foundation engineering eciv 4352 chapter (6) geometric...

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Foundation Engineering ECIV 4352

Chapter (6) Geometric Design of Shallow Foundations

First term

2010

Advantages of Shallow Foundations:

1. Minimum cost.

2. Simple construction procedure.

3. Materials mostly concrete.

4. Labor doesn’t need expertise.

Disadvantages of Shallow Foundations:

1. Settlement.

2. Limited bearing capacity.

3. Can’t be constructed on irregular ground surfaces.

Types of Shallow Foundations:

1) Isolated footing (Spread footing).

2) Combined footing.

3) Strip footing.

4) Mat foundation (Raft).

1. Geometric design of isolated footing:

This type of footings usually used when the loads on the footing are relatively small.

This type of footing can take three shapes: Square, Rectangular and Circular. It

always, preferable that the footing has the same shape as the column.

Sometimes even for a square column, the footing can be rectangular for the

following reasons :

a. We can’t extend the footing in one direction (too close to the next footing).

b. To increase the rigidity of the footing if the eccentricity is high.

1- Find the net allowable bearing capacity (Soil pressure):

scfcc

allunetu

netall

uall

hDhq

FS

qq

FS

qq

FS

qq

FS

qq

2- Find the required area of footing:

LD

netall

req

PPQ

q

QA

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First term

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Note: We can use gross allowable bearing capacity to find the Area of footing, but the

load Q should be also gross so that it includes the weight of foundation and soil above it.

scfsoil

ccfoot

soilfootLDGross

all

Grossreq

hDBLW

hBLW

WWPPQ

q

QA

3. In case of eccentricity, we increase the footing dimension in the direction of

eccentricity to ensure a uniform soil pressure.

LB

Qq

ew

CL

n

netall

22

2. Geometric design of combined footing: There are 3 types of combined footings:

A. Rectangular combined footings.

B. Trapezoidal combined footings.

C. Strap footings.

A. Geometric design of rectangular combined footings:

Used under closely spaced and heavily loaded columns where individual

footings, if they were provided, would be either very close or overlap each other.

Used as an alternative to an eccentrically loaded footing that has a property line

restriction so that the edge column is linked to an interior column.

Rectangular combined footing is more preferred than trapezoidal combined

footing due to its simplicity in both design and construction.

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First term

2010

Case I) No limitations:

1- Find the required area:

netall

reqq

QQA 21

2- Find the resultant force location (Xr):

RXLQ

QM

QQR

r

22

1

21

00.0@

3- To ensure uniform soil pressure, the resultant force (R) should be in the center of

rectangular footing:

L

AB

XLL

XLL

r

r

1

1

2

2

Case II) Limitation that we have known length or width of footing:

Find centerM @ :

1- if centerM @ =0.00, uniform soil pressure

And find the required area:

L

AB

q

QQA

netall

req

21

2- if 00.0@ centerM so that there is a value M

by means of eccentricity so that the soil pressure will not

be uniform.

Calculate the soil pressure:

L

e

LB

Rq

L

e

LB

Rq

61

61

min

max

Find the width of footing B, by equating maxq with netallq .

Check adequacy of footing width B that is 00.0min q

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First term

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B. Geometric design of trapezoidal combined footing:

Trapezoidal combined footing is used rather than rectangular combined footing and will

be more economical in the following two cases:

There are limitations on footing's longitudinal projection beyond the two

columns.

There is a large difference between the magnitudes of the columns loads.


netall

reqq

QQA 21

As the area of trapezoidal is given by ))((5.0 21 LBBA

So put AAreq to get an equation which is function of 21 & BB .

2- Determine the resultant force location by taking as example 0@ 1 QM

3- Put the resultant force location at the centroid of trapezoid to achieve uniform soil

pressure.

The centroid equation is:

21

21 2

3 BB

BBLX So we will have another equation of 21 & BB , solve them to get

21 & BB .

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First term

2010

C. Geometric design of strap footing (Cantilever):

Used when there is a property line which disables the footing to be extended

beyond the face of the edge column. In addition to that the edge column is

relatively far from the interior column so that the rectangular combined footing

will be too narrow and long which increases the cost.

