600

000,6006000,100

000,100

2'

'

BBA

Qq

IbFSQQ

IbQ

uu

allu

all

After arrangement of equation:

ftB

BBBBB

5.6

072.387208.11542571.113342.2451

5.0

000,60022

17



First term

2010

Example 5: An eccentrically loaded foundation is shown in the figure; determine the

ultimate load Qu that foundation can carry using Meyerhof’s equation.

Solution:


54.12 , 85.11 , 25.2226At NNN qc

ftL

ftB

8

7.665.028

'

'

Shape factors:

665.08

7.64.014.01

408.126tan8

7.61tan1

446.125.22

85.11

8

7.611

'

'

'

'

'

'

L

BF

L

BF

N

N

L

BF

s

qs

c

q

cs

Depth factors:

1

249.18

5.626sin126tan21

325.18

5.64.01

18/5.6/

2

d

qd

cd

f

F

F

F

BD

18



First term

2010

Inclination factors: Due to absence of inclined load, the inclination factor is 1 everywhere.

3'

2

/6.594.62122

/6.5384.621225.31103

ftIb

ftIbq

IbQ

ftA

ftIbq

q

u

u

u

4.18461706.5347.34443

6.5387.6

/47.34443

665.054.127.66.595.0249.1408.185.116.538325.1446.125.22500

2'

2'

'

19



First term

2010

Example 6: For the square foundation shown in the Figure, find the gross allowable

load that foundation can carry for the following cases:

A- No water table is observed.

B- Water table at depth 0.5m below the Bottom of the foundation.

Use FS = 3 with Meyerhof equation.

Solution: Case A: No Water Table

03.48 , 3.33 ,12.4653At NNN qc

2'

'

'

8.15.12.1

5.1

2.115.025.1

mA

mL

mB

Shape factors:

68.05.1

2.14.014.01

56.135tan5.1

2.11tan1

'

'

'

'

L

BF

L

BF

s

qs

Depth factors:

1

169.15.1

135sin135tan21

15.1/1/

2

d

qd

f

F

F

BD

20



First term

2010


18.035

2011

6049.090

201

901

22

22

i

qi

F

F

18.0168.003.482.1165.06049.0169.156.13.331610

5.0

'

''

u

idsqiqdqsqcicdcscu

q

FFFNBFFFqNFFFcNq

kNQmkNq uu 52.11598.1179.644/179.644 '2'

kNQall 5.3863/52.1159'

Case B: At water table at depth 0.5m below the bottom of foundation: Dw2 = 0.5m < (B = 1.5m) q = 16 x 1 = 16kN/m2

3'' /33.96165.1

5.01016 mKN

B

d

18.0168.003.482.133.95.06049.0169.156.13.33160

5.0

'

''

u

idsqiqdqsqcicdcscu

q

FFFNBFFFqNFFFcNq

kNQmkNq uu 17.11178.165.620/65.620 '2'

kNQall 39.3273/17.1117'

Bنس Bأخز '

انجاب يطق ، تأ انحم انر أصثحت تتحه انقاعدج ف جد

. أصثح أقم ي انحم انر تتحه ف حال غات Water table

21



First term

2010

Example 7: for the rectangular foundation shown below:

a) Compute the net allowable bearing capacity using general bearing capacity

equation (FS=3).

b) If the water table is lowered by 2m. What effect on bearing capacity would occur

due to water lowering?

Given data:

Dimensions of foundation ( 2m x 3m)

= 25o

C = 0.00

Solution:

Due to inclined load in the center of column, we have to translate it to the Center of

the footing, and so there will be moment exerted and so eccentricity develops.

mP

Me

mKNM

75.0700

525

.5255.1)60cos(700

The moment in the direction of L:

mLmB

mB

mL

used 2,5.1

2

5.1)75.0(23

''

'

'

،،انسؤال سابقزا الحظ انفشق ب eف را انسؤال نى عطا االحساف

نك قح انحم انائم شاح ،يثاشسج

انم، تحث ك انحم يائم عد

انعد، تانتان حسة انعصو اناشئ ع

قم انحم نهقاعدج ك االحساف

حاصم قسح انعصو عه انسكثح انسأسح

نهحم

22



First term

2010 Case I)

3'

2

/111021

/2010211185.0

mKN

mKNq

88.10 , 66.10 52At NNq


Shape factors:

7.02

5.14.01

3497.125tan2

5.11

s

qs

F

F

Depth factors:

1

233.12

5.1)25sin1(25tan21

12/5.1/D

2

f

d

qd

F

F

B


04.025

301

44.090

301

2

2

i

qi

F

F

./2.463

206.158/6.158

04.017.088.105.1115.044.0233.13497.166.1020

22'

'

mKNqmKNq

q

netallu

u

Case II) d = 1m<2m

3

2

/5.1411182

11021

/27185.1

mKN

mKNq

By similar calculations:

./35.623

2706.214 2mKNqnetall

23



First term

2010 Example 8: For the shown square footing (2.5m x 2.5m) if the allowable load P=800

KN, use FS=6 to determine the allowable resisting moment (M). Use Terzagi equations

( =35o)

Solution:

BNqNcNq qcu 4.03.1

mL

eB

5.2

25.2

'

'

41.45 , 44.41 35At NNq

3

2

/4.1610186.2

21020

/6.21182.1

mKN

mKNq

mKNPeM

mee

meme

ee

ee

AqP

ee

FS

qq

eq

eq

allall

uall

u

u

.3.421526.0800

526.0

526.0or 47.3

00.0829.14

25.25.2296.99305.273800

296.99305.2736

78.59583.1639

78.59583.1639

41.4525.24.164.044.416.21

min

2

'

24



First term

2010

P

eB

BL

eL

XY

P

BL

XY

MxMy

B

L

L1

eB

eL

Two way eccentricity

By

Lx

ePM

ePM


How to find A'?

