Problems on Foundation Design

8/13/2019 Problems on Foundation Design

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Eng . Haytham Besaiso2

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:

net all

reqq

QQ A 21

2- Find the resultant force location (Xr):

R X LQ

Q M

QQ R

r

22

1

21

00.0@

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

L

A B

X L L

X L L

r

r

1

1

2

2




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

Find center M @ :

1- If center M @ =0.00, uniform soil pressure

And find the required area:

L

A B

q

QQ A

net all

req

21

2- if 00.0@ center M 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

L B

Rq

L

e

L B

Rq

61

61

min

max

Find the width of footing B, by equating maxq

With net all q .

Check adequacy of footing width B that is 00.0min q




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 .


net all

reqq

QQ A 21

As the area of trapezoidal is given by ))((5.0 21 L B B A

So put A Areq to get an equation which is function of 21 & B B .

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

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

pressure.

The centroid equation is:

21

21 2

3 B B

B B L

X So we will have another equation of 21 & B B , solve them to

get 21 & B B where ,

B1 : the edge from which the centroid is measured .

B2: the other edge .

Please try to derive the centroid equation .




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:

R X Q

Q M

QQ R

r

Lc

00.0@

2

1

21

2- Assume the width of the exterior (edge) footing B1.

3- Find the distance X1 , X2 :

22

111

Bc X X r , X2 = Lc – Xr

4- Find the resultant of each soil pressure:

12

22112 100.0@

R R R

R Find X R X X R R M

5- Find the required area for each foot:

net all

net all

q

R A

q

R A

2

2

1

1




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

A X

~

Yg

~

Xg

2- Find the resultant force R:

iQ R

3- Find the location of the resultant force:

i

rii

R

i

rii

RQ

Y QY

Q

X Q X

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

i A : shape's area . i

X : distance between y-axis and the centroid of the shape

iY :

distance between x-axis and the centroid of the shapeiQ :The load of column i

ri X : 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.




4- Find the eccentricities:' Y Y e X X e R y R x

5- Find the following :

i y X Qe M ….. moment of the columns loads about x-axis

i x y Qe M ….. moment of the columns loads about y-axis

12

1

3 L B I x ….. moment of inertia of the mat's area about its centroidal x-axis

12

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

6- Find the stresses:

gravityof centerthe point tothefromDistances:,Y X

Y I

M X

I

M

A

Qq

X

X

Y

y

mat

i

Now check that:

00.0min

max

q

qqnet all

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:

ui X uY

ui yu X

X

u X

Y

uY

mat

ui

Qe M

Qe M

Y I

M X

I

M

A

Qq

Find the average stress:

2

F E avg

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 loadAverage

stripavg Stripui AqQ

Find the modified column loads:

stripui

uiuiQ

QQ

loadAverage

mod

Find the modified soil pressure:

stripavg u

avg uavg u Aq

qq

,

,mod,

loadAverage

Now we must have that:

StripuiQ = stripavg Aq

Draw SFD and BMD.




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 2mt q net all

Solution


2

21 33.1315

12080mq

QQ A

net all

req


m X X

Q M

tonsQQ R

r r 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:

m B

m L

L

76.16.7

333.13

6.78.32

2.06.3

2




Example 2. Design a rectangular combined footing, given that

net all q = 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

330255 ft

q

QQ A

net all

req


ft X X

Q M

kipsQQ R

r r 15.1058518330

00.0@

585330255

1

21

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

rectangular footing:

L = (0.7

5+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

= ksf 96.7

1175225.409

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

uq

* B = 7.96 * 5.31=42.3 k/ft




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

columns. The shear and moment diagrams are




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 2mt q net all

Solution


221 34.10

85.18

75120m

q

QQ A

net all

req

As the area of trapezoidal is given by ))((5.0 21 L B B A

So put A Areq to get an equation which is function of 21 & B B .

75.434.1035.45.0 2121 B B B B

...............1

2- Determine the resultant force

m X X

Q M

r r 55.1195475

0@ 1




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

pressure.