There is a "strap beam" which connects two separated footings. The edge

footing is eccentrically loaded and the interior footing is centrically loaded. The

purpose of the beam is to prevent overturning of the eccentrically loaded

footing.

a. Find the resultant force location:

RXdQ

QM

QQR

r

2

1

21

00.0@

b. Assume the length of any foot, let we assume L1.

c. Find the distance a:

22

1 wLXa r

d. Find the resultant of each soil pressure:

12

12 1 00.0@

RRR

RFindbRbaRRM

e. Find the required area for each footing:

netall

netall

q

RA

q

RA

22

11

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First term

2010

3. Geometric and structural design of mat Foundation:

When to use mat foundation:

1. The area of isolated footings (Isolated and combined) > 50 % of the total area of the

structure.

2. If the bearing capacity of soil < 15 ton/m².

3. Difficult soils, such as:

a. Compressible soils:

It contains a high content of organic material and not exposed to great pressure

during its geological history, so it will be exposed to a significant settlement; so mat

foundation is used to avoid differential settlement.

b. Expansive soils:

Expansive soils are characterized by clayey material that shrinks and swells as it dries

or becomes wet respectively. It is recognized from high values of Plasticity Index,

Plastic Limit and Shrinkage Limit.

c. Collapsible soils:

Collapsible soils are those that appear to be strong and stable in their natural (dry) state, but which rapidly consolidate under wetting, generating large and often unexpected settlements. This can yield disastrous consequences for structures unwittingly built on such deposits.

Types of Mat Foundations:

1. Uniform Thickness.

2. Flat plate thickened under columns.

3. Beams and Slab.

4. Slab with basement walls.

Compensated Foundations:

fDA

P

In order to make minimum, we increase the depth of foundation.

For fully compensated foundation A

PDD

A

Pff

0

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First term

2010

i. Geometric design (Service loads):

a. Find the center of gravity of mat footing:

i

ii

g

i

ii

gA

AYY

A

AXX

~

~

b. Find the resultant force R:

iQR

c. Find the location of the resultant force:

i

rii

R

i

rii

RQ

YQY

Q

XQX

That means, the moment of the resultant equals the sum of forces' moments

iA : shape’s area

iX : Distance between y-axis and the centroid of the shape

iY : Distance between x-axis and the centroid of the shape

iQ : The load of column i

riX : Distance between column's center and y-axis

riY : Distance between column's center and y-axis

Note: Xg , Yg ,XR and YR are all measured from the same axis.

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First term

2010

d. Find the eccentricities: '' YYeXXe RyRx

e. Find YX MM and :

ixy

iyX

QeM

QeM

f. Find the stresses:

22

33

: theoryaxis Parallel :Recall

12

1

12

1

gravity ofcenter thepoint to thefrom Distances : ,

)(

)(

xxoxyyoy

xy

X

X

Y

y

mat

i

AdIIAdII

LBIBLI

YX

Tension

nCompressio

YI

MX

I

M

A

Qq

Now check that:

00.0min

max

q

qqnetall

Otherwise increase the dimensions of the mat foundation

Be Careful that up till now service loads are use.

ii. Structural design (Ultimate loads):

We have to subdivide the mat foundation into strips in

both directions. The width of the strip is directly

proportion to the loads of the column included in this

strip.

For the previous mat let we take a strip of width B1 for the

columns 13-14-15-16

Locate the points E and F at the middle of strip edges.

Find the stresses at E and F and be careful that we use

ultimate loads:

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First term

2010

uiXuY

uiyuX

X

uX

Y

uY

mat

ui

QeM

QeM

YI

MX

I

M

A

Qq

Find the average stress:

2

FEavg

qqq

Find the summation of loads of the strip

StripuiQ .

Check that:

StripuiQ = stripavg Aq

If ok, draw the SFD and BMD and design the mat.

Otherwise;

We have to make adjustment for the loads as follow:

2 load Average

stripavgStripui AqQ

Find the modified column loads:

stripui

uiuiQ

QQ

load Average

mod

Find the modified soil pressure:

stripavgu

avguavguAq

qq

,

,mod,

load Average

Now we must have that:

StripuiQ = stripavg Aq

Draw SFD and BMD.

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First term

2010

Example1:

Determine the dimensions of a rectangular footing, given the following data:

Column (1): PD = 55 t, PL = 45 t and 40 40 cross section.

Column (2): PD = 90 t, PL = 70 t and 60 60 cross section.

