Foundation Examples Dr, Basma.pdf

DESIGN OF MAT (RAFT) FOUNDATIONS Design Steps and Equations For an example on Design of Mat Foundations click here 3D View of Mat (Raft) Foundation

y L1

L2

B

A Pu1

C Pu3

Pu2 ex

Column dimensions l i x bi

B1

Pu ey

B

E

D Pu4

Pu5

x

F Pu6 B2

Pu7 G

Pu8

H

Pu9 I

L Top View of Mat (Raft) Foundation 1 of 6

Foundation Engineering 2 Design of Mat (Raft) Footings Dr. Adnan A. Basma

STEP 1 – CHECK SOIL PRESSURE FOR SELECTED DIMENTION Allowable load P = !Pi Ultimate load

Piu = ! [1.4DLi + 1.7LLi]

Ultimate ratio

ru =

Pu , Ultimate pressure qu = qa x ru P

Locate the resultant load Pu In x- direction:

!My-axis = 0,

ex &

In y- direction:

!Mx-axis = 0,

ey &

"Pu3 $ Pu9 # L2

% "Pu1 $ Pu7 # L1 Pu

"Pu1 $ Pu3 # B1

% "Pu7 $ Pu9 # B2 Pu

,P My x Mx y ) ' Applied ultimate pressure, qu,applied = * u Iy Ix ' *A + ( Where

A = Area = BL Mx = Pu ey

and

My = Pu ex

Ix = ! B L

and

Iy = ! L B

3

3

For the mat shown in Top View, the following sign convention is used to estimate qu,applied x (+)

x (%) A

B

C y (+)

D

E

F

G

H

I

2 of 6

y (%)


The following values for x and y along with the sign conventions are used to estimate qu,applied Point A

B

C

D

E

F

G

H

I

x

L1

0

L2

L1

0

L2

L1

0

L2

y

B1

B1

B1

0

0

0

B2

B2

B2

For the dimensions L and B to be adequate, qu,max . qu

and

qu,min / 0

STEP 2 – DRAW SHEAR AND MOMENT DIAGRAMS (L - DIRECTION) The mat is divided into several strips in L-direction as shown below where B' = B/4. y L2

L1 B

A B'

C

I

K

J

B1

B' B D

II

E

x

F

B' B2 M

L B'

III

G

H

I

L

3 of 6


Calculations for Strip ABC: a) The average uniform soil reaction, quI =

q uA $ quC 2

b) Total soil reaction Q = quI x (B1 x L) where B1 = B' c) Total Column loads Pu, ABC = Pu1 + Pu2 + Pu3 d) Average load Pu, avg(ABC) =

Q $ Pu,

e) Load multiplying factor FABC =

ABC

2 Pu,

avg( ABC )

Pu,

ABC

f) The modified loads on this strip P'ui = (FABC) x (Pui) , Pu, avg( ABC ) ) g) Modified Average soil pressure qu, modified = quI x * ' Q *+ '( h) The pressure distribution along the length of the strip qu, L (ABC) =

0 P' ui L

Note that the same can be done for strips DEF (B2 = 2B') and GHI (B3 = B') where: quII = quIII =

q uD $ q uF 2 q uG $ q uI 2

Steps (b) to (h) are repeated as above

The shear and bending moment diagrams for strip ABC is shown below. Other strips will have similar plots.

4 of 6


P'u1

P'u3

P'u2 L1

L2 B

A

C qu, L (ABC)

+ V(kN)

%

Top Steel between Column 2 and 3


% M(kN.m)

+ Bottom Steel under Column 2

Moment drawn on tension side

5 of 6


Similar plots should be made for strips in B-direction as shown below A

B I

C III

II

D

E

F

G

H

I

STEP 3 – DEPTH OF CONCRETE, d' Estimate d' for: a) Column 1, 3, 7 and 9 by 2-way punching shear (p' = l + w). b) Column 2, 4, 6 and 8 by 3-way punching shear (p' = 2l + w). c) Column 5 by 4-way punching shear (p' = 2l + 2w). (Use Equations For Punching Shear or approximate d' by Structural Depth of Concrete table for punching shear failure). Select the largest d' from (a), (b) or (c)

STEP 4 – REINFORCEMENT The calculations below are repeated for every strip in L and B direction. a) Select the appropriate moments for each strip (refer to moment diagram) and estimate the moment per meter by Mui/m = Mu/Bi or Li b) Using Mui/m, d', fc' and fy estimate the reinforcement As (refer to Equations for Reinforcement or the percent reinforcement can be obtained directly from Percent Steel Tables).

For an example on Design of Mat Footings click here

6 of 6


FOUNDATION ENGINEERING 2 Mat Foundations (Design Equations) Design Example Design the mat foundation for the plan 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. Use fc’ = 24 Mpa and fy = 275 MPa

4.25 m

4.25 m

8m y

0.25 m M

B

3.75 m

DL= 200 kN LL= 200 kN

N DL = 250 kN LL = 250 kN

C DL=250 kN LL=200 kN

7m

M

D

N

DL=800 kN LL= 700 kN


E

DL=650 kN LL=550 kN

7m

A

F



H

DL=650 kN LL=550 kN

3.75 m

R

DL = 250 kN LL = 250 kN

J 0.25 m

O

K

8m

x

I

Q

DL=200 kN LL= 200kN

21.5 m

P

7m

7m

G

0.1 m

Pu

O

7m

0.44m

DL=200 kN LL=150 kN

P

8m

L 0.25 m

16.5 m

Plan of Mat Foundation with column loads and dimensions

1 of 9

Foundation Engineering 2 Example on Design of Mat Foundations Dr. Adnan A. Basma


Piu = ! [1.4DLi + 1.7LLi]

Ultimate ratio

ru =

Pu P

The table below shows the calculations for the loads: Column

DL, kN

A B C D E F G H I J K L

200 250 250 800 800 650 800 800 650 200 250 200

LL, kN

P, kN

200 250 200 700 700 550 700 700 550 200 250 150 Total Loads =

400 500 450 1500 1500 1200 1500 1500 1200 400 500 350 11000 ru =

Pu, kN 620 775 690 2310 2310 1845 2310 2310 1845 620 775 535 16945 1.54

Ultimate pressure qu = qa x ru = 60 x 1.54 = 92.4 kPa Location of the resultant load Pu In x- direction: ex '

!My-axis = 0,

"620 % 2310 % 2310 % 620# $ 8

&

"690 % 1845 % 1845 % 535# $ 8

16945

= 0.44 m In y- direction: ey '

!Mx-axis = 0,

"620 % 775 % 690#x10.5 % "2310 % 2310 % 1845#x3.5 - "2310 % 2310 % 1845#x3.5 & "620 % 775 % 535#x10.5 16945

= 0.1 m

2 of 9


-P My x Mx y * ( Applied ultimate pressure, qu,applied = + u . . Iy Ix ( +A , ) A = Area = BL = (16.5) x (21.5) = 354.75 m2

Where

Mx = Pu ey = (16945)(0.1) = 1694.5 kN.m My = Pu ex = (16945)(0.44) = 7455.8 kN.m 1

Ix =

Iy =

12 1 12

1

3

BL =

12 1

3

LB =

3

12

(16.5) (21.5) = 13665 m4 3

(21.5) (16.5) = 8050 m4

Therefore, 7455.8 x 1694.5 y * - 16945 qu,applied = + . . ( 8050 13665 )( ,+ 354.75

"

= 47.76 . 0.93 x . 0.124 y

#

Point

x

y

Sign for x Sign for y

A

8.25

10.75

+

+

56.77

B

0

10.75

+

+

49.09

C

8.25

10.75

&

+

41.42

J

8.25

10.75

+

&

54.10

K

0

10.75

+

&

46.43

L

8.25

10.75

&

&

38.75

qu,applied, kPa

For the dimensions L = 21.5 m and B = 16.5 m, qu,max = 56.77 kPa < qu = 92.4 kPa and qu,min = 38.75 kPa > 0 Therefore, the dimensions are adequate. 3 of 9


