Foundation engineering ECIV 4352site.iugaza.edu.ps/ashaat/wp-content/uploads/Ch6(2).pdf · Find the Dimensions of the trapezoidal combined footing for the columns A and B that spaced

1

Ch 6 Geometric design of shallow foundations and Mat footing

1) Geometric design of isolated footing: Most economical type .

Can be rectangular ,circular or square . It is preferred that the footing matches the

shape of its column .

Used in case of light columns loads and when columns are not closely spaced

1- Find the net allowable bearing capacity

(Soil pressure):

scfcc

all

unetu

netall

u

all

hDhq

FS

qq

FS

qq

FS

qq

FS

qq

2- Find the required area of footing:

LD

netall

req

PPQ

q

QA

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

Gross

req

hDBLW

hBLW

WWPPQ

q

QA

2

2) Geometric design of rectangular combined footing: 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 columns 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 .

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

3

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

4

3) 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 where ,

B1 : the edge from which the centroid is measured .

B2: the other edge .

Please try to derive the centroid equation .

5

4) 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 .

1- Find the resultant force location:

RXdQ

QM

QQR

r

2

1

21

00.0@

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

3- Find the distance a:

22

1 wLXa r

4- Find the resultant of each soil pressure:

12

12 1 00.0@

RRR

RFindbRbaRRM

5- Find the required area for each foot:

netall

netall

q

RA

q

RA

22

11

6

5) Geometric and structural design of Mat foundation:

Geometric design (Service loads):

1- Find the center of gravity of mat footing:

i

ii

i

ii

A

AY

A

AX

~

Yg

~

Xg

2- Find the resultant force R:

iQR

3- Find the location of the resultant force:

i

riiR

i

riiR

Q

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.

7

4- Find the eccentricities: ' YYeXXe RyRx

5- Find the following :

iyX QeM ….. moment of the columns loads about x-axis

ixy QeM ….. moment of the columns loads about y-axis

12

1 3 LBI x ….. moment of inertia of the mat's area about its centroidal x-axis

12

1 3BLI y ….. moment of inertia of the mat's area about its centroidal y-axis

6- Find the stresses:

gravity ofcenter thepoint to thefrom Distances : ,YX

YI

MX

I

M

A

Qq

X

X

Y

y

mat

i

Now check that:

00.0min

max

q

qqnetall

Otherwise increase the dimensions of the mat foundation

Structural design (Ultimate loads):

The mat foundation is divided 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:

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.

8

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.

9

Example1

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


mXX

QM

tonsQQR

rr 6.32006120

00.0@

20012080

1

21



mB

mL

L

76.16.7

333.13

6.78.32

2.06.32

10

Example 2.

Design a rectangular combined footing, given that netallq = 5 ksf , Df = 5 feet, the edge of

column 1 is at the property line, and the spacing between columns is 18 feet center-to-

center (c.c.).

Column1 (18'' *18'') Column 2 (24'' *24'')

DL 80 kips 130 kips

LL 175 kips 200 kips

Solution:


221 117

5

330255ft

q

QQA

netall

req


ftXX

QM

kipsQQR

rr 15.1058518330

00.0@

585330255

1

21



L = (0.75+10.15) * 2 = 21.8 ft =22 ft (approximately)

Note : This round off isn't required but it can be accepted

B = A/L = 117/22 = 5.31 ft = 5 ft- 4 in

4- Evaluate the net factored soil pressure :

uQ1 = 1.4 *80 + 1.7 *175 = 410 kips

uQ2 =1.4 *130 + 1.7*200 = 522 kips

net

uq

= ksf96.7

117

5225.409

So, the uniform soil pressure along footing's length q' = net

uq

* B = 7.96 * 5.31=42.3 k/ft

11

5- Now, The column loads are treated as concentrated loads acting at the centers of the

columns. The shear and moment diagrams are

12

Example3

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


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

2- Determine the resultant force

mXX

QM

rr 55.1195475

0@ 1

13

3- 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

14

Example 4

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


mXX

QM

KNQQR

rr 5400082500

00.0@

400025001500

1

21

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

15

3- Find the distance a:

ma 4.42

8.0

2

25


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

Example 5

Design a strap-footing for netallq = 2.5 ksf . The edge of column 1 is placed at the

property line, and the center of the columns are 25 feet center-to-center (c.c.).

Column1 (12'' *12'') Column 2(16'' *16'')

DL 80 kips 120 kips

LL 60 kips 110 kips

Solution:


ftXX

QM

KipsQQR

rr 54.1537025230

00.0@

370230140

1

21

2- Assume the length of any foot, let we assume L1=7ft.