Case I) 6

1&

6

1

B

e

L

e BL

'

''

11

'

1

1

11

'

),max(

35.1

35.1

2

1

L

AB

LBL

L

eLL

B

eBB

LBA

L

B

25



First term

2010

Case II) 6

10&

2

1

B

e

L

e BL

Case III) 2

10&

6

1

B

e

L

e BL

graph. eThrough th :

&B

B:axis-x

B

e :axis-y

P168Figure3.22at look B and B find To

2

1

21

B

21

''

21

'

L

e

B

B

L

AB

LBBA

L


&L

L:axis-x

L

e :axis-y

. P.167Figure3.21at look L and L find To

),max(

2

1

21

L

21

'

''

21

'

21

'

B

e

L

L

L

AB

LLL

BLLA

B

26



First term

2010

Case IV) 6

1&

6

1

B

e

L

e BL


&L

L:axis-x

B

e :axis-y

P169. 3.23 Figureat look B and L find To

2

1

22

B

22

''

222

'

L

e

B

B

L

AB

LLBBBLA

L

27



First term

2010

Qu=??

B

Myd=18KN/m3

sand

=35

d=18KN/m3

sand

=35

1 m

Mx

Example 9: Rectangular footing is subjected to two way eccentricity. Determine the

gross ultimate load that foundation can carry given that:

mD

mLmB

me

me

f

L

B

1

2,5.1

364.0

3.0

Solution:

1- Study the case of eccentricity:

ICase

B

e

L

e

B

L

6

12.0

5.1

3.0

6

1182.0

2

364.0

mL

AB

mLBA

mLBL

mL

mB

675.0908.1

2879.1

2879.1908.135.15.02

1

908.1),max(

908.1)182.0(35.12

35.1)2.0(35.15.1

'

''

2

11

'

11

'

1

1


Foe sand: C=0.00

03.48 , 3.33 53At NNq

28



First term

2010

Shape factors:

858.0908.1

675.04.01

248.135tan908.1

675.01

s

qs

F

F

Depth factors:

1

167.15.1

1)35sin1(35tan21

166.05.1/1/D

2

f

d

qd

F

F

B



KNQ

mKNq

ult

u

1.14492879.132.1123

./32.1123 2

29



First term

2010

eB

eL

6 ft

4 ft

eBeL

L2

L1

L

B

Example 10: Refer to the following figure. The shallow foundation measuring 4 ft 6 ft is

subjected to a centric load and a moment. If = 0.4 ,ft = 1.2 ft and the depth of the

foundation is 3 ft. Determine the allowable load the foundation can carry . Use a factor of

safety of 4 .Soil properties are: =115 Ib/ft³, = 35⁰, c = 0.0.

Solution: Study the case of eccentricity:

6

11.0

4

4.0

2

12.0

6

2.1

B

e

L

e

B

L

CaseIIB

e

L

e BL 6

10&

2

1

ftBftA

ftL

ftLL

ftLL

L

AB

LLL

BLLA

48.222.5

96.1296.12426.122.5

2

1

22.5

26.121.06

22.587.06

P167 3.21 figureat look L and L find To

),max(

2

1

'2'

'

22

11

21

'

''

21

'

21

'


KipsIbQ

ftIbq

N

N

ftIbq

all

u

q

1941941048.23

73.23479

/73.23479181.041.4548.21155.0191.133.13.333450

03.48

3.33

/3453115

2

2

30



First term

2010

Example 11: Refer to the following figure determine the gross ultimate bearing

capacity

Solution:

6

12.05.1/3.0/

6

133.05.1/5.0/

Le

Be

L

B

Case I

mL

AB

LBL

mLBA

mL

eLL

mB

eBB

L

B

375.035.1

50625.0

35.1),max(

50625.035.175.05.02

1

35.12.035.15.13

5.1

75.0333.035.15.13

5.1

'

''

11

'

2

11

'

1

1

At = 30:

Nq = 18.4 , = 22.4

Shape factors:

88.035.1

375.04.014.01

16.130tan35.1

375.01tan1

'

'

'

'

L

BF

L

BF

s

qs

Depth factors:

1

135.15.1

7.030sin130tan21

15.1/7.0/

2

d

qd

f

F

F

BD

31



First term

2010



2'

''

/76.3711188.04.22375.0185.01135.116.14.18187.00

5.0

mkNq

FFFNBFFFqNFFFcNq

u

idsqiqdqsqcicdcscu

foundation engineering eciv 4352 first term 2010site.iugaza.edu.ps/nhindi/files/chapter31.pdf · 2...

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