STEP 2 – DRAW SHEAR AND MOMENT DIAGRAMS (L - DIRECTION) The mat is divided into several strips in L-direction (see figure on page 1). The following strips are considered: AMOJ, MNPO and NCLP The following calculations are performed for every strip: a) The average uniform soil reaction, qu =

q u,

Edge 1

% qu

edge 2

2

Refer to table on page 3 for pressure values Strip AMOJ: Edge 1 is point A

and

Edge 2 is point J

Strip MNPO: Edge 1 is point B

and

Edge 2 is point K

Strip NCLP: Edge 1 is point C

and

Edge 2 is point L

b) Total soil reaction Qi = qu x (Bi x L) Strip AMOJ: B1 = 4.25 m Strip MNPO: B2 = 8.00 m Strip NCLP: B3 = 4.25 m For all strips L = 21.5 m c) Total Column loads Pu, total = !Pui d) Average load Pu, avg =

Q i % Pu,total 2

e) Load multiplying factor F =

Pu,

avg

Pu,

total

f) The modified loads on this strip P'ui = (F) x (Pui) - Pu, avg * g) Modified Average soil pressure qu, modified = qu x + ( +, Q i () h) The pressure distribution along the length of the strip qu, L =

/ P' ui L

4 of 9


The following table presents the calculations for the selected strips: Strip

Bi, m Point A J B K C L

AMOJ 4.25 MNPO 8.00 NCLP 4.25

qu, kPa 56.77 54.10 49.09 46.43 41.42 38.75

qu,avg kPa

Q, kN

Total Pu, kN

Pu, avg, qu, mod kN kPa

55.44

5065.37

5860.00 5462.69 59.47 0.932

47.76

8214.72

6170.00 7192.36 40.97 1.166

40.09

3662.77

4915.00 4288.88 45.94 0.873

F

Based on table above, the adjusted column loads and the pressure under each is strip are: Strip

Column

DL, kN

LL, kN

P, kN

Pu, kN

AMOJ F=0.932

P'u, kN qu,L, kN/m

A

200

200

400

620

D

800

700

1500

2310

2152.92

G

800

700

1500

2310

2152.92

J

200

200

400

620

577.84

Total =

2000

1800

3800

5860

5461.52

MNPO

B

250

250

500

775

F=1.166

E

800

700

1500

2310

2693.46

H

800

700

1500

2310

2693.46

K

250

250

500

775

903.65

Total =

2100

1900

4000

6170

7194.22

NCLP

C

250

200

450

690

F=0.873

F

650

550

1200

1845

1610.685

I

650

550

1200

1845

1610.685

L

200

150

350

535

467.055

Total =

1750

1450

3200

4915

4290.795

Total Loads =

5850

5150

11000

16945

577.84 254.02

903.65 334.61

602.37 199.57

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

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Strip AMOJ (B' = 4.25 m) 577.8 kN 0.25 m

2152.9 kN

2152.9 kN 7m

7m

7m

A

D

577.8 kN 0.25 m

G

J

254.02 kN/m

1500 1265.4 888.1

1000

510.8

500 63.4

+

0

V(kN)

0

& -500

5

10

15

20 -63.4

-510.8

-1000

-888 .1 -1265.4

-1500

Max. moment between columns Steel at the bottom 2660.5 2660.5

3000 2500 2000 1500 1000

+

500

M(kN.m)

&

1106.3 7.9

7.9

0

-500

0

5 -503.0

-1000

10

15

20

- 503.0

Max. moment under columns Steel at the top

Moment drawn on compression side 6 of 9


Strip MNPO (B' = 8.0 m) 903.7 kN 0.25 m

2693.5 kN

2693.5 kN 7m

7m

7m

B

E

903.7 kN 0.25 m

H

K

334.6 kN/m

2000 1524.9

1500

1169.8

1000

814.7

500 83.6

+

0

V(kN)

& -500

0

5

10

15

20 -83.6

-814.7

-1000

-1169.8

-1500

-1524.9

-2000 Max. moment between columns Steel at the bottom

3000

2496.2

2496.2

2500 2000 1500 1000 500

+ &

449.1

10.5

M(kN.m)

10.5

0

-500

0

5

10

15

20

-1000

-982.4 -1500

Max. moment under columns Steelon atcompression the top Moment drawn side

-982.4



Strip NCLP (B' = 4.25 m) 602.4 kN 0.25 m

1610.7 kN

1610.7 kN 7m

7m

7m

C

F

467.1 kN 0.25 m

I

L

199.6 kN/m

1000 845.0

632.6

500

420.4

+ V(kN)

50.1 0

&

0

-500

5

10

20 -49.7

15

-554.4 -766.6

-1000

-978.8

-1500

1500

1023.5

Max. moment between columns Steel at the bottom

1000

554.7 500

+ M(kN.m)

0

&

6.3

6.3 0

5

-500 -1000 -1500 -2000

10

15

20

-446.6 -762.3

-1842.1




STEP 3 – DEPTH OF CONCRETE, d' The table below was prepared for the columns and d’ was estimated using Structural Depth of Concrete table for punching shear failure (fc' = 24 MPa). Column

Punching

p', m

Pu, kN

d', m

A&J

2

1.00

620

0.29

B&K

3

1.50

775

0.27

C

2

1.00

690

0.31

D&G

3

1.50

2310

0.62

Maximum

E&H

4

2.00

2310

0.54

Use d' = 0.65 m

F&I

3

1.50

1845

0.53

L

2

1.00

535

0.25

STEP 4 – REINFORCEMENT L-direction Reinforcement: The table below is prepared by selecting the appropriate moments for each strip and estimate the moment per meter by Mui/m = Mu/Bi or Li . Using Mui/m, d' = 0.65, fc' = 24 MPa and fy = 275 MPa*, p (percent steel) and thus As can be estimated by Percent Steel Tables. Strip

B', m

Location

AMJO

4.25

0-5 m / top 5-17 m / bottom 17-21.5 m / top

503 2660.5 503

MNPO

8

0-5 m / top 5-16.5 m / bottom 16.5-21.5 m / top

MNPO 4.25

Mu, kN.m Mu, kN.m/m

+

p, %

AS (cm2/m)

Reinforcement

118.4 626.0 118.4

0.11** 0.62.. 0.11**

33.2 40.3 33.2

030@25 cm c-c 030@20 cm c-c 030@25 cm c-c

982.4 2496.2 982.4

122.8 312.0 122.8

0.11** 0.30** 0.11**

33.2 33.2 33.2

030@25 cm c-c 030@25 cm c-c 030@25 cm c-c

0-6 m / top 762.3 6-9 m / bottom 1023.5 9-13 m / top 446.6 1314.5 m / bottom 554.7 14.5-21.5 m / top 1842.1

179.4 240.8 105.1 130.5 433.4

0.18** 0.24** 0.10** 0.12** 0.43**

33.2 33.2 33.2 33.2 33.2

030@25 cm c-c 030@25 cm c-c 030@25 cm c-c 030@25 cm c-c 030@25 cm c-c

* For fc' = 24 MPa and fy = 275 MPa, p(min) = 0.51%, and p(max) = 2.99% ** p < p(min) so use p(min) = 0.51% + AS = p x d' x 1 = (p/100) x (65) x (100)

B-direction Reinforcement: The same can be done for the B-direction moments (step 2) by considering (refer to page 1) strips: ACNM (B' = 3.75 m), MNPO (B' = 7 m), OPRQ (B' = 7m) and QRLJ (B' = 3.75 m). 9 of 9


STRUCTURAL DEPTH OF CONCRETE FOOTINGS, d' (m) BY PUNCHING SHEAR *

w

w

l

w

l

l

EDGE COLUMN

EXTERIOR COLUMN

INTERIOR COLUMN

2-Way Punch at d'/2 from sides of column



p' = l + w

p' = 2l + 2w

p' = 2l + w

f'c = 21 MPa Dimension p'