3- Find the distance X1 , X2 :

ft54.122

7

2

154.15X1 , X2 = 25-15.54 = 9.46 ft

16


ftRRR

kipsRRRM

9.2101.159370

1.15946.93702200.0@

12

112

5- Find the required area for each footing:

ftBftA

ftLftA

2.936.8436.845.2

9.210

97

64.6364.63

5.2

1.159

2

2

1

2

1

1- Evaluate the net factored soil pressure :

uQ1 = 1.4 *80 + 1.7 *60 = 214 kips

uQ2=1.4 *120 + 1.7*110 = 355 kips

Repeat the usual steps to find out R1 = 243 kips and R2 = 326 kips

Edge footing : net

uq

1

= ksf85.39*7

243

So, the uniform soil pressure along footing's width q' = net

uq

* L = 3.85 * 9=34.65k/ft

Interior footing : net

uq

2 = ksf85.3

2.9*2.9

326

So, the uniform soil pressure along footing's length q' = net

uq

* B = 3.85 * 9=34.65 k/ft

17

18

Example 6

For the shown mat foundation:

Interior columns Edge columns Corner 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.

Check the adequacy of the foundation dimensions.

1) Find the center of gravity of mat footing:

Xg = 13.4/2 – 0.2 = 6.5 m , Yg = 17.4/2 – 0.2 = 8.5 m

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

19

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 the following :

mKNM

mKNM

y

X

.108,9200,1369.0

.224,4200,1332.0

43

43

627.882,5 4.134.1712

1

851.488,3 4.174.1312

1

mI

mI

x

y

6) Find the stresses:

2

min

2

max

/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

mKNq

mKNq

YX

YXq

OKq

OKqqnetall

00.0min

max

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 notice that we use ultimate loads:

20

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


2/2.1072

6.1168.97

2mKN

qqq FE

avg

KNQStripui 5400180029002 .

KNAq stripavg 76.37304.1722.107

StripuiQ stripavg Aq


KN4.45652

76.37305400 load Average


0.845by loadcolumn each

845.05400

4565.4mod

Multiply

QQQ uiuiui


2

mod, /2.13176.3730

4565.42.107 mKNq avgu

Draw SFD and BMD.

21

Example 7

For the mat foundation shown below. All column dimensions are 50 cm x 50 cm with the

load schedule shown below. The allowable soil pressure is qall =60 kPa.

1) Check the adequacy of the foundation dimensions

2) Draw shear and moment diagrams for the strip AMOJ

22

DL, kN LL, kN Column

200 200 A

250 250 B

250 200 C

800 700 D

800 700 E

650 550 F

800 700 G

800 700 H

650 550 I

200 200 J

250 250 K

200 150 L

Solution

1- Find the center of gravity of mat footing:

Xg = 8.25 m , Yg =10.75 m

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

2- Find the resultant force R:

KNQR i 000,11

3- Find the location of the resultant force:

mY

mX

R

R

85.10000,11

)450500400(25.21)120015002(25.14)120021500(25.7)350500400(25.0

81.7000,11

)12002450350(25.16)150025002(25.8)150024002(25.0

4- Find the eccentricities:

me

m.e

y

x

1.075.1085.10

44052.881.7

5- Find the following :

mKNM

mKNM

y

X

.840,,4000,1144.0

.100,1000,111.0

23

43

43

048,8 5.215.1612

1

665,13 5.165.2112

1

mI

mI

x

y

6- Find the stresses:

2

min

2

max

/62.2675.10048,8

110025.8

665,13

840,,4

5.21*5.16

000,11

/4.3575.10048,8

110025.8

665,13

840,,4

5.21*5.16

000,11

gravity ofcenter thepoint to thefrom Distances : ,

048,8

1100

665,13

840,,4

5.21*5.16

000,11

mKNq

mKNq

YX

YXq

OKq

OKqqnetall

00.0min

max

So, the mat dimensions are ok

Now, to draw shear and moment diagrams, factored forces are considered

2

2

/84.4875.10048,8

1694.5125.6

665,13

8.455,7

5.21*5.16

16945

/37.5375.10048,8

1694.5125.6

665,13

8.455,7

5.21*5.16

16945

048,8

1694.5

665,13

8.455,7

5.21*5.16

16945

1694.5169451.0

8.455,71694544.0

16945

mKNq

mKNq

YXq

KNM

KNM

KNQ

b

a

uX

uY

ui


2/105.512

84.4837.53

2mKN

qqq ba

avg

KNQStripui 5860 .

KNAq stripavg 46705.2125.4105.51

StripuiQ stripavg Aq


KN52652

46705860 load Average

24


0.9by loadcolumn each

9.05860

5265mod

Multiply

QQQ uiuiui


2

mod, /62.574670

5265105.51 mKNq avgu

The shear and bending moment diagrams for the selected strip in are shown below.

25

Note : This moment diagram could be mirrored about the horizontal axis to get

BMD we are used to (i.e, Moment drawn on tension side)

Special thanks for both Eng.Amr Qmbz and Eng. Eng. Haytham Besaiso

Documents

Foundation engineering ECIV 4352site.iugaza.edu.ps/ashaat/wp-content/uploads/Ch6(2).pdf · Find the Dimensions of the trapezoidal combined footing for the columns A and B that spaced