Allowable gross soil pressure = 2.0 kg/cm2 , mD f 5.1 and3/7.1 mtsoil

Solution:

Assume depth of footing = 60 cm

ccssgrossallnetall hhqq )()(

2/97.165.26.07.19.020 mts

mx

x

tPPR

mq

QA

netall

required

46.2

4)7090(260

260)7090()4555(

32.1597.16

70904555

21

2

)(

mL 32.5)2.046.2(2 taken as 5.4 m

mL

AB 84.2

4.5

32.15 taken as 2.9 m

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First term

2010

Example2:

Find the Dimensions of the combined footing for the columns A and B that spaced

6.0m center to center, column A is 40cm x 40cm carrying dead loads of 50tons and

30tons live load and column B is 40cm x 40cm carrying 70tons dead load and 50 tons

live loads.

./15 2mtqnetall

Solution


221 33.13

15

12080m

q

QQA

netall

req

2- Find the resultant force location (Xr):

mXX

QM

tonsQQR

rr 6.32006120

00.0@

20012080

1

21

3- To ensure uniform soil pressure, the resultant force (R) should be in the center of

rectangular footing:

mB

mL

L

76.16.7

333.13

6.78.32

2.06.32

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First term

2010

Example3:

A combined footing consists of four columns as shown in figure, determine:

1- Width of the combined footing B

2- Draw shear and moment diagram

Allowable gross soil pressure = 255 KN/m2 , 3/24 mKNc and

3/20 mKNsoil

Solution:

Weight of footing & soil = BBLB 15685.112)201245.1(28

KNBBQ )156811500(15682500400035001500

M (Moment taken about the centroid of the base)

mKNM .135005.11250025004350025005.111500

Eccentricity:

BQ

Me

156811500

13500

)156811500(28

1350061

28

156811500255

61

61

maxBB

B

L

e

BL

Qq

L

e

BL

Qq

046345000001162280000244633088

43904322000

43904403000)156811500(7140

43904322000

810001

28

156811500255

2

BB

B

BBB

BB

B

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First term

2010

m

mB

BB

58.2

334.72

945.184751.4751.4

0945.18751.4

2

2

Take B = 2.6 m and check qmax < (qall = 255 KN/m2)

0/174)6.2156811500(28

1350061

6.228

6.2156811500

/2557.253)6.2156811500(28

1350061

6.228

6.2156811500

2

max

2

max

mKNq

mKNq

Base pressure due to weight of soil & footing = 1.5x24 + 1x20 = 56 KN/m2

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First term

2010

Example4:

Find the Dimensions of the trapezoidal combined footing for the columns A and B that

spaced 4.0m center to center, column A is 40cm x 40cm carrying dead loads of 80tons

and 40tons live load and column B is 30cm x 30cm carrying 50tons dead load and 25

tons live loads.

./85.18 2mtqnetall

Solution:

a. Find the required area:

221 34.10

85.18

75120m

q

QQA

netall

req

As the area of trapezoidal is given by ))((5.0 21 LBBA

So put AAreq to get an equation which is function of 21 & BB .

75.434.1035.45.0 2121 BBBB ...............1

b. Determine the resultant force

mXX

QM

rr 55.1195475

0@ 1

c.

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First term

2010

d. Put the resultant force location at the centroid of trapezoid to achieve uniform soil

pressure.

The centroid equation is:

2121

21

21 2305.075.4

2

3

35.42

3BBX

BB

BB

BBLX

For uniform soil pressure:

1.55 + 0.2 = X

75.12305.0

75.1

21

BB

mX

75.52 21 BB .........................2

Solve 1 and 2:

mB 75.31

mB 12

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First term

2010

Example5:

Consider the weight of footing and determine L and B of a trapezoidal footing for a

uniform soil pressure of 300 KN/m2

Useful formula:

ba

bahy

2

3

Solution:

Weight of footing = BLLB

5.11224

2

5.1

BLBL

LB

BL

Q

5.1150)5.1(1210000

2

5.13005.112450025003000

0

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First term

2010

LY

LYLY

YBLet

13810000

1501210000

5.1

5.72LY ………………………………………………………………………(1)

0M (Moment taken about the centroid of the base)

maa

aaa

88.8888100

01000)14(4500700)8(2500500)2(3000

But :

LYLYLY

LYLLY

Y

YL

B

BLL

5.164.263

5.1)88.8(3

5.1

35.1

5.12

388.8

YLYL 64.2625.1 ……………………………………………………………….(2)

Substitute by LY from equation (1) in equation (2)

mBY

mL

LL

LL

YL

04.354.495.15

5.72

95.155.12

4.19315.14145145

04.19311455.1

5.7264.261455.1

64.265.7225.1

2

2

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First term

2010

Example 6:

Design a strap footing to support two columns, that spaced 4.0m center to center

exterior column is 80cm x 80cm carrying 1500 KN and interior column is 80cm x 80cm

carrying 2500KN.