Ultimate Load, Pu

1.00

1.20

1.40

1.60

1.80

2.00

2.20

2.40

2.60

2.80

3.00

300 400 500 600 700 800 900 1000 1100 1200 1300 1400 1500 1600 1700 1800 1900 2000 2100 2200 2300 2400 2500 2600 2700 2800 2900 3000 3100 3200 3300 3400 3500

0.17 0.22 0.26 0.29 0.33 0.36 0.39 0.42 0.45 0.48 0.50 0.53 0.55 0.58 0.60 0.62 0.65 0.67 0.69 0.71 0.73 0.75 0.77 0.79 0.81 0.82 0.84 0.86 0.88 0.89 0.91 0.93 0.94

0.15 0.20 0.23 0.27 0.30 0.33 0.36 0.39 0.42 0.45 0.47 0.50 0.52 0.54 0.57 0.59 0.61 0.63 0.65 0.67 0.69 0.71 0.73 0.75 0.77 0.79 0.80 0.82 0.84 0.86 0.87 0.89 0.90

0.14 0.18 0.21 0.25 0.28 0.31 0.34 0.37 0.39 0.42 0.44 0.47 0.49 0.51 0.54 0.56 0.58 0.60 0.62 0.64 0.66 0.68 0.70 0.71 0.73 0.75 0.77 0.79 0.80 0.82 0.84 0.85 0.87

0.13 0.16 0.20 0.23 0.26 0.29 0.31 0.34 0.37 0.39 0.42 0.44 0.46 0.48 0.51 0.53 0.55 0.57 0.59 0.61 0.63 0.65 0.66 0.68 0.70 0.72 0.74 0.75 0.77 0.79 0.80 0.82 0.83

0.12 0.15 0.18 0.21 0.24 0.27 0.29 0.32 0.34 0.37 0.39 0.41 0.44 0.46 0.48 0.50 0.52 0.54 0.56 0.58 0.60 0.62 0.63 0.65 0.67 0.69 0.70 0.72 0.74 0.75 0.77 0.79 0.80

0.11 0.14 0.17 0.20 0.22 0.25 0.27 0.30 0.32 0.35 0.37 0.39 0.41 0.43 0.45 0.47 0.49 0.51 0.53 0.55 0.57 0.59 0.61 0.62 0.64 0.66 0.67 0.69 0.71 0.72 0.74 0.75 0.77

0.10 0.13 0.16 0.18 0.21 0.23 0.26 0.28 0.30 0.33 0.35 0.37 0.39 0.41 0.43 0.45 0.47 0.49 0.51 0.53 0.54 0.56 0.58 0.60 0.61 0.63 0.65 0.66 0.68 0.69 0.71 0.72 0.74

0.09 0.12 0.14 0.17 0.20 0.22 0.24 0.27 0.29 0.31 0.33 0.35 0.37 0.39 0.41 0.43 0.45 0.47 0.48 0.50 0.52 0.54 0.55 0.57 0.59 0.60 0.62 0.64 0.65 0.67 0.68 0.70 0.71

0.08 0.11 0.14 0.16 0.18 0.21 0.23 0.25 0.27 0.29 0.31 0.33 0.35 0.37 0.39 0.41 0.43 0.45 0.46 0.48 0.50 0.51 0.53 0.55 0.56 0.58 0.60 0.61 0.63 0.64 0.66 0.67 0.69

0.08 0.10 0.13 0.15 0.17 0.20 0.22 0.24 0.26 0.28 0.30 0.32 0.34 0.35 0.37 0.39 0.41 0.43 0.44 0.46 0.48 0.49 0.51 0.53 0.54 0.56 0.57 0.59 0.60 0.62 0.63 0.65 0.66

0.07 0.10 0.12 0.14 0.16 0.18 0.21 0.23 0.25 0.26 0.28 0.30 0.32 0.34 0.36 0.37 0.39 0.41 0.42 0.44 0.46 0.47 0.49 0.50 0.52 0.54 0.55 0.57 0.58 0.59 0.61 0.62 0.64

* Neglecting ultimate soils pressure qu. Values of d' in the table are thus 10% - 30% greater than actual values. 1 of 4

Foundation Engineering 2 / Review Depth of Concrete Footings by Punching Shear Dr. Adnan A. Basma

STRUCTURAL DEPTH OF CONCRETE FOOTINGS, d' (m) BY PUNCHING SHEAR * f'c = 24 MPa Dimension p'

Ultimate Load, Pu

1.00

1.20

1.40

1.60

1.80

2.00

2.20

2.40

2.60

2.80

3.00

300 400 500 600 700 800 900 1000 1100 1200 1300 1400 1500 1600 1700 1800 1900 2000 2100 2200 2300 2400 2500 2600 2700 2800 2900 3000 3100 3200 3300 3400 3500

0.16 0.21 0.24 0.28 0.31 0.34 0.37 0.40 0.43 0.46 0.48 0.51 0.53 0.55 0.58 0.60 0.62 0.64 0.66 0.68 0.70 0.72 0.74 0.75 0.77 0.79 0.81 0.82 0.84 0.86 0.87 0.89 0.91

0.15 0.19 0.22 0.26 0.29 0.32 0.35 0.37 0.40 0.43 0.45 0.47 0.50 0.52 0.54 0.56 0.58 0.60 0.62 0.64 0.66 0.68 0.70 0.72 0.74 0.75 0.77 0.79 0.80 0.82 0.84 0.85 0.87

0.13 0.17 0.20 0.23 0.26 0.29 0.32 0.35 0.37 0.40 0.42 0.44 0.47 0.49 0.51 0.53 0.55 0.57 0.59 0.61 0.63 0.65 0.67 0.68 0.70 0.72 0.74 0.75 0.77 0.78 0.80 0.82 0.83

0.12 0.15 0.18 0.22 0.24 0.27 0.30 0.32 0.35 0.37 0.40 0.42 0.44 0.46 0.48 0.50 0.52 0.54 0.56 0.58 0.60 0.62 0.63 0.65 0.67 0.69 0.70 0.72 0.73 0.75 0.77 0.78 0.80

0.11 0.14 0.17 0.20 0.23 0.25 0.28 0.30 0.33 0.35 0.37 0.39 0.42 0.44 0.46 0.48 0.50 0.51 0.53 0.55 0.57 0.59 0.60 0.62 0.64 0.66 0.67 0.69 0.70 0.72 0.73 0.75 0.76

0.10 0.13 0.16 0.18 0.21 0.24 0.26 0.28 0.31 0.33 0.35 0.37 0.39 0.41 0.43 0.45 0.47 0.49 0.51 0.52 0.54 0.56 0.58 0.59 0.61 0.63 0.64 0.66 0.67 0.69 0.70 0.72 0.73

0.09 0.12 0.15 0.17 0.20 0.22 0.24 0.27 0.29 0.31 0.33 0.35 0.37 0.39 0.41 0.43 0.45 0.47 0.48 0.50 0.52 0.53 0.55 0.57 0.58 0.60 0.62 0.63 0.65 0.66 0.68 0.69 0.71

0.08 0.11 0.14 0.16 0.18 0.21 0.23 0.25 0.27 0.29 0.31 0.33 0.35 0.37 0.39 0.41 0.43 0.44 0.46 0.48 0.49 0.51 0.53 0.54 0.56 0.57 0.59 0.60 0.62 0.63 0.65 0.66 0.68

0.08 0.10 0.13 0.15 0.17 0.19 0.22 0.24 0.26 0.28 0.30 0.32 0.33 0.35 0.37 0.39 0.41 0.42 0.44 0.46 0.47 0.49 0.50 0.52 0.54 0.55 0.57 0.58 0.60 0.61 0.62 0.64 0.65

0.07 0.10 0.12 0.14 0.16 0.18 0.20 0.22 0.24 0.26 0.28 0.30 0.32 0.34 0.35 0.37 0.39 0.40 0.42 0.44 0.45 0.47 0.48 0.50 0.51 0.53 0.54 0.56 0.57 0.59 0.60 0.61 0.63

0.07 0.09 0.11 0.13 0.15 0.17 0.19 0.21 0.23 0.25 0.27 0.29 0.30 0.32 0.34 0.35 0.37 0.39 0.40 0.42 0.43 0.45 0.46 0.48 0.49 0.51 0.52 0.54 0.55 0.56 0.58 0.59 0.61

* Neglecting ultimate soils pressure qu. Values of d' in the table are thus 10% - 30% greater than actual values.