The centroid equation is:

21

21

21

21 2305.075.4

2

3

35.42

3 B B X

B B

B B

B B L X

For uniform soil pressure:

1.55 + 0.2 = X

75.12305.0

75.1

21

B B

m X

75.52 21 B B .........................2

Solve 1 and 2:

m B 75.31

m B 12

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 2m KN qnet all





m X X

Q M

KN QQ R

r r 5400082500

00.0@

400025001500

1

21

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

3- Find the distance X1 , X2 :

m4.42

2

2

8.05X1 , X2 = 8-5 = 3.0 m


KN R R R

KN R R R M

4.23786.16214000

6.1621340004.700.0@

12

112

5- Find the required area for each foot:

m B A

m Bm Aext

45.3892.11892.11200

4.2378

1.42

108.8108.8

200

6.1621

2

1

2




Example 5

Design a strap-footing for net all q = 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:


ft X X

Q M

KipsQQ R

r r 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 : ft 54.12

2

7

2

154.15X1 , X2 = 25-15.54 = 9.46 ft


ft R R R

kips R R R M

9.2101.159370

1.15946.93702200.0@

12

112

5- Find the required area for each footing:

ft B ft A

ft L ft A

2.936.8436.845.2

9.210

97

64.6364.63

5.2

1.159

22

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 R 1 = 243 kips and R 2 = 326 kips

Edge footing :net

uq

1 = ksf 85.3

9*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

= ksf 85.32.9*2.9

326

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

uq

* B = 3.85 * 9=34.65 k/ft







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 2m KN qnet all

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 foundati on 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.




2) Find the resultant force R:

KN Q R i 200,1312006180026004

3) Find the location of the resultant force:

mY

m X

R

R

18.8200,13

6002120017120021800121200218004

81.5

200,13

1312002136002518002512002

4) Find the eccentricities:

me

m.e

y

x

32.05.818.8

6905.681.5

5) Find the following :

m KN M

m KN M

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

m I

m I

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

gravityof centerthe point tothefromDistances:,

627.882,5

224,4

85.488,3

108,9

4.174.13

200,13

m KN q

m KN q

Y X

Y X q

OK q

OK qqnet all

00.0min

max

Calculate the modif ied soil pressure under the str ip 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:




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

800,19627.882,5

336,6

85.488,3

662,13

4.174.13

800,19

336,6800,1932.0

662,13800,1969.0

800,1927002180069004

m KN q

m KN q

Y X q

KN M

KN M

KN Q

F

E

u X

uY

ui


2/2.1072

6.1168.97

2m KN

qqq F E avg

KN QStripui 5400180029002 .

KN Aq stripavg 76.37304.1722.107

StripuiQ stripavg Aq


KN 4.45652

76.37305400 loadAverage


0.845 byloadcolumneach

845.05400

4565.4mod

Multiply

QQQ uiuiui


2

mod, /2.13176.3730

4565.42.107 m KN q avg u

Draw SFD and BMD.




Example 7For 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




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:

KN Q R i 000,11

3- Find the location of the resultant force:

Y

m X

R

R

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 :

m KN M

m KN M

y

X

.840,,4000,1144.0

.100,1000,111.0




43

43

048,8 5.215.1612

1

665,13 5.165.2112

1

m I

m I

x

y

6- Find the stresses:

2

min

2

max

/62.2675.10

048,8

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

gravityof centerthe point tothefromDistances:,

048,8

1100

665,13

5390

5.21*5.16

000,11

m KN q

m KN q

Y X

Y X q

OK q

OK qqnet all

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

m KN q

m KN q

Y X q

KN M

KN M

KN Q

b

a

u X

uY

ui


2/105.512

84.4837.53

2m KN

qqq baavg

KN QStripui 5860 .

KN Aq stripavg 46705.2125.4105.51

StripuiQ stripavg Aq


KN 52652

46705860 loadAverage





0.9 byloadcolumneach

9.05860

5265mod

Multiply

QQQ uiuiui


2

mod, /62.574670

5265105.51 m KN q avg u

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



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)

Documents

Problems on Foundation Design