./200 2mKNqnetall

Solution:

1- Find the resultant force location:

mXX

QM

KNQQR

rr 5400082500

00.0@

400025001500

1

21

2- Assume the length of any foot, let we assume L1=2m.

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First term

2010

3- Find the distance a:

ma 4.42

8.0

2

25

4- Find the resultant of each soil pressure:

KNRRR

KNRRRM

4.23786.16214000

6.1621340004.700.0@

12

112

5- Find the required area for each foot:

mBA

mBmAext

45.3892.11892.11200

4.2378

1.42

108.8108.8

200

6.1621

2

1

2

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First term

2010

Example 7:

For the shown mat foundation:

Interior columns Edge columns Exterior columns

Column dimension 60cm x 60cm 60cm x 40cm 40cm x 40cm

Service loads 1800KN 1200KN 600KN

Ultimate loads 2700KN 1800KN 900KN

./150 2mKNqnetall

Check the adequacy of the foundation dimensions.

Calculate the modified soil pressure under the strip ABCD which is 2m width.

Draw SFD and BMD for the strip.

Solution:

Check the adequacy of the foundation dimensions.

1- Find the center of gravity of mat footing:

mYmX gg 5.8 5.6

The distances are taken from (x-y) axes shown in the figure.

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First term

2010

2- Find the resultant force R:

KNQR i 200,1312006180026004

3- Find the location of the resultant force:

mY

mX

R

R

18.8200,13

6002120017120021800121200218004

81.5200,13

1312002136002518002512002

4- Find the eccentricities:

me

m.e

y

x

32.05.818.8

6905.681.5

5- Find YX MM and :

mKNM

mKNM

y

X

.108,9200,1369.0

.224,4200,1332.0

6- Find the stresses:

2

min

2

max

43

43

/87.327.8627.882,5

224,47.6

85.488,3

108,9

4.174.13

200,13

/35.807.8627.882,5

224,47.6

85.488,3

108,9

4.174.13

200,13

gravity ofcenter thepoint to thefrom Distances : ,

627.882,5

224,4

85.488,3

108,9

4.174.13

200,13

627.882,5 4.134.1712

1

851.488,3 4.174.1312

1

mKNq

mKNq

YX

TensionT

nCompressioc

YXq

mI

mI

x

y

OKq

OKqqnetall

00.0min

max

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First term

2010

Calculate the modified soil pressure under the strip ABCD which is 2m width.

Locate the points E and F at the middle of strip edges.

Find the stresses at E and F and be careful that we use ultimate loads:

2

2

/6.1167.8627.882,5

336,67.5

85.488,3

662,13

4.174.13

800,19

/8.977.8627.882,5

336,67.5

85.488,3

662,13

4.174.13

800,19

627.882,5

336,6

85.488,3

662,13

4.174.13

800,19

336,6800,1932.0

662,13800,1969.0

800,1927002180069004

mKNq

mKNq

YXq

KNM

KNM

KNQ

F

E

uX

uY

ui

Find the average stress:

2/2.1072

6.1168.97

2mKN

qqq FE

avg

KNQStripui 5400180029002 .

KNAq stripavg 76.37304.1722.107

StripuiQ stripavg Aq

We have to make adjustment for the loads as follow:

KN4.45652

76.37305400 load Average

Find the modified column loads:

0.845by loadcolumn each

845.05400

4565.4mod

Multiply

QQQ uiuiui

Find the modified soil pressure:

2

mod, /2.13176.3730

4565.42.107 mKNq avgu

Draw SFD and BMD.

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First term

2010

Example8:

Calculate the base pressures at all points beneath the mat foundation shown below

Weight of the mat will not be considered.

Total vertical force acting on the mat foundation is 26000 KN.

Solution:

2

2

2

2

433

2

/5.305)52.21(200),,,(

/3.263)32.21(200),(

/7.136)32.21(200),(

/5.94)52.21(200),,,(

2.2120033.923

19500

130

26000

.19500)575.5(26000.

33.9236512

1105

12

12

130)52(21015

mKNNLDBq

mKNHFq

mKNGEq

mKNMKCAq

XXq

mKNeQM

I

XM

A

Qq

mI

mA

yY

Y

Y

Y

foundation engineering eciv 4352 chapter (6) geometric...

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