2 of 4



Ultimate Load, Pu

1.00

1.20

1.40

1.60

1.80

2.00

2.20

2.40

2.60

2.80

3.00

300 400 500 600 700 800 900 1000 1100 1200 1300 1400 1500 1600 1700 1800 1900 2000 2100 2200 2300 2400 2500 2600 2700 2800 2900 3000 3100 3200 3300 3400 3500

0.15 0.19 0.23 0.26 0.30 0.33 0.35 0.38 0.41 0.43 0.46 0.48 0.50 0.53 0.55 0.57 0.59 0.61 0.63 0.65 0.67 0.68 0.70 0.72 0.74 0.75 0.77 0.79 0.80 0.82 0.83 0.85 0.86

0.14 0.17 0.21 0.24 0.27 0.30 0.33 0.35 0.38 0.40 0.43 0.45 0.47 0.49 0.51 0.53 0.55 0.57 0.59 0.61 0.63 0.65 0.67 0.68 0.70 0.72 0.73 0.75 0.77 0.78 0.80 0.81 0.83

0.12 0.16 0.19 0.22 0.25 0.28 0.30 0.33 0.35 0.38 0.40 0.42 0.44 0.46 0.48 0.50 0.52 0.54 0.56 0.58 0.60 0.61 0.63 0.65 0.67 0.68 0.70 0.71 0.73 0.75 0.76 0.78 0.79

0.11 0.14 0.17 0.20 0.23 0.26 0.28 0.30 0.33 0.35 0.37 0.39 0.42 0.44 0.46 0.48 0.49 0.51 0.53 0.55 0.57 0.58 0.60 0.62 0.63 0.65 0.67 0.68 0.70 0.71 0.73 0.74 0.76

0.10 0.13 0.16 0.19 0.21 0.24 0.26 0.28 0.31 0.33 0.35 0.37 0.39 0.41 0.43 0.45 0.47 0.49 0.50 0.52 0.54 0.56 0.57 0.59 0.60 0.62 0.64 0.65 0.67 0.68 0.70 0.71 0.73

0.09 0.12 0.15 0.17 0.20 0.22 0.24 0.27 0.29 0.31 0.33 0.35 0.37 0.39 0.41 0.43 0.44 0.46 0.48 0.50 0.51 0.53 0.55 0.56 0.58 0.59 0.61 0.62 0.64 0.65 0.67 0.68 0.70

0.09 0.11 0.14 0.16 0.18 0.21 0.23 0.25 0.27 0.29 0.31 0.33 0.35 0.37 0.39 0.40 0.42 0.44 0.46 0.47 0.49 0.50 0.52 0.54 0.55 0.57 0.58 0.60 0.61 0.63 0.64 0.65 0.67

0.08 0.10 0.13 0.15 0.17 0.19 0.21 0.23 0.25 0.27 0.29 0.31 0.33 0.35 0.37 0.38 0.40 0.42 0.43 0.45 0.47 0.48 0.50 0.51 0.53 0.54 0.56 0.57 0.59 0.60 0.61 0.63 0.64

0.07 0.10 0.12 0.14 0.16 0.18 0.20 0.22 0.24 0.26 0.28 0.30 0.31 0.33 0.35 0.36 0.38 0.40 0.41 0.43 0.44 0.46 0.47 0.49 0.50 0.52 0.53 0.55 0.56 0.58 0.59 0.60 0.62

0.07 0.09 0.11 0.13 0.15 0.17 0.19 0.21 0.23 0.25 0.26 0.28 0.30 0.31 0.33 0.35 0.36 0.38 0.39 0.41 0.42 0.44 0.45 0.47 0.48 0.50 0.51 0.53 0.54 0.55 0.57 0.58 0.59

0.06 0.09 0.10 0.12 0.14 0.16 0.18 0.20 0.22 0.23 0.25 0.27 0.28 0.30 0.32 0.33 0.35 0.36 0.38 0.39 0.41 0.42 0.44 0.45 0.46 0.48 0.49 0.50 0.52 0.53 0.54 0.56 0.57


3 of 4



Ultimate Load, Pu

1.00

1.20

1.40

1.60

1.80

2.00

2.20

2.40

2.60

2.80

3.00

300 400 500 600 700 800 900 1000 1100 1200 1300 1400 1500 1600 1700 1800 1900 2000 2100 2200 2300 2400 2500 2600 2700 2800 2900 3000 3100 3200 3300 3400 3500

0.14 0.18 0.21 0.24 0.27 0.30 0.33 0.35 0.38 0.40 0.42 0.45 0.47 0.49 0.51 0.53 0.55 0.57 0.58 0.60 0.62 0.64 0.65 0.67 0.69 0.70 0.72 0.73 0.75 0.76 0.78 0.79 0.81

0.12 0.16 0.19 0.22 0.25 0.28 0.30 0.33 0.35 0.37 0.39 0.42 0.44 0.46 0.48 0.50 0.51 0.53 0.55 0.57 0.58 0.60 0.62 0.63 0.65 0.67 0.68 0.70 0.71 0.73 0.74 0.76 0.77

0.11 0.14 0.17 0.20 0.23 0.25 0.28 0.30 0.32 0.35 0.37 0.39 0.41 0.43 0.45 0.47 0.48 0.50 0.52 0.54 0.55 0.57 0.59 0.60 0.62 0.63 0.65 0.66 0.68 0.69 0.71 0.72 0.74

0.10 0.13 0.16 0.18 0.21 0.23 0.26 0.28 0.30 0.32 0.34 0.36 0.38 0.40 0.42 0.44 0.46 0.47 0.49 0.51 0.52 0.54 0.56 0.57 0.59 0.60 0.62 0.63 0.65 0.66 0.67 0.69 0.70

0.09 0.12 0.14 0.17 0.19 0.22 0.24 0.26 0.28 0.30 0.32 0.34 0.36 0.38 0.40 0.41 0.43 0.45 0.46 0.48 0.50 0.51 0.53 0.54 0.56 0.57 0.59 0.60 0.62 0.63 0.64 0.66 0.67

0.08 0.11 0.13 0.16 0.18 0.20 0.22 0.24 0.26 0.28 0.30 0.32 0.34 0.36 0.37 0.39 0.41 0.42 0.44 0.46 0.47 0.49 0.50 0.52 0.53 0.55 0.56 0.57 0.59 0.60 0.62 0.63 0.64

0.08 0.10 0.12 0.15 0.17 0.19 0.21 0.23 0.25 0.26 0.28 0.30 0.32 0.34 0.35 0.37 0.39 0.40 0.42 0.43 0.45 0.46 0.48 0.49 0.51 0.52 0.53 0.55 0.56 0.58 0.59 0.60 0.62

0.07 0.09 0.11 0.14 0.16 0.18 0.19 0.21 0.23 0.25 0.27 0.28 0.30 0.32 0.33 0.35 0.37 0.38 0.40 0.41 0.43 0.44 0.46 0.47 0.48 0.50 0.51 0.52 0.54 0.55 0.56 0.58 0.59

0.07 0.09 0.11 0.13 0.15 0.16 0.18 0.20 0.22 0.24 0.25 0.27 0.29 0.30 0.32 0.33 0.35 0.36 0.38 0.39 0.41 0.42 0.43 0.45 0.46 0.48 0.49 0.50 0.51 0.53 0.54 0.55 0.57

0.06 0.08 0.10 0.12 0.14 0.16 0.17 0.19 0.21 0.22 0.24 0.25 0.27 0.29 0.30 0.32 0.33 0.35 0.36 0.37 0.39 0.40 0.42 0.43 0.44 0.45 0.47 0.48 0.49 0.51 0.52 0.53 0.54

0.06 0.08 0.09 0.11 0.13 0.15 0.16 0.18 0.20 0.21 0.23 0.24 0.26 0.27 0.29 0.30 0.32 0.33 0.34 0.36 0.37 0.38 0.40 0.41 0.42 0.44 0.45 0.46 0.47 0.49 0.50 0.51 0.52


− Equation (neglecting qu) : d' =

2 2 Pu p'  p'  + −  + 2 8.87 f 'c 2

2

(p' in m, Pu in kN, f'c in kPa)

4 of 4


Depth Factors for bearing capacity equations Values of dq (inside table) Df/B

0.2

0.4

0.6

0.8

1.0

1.2

1.4

1.6

1.8

2.0

1.00 1.01 1.01 1.02 1.02 1.03 1.03 1.04 1.04 1.05 1.05 1.05 1.05 1.06 1.06 1.06 1.06 1.06 1.06 1.06 1.06 1.06 1.06 1.06 1.06 1.06 1.06 1.06 1.06 1.06 1.06 1.06 1.06 1.05 1.05 1.05 1.05 1.05 1.05 1.04 1.04 1.04 1.04 1.04 1.04 1.03 1.08

1.00 1.01 1.03 1.04 1.05 1.06 1.07 1.08 1.08 1.09 1.10 1.10 1.11 1.11 1.11 1.12 1.12 1.12 1.12 1.13 1.13 1.13 1.13 1.13 1.13 1.12 1.12 1.12 1.12 1.12 1.12 1.11 1.11 1.11 1.10 1.10 1.10 1.10 1.09 1.09 1.09 1.08 1.08 1.08 1.07 1.07 1.16

1.00 1.02 1.04 1.06 1.07 1.09 1.10 1.11 1.12 1.14 1.14 1.15 1.16 1.17 1.17 1.18 1.18 1.18 1.19 1.19 1.19 1.19 1.19 1.19 1.19 1.19 1.18 1.18 1.18 1.18 1.17 1.17 1.17 1.16 1.16 1.15 1.15 1.14 1.14 1.13 1.13 1.12 1.12 1.11 1.11 1.10 1.24

1.00 1.03 1.05 1.08 1.10 1.12 1.13 1.15 1.17 1.18 1.19 1.20 1.21 1.22 1.23 1.24 1.24 1.24 1.25 1.25 1.25 1.25 1.25 1.25 1.25 1.25 1.25 1.24 1.24 1.24 1.23 1.23 1.22 1.22 1.21 1.20 1.20 1.19 1.18 1.18 1.17 1.16 1.16 1.15 1.14 1.14 1.32

1.00 1.03 1.05 1.07 1.10 1.11 1.13 1.15 1.16 1.18 1.19 1.20 1.21 1.22 1.23 1.23 1.24 1.24 1.24 1.25 1.25 1.25 1.25 1.25 1.25 1.24 1.24 1.24 1.24 1.23 1.23 1.22 1.22 1.21 1.21 1.20 1.19 1.19 1.18 1.17 1.17 1.16 1.15 1.15 1.14 1.13 1.31

1.00 1.03 1.06 1.08 1.11 1.13 1.15 1.17 1.18 1.20 1.21 1.22 1.23 1.24 1.25 1.26 1.26 1.27 1.27 1.27 1.28 1.28 1.28 1.28 1.27 1.27 1.27 1.27 1.26 1.26 1.25 1.25 1.24 1.24 1.23 1.22 1.22 1.21 1.20 1.19 1.19 1.18 1.17 1.17 1.16 1.15 1.35

1.00 1.03 1.06 1.09 1.12 1.14 1.16 1.18 1.20 1.21 1.23 1.24 1.25 1.26 1.27 1.28 1.29 1.29 1.29 1.30 1.30 1.30 1.30 1.30 1.30 1.30 1.29 1.29 1.28 1.28 1.27 1.27 1.26 1.26 1.25 1.24 1.23 1.23 1.22 1.21 1.20 1.20 1.19 1.18 1.17 1.16 1.38

1.00 1.03 1.07 1.10 1.12 1.15 1.17 1.19 1.21 1.23 1.24 1.26 1.27 1.28 1.29 1.30 1.30 1.31 1.31 1.32 1.32 1.32 1.32 1.32 1.32 1.31 1.31 1.31 1.30 1.30 1.29 1.29 1.28 1.27 1.27 1.26 1.25 1.24 1.23 1.23 1.22 1.21 1.20 1.19 1.18 1.17 1.40

1.00 1.04 1.07 1.10 1.13 1.16 1.18 1.20 1.22 1.24 1.26 1.27 1.28 1.30 1.30 1.31 1.32 1.33 1.33 1.33 1.34 1.34 1.34 1.34 1.33 1.33 1.33 1.32 1.32 1.31 1.31 1.30 1.29 1.29 1.28 1.27 1.26 1.25 1.25 1.24 1.23 1.22 1.21 1.20 1.19 1.18 1.43

1.00 1.04 1.07 1.10 1.13 1.16 1.19 1.21 1.23 1.25 1.27 1.28 1.30 1.31 1.32 1.33 1.33 1.34 1.34 1.35 1.35 1.35 1.35 1.35 1.35 1.34 1.34 1.34 1.33 1.33 1.32 1.31 1.31 1.30 1.29 1.28 1.27 1.26 1.26 1.25 1.24 1.23 1.22 1.21 1.20 1.19 1.44

φ° 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 dc

Prepared by Dr. Adnan Basma

STRUCTURAL DEPTH OF CONCRETE FOOTINGS, d' (m) PUNCHING SHEAR Punching shear failure occurs at a distance d'/2 from the sides of the column

INTERIOR COLUMN

EXTERIOR COLUMN b

l


b

l


For equilibrium,

!F = 0 , where:

Pu " Vc " Fs = 0 Pu = Ultimate load on column Vc = maximum shear force = Ac #c For interior column Ac = 2( ! +d’)d’ + 2(b + d’)d' For exterior column Ac = 2( ! +d’/2)d’ + (b + d’)d' # c = 8.87 fc ' (SI units) with fc’ and #c are in kN/m2

Fs = Soil pressure under failed portion = ASoil qu For interior column ASoil = ( ! +d’)(b + d’) For exterior column ASoil = ( ! +d’/2)(b + d’) Substitute in this equation and solve for d'. In many cases the effect of Fs is small and thus it is neglected and d' is estimated from Pu = Vc . For this solution refer to Structural Depth of Concrete for Punching Shear table for punching shear failure.

1 of 2

Foundation Engineering Equations for Concrete Depth of Footings Dr. Adnan A. Basma

WIDE BEAM FAILURE (Uniform footing width B) For wide beam shear, failure is assumed at a distance d' from the side of the column.

L d'

l

B

b

Lw Wide beam failure at d' from column

For equilibrium

!F = 0 , where

Fs - Vc = 0 Fs = qua . Lw.1 Vc = #c . d’ . 1 # c = 4.56

Thus

d’ =

fc ' (SI units) with fc’ and #c are in kN/m2

L " ! " 2d' qua $ # c d' , re-arranging term and solving 2 (L " ! )qua 2(# c % qua )

The value of d' is that by punching shear or wide beam whichever governs (larger value).

2 of 2

Foundation Engineering Equations for Concrete Depth of Footings Dr. Adnan A. Basma

ESTIMATION OF REINFORCEMENT FOR CONCRETE DESIGN OF FOOTINGS To resist tension in concrete we need reinforcements in such a way that a# & Mu ) ( A S fy $ d' ' ! 2" % where

(*)*0.9*for moment As = total area of steel required (m2 /meter) fy = yield strength of reinforcing bars (kN/m2) fc’ = 28-days ultimate compressive strength of concrete (kN/m2) d’ = footing thickness (m) a=

A s fy 0.85 fc ' + 1

Percent steel p =

where

and

pmin )

As d' +1

pmin , p , pmax

1400 fy

pmax . 0.5

indicates As / unit length (1m)

or 0.002 (0.2% )

fc ' 600 + 103 + fy ( fy - 600 + 103 )

fc’ and fy in kN/m2

Using the estimated value of AS, calculate number of bars / m = n = where Ab = Area of bar with diameter db where db =

Spacing of bars, s =

/db 4

As Ab

2

100 cm n'1

This gives the spacing of the bars per m (100 cm) and is repeated for every meter of the footing (L and B - direction)

Alternatively, the steel required can be obtained from Reinforcement Tables.

1 of 1

Foundation Engineering 2 / Review Equations for Reinforcement of Footings Dr. Adnan A. Basma

FOUNDATION ENGINEERING Footing Depth by Punching Shear Example 3a Determine the depth of the footing d' by punching shear for the following conditions: Loads: DL = 1110 kN, LL = 1022 kN Materials: fc'= 21 MPa, fy = 415 MPa Column: 45 cm ! 45 cm (at center of footing) Soil: Ultimate soil pressure qu = 365 kPa SOLUTION: Allowable load P = DL + LL = 1110 + 1022 = 2132 kN Ultimate load Pu = 1.4DL + 1.7LL = 1.4!1110 + 1.7!1022 = 3291.4 kN Footing Depth, d' by Punching Shear (4-way punch)

l = 0.45m, b = 0.45 m and

3 vc = 8.87 21x 10 = 1287.4 kPa

Pu = Vc + Fs Vc = vc!Ac Ac = 2!(0.45+d')d’ + 2!(0.45+d')d’ = 4!(0.45+d')d’ = (1.8 + 4d')d’ therefore, Vc = 1287.4!(1.8 + 4d')d’ Fs = qu!As with As = (0.45 + d')!(0.45 + d') = (0.45 + d')2 Fs = 365!(0.45 + d')2 Equating to Pu we get : 3291.4 = 1287.4!(1.8 + 4d')d’ + 365!(0.45 + d')2 Solving we find d' = 0.571m or 57.1 cm

Using Structural Depth of Concrete table for punching shear failure with Pu = 3291.4 kN, p' = 2(0.45)+2(0.45) = 1.8 m and fc' = 21 MPa we get d' = 0.75 m. This value however is determined after neglecting qu and is about 30% greater than the actual value. Thus if one is to calculate the actual value it will be 0.75/1.3 = 0.576 or 57.6 cm which compares well with the calculated value

1 of 1

Foundation Engineering Punching Shear / Example Dr. Adnan A. Basma

FOUNDATION ENGINEERING Reinforcement Example 3b Determine the reinforcement for the following footing conditions: Footing: L = 4.1 m and B = 2.2m, d' = 0.65m Loads: DL = 1110 kN, LL = 1022 kN Materials: fc'= 21 MPa, fy = 415 MPa Column: 45 cm ! 45 cm (at center of footing) Soil: Ultimate soil pressure qu = 365 kPa SOLUTION: Allowable load P = DL + LL = 1110 + 1022 = 2132 kN Ultimate load Pu = 1.4DL + 1.7LL = 1.4!1110 + 1.7!1022 = 3291.4 kN Ultimate moments in L and B directions

' 4.1 - 0.45 $ 365.0 % " 2 & # Mu (L) = 2

' 2.2 - 0.45 $ 365.0 % " 2 & # Mu (B) = 2

2

= 608.8 kN.m/m

2

= 139.9 kN.m/m

The moment in L-direction governs thus use Mu = 608.8 kN.m/m for steel calculations in both directions

a=

(415x 103 ) A s = 23.24 A s [0.85x(21x 103 )x1]

Substituting in ultimate moment versus steel area equation we get 23.24 A s * 0.9 (415x103 ) A s + 0.650 ( = 608.8 2 , ) Re-arranging the terms, the final result becomes 2 A s - 0.057 A s + 0.000143 = 0

1 of 2

Foundation Engineering Reinforcement / Example Dr. Adnan A. Basma

Solving we get As = 0.0026 m2/m or 26 cm2/m. This value gives a percent steel

pmax = 0.5x

pmin =

p=

21x 103 600x 103 x = 0.015 415x 103 (415x 103 + 600x 103 )

or 1.5%

1400 = 0.0033 or 0.002 whichever is larger 415x 103

0.0026 = 0.004 0.650x1

or

0.4%

Since the calculated percent area of steel is between pmin and pmax then use As = 26 cm2/m. Using Reinforcement Tables with Mu = 608.8 kN.m/m, d' = 0.65m fc'= 21 MPa and fy = 415 MPa the following are obtained

p(min) = 0.34 %

p(max) = 1.50%

and

p = 0.4%

These values compare well with calculated values. Note that As is calculated by: AS = p x d' x 1 =

0. 4 ! 0.65 ! 1 = 0.0026 m2/m or 26 cm2/m. 100

The following are possible bas sizes and spacing that can be used (note that the provided As must be equal to or greater than 26 cm2/m). These were obtained from page 10 of Reinforcement Tables.

Alternative 1:

[email protected] c-c

(provide As = 26.61 cm2/m)

Alternative 2:



Alternative 3:



The most ideal choice is alternative 3 (Alternative 2 will also work).

2 of 2

Advanced Foundation Engineering Reinforcement / Example Dr. Adnan A. Basma

FOUNDATION ENGINEERING Reinforcement Example #5 (Reinforcement) Determine the reinforcement for the following footing conditions: Footing: L = 4.1 m and B = 2.2m, d' = 0.65m Loads: DL = 1110 kN, LL = 1022 kN Materials: fc'= 21 MPa, fy = 415 MPa Column: 45 cm ! 45 cm (at center of footing) Soil: Ultimate soil pressure qu = 365 kPa SOLUTION: Allowable load P = DL + LL = 1110 + 1022 = 2132 kN Ultimate load Pu = 1.4DL + 1.7LL = 1.4!1110 + 1.7!1022 = 3291.4 kN Ultimate moments in L and B directions

' 4.1 - 0.45 $ 365.0 % " 2 & # Mu (L) = 2

' 2.2 - 0.45 $ 365.0 % " 2 & # Mu (B) = 2

2

= 608.8 kN.m/m

2

= 139.9 kN.m/m

The moment in L-direction governs thus use Mu = 608.8 kN.m/m for steel calculations in both directions

a=

(415x 103 ) A s = 23.24 A s [0.85x(21x 103 )x1]

Substituting in ultimate moment versus steel area equation we get 23.24 A s * 0.9 (415x103 ) A s + 0.650 ( = 608.8 2 , ) Re-arranging the terms, the final result becomes 2 A s - 0.057 A s + 0.000143 = 0

1 of 2

Foundation Engineering 2 / Review Example 5 - Reinforcement Dr. Adnan A. Basma

Solving we get As = 0.0026 m2/m or 26 cm2/m. This value gives a percent steel

pmax = 0.5x

pmin =

p=

21x 103 600x 103 x = 0.015 415x 103 (415x 103 + 600x 103 )

or 1.5%

1400 = 0.0033 or 0.002 whichever is larger 415x 103

0.0026 = 0.004 0.650x1

or

0.4%

Since the calculated percent area of steel is between pmin and pmax then use As = 26 cm2/m. Using Reinforcement Tables with Mu = 608.8 kN.m/m, d' = 0.65m fc'= 21 MPa and fy = 415 MPa the following are obtained

p(min) = 0.34 %

p(max) = 1.50%

and

p = 0.4%

These values compare well with calculated values. Note that As is calculated by: AS = p x d' x 1 =

0. 4 ! 0.65 ! 1 = 0.0026 m2/m or 26 cm2/m. 100

The following are possible bas sizes and spacing that can be used (note that the provided As must be equal to or greater than 26 cm2/m). These were obtained from page 10 of Reinforcement Tables.

Alternative 1:



Alternative 2:



Alternative 3:



The most ideal choice is alternative 3 (Alternative 2 will also work).

2 of 2

Foundation Engineering 2 / Review Example 5 - Reinforcement Dr. Adnan A. Basma

DESIGN OF MAT (RAFT) FOUNDATIONS Design Steps and Equations For an example on Design of Mat Foundations click here 3D View of Mat (Raft) Foundation

y L1

L2

B

A Pu1

C Pu3

Pu2 ex

Column dimensions l i x bi

B1

Pu ey

B

E

D Pu4

Pu5

x

F Pu6 B2

Pu7 G

Pu8

H

Pu9 I

L Top View of Mat (Raft) Foundation 1 of 6



Piu = ! [1.4DLi + 1.7LLi]

Ultimate ratio

ru =

Pu , Ultimate pressure qu = qa x ru P

Locate the resultant load Pu In x- direction:

!My-axis = 0,

ex &

In y- direction:

!Mx-axis = 0,

ey &

"Pu3 $ Pu9 # L2

% "Pu1 $ Pu7 # L1 Pu

"Pu1 $ Pu3 # B1

% "Pu7 $ Pu9 # B2 Pu

,P My x Mx y ) ' Applied ultimate pressure, qu,applied = * u Iy Ix ' *A + ( Where

A = Area = BL Mx = Pu ey

and

My = Pu ex

Ix = ! B L

and

Iy = ! L B

3

3

For the mat shown in Top View, the following sign convention is used to estimate qu,applied x (+)

x (%) A

B

C y (+)

D

E

F

G

H

I

2 of 6

y (%)


The following values for x and y along with the sign conventions are used to estimate qu,applied Point A

B

C

D

E

F

G

H

I

x

L1

0

L2

L1

0

L2

L1

0

L2

y

B1

B1

B1

0

0

0

B2

B2

B2

For the dimensions L and B to be adequate, qu,max . qu

and

qu,min / 0

STEP 2 – DRAW SHEAR AND MOMENT DIAGRAMS (L - DIRECTION) The mat is divided into several strips in L-direction as shown below where B' = B/4. y L2

L1 B

A B'

C

I

K

J

B1

B' B D

II

E

x

F

B' B2 M

L B'

III

G

H

I

L

3 of 6


Calculations for Strip ABC: a) The average uniform soil reaction, quI =

q uA $ quC 2

b) Total soil reaction Q = quI x (B1 x L) where B1 = B' c) Total Column loads Pu, ABC = Pu1 + Pu2 + Pu3 d) Average load Pu, avg(ABC) =

Q $ Pu,

e) Load multiplying factor FABC =

ABC

2 Pu,

avg( ABC )

Pu,

ABC

f) The modified loads on this strip P'ui = (FABC) x (Pui) , Pu, avg( ABC ) ) g) Modified Average soil pressure qu, modified = quI x * ' Q *+ '( h) The pressure distribution along the length of the strip qu, L (ABC) =

0 P' ui L

Note that the same can be done for strips DEF (B2 = 2B') and GHI (B3 = B') where: quII = quIII =

q uD $ q uF 2 q uG $ q uI 2

Steps (b) to (h) are repeated as above

The shear and bending moment diagrams for strip ABC is shown below. Other strips will have similar plots.

4 of 6


P'u1

P'u3

P'u2 L1

L2 B

A

C qu, L (ABC)

+ V(kN)

%



% M(kN.m)

+ Bottom Steel under Column 2

Moment drawn on tension side

5 of 6


Similar plots should be made for strips in B-direction as shown below A

B I

C III

II

D

E

F

G

H

I

STEP 3 – DEPTH OF CONCRETE, d' Estimate d' for: a) Column 1, 3, 7 and 9 by 2-way punching shear (p' = l + w). b) Column 2, 4, 6 and 8 by 3-way punching shear (p' = 2l + w). c) Column 5 by 4-way punching shear (p' = 2l + 2w). (Use Equations For Punching Shear or approximate d' by Structural Depth of Concrete table for punching shear failure). Select the largest d' from (a), (b) or (c)

STEP 4 – REINFORCEMENT The calculations below are repeated for every strip in L and B direction. a) Select the appropriate moments for each strip (refer to moment diagram) and estimate the moment per meter by Mui/m = Mu/Bi or Li b) Using Mui/m, d', fc' and fy estimate the reinforcement As (refer to Equations for Reinforcement or the percent reinforcement can be obtained directly from Percent Steel Tables).

For an example on Design of Mat Footings click here

6 of 6


FOUNDATION ENGINEERING 2 Mat Foundations (Design Equations) Design Example Design the mat foundation for the plan 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. Use fc’ = 24 Mpa and fy = 275 MPa

4.25 m

4.25 m

8m y

0.25 m M

B

3.75 m

DL= 200 kN LL= 200 kN

N DL = 250 kN LL = 250 kN

C DL=250 kN LL=200 kN

7m

M

D

N



E

DL=650 kN LL=550 kN

7m

A

F



H

DL=650 kN LL=550 kN

3.75 m

R

DL = 250 kN LL = 250 kN

J 0.25 m

O

K

8m

x

I

Q

DL=200 kN LL= 200kN

21.5 m

P

7m

7m

G

0.1 m

Pu

O

7m

0.44m

DL=200 kN LL=150 kN

P

8m

L 0.25 m

16.5 m

Plan of Mat Foundation with column loads and dimensions

1 of 9



Piu = ! [1.4DLi + 1.7LLi]

Ultimate ratio

ru =

Pu P

The table below shows the calculations for the loads: Column

DL, kN

A B C D E F G H I J K L

200 250 250 800 800 650 800 800 650 200 250 200

LL, kN

P, kN

200 250 200 700 700 550 700 700 550 200 250 150 Total Loads =

400 500 450 1500 1500 1200 1500 1500 1200 400 500 350 11000 ru =

Pu, kN 620 775 690 2310 2310 1845 2310 2310 1845 620 775 535 16945 1.54

Ultimate pressure qu = qa x ru = 60 x 1.54 = 92.4 kPa Location of the resultant load Pu In x- direction: ex '

!My-axis = 0,

"620 % 2310 % 2310 % 620# $ 8

&

"690 % 1845 % 1845 % 535# $ 8

16945

= 0.44 m In y- direction: ey '

!Mx-axis = 0,

"620 % 775 % 690#x10.5 % "2310 % 2310 % 1845#x3.5 - "2310 % 2310 % 1845#x3.5 & "620 % 775 % 535#x10.5 16945

= 0.1 m

2 of 9


-P My x Mx y * ( Applied ultimate pressure, qu,applied = + u . . Iy Ix ( +A , ) A = Area = BL = (16.5) x (21.5) = 354.75 m2

Where

Mx = Pu ey = (16945)(0.1) = 1694.5 kN.m My = Pu ex = (16945)(0.44) = 7455.8 kN.m 1

Ix =

Iy =

12 1 12

1

3

BL =

12 1

3

LB =

3

12

(16.5) (21.5) = 13665 m4 3

(21.5) (16.5) = 8050 m4

Therefore, 7455.8 x 1694.5 y * - 16945 qu,applied = + . . ( 8050 13665 )( ,+ 354.75

"

= 47.76 . 0.93 x . 0.124 y

#

Point

x

y

Sign for x Sign for y

A

8.25

10.75

+

+

56.77

B

0

10.75

+

+

49.09

C

8.25

10.75

&

+

41.42

J

8.25

10.75

+

&

54.10

K

0

10.75

+

&

46.43

L

8.25

10.75

&

&

38.75

qu,applied, kPa

For the dimensions L = 21.5 m and B = 16.5 m, qu,max = 56.77 kPa < qu = 92.4 kPa and qu,min = 38.75 kPa > 0 Therefore, the dimensions are adequate. 3 of 9


STEP 2 – DRAW SHEAR AND MOMENT DIAGRAMS (L - DIRECTION) The mat is divided into several strips in L-direction (see figure on page 1). The following strips are considered: AMOJ, MNPO and NCLP The following calculations are performed for every strip: a) The average uniform soil reaction, qu =

q u,

Edge 1

% qu

edge 2

2

Refer to table on page 3 for pressure values Strip AMOJ: Edge 1 is point A

and

Edge 2 is point J

Strip MNPO: Edge 1 is point B

and

Edge 2 is point K

Strip NCLP: Edge 1 is point C

and

Edge 2 is point L

b) Total soil reaction Qi = qu x (Bi x L) Strip AMOJ: B1 = 4.25 m Strip MNPO: B2 = 8.00 m Strip NCLP: B3 = 4.25 m For all strips L = 21.5 m c) Total Column loads Pu, total = !Pui d) Average load Pu, avg =

Q i % Pu,total 2

e) Load multiplying factor F =

Pu,

avg

Pu,

total

f) The modified loads on this strip P'ui = (F) x (Pui) - Pu, avg * g) Modified Average soil pressure qu, modified = qu x + ( +, Q i () h) The pressure distribution along the length of the strip qu, L =

/ P' ui L

4 of 9


The following table presents the calculations for the selected strips: Strip

Bi, m Point A J B K C L

AMOJ 4.25 MNPO 8.00 NCLP 4.25

qu, kPa 56.77 54.10 49.09 46.43 41.42 38.75

qu,avg kPa

Q, kN

Total Pu, kN

Pu, avg, qu, mod kN kPa

55.44

5065.37

5860.00 5462.69 59.47 0.932

47.76

8214.72

6170.00 7192.36 40.97 1.166

40.09

3662.77

4915.00 4288.88 45.94 0.873

F

Based on table above, the adjusted column loads and the pressure under each is strip are: Strip

Column

DL, kN

LL, kN

P, kN

Pu, kN

AMOJ F=0.932

P'u, kN qu,L, kN/m

A

200

200

400

620

D

800

700

1500

2310

2152.92

G

800

700

1500

2310

2152.92

J

200

200

400

620

577.84

Total =

2000

1800

3800

5860

5461.52

MNPO

B

250

250

500

775

F=1.166

E

800

700

1500

2310

2693.46

H

800

700

1500

2310

2693.46

K

250

250

500

775

903.65

Total =

2100

1900

4000

6170

7194.22

NCLP

C

250

200

450

690

F=0.873

F

650

550

1200

1845

1610.685

I

650

550

1200

1845

1610.685

L

200

150

350

535

467.055

Total =

1750

1450

3200

4915

4290.795

Total Loads =

5850

5150

11000

16945

577.84 254.02

903.65 334.61

602.37 199.57

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

5 of 9


Strip AMOJ (B' = 4.25 m) 577.8 kN 0.25 m

2152.9 kN

2152.9 kN 7m

7m

7m

A

D

577.8 kN 0.25 m

G

J

254.02 kN/m

1500 1265.4 888.1

1000

510.8

500 63.4

+

0

V(kN)

0

& -500

5

10

15

20 -63.4

-510.8

-1000

-888 .1 -1265.4

-1500

Max. moment between columns Steel at the bottom 2660.5 2660.5

3000 2500 2000 1500 1000

+

500

M(kN.m)

&

1106.3 7.9

7.9

0

-500

0

5 -503.0

-1000

10

15

20

- 503.0




Strip MNPO (B' = 8.0 m) 903.7 kN 0.25 m

2693.5 kN

2693.5 kN 7m

7m

7m

B

E

903.7 kN 0.25 m

H

K

334.6 kN/m

2000 1524.9

1500

1169.8

1000

814.7

500 83.6

+

0

V(kN)

& -500

0

5

10

15

20 -83.6

-814.7

-1000

-1169.8

-1500

-1524.9

-2000 Max. moment between columns Steel at the bottom

3000

2496.2

2496.2

2500 2000 1500 1000 500

+ &

449.1

10.5

M(kN.m)

10.5

0

-500

0

5

10

15

20

-1000

-982.4 -1500

Max. moment under columns Steelon atcompression the top Moment drawn side

-982.4



Strip NCLP (B' = 4.25 m) 602.4 kN 0.25 m

1610.7 kN

1610.7 kN 7m

7m

7m

C

F

467.1 kN 0.25 m

I

L

199.6 kN/m

1000 845.0

632.6

500

420.4

+ V(kN)

50.1 0

&

0

-500

5

10

20 -49.7

15

-554.4 -766.6

-1000

-978.8

-1500

1500

1023.5

Max. moment between columns Steel at the bottom

1000

554.7 500

+ M(kN.m)

0

&

6.3

6.3 0

5

-500 -1000 -1500 -2000

10

15

20

-446.6 -762.3

-1842.1




STEP 3 – DEPTH OF CONCRETE, d' The table below was prepared for the columns and d’ was estimated using Structural Depth of Concrete table for punching shear failure (fc' = 24 MPa). Column

Punching

p', m

Pu, kN

d', m

A&J

2

1.00

620

0.29

B&K

3

1.50

775

0.27

C

2

1.00

690

0.31

D&G

3

1.50

2310

0.62

Maximum

E&H

4

2.00

2310

0.54

Use d' = 0.65 m

F&I

3

1.50

1845

0.53

L

2

1.00

535

0.25

STEP 4 – REINFORCEMENT L-direction Reinforcement: The table below is prepared by selecting the appropriate moments for each strip and estimate the moment per meter by Mui/m = Mu/Bi or Li . Using Mui/m, d' = 0.65, fc' = 24 MPa and fy = 275 MPa*, p (percent steel) and thus As can be estimated by Percent Steel Tables. Strip

B', m

Location

AMJO

4.25

0-5 m / top 5-17 m / bottom 17-21.5 m / top

503 2660.5 503

MNPO

8

0-5 m / top 5-16.5 m / bottom 16.5-21.5 m / top

MNPO 4.25

Mu, kN.m Mu, kN.m/m

+

p, %

AS (cm2/m)

Reinforcement

118.4 626.0 118.4

0.11** 0.62.. 0.11**

33.2 40.3 33.2

030@25 cm c-c 030@20 cm c-c 030@25 cm c-c

982.4 2496.2 982.4

122.8 312.0 122.8

0.11** 0.30** 0.11**

33.2 33.2 33.2

030@25 cm c-c 030@25 cm c-c 030@25 cm c-c

0-6 m / top 762.3 6-9 m / bottom 1023.5 9-13 m / top 446.6 1314.5 m / bottom 554.7 14.5-21.5 m / top 1842.1

179.4 240.8 105.1 130.5 433.4

0.18** 0.24** 0.10** 0.12** 0.43**

33.2 33.2 33.2 33.2 33.2

030@25 cm c-c 030@25 cm c-c 030@25 cm c-c 030@25 cm c-c 030@25 cm c-c

* For fc' = 24 MPa and fy = 275 MPa, p(min) = 0.51%, and p(max) = 2.99% ** p < p(min) so use p(min) = 0.51% + AS = p x d' x 1 = (p/100) x (65) x (100)

B-direction Reinforcement: The same can be done for the B-direction moments (step 2) by considering (refer to page 1) strips: ACNM (B' = 3.75 m), MNPO (B' = 7 m), OPRQ (B' = 7m) and QRLJ (B' = 3.75 m). 9 of 9


Foundation Examples Dr, Basma.pdf

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